url stringlengths 14 2.42k | text stringlengths 100 1.02M | date stringlengths 19 19 | metadata stringlengths 1.06k 1.1k |
|---|---|---|---|
https://aptitude.gateoverflow.in/6806/nielit-2019-feb-scientist-d-section-c-17 | 228 views
A dinner party is to be fixed for a group consisting of $100$ persons. In this party, $50$ persons do not prefer fish, $60$ prefer chicken and $10$ do not prefer either chicken or fish. The number of persons who prefer both fish and chicken is:
1. $10$
2. $20$
3. $30$
4. $40$
$n(A\cup B) = n(A)+n(B)-n(A\cap B)$
n(persons who do not prefer either chicken or fish) = n(persons who do not prefer chicken) + n(persons who do not prefer fish) – n(persons who do not prefer both chicken and fish)
n(persons who do not prefer either chicken or fish) = (100 – 60) + 50 – 10 = 80
ׯ The number of persons who prefer both fish and chicken = 100 – 80 = 20
3.0k points
1 vote
1
359 views
2
287 views
3
264 views
4
380 views
1 vote | 2023-02-05 14:55:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1917678415775299, "perplexity": 3347.353034417138}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00637.warc.gz"} |
https://www.physicsforums.com/threads/work-and-energy-unit-joule.855617/ | Work and Energy Unit Joule
Tags:
1. Feb 4, 2016
Work unit is said to be Joule and Energy unit is also Joule. Work formulas in F x S ( Force x Displacement) .
We know
1 J = 1 F x `1 s
Now Energy is also measured in Joule(J)
So can we state :
E = 1 F x 1 s as 1 J = 1 F x 1 s
that is
E = W
Then why work and energy are different by definition and meaning
2. Feb 4, 2016
QuantumQuest
A small glitch, 1J = 1N x 1s.
What do you mean by that? A very "rough" statement back from high school times, is "Energy of a body, is the ability to do work". Now, what are the definitions you refer to?
3. Feb 4, 2016
nasu
I am not sure if you mean to correct a glitch or to present one. :)
The Joule is not Nxs (Newton times second).
The OP meant Force x distance when he/she wrote 1J= 1F x 1s (by s probably meaning distance, displacement).
4. Feb 4, 2016
Well is W = E ? I am not going by definition but rather its unit of measurement which in this case is Joule (for both Work & Energy)? So why we state that work is said to be done when a is force applied on a body and it displaces while energy is ability or capacity to do work - why are the unit of measurement same for both work / energy - JOULE?
5. Feb 4, 2016
Svein
Because work and energy is two sides of the same coin. If you lift a boulder (10kg) up 1m, it means that:
• You have performed an amount of work = m⋅g⋅h = 10⋅9.81⋅1J = 98J
• The boulder has gained an amount of potential energy = m⋅g⋅h = 10⋅9.81⋅1J = 98J
6. Feb 4, 2016
QuantumQuest
Yes, sorry for the typo, I meant N x m - of course, in SI units.
Work expresses energy transmission or transformation of energy, from one form to another. So, they have the same unit of measurement.
As for the energy concept itself, I recommend this great insight by DaleSpam https://www.physicsforums.com/insights/what-is-energy/.
Last edited: Feb 4, 2016
7. Feb 4, 2016
ZapperZ
Staff Emeritus
Work is a form of energy
Potential energy is a form of energy
Kinetic energy is a form of energy
They ALL have the same dimension and SI units.
The similar way as
Electrostatic force is a form of force
Gravitational force is a form of force
Spring force is a form of force
Tension is a form of force
Frictional force is a form of force
They all have have the same dimension and SI units of N.
Zz.
8. Feb 5, 2016 | 2017-08-20 16:03:24 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8248814344406128, "perplexity": 1621.5907837745567}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886106779.68/warc/CC-MAIN-20170820150632-20170820170632-00240.warc.gz"} |
http://www.lofoya.com/Solved/1315/a-train-covers-a-distance-in-100-min-if-it-runs-at-a-speed-of-48kmph | # Easy Time, Speed & Distance Solved QuestionAptitude Discussion
Q. A train covers a distance in 100 min, if it runs at a speed of 48kmph on an average. The speed at which the train must run to reduce the time of journey to 40 min will be:
✖ A. 30 kmph ✖ B. 50 kmph ✖ C. 80 kmph ✔ D. 120 kmph
Solution:
Option(D) is correct
Time = $\dfrac{100}{60}=\dfrac{5}{3}$hr
Speed =48 mph
Distance $=S \times T$
$= 48 \times \dfrac{5}{3}$
$=80 \text{ km}$
Now, as per the question, journey is to be reduced to 40 min.
So, new time,
$=40 \text{ min}= \dfrac{40}{60} \text{ hr}$
$=\dfrac{2}{3} \text{ hr}$
New speed,
$=\dfrac{\text{Distance}}{\text{New Time}}$
$=\dfrac{80}{2/3}$
$=80 \times \dfrac{3}{2}$
$= \textbf{120 kmph}$
Edit: For an alternative solution (without calculating the distance), check comment by Chirag Goyal.
Edit 2: For yet another alternative solution, check comment by Vejayanantham TR.
Edit 3: For yet another alternative solution, check comment by VENI.
Edit 4: For yet another alternative solution using Ratio method, check comment by Akshay.
## (10) Comment(s)
Akshay
()
#AKSHAY we can do simply without using pen and paper previously it was 100 min and now it is 40 min so the ratio is 2/5. So speed is increased by 5/2 of the old speed therefore 5/2 *48 = 120
Bharat Bhushan
()
Since distance is same use S1T1 = S2T2. hence 100/40 *48 = speed required.
VENI
()
$t=100$ min $v=48$ kmph
$t=40$ min $v=?$
For the same distance,
$v=\dfrac{(48*100)}{40}$
$v=\textbf{120 kmph}$
Vejayanantham TR
()
$\dfrac{S_1}{S_2} = \dfrac{T_2}{T_1}$
$T_1 = 100 \text{ min} = \dfrac{100}{60} = \dfrac{5}{3}$
$T_2 = 40\text{ min} = \dfrac{40}{60} = \dfrac{2}{3}$
So, as per formula,
$\dfrac{48}{S_2} = \dfrac{2}{5}$
$S_2 = 48* \dfrac{5}{2} = 120$
Chirag Goyal
()
Hello there,
We don't have to calculate distance Because it is same for both cases
and we can use this as following.
Let $S_1$ is speed of train which is $48km/hr$
$T_1$ is the time $(100\ min\ or\ \dfrac{5}{3}hr)$ to complete Certain Distance $D$
$S_2$ is required Speed of Train to cover the same distance $D$ in time $T_2$ which is $40\ min\ or\ \dfrac{2}{3}hr$.
$S_1\times{T_1}$=$S_2\times{T_2}$ as Distance $D$ is same
Solving this we get
$S_2=120\ km/hr$ is the required answer.
Saif Pathan
()
Initial speed =48kmph
Dist=?
Time = 100min
S=D/T
48=D/100
48*100=D
D=4800
Now
When Time=40kmph and D=4800
Using formula S= D/T
i.e. S= 4800/40
S= 120 kmph
Shubh
()
You have arrived at the result but I strongly advise against mixing the units.
You are multiplying 'kmph' with 'min' in the first step $(D=4800)$.
In the second step, its effect is nullified during the division operation $(S=D/T)$, when you have divided again by min'.
Very bad approach of solving questions this way though.
Rishabh Chaudha
()
the ans is 120 km/hr
Anujpachauri
()
How can you travel 80 km in 40 min with speed of 30 km?
sol is.
total distant is $\dfrac{100}{60} \times 48=80$ km
and
speed $=\dfrac{80}{40/60}$
$\dfrac{80}{40} \times 60=120$ kmph
hence and ans is $\textbf{120 kmph}$
Deepak
()
Thank you Anuj for pointing out the error. Solution is updated and correct option choice is pointed as a right answer now. | 2016-10-22 01:49:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6844271421432495, "perplexity": 2550.7335996861348}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718423.28/warc/CC-MAIN-20161020183838-00236-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://paperswithcode.com/paper/detecting-clitics-related-orthographic-errors | # Detecting Clitics Related Orthographic Errors in Turkish
For the spell correction task, vocabulary based methods have been replaced with methods that take morphological and grammar rules into account. However, such tools are fairly immature, and, worse, non-existent for many low resource languages... Checking only if a word is well-formed with respect to the morphological rules of a language may produce false negatives due to the ambiguity resulting from the presence of numerous homophonic words. In this work, we propose an approach to detect and correct the {}de/da{''} clitic errors in Turkish text. Our model is a neural sequence tagger trained with a synthetically constructed dataset consisting of positive and negative samples. The model{'}s performance with this dataset is presented according to different word embedding configurations. The model achieved an F1 score of 86.67{\%} on a synthetically constructed dataset. We also compared the model{'}s performance on a manually curated dataset of challenging samples that proved superior to other spelling correctors with 71{\%} accuracy compared to the second-best (Google Docs) with and accuracy of 34{\%}. read more
PDF Abstract
## Code Add Remove Mark official
No code implementations yet. Submit your code now
## Datasets
Add Datasets introduced or used in this paper
## Results from the Paper Add Remove
Submit results from this paper to get state-of-the-art GitHub badges and help the community compare results to other papers.
## Methods Add Remove
No methods listed for this paper. Add relevant methods here | 2021-12-04 23:44:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19026516377925873, "perplexity": 3845.1642694567176}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00059.warc.gz"} |
https://en.wikipedia.org/wiki/Hash-based_message_authentication_code | # HMAC
HMAC-SHA1 generation
In cryptography, an HMAC (sometimes expanded as either keyed-hash message authentication code or hash-based message authentication code) is a specific type of message authentication code (MAC) involving a cryptographic hash function and a secret cryptographic key. As with any MAC, it may be used to simultaneously verify both the data integrity and the authenticity of a message.
HMAC can provide message authentication using a shared secret instead of using digital signatures with asymmetric cryptography. It trades off the need for a complex public key infrastructure by delegating the key exchange to the communicating parties, who are responsible for establishing and using a trusted channel to agree on the key prior to communication.[1]
## Details
Any cryptographic hash function, such as SHA-2 or SHA-3, may be used in the calculation of an HMAC; the resulting MAC algorithm is termed HMAC-X, where X is the hash function used (e.g. HMAC-SHA256 or HMAC-SHA3-256). The cryptographic strength of the HMAC depends upon the cryptographic strength of the underlying hash function, the size of its hash output, and the size and quality of the key.
HMAC uses two passes of hash computation. The secret key is first used to derive two keys – inner and outer. The first pass of the algorithm produces an internal hash derived from the message and the inner key. The second pass produces the final HMAC code derived from the inner hash result and the outer key. Thus the algorithm provides better immunity against length extension attacks.
An iterative hash function breaks up a message into blocks of a fixed size and iterates over them with a compression function. For example, SHA-256 operates on 512-bit blocks. The size of the output of HMAC is the same as that of the underlying hash function (e.g., 256 and 512 bits in the case of SHA-256 and SHA-512, respectively), although it can be truncated if desired.
HMAC does not encrypt the message. Instead, the message (encrypted or not) must be sent alongside the HMAC hash. Parties with the secret key will hash the message again themselves, and if it is authentic, the received and computed hashes will match.
The definition and analysis of the HMAC construction was first published in 1996 in a paper by Mihir Bellare, Ran Canetti, and Hugo Krawczyk,[2] and they also wrote RFC 2104 in 1997. The 1996 paper also defined a nested variant called NMAC. FIPS PUB 198 generalizes and standardizes the use of HMACs. HMAC is used within the IPsec, SSH and TLS protocols and for JSON Web Tokens.
## Definition
This definition is taken from RFC 2104:
{\displaystyle {\begin{aligned}\operatorname {HMAC} (K,m)&=\operatorname {H} {\Bigl (}{\bigl (}K'\oplus opad{\bigr )}\parallel \operatorname {H} {\bigl (}\left(K'\oplus ipad\right)\parallel m{\bigr )}{\Bigr )}\\K'&={\begin{cases}\operatorname {H} \left(K\right)&K{\text{ is larger than block size}}\\K&{\text{otherwise}}\end{cases}}\end{aligned}}}
where
H is a cryptographic hash function
m is the message to be authenticated
K is the secret key
K' is a block-sized key derived from the secret key, K; either by padding to the right with 0s up to the block size, or by hashing down to less than or equal to the block size first and then padding to the right with zeros
‖ denotes concatenation
⊕ denotes bitwise exclusive or (XOR)
opad is the block-sized outer padding, consisting of repeated bytes valued 0x5c
ipad is the block-sized inner padding, consisting of repeated bytes valued 0x36
## Implementation
The following pseudocode demonstrates how HMAC may be implemented. Blocksize is 512 bits (64 bytes) when using one of the following hash functions: SHA-1, MD5, RIPEMD-128.[3]
function hmac is
input:
key: Bytes // Array on bytes
message: Bytes // Array of bytes to be hashed
hash: Function // The hash function to use (e.g. SHA-1)
blockSize: Integer // The block size of the hash function (e.g. 64 bytes for SHA-1)
outputSize: Integer // The output size of the hash function (e.g. 20 bytes for SHA-1)
// Keys longer than blockSize are shortened by hashing them
if (length(key) > blockSize) then
key ← hash(key) // key is outputSize bytes long
// Keys shorter than blockSize are padded to blockSize by padding with zeros on the right
if (length(key) < blockSize) then
key ← Pad(key, blockSize) // Pad key with zeros to make it blockSize bytes long
## Design principles
The design of the HMAC specification was motivated by the existence of attacks on more trivial mechanisms for combining a key with a hash function. For example, one might assume the same security that HMAC provides could be achieved with MAC = H(keymessage). However, this method suffers from a serious flaw: with most hash functions, it is easy to append data to the message without knowing the key and obtain another valid MAC ("length-extension attack"). The alternative, appending the key using MAC = H(messagekey), suffers from the problem that an attacker who can find a collision in the (unkeyed) hash function has a collision in the MAC (as two messages m1 and m2 yielding the same hash will provide the same start condition to the hash function before the appended key is hashed, hence the final hash will be the same). Using MAC = H(keymessagekey) is better, but various security papers have suggested vulnerabilities with this approach, even when two different keys are used.[2][4][5]
No known extension attacks have been found against the current HMAC specification which is defined as H(keyH(keymessage)) because the outer application of the hash function masks the intermediate result of the internal hash. The values of ipad and opad are not critical to the security of the algorithm, but were defined in such a way to have a large Hamming distance from each other and so the inner and outer keys will have fewer bits in common. The security reduction of HMAC does require them to be different in at least one bit.[citation needed]
The Keccak hash function, that was selected by NIST as the SHA-3 competition winner, doesn't need this nested approach and can be used to generate a MAC by simply prepending the key to the message, as it is not susceptible to length-extension attacks.[6]
## Security
The cryptographic strength of the HMAC depends upon the size of the secret key that is used. The most common attack against HMACs is brute force to uncover the secret key. HMACs are substantially less affected by collisions than their underlying hashing algorithms alone.[7][8] In particular, in 2006 Mihir Bellare proved that HMAC is a PRF under the sole assumption that the compression function is a PRF.[9] Therefore, HMAC-MD5 does not suffer from the same weaknesses that have been found in MD5.
RFC 2104 requires that “keys longer than B bytes are first hashed using H” which leads to a confusing pseudo-collision: if the key is longer than the hash block size (e.g. 64 bytes for SHA-1), then HMAC(k, m) is computed as HMAC(H(k), m).This property is sometimes raised as a possible weakness of HMAC in password-hashing scenarios: it has been demonstrated that it's possible to find a long ASCII string and a random value whose hash will be also an ASCII string, and both values will produce the same HMAC output.[10][11]
In 2006, Jongsung Kim, Alex Biryukov, Bart Preneel, and Seokhie Hong showed how to distinguish HMAC with reduced versions of MD5 and SHA-1 or full versions of HAVAL, MD4, and SHA-0 from a random function or HMAC with a random function. Differential distinguishers allow an attacker to devise a forgery attack on HMAC. Furthermore, differential and rectangle distinguishers can lead to second-preimage attacks. HMAC with the full version of MD4 can be forged with this knowledge. These attacks do not contradict the security proof of HMAC, but provide insight into HMAC based on existing cryptographic hash functions.[12]
In 2009, Xiaoyun Wang et al. presented a distinguishing attack on HMAC-MD5 without using related keys. It can distinguish an instantiation of HMAC with MD5 from an instantiation with a random function with 297 queries with probability 0.87.[13]
In 2011 an informational RFC 6151[14] was published to summarize security considerations in MD5 and HMAC-MD5. For HMAC-MD5 the RFC summarizes that – although the security of the MD5 hash function itself is severely compromised – the currently known "attacks on HMAC-MD5 do not seem to indicate a practical vulnerability when used as a message authentication code", but it also adds that "for a new protocol design, a ciphersuite with HMAC-MD5 should not be included".
In May 2011, RFC 6234 was published detailing the abstract theory and source code for SHA-based HMACs.
## Examples
Here are some non-empty HMAC values, assuming 8-bit ASCII or UTF-8 encoding:
HMAC_MD5("key", "The quick brown fox jumps over the lazy dog") = 80070713463e7749b90c2dc24911e275
HMAC_SHA1("key", "The quick brown fox jumps over the lazy dog") = de7c9b85b8b78aa6bc8a7a36f70a90701c9db4d9
HMAC_SHA256("key", "The quick brown fox jumps over the lazy dog") = f7bc83f430538424b13298e6aa6fb143ef4d59a14946175997479dbc2d1a3cd8
## References
1. ^ "How and when do I use HMAC?". Security Stack Exchange. April 2015. Retrieved 25 November 2020.
2. ^ a b Bellare, Mihir; Canetti, Ran; Krawczyk, Hugo (1996). "Keying Hash Functions for Message Authentication": 1–15. CiteSeerX 10.1.1.134.8430. Cite journal requires |journal= (help)
3. ^
4. ^ Preneel, Bart; van Oorschot, Paul C. (1995). "MDx-MAC and Building Fast MACs from Hash Functions". CiteSeerX 10.1.1.34.3855. Cite journal requires |journal= (help)
5. ^ Preneel, Bart; van Oorschot, Paul C. (1995). "On the Security of Two MAC Algorithms". CiteSeerX 10.1.1.42.8908. Cite journal requires |journal= (help)
6. ^ Keccak team. "Keccak Team – Design and security". Retrieved 31 October 2019. Unlike SHA-1 and SHA-2, Keccak does not have the length-extension weakness, hence does not need the HMAC nested construction. Instead, MAC computation can be performed by simply prepending the message with the key.
7. ^ Bruce Schneier (August 2005). "SHA-1 Broken". Retrieved 9 January 2009. although it doesn't affect applications such as HMAC where collisions aren't important
8. ^ IETF (February 1997). "Security". HMAC: Keyed-Hashing for Message Authentication. sec. 6. doi:10.17487/RFC2104. RFC 2104. Retrieved 3 December 2009. The strongest attack known against HMAC is based on the frequency of collisions for the hash function H ("birthday attack") [PV,BCK2], and is totally impractical for minimally reasonable hash functions.
9. ^ Bellare, Mihir (June 2006). "New Proofs for NMAC and HMAC: Security without Collision-Resistance". In Dwork, Cynthia (ed.). Advances in Cryptology – Crypto 2006 Proceedings. Lecture Notes in Computer Science 4117. Springer-Verlag. Retrieved 25 May 2010. This paper proves that HMAC is a PRF under the sole assumption that the compression function is a PRF. This recovers a proof based guarantee since no known attacks compromise the pseudorandomness of the compression function, and it also helps explain the resistance-to-attack that HMAC has shown even when implemented with hash functions whose (weak) collision resistance is compromised.
10. ^ "PBKDF2+HMAC hash collisions explained · Mathias Bynens". mathiasbynens.be. Retrieved 7 August 2019.
11. ^ "Aaron Toponce : Breaking HMAC". Retrieved 7 August 2019.
12. ^ Jongsung, Kim; Biryukov, Alex; Preneel, Bart; Hong, Seokhie (2006). "On the Security of HMAC and NMAC Based on HAVAL, MD4, MD5, SHA-0 and SHA-1" (PDF). Cite journal requires |journal= (help)
13. ^ Wang, Xiaoyun; Yu, Hongbo; Wang, Wei; Zhang, Haina; Zhan, Tao (2009). "Cryptanalysis on HMAC/NMAC-MD5 and MD5-MAC" (PDF). Retrieved 15 June 2015. Cite journal requires |journal= (help)
14. ^ "RFC 6151 – Updated Security Considerations for the MD5 Message-Digest and the HMAC-MD5 Algorithms". Internet Engineering Task Force. March 2011. Retrieved 15 June 2015.
Notes
• Mihir Bellare, Ran Canetti and Hugo Krawczyk, Keying Hash Functions for Message Authentication, CRYPTO 1996, pp. 1–15 (PS or PDF).
• Mihir Bellare, Ran Canetti and Hugo Krawczyk, Message authentication using hash functions: The HMAC construction, CryptoBytes 2(1), Spring 1996 (PS or PDF). | 2021-05-12 09:36:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.34672197699546814, "perplexity": 4916.31843085557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991685.16/warc/CC-MAIN-20210512070028-20210512100028-00473.warc.gz"} |
http://physics.stackexchange.com/questions/108258/why-does-the-induced-emf-oppose-the-change-in-magnetic-flux-lenzs-law-question | # Why does the induced EMF oppose the change in magnetic flux? Lenz's Law question
Can anyone explain to me why the induced magnetic field will oppose the change in magnetic flux? Is it an energy thing?
I know that the induced emf is $$emf= - \frac{d\phi}{dt}$$ but my book doesn't give a satisfactory (or any) explanation as to why this is the case. I recognize that an emf source supplies energy and does work on charge carriers, but I dont see what that has to do with the magnetic flux.
-
Yes a "conservation of energy thing". The wiki explains this quite well I think. en.wikipedia.org/wiki/Lenz%27s_law Also there is a good MIT demo video video.mit.edu/watch/… If you have a more specific question you can edit/ask them. – user6972 Apr 13 at 19:46
– Larry Harson Apr 13 at 22:44
Simply consider Maxwell equation :
$$\vec{\nabla}\wedge\vec{E}=-\frac{\partial\vec{B}}{\partial t}$$
If you interger this on a given closed surface $\Sigma$, it follows :
$$\oint_\Sigma \left(\vec{\nabla}\wedge\vec{E}\right) \cdot d\vec{S} =-\frac{\partial}{\partial t}\oint_\Sigma \vec{B}\cdot d\vec{S}$$
where $d\vec{S}=dS\,.\vec{n}$ with $dS$ the differential element of the surface $\Sigma$ and $\vec{n}$ the local normal direction of $\Sigma$, centred on $dS$. Here, $\phi=\oint_\Sigma \vec{B}\cdot d\vec{S}$ stands for the magnetic flux through the surface $\Sigma$.
At this point, Stockes theorem gives you :
$$\oint_\Sigma \left(\vec{\nabla}\wedge\vec{E}\right) \cdot d\vec{S}=\oint_\Gamma \vec{E}\cdot d\vec{l}$$ where $\Gamma$ is a given contour included in $\Sigma$, and $d\vec{l}=dl.\vec{r}$ is the differential length element along $\Gamma$. What is called emf is simply $e=\oint_\Gamma \vec{E}\cdot d\vec{l}$. Then directly follows :
$$e=-\frac{d\phi}{dt}$$
If you want to interprete the $-$ signe, keep in mind that it is linked magnetic flux conservation that follows a moderation law : "Effects are in opposition with their causes".
-
Maybe you have seen the experiment where a permanent magnet is dropped through a pipe made of conducting metal. The magnetic flux through a cross-section of the pipe will be changing, so a current is induced. The induced magnetic field is such as to oppose the change in magnetic flux, so it will slow down the falling magnet. Eventually an equilibrium is reached where the magnetic force and the gravitational force balance, and the magnet falls at constant velocity until the end of the pipe. You can find many demonstrations on YouTube by searching for "magnet in pipe" or something similar.
Now imagine if it were the opposite way, that is, the induced magnetic field was in the opposite direction, as to enhance the change in magnetic flux. Then the magnet would be accelerated, so the induced current would grow, accelerating the magnet more, and so on. This is absurd from conservation of energy! Not only would the kinetic energy of the magnet increase without a source, the pipe would get hotter and hotter from the increasing current.
(In the first case, the one that is realized in nature, the pipe does get heated. Instead of the magnets gravitational potential energy going to kinetic energy, it becomes heat. So conservation of energy is respected.)
-
That is the video I linked in the comment to his question. – user6972 Apr 13 at 23:39 | 2014-09-22 04:37:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6954094767570496, "perplexity": 161.9171172680275}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657136545.14/warc/CC-MAIN-20140914011216-00176-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://www.kofastudy.com/courses/ss3-maths-2nd-term/lessons/differentiation-of-algebraic-expressions-week-5-6/topic/differentiation-from-first-principle/ | Back to Course
0% Complete
0/0 Steps
• ## Do you like this content?
Lesson 5, Topic 2
In Progress
# Differentiation from First Principle
Lesson Progress
0% Complete
In general if, y =f(x) then f'(x) = $$\frac{dy}{dx}$$ is the limiting value of $$\frac{f(x + \delta x) – f(x)}{\delta x}$$
where $$\scriptsize \delta x \rightarrow 0$$
We assume that this limiting value will exist and can be found.
We write$$\frac{dy}{dx} = \scriptsize \lim\limits_{\delta x \to 0} \normalsize \frac{f(x + \delta x) – f(x)}{\delta x}$$
where $$\scriptsize \lim\limits_{\delta x \to 0}$$ means the limiting value as $$\scriptsize \delta x \rightarrow 0$$
Example:
Differentiate $$\scriptsize 6x^2 – 1$$ with respect to x from first principles.
We write $$\scriptsize 6x^2 – 1$$
Working from first principle, any increase in x to x + δx produces a corresponding increase in y to y + δy
Then $$\scriptsize y + \delta y = 6(x + \delta x)^2 – 1$$
$$\scriptsize \delta y = 6(x + \delta x)^2 – 1 – y$$
$$\scriptsize \delta y = 6[x^2 + 2x \delta x + (\delta x)^2] – 1 – (6x^2 – 1)$$
$$\scriptsize \delta y = 6x^2 + 12x \delta x + 6(\delta x)^2 – 1 – 6x^2 + 1$$
$$\scriptsize \delta y = 12x \delta x + 6(\delta x)^2$$
Divide through by δx, then
$$\frac{ \delta y }{\delta x} =\scriptsize 12x + 6(\delta x)$$
Now consider $$\scriptsize \delta x \rightarrow 0$$ and so
$$\frac{dy }{dx} = \scriptsize \lim\limits_{\delta x \to 0} \normalsize\frac{ \delta y }{\delta x} = \scriptsize 12x + 6(\delta x)$$
$$\frac{dy }{dx} = \scriptsize 12x$$
error: | 2022-01-20 17:31:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6323221325874329, "perplexity": 2762.546248963813}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320302355.97/warc/CC-MAIN-20220120160411-20220120190411-00278.warc.gz"} |
https://stats.stackexchange.com/questions/232718/engineering-features-using-pandas/232732 | # Engineering features using pandas
I have a dataframe with around 37,000 rows and 54 columns. Out of these 54 columns two columns namely 'user_id' and 'mail_id' are provided in avery creepy format as shown below:
user_id mail_id
AR+tMy3H/E+Re8Id20zUIz+amJkv6KU12o+BrgIDin0= DQ/4I+GIOz2ZoIiK0Lg0AkwnI35XotghgUK/MYc101I=
1P4AOvdzJzhDSHi7jJ3udWv4ajpKxOn4T/rCLv4PrXU= BL3z4RtiyfIDydaRYWX2ZXL6IX10QH1yG5ak1s/8Lls=
OEfFUcsTAGInCfsHuLZuIgdSNtuNsg8EdfN98VUZVTs= BL3z4RtiyfIDydaRYWX2ZXL6IX10QH1yG5ak1s/8Lls=
1P4AOvdzJzhDSHi7jJ3udWv4ajpKxOn4T/rCLv4PrXU= EHNBRbi6i9KO6cMHsuDPFjZVp2cY3RH+BiOKwPwzLQs=
CYRcuV0cR0algMZJ1N6+3uKcqi8iu+6tJNzmBbmgN7o= K0y/NW59TJkYc5y0HUwDeAXrewYT0JQlkcozz0s2V5Q=
After a detailed analysis of my data I figured out that I cannot drop these two columns from my dataframe as they are too importanct for prediction. I can hash these two features but there is one more interesting thing. There are only 2,000 types of user_ids and mail_ids. So doing one hot encoding can help a lot. My question is that if I convert this into one hot encoding using 'get_dummies' method in pandas with sparse=True, will it be memory efficient or is there any other efficient way to do it?
That "creepy" format is just a form of anonymization - in Python you can use base64 lib to b64encode('hi my name is derek') and get aGkgbXkgbmFtZSBpcyBkZXJlaw== as output. You'll notice the similarity to the above.
When I use hashlib and do base64.b64encode(hashlib.sha1('derek').hexdigest()). I get my name hashed and encoded as b64 - likely what you have above. Might be fun to experiment and see if you can b64.decode(user_name) and get anything useful out of it (unlikely since SHA1 and other popular hashes are one-way).
Yeah, you can hash those together and use pandas.get_dummies if you like. I usually use sklearn for this type of thing, and I like to work within that ecosystem more than with pandas. Either will be equal from a memory standpoint - both implementations use the sparse=True param to indicate that they want to use a numpy sparse matrix instead of a full featured numpy array under the hood. | 2021-07-28 00:44:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28113970160484314, "perplexity": 1689.2259891491333}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153515.0/warc/CC-MAIN-20210727233849-20210728023849-00284.warc.gz"} |
http://narimanfarsad.com/ | ### Latest News (Scroll Down For More)
• Feb 2017
Tutorial at WCNC
Chris Rose and I gave a tutorial at WCNC 2017 on molecular communication. The slides for my portion of the talk can be downloaded from here.
• Feb 2017
Patent Filed
I am the lead inventor on a provisional patent, which presents machine learning and deep learning detection algorithms for communication systems.
• Jan 2017
Journal Papers Accepted and Submitted
Our journal paper was acceped for publication at Elsevier Nano Communication Networks. We had another journal paper submitted to IEEE Access.
• Dec 2016
MIT Visit
I spent a week at Muriel Médard's reserach group at MIT working on one of the three ISIT 2017 submissions for which I am a co-author.
• Nov 2016
Media Coverage
My new experimental platform that uses household chemicals as carries of information has been covered by Stanford News.
• Nov 2016
Conference Presentation at Asilomar
I will present our paper "On the Capacity of Diffusion-Based Molecular Timing Channels with Diversity" at Asilomar Conference on Signals, Systems, and Computers.
• Sep 2016
Conference Presentation at ACM NANOCOM
Vahid Jamali presented our paper "Non-Coherent Multiple-Symbol Detection for Diffusive Molecular Communications" at ACM International Conference on Nanoscale Computing and Communication (NanoCom).
• Sep 2016
Conference Presentation at ACM NANOCOM
Yonathan Morin presented our paper "On Time-Slotted Communication over Molecular Timing Channels" at ACM International Conference on Nanoscale Computing and Communication (NanoCom).
• July 2016
News Coverage by IEEE Spectrum
The MIMO molecular communication platform we designed and built is covered by IEEE Spectrum.
• July 2016
Conference Presentation at ISIT
I presented our paper "On the Capacity of Diffusion-Based Molecular Timing Channels" at IEEE International Symposium on Information Theory (ISIT).
• July 2016
Conference Presentation at SPAWC
I presented our paper "A Molecular Communication System Using Acids, Bases and Hydrogen Ions" at IEEE International workshop on Signal Processing advances in Wireless Communications (SPAWC).
• June 2016
Invited Talk at University of British Columbia
I gave a talk titled "Molecular Communication: Theoretical Limits and Experimental Implementations" at University of British Columbia hosted by Victor Leung.
• June 2016
Invited Conference Talk at BSC
I gave a talk titled "Capacity Limits of Diffusion-Based Molecular Timing Channels" at the Canadian Biennial Symposium on Communications (BSC).
• May 2016
Conference Presentation at ICC
Birkan Yilmaz presented our paper titled "Energy Model for Vesicle-Based Active Transport Molecular Communication" at IEEE International Conference on Communications (ICC).
• May 2016
Invited Talk at Georgia Institute of Technology
I gave a talk titled "Molecular Communication: Theoretical Limits and Experimental Implementations" at Georgia Institute of Technology hosted by Faramaz Fekri.
• May 2016
Invited Talk at Carnegie Mellon University
I gave a talk titled "Molecular Communication: Theoretical Limits and Experimental Implementations" at Carnegie Mellon University hosted by Pulkit Grover.
• May 2016
Invited Talk at Boston University
I gave a talk titled "Molecular Communication: Theoretical Limits and Experimental Implementations" at Boston University hosted by Bobak Nazer.
• May 2016
Invited Talk at Massachusetts Institute of Technology
I gave a talk titled "Molecular Communication: Theoretical Limits and Experimental Implementations" at Massachusetts Institute of Technology hosted by Muriel Médard.
• April 2016
Invited Talk organized by IEEE Toronto Section at University of Toronto
I gave a talk titled "Molecular Communication: Theoretical Limits and Experimental Implementations" organized by IEEE Toronto Section at University of Toronto.
• April 2016
Invited Talk at Princton University
I gave a titled "Molecular Communication: Theoretical Limits and Experimental Implementations" at Princton University hosted by Vincent Poor.
• Feb 2016
Invited Workshop Presentation at ITA
I presented our work on "Capacity Limits of Molecular Timing Channels" at Information Theory and Applications (ITA) Workshop.
• Feb 2016
Journal Submission to IEEE T-IT
I submitted our paper "Capacity Limits of Diffusion-Based Molecular Timing Channels" to IEEE Transactions on Information Theory.
• Dec 2015
Conference Presentation at GLOBECOM
I presented our paper "Stable Distributions as Noise Models for Molecular Communication" at IEEE Global Communication Conference (GLOBCOM).
• Dec 2015
Invited Workshop Presentation at University of Southern California
I presented my work on "Molecular Communication using Acids and Bases" at USC's Communications, Inference, And Computing In Molecular And Biological Systems Workshop.
### Filter by type:
#### An Experimentally Validated Channel Model for Molecular Communication Systems
N. Kim, N. Farsad, A. W. Eckford, and C.-B. Chae
Journal Paper 19IEEE Access, submitted, 2017.
#### Abstract
In this paper, we present an experimentally validated end-to-end channel model for molecular communication systems with metal-oxide sensors. In particular, we focus on the recently developed tabletop molecular communication platform. Unlike previous work, this work separates the system into two parts-- the propagation and the sensing. Based on this separation, a more realistic channel model is derived. The coefficients in the derived models are estimated using a large collection of experimental data and it is shown how the coefficients change as a function of different system parameters such as distance, spraying duration, and initial condition. Finally, a noise model is derived for the system to complete an end-to-end system model for the tabletop platform that can be utilized with various system variables. Using this new channel model, we propose a multi-level modulation technique that represents different symbols with different spraying durations while still providing more feasibility and less computational complexity in practice.
#### Capacity Limits of Diffusion-Based Molecular Timing Channels
N. Farsad, Y. Murin, A. W. Eckford, and A. Goldsmith
Journal Paper 18IEEE Transactions on Information Theory, submitted, 2017.
#### Abstract
This work introduces capacity limits for molecular timing (MT) channels, where information is modulated in the release timing of small information particles, and decoded from the time of arrivals at the receiver. It is shown that the random time of arrival can be represented as an additive noise channel, and for the diffusion-based MT (DBMT) channel this noise is distributed according to the Levy distribution. Lower and upper bounds on the capacity of the DBMT channel are derived for the case where the delay associated with the propagation of the information particles in the channel is finite, namely, when the information particles dissipate after a finite time interval. For the case where a single particle is released per channel use, these bounds are shown to be tight. When the transmitter simultaneously releases a large number of particles, the detector at the receiver may not be able to precisely detect the arrival time of all the particles. Therefore, two alternative models are considered: detection based on the particle that arrives first, or detection based on the average arrival times. Tight lower and upper bounds are derived for these two models. It is shown that by controlling the maximal delay of the information particles, the capacity can increase poly-logarithmically with the number of released particles. As each particle takes a random independent path, this diversity of paths is analogous to receiver diversity and can be used to considerably increase the achievable data rates.
#### Time-Slotted Transmission over Molecular Timing Channels
Y. Murin,N. Farsad, M. Chowdhurya, and A. Goldsmith
Journal Paper 17Elsevier Nano Communication Networks, accepted, 2017.
#### Abstract
This work studies time-slotted communication over molecular timing (MT) channels. The transmitter, assumed to be perfectly synchronized in time with the receiver, is required to send K bits to the receiver using K information particles. It releases a single information particle in each time-slot, where the information is encoded in the time of release. The receiver decodes the transmitted information based on the random time of arrivals of the information particles during a finite-time observation window. The maximum-likelihood (ML) detector is derived in terms of the permanent of a matrix involving the arrival times, and shown to to have an exponential computational complexity, thus, rendering it impractical. Therefore, two additional (practical) detectors are presented: The first is a symbol-by-symbol detector. The second is a sequence detector which is based on the Viterbi algorithm (VA), yet, the VA is used differently than in its common application in electromagnetic communications where the channels are linear. Numerical simulations indicate that the proposed sequence detection algorithm significantly improves the performance compared to the symbol-by-symbol detector. For a short number of transmitted symbols, the numerical results indicate that the performance of the proposed sequence detector closely approaches the performance of the highly complicated ML detector. Finally, the proposed sequence detector is numerically compared with a one-shot transmission scheme that releases all K particles simultaneously to send a single symbol out of a constellation of size 2^K. It is shown that while for a small number of bits the one-shot scheme is better, when the number of bits is medium to large, the sequence detector achieves significantly better performance.
#### SMIET: Simultaneous Molecular Information and Energy Transfer
W. Guo, Y. Deng, B. H. Yilmaz, N. Farsad, M. Elkashlan, C.-B. Chae, A. Eckford, and A. Nallanathan
Journal Paper 16IEEE Wireless Communications, under revision to address reviewers’ comments, 2016.
#### Abstract
The performance of communication systems is fundamentally limited by the loss of energy through propagation and circuit inefficiencies. The emergence of Internet of Nano Things ecosystem means there is need to design and build nanoscale energy efficient communication subsystems. In this article, we show that it is possible to achieve ultra low energy communications at the nanoscale, if diffusive molecules are used for carrying data. While the energy of electromagnetic waves will inevitably decays as a function of transmission distance and time, the energy in individual molecules does not. Over time, the receiver has an opportunity to recover some, if not all of the molecular energy transmitted. The article demonstrates the potential of ultra-low energy simultaneous molecular information and energy transfer (SMIET) through the design of two different nano-relay systems. It also discusses the benefits of crowd energy harvesting compared to traditional wave-based systems.
#### A Comprehensive Survey of Recent Advancements in Molecular Communication
N. Farsad, H. B. Yilmaz, A. W. Eckford, C.-B. Chae, and W. Guo
Journal Paper 15IEEE Communications Surveys & Tutorials, Volume 18, Issue 3, Pages 1887-1919, 2016.
#### Abstract
In molecular communication, information is conveyed through chemical messages. With much advancement in the field of nanotechnology, bioengineering and synthetic biology over the past decade, micro- and nano-scales devices are becoming a reality. Yet the problem of engineering a reliable communication system between tiny devices is still an open problem. At the same time, despite the prevalence of radio communication, there are still areas where traditional electromagnetic waves find it difficult or expensive to reach. Points of interest in industry, cities, medical, and military applications often lie in embedded and entrenched areas, accessible only by ventricles at scales too small for conventional radio- and micro-waves, or they are located in such a way that directional high frequency systems are ineffective. Molecular communication is a biologically inspired communication scheme that could be employed for solving these problems. Although biologists have studied molecular communication, it is poorly understood from a telecommunication perspective. In this paper, we highlight the recent advancements in the field of molecular communication engineering.
#### Molecular Communications: Channel Model and Physical Layer Techniques
W. Guo, T. Asyhari, N. Farsad, H. B. Yilmaz, B. Li, A. Eckford, C.-B. Chae
Journal Paper 14IEEE Wireless Communications, Volume 23, Number 4, Pages 120-127, 2016.
#### Abstract
This article examines recent research in molecular communications from a telecommunications system design perspective. In particular, it focuses on channel models and state-of-the-art physical layer techniques. The goal is to provide a foundation for higher layer research and motivation for research and development of functional prototypes. In the first part of the article, we focus on the channel and noise model, comparing molecular and radio-wave pathloss formulae. In the second part, the article examines, equipped with the appropriate channel knowledge, the design of appropriate modulation and error correction coding schemes. The third reviews transmitter and receiver side signal processing methods that suppress inter-symbol-interference. Taken together, the three parts present a series of physical layer techniques that are necessary to producing reliable and practical molecular communications.
#### Molecular MIMO: From Theory to Prototype
B. Koo, C. Lee, H. B. Yilmaz, N. Farsad, A. W. Eckford, and C.-B. Chae
Journal Paper 13IEEE Journal on Selected Areas in Communication, Volume 34, Number 3, Pages 600-614, 2016.
#### Abstract
In diffusion-based molecular communications, the channel is governed by diffusion through a fluid medium, which leads to extremely low data rates compared to the radio frequency communication system. To mitigate this problem, we propose a novel design for molecular communication that utilizes multiple bulges (in RF communication this corresponds to antenna) at both the transmitter and molecular detectors at the receiver. We simulate the system with a one-shot signal to obtain the channel's finite impulse response. We then incorporate this result within our mathematical analysis to determine inter-link interference (ILI) and inter-symbol interference (ISI). Low complexity symbol detection methods are needed for the cases of incomplete information regarding the system and the channel state, since the receiver is supposed to be a small and simple node. Thus we propose four detection algorithms, namely adaptive thresholding, practical zero forcing with channel models excluding/including the ILI and ISI, and Genie-aided zero forcing. We verify the proposed system via extensive numerical/analytical evaluations and a novel macro-scale testbed.
#### Molecular Versus Electromagnetic Wave Propagation Loss in Macro-Scale Environments
W. Guo, C. Mias, N. Farsad, and J.–L. Wu
Journal Paper 12IEEE Transactions on Molecular, Biological, and Multi-Scale Communications, Volume 1, Number 1, Pages 18–25, 2015.
#### Abstract
Molecular communications (MC) has been studied as a bio-inspired information carrier for micro-scale and nanoscale environments. On the macro-scale, it can also be considered as an alternative to electromagnetic (EM) wave based systems, especially in environments where there is significant attenuation to EM wave power. This paper goes beyond the unbounded free space propagation to examine three macro-scale environments: the pipe, the knife edge, and the mesh channel. Approximate analytical expressions shown in this paper demonstrate that MC has an advantage over EM wave communications when: (i) the EM frequency is below the cut-off frequency for the pipe channel, (ii) the EM wavelength is considerably larger than the mesh period, and (iii) when the receiver is in the high diffraction loss region of an obstacle.
#### Design and Optimizing of On-Chip Kinesin Substrates for Molecular Communication
N. Farsad, A. W. Eckford, and S. Hiyama
Journal Paper 11IEEE Transactions on Nanotechnology, Volume 14, Number 4, Pages 699–708, 2015.
#### Abstract
Lab-on-chip devices and point-of-care diagnostic chip devices are composed of many different components, such as nanosensors that must be able to communicate with other components within the device. Molecular communication is a promising solution for on-chip communication. In particular, kinesin driven microtubule motility is an effective means of transferring information particles from one component to another. However, finding an optimal shape for these channels can be challenging. In this paper, we derive a mathematical optimization model that can be used to find the optimal channel shape and dimensions for any transmission period. We derive three specific models for the rectangular channels, regular polygonal channels, and regular polygonal ring channels. We show that the optimal channel shapes are the square-shaped channel for the rectangular channel, and circular-shaped channel for the other classes of shapes. Finally, we show that among all 2-D shapes the optimal design choice that maximizes information rate is the circular-shaped channel.
#### Channel and Noise Models for Nonlinear Molecular Communication Systems
N. Farsad, N.-R. Kim, A. W. Eckford, and C.-B. Chae
Journal Paper 10IEEE Journal on Selected Areas in Communications, Volume 32, Number 12, Pages 2392–2401, 2014.
#### Abstract
Recently, a tabletop molecular communication platform has been developed for transmitting short text messages across a room. The end-to-end system impulse response for this platform does not follow previously published theoretical works because of imperfect receiver, transmitter, and turbulent flows. Moreover, it is observed that this platform resembles a nonlinear system, which makes the rich body of theoretical work that has been developed by communication engineers not applicable to this platform. In this work, we first introduce corrections to the previous theoretical models of the end-to-end system impulse response based on the observed data from experimentation. Using the corrected impulse response models, we then formulate the nonlinearity of the system as noise and show that through simplifying assumptions it can be represented as Gaussian noise. Through formulating the system's nonlinearity as the output a linear system corrupted by noise, the rich toolbox of mathematical models of communication systems, most of which are based on linearity assumption, can be applied to this platform.
#### A Markov Chain Channel Model for Active Transport Molecular Communication
N. Farsad, A. W. Eckford, and S. Hiyama
Journal Paper 9IEEE Transactions on Signal Processing, Volume 62, Number 9, Pages 2424–2436, 2014.
#### Abstract
In molecular communication, small particles such as molecules are used to convey information. These particles are released by a transmitter into a fluidic environment, where they propagate freely (e.g. through diffusion) or through externals means (e.g. different types of active transport) until they arrive at the receiver. Although there are a number of different mathematical models for the diffusion-based molecular communication, active transport molecular communication (ATMC) lacks the necessary theoretical framework. Previous works had to rely almost entirely on full Monte Carlo simulations of these systems. However, full simulations can be time consuming because of the computational complexities involved. In this paper, a Markov channel model has been presented, which could be used to reduce the amount of simulations necessary for studying ATMC without sacrificing accuracy. Moreover, a mathematical formula for calculating the transition probabilities in the Markov chain model is derived to complete our analytical framework. Comparing our proposed models with full simulations, it is shown that these models can be used to calculate parameters such channel capacity accurately in a timely manner.
#### Nanoparticle Communications: from Chemical Signals in Nature to Wireless Sensor Networks
S. Qiu, W. Guo, M. Leeson, S. Wang, N. Farsad, and A. W. Eckford
Journal Paper 8Nanotechnology Perceptions, Volume 10, Number 1, Pages 1–13, 2014.
#### Abstract
The need to convey information has always existed in both the animal and human kingdoms. The article offers a review of the latest developments in transporting information using nanosized particles. The article begins by examining the usage of chemical signalling in nature, and goes on to discuss the recent advances in mimicking this in bio-inspired engineering. The article then distinguishes the important difference between signalling and general communications, and explains why the latter is a more challenging problem. The article then goes on to examine existing research on mimicking chemical signalling in nature, which is a precurser to research in general chemical communications. A review of the latest theoretical research in general chemical communications is presented, along with the practical developments of the world’s first nanoparticle communications test-bed. In the end, the authors discuss the potential research challenges and name three important areas for future development: robustness, miniaturization, and scalability.
#### Tabletop Molecular Communication: Text Messages through Chemical Signals
N. Farsad, W. Guo, and A. W. Eckford
Journal Paper 7PLOS ONE, 2013.
#### Abstract
In this work, we describe the first modular, and programmable platform capable of transmitting a text message using chemical signalling – a method also known as molecular communication. This form of communication is attractive for applications where conventional wireless systems perform poorly, from nanotechnology to urban health monitoring. Using examples, we demonstrate the use of our platform as a testbed for molecular communication, and illustrate the features of these communication systems using experiments. By providing a simple and inexpensive means of performing experiments, our system fills an important gap in the molecular communication literature, where much current work is done in simulation with simplified system models. A key finding in this paper is that these systems are often nonlinear in practice, whereas current simulations and analysis often assume that the system is linear. However, as we show in this work, despite the nonlinearity, reliable communication is still possible. Furthermore, this work motivates future studies on more realistic modelling, analysis, and design of theoretical models and algorithms for these systems.
#### On-Chip Molecular Communication: Analysis and Design
N. Farsad, A. W. Eckford, S. Hiyama, and Y. Moritani
Journal Paper 6IEEE Transactions on NanoBioscience, Volume 11, Number 3, Pages 304–314, 2012.
#### Abstract
We consider a confined space molecular communication system, where molecules or information carrying particles are used to transfer information on a microfluidic chip. Considering that information-carrying particles can follow two main propagation schemes: passive transport, and active transport, it is not clear which achieves a better information transmission rate. Motivated by this problem, we compare and analyze both propagation schemes by deriving a set of analytical and mathematical tools to measure the achievable information rates of the on-chip molecular communication systems employing passive to active transport. We also use this toolbox to optimize design parameters such as the shape of the transmission area, to increase the information rate. Furthermore, the effect of separation distance between the transmitter and the receiver on information rate is examined under both propagation schemes, and a guidepost to design an optimal molecular communication setup and protocol is presented.
#### Resource Allocation via Linear Programming for Fractional Cooperation
N. Farsad, and A. W. Eckford
Journal Paper 5IEEE Transactions on Wireless Communications, Volume 11, Number 5, Pages 1633–1637, 2012.
#### Abstract
In this letter, resource allocation is considered for large multi-source, multi-relay networks employing fractional cooperation, in which each potential relay only allocates a fraction of its resources to relaying. Using a Gaussian approximation, it is shown that the optimization can be posed as a linear program, where the relays use a demodulate-and-forward (DemF) strategy, and where the transmissions are protected by low-density parity-check (LDPC) codes. This is useful since existing optimization schemes for this problem are nonconvex.
#### Quick System Design of Vesicle-Based Active Transport Molecular Communication by Using a Simple Transport Model
N. Farsad, A. W. Eckford, S. Hiyama, and Y. Moritani
Journal Paper 4Nano Communication Networks, Volume 2, Number 4, Pages 175–188, 2011.
#### Abstract
This paper will provide a guidepost to design an optimal molecular communication setup and protocol. A barrier to the design of vesicle-based molecular communication nanonetworks is the computational complexity of simulating them. In this paper, a computationally efficient transport model is presented, which could be employed to design active transport molecular communication systems, particularly to optimize the shape of the transmission zone. Furthermore, a vesicular encapsulation model is presented as an addition to the transport model, and it is shown that there exists an optimal vesicle size for each molecular communication channel. As an application, our transport model is used to estimate the channel capacity of a molecular communication nanonetwork in a computationally efficient manner compared to traditional Monte Carlo techniques. Moreover, it is shown that the derived optimal vesicle size maximizes channel capacity.
#### Bit Error Rates in Stable Distributed Molecular Timing Channels
N. Farsad, Y. Murin, W. Guo, C.-B. Chae, A. W. Eckford, and A. Goldsmith
Journal Paper 3IEEE Transactions on Signal Processing, in preparation for submission, 2017.
#### Abstract
This work studies modulation techniques for molecular timing (MT) channels. Three different modulation techniques are proposed: 1) Modulating information on the release timing of information particles, 2) Modulating information on the time between two consecutive releases of {\em indistinguishable} information particles, and 3) Modulating information on the time between two consecutive releases of {\em distinguishable} information particles. While the first modulation scheme requires transmitter-receiver synchronization, the latter two are asynchronous. We show that for all three modulation techniques the channel can be represented as an additive noise channel, where for diffusion-based MT (DBMT) channels the noise follows a stable distribution with characteristic exponent $1/2$. For DBMT channels, we provide expressions for the probability density function of the additive noise in terms of the Voigt functions, which can be numerically calculated efficiently. Next, we focus on the binary communication and derive the optimal detection rules for each modulation. To compare the performance of the different modulations, we first derive an expression for the geometric power of almost all stable distributions, and then use this results to obtain the geometric SNR (G-SNR) for each modulation scheme. Numerical evaluations indicate that the bit error rate is constant for a given G-SNR. Moreover, it is shown that synchronized communication in DBMT channels provides a significant performance gain. Yet, by using two {\em distinguishable} particles per bit instead of one, the probability of error using asynchronous communication in the third modulation technique can approach the probability of error obtained in synchronized communication of the first modulation scheme.
#### Non-Coherent Detection for Diffusive Molecular Communications,
V. Jamali, N. Farsad, R. Schober, and A. Goldsmith
Journal Paper 2IEEE Transactions on Communication, in preparation for submission, 2017.
#### Abstract
In diffusive molecular communication (MC), the channel state information (CSI) is defined as the probability that a molecule released by the transmitter is observed at the receiver as well as the expected number of the interfering molecules observed at the receiver. Most of the available works on MC assume that the CSI is perfectly known at the receiver for data detection, e.g., to determine the detection threshold. In contrast, in this paper, we study non-coherent detection schemes which do not require knowledge of the CSI. In particular, we first derive the optimal maximum likelihood (ML) multiple-symbol (MS) detector and the decision-feedback (DF) detector. As a special case of the optimal MS detector, we show that the optimal ML symbol-by-symbol (SS) detector can be equivalently written in the form of a threshold-based detector where the optimal threshold is constant and depends only on the statistics of the MC channel. Having stated that the complexity of the DF detector is lower than that of the optimal MS detector, the main challenge of both of these detectors is the complexity associated with the calculation of the their detection metrics. To cope with this issue, we propose approximated MS and DF detection metrics as well as a suboptimal blind detector with a complexity of much lower than the MS and DF detectors. Finally, we derive analytical expressions for the bit error rate (BER) of the optimal SS detector, and an upper bound and a lower bound for the BER of the optimal MS detector. Simulation results confirm our analyses and reveal the effectiveness of the proposed optimal and suboptimal detection schemes with respect to a baseline scheme which assumes perfect CSI knowledge, particularly when the number of observations used for detection is sufficiently large.
#### Optimal Detection for Diffusion-Based Molecular Timing Channels,
Y. Murin, N. Farsad, M. Chowdhury, and A. Goldsmith
Journal Paper 1IEEE Transactions on Signal Processing, in preparation for submission, 2017.
#### Abstract
This work studies optimal detection for communication over diffusion-based molecular timing (DBMT) channels. The transmitter simultaneously releases multiple small information particles, where the information is encoded in the time of release. The receiver decodes the transmitted information based on the random time of arrival of the information particles, which is represented as an additive noise channel. For a diffusion-based MT channel, without flow, this noise follows the L ́evy distribution. Under this channel model, the maximum-likelihood (ML) detector is derived and shown to have high computational complexity. It is further shown that for any additive channel with α-stable noise, α < 1, such as the DBMT channel, a linear receiver is not able to take advantage of the release of multiple information particles. Thus, instead of the common low complexity linear approach, a new detector, which is based on the first arrival (FA) among all the transmitted particles, is derived. It is shown that for small number of released particles the performance of the FA detector is very close to that of the ML detector. On the other hand, via error exponent analysis, it is shown that the performance of the two detectors differ when the number of released particles is large. Thus, in the regime of small to medium number of sent particles, the FA detector is an attractive alternative to the relatively complicated ML detector.
#### On the Impact of Time-Synchronization in Molecular Timing Channels
N. Farsad, Y. Murin, W. Guo, C.-B. Chae, A. Eckford and A. Goldsmith
Conference Paper 24IEEE Global Communications Conference (GLOBECOM), To be presented,2016.
#### Abstract
This work studies the impact of time-synchronization in molecular timing (MT) channels by analyzing three different modulation techniques. The first requires transmitter-receiver synchronization and is based on modulating information on the release timing of information particles. The other two are asynchronous and are based on modulating information on the relative time between two consecutive releases of information particles using indistinguishable or distinguishable particles. All modulation schemes result in a system that relate the transmitted and the received signals through an additive noise, which follows a stable distribution. As the common notion of the variance of a signal is not suitable for defining the power of stable distributed signals (due to infinite variance), we derive an expression for the geometric power of a large class of stable distributions, and then use this result to characterize the geometric signal-to-noise ratio (G-SNR) for each of the modulation techniques. In addition, for binary communication, we derive the optimal detection rules for each modulation technique. Numerical evaluations indicate that the bit error rate (BER) is constant for a given G-SNR, and the performance gain obtained by using synchronized communication is significant. Yet, it is also shown that by using two distinguishable particles per bit instead of one, the BER of the asynchronous technique can approach that of the synchronous one.
#### Communication over Diffusion-Based Molecular Timing Channels
Y. Murin, N. Farsad, M. Chowdhury, and A. Goldsmith
Conference Paper 23IEEE Global Communications Conference (GLOBECOM), To be presented,2016.
#### Abstract
This work studies communication over diffusion- based molecular timing (DBMT) channels. The transmitter si- multaneously releases multiple small information particles, where the information is encoded in the time of release. The receiver decodes the transmitted information based on the random time of arrival of the information particles, which is represented as an additive noise channel. For a DBMT channel, without flow, this noise follows the Lévy distribution. Under this channel model, the maximum-likelihood (ML) detector is derived and shown to have high computational complexity. It is further shown that for any additive noise channel with α-stable noise, α < 1, such as the DBMT channel, a linear receiver is not able to take advantage of the release of multiple information particles. Thus, instead of the common low complexity linear approach, a new detector, which is based on the first arrival (FA) among all the transmitted particles, is derived. Numerical simulations indicate that for a small to medium number of released particles, the performance of the FA detector is very close to the performance of the ML detector.
#### On the Capacity of Diffusion-Based Molecular Timing Channels with Diversity
N. Farsad, Y. Murin, M. Rao, and A. Goldsmith
Conference Paper 22Asilomar Conference on Signals, Systems, and Computers (Asilomar), 2016.
#### Abstract
This work introduces bounds on the capacity of molecular timing (MT) channels, where information is modulated on the release timing of multiple indistinguishable information particles, and decoded from the times of arrival at the receiver. It is shown that for diffusion-based MT channels, the capacity scales linearly in the number of particles. This is analogous to receiver diversity as each particle takes a random independent path. However, unlike receiver diversity in wireless channels, which mitigates fading, this form of diversity in MT channels can be used to significantly increase data rate.
#### Non-Coherent Multiple-Symbol Detection for Diffusive Molecular Communications
V. Jamali, N. Farsad, R. Schober, and A. Goldsmith
Conference Paper 21ACM International Conference on Nanoscale Computing and Communication (NANOCOM), Pages 1-7, 2016.
#### Abstract
Most of the available works on molecular communication (MC) assume that the channel state information (CSI) is perfectly known at the receiver for data detection. In contrast, in this paper, we study non-coherent multiple-symbol detection schemes which do not require knowledge of the CSI. In particular, we derive the optimal maximum likelihood (ML) multiple-symbol (MLMS) detector. Moreover, we propose an approximated detection metric and a sub-optimal detector to cope with the high complexity of the optimal MLMS detector. Numerical results reveal the effectiveness of the proposed optimal and suboptimal detection schemes with respect to a baseline scheme which assumes perfect CSI knowledge, particularly when the number of observations used for detection is sufficiently large.
#### On Time-Slotted Communication over Molecular Timing Channels
Y. Murin, N. Farsad, M. Chowdhury, and A. Goldsmith
Conference Paper 20ACM International Conference on Nanoscale Computing and Communication (NANOCOM), Pages 1-6, 2016.
#### Abstract
This work studies time-slotted communication over molecular timing (MT) channels. The transmitter, assumed to be perfectly synchronized in time with the receiver, releases a single information particle in each time slot, where the information is encoded in the time of release. The receiver decodes the transmitted information based on the random time of arrivals of the information particles during a finite-time reception window. The maximum-likelihood (ML) detector is derived and shown to have an exponential computational complexity, thus, rendering it impractical. In addition, two practical detectors are presented: The first is a symbol-by-symbol detector. The second is a sequence detector which is based on the Viterbi algorithm (VA), yet, the VA is used differently than in its common application in ML detection where information is transmitted over linear channels with memory. Numerical simulations indicate that the proposed sequence detection algorithm significantly improves the performance compared to the symbol-by-symbol detector. Furthermore, for a short number of transmitted symbols it closely approaches the highly complicated ML detector.
#### On the Capacity of Diffusion-Based Molecular Timing Channels
N. Farsad, Y. Murin, A. W. Eckford, and A. Goldsmith
Conference Paper 19IEEE International Symposium on Information Theory (ISIT), Pages 1023-1027, 2016.
#### Abstract
This work introduces capacity limits for molecular timing (MT) channels, where information is modulated on the release timing of small information particles, and decoded from the time of arrival at the receiver. It is shown that the random time of arrival can be represented as an additive noise channel, and for the diffusion-based MT (DBMT) channel this noise is distributed according to the Levy distribution. Lower and upper bounds on capacity of the DBMT channel are derived for the case where the delay associated with the propagation of information particles in the channel is finite. These bounds are also shown to be tight.
#### A Molecular Communication System Using Acids, Bases and Hydrogen Ions
Conference Paper 18 International Workshop on Signal Processing Advances in Wireless Communications (SPAWC), Pages 1-6, 2016.
#### Abstract
Concentration modulation, whereby information is encoded in the concentration level of chemicals, is considered. One of the main challenges with such systems is the limited control the transmitter has on the concentration level at the receiver. For example, concentration cannot be directly decreased by the transmitter, and the decrease in concentration over time occurs solely due to transport mechanisms such as diffusion. This can result in inter-symbol interference (ISI), which degrades performance. In this work, a novel scheme is proposed that uses the transmission of acids, bases, and the concentration of hydrogen ions for carrying information. By employing this technique, the concentration of hydrogen ions at the receiver can be both increased and decreased through the sender's transmissions. This enables ISI mitigation as well as the ability to form a wider array of signal patterns at the receiver.
#### Energy Model for Vesicle-Based Active Transport Molecular Communication
N. Farsad, H. B. Yilmaz, C.-B. Chae and A. Goldsmith
Conference Paper 17IEEE International Conference on Communications (ICC), accepted, Pages 1-6, 2016.
#### Abstract
In active transport molecular communication (ATMC), information particles are actively transported from a transmitter to a receiver using special proteins. Prior work has demonstrated that ATMC can be an attractive and viable solution for on-chip applications. The energy consumption of an ATMC system plays a central role in its design and engineering. In this work, an energy model is presented for ATMC and the model is used to provide guidelines for designing energy efficient systems. The channel capacity per unit energy is analyzed and maximized. It is shown that based on the size of the symbol set and the symbol duration, there is a vesicle size that maximizes rate per unit energy. It is also demonstrated that maximizing rate per unit energy yields very different system parameters compared to maximizing the rate only.
#### Stable Distributions as Noise Models for Molecular Communication
N. Farsad, W. Guo, C.-B. Chae,and A. W. Eckford
Conference Paper 16IEEE Global Communications Conference (GLOBCOM), Pages 1-6, 2015.
#### Abstract
In this work, we consider diffusion-based molecular communication timing channels. Three different timing channels are presented based on three different modulation techniques, i.e., i) modulation of the release timing of the information particles, ii) modulation on the time between two consecutive information particles of the same type, and iii) modulation on the time between two consecutive information particles of different types. We show that each channel can be represented as an additive noise channel, where the noise follows one of the subclasses of stable distributions. We provide expressions for the probability density function of the noise terms, and numerical evaluations for the probability density function and cumulative density function. We also show that the tails are longer than Gaussian distribution, as expected.
#### Molecular MIMO with Drift
C. Lee, B. Koo, N.-R. Kim, H. B. Yilmaz, N. Farsad, A. W. Eckford, and C.-B. Chae
Conference Paper 15Proceedings of the 21st Annual International Conference on Mobile Computing and Networking (MobiCom), Pages 201–203, 2015.
#### Abstract
In molecular communication information is transferred with the use of molecules. Molecular multiple-input multiple- output (MIMO) system with drift (positive velocity) at macro- scale will be presented and the improvement against single- input single-output (SISO) molecular communication systems will be verified via our testbed. Until now it was unclear whether MIMO techniques, which are extensively used in modern radio frequency (RF) communications, could be applied to molecular communication. In the demonstration, using our MIMO testbed we will show that we can achieve nearly 1.7 times higher data rate than SISO molecular communication systems. Moreover, signal-to-inter-link-interfeence metric for one-shot signal will be depicted for a given symbol duration.
#### Molecular Barcodes: Information Transmission via Persistent Chemical Tags
L. Wang, N. Farsad, W. Guo, S. Magierowski, and A. W. Eckford
Conference Paper 14Proceedings of IEEE International Conference on Communications (ICC), Pages 1097–1102, 2015.
#### Abstract
In molecular communication information is conveyed through chemical signals. In this work, we have considered a novel communication scheme where information is encoded in chemical barcodes, through use of persistent chemical tags. We have assumed that this information is already encoded in the environment, and we have devised a robotic platform for reading the chemical tag. We have performed many experiments to find the optimal encoding scheme and an algorithm for reading and decoding the chemically tagged information. We have demonstrated that chemical tags can be decoded using simple algorithms and inexpensive, off-the-shelf sensors. Finally, we have evaluated and presented the bit error rate performance of our devised algorithm.
#### Under-Water Molecular Signalling: a Hidden Transmitter and Absent Receivers Problem
S. Qiu, N. Farsad, T. Dong, A. W. Eckford, and W. Guo
Conference Paper 13Proceedings of IEEE International Conference on Communications (ICC), Pages 1085–1090, 2015.
#### Abstract
Wave-based signals have been successful in reliably and efficiently transferring data between two or more well defined points (e.g., known location area). However, it is challenged when the transmitter is hidden and the receivers are absent. Essentially, the transmitter and the receivers have no location knowledge of each other. We demonstrate that unlike wave-based transmissions, the total molecular energy doesn't monotonically degrade as a function of time. This paper uses a bio-inspired method of communicating data from a hidden transmitter to a group of absent receivers. A specialized molecular communication system is designed, including how to embed vital location information in the structure of a heterogeneous biochemical molecule. Like message in a bottle, there is a growing probability of receiving the location message over a period of several years. The only caveat is that there is an initial delay of a few hours to days, depending on the proximity of the rescue team to the crash site. This will provide an attractive alternative to current wave-based communications for delay-tolerant crash recovery.
#### A Universal Channel Model for Molecular Communication Systems with Metal-Oxide Detectors
N.-R. Kim, N. Farsad, C.-B. Chae, and A. W. Eckford
Conference Paper 12Proceedings of IEEE International Conference on Communications (ICC), Pages 1054–1059, 2015.
#### Abstract
In this paper, we propose an end-to-end channel model for molecular communication systems with metal-oxide sensors. In particular, we focus on the recently developed table top molecular communication platform. The system is separated into two parts: the propagation and the sensor detection. There is derived, based on this, a more realistic end-to-end channel model. However, since some of the coefficients in the derived models are unknown, we collect a great deal of experimental data to estimate these coefficients and evaluate how they change with respect to the different system parameters. Finally, a noise model is derived for the system to complete an end-to-end system model for the tabletop platform.
C. Lee, B. Koo, N.-R. Kim, H. B. Yilmaz, N. Farsad, A. W. Eckford, and C.-B. Chae
Conference Paper 11Proceedings of IEEE International Conference on Computer Communications (INFOCOM), Pages 13–14, 2015.
#### Abstract
In this demonstration, we will present the world's first molecular multiple-input multiple-output (MIMO) communication link to deliver two data streams in a spatial domain. We show that chemical signals such as concentration gradients could be used in MIMO fashion to transfer sequential data. Until now it was unclear whether MIMO techniques, which are used extensively in modern radio communication, could be applied to molecular communication. In the demonstration, using our devised MIMO apparatus and carefully designed detection algorithm, we will show that we can achieve about 1.7 times higher data rate than single input single output (SISO) molecular communication systems.
#### A Realistic Channel Model for Molecular Communication with Imperfect Receivers
N.-R. Kim, N. Farsad, C.-B. Chae, and A. W. Eckford
Conference Paper 10Proceedings of IEEE International Conference on Communications (ICC), Pages 3987–3992, 2014.
#### Abstract
In this paper, we propose a realistic channel model for a table-top molecular communication platform that is capable for transmitting short text messages across a room. The observed system response for this experimental platform does not match the theoretical results in the literature. This is because many simplifying assumptions regarding the flow, the sensor, and environmental conditions, which were used in derivations of previous theoretical models do not hold in practice. Therefore, in this paper, based on experimental observations, theoretical models are modified to create more realistic channel models.
#### A Molecular Communication Link for Monitoring in Confined Environments
S. Qiu, W. Guo, S. Wang, N. Farsad, and A. W. Eckford
Conference Paper 9Proceedings of IEEE International Conference on Communications Workshops (ICC), Pages 718–723, 2014.
#### Abstract
In this paper, we consider a molecular diffusion based communications link that can reliably transport data over-the-air. We show that the system can also reliably transport data across confined structural environments, especially in cases where conventional electromagnetic (EM) wave based systems may fail. In particular, this paper compares the performance of our proprietary molecular communication test-bed with Zigbee wireless sensors in a metal pipe network that does not act as a radio wave-guide. The paper first shows that a molecular-based communication link's performance is determined primarily by the delay time spread of the pulse response. The paper go on to show that molecular-based systems can transmit more reliably in complex and confined structural environments than conventional EM-based systems. The paper then utilizes empirical data to find relationships between the received radio signal strength, the molecular pulse spread, data rate (0.1 bits/s) and the structural propagation environment.
N. Farsad, W. Guo, and A. W. Eckford
Conference Paper 8Proceedings of IEEE International Conference on Computer Communications (INFOCOM), Pages 107–108 , 2014.
#### Abstract
This demonstration will present the world's first macroscale molecular communication link to reliably transmit a continuous data stream. The system modulates alcohol molecules, which are then diffused via ambient and induced air currents to carry information to a receiver. The communication distance is several meters and the propagation channel we will demonstrate consists of both free space and tunnel environments. The goal is to show that molecules can be used as an alternative to electromagnetic (EM) waves in challenging environments where EM waves do not perform well.
#### Modelling and Design of Polygon-Shaped Kinesin Substrates for Molecular Communication
N. Farsad, A. W. Eckford, and S. Hiyama
Conference Paper 7Proceedings of IEEE International Conference on Nanotechnology (NANO), Pages 1–5, 2012.
#### Abstract
One of the most prominent forms of information transmission between nano- or micro-scale devices is molecular communication, where molecules are used to transfer information inside a fluidic channel. The effects of channel shape on achievable information transmission rates is considered in this work. Specifically, regular convex polygons are studied. A mathematical framework for finding the optimal channel among this class of geometric shapes is derived. Using this framework it is shown that the optimal channel tends to be circular. This result is verified using computer simulations.
#### A Mathematical Channel Optimization Formula for Active Transport Molecular Communication
N. Farsad, A. W. Eckford, and S. Hiyama
Conference Paper 6Proceedings of IEEE International Conference on Communications (ICC), Pages 6137–6141, 2012.
#### Abstract
In this paper, a mathematical optimization formula for estimating the optimal channel dimensions of active transport molecular communication is presented. More specifically, rectangular channels with constant microtubule (MT) concentration are considered. It is shown, both using our formula and using Monte Carlo simulations, that square-shaped channels are optimal. Furthermore, when the value of time per channel use is on the order of a few minutes, which is the range of interest for a lot of potential applications such as diagnostic chips for healthcare, it is shown that our optimization formula can quickly and accurately estimate the optimal channel dimensions.
#### An experimental study of fractional cooperation in wireless mesh networks
A. Calce, N. Farsad, and A. W. Eckford
Conference Paper 5Proceedings of IEEE Symposium on Personal Indoor Mobile Radio Communications (PIMRC), Pages 990–994, 2011.
#### Abstract
Fractional cooperation is a decentralized, low-complexity wireless networking protocol in which nodes have the ability to dynamically select a fraction of its resources to commit to forwarding, and where sources may use more than one relay to convey information to the destination. In this paper, an implementation and a series of experiments are presented to demonstrate the practical performance and effectiveness of fractional cooperation. A low-complexity MAC layer protocol is used, which employs fractional cooperation using LT codes in the absence of central coordination. Experimental results from real-world trials are given, which show that this protocol can maintain a reasonable throughput when nodes are abruptly entering and leaving, making it ideal for a dynamically changing system, such as an ad-hoc network. The redundancy of information seen in the network makes this scheme robust to unfavourable channel conditions.
#### A simple mathematical model for information rate of active transport molecular communication
N. Farsad, A. W. Eckford, S. Hiyama, and Y. Moritani
Conference Paper 4Proceedings of IEEE International Conference on Computer Communications Workshops (INFOCOM), Pages 473–478, 2011.
#### Abstract
In molecular communication, gaps in the underlying theoretical and mathematical framework create numerous challenges. Currently, most researchers rely on simulations to study these systems. However, simulations can be time consuming and impractical. Moreover, due to the complexity and dependencies present in these systems, deriving a mathematical framework that can capture the essence of molecular communication systems is also challenging. In this work, we derive a simple mathematical model, based on some independence assumptions, to estimate the information rate of a molecular communication system employing active transport propagation. We show that the presented model estimates the simulated information rate closely for small communication time intervals. We also use the derived mathematical model to design and verify an optimal loading area that would maximize the information rate.
#### Microchannel Molecular Communication with Nanoscale Carriers: Brownian Motion Versus Active Transport
A. W. Eckford, N. Farsad, S. Hiyama, and Y. Moritani
Conference Paper 3Proceedings of IEEE International Conference on Nanotechnology (NANO), Pages 854–858, 2010.
#### Abstract
In molecular communication, information is encoded and transmitted as a pattern of molecules or other very small information carriers (in this paper, vesicles are used). Nanoscale techniques, such as molecular motors or Brownian motion, are used to convey the vesicles from the transmitter to the receiver, where the transmitted message is deciphered. In this paper, the microchannel environment is considered, and the achievable information rates are compared between the use of Brownian motion and molecular motors, which are evaluated through simulation. Communication is viewed as a mass transfer problem, where messages are sent by transporting a number of vesicles from transmitter to receiver. Results are provided which suggest that active transport is best when the available number of vesicles is small, and Brownian motion is best when the number of vesicles is large.
#### Resource Allocation via Linear Programming for Multi-Source, Multi-Relay Wireless Networks
N. Farsad, and A. W. Eckford
Conference Paper 2Proceedings of IEEE International Conference on Communications (ICC), Pages 1–5, 2010.
#### Abstract
In a cooperative wireless network, there may be many potential relays within radio range of a source; similarly, there may be many potential sources seeking to use relays. Allocating these resources is a non-trivial optimization problem. In this paper, fractional cooperation is considered, where each potential relay only allocates a fraction of its resources to relaying. It is shown that linear programming can be used to optimally allocate resources in multi-source, multi-relay net- works, where the relays use a demodulate-and-forward (DemF) strategy, and where the transmissions are protected by low-density parity-check (LDPC) codes. Compared with existing optimization schemes, this method is particularly suitable for very large networks with numerous sources and relays. Simulation results are presented to demonstrate the performance of this scheme.
#### Low-Complexity Cooperation with Correlated Sources: Diversity Order Analysis
N. Farsad, and A. W. Eckford
Conference Paper 1Proceedings of Annual Conference on Information Sciences and Systems (CISS), Pages 663–668, 2009.
#### Abstract
Wireless sensor networks, which consist of numerous devices that take measurements of a physical phenomenon, are commonly used to observe phenomena that are correlated in space. In this paper, we devise a low-complexity coding scheme for correlated sources based on Slepian-Wolf compression, and analyze its performance in terms of diversity order. The main idea of this scheme is to use the correlated measurements as a substitute for relay links. Although we show that the asymptotic diversity order is limited by the constant correlation factor, we give experimental results that show excellent performance over practical ranges of SNR.
#### Channel Design and Optimization of Active Transport Molecular Communication
N. Farsad, A. W. Eckford, and S. Hiyama
Book Chapter 2 Springer | Pages 213–223 | 2012 | ISBN-13: 978-3-642-32710-0
#### Abstract
In this paper, a mathematical optimization formula for estimating the optimal channel dimensions of active transport molecular communication is presented. More specifically, rectangular channels with constant microtubule (MT) concentration are considered. It is shown, both using our formula and using Monte Carlo simulations, that square-shaped channels are optimal. Furthermore, when the value of time per channel use is on the order of a few minutes, which is the range of interest for a lot of potential applications such as diagnostic chips for healthcare, it is shown that our optimization formula can quickly and accurately estimate the optimal channel dimensions.
#### Information Rates of Active Propagation in Microchannel Molecular Communication
N. Farsad, A. W. Eckford, S. Hiyama, and Y. Moritani
Book Chapter 1 Springer | Pages 16–21 | 2012 | ISBN-13: 978-3-642-32614-1
#### Abstract
Molecular communication is a promising technique for microchannel systems. In this paper, various microchannel molecular communication schemes are simulated and analyzed using information theory, including molecular motors and Brownian motion with drift. Results suggest Brownian motion with drift can deliver excellent performance, depending on the drift velocity.
#### Optimal Channel Design and Markov Chain Channel Model for Active Transport Molecular Communication
N. Farsad, and A. W. Eckford
Technical Report 4NTT DOCOMO Inc., Yokosuka, Kanagawa, Japan | March 2013.
#### Channel Design and Optimization in Active Transport Molecular Communication
N. Farsad, and A. W. Eckford
Technical Report 3NTT DOCOMO Inc., Yokosuka, Kanagawa, Japan | March 2012.
#### Information transfer in Microchannel Systems: Effects of Flow and Mass Transport
N. Farsad, and A. W. Eckford
Technical Report 2NTT DOCOMO Inc., Yokosuka, Kanagawa, Japan | March 2011.
#### Mathematical Models of Information Transfer in Molecular Active Transport Systems
N. Farsad, and A. W. Eckford
Technical Report 1NTT DOCOMO Inc., Yokosuka, Kanagawa, Japan | March 2010.
## Research Summary
My research is currently focused on how bio-inspired forms of communication, such as use of chemical signals or exchange of molecules could be used to create networks in environments that are harsh to radio propagation. This technique, which is called molecular communication in the literature (this is a good introduction) has attracted a lot of attention in recent years. This new multidisciplinary field can be used for in-body communication, robotics, secrecy, networking microscale and nanoscale devices, infrastructure monitoring in smart cities and industrial complexes, as well as for underwater communications. Therefore, molecular communication, as a technology, can have a disruptive effect by unlocking many futuristic applications much like Guglielmo Marconi’s pioneering work in radio that laid the foundation for modern telecommunications and the wealth of applications it supports.
Much of my past and current research has been focused on studying, designing, and building point-to-point molecular communication links at both macroscales and microscales, and improving the performance of these links. Specifically, my current research spans evaluating the theoretical limits of molecular communications, system design, analysis, and optimization, as well as building experimental platforms. I use tools and models from several disciplines including communication and network engineering, machine learning, information theory, signal processing, optimization, chemistry, biology, bioengineering, and medical sciences. I also design and build experimental systems to support, validate, and complement my theoretical work (e.g., world's first experimental demonstrator for molecular communication). Below, I highlight some of the applications of molecular communication at microscales and macroscales, which I strive to explore, and why molecular communication is central to these applications.
#### Microscale Communication
At microscale, I am solving the communication problem among tiny devices (smaller than a few micrometers). Engineering radio-based communication networks at these scales can be very challenging, since 1) we need to use higher frequencies to have small antennas that fit inside the small device; and 2) High frequency radio does not propagate very well in ionic fluids (e.g. inside human body). Nevertheless, there are research group that are trying to adapt the radio technology to these small dimensions using novel materials such as carbon nanotube or graphene. However to date, there have not been any practical demonstrations of feasibility of this approach. On the other hand, molecular communication is already used in nature to solve this problem and it can be biocompatible, which would make it very suitable for many potential applications, especially in medicine.
Why is this important? Engineering micro- and nano-scale systems are the key to unlocking many futuristic applications such as nanomedicine, microrobotics, nanorobotics, lab-on-a-chip devices, diagnostic chip devices, targeted drug delivery and biological Computation. Most of these futuristic and transformative applications have one feature in common: they involve not just single devices working independently, but swarms of devices working in concert.
#### Macroscale Communication
At macroscale, chemical signals open a new mode of communication (e.g., using chemical tags) for connecting devices such as robots. Although molecular communication cannot outperform radio-based communication in terms of delay and throughput, there are environments that are harsh to electromagnatic waves. For example, inside networks of tunnels or pipes as we show here. However, chemical communication can be used instead to transfer small amounts of data with a delay in these situations.
Why is this important? One of the requirements of smart cities is infrastructure monitoring. However, there are a number of challenges for radio-based solutions that needs to be overcome (for example see here). Chemical communication can be used in conjunction with radio based systems to overcome some of these challenges. Another area that chemical communication can prove useful is inside pipelines, where applications of interest would be in different industries such as oil and gas. Finally, employing chemical communication can add a new and exciting dimension to the way robots and devices in general communicate.
### Interests
• Bio-inspired Communication Networks
• Communication & Network Engineering
• Machine Learning & Deep Learning
• Information Theory
• Molecular Communication
• Chemical Signaling
• Bioengineering
### Experience as Lecturer
• Winter 2015
University of Ontario Institute of Technology, Oshawa, Canada
Course offered by the Departments of Electrical Engineering/Computer Sciece/Information Technology
• Fall2012
INFR3710U: Signals and Random Processes (Third Year Undergraduate Course)
University of Ontario Institute of Technology, Oshawa, Canada
Department of Business and Information Technology
• Fall2011
INFR3710U: Signals and Random Processes (Third Year Undergraduate Course)
University of Ontario Institute of Technology, Oshawa, Canada
Department of Business and Information Technology
### Experience as Teaching Assistant
I have served as a teaching assistant for various courses in electrical engineering and computer science, almost every fall and winter throughout my Masters and Ph.D. Most courses that I have been assigned have had dedicated lab sessions as well, where I was the lab instructor. Typically, the number of students in the lab ranged from 10 to 30 students, where they designed and implemented software and hardware solutions to a set of problems and projects. The following are the list of courses:
• EECS1021: Object Oriented Programming from Sensors to Actuators (Winter 2015)
• EECS3451: Signals and Systems (Fall 2014)
• CSE4215/CSE5431: Mobile Communications (Winter 2014)
• CSE2011: Fundamentals of Data Structures (Summer 2013)
• CSE4214: Digital Communications (Fall 2012)
• CSE3215: Embedded Systems (Winter 2012)
• CSE3451: Signals and Systems (Fall 2011)
• CSE4411: Database Management Systems (Summer 2011)
• CSE1560: Introduction to Computing for Mathematics and Statistics (Winter 2011)
• CSE4214: Digital Communications (Fall 2010)
• CSE3451: Signals and Systems (Fall 2009)
• CSE4214: Digital Communications (Fall 2009)
• CSE1520: Computer Use: Fundamentals (Winter 2009)
• CSE1020: Introduction to Computer Science I (Winter 2009)
• CSE4471: Introduction to Virtual Reality (Fall 2008)
• CSE3215: Embedded Systems (Winter 2008)
• CSE1020: Introduction to Computer Science I (Fall 2007)
### Honors and Awards
• 2015-2017
NSERC Postdoctoral Fellowship Award
Receipient of the Natural Sciences and Engineering Research Council of Canada (NSERC) Postdoctoral Fellowship Award provided by the government of Canada.
• 2015
INFOCOM Best Demo Award
We demoed the "Molecular MIMO Communication Link" at IEEE International Conference on Computer Communications (IEEE-INFOCOM), and won the best demo award from the conference.
• 2014
Finalist at the Bell Labs Prize Competition
One of the seven finalists (from about 500 applicants) in the 2014 Bell Labs Prize competition.
• 2014
Second Prize in IEEE ComSoc Student Competition
• 2011-2014
Awarded the Ontario Graduate Scholarship for two consecutive years (2012-2014), and the Queen Elizabeth II Graduate Scholarship in Science & Technology during the 2011-2012 school year. These awards were all provided by the government of the province on Ontario in Canada.
### Professional Service
• Present 2014
#### Technical Program Committee
IEEE Global Communications Conference (GLOBCOM'2015,2016)
IEEE International Conference on Communications (ICC'2015)
International Conference on Bio-inspired Information and Communications Technologies (BICT'2015,2016)
• Present 2011
#### Member of Standardization Project
P1906.1 - Recommended Practice for Nanoscale and Molecular Communication Framework
• 2013 2012
#### Area Associate Editor
EEE Journal of Selected Areas of Communication (JSAC) - Special Issue on Emerging Technologies in Communications
• Present 2009
#### Technical Reviewer
IEEE Transactions on Molecular, Biological, and Multi-Scale Communications
IEEE Journal of Selected Areas of Communication (JSAC) - Special Issue on Molecular, Biological, and Multi-Scale Communications
IEEE Transactions on Information Theory
IEEE Transactions on Signal Processing
IEEE Transactions on Communications
IEEE Transactions on NanoBioscience
IEEE Transactions on Nanotechnology
IEEE Wireless Communications Letters
My newest experimental platform uses acids and bases for communication and is 10 times faster than any of my previous platforms.
The world's first molecular communication system, capable of transferring short text messages with alcohol.
A MIMO implementation of this work demonstrates improvment in data rate.
Some of my work has been covered by these media outlets. Click the logos to go to the corresponding site. I recommend the article by the ChemistryWorld.
I am always interested in discussing research opportunities, assisting research enthusiasts, and collaborating on new projects. Feel free to contact me at nfarsad@stanford.edu or meet me in my office:
350 Serra Mall
Packard Building, Room 372
Stanford, CA 94305 | 2017-03-27 22:27:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45286503434181213, "perplexity": 1481.2852722700375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189534.68/warc/CC-MAIN-20170322212949-00434-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://wsw.academickids.com/encyclopedia/index.php/Proper_time | # Proper time
Proper time is time as measured by the clock for an observer who is traveling through spacetime. The concept of proper time is necessary in Einstein's theories of relativity because of effects such as time dilation, which result in observers travelling between the same positions along different paths/world lines through spacetime experiencing time differently.
Contents
## Mathematical formalism
The formal definition of proper time involves describing the path that a clock, observer, or test particle is taking through spacetime and the metric structure of that spacetime.
Using tensor calculus, proper time is defined as follows: Given a spacetime which is a pseudo-Riemannian manifold mapped with a coordinate system [itex]x^\mu[itex] and equipped with a corresponding metric tensor [itex]g_{\mu\nu}[itex], the proper time [itex]\tau\ [itex] experienced in moving between two events along a timelike path P is given by the path integral
[itex]\tau = \int_P \;\! d\tau [itex]
where
[itex] d\tau = \sqrt{dx_\mu \; dx^\mu} = \sqrt{g_{\mu\nu} \; dx^\mu \; dx^\nu}[itex]
### Derivation
For any spacetime, there is an incremental invariant interval ds between events with an incremental coordinate separation dxμ of
[itex]ds^2 = g_{\mu\nu} \; dx^\mu \; dx^\nu[itex].
This is refered to as the line element of the spacetime. s may be spacelike, lightlike, or timelike. Spacelike paths cannot be physically traveled (as they require moving faster than light). Lightlike paths can only be followed by light beams, for which there is no passage of proper time. Only timelike paths can be traveled by massive objects, in which case the invariant interval becomes the proper time [itex]\tau\ [itex]. So for our purposes [itex]\tau\ \equiv[itex] s.
Taking the square root of each side of the line element gives the above definition of [itex]d\tau\ [itex]. After that, take the path integral of each side to get [itex]\tau\ [itex] as described by the first equation.
## Usage in special relativity
In special relativity spacetime is mapped with a four-vector coordinate system [itex]x^\mu = (t,x,y,z)[itex] where
t is a temporal coordinate and
x,y, and z are orthogonal spatial coordinates.
This spacetime and mapping are described with the Minkowski metric:
[itex] g_{\mu\nu} = \left ( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -\frac{1}{c^2} & 0 & 0 \\ 0 & 0 & -\frac{1}{c^2} & 0 \\ 0 & 0 & 0 & -\frac{1}{c^2} \end{matrix} \right ) . [itex]
(Note: The +--- metric signature is used in this article so that [itex]d\tau\ [itex] will always be positive definite for timelike paths.)
In special relativity, the proper time equation becomes
[itex]\tau = \int_P \sqrt {dt^2 - dx^2/c^2 - dy^2/c^2 - dz^2/c^2}[itex].
To make things even easier, inertial motion in special relativity is where the spatial coordinates change at a constant rate with respect to the temporal coordinate. This further simplifies the proper time equation to
[itex]\tau = \sqrt{\Delta t^2 - \Delta x^2/c^2 - \Delta y^2/c^2 - \Delta z^2/c^2}[itex],
where [itex]\Delta[itex] means "the change in" between two events.
### Example 1: The twin "paradox"
For a twin "paradox" scenario, let there be an observer A who moves between the coordinates (0,0,0,0) and (10 years, 0, 0, 0) inertially. This means that A stays at [itex]x=y=z=0[itex] for 10 years of coordinate time. The proper time for A is then
[itex]\tau = \sqrt{(10\; \mathrm{years})^2} = 10\; \mathrm{years}[itex]
So we find that being "at rest" in a special relativity coordinate system means that proper time and coordinate time are the same.
Let there now be another observer B who travels in the x direction from (0,0,0,0) for 5 years of coordinate time at 0.866c to (5 years, 4.33 light-years, 0, 0). Once there, B accelerates, and travels in the other spatial direction for 5 years to (10 years, 0, 0, 0). For each leg of the trip, the proper time is
[itex]\tau = \sqrt{(5\;\mathrm{years})^2 - (4.33\;\mathrm{years})^2} = \sqrt{6.25\;\mathrm{years}^2} = 2.5 \; \mathrm{years}.[itex]
So the total proper time for observer B to go from (0,0,0,0) to (5 years, 4.33 light-years, 0, 0) to (10 years, 0, 0, 0) is 5 years. Thus is it is shown that the proper time equation incorporates the time dilation effect. In fact, for an object in a SR spacetime traveling with a velocity of v for a time T, the proper time experienced is
[itex]\tau = \sqrt{T^2 - (v_x T/c)^2 - (v_y T/c)^2 - (v_z T/c)^2 } = T \sqrt{1 - v^2/c^2} [itex],
which is the SR time dilation formula.
### Example 2: The rotating disk
An observer rotating around another, inertial observer is in an accelerated frame of reference. For such an observer, the incremental ([itex]d\tau\ [itex]) form of the proper time equation is needed, along with a parameterized description of the path being taken, as shown below.
Let there be an observer C on a disk rotating in the xy plane at an coordinate angular rate of [itex]\omega[itex] and who is at a distance of r from the center of the disk with the center of the disk at x=y=z=0. The path of observer C is given by [itex](T, \;\, r\sin(\omega T),\;\, r\cos(\omega T), \;\, 0)[itex], where [itex]T [itex] is the current coordinate time. When r and [itex]\omega[itex] are constant, [itex]dx = -r \omega \cos(\omega T) \; dT[itex] and [itex]dy = r \omega \sin(\omega T) \; dT[itex]. The incremental proper time formula then becomes
[itex]d\tau = \sqrt{dT^2 - (r \omega /c)^2 \cos^2(\omega T)\; dT^2 - (r \omega /c)^2 \sin^2(\omega T) \; dT^2} = dT\sqrt{1 - \left ( \frac{r\omega}{c} \right )^2}[itex].
So for an observer rotating at a constant distance of r from a given point in spacetime at an constant angular rate of ω between coordinate times [itex]T_1[itex] and [itex]T_2[itex], the proper time experienced will be
[itex]\int_{T_1}^{T_2} d\tau = (T_2 - T_1) \sqrt{ 1 - \left ( \frac{r\omega}{c} \right )^2}[itex].
As v= for an rotating observer, this result is as expected given the time dilation formula above, and shows the general application of the integral form of the proper time formula.
## Usage in general relativity
The difference between SR and general relativity (GR) is that in GR you can use any metric which is a solution of the Einstein field equations, not just the Minkowski metric. Because inertial motion in curved spacetimes lacks the simple expression it has in SR, the path integral form of the proper time equation must always be used.
### Example 3: The rotating disk (again)
An appropriate coordinate conversion done against the Minkowski metric creates coordinates where an object on a rotating disk stays in the same spatial coordinate position. The new coordinates are [itex]r=\sqrt{x^2 + y^2}[itex] and [itex]\theta = \arctan(x/y) - \omega t[itex]. The t and z coordinates remain unchanged. In this new coordinate system, the incremental proper time equation is
[itex]d\tau^2 = \sqrt{\left [ 1 - \left (\frac{r \omega}{c} \right )^2 \right ] dt^2 - \frac{dr^2}{c^2} - \frac{r^2\, d\theta^2}{c^2} - \frac{dz^2}{c^2} - 2 \frac{r^2 \omega \, dt \, d\theta}{c^2}}[itex]
With r, θ, and z being constant over time, this simplifies to [itex]d\tau = dt \sqrt{ 1 - \left (\frac{r \omega}{c} \right )^2 }[itex], which is the same as in Example 2.
Now let there be an object off of the rotating disk and at inertial rest with respect to the center of the disk and at a distance of R from it. This object has a coordinate motion described by dθ = -ω dt, which describes the inertially at-rest object of counter-rotating in the view of the rotataing observer. Now the proper time equation becomes
[itex]d\tau = \sqrt{\left [ 1 - \left (\frac{R \omega}{c} \right )^2 \right] dt^2 - \left (\frac{R\omega}{c} \right )^2 \,dt^2 + 2 \left ( \frac{R \omega}{c} \right )^2 \,dt^2} = dt [itex].
So for the inertial at-rest observer, coordinate time and proper time are once again found to pass at the same rate, as expected and required for the internal self-consistency of relativity theory.
### Example 4: The Schwarzschild solution - Time on Planet Earth
The Schwarzschild solution has an incremental proper time equation of
[itex]d\tau = \sqrt{\left( 1 - 2m/r \right ) dt^2 - \frac{1}{c^2}\left ( 1 - 2m/r \right )^{-1} dr^2 - \frac{r^2}{c^2} d\theta^2 - \frac{r^2}{c^2} \sin^2 \theta \; d\phi^2}[itex],
where
t is time as calibrated with a clock distant from and at inertial rest with respect to the Earth,
r is a radial coordinate (which is effectively the distance from the Earth's center),
θ is the latitudinal coordinate, being the angular separation from the north pole in radians.
[itex]\phi\ [itex] is a longitudinal coordinate, analogous to the latitude on the Earth's surface but independent of the Earth's rotation. This is also given in radians.
m is the geometrized mass of a central massive object, being m=MG/c2,
M is the mass of the object,
G is the gravitational constant.
To demonstrate the use of the proper time relationship, several sub-examples involving the Earth will be used here. It should be noted that the use of the Schwarzschild solution for the Earth is not entirely correct for the following reasons:
• Due to its rotation, the Earth is an oblate spheroid instead of being a true sphere. This results in the gravitational field also being oblate instead of spherical.
• In GR, a rotataing object also drags spacetime along with itself. This is described by the Kerr solution. However, the amount of frame dragging that occurs for the Earth is so small that it can be ignored.
For the Earth, [itex]M=5.9736 \times 10^{24}[itex]kg, meaning that [itex]m = 4.436 \times 10^{-3}[itex]m. When standing on the north pole, we can assume [itex]dr = d\theta\ = d\phi\ [itex] (meaning that we are neither moving up or down or along the surface of the Earth). In this case, the Schwarzschild solution proper time equation becomes [itex]d\tau = dt \sqrt{1 - 2m/r}[itex]. Then using the polar radius of the Earth as the radial coordinate (or r = 6356780 meters), we find that
[itex]d\tau = \sqrt{\left ( 1 - 1.3957 \times 10^{-9} \right ) \;dt^2} = \left (1 - 6.978 \times 10^{-10} \right ) dt[itex].
At the equator, the radius of the Earth is r = 63781400 meters. In addition, the rotation of the Earth needs to be taken into account. This imparts on an observer an angular velocity of dφ/dt of 2π divided by the sidereal period of the Earth's rotation, 86162.4 seconds. So [itex]d\phi\ [itex] = 7.2922×10^{-5} dt. The proper time equation then produces
[itex]d\tau = \sqrt{\left ( 1 - 1.39093 \times 10^{-9} \right ) dt^2 - 2.40 \times 10^{-12} dt^2} = \left( 1 - 6.9667 \times 10^{-10}\right )[itex].
This should have been the same as the previous result, but as noted above the Earth is not spherical as assumed by the Schwarzschild solution. Even so this demonstrates how the proper time equation is used.
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy | 2021-05-07 10:43:39 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9832170009613037, "perplexity": 1148.004225804623}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988775.80/warc/CC-MAIN-20210507090724-20210507120724-00383.warc.gz"} |
https://or.stackexchange.com/questions/3781/is-there-a-better-way-to-formulate-this-constraint/3783 | # Is there a better way to formulate this constraint?
Let $$x_{r}^{j}=1\iff$$ the machine schedules job $$j$$ using resource $$r$$. My constraint says that: a resource cannot be used twice, i.e., if $$x_{r}^{j}=1$$, then $$x_{r}^{j'}=0$$ for $$j'\neq j$$. I write this as: $$x_{r}^{j}+x_{r}^{j'}\leq1,\forall j\neq j', \forall r.$$
Is there a better way to formulate this?
You can strengthen your "conflict" constraint to a "clique" constraint: $$\sum_j x_r^j \le 1$$ for all $$r$$. There are fewer of these, and they dominate the conflict constraints.
• What is the difference between your answer and the answer that was given before yours? I am asking because I can't see any difference in the both approaches. Am I missing something? Mar 27, 2020 at 22:51
• They were entered simultaneously. Mar 27, 2020 at 22:52
Same idea, but typically formulated as $$\sum_j x_r^j \leq 1, \: \forall r$$ For binary $$x$$
That seems like the right way to formulate it. There are lots of problems that use that sort of approach to ensure that at most one of two binary variables equal 1.
• RobPratt's and @JBL's approach is better than mine. :) Mar 27, 2020 at 23:33
• Those approaches dominate algebraically, but they also result in denser constraint matrices (albeit with fewer rows), so I'm not sure it's a lock that they are always preferable.
– prubin
Mar 28, 2020 at 19:00 | 2022-12-04 15:17:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.47172650694847107, "perplexity": 415.0004257936566}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710974.36/warc/CC-MAIN-20221204140455-20221204170455-00259.warc.gz"} |
https://sqlpete.wordpress.com/2017/02/ | # Archive for February, 2017
### Brute-force packing
Because I’m a huge music fan, I own quite a few CDs; rather more than I’d readily admit to! A few years ago, I started ripping my collection to FLAC, a lossless audio format, and backing everything up to DVD-R. In order to be efficient, I want to get as many albums as I can on each DVD-R disc, minimising the space wasted — this is known as a Packing Problem.
Now, I’d love to be able to present you with a highly-optimised algorithm that I wrote, but that’s not what I did: I brute-forced it. Processor cycles are very cheap, and if it’s going to be orders of magnitude quicker to iterate a few million times than it will be to research a whole new area of maths, then iterating it’ll be. My original code was a VB app (so I could drag folders onto a GUI), but here’s a similar version of the code in R:
set.seed(1234);
containerSize <- 4500; # roughly DVD size in MB
itemSize <- c(1641,1498,1747,751,1090,164,1602,1020,1126,553); # album sizes in MB
cat(sprintf("No. containers needed (no partitioning): %5.2f\n", sum(itemSize) / containerSize));
Z <- 1000; # Number of iterations
# To keep track of the best partition
best.remainder <- 1.0;
best.partition <- NULL;
for(i in 1:Z) {
working <- sample(itemSize); # randomly re-order our list of sizes
partition <- list();
k <- 1;
# Using the order as per 'working', partition the items
# such that the container size isn't exceeded:
while (length(working) > 0) {
this.partition.indexes <- which( cumsum(working) <= containerSize );
partition[[k]] <- working[this.partition.indexes];
working <- working[-(this.partition.indexes)];
k <- k+1;
}
npm1 <- length(partition) - 1; # Number of partitions minus 1
partition.totals <- unlist(lapply(partition, sum));
remainder <- (sum(rep(containerSize, npm1) - partition.totals[1:npm1]))
/ (npm1 * containerSize);
if (remainder < best.remainder) {
best.remainder <- remainder;
best.partition <- partition;
totals.str <- paste("(", paste(partition.totals, collapse=","), ")", sep="");
partition.str <- paste(unlist(lapply(partition,
function(x) paste("(",paste(x,collapse=","),")",sep=""))),collapse=",")
cat(sprintf("i = %3d, rem. = %5.2f%%; totals = %s; partition = %s\n", i,
remainder * 100.0), totals.str, partition.str));
}
} # end for loop
This code (1000 iterations) runs in the blink of an eye:
i = 1, rem. = 19.00%; totals = (3772,3518,3902);
partition = (1498,164,1090,1020),(1126,751,1641),(1602,553,1747)
i = 2, rem. = 13.56%; totals = (4439,3341,3412);
partition = (1602,1090,1747),(553,1498,1126,164),(1641,1020,751)
i = 4, rem. = 13.18%; totals = (3963,3851,3378);
partition = (1090,1747,1126),(751,1498,1602),(1641,553,164,1020)
i = 6, rem. = 4.78%; totals = (4303,4267,2622);
partition = (1641,1747,164,751),(553,1126,1498,1090),(1020,1602)
i = 13, rem. = 4.04%; totals = (4301,4335,2556);
partition = (1020,1126,553,1602),(1747,1498,1090),(751,1641,164)
i = 23, rem. = 0.26%; totals = (4478,4499,2215);
partition = (1090,1641,1747),(1020,1126,1602,751),(1498,553,164)
i = 524, rem. = 0.02%; totals = (4499,4499,2194);
partition = (1126,1602,751,1020),(1747,1498,1090,164),(553,1641)
The figure rem. is the percentage of space wasted on the discs that could be full — clearly, not all the discs can be 100% full. So in this case, I knew I was going to be burning three DVD-Rs, but there’s only 1 MB of unused space on each of the first two discs; for the third, I can either find some other files to backup, or keep those two albums to burn later — which is what I usually do; saving even more space, by repeatedly putting off burning the least-full disc.
### Exact p-value versus Information Value
As I think I’ve mentioned before, one of the ‘go-to’ stats in my scorecard-building toolkit is the p-value that results from performing Fisher’s Exact Test on contingency tables. It’s straightforward to generate (in most cases), and directly interpretable: it’s just a sum of the probabilities of ‘extreme’ tables. When I started building credit risk scorecards, and using the Information Value (IV) statistic, I had to satisfy myself that there was a sensible relationship between the two values. Now, my combinatoric skills are far too lacking to attempt a rigorous mathematical analysis, so naturally I turn to R and the far easier task of simulating lots of data!
I generated 10,000 2-by-2 tables at random, with cell counts between 5 and 100. Here’s a plot of the (base e) log of the resulting exact p-value, against the log of the IV:
(I’ve taken logs as the relationship is clearer.) As you can see, I’ve drawn in some lines for the typical levels of p-value that people care about (5%, 1% and 0.1%), and the same for the IV (0.02, 0.1, 0.3 and 0.5). In the main, it looks like you’d expect, no glaring outliers.
For fun, I’ll look at those that fall into the area (p_exact > 0.05) and (0.3 < IV < 0.5):
6 29 10 15
p = 0.0751, IV = 0.332
5 84 7 37
p = 0.0613, IV = 0.321
In both cases, the exact p-value says there’s not much evidence that the row/column categories are related to each other — yet the IV tells us there’s “strong evidence”! Of course, the answer is that there’s no one single measure of independence that covers all situations; see, for instance, the famous Anscombe’s Quartet for a visual representation.
Practically, for the situations in which I’m using these measures, it doesn’t matter: if I have at least one indication of significance, I may as well add another candidate variable to the logistic regression that’ll form the basis of my scorecard. If the model selection process doesn’t end up using it, that’s fine.
Anyway, I end with a minor mystery. In my previous post, I came up with an upper bound for the IV, which means I can scale my IV to be between zero and one. I presumed that this new scaled version would be more correlated with the exact p-value; after all, how can a relationship with an IV of 0.25, but an upper bound of 5, be less significant than one with an IV of 0.375, but an upper bound of 15 (say)? Proportionally, the former is twice as strong as the latter, no?
What I found was that the scaled version was consistently less correlated! Why would this be? Surely, the scaling is providing more information? I have some suspicions, but nothing concrete at present — hopefully, I can clear this up in a future post.
### Information Value
Despite having worked with it for years, it has always irked me that I don’t know the derivation of the Information Value (IV) statistic.
It’s used liberally throughout credit risk work, but the background to its invention seems somewhat hazy. Clearly it’s related to Shannon Entropy, via the $\sum p \log(p)$ construct. In Naeem Siddiqi’s well-known book Credit Risk Scorecards, he writes “Information Value, […] comes from information theory” and references Kulback’s 1959 book Information Theory and Statistics, which I don’t have. Someone else suggested that it stems from the work of I.J. Good, but I can’t find an explicit definition in any of his papers I’ve managed to look at. (I bought his book Good Thinking, about the foundations of probability and statistical interference, but it’s waaaay too complex for me!)
The Information Value (IV) is defined as:
$\mathrm{IV} = \sum_{i=1}^{k} (g_{i} - b_{i}) \log_e (g_{i} / b_{i})$
, where $g_{i}$ is the number of ‘goods’ in category i, and $b_{i}$ is the number of ‘bads’.
In his book, Siddiqi gives the following rule of thumb regarding the value of IV:
< 0.02 unpredictive 0.02 to 0.1 weak 0.1 to 0.3 medium 0.3 to 0.5 strong 0.5+ “should be checked for over-predicting”
For an independent variable with an IV over 0.5, it might be somehow related to the dependent variable, and you might want to consider leaving it out. (If you build a scorecard that has a bureau score as one of your variables, then you’ll almost certainly see this.)
[See these two links for more about Information Value, and an example or two of its use: All about “Information Value” and Information Value (IV) and Weight of Evidence (WOE).]
### Upper Bound
The lower bound of the IV is fairly obviously zero: if $g_{i} \equiv b_{i}$ for all the categories, then the difference is zero, so their sum is zero times $\log_{e}(1)$, which is also zero. But what about the upper bound?
I’ve put together this small PDF document: Upper bound of the Information Value (IV), in which (I think!) I show that the upper bound is very close to $\log_{e}(N_{G}) + \log_{e}(N_{B})$, where $N_G$ is the total number of goods, and $N_B$ is the total number of bads.
Of course, it’s wise to at least check the result with some code — so in R, let’s create a million tables at random, and look at the actual figures that are produced:
Z <- 1000000; # number of iterations
IV <- rep(0, Z); # array of IVs
lGB <- rep(0, Z); # array of (log(n_g) + log(n_b))
for (i in 1:Z)
{
k <- sample(2:20, 1); # number of categories
g <- sample(1:100, k, replace=T); # good
b <- sample(1:100, k, replace=T); # bad
ng <- sum(g);
nb <- sum(b);
IV[i] <- sum( ((g/ng)-(b/nb)) * log((g/ng)/(b/nb)) );
lGB[i] <- log(ng) + log(nb);
}
plot(IV, lGB, xlab="IV", ylab="log(N_G)+log(N_B)",
main="IV vs log(N_G)+log(N_B)", pch=19,col="blue",cex=0.5);
abline(a=0,b=1,col="red",lwd=2); # draw the line x=y
As you can see, there are no points below the red ‘x=y’ line; in other words, the IV is always less than $\log_{e}(N_{G}) + \log_{e}(N_{B})$. There are a few points that are close; the closest is:
min(lGB-IV)
[1] 0.2161227
I know that $\log_{e}(N_{G}) + \log_{e}(N_{B})$ is not the best possible upper bound — a closer, but more complex answer is reasonably obvious from the document — but “log(number of goods) plus log(number of bads)” is (a) memorable, and (b) close enough for me!
### Inexact date grouping using islands
A few years ago at SQLBits, I was fortunate enough to attend a fascinating lecture given by SQL Server MVP Itzik Ben-Gan on Gaps and islands:
Gaps and islands problems involve finding a range of missing values (gaps) or a range of consecutive values (islands) in a sequence of numbers or dates.
Recently, I had to produce a dataset which showed how our interest rates varied over time, by product; for example, products A to E started at 10% through to 50%, but have been adjusted periodically to where they are today, a few points different. Practically, most of the changes have been made at or around the same time — but not exactly. For technical reasons, the specified rates aren’t stored in the database, so there’s no InterestRate table, or a parent table called InterestRateSet that links them together. However, the final result of an application is stored, so we know that a product has been sold, and what the corresponding interest rate was on that day.
The challenge was to work out how many sets of interest rates we’ve had since day 1; but because not every product is purchased every day, if we group by product/day, then it looks like our rates change more often than they do. This is where the ‘gaps and islands’ concept comes in, and luckily I remembered the lecture from a few years before. I found and tweaked some of Itzik’s code from a 2012 article Solving Gaps and Islands with Enhanced Window Functions (SQL Server Pro website) to accept a UDT (in this case, a list of dates). [See previous post, Passing structured data to stored procedures.]
Here it is:
-- Drop/Create our UDT:
IF EXISTS (SELECT * FROM sys.types WHERE is_table_type = 1 AND name = 'DateListType')
DROP TYPE DateListType
GO
CREATE TYPE DateListType AS TABLE ([Date] DATE)
GO
-- Drop/Create our function:
IF OBJECT_ID('dbo.fn_GetIslandsFromDateList') IS NOT NULL
DROP FUNCTION dbo.fn_GetIslandsFromDateList
GO
CREATE FUNCTION dbo.fn_GetIslandsFromDateList
(
,@GapSize INT
)
RETURNS TABLE
AS
RETURN
(
WITH cte_Distinct AS
(
SELECT
[Date]
FROM @dateListType
GROUP BY [Date]
), cte_Part1 AS
(
SELECT
[Date]
,CASE
WHEN DATEDIFF(day, LAG([Date]) OVER(ORDER BY [Date]), [Date]) <= @GapSize
THEN 0 ELSE 1 END AS IsStart
,CASE
WHEN DATEDIFF(day, [Date], LEAD([Date]) OVER(ORDER BY [Date])) <= @GapSize
THEN 0 ELSE 1 END AS IsEnd
FROM cte_Distinct
)
, cte_Part2 AS
(
SELECT
[Date] AS RangeStart
,CASE
WHEN IsEnd = 1
THEN [Date] ELSE LEAD([Date], 1) OVER(ORDER BY [Date]) END AS RangeEnd
,IsStart
FROM cte_Part1
WHERE IsStart = 1 OR IsEnd = 1
)
SELECT
ROW_NUMBER() OVER (ORDER BY RangeStart) AS ID
,RangeStart
,RangeEnd
FROM cte_Part2
WHERE IsStart = 1
)
GO
Some things to note:
• dbo.fn_GetIslandsFromDateList is an inline function, which you can almost think of as ‘a view that takes parameters’ (a parameterised view).
• You can use CTEs (Common Table Expressions) in inline functions. I love using CTEs, they can make the code very readable. Often, the parser turns them into standard sub-queries, so there’s no performance hit.
• The @GapSize parameter controls how far apart our islands can be — see the examples below.
• If you follow the code through, and break it down in to its component parts, you can see how it works — like all the best code, it’s very neat and compact.
• To re-emphasise, this isn’t my algorithm, it’s Itzik Ben-Gan’s; I’ve done little more than re-format it for my own use.
Let’s feed some dates into our function:
DECLARE @myList DateListType
INSERT @myList([Date])
VALUES('2017Feb01'),('2017Feb02'),('2017Feb03')
,('2017Feb06'),('2017Mar01'),('2017Mar02')
SELECT * FROM dbo.fn_GetIslandsFromDateList(@myList, 2)
GO
ID RangeStart RangeEnd
---- ---------- ----------
1 2017-02-01 2017-02-03
2 2017-02-06 2017-02-06
3 2017-03-01 2017-03-02
With a @GapSize of 2, we get 3 ranges (islands). With a @GapSize of 3:
SELECT * FROM dbo.fn_GetIslandsFromDateList(@myList, 2)
GO
ID RangeStart RangeEnd
---- ---------- ----------
1 2017-02-01 2017-02-06
2 2017-03-01 2017-03-02
, we get 2 ranges, because the difference in days between 2017-02-06 and 2017-02-03 is less than or equal to 3.
So this code did the trick, and allowed us to work out exactly how many different sets of rates we’d actually had live. Yes, we could’ve worked it out by hand; but now we’ve got some reproducible code that can drive various different reports, that’ll show us exactly how our changes have affected the business.
A final thought: Quite often, solving a problem comes down to just knowing the right phrase to google!
### Passing structured data to stored procedures
As I wrote about here, I like to pass structured data around using XML (where applicable). While I think it’s the most flexible way, it’s understandable that people might not want to learn XML-parsing syntax, if they don’t have to. Another way of passing data into stored procedures (sprocs) is to use user-defined types (UDTs).
Here’s a simple example:
-- If the UDT already exists, delete it:
IF EXISTS (
SELECT *
FROM sys.types
WHERE is_table_type = 1
AND [name] = 'ListOfDates'
AND [schema_id] = SCHEMA_ID('dbo')
)
BEGIN
DROP TYPE dbo.ListOfDates
END
GO
-- Create our UDT:
CREATE TYPE dbo.ListOfDates AS TABLE
(
DateTypeID TINYINT NOT NULL
,[Date] DATE NOT NULL
)
GO
-- Let's test it out:
SET DATEFORMAT YMD
DECLARE @MyListOfDates AS dbo.ListOfDates
INSERT @MyListOfDates(DateTypeID, [Date])
VALUES(1, '2016Jan01'),(1, '2016Jan02'),(1, '2016Jan03')
,(2, '2016Oct10'),(2, '2017Jan01')
SELECT * FROM @MyListOfDates
GO
DateTypeID Date
---------- ----------
1 2016-01-01
1 2016-01-02
1 2016-01-03
2 2016-10-10
2 2017-01-01
Hopefully, that looks straightforward; the way we’ve used it here is not dissimilar to a table variable (e.g. DECLARE @MyTable AS TABLE...), but we can’t pass a table variable to a sproc. We’ll create a test sproc and pass our new user-defined type to it:
-- If the sproc already exists, delete it:
IF OBJECT_ID('dbo.usp_MySproc') IS NOT NULL
BEGIN
DROP PROCEDURE dbo.usp_MySproc
END
GO
-- Create our sproc:
CREATE PROCEDURE dbo.usp_MySproc
(
)
AS
BEGIN
SELECT
DateTypeID
,MIN([Date]) AS MinDate
,MAX([Date]) AS MaxDate
FROM @ListOfDates
GROUP BY DateTypeID
ORDER BY DateTypeID
END
GO
-- Test it out:
DECLARE @MyListOfDates AS dbo.ListOfDates
INSERT @MyListOfDates(DateTypeID, [Date])
VALUES(1, '2016Jan01'),(1, '2016Jan02'),(1, '2016Jan03')
,(2, '2016Oct10'),(2, '2017Jan01')
,(3, '2017Feb01'),(3, '2017Feb03'),(3, '2017Feb28')
EXEC dbo.usp_MySproc @ListOfDates = @MyListOfDates
GO
DateTypeID MinDate MaxDate
---------- ---------- ----------
1 2016-01-01 2016-01-03
2 2016-10-10 2017-01-01
3 2017-02-01 2017-02-28
See the READONLY in the CREATE PROCEDURE? Here’s what happens when you omit it:
The table-valued parameter "@ListOfDates" must be declared with the READONLY option.
This is one of the limitations of user-defined types: they can’t be used to pass data back. So we couldn’t make any changes to @ListOfDates in the sproc, and see those changes reflected outside of the sproc.
There’s another limitation: UDTs can’t be used cross-database. Here’s what happens:
The type name 'MyOtherDatabase.dbo.ListOfDates' contains more than the
maximum number of prefixes. The maximum is 1.
Even if you create the exact same type in another database, with the same name, it won’t work.
(Note that UDTs don’t have to be tables: go here [MSDN] for more info.)
Just for kicks, here’s how I’d replicate the above functionality, but using XML:
-- If the sproc already exists, delete it:
IF OBJECT_ID('dbo.usp_MySproc') IS NOT NULL
BEGIN
DROP PROCEDURE dbo.usp_MySproc
END
GO
-- Create the sproc:
CREATE PROCEDURE dbo.usp_MySproc
(
@ListOfDates XML
)
AS
BEGIN
WITH cte_ExtractFromXML AS
(
SELECT
DateTypeID = d.i.value('(@type)[1]', 'INT')
,[Date] = d.i.value('.[1]', 'DATE')
FROM @ListOfDates.nodes('//date') d(i)
)
SELECT
DateTypeID
,MIN([Date]) AS MinDate
,MAX([Date]) AS MaxDate
FROM cte_ExtractFromXML
GROUP BY DateTypeID
ORDER BY DateTypeID
END
GO
-- Test it with some XML data:
DECLARE @MyListOfDates XML
SET @MyListOfDates = '
<listOfDates>
<date type="1">2016Jan01</date>
<date type="1">2016Jan02</date>
<date type="1">2016Jan03</date>
<date type="2">2016Oct10</date>
<date type="2">2017Jan01</date>
<date type="3">2017Feb01</date>
<date type="3">2017Feb03</date>
<date type="3">2017Feb28</date>
</listOfDates>'
EXEC dbo.usp_MySproc @ListOfDates = @MyListOfDates
GO
DateTypeID MinDate MaxDate
----------- ---------- ----------
1 2016-01-01 2016-01-03
2 2016-10-10 2017-01-01
3 2017-02-01 2017-02-28
If we wanted to, we could declare our @ListOfDates parameter as OUTPUT, and make changes to it in the sproc, e.g.:
Drop the sproc if it exists:
IF OBJECT_ID('dbo.usp_MySproc') IS NOT NULL
BEGIN
DROP PROCEDURE dbo.usp_MySproc
END
GO
-- Create our (new and improved) sproc:
CREATE PROCEDURE dbo.usp_MySproc
(
@ListOfDates XML OUTPUT
)
AS
BEGIN
DECLARE @Aggregated XML
;WITH cte_ExtractFromXML AS
(
SELECT
DateTypeID = d.i.value('(@type)[1]', 'INT')
,[Date] = d.i.value('.[1]', 'DATE')
FROM @ListOfDates.nodes('//date') d(i)
)
SELECT @Aggregated = (
SELECT
DateTypeID AS '@id'
,MIN([Date]) AS '@MinDate'
,MAX([Date]) AS '@MaxDate'
FROM cte_ExtractFromXML
GROUP BY DateTypeID
ORDER BY DateTypeID
FOR XML PATH('DataType'), ROOT('Aggregated')
)
SELECT @ListOfDates = (
SELECT @ListOfDates, @Aggregated FOR XML PATH('Output'), TYPE
)
END
GO
-- Run it:
DECLARE @MyListOfDates XML
SET @MyListOfDates = '
<listOfDates>
<date type="1">2016Jan01</date>
<date type="1">2016Jan02</date>
<date type="1">2016Jan03</date>
<date type="2">2016Oct10</date>
<date type="2">2017Jan01</date>
<date type="3">2017Feb01</date>
<date type="3">2017Feb03</date>
<date type="3">2017Feb28</date>
</listOfDates>'
EXEC dbo.usp_MySproc @ListOfDates = @MyListOfDates OUTPUT
SELECT @MyListOfDates AS MyListOfDates
GO
<Output>
<listOfDates>
<date type="1">2016Jan01</date>
<date type="1">2016Jan02</date>
<date type="1">2016Jan03</date>
<date type="2">2016Oct10</date>
<date type="2">2017Jan01</date>
<date type="3">2017Feb01</date>
<date type="3">2017Feb03</date>
<date type="3">2017Feb28</date>
</listOfDates>
<Aggregated>
<DataType id="1" MinDate="2016-01-01" MaxDate="2016-01-03" />
<DataType id="2" MinDate="2016-10-10" MaxDate="2017-01-01" />
<DataType id="3" MinDate="2017-02-01" MaxDate="2017-02-28" />
</Aggregated>
</Output>
As you can see, we’ve returned our original list, along with the aggregated data, both under a new parent node. | 2018-03-18 17:12:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4176166355609894, "perplexity": 5111.1155286923395}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645830.10/warc/CC-MAIN-20180318165408-20180318185408-00207.warc.gz"} |
http://www.cs.cmu.edu/~yandongl/optimization.html | ### Optimization for machine learning
#### Estimator choice:
First we need to write out the function to be optimized.
• Point estimate: Maximum Likelihood. Obtain the single point (model parameters $\theta$) that maximizes the probability data is generated from your model $p(Data; \theta)$.
• Bayesian point estimate: Maximum A Posteriori. Obtain the single point (model paramters $\theta$) that maximizes the posterior probability $p(\theta | Data)$
• Fully Bayesian approach: estimate the mean and variance
To minimize/maximize a function $F$, there are a few choices:
• only needs first derivative of $F$. simple to implement
• each iteration is cheap
• has an extra parameter (learning rate)
• features better be rescaled
• for linear regression, if use normal equaiton, no need to resclae
• When objective is convex but not smooth (e.g. hinge loss)
#### Newton's method:
• needs first and second derivatives of $F$ (Hessian matrix)
• each iteration is expensive
• but converges much faster
• $\theta = \theta - H^{-1}\bigtriangledown_{\theta}l(\theta)$. $\bigtriangledown_{\theta}l(\theta)$ is the partial derivative wrt $\theta$. $H$ is the Hessian matrix. | 2017-01-19 07:46:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9816906452178955, "perplexity": 2499.0549994823778}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280485.79/warc/CC-MAIN-20170116095120-00081-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://www.numdam.org/numdam-bin/fitem?id=ITA_2001__35_3_287_0 | ### Recherche et téléchargement d’archives de revues mathématiques numérisées
Table des matières de ce fascicule | Article précédent
Mishra, Sounaka; Sikdar, Kripasindhu
On the hardness of approximating some NP-optimization problems related to minimum linear ordering problem. RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications, 35 no. 3 (2001), p. 287-309
Texte intégral djvu | pdf | Analyses MR 1869219 | Zbl 1014.68063
Class. Math.: 68Q17, 68R01, 68W25
Mots clés: NP-optimization problems, minimaximal and maximinimal NP-optimization problems, approximation algorithms, hardness of approximation, APX-hardness, AP-reduction, L-reduction, S-reduction
URL stable: http://www.numdam.org/item?id=ITA_2001__35_3_287_0
Résumé
We study hardness of approximating several minimaximal and maximinimal NP-optimization problems related to the minimum linear ordering problem (MINLOP). MINLOP is to find a minimum weight acyclic tournament in a given arc-weighted complete digraph. MINLOP is APX-hard but its unweighted version is polynomial time solvable. We prove that MIN-MAX-SUBDAG problem, which is a generalization of MINLOP and requires to find a minimum cardinality maximal acyclic subdigraph of a given digraph, is, however, APX-hard. Using results of Håstad concerning hardness of approximating independence number of a graph we then prove similar results concerning MAX-MIN-VC (respectively, MAX-MIN-FVS) which requires to find a maximum cardinality minimal vertex cover in a given graph (respectively, a maximum cardinality minimal feedback vertex set in a given digraph). We also prove APX-hardness of these and several related problems on various degree bounded graphs and digraphs.
Bibliographie
[1] P. Alimonti and V. Kann, Hardness of approximating problems on cubic graphs, in Proc. 3rd Italian Conf. on Algorithms and Complexity. Springer-Verlag, Lecture Notes in Comput. Sci. 1203 (1997) 288-298. MR 1488983
[2] G. Ausiello, P. Crescenzi and M. Protasi, Fundamental Study: Approximate solution of NP optimization problems. Theoret. Comput. Sci. 150 (1995) 1-55. MR 1357119 | Zbl 0874.68145
[3] G. Ausiello, P. Crescenzi, G. Gambosi, V. Kann, A. Marchetti–Spaccamela and M. Protasi, Complexity and Approximation: Combinatorial Optimization Problems and Their Approximability Properties. Springer-Verlag, Berlin Heidelberg (1999). Zbl 0937.68002
[4] V. Bafna, P. Berman and T. Fujito, Constant ratio approximations of feedback vertex sets in weighted undirected graphs, in 6th Annual International Symposium on Algorithms and Computation (1995).
[5] A. Brandstädt, V.D. Chepoi and F.F. Dragan, The algorithmic use of hypertree structure and maximum neighborhood orderings. Discrete Appl. Math. 82 (1998) 43-77. MR 1610005 | Zbl 0893.05018
[6] A. Brandstädt and D. Kratsch, On domination problems for permutation and other graphs. Theoret. Comput. Sci. 54 (1987) 181-198. MR 919590 | Zbl 0641.68100
[7] S. Chanas and P. Kobylański, A new heuristic algorithm solving the linear ordering problem. Comput. Optim. Appl. 6 (1996) 191-205. MR 1398267 | Zbl 0860.90100
[8] M.S. Chang, Efficient algorithms for the domination problems on interval and circular-arc graphs. SIAM J. Comput. 27 (1998) 1671-1694. MR 1622646 | Zbl 0911.05051
[9] A. Chaudhary and S. Vishwanathan, Approximation algorithms for achromatic number, in Proc. 8th Ann. ACM-SIAM Symp. on Discrete Algorithms. ACM-SIAM (1997) 558-563. MR 1447703
[10] G.A. Cheston, G. Fricke, S.T. Hedetniemi and D.P. Jacobs, On the computational complexity of upper fractional domination. Discrete Appl. Math. 27 (1990) 195-207. MR 1058945 | Zbl 0717.05068
[11] P. Crescenzi, V. Kann, R. Silvestri and L. Trevisan, Structures in approximation classes, in 1st. Annu. Int. Conf. on Computing and Combinatorics. Springer-Verlag, Lecture Notes in Comput. Sci. 959 (1995) 539-548. MR 1450157
[12] U. Feige and J. Kilian, Zero knowledge and the chromatic number. Proc. Comp. Complexity (1996).
[13] M. Fraber, Independent domination in chordal graphs. Oper. Res. Lett. 1 (1982) 134-138. MR 687354 | Zbl 0495.05053
[14] M. Fraber and J.M. Keil, Domination in permutation graphs. J. Algorithms 6 (1985) 309-321. MR 800722 | Zbl 0598.05056
[15] T. Fujito, Personal communication (1999).
[16] M. Grötschel, M. Jünger and G. Reinelt, A cutting plane algorithm for linear ordering problem. Oper. Res. 32 (1984) 1195-1220. MR 775257 | Zbl 0554.90077
[17] M. Grötschel, M. Jünger and G. Reinelt, On the acyclic subgraph polytope. Math. Programming 33 (1985) 28-42. MR 809747 | Zbl 0577.05034
[18] J. Håstad, Clique is hard to approximate within $n^{1-\epsilon }$, in Proc. 37th IEEE Sympo. on Foundation of Comput. Sci. (1996) 627-636. MR 1450661
[19] F. Harary, Graph Theory. Addition-Wesley, Reading, MA (1969). MR 256911 | Zbl 0182.57702
[20] F. Harary, Maximum versus minimum invariants for graphs. J. Graph Theory 7 (1983) 275-284. MR 710904 | Zbl 0515.05053
[21] F. Harary and S. Hedetniemi, The achromatic number of a graph. J. Combin. Theory 8 (1970) 154-161. MR 253930 | Zbl 0195.25702
[22] M.M. Halldórsson, Approximating the minimum maximal independence number. Inform. Process. Lett. 46 (1993) 169-172. MR 1229204 | Zbl 0778.68041
[23] R.W. Irving, On approximating the minimum independent dominating set. Inform. Process. Lett. 37 (1991) 197-200. MR 1095707 | Zbl 0713.68033
[24] V. Kann, On the Approximability of NP-complete Optimization Problems, Ph.D. Thesis. Department of Numerical Analysis and Computing Science, Royal Institute of Technology, Stockholm (1992).
[25] V. Kann, Polynomially bounded minimization problems that are hard to approximate. Nordic J. Comput. 1 (1994) 317-331. MR 1335251 | Zbl 0817.68082
[26] S. Khanna, R. Motwani, M. Sudan and U. Vazirani, On syntactic versus computational views of approximability, in Proc. 35th Ann. IEEE Symp. on Foundations of Computer Science (1994) 819-836.
[27] C. Lund and M. Yannakakis, On the hardness of approximating minimization problems. J. ACM 41 (1994) 960-981. MR 1371491 | Zbl 0814.68064
[28] C.H. Papadimitriou and M. Yannakakis, Optimization, Approximation, and Complexity Classes. J. Comput. System Sci. 43 (1991) 425-440. MR 1135471 | Zbl 0765.68036
[29] K. Peters, R. Laskar and S.T. Hedetniemi, Maximinimal/Minimaximal connectivity in graphs. Ars Combinatoria 21 (1986) 59-70. MR 846681 | Zbl 0602.05044
[30] D.F. Manlove, Minimaximal and maximinimal optimization problems: A partial order-based approach, Ph.D. Thesis. University of Glasgow (1998).
[31] S. Mishra and K. Sikdar, On approximation solutions of linear ordering and related NP-Optimization problems on graphs (Extended Abstract), Electronic Notes in Discrete Mathematics, Vol. 8, edited by H. Broersma, U. Faigle, J. Hurink and S. Pickl. Elsevier Science Publishers (2001), Full version submitted for publication.
[32] S. Mishra and K. Sikdar, On the hardness of approximating some NP-optimization problems related to minimum linear ordering problem (extended abstract), edited by J. van Leeuwen et al., IFIP TCS 2000. Lecture Notes in Comput. Sci. 1872 (2000) 186-199. MR 1869219 | Zbl 1010.90523
[33] S. Ueno, Y. Kajtani and S. Gotoh, On the nonseparating independent set problem and feedback set problem for graphs with no vertex exceeding three. Discrete Math. 72 (1988) 355-360. MR 975556 | Zbl 0678.05026
Copyright Cellule MathDoc 2015 | Crédit | Plan du site | 2015-03-01 04:26:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.852276623249054, "perplexity": 4569.354568548653}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462206.12/warc/CC-MAIN-20150226074102-00311-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://canonij.co/canon-ij-network-tools/ | Canon Ij Network Tools
Canon Ij Network Tools
Canon Ij Network Tools
Canon IJ Network Tools Download Support for OS Windows and Mac – Canon IJ Network Tools Setup device is a utility that allows you to screen and modify the community settings within the instrument. Canon Network Tools It’s put in in the event the machine is ready
Canon Ij Network Tools that enable you to print and Scan from the wireless Canon IJ Networkprinter that is connected through a network
IJ Network Tool. IJ Network Tool is a utility that enables you to display and modify the machinenetwork settings. It is installed when the machine is set up.. Important. To use the machine over LAN, make sure you have the equipment necessary for the connection type, such as an access point or a LAN cable.
Canon IJ Network Tools is a Canon software that helps you to adjust the network printing configuration
Canon IJ Network Tools is a free application that allows you to install on your laptop or computer, view or configure the network settings of your printer is connected via a network.
Canon IJ Network Tools The Canon IJ Network Tool is a utility that enables you to display and modify the machine network settings. It is installed when the machine is set up.
canon ij network tools free download – Canon IJ Printer Driver Canon iP4200, Canon IJ Printer Driver Canon iP5200, Canon IJ Printer Driver Canon iP6600D, and many more programs.
canon ij network tools free download – Canon IJ Printer Driver Canon iP4200, Canon IJ Printer Driver Canon iP5200, Canon IJ Printer Driver Canon iP6600D, and many more programs.
Canon Ij Network Tools
Canon Ij Network Tools | 2019-06-15 20:39:49 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.803802490234375, "perplexity": 9960.81914211789}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627997335.70/warc/CC-MAIN-20190615202724-20190615224724-00556.warc.gz"} |
https://dakep.github.io/examinr/articles/grading.html | Grading is arguably one of the most time-consuming tasks of conducting an exam. Exams created with examinr come with a grading interface to support you in this task. Moreover, some question types (numeric questions and multiple-choice questions) are automatically graded, but still give you the opportunity to override these grades.
The grading interface is available under the same path as the exam document, but you need to append display=feedback to the query string. For instance, if the exam is available under http://127.0.0.1:8181/exam.Rmd, the grading interface would be accessible under http://127.0.0.1:8181/exam.Rmd?display=feedback. If the user accessing this page does not have grading permissions, the user will see only their individual feedback (if the feedback has been made public already). Only users with grading permissions are able to access the grading interface.
At the top of the grading interface is the control bar for selecting the learner (center) and the learner’s attempt:
Attempts are identified and ordered by the time they have been finished (i.e., the last section was submitted). By default, the learner’s latest finished attempt is selected. You can also select un-finished attempts Selecting the learner/attempt will re-render the page in the state as shown to the learner for this attempt, and update the solutions according to the realized values. The button on the far right exports grades for all learners. See below for details on the export.
exam id exam version learner id attempt id question label date-time of submission (NA for un-finished attempts) points awarded total points available for the question | 2021-01-16 22:02:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17744821310043335, "perplexity": 2109.726379804296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703507045.10/warc/CC-MAIN-20210116195918-20210116225918-00595.warc.gz"} |
https://houxianxu.github.io/2015/03/31/linear-regression-post/ | This post dicusses how to come up with linear regression algorithm, specifically how to define the loss function and minimize the loss with gradient decent algorithm. I also implement the linear regression using Python (numpy) to do experiment with a datasets, and the result can be found in this IPython notebook.
## 1. Problem setting
We want to use a predictor variable X to predict a quantitative response Y, such as using living area (X) to predict the price (Y) of house.
## 2. Basic idea
Becuase of just two variables, we can simply visualize the data on a scatter plot, then we can predict Y by the structure of the plot (see following figure)
After getting the scatter plot, we can estimate the position of potiential point when given the x value and get the corresponding value y.
## 3. Can we do better
So far it seems that the problem can be solved, however, we shold always ask the quesion, i.e., “Can we do better?”
What’s the shortcoming of the above solution?
• We have to firstly get the scatter plot, which is a problem when it scales to high dimension prediction problems.
• We need our human’s eyes to find the position of the position of potential point. We humans are not happy with that, instead we want the computer to do all the work. In addition, we can easily get overwhelmed when amounts of prediction needed.
## 4. Better idea
If we look at the structure of the scatter plot above, it is not hard to figure that the Y value is increasing when X gets bigger. So it is possible to find a model to fit all the data, then use the model instead for prediction.
Then what’s kind of model we should use? Linear model may be a good choice because of its simplicity and ability to show the general trend.
The next task is how to find the “best” line, such as $$y$$ $$\approx$$ $$f(x, w, b) = w x + b,$$ where w and b are the parameters of the function, in which w is called the weight and b is called bias, which doesn’t interact with the actual data $$x_i$$. In order to find the “best” line, we need to define “best” and measure it. Ideally we hope $y_i == f_i$ for every sample i, so we can use the difference or loss ($|y_i - f_i|$, also called L1 distance) between the target $y_i$ and predicted $f_i$ for measurement for a single sample. When considering all the samples, we want to minimize the average loss $$\frac{1}{m} \sum_{i=1}^m|f_i - y_i|$$ for all samples (m is the number of samples). Alternatively L2 distance can be used as well and the average loss is $L_2 = \frac{1}{m} \sum_i^m(f_i - y_i)^2,$ of course other measurement could be used as well.
Next we need to find the w and b to minimize L_2(w, b), i.e. least-squares loss, which is an optimization problem. Because of quadratic formula, we can guess $L_2$ is has bowl-shaped appearance in 3-dimension that $L_2$ in fact is a convex function. So based on college caculus，we can compute the partial derivative of w and b, then set them to be zero and compute the w and b (the bottom of the shape).
Above approach directly compute the best w and b based on the property of convex function, and we could ask ourselves is there other ways (say indirectly) to get w and b? Maybe we could firstly inilize w and b randomly, and then try to make it better little by little. By analogy, a blind hiker tries his best to reach the bottom of a hill, specifically try to take a step at every point.
• The first approach (Random Local Search) could be try to extend on foot in a random direction and take a step only if it leads down hill.
• Another better way is to follow the direction of steepest decend, which is the gradient or derivative of loss function at one point, and the w and b can be updated by following the best direction (gradient) and a given step (known as learning rate). Obviously the learning rate will have big impact on our algorithm; we can only get a very small progress if learing rate is too small, however when making a bigger step, we may get a higher loss because the point maybe jump to the other side of the bowl-shaped line. So we could do some research here, e.g. how to decide the learning rate (try different values with validation method), maybe we could make it dynamically. Another potential problem here is that we use all the samples to complish just one update when taking compulation complex into account. One solution is to update the parameters according to the gradient of the error with respect to one single training example only. This alogrithm is called stochastic gradient descent or online gradient decent, and batch gradient descent for previous one. SGD often gets “close” to the minimum much faster than BGD, however it may never “converge” to the minimum. Another bonus is that it is possible to ensure that the parameters will converge to the global minimum rather then merely oscillate around the minimum.
## 5. Genralization for high dimension data
When there are more than 1 predictor variable, we just need to change the model as $y \approx f(x, w, b) = \sum_{j=1}^n x_j w_j + b = w^T x + b$ (w and x are vector, and n is the number of features), in fact we can make the expression more compact by setting b = $w_0$ and $x_0 = 1$, then $f(x, w) = w^T x.$
The loss $L = \frac{1}{2m} \sum_{i=1}^m(f(x^{(i)}, w) - y^{(i)})^2$, where $x^{(i)}$ is a vector for all features $x_j^{(i)}$ (j=0,1, … , n) for single sample i, and $y^{(i)}$ is the target value for this example.
Compute the gradient for all w:
• Analytic gradient, using calculus to compute the gradient directly
• Numerical gradient, which is an approximation approach based on the definition of derivatives (or gradient). The derivative of a 1-D function is the limit of the function with respect its input. When the function takes more than one parameters, the derivatives are called partial derivatives.
Update w by gradient decient: $w_j := w_j - \alpha \frac{\partial}{\partial w_j} L(w, x)$, this regression is also called LMS standing for “least mean squares”.
Compared with two dimension model that is a line in 2-D space, we can look on n-dimension model as a n-hyperplane (subspace) in (n+1)-D space, e.g. a plane in a 3-D space.
## 6. Probabilistic interpretation
After understanding LMS regression above, we should again ask “Can we do better?”. I think the key or foundation of above is the least-square loss function $L_2 = \frac{1}{m} \sum_i^m(f_i - y_i)^2$. So why this is a reasonable choice?
Because we just use f(w, x) to estimate the target y and expectation is often used for estimation, so we can interpret $f(w, x^{(i)}) = w^T x^{(i)}$ as the expectation of estimation. So we could add an error term $\epsilon^{(i)}$ to previous experession, as a result $y = w^T x^{(i)} + \epsilon^{(i)}$. Because the expectation could be higher or less than the target value, we could even assume all $\epsilon^{(i)}$ are distributed IID (independently and identically distributed) according to a Gaussian Distribution (also called a Normal distribution) with mean zero and some variance $\sigma^2$, i.e., $\epsilon^{(i)} \sim \mathcal{N} (0,\sigma^2)$, so $y^{(i)} \sim \mathcal{N} (w^T x^{(i)},\sigma^2)$
So far when given a vector w and all $x_j$, we can compute the probability of $y^{(i)}$ from the Gaussian Distribution. Naturally we want the maximize all the probability of $y^{(i)}$ at the same time, and this method is called maximum likelihood. The corresponding likelihood function is
Instead of maximizing $L(w; x)$, we can also maximize any strictly increasing function of $L(w; x)$, naturally we can instead maximize likelihood $l(w; x)$
Because w are the only unknown parameters (assume $\sigma$ is known), we need only to minimize the second item $\frac{1}{\sigma^2} \frac{1}{2} \sum_{i=1}^m (y^{(i)} - w^T x^{(i)})^2$, which could be viewed as less-square loss.
When we see $\sigma$ as the unknown parameter, we could also calculate the “best” $\sigma$ to maximize the likelihood. I think the assumption that all the point have the same $\sigma$ is too strong to some degree, so if they are not the same and depend on X, we can get a different loss function.
## 7. Get your hands dirty and have fun
• Data: I use the data from linear regression exercise from Andrew Ng’s Machine learning on Coursera.
• Setup: I choose Python (IPython, numpy etc.) on Mac for implementation, and the results are published in a IPython notebook, click here for the details
• Following is code to implement the batch and stochastic gradient decent algorithms.
import numpy as np
class LinearRegression:
def __init__(self):
self.W = None # set the weight vector
def train(self, X, y, method='bgd', learning_rate=1e-2, num_iters=100, verbose=False):
"""
Train linear regression using batch gradient descent or stochastic gradient descent
Parameters
----------
X: (D x N) array of training data, each column is a training sample with D-dimension.
y: (N, ) 1-dimension array of target data with length N.
method: (string) determine wheter use 'bgd' or 'sgd'
learning_rate: (float) learning rate for optimization
num_iters: (integer) number of steps to take when optimization
verbose: (boolean) if True, print out the progress when optimization
Returns
-------
losses_history: (list) of losses at each training iteration
"""
dim, num_train = X.shape
if self.W is None:
# initilize weights with small values
self.W = np.random.randn(1, dim) * 0.001 # [1, D]
losses_history = []
for i in xrange(num_iters):
if method == 'sgd':
# randomly choose a sample
idx = np.random.choice(num_train)
loss, grad = self.loss_grad( X[:, idx, np.newaxis], y[idx, np.newaxis])
else:
loss, grad = self.loss_grad(X, y)
losses_history.append(loss)
# Update weights using matrix computing (vectorized)
self.W -= learning_rate * grad
if verbose and i % (num_iters / 10) == 0:
print 'iteration %d / %d : loss %f' %(i, num_iters, loss)
return losses_history
def predict(self, X):
"""
Predict value of y using trained weights
Parameters
----------
X: (D x N) array of data, each column is a sample with D-dimension.
Returns
-------
pred_ys: (N, ) 1-dimension array of y for N sampels
"""
pred_ys = self.W.dot(X)
return pred_ys
def loss_grad(self, X, y, vectorized=True):
"""
Compute the loss and gradients
Parameters
----------
The same as self.train function
Returns
-------
a tuple of two items (loss, grad)
loss: (float)
grad: (array) with respect to self.W
"""
if vectorized:
return linear_loss_grad(self.W, X, y)
else:
return linear_loss_grad_naive(self.W, X, y)
def linear_loss_grad(W, X, y):
"""
Compute the loss and gradients with weights, vectorized version
Parameters and Returns are the same as LinearRegression.loss_grad, except including W as parameter
"""
# vectorized implementation
num_train = X.shape[1]
f_mat = W.dot(X) # [1, D] * [D, N]
diff = f_mat - y # [1, N] - [1, N]
loss = 1.0 / (2 * num_train) * np.sum(diff * diff) # [1, N] * [N, 1]
grad = 1.0 / num_train * diff.dot(X.T) # [1, N] * [N, D]
return (loss, grad)
def linear_loss_grad_naive(W, X, y):
"""
Compute the loss and gradients with weights, for loop version
"""
num_train = X.shape[1]
loss = 0
grad = np.zeros_like(W) # [1, D]
for i in xrange(num_train):
sample_X = X[:, i] # a vector
f = 0
for j in xrange(W.shape[1]):
f += sample_X[j] * W[0, j]
diff = f - y[i]
loss += diff ** 2
for j in xrange(W.shape[1]):
grad[0, j] += diff * sample_X[j]
loss = 1.0 / (2 * num_train) * loss
grad = 1.0 / (num_train) * grad
return (loss, grad)
## 8. Summary
• We all should keep it in mind that linear regression is based on the assumption that the true model is linear or close to linear, so we should be very careful if we don’t known the true model in advance.
• Most people use least square error to indicate the loss of the linear model and it can be interpretated from probabilistic aspect, i.e., assuming that the errors are distributed IID according to a Gaussian Distribution, the probability of y based on x (p(y|x)) for all the samples can be maximized to minimize the least square error. Of course, we can choose other loss function as long as it makes sense to measure the agreement between the predicted scores and the ground truth value.
• We can use normal equation $W = (X^T X)^{-1} X^T y$ to compute W directly based on calculus, however it works slow when n is large, instead, gradient decent algorithm is more practical based on the bowl-shape of loss function. The basic idea is to reduce the loss step by step.
• For implementation, it is critical to use matrix calculation. Not only can it speed up the computation, but also can make code simpler and conciser when compared to naive loop version. | 2017-10-16 21:57:33 | {"extraction_info": {"found_math": true, "script_math_tex": 36, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.8020723462104797, "perplexity": 969.1227020012495}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187820466.2/warc/CC-MAIN-20171016214209-20171016234209-00396.warc.gz"} |
https://math.stackexchange.com/questions/1487046/infinite-coin-flips-set-notation-language | # Infinite Coin Flips- Set Notation language
Suppose a coin is flipped infinitely many times and we take as our sample space $S$ all possible infinite length sequences of heads and tails. For $n\geq1$, let $A_n$ be the event that in the first $n$ flips the proportion of heads is at least $1/2$.
(a) Express in terms of the events $A_n$ using set operations the event $A$, which is defined as the set of all sequences in $S$ with the property that for some $k$, the proportion of heads in the first $n$ flips is at least $1/2$ for all $n\geq k$.
Could someone help me parse this out a bit. I don't have trouble calculating actual probabilities but the set operation language and notation is a bit confusing for me. Any help would be appreciated.
• Hint: Assuming the $A_n$ are as described, given some integer $k$, what is $\bigcap_{n \ge k} A_n$? – BrianO Oct 19 '15 at 6:12
Let's start from the end. We want $A_n$ to occur for all $n\ge k$, i.e., $A_k$ and $A_{k+1}$ and ... . "And" translates to intersection, so this event is $A_k\cap A_{k+1}\cap \ldots$ or more succinctly $B_k:=\bigcap _{n\ge k} A_n$. This again shall happen "for some $k$", so for $k=1$ or $k=2$ or $k=3$ or ... . "Or" translates to union, so this event is $B_1\cup B_2\cup\ldots$, or in short $\bigcup_{k\in\Bbb N}B_k$. In summary, $$\bigcup_{k\in\Bbb N}\bigcap_{n\ge k}A_n.$$ | 2019-12-08 13:32:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9277450442314148, "perplexity": 128.34132540909667}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540510336.29/warc/CC-MAIN-20191208122818-20191208150818-00550.warc.gz"} |
https://www.calctool.org/thermodynamics/biot-numer | # Biot Number Calculator
Created by Krishna Nelaturu
Last updated: Nov 21, 2022
Our Biot number calculator will help you determine the Biot number of a body. The Biot number will help you understand the heat distribution in a body when you heat a part of its surface. In this article, we shall briefly discuss the Biot number, including:
• What is Biot number?
• Biot number equation
• Biot number for spheres and cylinders.
Is your goal to heat a body to a specific temperature? Our heat capacity calculator and specific heat calculator will help you with calculations on how much heat is needed to reach the desired temperature!
## What is Biot number?
The Biot number (Bi) is a dimensionless number used to predict how well a solid material conducts heat. It is named after French physicist Jean Baptiste Biot.
The Biot number is a ratio of the rate of heat transfer by conduction to the rate of heat transfer by convection. If you're heating a rod in a furnace, you can use this number to predict the heat distribution within the rod. Further, we can understand whether the rod will achieve thermal equilibrium quickly.
Objects with a large Biot number require more time to reach thermal equilibrium. However, this increased insulation can also lead to increased energy costs, as it takes longer for heat to escape from the system. As such, engineers must often strike a balance between different competing factors when designing systems that involve heat transfer.
## Biot number equation
The formula for the Biot number is:
$\text{Bi} = \frac{h}{k}L_c$
Where:
• $\text{Bi}$ - Biot number;
• $h$ - Heat transfer coefficient at the body surface in $\rm{W/m^2 \cdot K}$;
• $k$ - Thermal conductivity of the body in $\rm{W/m \cdot K}$; and
• $L_c$ - Characteristic length of the body.
We can calculate the body's characteristic length using the equation:
$L_c = \frac{V}{A}$
Where:
• $V$ - Volume of the body; and
• $A$ - Surface area through which the body is heated.
In other words, the characteristic length depends on the body geometry. Let's look at two common geometries - cylinders and spheres.
• Biot number for cylinder - The characteristic length of the cylinder with radius $r$ and height $H$ would be:
$\qquad L_c = \frac{V}{A} = \frac{\pi r^2 H}{2\pi r (r + H)}\\ [1em] \qquad L_c = \frac{ r H}{2 (r + H)}$
Using this characteristic length in the formula for Biot number, we would get the following:
$\qquad \text{Bi} = \frac{h}{k}\left( \frac{ r H}{2 (r + H)} \right)$
• Biot number for sphere - The characteristic length of the sphere with radius $r$ would be:
$\qquad L_c = \frac{V}{A} = \frac{\frac{4}{3}\pi r^3}{4\pi r^2}\\ [1em] \qquad L_c = \frac{ r}{3}$
Using this characteristic length in the formula for the Biot number, we would get the following:
$\qquad \text{Bi} = \frac{rh}{3k}$
## Using this tool to calculate Biot number
Our Biot number calculator is simple to use:
• Enter the body's surface area and volume. The calculator will determine the characteristic length using these values. If you know the characteristic length, you can enter it directly in its corresponding field.
• Provide the heat transfer coefficient value.
• Enter the thermal conductivity of the material. The calculator will use all the given data to calculate the Biot number.
Krishna Nelaturu
Characteristic length
Surface area
Volume
Biot number
Characteristic length
m
Heat transfer coefficient
W/(m² * K)
Thermal conductivity
W/(m⋅K)
Biot number
People also viewed…
### Gay-Lussacs's law
Learn how to find the missing variable in an isochoric process with our Gay-Lussac's law calculator.
### Gibbs free energy
With our Gibbs free energy calculator, you will calculate the delta G in any chemical reaction with a few clicks! | 2022-12-05 14:40:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 17, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7415981888771057, "perplexity": 806.184875290538}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711017.45/warc/CC-MAIN-20221205132617-20221205162617-00661.warc.gz"} |
https://www.nature.com/articles/nphys781?error=cookies_not_supported&code=be887b8a-d455-4ce5-96e4-37bae20f503a | Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.
# Observation of electron–hole puddles in graphene using a scanning single-electron transistor
## Abstract
The electronic structure of graphene causes its charge carriers to behave like relativistic particles. For a perfect graphene sheet free from impurities and disorder, the Fermi energy lies at the so-called ‘Dirac point’, where the density of electronic states vanishes. But in the inevitable presence of disorder, theory predicts that equally probable regions of electron-rich and hole-rich puddles will arise. These puddles could explain graphene’s anomalous non-zero minimal conductivity at zero average carrier density. Here, we use a scanning single-electron transistor to map the local density of states and the carrier density landscape in the vicinity of the neutrality point. Our results confirm the existence of electron–hole puddles, and rule out extrinsic substrate effects as explanations for their emergence and topology. Moreover, we find that, unlike non-relativistic particles the density of states can be quantitatively accounted for by considering non-interacting electrons and holes.
This is a preview of subscription content
## Access options
\$32.00
All prices are NET prices.
## References
1. Brandt, N. B., Chudinov, S. M. & Ponomarev, Y. G. Semimetals 1, Graphite and Its Compounds (North-Holland, Amsterdam, 1988).
2. Semenoff, G. W. Condensed-matter simulation of a three dimensional anomaly. Phys. Rev. Lett. 53, 2449–2452 (1984).
3. Haldane, F. D. M. Model for a quantum Hall effect without Landau levels: Condensed-matter realization of the “parity anomaly”. Phys. Rev. Lett. 61, 2015–2018 (1988).
4. Zhou, S. Y. et al. First direct observation of Dirac fermions in graphite. Nature Phys. 2, 595–599 (2006).
5. Bostwick, A., Ohta, T., Seyller, T., Horn, K. & Rotenberg, E. Quasiparticle dynamics in graphene. Nature Phys. 3, 36–40 (2007).
6. Fetter, A. L. & Walecka, J. D. Quantum Theory of Many Particle Systems (McGraw-Hill, New York, 1971).
7. Bello, M. S., Levin, E. I., Shklovskii, B. I. & Efros, A. L. Density of localized states in the surface impurity band of a metal-insulator-semiconductor structure. Sov. Phys. JETP 53, 822–829 (1981).
8. Eisenstein, J. P., Pfeiffer, L. N. & West, K. W. Compressibility of the two-dimensional electron gas: Measurements of the zero-field exchange energy and fractional quantum Hall gap. Phys. Rev. B 50, 1760–1778 (1994).
9. Peres, N. M. R., Guinea, F. & Castro Neto, A. H. Coulomb interactions and ferromagnetism in pure and doped graphene. Phys. Rev. B 72, 174406 (2005).
10. Barlas, Y., Pereg-Barnea, T., Polini, M., Asgari, R. & MacDonald, A. H. Chirality and correlations in graphene. Phys. Rev. Lett. 98, 236601–236604 (2007).
11. Hwang, E. H., Hu, B. Y.-K. & Das Sarma, S. Density dependent exchange contribution to δμ/δn in extrinsic graphene. Preprint at <http://www.arxiv.org/cond-mat/0703499v1> (2007).
12. Eisenstein, J. P., Pfeiffer, L. N. & West, K. W. Negative compressibility of interacting two-dimensional electron and quasiparticle gases. Phys. Rev. Lett. 68, 674–677 (1992).
13. Millard, S. et al. Effect of finite quantum well width on the compressibility of a two-dimensional electron gas. Phys. Rev. B 55, 6715–1618 (1997).
14. Shapira, S. et al. Thermodynamics of a charged fermion layer at high r s values. Phys. Rev. Lett. 77, 3181–3184 (1996).
15. Kravchenko, S. V., Ringberg, D. A., Semenchinsky, S. G. & Pudalov, V. M. Evidence for the influence of electron–electron interaction on the chemical potential of the two-dimensional electron gas. Phys. Rev. B 42, 3741–3744 (1990).
16. Novoselov, K. S. et al. Electric field effect in atomically thin carbon films. Science 306, 666–669 (2004).
17. Novoselov, K. S. et al. Two dimensional gas of massless Dirac fermions in graphene. Nature 438, 197–200 (2005).
18. Zhang, Y., Tan, Y.-W., Stormer, H. L. & Kim, P. Experimental observation of the quantum Hall effect and Berry’s phase in graphene. Nature 438, 201–204 (2005).
19. Peres, N. M. R., Guinea, F. & Castro Neto, A. H. Electronic properties of disordered two-dimensional carbon. Phys. Rev. B 73, 125411 (2006).
20. Sheng, D. N., Sheng, L. & Sheng, Z. Y. Quantum Hall effect in graphene: disorder effect and phase diagram. Phys. Rev. B 73, 233406 (2006).
21. Khveschenko, D. V. Electron localization properties in graphene. Phys. Rev. Lett. 97, 036802–036805 (2006).
22. Ziegler, K. Robust transport properties in graphene. Phys. Rev. Lett. 97, 266802–266805 (2006).
23. Ostrovsky, P. M., Gornyi, I. V. & Mirlin, A. D. Electron transport in disordered graphene. Phys. Rev. B 74, 235443 (2006).
24. Aleiner, I. L. & Efetov, K. B. Effect of disorder on transport in graphene. Phys. Rev. Lett. 97, 236801–236804 (2006).
25. Hwang, E. H., Adam, S. & Das Sarma, S. Carrier transport in two-dimensional graphene layers. Phys. Rev. Lett. 98, 186806–186809 (2007).
26. Nomura, K. & MacDonald, A. H. Quantum transport of massless Dirac fermions in graphene. Phys. Rev. Lett. 98, 076602–076605 (2007).
27. Cheianov, V. & Fal’ko, V. I. Friedel oscillations, impurity scattering and temperature dependence of resistivity in graphene. Phys. Rev. Lett. 97, 226801–226804 (2006).
28. McCann, E. et al. Weak-localization magnetoresistance and valley symmetry in graphene. Phys. Rev. Lett. 97, 146805–146809 (2006).
29. Gonzales, J., Guinea, F. & Vozmediano, M. A. H. Electron–electron interactions in graphene sheets. Phys. Rev. B. 63, 134421 (2001).
30. Castro Neto, A. H. & Kim, E.-A. Charge inhomogeneity and the structure of graphene sheets. Preprint at <http://www.arxiv.org/cond-mat/0702562v1> (2007).
31. Novoselov, K. S. et al. Two-dimensional atomic crystals. Proc. Natl Acad. Sci. USA 102, 10451–10453 (2005).
32. Schedin, F. et al. Detection of individual gas molecules adsorbed on graphene. Nature Mater. 6, 652–655 (2007).
33. Matsui, T. et al. STS observations of Landau levels at graphite surfaces. Phys. Rev. Lett. 94, 226403–226406 (2005).
34. Niimi, Y., Kambara, H., Matsui, T., Yoshioka, D. & Fukuyama, H. Real-space imaging of alternate localization and extension of quasi-two-dimensional electronic states at graphite surfaces in magnetic fields. Phys. Rev. Lett. 97, 236804–236807 (2006).
35. Niimi, Y. et al. Scanning tunneling microscopy and spectroscopy of the electronic local density of states of graphite surfaces near monoatomic step edges. Phys. Rev. B 73, 085421 (2006).
36. Ishigami, M., Chen, J. H., Cullen, W. G., Fuhrer, M. S. & Williams, E. D. Atomic structure of graphene on SiO2 . Nano Lett. 7, 1643–1648 (2007).
37. Rutter, G. M. et al. Scattering and interference in epitaxial graphene. Science 317, 219–222 (2007).
38. Yoo, M. J. et al. Scanning single-electron transistor microscopy: Imaging individual charges. Science 276, 579–582 (1997).
39. Yacoby, A., Hess, H. F., Fulton, T. A., Pfeiffer, L. N. & West, K. W. Electrical imaging of the quantum Hall state. Solid State Commun. 111, 1–13 (1999).
40. Jiang, Z. et al. Infrared spectroscopy of Landau levels of graphene. Phys. Rev. Lett. 98, 197403–197407 (2007).
41. Cheianov, V. V., Fal’ko, V. I., Altschuler, B. L. & Aleiner, I. L. Random resistor network model of minimal conductivity in graphene. Phys. Rev. Lett. 99, 176801 (2007).
42. Geim, A. K. & Novoselov, K. S. The rise of graphene. Nature Mater. 6, 183–191 (2007).
43. Katnelson, M. I., Novoselov, K. S. & Geim, A. K. Chiral tunnelling and the Klein paradox in graphene. Nature Phys. 2, 620–625 (2006).
44. Cho, S. & Fuhrer, M. S. Charge transport and inhomogeneity near the charge neutrality point in graphene. Preprint at <http://www.arxiv.org/abs/0705.3239> (2007).
45. Prange, R. E. & Girvin, S. M. The Quantum Hall Effect (Springer, New York, 1990).
46. Ilani, S. et al. The microscopic nature of localization in the quantum Hall effect. Nature 427, 328–332 (2004).
## Acknowledgements
We acknowledge helpful discussions on the preparation of graphene flakes with K. Novoselov and A. Geim. We would also like to acknowledge fruitful discussions with F. von Oppen.
## Author information
Authors
### Corresponding author
Correspondence to A. Yacoby.
## Supplementary information
### Supplementary Information
Supplementary Material (PDF 58 kb)
## Rights and permissions
Reprints and Permissions
Martin, J., Akerman, N., Ulbricht, G. et al. Observation of electron–hole puddles in graphene using a scanning single-electron transistor. Nature Phys 4, 144–148 (2008). https://doi.org/10.1038/nphys781
• Accepted:
• Published:
• Issue Date:
• DOI: https://doi.org/10.1038/nphys781
• ### Improving stability in two-dimensional transistors with amorphous gate oxides by Fermi-level tuning
• Theresia Knobloch
• Burkay Uzlu
• Tibor Grasser
Nature Electronics (2022)
• ### Selective hydrogenation improves interface properties of high-k dielectrics on 2D semiconductors
• Yulin Yang
• Tong Yang
• Ming Yang
Nano Research (2022)
• ### Weak ferroelectric charge transfer in layer-asymmetric bilayers of 2D semiconductors
• Fábio Ferreira
• Samuel J. Magorrian
Scientific Reports (2021)
• ### Nanoscale electric-field imaging based on a quantum sensor and its charge-state control under ambient condition
• Ke Bian
• Wentian Zheng
• Ying Jiang
Nature Communications (2021)
• ### Systematic THz study of the substrate effect in limiting the mobility of graphene
• Samantha Scarfe
• Wei Cui
• Jean-Michel Ménard
Scientific Reports (2021) | 2022-06-30 15:03:00 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8876703977584839, "perplexity": 10420.750204219774}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00005.warc.gz"} |
https://gianluca--gianlucabaio.netlify.app/project/football/ | # Bayesian hierarchical model for the prediction of football results
## Introduction
The problem of modelling football data has become increasingly popular in the last few years and many different models have been proposed with the aim of estimating the characteristics that bring a team to lose or win a game, or to predict the score of a particular match. We have worked within a Bayesian hierarchical framework to fulfil both these aims and test the predictive strength of our resulting models, based on several observed datasets. In our first paper , we have used data about the Italian Serie A 1991–1992 and 2007–2008 championships.
We have then applied a similar version of this model to the FIFA World Cup 2014. In that case, we have used historical data on friendly, qualifiers and major competition games at the international level to model the results of the World Cup. We have run the model dynamically to continuously update the predictions based on the results that were observed during the competition and our results have been reported in a series of posts, starting from here. Finally, I have used a more complex version of the model, including a few relevant covariates, to model data on a variety of leagues and seasons. This work has then been written up in .
## Modelling
The basic structure of the model is represented in the following graph.
Here, the two observed variables ($$y_{g1}$$ and $$y_{g2}$$) represent the number of goals scored by teams $$t=1,2$$ in game $$g=1,\ldots,G$$. We model these using two Poisson distributions, $$y_{gt}\sim\mbox{Poisson}(\theta_{gt})$$ and then we impose a structure on the joint distribution of the scoring rates’’ $$\theta_{gt}$$. These depend on some common effects and — here the nested indices $$h(g)$$ and $$a(g)$$ indicate the team playing home or away in game $$g$$. These are in turn assumed to be exchangeable across the teams in a given season/league and thus we imply some correlation across the observable variables.
Our original model was fitted using MCMC in OpenBUGS. In we also discuss issues with overshrinkage’’ produced by the Bayesian hierarchical model, where predictions for the most extreme teams (either the very good or the very bad ones) were pooled too much towards the grand mean, thus diluting some of the differences in the league. To alleviate this problem, we specify a more complex mixture model that results in a better fit to the observed data.
Our later versions of the model have been constructed in INLA, to speed up the analysis and deal with much larger datasets. The basic structure of the model has not been changed.
Last updated: Friday 19 November 2021 | 2022-01-24 01:22:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40319085121154785, "perplexity": 477.88111644720834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00208.warc.gz"} |
https://codereview.stackexchange.com/questions/121872/number-guessing-game-with-betting | # Number-guessing game with betting
I'm really new to programming, currently following several tutorials and guidelines. After a couple of days of studying, I've learnt the basics of Java and conditionals and took on my first program. It was written all out of my head.
The program is a sort of gambling machine. Each turn, the player has to guess a number out of 10 and if his number gets drawn he wins 50 credits. He must choose how much money to play, for how many turns and the stake for each turn.
I would like to receive some feedback regarding my code. What did I do wrong or some helpful tips/suggestions. Once I completed it, I keep having this feeling that the code is too long considering the simple task it does. How could I make it more compact?
import java.util.Random;
import java.util.Scanner;
/**
* Created by doublin on 3/3/16.
*/
public class Lottery {
public static void main(String[] args) {
System.out.println("Welcome!");
System.out.println("In this game you must guess a number between 1 and 10.");
System.out.println("How much money you want to load in to the machine?");
int currentMoney;
Scanner moneyInput = new Scanner(System.in);
while (!moneyInput.hasNextInt()) {
moneyInput.next();
System.out.println("Invalid amount.");
}
currentMoney = moneyInput.nextInt();
//******************************(START) NUMBER OF PLAYS **********************************************
System.out.println("How many times you want to try?");
int maxTries;
int currentTries = 0;
Scanner userTries = new Scanner(System.in);
while (!userTries.hasNextInt()) {
userTries.next();
System.out.println("Invalid choice.");
}
maxTries = userTries.nextInt();
//*******************************(END) NUMBER OF PLAYS **********************************************
//*******************************(START) STAKE CHOICE ***********************************************
System.out.println("Choose a stake:");
int stake;
Scanner stakeInput = new Scanner(System.in);
while (!stakeInput.hasNextInt()) {
stakeInput.next();
System.out.println("Invalid stake.");
}
stake = stakeInput.nextInt();
//***************************************(END) STAKE CHOICE *************************************************
while(currentTries < maxTries && currentMoney > 0) {
currentMoney = currentMoney - stake;
int userNumber;
Scanner userChoice = new Scanner(System.in);
while (!userChoice.hasNextInt()) {
userChoice.next();
System.out.println("Only numbers from 1 to 10 please.");
}
userNumber = userChoice.nextInt();
Random random = new Random();
int extractedNumber = random.nextInt((10-1) + 1) + 1;
System.out.println("The extracted number is: " + extractedNumber);
if (userNumber == extractedNumber) {
currentMoney = currentMoney + 50;
System.out.println("Congratulations! You've won 50 bucks!");
System.out.println("Now you have " + currentMoney);
System.out.println();
}
else {
System.out.println("You were unlucky.");
System.out.println("You have left " + currentMoney);
System.out.println();
}
currentTries++;
if (currentMoney <= 0) {
System.out.println("*** GAME OVER ***");
System.out.println("Sorry, you lost all your money.");
}
if (currentTries == maxTries) {
System.out.println("*** GAME OVER ***");
System.out.println("You reached " + gameNumber + " games.");
System.out.println("Money left: " + currentMoney);
}
}
}
}
• Welcome to Code Review! This question is incomplete. To help reviewers give you better answers, please add sufficient context to your question. The more you tell us about what your code does and what the purpose of doing that is, the easier it will be for reviewers to help you. See also this meta question. Mar 4 '16 at 9:33
It's not a very fair betting machine, because the stake does nothing. You could bet 0 money and the machine would still payout 50 if you were correct. Similarly, you could stake 500 per round, and you'd lose even if you won!
Aside from that, you should split your code up into separate functions.
Take getting a number, for instance.
You managed to spot that they're three separate segments, by labelling them with a comment. But what you could do is just make them separate function calls:
System.out.println("How much money you want to load in to the machine?");
int currentMoney;
Scanner moneyInput = new Scanner(System.in);
while (!moneyInput.hasNextInt()) {
moneyInput.next();
System.out.println("Invalid amount.");
}
currentMoney = moneyInput.nextInt();
System.out.println("How many times you want to try?");
int maxTries;
int currentTries = 0;
Scanner userTries = new Scanner(System.in);
while (!userTries.hasNextInt()) {
userTries.next();
System.out.println("Invalid choice.");
}
maxTries = userTries.nextInt();
These two snippets differ in that currentTries is wedged in the second one, but we don't need it here - it's not useful for doing things to maxTries. They also differ in the variable written to, which we can use as return value and the strings displayed.
I'd make a function getNumber(Scanner inputSource, String initialMessage, String errorMessage).
public int getNumber(Scanner inputSource, String initialMessage, String errorMessage){
System.out.println(initialMessage);
while (!inputSource.hasNextInt()) {
inputSource.next();
System.out.println(errorMessage);
}
return inputSource.nextInt();
}
Directly integrating with your code, it would look like so:
//******************************(START) MONEY LOADING ***********************************************
int currentMoney = getNumber(new Scanner(System.in), "How much money you want to load in to the machine?", "Invalid amount.");
//******************************(START) NUMBER OF PLAYS **********************************************
int currentTries = 0;
int maxTries = getNumber(new Scanner(System.in), "How many times you want to try?", "Invalid choice.");
//*******************************(END) NUMBER OF PLAYS **********************************************
//*******************************(START) STAKE CHOICE ***********************************************
int stake = getNumber(new Scanner(System.in), "Choose a stake:", "Invalid stake.");
//***************************************(END) STAKE CHOICE *************************************************
As a direct result, you no longer need all the //****** anymore.
int currentMoney = getNumber(new Scanner(System.in), "How much money you want to load in to the machine?", "Invalid amount.");
int currentTries = 0;
int maxTries = getNumber(new Scanner(System.in), "How many times you want to try?", "Invalid choice.");
int stake = getNumber(new Scanner(System.in), "Choose a stake:", "Invalid stake.");
You might want to fix it so that they all use the same scanner, but you'll have to test carefully there, because users might be capable of entering things like "10 20" which is two separate ints (they'd answer 2 questions at the same time). | 2022-01-28 17:30:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22572894394397736, "perplexity": 5762.8895598650815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00245.warc.gz"} |
https://supportforums.cisco.com/t5/other-network-infrastructure/cpu-and-bw-monitoring-for-3550-series/td-p/299524 | cancel
Showing results for
Did you mean:
New Member
## CPU and BW monitoring for 3550 Series
Hi,
I am trying to monitor a couple of 3550 Switches for CPU and BW Utilization.
I know that the CPU usage didn't mean a lot for 2900\3500 series switches. Is that still true for 3550 switches also ?
What is the best way to monitor the Switch BW Utilization (not using CMS), Is there an OID for this ?
Regards \\ Naman
3 REPLIES
Silver
New Member
New Member
## Re: CPU and BW monitoring for 3550 Series
OID for CPU utilization of CISCO devices is
1.3.6.1.4.1.9.2.1.58.0
95
Views
0 | 2017-12-15 11:17:08 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9307606816291809, "perplexity": 11706.754910624648}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00651.warc.gz"} |
http://sun.aos.wisc.edu/research | ## Research
Posted by Alex • Comments •
Atmospheric aerosols - airborne liquid and/or solid particles - are a critical source of uncertainty in climate science. Aerosols can change Earth's radiation balance by scattering and absorbing sunlight. Whereas most aerosols scatter sunlight which cools the atmosphere, some aerosols (including soot and dust) absorb sunlight which warms the atmosphere.
The behavior of aerosols affects the world around us every day. For example, the scattering properties of aerosols can make a hazy sky appear brighter than a clear sky. Since much of the sunlight is scattered back to space, aerosols can greatly reduce the intensity of sunlight reaching the surface. In heavily polluted areas, aerosols can reduce sunlight reaching the surface by over ten percent.
The changes in solar radiation due to the presence of aerosols are known as aerosol direct radiative effects. Global estimates of aerosol radiative direct effects may be obtained by combining multi-sensor satellite data including CloudSat, CALIPSO, and MODIS. The figures below display global mean aerosol radiative direct effects from 2006 through 2011. Results are presented at several height levels, including Earth's surface and the top of the atmosphere. Figures also display aerosol forcing due to atmospheric absorption (TOA minus SFC). As shown, the largest aerosol direct effects are found at locations near the source regions of the aerosols.
For more information about the CloudSat 2B-FLXHR-LIDAR product dataset, please refer to the official documentation from the CloudSat Data Processing Center. A complete description of this work and key results are provided in my master's thesis, "A Global Survey of Aerosol Direct Effects".
# Liquid Water Path Algorithm
Satellite-borne microwave radiometers detect faint microwave emissions from Earth's surface and atmosphere, using a metric known as brightness temperature. From these measurements various geophysical parameters may be retrieved, such as column integrated liquid water path. These measurements provide valuable information about atmospheric conditions in locations otherwise poorly sampled by in-situ techniques.
AMSR-E is a dual-polarized microwave radiometer aboard the Aqua satellite platform. AMSR-E brightness temperature measurements may be used to retrieve liquid water path in the atmosphere. To improve the current liquid water path algorithm, quartic regression analysis was performed to reduce systematic error. The result is improved performance in the AMSR-E retrieval of liquid water path. The figures below show results for three AMSR-E channels. Binned averages of brightness temperature as a function of column water vapor and surface wind speed. Empirical regression fits are displayed, as well as the difference between observed and estimated values. The relationship of brightness temperature to retrieved wind speed and column water vapor is given by the following quartic equation.
\begin{eqnarray} T_B \left (U,V \right ) &=& T_{B0} + aV + bV^2 + cV^3 + dV^4 + eU + fUV + gUV^2 + \\ && hUV^3 + iU^2 + jU^2V + kU^2V^2 + lU^3 + mU^3V + nU^4 \end{eqnarray}
The terms included are as follows: TB is the estimated brightness temperature, TB0 is the computed brightness temperature, U is surface wind speed, V is column water vapor, and a-n are regression coefficients. Surface wind speed and column water vapor are computed a priori using brightness temperatures from a combination of AMSR-E channels. Differences between observed and estimated brightness temperatures provide a metric for evaluating the performance of the new algorithm at various combinations of surface wind speed and column water vapor.
# Volcanic Ash Dispersion
Aircraft in flight routes over Alaska are periodically faced with the threat of volcanic eruptions. Air traffic within close proximity to an active volcano are most at risk for encountering ash. Ash-related aircraft damages have been reported as far as 1800 miles from an eruption. Potential damages include windscreen scouring, communication disruption, and engine failure.
The economic impact to the aviation industry includes not only damages to aircraft but also the costs of rerouting, delaying, and canceling flights. The 1989 eruption of Mt. Redoubt was especially costly due after a commercial flight encountered volcanic ash. KLM flight 867 was able to recover from a 13,000-foot powerless descent over mountainous terrain, however the structural and mechanical damages to the aircraft exceeded $80 million. The total cost to the aviation industry during that year was estimated at$101 million, according to a 1998 economic impacts study.
A case study of the 2009 eruption of Mt. Redoubt was performed to examine the vulnerability posed to aircraft by volcanic ash clouds. A series of eruptions occurred during late March and early April 2009. The first major volcanic ash eruption, on 23 March 2009, produced an ash plume exceeding 60,000 feet according to NWS radar observations. The dispersion of volcanic ash was monitored using satellite retrievals, including OMI column SO2 and CALIPSO backscatter data, and model simulations, including HYSPLIT forward trajectories and NCEP/NCAR reanalyses. The figures below show Google Earth visualizations of satellite and model data for three days following an explosive eruption on 23 March 2009. | 2017-06-25 12:03:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6121899485588074, "perplexity": 4077.8620270996435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320491.13/warc/CC-MAIN-20170625115717-20170625135717-00564.warc.gz"} |
https://infoscience.epfl.ch/record/53056 | ## Communication Infrastructure Design for Wide-Area Mobile Computation: Specification in Nomadic Pict
We review an example of wide-area mobile agent applications: video-on-demand, long-lived scientific computation, and collaborative work, and the design of a distributed infrastructure required in each of these applications for location-independent communication. For the latter application, we propose an infrastructure algorithm that assumes two kinds of collaboration: (1) within a group of mobile'' individuals, who can communicate frequently using different computers connected to a local-area network (possibly via a wireless medium), and (2) some individuals may also communicate outside their groups using the global network. The algorithm has been specified formally, as an executable encoding in Nomadic Pict. The formal specification is concise but gives enough details to be directly translated by application programmers using their language of choice.
Year:
2005
Keywords:
Laboratories: | 2018-02-18 21:42:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27820485830307007, "perplexity": 3143.9523957811302}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812259.30/warc/CC-MAIN-20180218212626-20180218232626-00650.warc.gz"} |
https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-4-section-4-5-summary-of-curve-sketching-4-5-exercises-page-321/9 | ## Calculus: Early Transcendentals 8th Edition
A. The domain is $(-\infty,1) \cup(1, \infty)$ B. The y-intercept is $0$ The x-intercept is $0$ C. The function is not an odd function or an even function. D. $\lim\limits_{x \to -\infty} (\frac{x}{x-1}) = 1$ $\lim\limits_{x \to \infty} (\frac{x}{x-1}) = 1$ $y = 1$ is a horizontal asymptote. $x = 1$ is a vertical asymptote. E. The function is decreasing on the intervals $(-\infty, 1)\cup (1,\infty)$ F. There is no local maximum or local minimum. G. The graph is concave down on the interval $(-\infty, 1)$ The graph is concave up on the interval $(1, \infty)$ There no points of inflection. H. We can see a sketch of the curve below.
$y = \frac{x}{x-1}$ A. The function is defined for all real numbers except $x = 1$. The domain is $(-\infty,1) \cup(1, \infty)$ B. When $x = 0,$ then $y = \frac{0}{0-1}= 0$ The y-intercept is $0$ When $y = 0$: $\frac{x}{x-1} = 0$ $x = 0$ The x-intercept is $0$ C. The function is not an odd function or an even function. D. $\lim\limits_{x \to -\infty} (\frac{x}{x-1}) = 1$ $\lim\limits_{x \to \infty} (\frac{x}{x-1}) = 1$ $y = 1$ is a horizontal asymptote. $x = 1$ is a vertical asymptote. E. We can try to find the values of $x$ such that $y' = 0$: $y' = \frac{(x-1)-(x)}{(x-1)^2} = -\frac{1}{(x-1)^2} = 0$ There are no values of $x$ such that $y'=0$ When $x \lt 1$ or $x \gt 1$, then $y' \lt 0$ The function is decreasing on the intervals $(-\infty, 1)\cup (1,\infty)$ F. There is no local maximum or local minimum. G. We can find the values of $x$ such that $y'' = 0$: $y'' = -\frac{-2(x-1)}{(x-1)^4} = \frac{2}{(x-1)^3}$ There are no values of $x$ such that $y''=0$ When $x \lt 1~~$, then $y'' \lt 0$ The graph is concave down on the interval $(-\infty, 1)$ When $x \gt 1$, then $y'' \gt 0$ The graph is concave up on the interval $(1, \infty)$ There no points of inflection. H. We can see a sketch of the curve below. | 2019-11-19 08:55:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8606823086738586, "perplexity": 87.56811092667883}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670036.23/warc/CC-MAIN-20191119070311-20191119094311-00491.warc.gz"} |
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=dan&paperid=8220&option_lang=eng | RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
General information Latest issue Archive Search papers Search references RSS Latest issue Current issues Archive issues What is RSS
Dokl. Akad. Nauk: Year: Volume: Issue: Page: Find
Dokl. Akad. Nauk SSSR, 1987, Volume 293, Number 2, Pages 343–345 (Mi dan8220)
MATHEMATICAL PHYSICS
Wave scattering by an unclosed cylindrical screen of arbitrary profile with the Neumann boundary condition
Yu. A. Tuchkin
Institute of Radiophysics and Electronics of the Academy of Sciences of the Ukrainian SSR, Kharkov
Full text: PDF file (369 kB)
Bibliographic databases:
UDC: 517.9:535.4
Presented: Þ. À. Ìèòðîïîëüñêèé
Citation: Yu. A. Tuchkin, “Wave scattering by an unclosed cylindrical screen of arbitrary profile with the Neumann boundary condition”, Dokl. Akad. Nauk SSSR, 293:2 (1987), 343–345
Citation in format AMSBIB
\Bibitem{Tuc87} \by Yu.~A.~Tuchkin \paper Wave scattering by an unclosed cylindrical screen of arbitrary profile with the Neumann boundary condition \jour Dokl. Akad. Nauk SSSR \yr 1987 \vol 293 \issue 2 \pages 343--345 \mathnet{http://mi.mathnet.ru/dan8220} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=884044}
• http://mi.mathnet.ru/eng/dan8220
• http://mi.mathnet.ru/eng/dan/v293/i2/p343
SHARE:
Citing articles on Google Scholar: Russian citations, English citations
Related articles on Google Scholar: Russian articles, English articles
This publication is cited in the following articles:
1. P. A. Krutitskii, “Neumann's problem for the Helmholtz equation outside cuts in the plane”, Comput. Math. Math. Phys., 34:11 (1994), 1421–1431
2. P. A. Krutitskiǐ, “Dirichlet's problem for the Helmholtz equation outside cuts in a plane”, Comput. Math. Math. Phys., 34:8-9 (1994), 1073–1090
3. P. A. Krutitskiǐ, “The mixed problem for the Helmholtz equation in a multiply connected region”, Comput. Math. Math. Phys., 36:8 (1996), 1087–1095 | 2019-12-14 01:48:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22560231387615204, "perplexity": 10772.856443558758}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00247.warc.gz"} |
https://www.trustudies.com/question/1033/the-following-number-of-goals-were-sc/ | 3 Tutor System
Starting just at 265/hour
# The following number of goals were scored by a team in a series of 10 matches: 2, 3, 4, 5, 0, 1, 3, 3, 4, 3
The number of goals scored by the team is
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
We know that,
Mean of data
= $$\frac{Sum of all observations}{Total number of observations}$$
Thus, Mean score
= $$\frac{2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3}{10} = \frac{28}{10}$$
= 2.8 goals
Now, Arranging the number of goals in ascending order, we get,
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Hence, the number of observations is 10, which is an even number.
Therefore, median Score will be the mean of $$\frac{10}{2}$$ i.e.,5th and $$\frac{10}{2} + 1$$ i.e,6th observation while arranged in ascending or descending order.
i.e., Median score
= $$\frac{5th observation + 6th observation}{2}$$
$$= \frac{3 + 3}{2}$$
$$= \frac{6}{2}$$
$$= 3$$
Mode of data is the observation With the maximum frequency in data.
Therefore, the mode score of data is 3 as has the maximum frequency as 4 in the data. | 2023-03-24 06:46:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.34018707275390625, "perplexity": 643.4358417187902}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00246.warc.gz"} |
http://tlng.medport24.ru/20307/560 | # 0.025 as a fraction
## Fraction
Add: wenaf17 - Date: 2020-12-18 18:03:26 - Views: 6866 - Clicks: 666
It is invaluable when: - Helping kids do the math homework - Adjusting the recipe ingredients to the number of servings you need - Doing the calculations for your craft or even construction project and more. E-LDP makes funds available within a couple of days. This means it is still under.
Log into Facebook to start sharing and connecting with your friends, family, and people you know. "1" and "2" are the numerators. The number on the top of the line is called the numerator.
More As images. 9/20) Club: 132: Statistiques : pourcentage/fraction d'un nombre (2) vero7000: 8548: 61% (12. 5% as a fraction? Superimpose fractions upon each other to compare fractions or see equal parts. Practice what you learn with games and quizzes. 025 liters to milliliters = 25 ml. Principes actifs : C&233;thexonium Excipients : Sodium chlorure, Eau pour pr&233;parations injectables. Over 50 safe and hassle-free payment options (prepaid / post-paid) available!
Wide choice of products. It is called an improper fraction. 01% atropine compared with placebo over 1 year based on the Low-Conc. Adding fractions is a very handy skill to know. De im Lager verf&252;gbar und in k&252;rzester Zeit bei Ihnen. Qu'est ce qu'une puissance? Choose the fraction model and number of equal parts.
You can express an integer as a fraction by simply dividing by 1, or you can express any integer as a fraction by simply choosing a numerator and denominator so that the overall value is equal to the integer. These MIG welding contact tips are designed for improved contact and increased conductivity. Related Products:. Pour plus d'information, veuillez contacter le. 025" Thickness (48KU44)?
When starting with fractions, begin by focusing on 1/2 and then a 1/4 before moving to equivalent fractions and using the 4 operations with. 25, we get; 0. The numerator is the number on top of the fraction. Simplify the fractions. First, note that a Z Score of 0.
Lifetime membership to a Kroma Pro Plan, now 89 percent off, empowers you to plug and pitch just like Samsung, IBM and AT&T. 003 Grammes: 30 Milligrammes = 0. A fraction (frae Laitin: fractus, "broken") represents a pairt o a whole or, mair generally, ony nummer o equal pairts. Fr: capsa&239;cine.
The most common examples of fractions from real life are equal slices of pizza, fruit, cake, a bar of chocolate, etc. As a standalone company and as part of Maersk we have been providing safety and support at sea since 1833. Another digital impact is that has now become easy to access sundry medical products available worldwide without wasting time and other resources. Le tableau suivant fournit les valeurs de certains quantiles de la loi de Fisher pour diff&233;rents param&232;tres ν 1 et ν 2. Recognize that each part has size 1/b and that the endpoint of the part based at 0 locates the number 1/b on the number line. We have to willingly or unwillingly share that yummy pizza amongst our friends and families. Aibel is a leading service company within the oil, gas and offshore wind industries. 01 Grammes: 2500 Milligrammes = 2.
From Middle English fraccioun (“a breaking”), from Anglo-Norman, from Old French fraction, from Medieval Latin fractio (“a fragment, portion”), from earlier Latin fractio (“a breaking, a breaking into pieces”), from fractus (English fracture), past participle of frangere (“to break”) (whence English frangible), from Proto-Indo-European *bʰreg- (English break). Miley Cyrus Midnight Sky. See the formulas below. Find Tutor For Maths. 025 uF Capacitors are available at Mouser Electronics. Another word for as. בית השקעות אלטשולר שחם נמנה על בתי ההשקעות המובילים בשוק ההון הישראלי, ומנהל תיקי השקעות, קופות גמל, קרנות פנסיה וקרנות נאמנות בהיקף של מיליארדי שקלים. 0186 into focus, with the eventual estimated bullish target at .
Aker Solutions helps the world meet its energy needs. After a workflow solution has been activated for a site collection, it is available as a feature for all sub-sites. In writing a fraction, e.
However many fractions you have, if they have the same bottom numbers, add up all the top numbers. Example 1. The microsomal fraction is. Specialties: If you ask one of our regulars what makes Dizz&39;s As Is a special place, they&39;ll probably mention a few things.
Ingredient matches for Acnetin A 0. That is called Simplifying, or Reducing the Fraction. The heart contracts and relaxes when it beats. How do you write an integer as a fraction? Sirius har via kontinuerlig produktudvikling og samarbejde med anerkendte designere, skabt Europas st&230;rkeste sortiment inden for dekorationsbelysning.
In common usage a fraction is any part of a unit. , eczema, dermatitis, allergies, rash). The denominator is the number that the numerator is being divided by.
Pour chaque param&232;tre, le quantile donn&233; est tel que la probabilit&233; pour qu'une variable suivant une loi de Fisher lui soit inf&233;rieur est de −. Qu'il s'agisse d'un emm&233;nagement, d’un d&233;m&233;nagement ou d'une nouvelle installation, vous devez informer le gestionnaire du service de l'eau de la commune concern&233;e qui proc&232;dera au changement de titulaire de l’abonnement ou &224; la r&233;siliation administrative de celui-ci. &92;(&92;frac14&92;) is the same as &92;(1 &92;div 4. In stock and ready to ship. Ejection fraction (EF) measures the amount of blood pumped out of your heart ’s lower chambers, or ventricles. People can view the slides even if they don’t have PowerPoint, but they can’t make changes to it. We can&39;t wait for you to join us for theyear.
The quotient of two quantities. 025% may be available in the countries listed below. 025% in the following countries: Egypt; Important Notice: The Drugs. 0270 resistance zone against the US Dollar. 3 out of 5 are girls.
We call the top number the Numerator, it is the number of parts we have. See the. 5% two or three times daily depending on the severity of the condition.
I had occasion. How to use fraction in a sentence. Internationalized Domain Name Support.
Ma et ne pas &234;tre disponibles sur le march&233; marocain. 25 degr&233;s) jusqu'&224; 240h. Calculer la puissance d'un nombre. Hello Friends,Check out our new video on "What is Fraction? Vous &234;tes en copropri&233;t&233;? During the contraction, it pushes blood out of large chambers. 025 Secondes: 5 Microsecondes = 5.
Sign up for news. ASUS and our third party partners use cookies (cookies are small text files placed on your products to personalize your user experience on ASUS products and services) and similar technologies such as web beacons to provide our products and services to you. 05ng/ml, 6mois plus tard toujours en m&234;mes termes -0. Tretinoin Cream. Forum The best place to ask for advice, get help using the site, or just talk about WoW. A fraction in which both the numerator and the denominator are whole numbers.
Fractionnel, elle, adj. The top number is called the numerator, and the bottom number is called the denominator. Mister A&39;s is just minutes from Downtown and an eternity from the ordinary dining experience.
Meanwhile, if . Decimal to Fraction Calculator. Fractions represent equal partsof a whole or a collection. Notation for Fractions. Fractions represent equal parts of a whole or a collection.
Example 1: Becky, Merry and John want to share a chocolate bar evenly. Fractions - Topics. A fraction has two parts. Engaging questions and fun visuals motivate students to master new concepts. They are: 1. 5 Grammes: 2 Milligrammes = 0. Like fractions 4. Addon Get our in-game addon for quickly loading your character, managing gear, and more.
025 g * * par dose unitaire. The numerator (20) is less than the denominator (23), so this fraction is a proper fraction. 25 are different ways of representing the same fraction. Divide 14 by 25 to get 0. · Ejection fraction is a measure of how well.
We engineer the products, systems and services required to unlock energy. La finition en aluminium donne &224; chaque feuille d&233;corative un aspect unique et prot&232;ge contre la corrosion. But even with a normal ejection fraction, your overall heart function may not be normal. Item Information. This class only represents fractions with a determined value, so fractions with a zero in the denominator are not allowed (including 0/0).
When you save presentation as a PDF file it freezes the formatting and layout. The latest tweets from The official website of professional Italian football club AS Roma. 025 Aayush Jindal &183; Novem &183; 4:28 am Tron price started a fresh decline from the .
ASKinard, A&S Services Group LLC is the transportation solution for all of your surface transportation, logistic, Intermodal Transportation and warehouse needs. Your medication may look. 02 Grammes: 5000 Milligrammes = 5 Grammes: 3 Milligrammes = 0. Jobs: Contact: About Nordal: F&248;devarestyrelsens kontrolrapporter.
This medication is used to treat a variety of skin conditions (e. What&39;s Going On. Nos termes & conditions peuvent &234;tre consult&233;s ici : termes et conditions, politique de confidentialit&233;. The first time kids discover that there’s more to math than whole numbers, they are likely to be a tad confused. La minorit&233; s'efforce de constituer dans toutes les entreprises des comit&233;s syndicalistes r&233;volutionnaires, &224; la fois groupes fractionnels et futurs Soviets (Reynaud, Syndic. An improper fraction is a fraction where the numerator is larger than the denominator. 5k Followers, 579 Following, 21k Posts - See Instagram photos and videos from Diario AS Floor Proceedings Saturday, 11:00 a. K) pour la.
Our parent educators use an evidence-based home visiting model with parents and caregivers during a child’s earliest years in life, from prenatal through kindergarten. ,=breaking, in arithmetic, an expression representing a part, or several equal parts, of a unit. 提供眾多AS官方旗艦館商品，讓您輕鬆選購：新品上市、品牌快速搜尋、款式快速搜尋。AS官方旗艦館盡在 Yahoo 奇摩購物中心。.
At BASF, we create chemistry for a sustainable future. Two Astronomy Teams at UArizona Awarded Astrobiology Interdisciplinary Consortia. The State Corporation Commission’s (SCC) Bureau of Insurance (Bureau) is changing the duration of agent licenses from a perpetual basis to a biennial basis based on their birth month and year.
Fraction workshop allows users to practice ordering, reducing, adding, subtracting, multiplying, and dividing fractions and mixed numbers. So let&39;s say that I have this square. It is a complete square. For example, 4/6 = 2/3. L’isolation de la toiture permet donc d’&233;pargner beaucoup d’&233;nergie et d’argent. 025 But you can&39;t get it to print it exactly the way you declared it. Vi er distribut&248;r av Eika Kapitalforvaltning, distribut&248;r av Eika Kredittbank, distribut&248;r av Eika Forsikring.
Cc | &220;bersetzungen f&252;r 'residual fraction' im Isl&228;ndisch-Deutsch-W&246;rterbuch, mit echten Sprachaufnahmen, Illustrationen, Beugungsformen,. Methods: A single center, randomized study was conducted between April and March. 025 D&233;tails Pays d'origine CN-China Garanties et certifications. 025 Z Score probability, percentile, and explanations for all three. 025 per share, based on the closing market price on J. Aside from many potential performance issues, external hard drives may connect or disconnect from the computer at inopportune times as part of their normal operations. This method involves cross multiplication of the fractions.
Its headquarters is located in Ban. It’s the percentage of blood that leaves your left ventricle when your heart. If you have a graphing calculator with a fraction command, then you can enter the fraction as a division (because 4 / 8 means "four divided by eight"), and then convert to fraction form. The fraction 1 / 2, which means 1 divided by 2, can represent such things as 10 pencils out of a box of 20, or 50 cents out of a dollar.
Updated travel information. Any fraction with the value zero is expressed as 0/1 in lowest terms. Add to Cart + Add to. The NASA Astrobiology Program has selected eight new interdisciplinary research teams to inaugurate its Interdisciplinary Consortia for Astrobiology Research program, including two teams at the University of Arizona. Antimyopic effects of low-concentration atropine act mainly on reducing AL elongation, and therefore could reduce the risk of subsequent myopia complications. Du bygger – Vi tar oss av resten.
The velocity along the top surface is given in Figure 6, where the chord c = 10 cm and the free-stream. Multiply the numerators for 4 x 1 = 4. 025 mg tablet. Site also includes information about the photographer&39;s instructional photo-tours, books, and recent magazine articles.
Solid copper alloy construction for high conductivity and spatter resistance. A rebound from this area will bring the x-axis of the triangle at . Fractions form an important part of our daily lives. Fire Alarm and Sprinkler System Installation, monitoring, Inspection & Repair Service.
1 Milligrammes = 0. Com Associates Program and start earning money today. M&233;t&233;ociel propose le mod&232;le GFS &224; maille moyenne (0.
Microsomal fraction: a collection of tiny subcellular particles, invisible with the light MICROSCOPE, that are produced during DIFFERENTIAL CENTRIFUGATION of eukaryotic cells. Define simple fraction. For examples,. Like fractions are fractions that have the same denominator. Une puissance est la multiplication d'un nombre par rapport &224;. W&246;rterbuch - Synonyme - Deutsch-Englisch &220;bersetzungen f&252;r fraction. Thickness: 0.
4 4 : Four fourths are yellow. If you learn and visualize fractions the easy way, it will be more fun and exciting. VET, 4-hour chart. : Columbia University Press, →ISBN, page vii: 1.
The aluminum material is non-corrosive. 1/40 is the fraction representation of 2. Synonyms of fraction a broken or irregular part of something that often remains incomplete if even a tiny fraction of that cookie broke off and fell into the delicate watch works, it could mess things up. Administrative revendique avec succ&232;s dans l'application des r&232;glements le mono&173; pole de la contrainte physique l&233;gitimel &187;. External Hard Drives.
So the fraction here, x/4, defines 1/4th of number x. AS Domain Registry supports the registration of Internationalized Domain Names (IDN). No, It is not a proper fraction because the numerator 6 in this fraction is more than the denominator 5. Change $$\frac1425$$ to a decimal. External hard drives are not recommended for use with Steam or Steam's games. VES & Viking Dealerships. Elkem also has strong positions in specialty foundry alloys and carbon materials.
025 means that your statistic is 0. It is suggested. So, the fraction of boys is two-fifths ( 2⁄5).
Activate the solution by clicking the Activate button as shown in the figure. Find all the information for your next step. Experience Call of Duty: the world’s best-selling video game franchise. How do you find the product of a fraction? Trygg og rask levering! 0537 In Microsoft Excel or Google Sheets, you write this function as. 0001 Secondes: 100000 Microsecondes = 0. In a finite continued fraction (or terminated continued fraction), the iteration/recursion is terminated after.
Le pelage est facile, la mise &224; la cote est instantan&233;e. The numerator (75) is greater than the denominator (51), so this fraction is an improper fraction. 025 L in ml = 25 milliliters. Proper fractions 2. The Amazon Associates Program is one of the largest and most successful online affiliate programs, with over 900,000 members joining worldwide.
NumeratorDenominator You just have to remember those names! Subtract Fractions 2. We deliver unique experiences that unite people, and we bring stories to life in films, series, gaming, cinemas and digital gifting. Here is a Bell Curve so you can visualize where 0. More Fraction videos. Veidekke &228;r Skandinaviens 4:e st&246;rsta bygg- och anl&228;ggningsf&246;retag.
Stjerne Autoco - en av de mest profesjonelle selskapene i Mercedes-Benz utvalg i Norge. How do you divide a fraction by a fraction? For example, Fraction 1/2 can be written in decimal form as 0. Pensionsordninger der er tilpasset dit liv, hvor opsparing og sundhedsforsikring g&229;r h&229;nd i h&229;nd. Slice an apple, and we get fractions. Informed, written consent was obtained from each patient prior to study enrollment. You can divide or multiply the top and bottom of any fraction by any.
There are a lot of options for printf including leading spaces as well:. Synthroid is among the actual couple of drugs that nearly anybody can take, as it's maternity category A and is anticipated to be secure even when taken by expectant or breastfeeding ladies. 10 = 2 remainder 4 ). Learn more about fraction, decimal, percentage. 025 in the Fraction is 0. Learn More. Blog Read Mr. As we already learned enough about fractions, which are the part of a whole.
The quarterly stock dividend will have a value of . Value in inch = 0. To divide or break into fractions. Not only is it an important part of school — from elementary school all the way up to high school — it&39;s also a really practical skill to know. Ejection fraction is just one of many tests your doctor may use to determine how your heart works. In contrast, an infinite continued fraction is an infinite expression. Figure: Upload Solution button. (If you forget just think "Down"-ominator).
011" 5 psi @ 72° F: White: 2, 5, 10, 25, 50. Thick wire. Infor is a global software company that builds SMB and Enterprise ERP software cloud products for industries including Manufacturing, Healthcare, Retail, Hospitality and Services. Fraction Matcher - PhET Interactive Simulations. Ces valeurs peuvent varier selon les. What type of fraction is 1/4? Fractions - Table of Contents.
TopSearch Can Help You Find Multiples Results Within Seconds. Funding for Cyberchase is provided by The JPB Foundation, the National Science Foundation and Ernst & Young LLP. 05 Secondes: 6 Microsecondes = 6. We can also: 1.
3 4 : Three fourths are yellow. For example, in the fraction one-half, 1 is the numerator, 2 is the denominator. The convert fraction to decimal calculator will take any fraction and change it to a decimal. 12 M − 1 (ng − 1) Δg (cm) Created Date: 8:20:10 PM. 1898, Winston Churchill, chapter 2, in The Celebrity: 5. You can quickly add fractions with the same denominator. Grainger&39;s got your back. Com is the 1 question answering service that delivers the best answers from the web and real people - all in one place.
1200/10000 = 13012/100 Question 3: Add 3/ /15. , 2-5 or 2/5, the number after or below the bar represents the total number of parts into which the unit has been divided. Statistics Inference with the z and t Distributions One-sample z test. 025% two to four times daily; Triamcinolone Acetonide Cream USP, 0.
In a eucharistic service, the breaking of the host. Parents as Teachers builds strong communities, thriving families and children that are healthy, safe and ready to learn. The best part of decimals are they can be easily used for any arithmetic operations such as addition, subtraction, etc. Her kan du finne &229;rets jule-reklamefilm, strikkeoppskrift p&229; TINE-nissens genser, juleoppskrifter og. &0183;&32;Make professional presentations in a fraction of the time. Die Suche im W&246;rterbuch ergab folgende Treffer f&252;r "fraction":.
But what about when the denominators(the bottom numbers) are not the same? Please feel free to browse our terms and definitions free of charge. Fraction Formulas There is a way to add or subtract fractions without finding the least common denominator (LCD). &0183;&32;Tron (TRX) Price Analysis: Key Breakout Zone Near . Nous utilisons des cookies et des outils similaires pour faciliter vos achats, fournir nos services, pour comprendre comment les clients utilisent nos services afin de pouvoir apporter des am&233;liorations, et pour pr&233;senter des annonces.
Join the Amazon. You can do this on a calculator or manually using long division. Please Enter Zero Before The Decimal Number Like (0. Rule 8: Any Integer Can Be Written as a Fraction. Solid copper alloy construction For use with 0. Kostenlose Musikdownloads.
Consulta la clasificación de los equipos de la LaLiga SmartBank /, todos los datos de la LaLiga SmartBank / en AS. Color orange shape No data. Som Norges st&248;rste akt&248;r innen byggevarer, trelast og interi&248;r er vi i Optimera opptatt av &229; hjelpe deg med &229; forenkle og effektivisere arbeidsdagen din. Si certaines conditions sont v&233;rifi&233;es, on peut approcher une loi binomiale de param&232;tres n et p par une loi normale. Cartes de pr&233;vision des pr&233;cipitations, des orages, de la neige, du vent, des temp&233;ratures &224; tous les niveaux. 001 Grammes: 10 Milligrammes = 0. Flexit utvikler, produserer og markedsf&248;rer energieffektive produkter, tjenester og l&248;sninger for et bedre inneklima i boliger og yrkesbygg.
Fraction A mathematical expression representing the division of one whole number by another. JUMBO Stillads A/S er Danmarks st&248;rste totalleverand&248;r af l&248;sninger, n&229;r toppen skal n&229;s, hvad enten det g&230;lder stiger, stilladser eller lifte. Vi leverer alt fra smarte byggeplasscontainere til prosjektering av kj&248;kken, beregninger og gode logistikkl&248;sninger. Honda TRX 420 Rancher 4-Stroke. The number below the line is called the denominator.
Gj&248;r juletiden smakfull med TINE-produkter TINE og TINE-nissen &248;nsker god jul. When spoken in everyday English, a fraction describes how many parts of a certain size there are, for example, one-half, eight-fifths, three-quarters. For K-12 kids, teachers and parents. Test av metadesription. 025 x 4 x 10 In, PK3 (5MWN3)? Ceci peut servir pour obtenir une approximation d'une probabilit&233; de la forme p\left(c\leqslant X \leqslant d\right).
Asiasoft is an online game operator in Thailand, Vietnam, Indonesia, Philippines, Singapore and Malaysia. Com, a free online dictionary with pronunciation, synonyms and translation. Covid19 Operational Plan. That is called Simplifying, or Reducingthe Fraction. 3) Le flux maximum sous un p&244;le &233;tant de 0,025 Wb, le coefficient de Kapp valant 2,08 et le nombre de conducteurs actifs par phase 1620, calculer la f.
A fraction consist of two parts, a numerator and a denominator. Each order includes a 1/8" drill bit and a pair of nitrile coated gloves to protect your hands while working on your project. Under the ELECTRON MICROSCOPE, these microsomes can be seen to consist mainly of membranes and RIBOSOMES from the ENDOPLASMIC reticulum. Com conna&238;tre les tests de significativit&233; pour interpr&233;ter vos r&233;sultats statistiques. La cale pelable Jicey permet un montage d'une grande pr&233;cision.
Question 2: Convert 130. Fraction (third-person singular simple present fractions, present participle fractioning, simple past and past participle fractioned) 1. E-LDP: CMAS Warehouse Code 8-8663 The USDA FSA offers online LDP application. Suppose a number has to be divided into four parts, then it is represented as x/4. . This study was approved by the Medical Ethics Committee at our hospital. A fraction represents either a part of a whole or any number of equal parts.
The anxiety and depressive subscales are also valid measures of severity of the emotional disorder. And we&39;ll see there&39;s many ways to think about a fraction. Compare to. Leverand&248;r av utstyr for n&248;d- og reservekraft fra 10 kVA til 2500 KVA. There are two parts to a fraction: The number on top shows how many parts there are; The number on the bottom shows how. View full product details.
Stock Price: AS. The dividend reflects an aggregate distribution of approximately 62,632. A fraction represents a numerical value, which defines the parts of a whole. Now we just have to solve the simple equation, and we will get the solution we are looking for. 25 mg, 1 mg, or 5 mg triamcinolone acetonide respectively, in a washable cream base of cetyl alcohol, cetyl esters wax, glycerin, glyceryl monostearate, isopropyl palmitate, polysorbate-60, propylene glycol, purified water, sorbic acid, and sorbitan monostearate.
Just divide the numerator by the denominator. How to use your in a sentence. Printable fraction games and printable worksheets; Manipulative fraction strips, printable fraction pizzas, a memory-matching game, and more. 5% What is the percentage form of 0. So if we cut a cake and take half of it, it can be represented as one part of two equal parts which can be written as 1/2. 025% Isotretinoin.
Click Here if you want to buy Tretinoin online. 025" wire, including over 50 products in a wide range of styles and sizes. Math Help for Fractions: Easy-to-understand lessons for kids, parents and teachers. See full list on byjus. En fonction du type d'huile utilis&233;e le pourcentage d'huile dans le m&233;lange n'est pas le m&234;me. Click the Upload Solution button to upload the solution as shown in the figure. So a fraction is the number of shaded parts divided by the number of equal parts as shown below:.
(chemistry) A component of a mixture, separated by fractionation. GDPR Update. Usually written as two numbers separated by a horizontal or diagonal line, fractions are also used to indicate a part of a whole number or a ratio between two numbers. ) calculator that can handle fractions, then you can enter the fraction and then hit the "equals" button to get the reduced fraction. 1 Secondes: 7 Microsecondes = 7.
Whereas it is difficult sometimes to perform operations on fractions. An expression that indicates the quotient of two quantities, such as13. The dividend reflects an aggregate distribution of approximately 62,632. Visit The Home Depot to buy 12 in.
5) Calculer la puissance nominale de l'alternateur et le rendement. Valeurs du coefficient n de Manning: Nature des surfaces: Etats des parois: Parfait: Bon: Assez bon: Mauvais: A) Canaux artificiels: Ciment liss&233;: 0,01: 0,011: 0,012. Marnar Bruk Royal er mer milj&248;vennlig og krever mindre vedlikehold enn annet impregnert trevirke. We can assure that we have taken all precautions regarding hygiene and adapted to this on all our products so that you as a guest will have a worry-free experience. 5/100 = 25/1000 And finally we have:.
Specialties: The most spectacular views in San Diego. Let us take an example to understand; Example: Add 1/6 and 1/4. How do you calculate fractions on a calculator? En France, le taux est jug&233; normal s’il est compris entre 0,5 et 4,5 UI/L. Semi-Clear Tubing. Elkem is one of the world’s leading providers of silicones and silicon solutions and uniquely positioned in the market through full ownership of the value chain from quartz to specialty silicones. If you need to simplify fractions, this fraction calculator can do the work for you by entering a regular fraction, mixed fraction or improper fraction then multiply the value by one. It can also be defined as a part of n equal parts.
A fraction (from Latin fractus, "broken") represents a part of a whole or, more generally, any number of equal parts. Diphenoxylate-atropine 2. NET provides detailed descriptions, pictures, and directions to local estate sales, tag sales, and auctions in your area. Mixed fractions. 01%) have no clinical effect on corneal or lens power. Com, the world&39;s most trusted free thesaurus.
It tells how many equal parts of the whole or collection are taken. Divide Fractions Visit the Fractions Indexto find out even more. The top number of a fraction is called its numerator and the bottom part is its denominator. Looking for GRAINGER APPROVED Copper Sheet, 0. Fraction and whole number division in contexts Get 3 of 4 questions to level up! The USLegal Dictionary.
The same amount is green on both so. Fraction definition at Dictionary. 5mm Solvent & UV Resistant Temperature resistant to 350F. Facebook gibt Menschen die M&246;glichkeit, Inhalte zu teilen und die. Complete your Fraction collection.
03 Grammes: 10000 Milligrammes = 10 Grammes: 4 Milligrammes = 0. 2,5 km) et 0,01&176; (env 1,3km) Format : GRIB V2 ou GEOTIFF pour les donn&233;es en ligne, GRIB 1 ou 2 pour les donn&233;es hors ligne. P&229; brreg. 11 / 100 is called as vulgar or simple fractions. 17 and 1/4 = 0.
We provide our customers with optimal and innovative solutions within engineering, construction, modifications and maintenance throughout a project's entire life cycle. You would have to change the precision at different stages of the loop (%. Fraction Division Worksheets Mixed number to improper fraction worksheets Improper Fractions and Mixed number worksheets Comparing fractions worksheets Fractions on Number Line Worksheets Ordering fractions worksheets Fractions Worksheets: Addition, subtraction, multiplication and Division Simplify Fractions Worksheets Picture Fraction Worksheets. Amid a coronavirus outbreak in the United States, the US Centers for Disease Control and Prevention is encouraging older people and people with severe chronic medical conditions to "stay at home.
1f for the first 2 iterations and then %. Multiplying by 1 in the form of 3/3 turns 1/3 into the equivalent fraction 3/9 fact five Add and Subtract Equal Sized Parts. 975 Our critical z value is 2. آس آرابيا هو النسخة العربية من موقع آس الإسباني، ويضم مواضيع يومية ومقالات رأي ومقابلات. Maybe they&39;ll tell you about the eclectic decor and innovative cuisine. The numerator represents the number of equal parts of a whole, while the denominator is the total number of parts that make up said whole.
Quelle est alors l'intensit&233; du courant d'excitation? Possibilit&233; de bi-composition (feuilles d'&233;paisseurs diff&233;rentes sur chaque face). 004 Grammes: 40 Milligrammes = 0. Ll convient g&233;n&233;ralement de vous mettre en relation avec votre syndic.
2 days ago · The first version requires that numerator and denominator are instances of numbers. Google Scholar provides a simple way to broadly search for scholarly literature. Bonjour, J' ai &233;t&233; op&233;r&233; d' un cancer de la prostate en juillet, depuis je fais des contr&244;les tous les 6 mois &224;, la premi&233;re prise de sang mon taux d' antig&232;ne prostatique sp&233;cifique total &233;tait de -0. 25 Secondes: 8.
Figure: Activate Solution dialog and button. A list of US medications equivalent to Acretin 0. Yes, It is a proper fraction because the numerator 5 in this fraction is less than the denominator 9. " | Introduction to Fractions by LetstuteIn this video, we have discussed the following points on. Low-concentrations of atropine (0. 1200 = 130.
025 mg/5 mL oral liquid. Solution: 3 /5 + 10/15 LCM of is 15 =/15 = 19/15 A lot of word problems on fractions appear in our textbooks to compare fractions, multiply mixed numbers and improper fractions, decimals, and fractions. And sometimes that confusion extends throughout the entirety of elementary school, where an initial introduction to concepts like numerators and denominators is followed by comparing fractions, adding and subtracting fractions, multiplying fractions, simplifying fractions and so on. Reduce the fraction by dividing the numerator and the denominator with the gcd. 025 L to ml = 25 milliliters. 1/40 is an equivalent fraction form for decimal 0. Imprint No data. Get instant access to search and more every time you open your browser by setting your homepage to Google.
Recharge CC (our virtual credits) in your Cherry e-wallet! · Take the two numerators (top numbers) and add them up. 065 The 2D cross-section of an airfoil of width b into the page shown in Figure 5 passes through air at atmospheric pressure with steady speed U = 30 m. Some fractions may look different, but are really the same, for example: It is usually best to show an answer using the simplest fraction ( 1/2 in this case ). Fraction may also mean:. App calculates the Greatest Common Factor showing how the lowest term answer is derived. 4) En utilisant le diagramme de Behn-Eschenburg, retrouver cette f.
A small amount. Unlike fractions are fractions that are different. 1200 into fraction. 025 in to mm conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result). Velkommen til Hoyer. Aucun excipient &224; effet notoire? Free Domestic Shipping over 0. 2.
Check out our open positions. Since α = 0. A naturally sustainable, convenient and consumer preferred alternative to plastic bottles. Krigsvoll AS er import&248;r/grossist av verkt&248;y, maskiner og produkter for hjem/hus/hytte og fritid. No kan du registrere og endre opplysninger p&229; bedrift, finne bedriftsinformasjon og kunngj&248;ringer, sjekke heftelser i bil og stoppe telefonsalg. Given α = 0.
025 Profondeur de l’emballage (en pouces) 0. Check out our shop supply site for additional products www. Los resultados de la Quiniela 1X2 actualizados en directo. The Associated Students of the University of California. Yes, it’s a unit fraction.
002 Grammes: 20 Milligrammes = 0. See full list on en. S&228;mtliche der im Folgenden vorgestellten Bell Fraction sind direkt bei Amazon. Paulin T&244;le d'acier 24 x 36 po x 36 po de calibre 26 - galvanis&233;e. Diamond Tread Aluminum Sheet 57306. Each roll includes (10) 1/8" aluminum rivets. Les meilleures offres pour PLASTIGAGE 0,025-0. La conductivit&233; thermique de la laine de verre se rapproche de celle de l’air immobile.
Com international database is in BETA release. Bell Fraction - Die TOP Favoriten unter der Vielzahl an analysierten Bell Fraction. Just add the numerator, keeping the denominator same. RICHMOND – Changes in licensing, fees and continuing education due dates for insurance agents in Virginia will take effect Janu. For Mercedes-Benz salg og tjeneste p&229; L&248;renskog og Skedsmo. &233;l&233;mentaires de 0,025 &224; 0,2 mm. Click on the following link to learn more and sign up. What Fraction of the square is Blue?
Bien que tous les efforts aient &233;t&233; faits pour assurer l'exactitude des calculatrices et des graphiques figurant sur ce site, nous ne pouvons offrir aucune garantie ou &234;tre tenus pour responsable des &233;ventuelles erreurs commises. You are visiting enjoy your treatment experience just if you are sure regarding two things: that your Tretinoin Cream is of the very best top quality possible which you did not overpay for it. Retin-A Cream 0. Information on COVID-19. Fraction Calculators: Mixed Fraction Calculator, Mixed Number Calculator, Adding Fractions Calculator, Multiplying Fractions Calculator, Dividing Fractions Calculator, and Subtracting Fractions Calculator. Daily prayer whenever you need it. Farlex Partner Medical Dictionary. Final konnte sich beim Bell Fraction Test der Sieger behaupten.
Isolation toiture : Aper&231;u des mat&233;riaux isolants. So, the fraction of girls is three-fifths ( 3⁄5). 025% Ophthalmic Solution, 5 mL Antihistamine eye drops work in minutes to help temporarily relieve itchy eyes due to pollen, ragweed, grass, animal hair and dander Lasts up to 12 hours. &201;conomies et confort assur&233;s avec nos mousses expansives. We operate a state-of-the-art late model fleet with the latest technology for tracking, storing and transporting goods. Le taux moyen est une moyenne de proportions et la part d. The Fractions app lets students use a bar or circle to represent, compare, and perform operations with fractions with denominators from 1 to 100. And we can consider this a whole.
Fraction Lat. To write 0. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Welcome on board of the. The bottom surface is flat and the flow is relatively undisturbed.
Fractions in maths are painful if you do not grasp the underlying concept behind it. Solution: Yes, it is. Aim: To compare the clinical outcomes between 0. To make a s.
A fraction represents number of equal parts of a whole. When you want to give a high-fidelity version of your presentation to colleagues or customers (either as an e-mail attachment, published to the web, on a CD or DVD), save it and let it play as a video. Simplifying Fraction. Common Denominator (They both work nicely, use the one you prefer. Find more ways to say as, along with related words, antonyms and example phrases at Thesaurus. With fresh material, taxonomic conclusions are leavened by recognition that the material examined reflects the site it occupied; a herbarium packet gives one only a small fraction of the data desirable for sound conclusions.
. Motofen 1 mg-0. Decimal to fraction conversion. Occlusive dressings may be used for the management of psoriasis or recalcitrant conditions. Be the first to write a review. It shows the total divisible number of equal parts the whole into or the total number of equal parts which are there in a collection.
One half is yellow. What is the z-value at alpha = 0. Through science and innovation we enable our customers in nearly every industry to meet the current and future needs of society.
Infoclimat, la m&233;t&233;o en France en temps r&233;el - Weather in realtime Europe, Canada. Applies to. In mathematics, a fraction is a number that represents a part of a whole. Attention : ne s’applique qu’&224; une proportion. :| See more videos for Fraction. 025% is available on the Drugs. Xyz values). MX Power Play.
It can also be use to express a ratio of two values. N'est pr&233;sent dans la composition de ce m&233;dicament. This is a special time, and one may wonder if it is safe to travel or not.
025 standard deviation to the right of the mean on a bell curve. A number that compares part of an object or a set with the whole, especially the quotient of two whole numbers written in the form a/b. BK (Thailand) THB5. Fraction Calculator Plus is your best way to deal with everyday fraction problems or even more complex woodworking calculations.
As a verb, to separate into portions. This SQL tutorial explains how to use the SQL CREATE TABLE AS statement with syntax and examples. In either case, all integers in the sequence, other than the first, must be positive. This medicine is a orange, clear, cherry, liquid ‹ Back to Gallery. Malgr&233; tous les efforts fournis par notre &233;quipe, certains m&233;dicaments peuvent figurer sur la base medicament. Details about PROX PISTON TRX420 RANCHER 0. Craft cocktails might come to mind, or. The denominator shows how many equal parts make up a whole, and the numerator shows how many of these parts we have in mind.
But how do you add 2/4 + 3/9? Is &92;(&92;frac59&92;) a proper fraction? Conserving liquid foods in cartons instead of plastic bottles is better for everyone, in every way. 025 - fonte / cast iron. 25000 Microsecondes = 0. Search and save now! In a finite continued fraction (or terminated continued fraction), the iteration/recursion is terminated after finitely many steps by using an integer in lieu of another continued fraction.
Se v&229;rt store og eksklusive utvalg av kl&230;r, sko og tilbeh&248;r til dame og herre. We used three different distribution tables, and we will give you the 0. 2/20) Club: 133: Transformation de fractions en nombres d&233;cimaux: anonyme: 27717: 34% (6.
025 as a fraction you have to write 0. In general, where a0, a1, a2,. Question 1: Is 12/6 a fraction? Answers are fractions in lowest terms or mixed numbers in reduced form. Isotretinoin is reported as an ingredient of Acnetin A 0. Qui tend &224; cr&233;er une fraction dans un parti politique (cf. If the Denominator is the same, adding and subtracting fractions is an easy task. Statistiques : pourcentage/fraction d'un nombre (1) vero7000: 9285: 54.
FastQuickSearch Provides Comprehensive Information About Your Query. For information on how Viking Supply Ships treat the COVID-19 situation, please click below link. Sur Terre, naturellement, le meilleur isolant est l’air sec immobile &224; 10&176;C : son coefficient de conductivit&233; thermique, exprim&233; en lambda λ, est de 0,025 W/(m.
You&39;ll be spinning with knowledge in just a few minutes. Look it up now! Our goal is to maximize recovery and efficiency of oil and gas assets, while using our expertise to develop the sustainable solutions of the future. Apply Now to Join AS! Use this fraction calculator for adding, subtracting, multiplying and dividing fractions.
PROX PISTON TRX420 RANCHER 0. Among patients with heart failure and a reduced ejection fraction, those who received the SGLT2 inhibitor dapagliflozin had a lower risk of worsening heart failure or death from cardiovascular. Herbarium material does not, indeed, allow one to extrapolate safely: what you see is what you get.
Talk with your doctor if you have concerns about your heart. Find more ways to say fraction, along with related words, antonyms and example phrases at Thesaurus. Rational and returns a new Fraction instance with value numerator/denominator. Examples of fractions.
A self-assessment scale has been developed and found to be a reliable instrument for detecting states of depression and anxiety in the setting of an hospital medical outpatient clinic. There are many examples of fractions you will come across in real life. Fraction (plural fractions) 1. We call the bottom number the Denominator, it is the number of parts the whole is divided into. Ce calculateur permet de conna&238;tre le r&233;sultat de la puissance d'un nombre.
05ng/ml 6mois plus tard toujours en m&234;mes termes 0. 00025 Secondes: 250000 Microsecondes = 0. D&233;couvrez notre large s&233;lection de produits dans notre rubrique Condensateurs &233;lectrolytiques! Collection of North American bird photographs. 11 is numerator and 100 is denominator. In other words, etc.
Definition: A fraction names part of a region or part of a group. 8/20) Club: 134: Transformer des fractions en pourcentage: mmulet: 6391: 68% (13. Solution: 1/6 = 0. Calculator to reduce a fraction to its simplest form. Fraction Workshop - Online. If two of you split the take from a bank heist, you would each get half. 5% en huile 100% synth&233;tique ( + ou - selon la configuration moteur ). Lorsque l’on travaille avec un logiciel de statistiques g&233;n&233;raliste tel que r, sas, spss, stata ou spad, les tests de significativit&233; (souvent Vu sur market-audit.
Rational and returns a Fraction instance with the same value. It is an improper fraction. The number 1 can be represented as a fraction because any number divided by itself is equal to 1 (remember that the fraction notation means the same thing as division). 10,90 € ttc quantit&233;AS Monaco uses cookies on this website.
025 = 025 / 1000 Below is the Representation of. Multiply Fractions 3. Vi k&228;nnetecknas av en stark f&246;retagskultur som tror p&229; att engagemang och involvering &228;r grunden f&246;r all framg&229;ng. Les feuilles d'acier galvanis&233; ont des applications pour la r&233;paration de goutti&232;res, la r&233;paration automobile, les solins et.
Inch Definition of Millimeter A millimeter (mm) is a decimal fraction of the meter, The international standard unit of length, approximately equivalent to 39. You'll see more of this in the Equivalent Fractions Part 2 lesson, but here's a preview: If you multiply the numerator and denominator of by 3, you get. ICYNENE r&233;alise l'isolation globale de votre maison avec une mousse expansive tr&232;s performante. 05% Trypsin - EDTA (1x), phenol red Cat No:,,,,. See more videos for 0. 3f for the last iteration). It is equivalent to 11 / 100. We can multiply fractions, divide fractions, divide fractions by fractions and so much more.
Now checking fraction math takes just seconds. This fraction calculator will automatically simplify results. BYJU’S provides students with maths concepts as well so that students can practice and solve problems easily. Why turn your presentation into a video? Fraction (mathematics), a quotient of numbers, e. But first, we&39;ll think about the most fundamental. Fractions lets students use a bar or circle to represent, compare, and perform operations with fractions with denominators from 1 to 100.
Are all integers. Based on the properties of numerator and denominator, fractions are sub-divided into different types. Uk has been visited by 10K+ users in the past month. I'm Fraction Calculator Plus and I'm the best and easiest way to deal with everyday fraction problems.
The top number, called the numerator, represents the number of parts you&39;re working with. 025 What is 2. 230mm vert bleu, rouge 3 Rayures a 30cm Paliers lisses sont sur eBay Comparez les prix et les sp&233;cificit&233;s des produits neufs et d'occasion Pleins d'articles en livraison gratuite! The 2D cross-section of an airfoil of width b into the page shown in Figure 5 passes through air at atmospheric pressure with steady speed U = 30. 5% Triamcinolone Acetonide Cream USP contains 0.
Some fractions may look different, but are really the same, for example: It is usually best to show an answer using the simplest fraction (1/2 in this case). See full list on calculator. Fraction definition is - a numerical representation (such as 3/4, 5/8, or 3. 025 as numerator and put 1 as the denominator. Every display form has its own advantages and in different situations particular form is more convenient than another. From working with a number line to comparing fraction quantities, converting mixed numbers, and even using fractions in addition and subtraction problems, the fractions games below introduce your students to their next math challenge as they play to rack up points and win the game.
For example usage of. Les contributions conventionnelles sont mutualis&233;es d&232;s r&233;ception entre les entreprises. With your agreement, we use them to access how this website is used (analytic cookies) and to adapt it to your needs and interests (customization cookies depend on your browsing and your browser). The velocity along the top surface is given in Figure 6, where the chord c = 10 cm and the free-stream speed U are used as the. 025 uF Capacitors.
When adding and subtracting fractions the denominators must be the same. If you had a getaway driver, then the money would be split three ways, into fractions of one-third each. A fraction is a part of a whole. Fractions on a number line: Fractions can be represented on a number line, as shown below. Plus de temps mort au montage. Oslo Taxi - 02323. L'utilisation du format GRIB V2 n&233;cessite un outil sp&233;cifique (par exemple, celui du Centre Europ&233;en de Pr&233;vision &224; Moyenne Ech&233;ance). An aliquot portion or any portion.
Comprueba al instante tu boleto de quiniela de la liga espa&241;ola as&237; como la quiniela de las pr&243;ximas jornadas. You will notice once a term is defined there will be associated news and or court cases where the defined term is applicable. 0,025 % de la masse salariale due par les entreprises de 1 &224; 49 salari&233;s.
Solution for what is 0. Write the decimal fraction as a fraction of the digits to the right of the decimal period (numerator) and a power of 10 (denominator). The decimals are the numbers expressed in a decimal form which represents fractions, after division. David Guetta & Sia Let&39;s Love. We combine economic success with environmental protection and social responsibility. Equivalent Fractions & Simplifying. Bell Fraction - Der Testsieger der Redaktion. Continued fraction, expression of a number as the sum of an integer and a quotient, the denominator of which is the sum of an integer and a quotient, and so on.
Keynproject - polar fraction | Track anh&246;ren und downloaden, den K&252;nstler auf Jamendo Music unterst&252;tzen. Know that a fraction is a way of indicating parts of a whole. A small part of something, or a small amount: Although sexual and violent crimes have increased by 13 percent, they remain only a tiny / small fraction of the total number of crimes committed each year.
025&176; R&233;solution : 0,025&176; (env. We fly to 70+ destinations from Riga, Vilnius and Tallinn. We welcome all our guests to Fl&229;m.
Please select Country. Diamond Tread Aluminum Sheet 57306We would like to show you a description here but the site won’t allow us. We sell many types. Kids will learn to make friends with fractions in these engaging and interactive fractions games. IXL brings learning to life with over 200 different fraction skills. 24kGoldn Mood.
Reduce the proper fraction, improper fraction and mixed numbers to its lowest term. In mathematics, a continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number, then writing this other number as the sum of its integer part and another reciprocal, and so on. A part of a whole, especially a comparatively small part. 025% Cream helps to treats of Acne, Anti-aging, Diminishes Wrinkles Visibly, Blemishes Dark Spots and Skin Imperfections. Basic Fractions; Adding Fractions; Comparing Fractions. 025, the area under the curve is 1 - α → 1 - 0. : Convene and proceed to executive session to resume consideration of the nomination Thompson Michael Dietz, of New Jersey, to be a Judge of the United States Court of Federal Claims.
No you will find timetables, a Journey Planner and ticket information for public transport services in Oslo and Akershus. Return to Top. Or how will you solve 6/12- 3/27? Elopak is the owner of the Pure-Pak&174; carton brand. Com Vu sur memoireonline. Three people, four slices.
A) En parlant d'un ensemble de pers. 32 to fraction: 0. 0&215;10-6 Secondes: 250 Microsecondes = 0. يهتم الموقع بأخبار الرياضة، وعلى رأسها كرة القدم والسلة وفورمولا 1 والتنس. Thus any fractional number can be represented in decimal notation. What we&39;re going to talk about in this video is the idea of a fraction. Produits connexes.
Provides the mixed number solution for improper fractions. You can also use the SQL CREATE TABLE AS statement to create a table from an existing table by copying the existing table's columns. A fraction is a mathematical way of expressing a value that is less than a whole unit or one. Be pampered with an extremely attentive staff and enticed with the finest cuisine. Results, fixtures, interviews, information, tickets and more. So let me write that down. Entre phases.
See full list on mathsisfun. The second version requires that other_fraction is an instance of numbers. Svitzer is a global market leader providing towage and sustainable marine solutions to customers in 30+ countries across four regions. Tritt Facebook bei, um dich mit Marsha Fraction und anderen Nutzern, die du kennst, zu vernetzen. 025-inch and 0. Each gram of 0. Simple fraction synonyms, simple fraction pronunciation, simple fraction translation, English dictionary definition of simple fraction. 0&215;10-6 Secondes: 100 Microsecondes = 0.
B) En parlant d'un attribut, d'une manifestation de l. A fraction means a part of something or a number of parts of something. Ejection fraction is a measure of how well the heart is pumping blood around the body. Our VR headsets redefine digital gaming & entertainment. 025 fl oz to ml conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result). Eagle 1 embossed 0. 0&215;10-6 Secondes: 50 Microsecondes = 5. This is a fraction calculator with steps shown in the solution.
Cette contribution conventionnelle compl&232;te la contribution l&233;gale. 025% where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that: 100/x=100%/0. Discover the latest updates to this first person shooter series all in one place. A&S primarily serves the Northeast and mid-Atlantic United States; view our coverage map for more details. When spoken in everyday Inglis, a fraction describes hou mony pairts o a certain size thare are, for example, ane-hauf, aicht-fifts, three-quarters.
Fraction B 1 b). The numerator (8) is less than the denominator (9), so this fraction is a proper fraction. 234) indicating the quotient of two numbers.
- Need to adjust recipe quantities for a larger guest list? In the above section containing the definitions, a liter has been defined as being equivalent to 1,000 milliliters. Fluid Mechanics: y* U 0 0 0.
Condition. Unser Team hat den Markt von Bell Fraction verglichen und hierbei die relevantesten Informationen verglichen. Fraction represents a part of the whole unit. We must somehowmake the denominators the same.
Improper fractions 3. Fraction of a whole: When we divide a whole into equal parts, each part is a fraction of the whole. Is &92;(&92;frac65&92;) a proper fraction? Simply, a fraction is used to indicate a part of a whole. Com has been visited by 1M+ users in the past month. 025 aluminum tread plate rolls are perfect for protecting and decorating. 0537 In Microsoft Excel or Google Sheets, you write this function as =NORMSINV (0.
GigaPromo is the website to compare Tutor For Maths. Kontakt vores r&229;dgivere, s&229; I kan finde l&248;sningen for dig. The photos shown are samples only Not all photos of the drug may be displayed. In a simple continued fraction (SCF), all the bi are equal to 1 and. A fraction is part of an entire object.
The Difference Between GMT and UTC. 06ng/ml et l&224; je viens de faire une nouvelle prise de sang ou c' est. It will become easy if you visualize the number fractions and then solve systematically. What does fraction mean? 25 Hence, on adding 0. The fraction calculator will simply the answer for you. Please see GDPR for details regarding our compliance with GDPR. &0183;&32;To evaluate changes in ocular biometrics in groups receiving 0.
Additional funding is provided by the Tiger Baron Foundation, The V & L Marx Foundation in Memory of Virginia and Leonard Marx, Lynne and Marc Benioff, and Epstein Teicher Philanthropies. Robot's latest theorycraft articles. Fractions show parts of whole numbers, for example, the fraction &92;(&92;frac14&92;) shows a number that is 1 part out of 4, or a quarter. If denominator is 0, it raises a ZeroDivisionError. Ainsi, pour − =, et = et =, si X suit une loi de Fisher avec ces param&232;tres, on lit dans la table que (⩽,) =,. Vi har fokus p&229; specialmedier og leverer kvalitetsindhold og underholdning til vores kunder. Solution for Converting.
Using fractions. Find the greatest common divisor (gcd) of the numerator and the denominator. Bonnier Publications koncernen st&229;r bag nogle af Nordens mest kendte mediebrands, fx Illustreret Videnskab, Historie og Bo Bedre. Solution: Here will use the concept of how to convert decimals into fractions 130.
M&233;canisme d’action. 1992, Rudolf Mathias Schuster, The Hepaticae and Anthocerotae of North America: East of the Hundredth Meridian, volume V, New York, N. Math explained in easy language, plus puzzles, games, quizzes, videos and worksheets.
【発音】frǽkʃən【カナ】フラクション【変化】《複》fractions - アルクがお届けするオンライン英和・和英辞書検索サービス。. And b0, b1, b2,. Choisissez la taille et la finition qui correspondent le mieux &224; vos besoins. Looking for GRAINGER APPROVED Plate Stock, SS, 0. Apply to the affected area as a thin film as follows: Triamcinolone Acetonide Cream USP, 0. Some things that may cause a reduced ejection fraction are:. Hydroelectric Power Stations Wind Power Plants Solar Power Plants Nuclear Power Plants Coal Power Plants Biomass CCGT and small combined heat and power units Environment Market Transparency (REMIT) Map of Power Plants Industrial tourism. 035-inch guide wires (GWs) when used in wire-guided cannulation (WGC).
If you have a regular (scientific, business, etc. Search across a wide variety of disciplines and sources: articles, theses, books, abstracts and court opinions. Also Fit's. Rewrite each of the following fractions as a whole number. By Anne Buckle. Greenwich Mean Time (GMT) is often interchanged or confused with Coordinated Universal Time. Discover releases, reviews, credits, songs, and more about Fraction - Moon Blood at Discogs.
Any number that is represented in decimal notation is called as decimal number. 04 Grammes: 25000. Kj&230;rulf Pedersen a/s | Taastrupgaardsvej 8-10 | Postbox 58 | DK-2630 Taastrup | Tlf.
Whether you're checking homework, preparing recipes, or working on craft or even construction projects, I can help: - Wish you could find the time to check your kids' math homework? In this article. Nr:.
Least Common Denominator, or 2. Unlike fractions 5. UCAS connects people to University, post Uni studies including teacher training, apprenticeships & internships. Cc | &220;bersetzungen f&252;r 'residual fraction' im Latein-Deutsch-W&246;rterbuch, mit echten Sprachaufnahmen, Illustrationen, Beugungsformen,. MIG Gun Liner.
The method to change a fraction to a decimal is quite simple. Not only is. Support Get help with account-related or technical issues. &0183;&32;Defy reality with Oculus. 025, calculate the right-tailed and left-tailed critical value for Z Calculate right-tailed value: Since α = 0. Com website.
(arithmetic) A ratio of two numbers, the numerator and the denominator, usually written one above the other and separated by a horizontal bar. Easy online ordering for the ones who get it done along with 24/7 customer service, free technical support & more. La majorit&233; des pertes de chaleurs dans une habitation se font par la toiture. Reveal or hide numeric labels as needed. = 0. 42%) Mon,, 4:35AM EST. Another word for fraction. Frac·tion (frak&39;shŭn), 1.
Cc | &220;bersetzungen f&252;r 'solid fraction' im Franz&246;sisch-Deutsch-W&246;rterbuch, mit echten Sprachaufnahmen, Illustrationen, Beugungsformen,. Represent a fraction 1/b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. 025" Hex (625 microns) Tip Size: 2. &0183;&32;There are over 100 free fraction worksheets in PDFs below to support the many concepts encountered with fractions. Search For Info About Tutor maths. Son lambda varie en effet de 0,030 W/(m. Available 24/7. 0,025 Estampage es x x x - Forgeage fo x Grenaillage gn x x x x x x - Laminage filage - extrusion &224; chaud lac - x x - tr&233;filage - &233;tirage &224; froid laf - x x x - Matri&231;age &224; chaud ma -x: x: x-&224; froid Moulage au sable mo -x: x-cire perdue - proc&233;d&233; Schaw -x: x-en coquille par gravit&233; x: x: x: en coquille sous pression x: x: x: Moulage plastique mo Sablage sa Proc&233;d&233; Symbole.
Identify a proper fraction, improper fraction, mixed number, unit fraction, like fractions, unlike fractions like a pro with these printable types of fractions worksheets.
### 0.025 as a fraction
email: upitezu@gmail.com - phone:(499) 372-6301 x 6939
### Calmate in english - Eyebrow anti
-> Male dog nipples | 2021-05-06 19:13:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45451074838638306, "perplexity": 8355.398694431928}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988759.29/warc/CC-MAIN-20210506175146-20210506205146-00475.warc.gz"} |
http://mathhelpforum.com/calculus/114932-five-problems-derivatives-rates.html | # Math Help - Five problems, derivatives and rates
1. ## Five problems, derivatives and rates
I'm reviewing so I can test out of calculus. I was given a study guide with answers included. These five I don't get. In numerical order, the answers are B, B, B, A, and B respectively. But why?
24. ...What?
31. ...What?
41. I keep getting 1/13.
43. ...What?
44. I keep getting 16pi/25.
Thanks
2. Originally Posted by Sir Cumphrence
In numerical order, the answers are B, B, B, A, and B respectively.
I get B B A B A...
24. Lay your ruler through the points on the curve where x = -2 and x = 4. Which tangent might be parallel?
31. d/dr(r^3) must = 12. Why?
41. I agree with you.
43. Separate variables, giving 1/F * dF/dt = -2. Now integrate each side with respect to t. Call the constant of integration ln(A). Then take the exp of each side and simplify. See what occurs when t = 0.
44. Is your fraction flipped? Agreeing with you again, then.
Edit:
Just in case a picture helps with 43...
Spoiler:
... where
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to t, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
And for 31... Related rates usually involves the same shapes, but travelling down instead of up...
Spoiler:
__________________________________________
Don't integrate - balloontegrate!
http://www.ballooncalculus.org/examples/gallery.html
http://www.ballooncalculus.org/asy/doc.html | 2015-08-02 15:09:57 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8768779635429382, "perplexity": 1753.6749599088198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042989126.22/warc/CC-MAIN-20150728002309-00150-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/tags/superscripts/new | # Tag Info
3
The \ssp definition in the cited answer, How do you get a scriptscriptstyle-sized prime?, was specifically for \scriptscriptstyle only. To adapt a macro (I kept the same name, but you can change it to suit) for all math styles, just use \mathchoice. In most math styles, it is just #1^\prime. Only in \scriptscriptstyle does the alternate definition kick in ...
3
You are talking about a math environment I guess (because you mention $^$). I think the answer depends on the packages being loaded. In the example I think the package unicode-math actually makes $²$ to expand to and $^2$ (or something similar). Once this substitution is made the result (in math mode) is independent of the font selected. ...
1
You can make TeX ignore the height of a formula with the \smash command. Try ... some text $\smash{A_{b/c}^{dd}}$ more text ... The \smash macro is provided by plain TeX and it sets the height and depth of a maths subformula to zero. Beware that this might make your subscripts and superscripts overprint the lines above or below.
1
The parenthesis in the example provided is not really necessary. The problem here was the fact of forgetting to group the superscripts/subscripts in braces. This is a good practice. Hence the right approach is illustrated below: \documentclass[letterpaper]{article} \begin{document} $p= \sqrt{2}^{(\sqrt{2})}$ $p= \sqrt{2}^{\sqrt{2}}$ \end{document}
6
The font is respected if you use a better syntax: \documentclass{book} \usepackage{tipa} \newcommand{\upschwa}{\textsuperscript{\textschwa}} \begin{document} \textit{ahmiia zaoth etc\upschwa} ahmiia zaoth etc\upschwa \end{document} Note though that the TIPA fonts don't have italics, but only slanted. Loading tipa with noenc would give infinite ...
4
The prime symbols are not too low, they're too small: you're requesting that first level sub/superscripts are 6pt, which is too small next to an 11pt size symbol. Computing 80% of 11pt gives 8.8pt, and the standard first level sub/superscript size for 11pt is 8pt. Of course you need arbitrarily scalable fonts for this to work (\usepackage{lmodern}, for ...
5
Please always post complete documents that show the problem. It should work: \documentclass{article} \begin{document} 8\textsuperscript{th} 8th \end{document}
Top 50 recent answers are included | 2014-04-20 13:59:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9794769287109375, "perplexity": 3527.3596912633398}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00387-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://gmatclub.com/forum/in-the-figure-above-lines-l-and-p-are-parallel-the-segment-ad-is-the-202159.html | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 20 Oct 2019, 15:28
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# In the figure above, lines L and P are parallel. The segment AD is the
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 58434
In the figure above, lines L and P are parallel. The segment AD is the [#permalink]
### Show Tags
22 Jul 2015, 02:02
1
4
00:00
Difficulty:
45% (medium)
Question Stats:
74% (02:25) correct 26% (02:14) wrong based on 182 sessions
### HideShow timer Statistics
In the figure above, lines L and P are parallel. The segment AD is the same length as the segment DC; AD is parallel to BC. If the length of AD is equal to 4, and angle ADC is equal to 60º, what is the area of ABCD?
A. 4√2
B. 4√3
C. 8√2
D. 8√3
E. 16√2
Kudos for a correct solution.
Attachment:
parallel.gif [ 2.38 KiB | Viewed 5231 times ]
_________________
Intern
Joined: 16 Dec 2013
Posts: 32
Location: United States
GPA: 3.7
Re: In the figure above, lines L and P are parallel. The segment AD is the [#permalink]
### Show Tags
22 Jul 2015, 02:39
2
Bunuel wrote:
In the figure above, lines L and P are parallel. The segment AD is the same length as the segment DC; AD is parallel to BC. If the length of AD is equal to 4, and angle ADC is equal to 60º, what is the area of ABCD?
A. 4√2
B. 4√3
C. 8√2
D. 8√3
E. 16√2
Kudos for a correct solution.
Attachment:
parallel.gif
Drawing a line from A to C, we find that Triangle ADC is equilateral triangle.
Angle ADP is 120 deg=DCB=120. therefore ACB=60 deg
CBL=120 deg=ABC=60 deg. there fore triangle ABC is equilateral triangle.
Total area=area of 2 equilateral triangle=D
CEO
Joined: 20 Mar 2014
Posts: 2596
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: In the figure above, lines L and P are parallel. The segment AD is the [#permalink]
### Show Tags
22 Jul 2015, 03:51
3
Bunuel wrote:
In the figure above, lines L and P are parallel. The segment AD is the same length as the segment DC; AD is parallel to BC. If the length of AD is equal to 4, and angle ADC is equal to 60º, what is the area of ABCD?
A. 4√2
B. 4√3
C. 8√2
D. 8√3
E. 16√2
Kudos for a correct solution.
Attachment:
parallel.gif
Line L||P and AD||BC ---> $$\angle {ADC} = \angle{BCD} = 60$$
As AD = CD and AD || BC ---> ABCD is a rhombus with sides 4 inches each.
Thus the area = base X height
For height, draw a perpendicular line from A to CD meeting CD at O. Thus in right triangle AOD (rigth angled at O), we can apply the property of 30-60-90 triangle with the hypotenuse = AD = 4
Thus we get the height = $$2\sqrt{3}$$
Finally, the area = base X height = $$4*2\sqrt{3}$$ = $$8\sqrt{3}$$. D is the correct answer.
Math Expert
Joined: 02 Sep 2009
Posts: 58434
Re: In the figure above, lines L and P are parallel. The segment AD is the [#permalink]
### Show Tags
26 Jul 2015, 11:51
2
1
Bunuel wrote:
In the figure above, lines L and P are parallel. The segment AD is the same length as the segment DC; AD is parallel to BC. If the length of AD is equal to 4, and angle ADC is equal to 60º, what is the area of ABCD?
A. 4√2
B. 4√3
C. 8√2
D. 8√3
E. 16√2
Kudos for a correct solution.
Attachment:
parallel.gif
800score Official Solution:
AB is parallel to DC and AD is parallel to BC, so ABCD is parallelogram. We know how to find the area of a parallelogram we multiply the height of parallelogram by its base length.
We are told that the angle ADC is a 60º angle. If we draw an altitude from A straight down and perpendicular to line P, the length of that altitude will be equal to the height of the parallelogram. Moreover, it forms a 60-30-90 triangle, so we can easily find its length. Since the hypotenuse of that triangle is equal to 4, the second longest side must be equal to 2√3 (since the proportions for the triangle run x, x√3, and 2x).
To find the area of a parallelogram , multiply the height of the parallelogram by its base length: 4 × 2√3 = 8√3, or answer choice (D).
_________________
Retired Moderator
Joined: 04 Aug 2016
Posts: 477
Location: India
GPA: 4
WE: Engineering (Telecommunications)
Re: In the figure above, lines L and P are parallel. The segment AD is the [#permalink]
### Show Tags
12 Apr 2017, 21:24
Hello Bunuel,
Is below approach correct?
$$\angle$$ ADC = 60, AD=DC; so ADC will be an equilateral triangle with sides 4 each.
Thus height of the triangle will be the height of parallelogram = ($$\sqrt{3}$$ /2 )* side 4
Thus it will 8 * $$\sqrt{3}$$
Non-Human User
Joined: 09 Sep 2013
Posts: 13316
Re: In the figure above, lines L and P are parallel. The segment AD is the [#permalink]
### Show Tags
02 Nov 2018, 02:09
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: In the figure above, lines L and P are parallel. The segment AD is the [#permalink] 02 Nov 2018, 02:09
Display posts from previous: Sort by | 2019-10-20 22:28:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.825915515422821, "perplexity": 2676.513098962614}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00370.warc.gz"} |
http://www.tricki.org/article/Make_your_method_keep_the_symmetries_of_the_problem | Tricki
## Make your method keep the symmetries of the problem
### Quick description
Making your method keep the symmetries of the problem means that it is immune to errors that break symmetries. This can have two important effects:
• reducing the errors in the results, and particularly eliminating entire classes of errors that have unphysical or otherwise harmful effects,
• and reducing the amount of computational work.
### Prerequisites
Linear algebra, calculus.
### Example 1
For symmetric positive definite matrices, use the Cholesky factorization rather than the LU factorization.
The LU factorization of a matrix is , while the Cholesky factorization is where in each case is lower triangular and is upper triangular. Using the Cholesky factorization (which preserves the symmetry of ) roughly halves the time to compute the factorization, and avoids the problems of swapping rows and/or columns of to preserve numerical stability.
### Example 2
Use symplectic methods to solve Hamiltonian differential equations.
A Hamiltonian differential equation has the form
where . Symplectic methods preserve the two-form where . Such methods also nearly preserve a "numerical energy" function (which depends on the step-size), and are much better for long-time integration of mechanical systems such as arise in celestial mechanics. | 2019-03-23 04:39:02 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.847111165523529, "perplexity": 1029.543287358581}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202723.74/warc/CC-MAIN-20190323040640-20190323062640-00533.warc.gz"} |
https://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition-anton/chapter-0-before-calculus-0-1-functions-exercises-set-0-1-page-13/18 | Calculus, 10th Edition (Anton)
When $x = 0$, $y = 5$ and $y = -5$. As there is a value of x for which there is more than one corresponding value for y, this graph does not define a function | 2019-12-10 16:26:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8907313942909241, "perplexity": 114.81282198100651}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00351.warc.gz"} |
https://www.physicsforums.com/threads/qft-question-what-is-meant-by-dipole-form.650063/ | # QFT Question: What is meant by dipole form ?
1. Nov 6, 2012
### spooky51
QFT Question: What is meant by "dipole form"?
Hello physics people! Probably a very basic question, but here goes.
I'm taking a course on QFT based on Ryder. I've heard my professor refer to propagators as having a "polar" or "dipole" form. Things like (k^2 - m^2 + ie)^(-1)
For anyone who has a copy of Ryder hand, the specific case I'm puzzling over at at the moment is equation 1.20 on page 16, where the text above says that the form factor G_M has a dipole form. ( G_M = (1+q^2/M_q^2)^(-1)).
I'm not sure what is meant by something having a polar/dipole form. I can't seem to find anything on google or wikipedia, but I might not be digging deep enough. Can someone shed some light on this?
2. Nov 6, 2012
### Bill_K
Re: QFT Question: What is meant by "dipole form"?
The form factor is the Fourier transform of the charge distribution,
F(q) = ∫ρ(x) eiq·x d3x
For small q, (long wavelength photons) expand the exponential,
F(q) = ∫ρ(x)(1 + iq·x -(q·x)2/2 + ...) d3x
Keeping only these leading terms is called the dipole approximation. Assuming the charge distribution is spherically symmetric, you get a form factor,
F(q) = 1 + <r2> q2 + ...
where <r2> is the mean square radius of the charge cloud.
Whether you write it this way or in the denominator, (1 - <r2> q2)-1, the idea is the same. Keeping only the leading term in q2 is the dipole approximation.
Last edited: Nov 6, 2012
3. Nov 7, 2012
### andrien
Re: QFT Question: What is meant by "dipole form"?
what you say is dipole approximation which is essentially the point that atomic dimensions are small compared with wavelength.Moreover,the form factor also arises in rosenbluth formula where it has to do with the structured form of proton where form factors really has it's connection to anomalous magnetic moment and also the structure of charge distribution(there are two).I don't know what it has to do with the propagators of a fermion. | 2017-08-22 09:54:00 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8483472466468811, "perplexity": 1273.2528405445}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886110573.77/warc/CC-MAIN-20170822085147-20170822105147-00388.warc.gz"} |
https://electronics.stackexchange.com/questions/268478/op-amp-output-load | I'm using an audio op amp (LME49726) to combine two audio sources together, to be listened to with headphones at its output. Whenever I plug in the headphones, there is a large voltage drop in the supply voltage and it doesn't work. What could be the problem with having a headphone load at the output? What op amp parameters would this involve? I have also tried a general purpose op amp with the same result. This is datasheet of the audio op amp: http://www.ti.com/lit/ds/symlink/lme49726.pdf Here's the simple circuit:
• Second line of the datasheet : "Easily drives 2 kilohm loads". What impedance are your headphones? If they are less than 2 kilohms ... no go. Nov 9 '16 at 18:33
• I didn't realize that meant you cannot drive a load below 2k...is there an actual parameter in the characteristics that indicates this? Nov 9 '16 at 18:39
• What are you using for a power supply? And a schematic of your circuit would be helpful too. Nov 9 '16 at 18:41
• *is there an actual parameter * Forget the parameters, just read what this opamp is designed for. Does it say "headphone amplifier" anywhere ? I could not find it. Obviously this amplifier is not meant for driving headphones. Nov 9 '16 at 18:49
• And next time, include the schematic from the start as then we could have immediately pointed out that you need a DC blocking capacitor at the output ! Nov 10 '16 at 6:58
That particular op-amp is capable of hundreds of mA output current. I think you should be able to deafen yourself if the headphones are the normal rare-earth ~30$\Omega$ variety.
You seem to have identified part of the problem- the supply voltage drops. You need to capacitively couple the headphones. What you have now attempts to put 2.5VDC across the headphones, which is at least 83mA, thereby dragging down your supply.
Try a 100uF electrolytic in series with the output (+ side to the op-amp, - to the headphones).
• Ah that's the problem, I forgot to do that, thanks!! Nov 9 '16 at 19:05
• @cheeto if that kind of cap is too bulky or you don't have one at hand, you can also stabilise the 2.5 V level with another OP and use it as the headphone's ground. Or implement a proper push-pull by inverting the signal once more. But that's arguably overkill for headphones. Nov 9 '16 at 20:45
• Why does the output coupling capacitor need to be as high as 100 uF and be electrolytic? What about a 10 uF ceramic cap? Nov 10 '16 at 5:09
• It does not need to be electrolytic, but 100uF is pretty small and maintains bass response. You can work out the -3dB frequency. 10uF ceramic would limit the low end response and probably add a lot of audible distortion, but if you can't hear any difference why not. Nov 10 '16 at 5:18
• Would you calculate the output coupling capacitor you need for your desired frequency response through the 1/(2*pi*RC) formula, with R being the headphone impedance? Dec 21 '16 at 21:53
If your power supply voltage is dropping then it indicates that the power supply is not capable of supplying the current to the load.
Many headphones are only 8 ohms. Suppose you had an output signal level into the headphones of 1V you would need to supply up to 1V/8 ohms = 125mA. If your signal level is higher you will need correspondingly more current. For example at 4V you would need 0.5A.
The LME49726 is capable of driving an 8 ohm load at signal levels up to 2.8V with a 5V supply and is easily capable of driving typical headphones.
The parameters that would typically be considered when driving headphones from an op-amp would be...
1) Voltage headroom from each supply vs load current (also referred to output swing). | 2022-01-24 05:00:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.488526314496994, "perplexity": 1773.6089288110163}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00257.warc.gz"} |
https://zbmath.org/serials/?q=se%3A1331 | ## Random Structures & Algorithms
Short Title: Random Struct. Algorithms Publisher: Wiley, New York, NY ISSN: 1042-9832; 1098-2418/e Online: http://onlinelibrary.wiley.com/journal/10.1002/(ISSN)1098-2418/issues Comments: Indexed cover-to-cover
Documents Indexed: 1,476 Publications (since 1990) References Indexed: 1,103 Publications with 22,020 References.
all top 5
### Latest Issues
57, No. 4 (2020) 57, No. 3 (2020) 57, No. 2 (2020) 57, No. 1 (2020) 56, No. 4 (2020) 56, No. 3 (2020) 56, No. 2 (2020) 56, No. 1 (2020) 55, No. 4 (2019) 55, No. 3 (2019) 55, No. 2 (2019) 55, No. 1 (2019) 54, No. 4 (2019) 54, No. 3 (2019) 54, No. 2 (2019) 54, No. 1 (2019) 53, No. 4 (2018) 53, No. 3 (2018) 53, No. 2 (2018) 53, No. 1 (2018) 52, No. 4 (2018) 52, No. 3 (2018) 52, No. 2 (2018) 52, No. 1 (2018) 51, No. 4 (2017) 51, No. 3 (2017) 51, No. 2 (2017) 51, No. 1 (2017) 50, No. 4 (2017) 50, No. 3 (2017) 50, No. 2 (2017) 50, No. 1 (2017) 49, No. 4 (2016) 49, No. 3 (2016) 49, No. 2 (2016) 49, No. 1 (2016) 48, No. 4 (2016) 48, No. 3 (2016) 48, No. 2 (2016) 48, No. 1 (2016) 47, No. 4 (2015) 47, No. 3 (2015) 47, No. 2 (2015) 47, No. 1 (2015) 46, No. 4 (2015) 46, No. 3 (2015) 46, No. 2 (2015) 46, No. 1 (2015) 45, No. 4 (2014) 45, No. 3 (2014) 45, No. 2 (2014) 45, No. 1 (2014) 44, No. 4 (2014) 44, No. 3 (2014) 44, No. 2 (2014) 44, No. 1 (2014) 43, No. 4 (2013) 43, No. 3 (2013) 43, No. 2 (2013) 43, No. 1 (2013) 42, No. 4 (2013) 42, No. 3 (2013) 42, No. 2 (2013) 42, No. 1 (2013) 41, No. 4 (2012) 41, No. 3 (2012) 41, No. 2 (2012) 41, No. 1 (2012) 40, No. 4 (2012) 40, No. 3 (2012) 40, No. 2 (2012) 40, No. 1 (2012) 39, No. 4 (2011) 39, No. 3 (2011) 39, No. 2 (2011) 39, No. 1 (2011) 38, No. 4 (2011) 38, No. 3 (2011) 38, No. 1-2 (2011) 37, No. 4 (2010) 37, No. 3 (2010) 37, No. 2 (2010) 37, No. 1 (2010) 36, No. 4 (2010) 36, No. 3 (2010) 36, No. 2 (2010) 36, No. 1 (2010) 35, No. 4 (2009) 35, No. 3 (2009) 35, No. 2 (2009) 35, No. 1 (2009) 34, No. 4 (2009) 34, No. 3 (2009) 34, No. 2 (2009) 34, No. 1 (2009) 33, No. 4 (2008) 33, No. 3 (2008) 33, No. 2 (2008) 33, No. 1 (2008) 32, No. 4 (2008) ...and 114 more Volumes
all top 5
### Authors
69 Frieze, Alan Michael 49 Janson, Svante 44 Krivelevich, Michael 41 Bollobás, Béla 37 Sudakov, Benny 31 Rodl, Vojtech 26 Alon, Noga M. 25 Wormald, Nicholas Charles 23 Cooper, Colin 21 Pittel, Boris G. 20 Devroye, Luc P. J. A. 20 Łuczak, Tomasz 20 Riordan, Oliver Maxim 19 Kohayakawa, Yoshiharu 19 McDiarmid, Colin J. H. 18 Spencer, Joel H. 18 Vu, Van H. 17 Dyer, Martin E. 17 Steger, Angelika 14 Kahn, Jeff D. 14 Kang, Mihyun 13 Bohman, Tom 13 Ferber, Asaf 13 Kostochka, Aleksandr Vasil’evich 12 Balogh, József 12 Hwang, Hsien-Kuei 11 Balister, Paul N. 11 Chayes, Jennifer Tour 11 Feige, Uriel 11 Kim, Jeong Han 11 Reed, Bruce Alan 11 Vigoda, Eric 10 Borgs, Christian 10 Coja-Oghlan, Amin 10 Drmota, Michael 10 Gamarnik, David 10 Linial, Nathan 10 Lubetzky, Eyal 10 Mossel, Elchanan 10 Osthus, Deryk 10 Schacht, Mathias 9 Broutin, Nicolas 9 Goldreich, Oded 9 Luczak, Malwina J. 9 Peres, Yuval 9 van der Hofstad, Remco W. 8 Böttcher, Julia 8 Brightwell, Graham R. 8 Chung, Fan 8 Fountoulakis, Nikolaos 8 Greenhill, Catherine S. 8 Kühn, Daniela 8 Morris, Robert D. 8 Szabó, Tibor 8 Szpankowski, Wojciech 8 Tetali, Prasad 8 Winkler, Peter M. 7 Aldous, David John 7 Benjamini, Itai 7 Fill, James Allen 7 Flajolet, Philippe 7 Gao, Pu 7 Jaworski, Jerzy 7 Loh, Po-Shen 7 Louchard, Guy 7 Marckert, Jean-François 7 McKay, Brendan D. 7 Nenadov, Rajko 7 Panholzer, Alois 7 Pegden, Wesley 7 Pippenger, Nicholas J. 7 Prałat, Paweł 7 Prodinger, Helmut 7 Stojaković, Miloš 7 van den Berg, Rob 6 Conlon, David 6 Czumaj, Artur 6 Erdős, Pál 6 Fernandez de la Vega, Wenceslas 6 Flaxman, Abraham D. 6 Fox, Jacob 6 Friedgut, Ehud 6 Häggström, Olle 6 Håstad, Johan Torkel 6 Hefetz, Dan 6 Hitczenko, Paweł 6 Karp, Richard Manning 6 Mubayi, Dhruv 6 Nagle, Brendan 6 Neininger, Ralph 6 Ron, Dana 6 Ruciński, Andrzej 6 Schmutz, Eric 6 Shapira, Asaf 6 Smythe, Robert T. 6 Sorkin, Gregory B. 6 Srinivasan, Aravind 6 Warnke, Lutz 5 Barbour, Andrew David 5 Chern, Huahuai ...and 1,326 more Authors
all top 5
### Fields
1,053 Combinatorics (05-XX) 559 Probability theory and stochastic processes (60-XX) 302 Computer science (68-XX) 62 Operations research, mathematical programming (90-XX) 55 Statistical mechanics, structure of matter (82-XX) 43 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 34 Information and communication theory, circuits (94-XX) 27 Statistics (62-XX) 26 Linear and multilinear algebra; matrix theory (15-XX) 24 Convex and discrete geometry (52-XX) 24 Numerical analysis (65-XX) 22 Mathematical logic and foundations (03-XX) 21 Order, lattices, ordered algebraic structures (06-XX) 19 Number theory (11-XX) 16 General and overarching topics; collections (00-XX) 12 Biology and other natural sciences (92-XX) 9 Group theory and generalizations (20-XX) 6 Geometry (51-XX) 6 Algebraic topology (55-XX) 5 Dynamical systems and ergodic theory (37-XX) 3 History and biography (01-XX) 3 Special functions (33-XX) 3 Functional analysis (46-XX) 3 Calculus of variations and optimal control; optimization (49-XX) 2 Field theory and polynomials (12-XX) 2 Real functions (26-XX) 2 Approximations and expansions (41-XX) 2 Operator theory (47-XX) 2 Manifolds and cell complexes (57-XX) 1 Algebraic geometry (14-XX) 1 Measure and integration (28-XX) 1 Potential theory (31-XX) 1 Partial differential equations (35-XX) 1 Difference and functional equations (39-XX) 1 Harmonic analysis on Euclidean spaces (42-XX) 1 Global analysis, analysis on manifolds (58-XX) 1 Quantum theory (81-XX)
### Citations contained in zbMATH Open
1,291 Publications have been cited 14,125 times in 8,259 Documents Cited by Year
A critical point for random graphs with a given degree sequence. Zbl 0823.05050
Molloy, Michael; Reed, Bruce
1995
Exact sampling with coupled Markov chains and applications to statistical mechanics. Zbl 0859.60067
Propp, James Gary; Wilson, David Bruce
1996
The phase transition in inhomogeneous random graphs. Zbl 1123.05083
Bollobàs, Bela; Janson, Svante; Riordan, Oliver
2007
The degree sequence of a scale-free random graph process. Zbl 0985.05047
Bollobás, Béla; Riordan, Oliver; Spencer, Joel; Tusnády, Gábor
2001
The Ramsey number $$R(3,t)$$ has order of magnitude $$t^ 2 /\log t$$. Zbl 0832.05084
Kim, Jeong Han
1995
Acyclic coloring of graphs. Zbl 0735.05036
Alon, Noga; McDiarmid, Colin; Reed, Bruce
1991
An elementary proof of a theorem of Johnson and Lindenstrauss. Zbl 1018.51010
Dasgupta, Sanjoy; Gupta, Anupam
2003
Random walks in a convex body and an improved volume algorithm. Zbl 0788.60087
Lovász, László; Simonovits, M.
1993
Simple constructions of almost $$k$$-wise independent random variables. Zbl 0755.60002
Alon, Noga; Goldreich, Oded; Håstad, Johan; Peralta, René
1992
Regularity lemma for $$k$$-uniform hypergraphs. Zbl 1046.05042
Rödl, Vojtěch; Skokan, Jozef
2004
The birth of the giant component. Zbl 0795.05127
Janson, Svante; Knuth, Donald E.; Łuczak, Tomasz; Pittel, Boris
1993
The counting lemma for regular $$k$$-uniform hypergraphs. Zbl 1093.05045
Nagle, Brendan; Rödl, Vojtěch; Schacht, Mathias
2006
A very simple algorithm for estimating the number of $$k$$-colorings of a low-degree graph. Zbl 0833.60070
Jerrum, Mark
1995
Component behavior near the critical point of the random graph process. Zbl 0745.05048
Łuczak, Tomasz
1990
A general model of web graphs. Zbl 1018.60007
Cooper, Colin; Frieze, Alan
2003
Balls and bins: A study in negative dependence. Zbl 0964.60503
Dubhashi, Devdatt; Ranjan, Desh
1998
Poisson approximation for large deviations. Zbl 0747.05079
Janson, Svante
1990
The complexity of counting graph homomorphisms. Zbl 0965.68073
Dyer, Martin; Greenhill, Catherine
2000
Nonrepetitive colorings of graphs. Zbl 1018.05032
Alon, Noga; Grytczuk, Jarosław; Hałuszczak, Mariusz; Riordan, Oliver
2002
Improved bounds and algorithms for hypergraph 2-coloring. Zbl 0942.05024
Radhakrishnan, Jaikumar; Srinivasan, Aravind
2000
Homological connectivity of random $$k$$-dimensional complexes. Zbl 1177.55011
Meshulam, R.; Wallach, N.
2009
Extremal problems on set systems. Zbl 0995.05141
Frankl, Peter; Rödl, Vojtěch
2002
Random Cayley graphs and expanders. Zbl 0798.05048
Alon, Noga; Roichman, Yuval
1994
Problems and results on judicious partitions. Zbl 1013.05059
Bollobás, B.; Scott, A. D.
2002
Randomized broadcast in networks. Zbl 0712.68011
Feige, Uriel; Peleg, David; Raghavan, Prabhakar; Upfal, Eli
1990
Almost all regular graphs are Hamiltonian. Zbl 0795.05088
Robinson, R. W.; Wormald, N. C.
1994
Random walks and an $$O^*(n^5)$$ volume algorithm for convex bodies. Zbl 0895.60075
Kannan, Ravi; Lovász, László; Simonovits, Miklós
1997
The $$\zeta(2)$$ limit in the random assignment problem. Zbl 0993.60018
Aldous, David J.
2001
The infamous upper tail. Zbl 0996.60023
Janson, Svante; Ruciński, Andrzej
2002
Note on the heights of random recursive trees and random $$m$$-ary search trees. Zbl 0790.05077
Pittel, Boris
1994
Asymptotic analysis of a random walk on a hypercube with many dimensions. Zbl 0723.60085
Diaconis, Persi; Graham, R. L.; Morrison, J. A.
1990
Survey propagation: an algorithm for satisfiability. Zbl 1087.68126
Braunstein, A.; Mézard, M.; Zecchina, R.
2005
Invertibility of symmetric random matrices. Zbl 1291.15088
Vershynin, Roman
2014
Bootstrap percolation on the random regular graph. Zbl 1106.60076
Balogh, József; Pittel, Boris G.
2007
Spectral techniques applied to sparse random graphs. Zbl 1076.05073
Feige, Uriel; Ofek, Eran
2005
The transitive closure of a random digraph. Zbl 0712.68076
Karp, Richard M.
1990
Almost all cubic graphs are Hamiltonian. Zbl 0755.05075
Robinson, R. W.; Wormald, N. C.
1992
Local resilience of graphs. Zbl 1182.05114
Sudakov, Benny; Vu, V. H.
2008
On random $$\pm 1$$ matrices: Singularity and determinant. Zbl 1086.60008
Tao, Terence; Vu, Van
2006
Sparse random graphs: eigenvalues and eigenvectors. Zbl 1257.05089
Tran, Linh V.; Vu, Van H.; Wang, Ke
2013
The geometry of logconcave functions and sampling algorithms. Zbl 1122.65012
Lovász, László; Vempala, Santosh
2007
Finding a large hidden clique in a random graph. Zbl 0959.05082
Alon, Noga; Krivelevich, Michael; Sudakov, Benny
1998
An algorithmic approach to the Lovász local lemma. I. Zbl 0756.05080
Beck, József
1991
Large cliques elude the Metropolis process. Zbl 0754.60018
Jerrum, Mark
1992
Random maps, coalescing saddles, singularity analysis, and Airy phenomena. Zbl 1016.68179
Banderier, Cyril; Flajolet, Philippe; Schaeffer, Gilles; Soria, Michèle
2001
Excluding induced subgraphs. III: A general asymptotic. Zbl 0751.05041
Prömel, H. J.; Steger, A.
1992
A few logs suffice to build (almost) all trees. I. Zbl 0945.60004
Erdős, Péter L.; Steel, Michael A.; Székely, László A.; Warnow, Tandy J.
1999
Factors in random graphs. Zbl 1146.05040
Johansson, Anders; Kahn, Jeff; Vu, Van
2008
Approximating the unsatisfiability threshold of random formulas. Zbl 0936.68038
Kirousis, Lefteris M.; Kranakis, Evangelos; Krizanc, Danny; Stamatiou, Yannis C.
1998
A tight upper bound on the cover time for random walks on graphs. Zbl 0811.60060
Feige, Uriel
1995
Systems of functional equations. Zbl 0869.39010
Drmota, Michael
1997
Three theorems regarding testing graph properties. Zbl 1048.68062
Goldreich, Oded; Trevisan, Luca
2003
Random cutting and records in deterministic and random trees. Zbl 1120.05083
Janson, Svante
2006
The scaling window of the 2-SAT transition. Zbl 0979.68053
Bollobás, Béla; Borgs, Christian; Chayes, Jennifer T.; Kim, Jeong Han; Wilson, David B.
2001
Testing subgraphs in large graphs. Zbl 1027.68095
Alon, Noga
2002
A new approach to the giant component problem. Zbl 1177.05110
Janson, Svante; Luczak, Malwina J.
2009
Degrees and choice numbers. Zbl 0958.05049
Alon, Noga
2000
A tight lower bound on the cover time for random walks on graphs. Zbl 0832.60016
Feige, Uriel
1995
An exponential lower bound to the size of bounded depth Frege proofs of the pigeonhole principle. Zbl 0843.03032
Krajíček, Jan; Pudlák, Pavel; Woods, Alan
1995
Nearest neighbor and hard sphere models in continuum percolation. Zbl 0866.60088
Häggström, Olle; Meester, Ronald
1996
Dependent random choice. Zbl 1215.05159
Fox, Jacob; Sudakov, Benny
2011
Random trees and general branching processes. Zbl 1144.60051
Rudas, Anna; Tóth, Bálint; Valkó, Benedek
2007
2+p-SAT: Relation of typical-case complexity to the nature of the phase transition. Zbl 0931.68056
Monasson, Rémi; Zecchina, Riccardo; Kirkpatrick, Scott; Selman, Bart; Troyansky, Lidror
1999
Distances in random graphs with finite variance degrees. Zbl 1074.05083
van der Hofstad, Remco; Hooghiemstra, Gerard; Van Mieghem, Piet
2005
On tree census and the giant component in sparse random graphs. Zbl 0747.05080
Pittel, Boris
1990
A sharp concentration inequality with applications. Zbl 0954.60008
Boucheron, Stéphane; Lugosi, Gábor; Massart, Pascal
2000
On the structure of random plane-oriented recursive trees and their branches. Zbl 0773.05040
Mahmoud, Hosam M.; Smythe, R. T.; Szymański, Jerzy
1993
Quasi-random hypergraphs. Zbl 0708.05044
Chung, F. R. K.; Graham, R. L.
1990
Random disease on the square grid. Zbl 0964.60006
Balogh, József; Pete, Gábor
1998
Cores in random hypergraphs and Boolean formulas. Zbl 1068.05063
Molloy, Michael
2005
Random subgraphs of finite graphs. I: The scaling window under the triangle condition. Zbl 1076.05071
Borgs, Christian; Chayes, Jennifer T.; van der Hofstad, Remco; Slade, Gordon; Spencer, Joel
2005
The strong chromatic index of $$C_4$$-free graphs. Zbl 0961.05022
2000
Maximum hitting time for random walks on graphs. Zbl 0744.05044
Brightwell, Graham; Winkler, Peter
1990
Simple Markov-chain algorithms for generating bipartite graphs and tournaments. Zbl 0933.05145
Kannan, Ravi; Tetali, Prasad; Vempala, Santosh
1999
A sharp threshold for $$k$$-colorability. Zbl 0962.05055
Achlioptas, Dimitris; Friedgut, Ehud
1999
Random regular graphs of high degree. Zbl 0996.05106
Krivelevich, Michael; Sudakov, Benny; Vu, Van H.; Wormald, Nicholas C.
2001
Asymptotic degree distribution in random recursive trees. Zbl 1059.05094
Janson, Svante
2005
Properly colored subgraphs and rainbow subgraphs in edge-colorings with local constraints. Zbl 1037.05033
Alon, Noga; Jiang, Tao; Miller, Zevi; Pritikin, Dan
2003
Large deviations for sums of partly dependent random variables. Zbl 1044.60021
Janson, Svante
2004
A parallel algorithmic version of the local lemma. Zbl 0768.05086
Alon, Noga
1991
Randomly coloring sparse random graphs with fewer colors than the maximum degree. Zbl 1115.05030
Dyer, Martin; Flaxman, Abraham D.; Frieze, Alan M.; Vigoda, Eric
2006
Properly colored Hamilton cycles in edge-colored complete graphs. Zbl 0882.05084
Alon, N.; Gutin, Gregory
1997
On replica symmetry of large deviations in random graphs. Zbl 1348.05195
Lubetzky, Eyal; Zhao, Yufei
2015
Counting without sampling: Asymptotics of the log-partition function for certain statistical physics models. Zbl 1157.82006
Bandyopadhyay, Antar; Gamarnik, David
2008
Tail bounds for occupancy and the satisfiability threshold conjecture. Zbl 0834.68051
Kamath, Anil; Motwani, Rajeev; Palem, Krishna; Spirakis, Paul
1995
Mixing in time and space for lattice spin systems: a combinatorial view. Zbl 1126.82021
Dyer, Martin; Sinclair, Alistair; Vigoda, Eric; Weitz, Dror
2004
Asymptotic random graph intuition for the biased connectivity game. Zbl 1198.91050
Gebauer, Heidi; Szabó, Tibor
2009
Concentration and regularization of random graphs. Zbl 1373.05179
Le, Can M.; Levina, Elizaveta; Vershynin, Roman
2017
Ramsey properties of random discrete structures. Zbl 1228.05284
Friedgut, Ehud; Rödl, Vojtěch; Schacht, Mathias
2010
Local resilience of almost spanning trees in random graphs. Zbl 1215.05154
Balogh, József; Csaba, Béla; Samotij, Wojciech
2011
An algorithmic version of the blow-up lemma. Zbl 0917.05071
Komlós, János; Sárközy, Gabor N.; Szemerédi, Endre
1998
A simple and linear time randomized algorithm for computing sparse spanners in weighted graphs. Zbl 1118.68582
Baswana, Surender; Sen, Sandeep
2007
Applications of the regularity lemma for uniform hypergraphs. Zbl 1087.05031
Rödl, Vojtěch; Skokan, Jozef
2006
On variants of the Johnson-Lindenstrauss Lemma. Zbl 1154.51002
Matoušek, Jiří
2008
Algorithms for graph partitioning on the planted partition model. Zbl 0972.68129
Condon, Anne; Karp, Richard M.
2001
Avoiding a giant component. Zbl 0986.05091
Bohman, Tom; Frieze, Alan
2001
The vertex degree distribution of random intersection graphs. Zbl 1042.05091
Stark, Dudley
2004
Limit theorems for combinatorial structures via discrete process approximations. Zbl 0758.60009
Arratia, Richard; Tavaré, Simon
1992
Critical percolation on random regular graphs. Zbl 1209.05228
Nachmias, Asaf; Peres, Yuval
2010
New approach to nonrepetitive sequences. Zbl 1349.68131
Grytczuk, Jarosław; Kozik, Jakub; Micek, Piotr
2013
Permutations with fixed pattern densities. Zbl 1442.05008
Kenyon, Richard; Král’, Daniel; Radin, Charles; Winkler, Peter
2020
Powers of Hamiltonian cycles in randomly augmented graphs. Zbl 1444.05080
Dudek, Andrzej; Reiher, Christian; Ruciński, Andrzej; Schacht, Mathias
2020
A graphon approach to limiting spectral distributions of Wigner-type matrices. Zbl 1442.05214
Zhu, Yizhe
2020
Rainbow structures in locally bounded colorings of graphs. Zbl 1450.05070
Kim, Jaehoon; Kühn, Daniela; Kupavskii, Andrey; Osthus, Deryk
2020
Condensation in preferential attachment models with location-based choice. Zbl 1455.05070
Haslegrave, John; Jordan, Jonathan; Yarrow, Mark
2020
Swendsen-Wang dynamics for general graphs in the tree uniqueness region. Zbl 1436.05104
Blanca, Antonio; Chen, Zongchen; Vigoda, Eric
2020
Almost all trees are almost graceful. Zbl 1456.05134
Adamaszek, Anna; Allen, Peter; Grosu, Codruţ; Hladký, Jan
2020
Size of nodal domains of the eigenvectors of a $$G(n,p)$$ graph. Zbl 1453.05060
Huang, Han; Rudelson, Mark
2020
Invertibility via distance for noncentered random matrices with continuous distributions. Zbl 1456.60028
Tikhomirov, Konstantin
2020
Rapid mixing of the switch Markov chain for strongly stable degree sequences. Zbl 1456.60175
Amanatidis, Georgios; Kleer, Pieter
2020
Coloring triangle-free graphs with local list sizes. Zbl 1464.05148
Davies, Ewan; de Verclos, Rémi de Joannis; Kang, Ross J.; Pirot, François
2020
Finding cliques using few probes. Zbl 1442.05207
Feige, Uriel; Gamarnik, David; Neeman, Joe; Rácz, Miklós Z.; Tetali, Prasad
2020
Semi-random graph process. Zbl 1442.05203
Ben-Eliezer, Omri; Hefetz, Dan; Kronenberg, Gal; Parczyk, Olaf; Shikhelman, Clara; Stojaković, Miloš
2020
Random graphs with given vertex degrees and switchings. Zbl 1451.05211
Janson, Svante
2020
A rainbow blow-up lemma. Zbl 1458.05067
Glock, Stefan; Joos, Felix
2020
Resolution of a conjecture on majority dynamics: rapid stabilization in dense random graphs. Zbl 1454.05109
Fountoulakis, Nikolaos; Kang, Mihyun; Makai, Tamás
2020
On the weight distribution of random binary linear codes. Zbl 1475.94193
Linial, Nati; Mosheiff, Jonathan
2020
Spanning trees in randomly perturbed graphs. Zbl 1444.05132
Joos, Felix; Kim, Jaehoon
2020
The combinatorics of the colliding bullets. Zbl 1446.60008
Broutin, Nicolas; Marckert, Jean-François
2020
Limiting shape of the depth first search tree in an Erdős-Rényi graph. Zbl 1436.05096
Enriquez, Nathanaël; Faraud, Gabriel; Ménard, Laurent
2020
Quasi-polynomial mixing of critical two-dimensional random cluster models. Zbl 1453.82043
Gheissari, Reza; Lubetzky, Eyal
2020
Phase transitions of the Moran process and algorithmic consequences. Zbl 1444.92072
Goldberg, Leslie Ann; Lapinskas, John; Richerby, David
2020
Geometry of the vacant set left by random walk on random graphs, Wright’s constants, and critical random graphs with prescribed degrees. Zbl 1445.05096
Bhamidi, Shankar; Sen, Sanchayan
2020
Further results on random cubic planar graphs. Zbl 1445.05095
Noy, Marc; Requilé, Clément; Rué, Juanjo
2020
Eigenvector delocalization for non-Hermitian random matrices and applications. Zbl 1448.15046
Luh, Kyle; O’Rourke, Sean
2020
Modularity of Erdős-Rényi random graphs. Zbl 1452.05168
McDiarmid, Colin; Skerman, Fiona
2020
Sharp thresholds for nonlinear Hamiltonian cycles in hypergraphs. Zbl 1472.05088
Narayanan, Bhargav; Schacht, Mathias
2020
Shotgun assembly of random jigsaw puzzles. Zbl 1450.05023
Bordenave, Charles; Feige, Uriel; Mossel, Elchanan
2020
Tight bounds for popping algorithms. Zbl 1453.62542
Guo, Heng; He, Kun
2020
Diameter of P.A. random graphs with edge-step functions. Zbl 1462.05322
Alves, Caio; Ribeiro, Rodrigo; Sanchis, Rémy
2020
Every planar graph with the Liouville property is amenable. Zbl 1453.31015
Carmesin, Johannes; Georgakopoulos, Agelos
2020
Vertex Ramsey properties of randomly perturbed graphs. Zbl 1454.05075
Das, Shagnik; Morris, Patrick; Treglown, Andrew
2020
On connectivity, conductance and bootstrap percolation for a random $$K$$-out, age-biased graph. Zbl 1444.05130
Acan, Hüseyin; Pittel, Boris
2020
The genus of the Erdős-Rényi random graph and the fragile genus property. Zbl 1443.05171
Dowden, Chris; Kang, Mihyun; Krivelevich, Michael
2020
Vanishing of cohomology groups of random simplicial complexes. Zbl 1436.05095
Cooley, Oliver; Del Giudice, Nicola; Kang, Mihyun; Sprüssel, Philipp
2020
Slightly subcritical hypercube percolation. Zbl 1436.60081
Hulshof, Tim; Nachmias, Asaf
2020
Connectivity of a general class of inhomogeneous random digraphs. Zbl 1453.82005
Cao, Junyu; Olvera-Cravioto, Mariana
2020
Random tree recursions: which fixed points correspond to tangible sets of trees? Zbl 1445.05028
Johnson, Tobias; Podder, Moumanti; Skerman, Fiona
2020
The chromatic number of random Borsuk graphs. Zbl 1442.05210
Kahle, Matthew; Martinez-Figueroa, Francisco
2020
Minimalist designs. Zbl 1451.05159
Barber, Ben; Glock, Stefan; Kühn, Daniela; Lo, Allan; Montgomery, Richard; Osthus, Deryk
2020
New bounds for the Moser-Tardos distribution. Zbl 1451.05240
Harris, David G.
2020
On Hamilton cycles in Erdős-Rényi subgraphs of large graphs. Zbl 1451.05132
Johansson, Tony
2020
Asymptotics in percolation on high-girth expanders. Zbl 1455.05041
Krivelevich, Michael; Lubetzky, Eyal; Sudakov, Benny
2020
Asymptotic normality in random graphs with given vertex degrees. Zbl 1451.05212
Janson, Svante
2020
Phase transitions in graphs on orientable surfaces. Zbl 1451.05213
Kang, Mihyun; Moßhammer, Michael; Sprüssel, Philipp
2020
Judiciously 3-partitioning 3-uniform hypergraphs. Zbl 1450.05062
Spink, Hunter; Tiba, Marius
2020
1-factorizations of pseudorandom graphs. Zbl 1453.05100
Ferber, Asaf; Jain, Vishesh
2020
The real tau-conjecture is true on average. Zbl 07279068
Briquel, Irénée; Bürgisser, Peter
2020
Recursive functions on conditional Galton-Watson trees. Zbl 07279069
Broutin, Nicolas; Devroye, Luc; Fraiman, Nicolas
2020
Hyperuniform and rigid stable matchings. Zbl 1456.60121
Klatt, Michael Andreas; Last, Günter; Yogeshwaran, D.
2020
Constrained percolation, Ising model, and XOR Ising model on planar lattices. Zbl 1453.82036
Li, Zhongyang
2020
Projections of the Aldous chain on binary trees: intertwining and consistency. Zbl 1453.05134
Forman, Noah; Pal, Soumik; Rizzolo, Douglas; Winkel, Matthias
2020
Information percolation and cutoff for the random-cluster model. Zbl 1453.82077
Ganguly, Shirshendu; Seo, Insuk
2020
On-line balancing of random inputs. Zbl 1454.91051
Bansal, Nikhil; Spencer, Joel H.
2020
Characterization of quasirandom permutations by a pattern sum. Zbl 1454.05007
Chan, Timothy F. N.; Král’, Daniel; Noel, Jonathan A.; Pehova, Yanitsa; Sharifzadeh, Maryam; Volec, Jan
2020
Short proofs of some extremal results. III. Zbl 1454.05056
Conlon, David; Fox, Jacob; Sudakov, Benny
2020
Ramsey, paper, scissors. Zbl 1454.91048
Fox, Jacob; He, Xiaoyu; Wigderson, Yuval
2020
Asymptotic normality of the number of corners in tableaux associated with the partially asymmetric simple exclusion process. Zbl 1454.05123
Hitczenko, Paweł; Yaroslavskiy, Aleksandr
2020
Sharp bounds for the variance of linear statistics on random permutations. Zbl 1456.60034
Manstavičius, Eugenijus
2020
Hypergraph coloring up to condensation. Zbl 1417.05057
Ayre, Peter; Coja-Oghlan, Amin; Greenhill, Catherine
2019
The Johansson-Molloy theorem for DP-coloring. Zbl 1417.05059
Bernshteyn, Anton
2019
Resilience of perfect matchings and Hamiltonicity in random graph processes. Zbl 1417.05189
Nenadov, Rajko; Steger, Angelika; Trujić, Miloš
2019
Powers of tight Hamilton cycles in randomly perturbed hypergraphs. Zbl 1433.05178
Bedenknecht, Wiebke; Han, Jie; Kohayakawa, Yoshiharu; Mota, Guilherme O.
2019
Patterns in random permutations avoiding the pattern 321. Zbl 1423.60026
Janson, Svante
2019
Rainbow matchings in Dirac bipartite graphs. Zbl 1426.05137
Coulson, Matthew; Perarnau, Guillem
2019
Universality for bounded degree spanning trees in randomly perturbed graphs. Zbl 1433.05275
Böttcher, Julia; Han, Jie; Kohayakawa, Yoshiharu; Montgomery, Richard; Parczyk, Olaf; Person, Yury
2019
Monochromatic paths in random tournaments. Zbl 1405.05158
Bucić, Matija; Letzter, Shoham; Sudakov, Benny
2019
Powers of Hamilton cycles in random graphs and tight Hamilton cycles in random hypergraphs. Zbl 1405.05167
Nenadov, Rajko; Škorić, Nemanja
2019
Multicolor containers, extremal entropy, and counting. Zbl 1417.05104
Falgas-Ravry, Victor; O’Connell, Kelly; Uzzell, Andrew
2019
A counterexample to the DeMarco-Kahn upper tail conjecture. Zbl 1433.05285
Šileikis, Matas; Warnke, Lutz
2019
Sharp asymptotic for the chemical distance in long-range percolation. Zbl 1428.05164
Biskup, Marek; Lin, Jeffrey
2019
Component structure of the configuration model: barely supercritical case. Zbl 1425.05145
van der Hofstad, Remco; Janson, Svante; Luczak, Malwina
2019
Uniform random colored complexes. Zbl 1428.05098
Carrance, Ariane
2019
On polynomial approximations to $$\mathrm{AC}^0$$. Zbl 1421.68055
Harsha, Prahladh; Srinivasan, Srikanth
2019
The size Ramsey number of short subdivisions of bounded degree graphs. Zbl 1409.05136
Kohayakawa, Yoshiharu; Retter, Troy; Rödl, Vojtěch
2019
Random Čech complexes on Riemannian manifolds. Zbl 1414.05318
Bobrowski, Omer; Oliveira, Goncalo
2019
On densities for solutions to stochastic fixed point equations. Zbl 1454.60017
Leckey, Kevin
2019
Most edge-orderings of $$K_{n}$$ have maximal altitude. Zbl 1414.05268
Martinsson, Anders
2019
Rapid mixing of hypergraph independent sets. Zbl 1417.05147
Hermon, Jonathan; Sly, Allan; Zhang, Yumeng
2019
Asymptotically optimal bound on the adjacent vertex distinguishing edge choice number. Zbl 1417.05072
Kwaśny, Jakub; Przybyło, Jakub
2019
Fractional clique decompositions of dense graphs. Zbl 1416.05232
Montgomery, Richard
2019
The continuum random tree is the scaling limit of unlabeled unrooted trees. Zbl 1426.05153
Stufler, Benedikt
2019
Spatial mixing and nonlocal Markov chains. Zbl 1427.60150
Blanca, Antonio; Caputo, Pietro; Sinclair, Alistair; Vigoda, Eric
2019
Enumeration and randomized constructions of hypertrees. Zbl 1433.05340
Linial, Nati; Peled, Yuval
2019
Geometry of large Boltzmann outerplanar maps. Zbl 1428.05286
Stefánsson, Sigurdur Örn; Stufler, Benedikt
2019
Perfect matchings and Hamiltonian cycles in the preferential attachment model. Zbl 1417.05188
Frieze, Alan; Pérez-Giménez, Xavier; Prałat, Paweł; Reiniger, Benjamin
2019
A new combinatorial representation of the additive coalescent. Zbl 1409.05103
Marckert, Jean-François; Wang, Minmin
2019
Law of the iterated logarithm for random graphs. Zbl 1405.05159
Ferber, Asaf; Montealegre, Daniel; Vu, Van
2019
Swendsen-Wang algorithm on the mean-field Potts model. Zbl 1412.82009
Galanis, Andreas; Štefankovič, Daniel; Vigoda, Eric
2019
Graph limits of random graphs from a subset of connected $$k$$-trees. Zbl 1423.05152
Drmota, Michael; Jin, Emma Yu; Stufler, Benedikt
2019
Monotonicity and phase diagram for multirange percolation on oriented trees. Zbl 1423.05044
De Lima, Bernardo N. B.; Rolla, Leonardo T.; Valesin, Daniel
2019
Tunneling of the hard-core model on finite triangular lattices. Zbl 1423.82018
Zocca, Alessandro
2019
An optimal $$(\varepsilon,\delta)$$-randomized approximation scheme for the mean of random variables with bounded relative variance. Zbl 1423.60007
Huber, Mark
2019
Asymmetry and structural information in preferential attachment graphs. Zbl 1440.05071
Łuczak, Tomasz; Magner, Abram; Szpankowski, Wojciech
2019
Fluctuations in a general preferential attachment model via Stein’s method. Zbl 1433.05273
Betken, Carina; Döring, Hanna; Ortgiese, Marcel
2019
On the threshold problem for Latin boxes. Zbl 1439.05040
Luria, Zur; Simkin, Michael
2019
Competition in growth and urns. Zbl 1423.60148
Ahlberg, Daniel; Griffiths, Simon; Janson, Svante; Morris, Robert
2019
Diameter in ultra-small scale-free random graphs. Zbl 1414.05098
Caravenna, Francesco; Garavaglia, Alessandro; van der Hofstad, Remco
2019
Packing and counting arbitrary Hamilton cycles in random digraphs. Zbl 1414.05265
Ferber, Asaf; Long, Eoin
2019
Water transport on infinite graphs. Zbl 1414.05204
Häggström, Olle; Hirscher, Timo
2019
...and 1191 more Documents
all top 5
### Cited by 8,304 Authors
103 Sudakov, Benny 91 Frieze, Alan Michael 88 Krivelevich, Michael 87 Janson, Svante 70 Alon, Noga M. 61 van der Hofstad, Remco W. 60 Bollobás, Béla 58 Rodl, Vojtech 54 Balogh, József 49 Kohayakawa, Yoshiharu 49 Shabanov, Dmitry A. 46 Mahmoud, Hosam M. 44 Wormald, Nicholas Charles 42 Fox, Jacob 41 Osthus, Deryk 41 Vu, Van H. 40 Coja-Oghlan, Amin 40 Pittel, Boris G. 38 Cooper, Colin 38 Kang, Mihyun 38 Schacht, Mathias 37 Ferber, Asaf 36 Kühn, Daniela 36 Mubayi, Dhruv 36 Peres, Yuval 35 Goldberg, Leslie Ann 33 Prałat, Paweł 33 Sly, Allan 32 Goldreich, Oded 31 Bhamidi, Shankar 31 Conlon, David 30 Devroye, Luc P. J. A. 30 Hou, Jianfeng 30 Łuczak, Tomasz 30 Yuster, Raphael 29 Mossel, Elchanan 29 Steger, Angelika 28 Lubetzky, Eyal 27 Addario-Berry, Louigi 27 Dudek, Andrzej 27 Kostochka, Aleksandr Vasil’evich 27 McDiarmid, Colin J. H. 27 Morris, Robert D. 27 Riordan, Oliver Maxim 26 Person, Yury Aleksandrovic 26 Tao, Terence 25 Kang, Ross J. 25 Shapira, Asaf 24 Jerrum, Mark R. 24 Panholzer, Alois 24 Samotij, Wojciech 23 Böttcher, Julia 23 Chung, Fan 23 Dyer, Martin E. 23 Greenhill, Catherine S. 23 Panagiotou, Konstantinos D. 23 Prodinger, Helmut 23 Sauerwald, Thomas 22 Drmota, Michael 22 Mitsche, Dieter 22 Müller, Tobias M. 22 Noy, Marc 22 Sarkozy, Gabor N. 22 Taraz, Anusch 21 Barvinok, Alexander I. 21 Diaconis, Persi Warren 21 Ruciński, Andrzej 21 Vigoda, Eric 20 Broutin, Nicolas 20 Fountoulakis, Nikolaos 20 Gamarnik, David 20 Kahn, Jeff D. 20 Perarnau, Guillem 20 Rué, Juanjo 20 Zhao, Yufei 19 Hwang, Hsien-Kuei 19 Penrose, Mathew D. 19 Pikhurko, Oleg 19 Ron, Dana 19 Scott, Alexander D. 19 Sinclair, Alistair 18 Benjamini, Itai 18 Bertoin, Jean 18 Bohman, Tom 18 Galanis, Andreas 18 Iksanov, Aleksander M. 18 Kwan, Matthew 18 Linial, Nathan 18 Loh, Po-Shen 18 Neininger, Ralph 18 Nenadov, Rajko 18 Pach, János 18 Reed, Bruce Alan 18 Spencer, Joel H. 18 Warnke, Lutz 17 Borgs, Christian 17 Chayes, Jennifer Tour 17 Curien, Nicolas 17 Friedrich, Tobias 17 Hefetz, Dan ...and 8,204 more Authors
all top 5
### Cited in 581 Journals
479 Random Structures & Algorithms 315 Discrete Mathematics 306 Combinatorics, Probability and Computing 291 Theoretical Computer Science 217 The Annals of Applied Probability 210 Discrete Applied Mathematics 199 The Electronic Journal of Combinatorics 190 The Annals of Probability 188 European Journal of Combinatorics 181 Journal of Statistical Physics 179 Journal of Combinatorial Theory. Series B 158 SIAM Journal on Discrete Mathematics 135 Probability Theory and Related Fields 108 Journal of Applied Probability 107 SIAM Journal on Computing 106 Algorithmica 106 Stochastic Processes and their Applications 105 Combinatorica 104 Advances in Applied Probability 97 Graphs and Combinatorics 94 Journal of Combinatorial Theory. Series A 87 Information Processing Letters 79 Journal of Graph Theory 76 Electronic Journal of Probability 73 Journal of Computer and System Sciences 70 Israel Journal of Mathematics 68 Annales de l’Institut Henri Poincaré. Probabilités et Statistiques 67 Statistics & Probability Letters 65 The Annals of Statistics 61 Advances in Mathematics 61 Advances in Applied Mathematics 61 Linear Algebra and its Applications 58 Discrete & Computational Geometry 54 Bernoulli 54 Journal of Combinatorial Optimization 52 Communications in Mathematical Physics 51 Transactions of the American Mathematical Society 48 Computational Complexity 46 Journal of Statistical Mechanics: Theory and Experiment 38 Proceedings of the American Mathematical Society 37 Physica A 35 Information and Computation 34 Journal of Theoretical Probability 34 Distributed Computing 34 Annals of Combinatorics 32 Electronic Communications in Probability 32 Theory of Computing Systems 31 Journal of Mathematical Sciences (New York) 31 Journal of Machine Learning Research (JMLR) 31 Electronic Journal of Statistics 30 ALEA. Latin American Journal of Probability and Mathematical Statistics 28 Geometric and Functional Analysis. GAFA 28 Methodology and Computing in Applied Probability 27 Journal of Mathematical Physics 26 Journal of Mathematical Biology 23 Applied Mathematics and Computation 23 Doklady Mathematics 23 Annals of Mathematics. Second Series 23 Journal of Discrete Algorithms 21 Duke Mathematical Journal 21 Journal of Computational and Applied Mathematics 21 Journal of the American Mathematical Society 20 Internet Mathematics 19 European Journal of Operational Research 19 Mathematical Programming. Series A. Series B 18 Mathematical Biosciences 18 Journal of Functional Analysis 17 Mathematics of Operations Research 17 Journal of Theoretical Biology 17 Discrete Analysis 16 Bulletin of the American Mathematical Society. New Series 15 Mathematical Notes 15 Annals of Pure and Applied Logic 15 Discussiones Mathematicae. Graph Theory 15 Acta Mathematica Sinica. English Series 15 Computer Science Review 14 Acta Mathematica Hungarica 14 Probability in the Engineering and Informational Sciences 14 Stochastic Models 14 Probability Surveys 14 Random Matrices: Theory and Applications 14 Forum of Mathematics, Sigma 13 Artificial Intelligence 13 Inventiones Mathematicae 13 Queueing Systems 13 Annals of Operations Research 13 International Journal of Foundations of Computer Science 13 Designs, Codes and Cryptography 13 Proceedings of the National Academy of Sciences of the United States of America 13 Computational Statistics and Data Analysis 13 Chaos 13 Statistics and Computing 12 Journal of Mathematical Analysis and Applications 12 Linear and Multilinear Algebra 12 Bulletin of Mathematical Biology 12 Journal of Number Theory 12 The Journal of Symbolic Logic 12 Operations Research 12 Physica D 12 Journal of the European Mathematical Society (JEMS) ...and 481 more Journals
all top 5
### Cited in 62 Fields
4,690 Combinatorics (05-XX) 2,692 Probability theory and stochastic processes (60-XX) 1,964 Computer science (68-XX) 687 Statistical mechanics, structure of matter (82-XX) 560 Operations research, mathematical programming (90-XX) 515 Statistics (62-XX) 373 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 309 Numerical analysis (65-XX) 289 Linear and multilinear algebra; matrix theory (15-XX) 288 Information and communication theory, circuits (94-XX) 286 Biology and other natural sciences (92-XX) 243 Number theory (11-XX) 187 Convex and discrete geometry (52-XX) 141 Mathematical logic and foundations (03-XX) 127 Group theory and generalizations (20-XX) 116 Dynamical systems and ergodic theory (37-XX) 88 Order, lattices, ordered algebraic structures (06-XX) 87 Functional analysis (46-XX) 78 Measure and integration (28-XX) 78 Quantum theory (81-XX) 56 Algebraic topology (55-XX) 41 Manifolds and cell complexes (57-XX) 41 Systems theory; control (93-XX) 38 Geometry (51-XX) 35 Partial differential equations (35-XX) 29 Operator theory (47-XX) 29 Differential geometry (53-XX) 25 Ordinary differential equations (34-XX) 24 Approximations and expansions (41-XX) 22 Calculus of variations and optimal control; optimization (49-XX) 20 Harmonic analysis on Euclidean spaces (42-XX) 20 Global analysis, analysis on manifolds (58-XX) 19 Real functions (26-XX) 17 Algebraic geometry (14-XX) 17 Special functions (33-XX) 16 Difference and functional equations (39-XX) 16 General topology (54-XX) 15 General and overarching topics; collections (00-XX) 12 Commutative algebra (13-XX) 12 Topological groups, Lie groups (22-XX) 12 Functions of a complex variable (30-XX) 11 History and biography (01-XX) 11 Abstract harmonic analysis (43-XX) 10 Fluid mechanics (76-XX) 9 Field theory and polynomials (12-XX) 9 Mechanics of deformable solids (74-XX) 8 Relativity and gravitational theory (83-XX) 7 Potential theory (31-XX) 5 Associative rings and algebras (16-XX) 5 Sequences, series, summability (40-XX) 5 Integral transforms, operational calculus (44-XX) 5 Geophysics (86-XX) 4 General algebraic systems (08-XX) 4 Integral equations (45-XX) 4 Mechanics of particles and systems (70-XX) 3 Nonassociative rings and algebras (17-XX) 3 Category theory; homological algebra (18-XX) 2 Several complex variables and analytic spaces (32-XX) 1 Optics, electromagnetic theory (78-XX) 1 Classical thermodynamics, heat transfer (80-XX) 1 Astronomy and astrophysics (85-XX) 1 Mathematics education (97-XX) | 2022-10-05 14:38:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5285788178443909, "perplexity": 7673.001374402543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337631.84/warc/CC-MAIN-20221005140739-20221005170739-00127.warc.gz"} |
https://www.cuemath.com/ncert-solutions/q-1-exercise-2-5-linear-equations-in-one-variable-class-8-maths/ | In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!
# Ex.2.5 Q1 Linear Equations in One Variable Solution - NCERT Maths Class 8
Go back to 'Ex.2.5'
## Question
Solve the linear equation \begin{align}\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\end{align}
Video Solution
Linear Equations
Ex 2.5 | Question 1
## Text Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.
Steps:
\begin{align}\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\end{align}
LCM of the denominators, $$2, 3, 4,$$ and $$5$$, is $$60.$$
Multiplying both sides by $$60,$$ we obtain
\begin{align}60\left( {\frac{x}{2} - \frac{1}{5}} \right) = 60\left( {\frac{x}{3} + \frac{1}{4}} \right)\end{align}
Opening the brackets, we get,
\begin{align}30x - 12 &= 20x + 15 \\30x - 20x &= 15 + 12\\10x &= 27\quad \quad \\\,\,\,\,\,x &= \frac{{27}}{{10}}\end{align}
Learn from the best math teachers and top your exams
• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school | 2020-04-08 09:36:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 8, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9983733296394348, "perplexity": 5303.934475045107}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371810807.81/warc/CC-MAIN-20200408072713-20200408103213-00070.warc.gz"} |
https://tutorialspoint.com/converting_fractions_to_decimals/converting_fraction_with_denominator_10_100_decimal_online_quiz.htm | # Converting a Fraction With a Denominator of 10 or 100 to a Decimal Online Quiz
#### Complete Python Prime Pack
9 Courses 2 eBooks
#### Artificial Intelligence & Machine Learning Prime Pack
6 Courses 1 eBooks
#### Java Prime Pack
9 Courses 2 eBooks
Following quiz provides Multiple Choice Questions (MCQs) related to Converting a Fraction With a Denominator of 10 or 100 to a Decimal. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Write $\mathbf {\frac{18}{10}}$ as a decimal.
### Explanation
Step 1:
We rewrite 18 as 18.0
Step 2:
We shift the decimal one place to the left in 18.0 as there is only one zero in 10.
Step 3:
So, $\frac{18}{10} = 1.8$
Q 2 - Write $\mathbf {\frac{49}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 49 as 49.0
Step 2:
We shift the decimal two places to the left in 49.0 as there are two zeros in 100.
Step 3:
So, $\frac{49}{100} = 0.49$
Q 3 - Write $\mathbf {\frac{29}{10}}$ as a decimal.
### Explanation
Step 1:
We rewrite 29 as 29.0
Step 2:
We shift the decimal one place to the left in 29.0 as there is only one zero in 10.
Step 3:
So, $\frac{29}{10} = 2.9$
Q 4 - Write $\mathbf {\frac{78}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 78 as 78.0
Step 2:
We shift the decimal two places to the left in 78.0 as there are two zeros in 100.
Step 3:
So, $\frac{78}{100} = 0.78$
Q 5 - Write $\mathbf {\frac{46}{10}}$ as a decimal.
### Explanation
Step 1:
We rewrite 46 as 46.0
Step 2:
We shift the decimal one place to the left in 46.0 as there is only one zero in 10.
Step 3:
So, $\frac{46}{10} = 4.6$
Q 6 - Write $\mathbf {\frac{543}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 543 as 543.0
Step 2:
We shift the decimal two places to the left in 543.0 as there are two zeros in 100.
Step 3:
So, $\frac{543}{100} = 5.43$
Q 7 - Write $\mathbf {\frac{124}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 124 as 124.0
Step 2:
We shift the decimal two places to the left in 124.0 as there are two zeros in 100.
Step 3:
So, $\frac{124}{100} = 1.24$
Q 8 - Write $\mathbf {\frac{237}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 237 as 237.0
Step 2:
We shift the decimal two places to the left in 237.0 as there are two zeros in 100.
Step 3:
So, $\frac{237}{100} = 2.37$
Q 9 - Write $\mathbf {\frac{99}{10}}$ as a decimal.
### Explanation
Step 1:
We rewrite 99 as 99.0
Step 2:
We shift the decimal one place to the left in 99.0 as there is only one zero in 10.
Step 3:
So, $\frac{99}{10} = 9.9$
Q 10 - Write $\mathbf {\frac{161}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 161 as 161.0
Step 2:
We shift the decimal two places to the left in 161.0 as there are two zeros in 100.
Step 3:
So, $\frac{161}{100} = 1.61$
converting_fraction_with_denominator_10_100_decimal.htm | 2022-09-29 20:50:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6670805811882019, "perplexity": 3057.941467672868}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335365.63/warc/CC-MAIN-20220929194230-20220929224230-00263.warc.gz"} |
https://www.projecteuclid.org/euclid.jam/1394808069 | ## Journal of Applied Mathematics
### A Phantom-Node Method with Edge-Based Strain Smoothing for Linear Elastic Fracture Mechanics
#### Abstract
This paper presents a novel numerical procedure based on the combination of an edge-based smoothed finite element (ES-FEM) with a phantom-node method for 2D linear elastic fracture mechanics. In the standard phantom-node method, the cracks are formulated by adding phantom nodes, and the cracked element is replaced by two new superimposed elements. This approach is quite simple to implement into existing explicit finite element programs. The shape functions associated with discontinuous elements are similar to those of the standard finite elements, which leads to certain simplification with implementing in the existing codes. The phantom-node method allows modeling discontinuities at an arbitrary location in the mesh. The ES-FEM model owns a close-to-exact stiffness that is much softer than lower-order finite element methods (FEM). Taking advantage of both the ES-FEM and the phantom-node method, we introduce an edge-based strain smoothing technique for the phantom-node method. Numerical results show that the proposed method achieves high accuracy compared with the extended finite element method (XFEM) and other reference solutions.
#### Article information
Source
J. Appl. Math., Volume 2013 (2013), Article ID 978026, 12 pages.
Dates
First available in Project Euclid: 14 March 2014
https://projecteuclid.org/euclid.jam/1394808069
Digital Object Identifier
doi:10.1155/2013/978026
Mathematical Reviews number (MathSciNet)
MR3082126
Zentralblatt MATH identifier
1271.74418
#### Citation
Vu-Bac, N.; Nguyen-Xuan, H.; Chen, L.; Lee, C. K.; Zi, G.; Zhuang, X.; Liu, G. R.; Rabczuk, T. A Phantom-Node Method with Edge-Based Strain Smoothing for Linear Elastic Fracture Mechanics. J. Appl. Math. 2013 (2013), Article ID 978026, 12 pages. doi:10.1155/2013/978026. https://projecteuclid.org/euclid.jam/1394808069
#### References
• T. Belytschko and T. Black, “Elastic crack growth in finite elements with minimal remeshing,” International Journal for Numerical Methods in Engineering, vol. 45, no. 5, pp. 601–620, 1999.
• A. Hansbo and P. Hansbo, “A finite element method for the simulation of strong and weak discontinuities in solid mechanics,” Computer Methods in Applied Mechanics and Engineering, vol. 193, no. 33–35, pp. 3523–3540, 2004.
• J. Mergheim, E. Kuhl, and P. Steinmann, “A finite element method for the computational modelling of cohesive cracks,” International Journal for Numerical Methods in Engineering, vol. 63, no. 2, pp. 276–289, 2005.
• J.-H. Song, P. M. A. Areias, and T. Belytschko, “A method for dynamic crack and shear band propagation with phantom nodes,” International Journal for Numerical Methods in Engineering, vol. 67, no. 6, pp. 868–893, 2006.
• P. M. A. Areias and T. Belytschko, “A comment on the article “A finite element method for simulation of strong and weak discontinuities in solid mechanics”,” Computer Methods in Applied Mechanics and Engineering, vol. 195, no. 9–12, pp. 1275–1276, 2006.
• J. Shi, D. Chopp, J. Lua, N. Sukumar, and T. Belytschko, “Abaqus implementation of extended finite element method using a level set representation for three-dimensional fatigue crack growth and life predictions,” Engineering Fracture Mechanics, vol. 77, no. 14, pp. 2840–2863, 2010.
• M. Duflot, “Industrial applications of XFEM for 3D crack propagation with Morfeo/Crack and Abaqus,” in ECCOMAS Thematic Conference on XFEM, Cardiff, UK, June 2011.
• T. Menouillard, J. Réthoré, A. Combescure, and H. Bung, “Efficient explicit time stepping for the eXtended Finite Element Method (X-FEM),” International Journal for Numerical Methods in Engineering, vol. 68, no. 9, pp. 911–939, 2006.
• T. Menouillard, J. Réthoré, N. Moës, A. Combescure, and H. Bung, “Mass lumping strategies for X-FEM explicit dynamics: application to crack propagation,” International Journal for Numerical Methods in Engineering, vol. 74, no. 3, pp. 447–474, 2008.
• T. Chau-Dinh, G. Zi, P.-S. Lee, T. Rabczuk, and J.-H. Song, “Phantom-node method for shell models with arbitrary cracks,” Computers and Structures, vol. 92-93, pp. 242–246, 2012.
• P. Laborde, J. Pommier, Y. Renard, and M. Salaün, “High-order extended finite element method for cracked domains,” International Journal for Numerical Methods in Engineering, vol. 64, no. 3, pp. 354–381, 2005.
• E. Béchet, H. Minnebo, N. Moës, and B. Burgardt, “Improved implementation and robustness study of the X-FEM for stress analysis around cracks,” International Journal for Numerical Methods in Engineering, vol. 64, no. 8, pp. 1033–1056, 2005.
• G. Ventura, R. Gracie, and T. Belytschko, “Fast integration and weight function blending in the extended finite element method,” International Journal for Numerical Methods in Engineering, vol. 77, no. 1, pp. 1–29, 2009.
• R. Gracie, H. Wang, and T. Belytschko, “Blending in the extended finite element method by discontinuous Galerkin and assumed strain methods,” International Journal for Numerical Methods in Engineering, vol. 74, no. 11, pp. 1645–1669, 2008.
• S. P. A. Bordas, T. Rabczuk, N.-X. Hung et al., “Strain smoothing in FEM and XFEM,” Computers and Structures, vol. 88, no. 23-24, pp. 1419–1443, 2010.
• T. Rabczuk, G. Zi, A. Gerstenberger, and W. A. Wall, “A new crack tip element for the phantom-node method with arbitrary cohesive cracks,” International Journal for Numerical Methods in Engineering, vol. 75, no. 5, pp. 577–599, 2008.
• G. R. Liu, K. Y. Dai, and T. T. Nguyen, “A smoothed finite element method for mechanics problems,” Computational Mechanics, vol. 39, no. 6, pp. 859–877, 2007.
• G. R. Liu, T. T. Nguyen, K. Y. Dai, and K. Y. Lam, “Theoretical aspects of the smoothed finite element method (SFEM),” International Journal for Numerical Methods in Engineering, vol. 71, no. 8, pp. 902–930, 2007.
• N. Nguyen-Thanh, T. Rabczuk, H. Nguyen-Xuan, and S. P. A. Bordas, “A smoothed finite element method for shell analysis,” Computer Methods in Applied Mechanics and Engineering, vol. 198, no. 2, pp. 165–177, 2008.
• S. P. A. Bordas and S. Natarajan, “On the approximation in the smoothed finite element method (SFEM),” International Journal for Numerical Methods in Engineering, vol. 81, no. 5, pp. 660–670, 2010.
• T. Rabczuk, P. M. A. Areias, and T. Belytschko, “A meshfree thin shell method for non-linear dynamic fracture,” International Journal for Numerical Methods in Engineering, vol. 72, no. 5, pp. 524–548, 2007.
• T. Rabczuk and T. Belytschko, “Application of particle methods to static fracture of reinforced concrete structures,” International Journal of Fracture, vol. 137, no. 1-4, pp. 19–49, 2006.
• T. Rabczuk and T. Belytschko, “A three-dimensional large deformation meshfree method for arbitrary evolving cracks,” Computer Methods in Applied Mechanics and Engineering, vol. 196, no. 29-30, pp. 2777–2799, 2007.
• S. Bordas, P. V. Nguyen, C. Dunant, A. Guidoum, and H. Nguyen-Dang, “An extended finite element library,” International Journal for Numerical Methods in Engineering, vol. 71, no. 6, pp. 703–732, 2007.
• G. R. Liu, T. Nguyen-Thoi, and K. Y. Lam, “An edge-based smoothed finite element method (ES-FEM) for static, free and forced vibration analyses of solids,” Journal of Sound and Vibration, vol. 320, no. 4-5, pp. 1100–1130, 2009.
• G. R. Liu, “A G space theory and a weakened weak (W$^{2}$) form for a unified formulation of compatible and incompatible methods: part I theory,” International Journal for Numerical Methods in Engineering, vol. 81, no. 9, pp. 1093–1126, 2010.
• L. Chen, T. Rabczuk, S. P. A. Bordas, G. R. Liu, K. Y. Zeng, and P. Kerfriden, “Extended finite element method with edge-based strain smoothing (ESm-XFEM) for linear elastic crack growth,” Computer Methods in Applied Mechanics and Engineering, vol. 209–212, pp. 250–265, 2012.
• N. Vu-Bac, H. Nguyen-Xuan, L. Chen et al., “A node-based smoothed extended finite element method (NS-XFEM) for fracture analysis,” CMES: Computer Modeling in Engineering and Sciences, vol. 73, no. 4, pp. 331–355, 2011.
• F. Z. Li, C. F. Shih, and A. Needleman, “A comparison of methods for calculating energy release rates,” Engineering Fracture Mechanics, vol. 21, no. 2, pp. 405–421, 1985.
• B. Moran and C. F. Shih, “Crack tip and associated domain integrals from momentum and energy balance,” Engineering Fracture Mechanics, vol. 27, no. 6, pp. 615–642, 1987.
• S. S. Wang, J. F. Yau, and H. T. Corten, “A mixed-mode crack analysis of rectilinear anisotropic solids using conservation laws of elasticity,” International Journal of Fracture, vol. 16, no. 3, pp. 247–259, 1980.
• F. Erdogan and G. Sih, “On the crack extension in sheets under plane loading and transverse shear,” Journal Basic Engineering, vol. 85, no. 6, pp. 519–527, 1963.
• A. Menk and S. Bordas, “Crack growth calculations in solder joints based on microstructural phenomena with X-FEM,” Computational Materials Science, vol. 50, no. 3, pp. 1145–1156, 2011.
• D. F. Li, C. F. Li, S. Q. Shu, Z. X. Wang, and J. Lu, “A fast and accurate analysis of the interacting cracks in linear elastic solids,” International Journal of Fracture, vol. 151, pp. 169–185, 2008. | 2019-11-16 02:45:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2485668808221817, "perplexity": 3552.739439544619}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668716.69/warc/CC-MAIN-20191116005339-20191116033339-00064.warc.gz"} |
https://mathzsolution.com/lplp-and-lqlq-space-inclusion/ | # L^pL^p and L^qL^q space inclusion
Let $(X, \mathcal B, m)$ be a measure space. For $1 \leq p < q \leq \infty$, under what condition is it true that $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)$ and what is a counterexample in the case the condition is not satisfied?
Theorem Let $$XX$$ be a finite measure space. Then, for any $$1\leq p< q\leq +\infty1\leq p< q\leq +\infty$$ $$L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).$$ The proof follows from Hölder inequality. Note that $$\frac{1}{p}=\frac{1}{q}+\frac{1}{r}\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$$, with $$r>0r>0$$. Hence $$\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.$$
The case reported on the Wikipedia link of commenter answer follows from this, since of course, if $$XX$$ does not contain sets of arbitrary large measure, $$XX$$ itself can't have an arbitrary large measure.
For the counterexample: $$f(x)=\frac{1}{x}f(x)=\frac{1}{x}$$ belongs to $$L^2([1,+\infty))L^2([1,+\infty))$$, but clearly it does not belong to $$L^1([1,+\infty)).L^1([1,+\infty)).$$
I would like to add other lines to this interesting topic. Namely, I would like to prove what is mentioned in Wikipedia, hope it is correct:
Theorem Suppose $$(X,\mathcal B,m)(X,\mathcal B,m)$$ is a measure space such that, for any $$1\leq p $$L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).$$ Then $$XX$$ doesn't contain sets of arbitrarily large measure.
Indeed it is well defined the embedding operator $$G:L^q(X, \mathcal B, m) \to L^p(X, \mathcal B, m)G:L^q(X, \mathcal B, m) \to L^p(X, \mathcal B, m)$$, and it is bounded.
Indeed the inclusion $$L^q(X,\mathcal B,m)\subset L^p(X,\mathcal B,m)L^q(X,\mathcal B,m)\subset L^p(X,\mathcal B,m)$$ is continuous. Convergence in $$L^pL^p$$ and in $$L^qL^q$$ imply convergence almost everywhere and we can conclude by the closed graph theorem.
By Hölder inequality, $$\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.$$ This means $$\|G\|\leq \text{meas}(X)^{1/r}=\text{meas}(X)^{1/p-1/q}.\|G\|\leq \text{meas}(X)^{1/r}=\text{meas}(X)^{1/p-1/q}.$$ But, considering $$f(x)=\chi_X(x)f(x)=\chi_X(x)$$, one sees that $$\|G\|=\text{meas }(X)^{1/r}<+\infty \Leftrightarrow \text{meas }(X)<+\infty.\|G\|=\text{meas }(X)^{1/r}<+\infty \Leftrightarrow \text{meas }(X)<+\infty.$$ Now we can proceed by density of the vector space of the simple functions in both $$L^p(X,\mathcal B,m)L^p(X,\mathcal B,m)$$ and $$L^q(X,\mathcal B,m).L^q(X,\mathcal B,m).$$
Theorem Let $$(X,\mathcal B,m)(X,\mathcal B,m)$$ be a measure space. Then $$XX$$ doesn't contain sets of arbitrarily small measure if and only if for any $$1\leq p, one has $$L^p(X, \mathcal B, m) \subset L^q(X, \mathcal B, m).L^p(X, \mathcal B, m) \subset L^q(X, \mathcal B, m).$$
Let us suppose that, for any subset $$Y\subseteq X,\quad Y\in\mathcal BY\subseteq X,\quad Y\in\mathcal B$$, we have $$0<\alpha\leq\text{meas}(Y)0<\alpha\leq\text{meas}(Y)$$.
It sufficies to prove the statement for simple functions. Pick now $$f(x) =\sum_{j=1}^n a_j\chi_{E_j},f(x) =\sum_{j=1}^n a_j\chi_{E_j},$$ where $$\{E_j\}_{j=1,\dots,n}\{E_j\}_{j=1,\dots,n}$$ is a collection of disjoint subsets of $$\mathcal B.\mathcal B.$$ Then $$\|f\|_{L^q} \le \sum_{j=1}^n \|a_j\text{meas}(E_j)\|_{L^q} = \sum_{j=1}^n a_j\text{meas}(E_j)^{1/q}=\sum_{j=1}^n a_j\text{meas}(E_j)^{1/q-1/p+1/p}\leq\frac{1}{\alpha^{1/p-1/q}}\|f\|_{L^p}.\|f\|_{L^q} \le \sum_{j=1}^n \|a_j\text{meas}(E_j)\|_{L^q} = \sum_{j=1}^n a_j\text{meas}(E_j)^{1/q}=\sum_{j=1}^n a_j\text{meas}(E_j)^{1/q-1/p+1/p}\leq\frac{1}{\alpha^{1/p-1/q}}\|f\|_{L^p}.$$
The first inequality is due to Minkowski inequality.
For the converse of the theorem note that again it is well defined the embedding operator $$G:L^p(X,\mathcal B,m)\to L^q(X,\mathcal B,m)G:L^p(X,\mathcal B,m)\to L^q(X,\mathcal B,m)$$, and the operator is bounded. Now consider that, for any subset $$Y\subset XY\subset X$$, $$Y\in\mathcal BY\in\mathcal B$$, the function $$g_Y(x)=\frac{\chi_Y(x)}{(\text{meas(Y)})^{1/p}}g_Y(x)=\frac{\chi_Y(x)}{(\text{meas(Y)})^{1/p}}$$ satisfies $$\|g_Y\|_{L^q}= \frac{1}{(\text{meas}(Y))^{1/p-1/q}}. \|g_Y\|_{L^q}= \frac{1}{(\text{meas}(Y))^{1/p-1/q}}.$$
But then, for any $$Y\subset XY\subset X$$, $$Y\in\mathcal BY\in\mathcal B$$, we have $$\frac{1}{(\text{meas}(Y))^{1/p-1/q}}\leq \|G\|,\frac{1}{(\text{meas}(Y))^{1/p-1/q}}\leq \|G\|,$$ which means $$0<\frac{1}{\|G\|^{1/p-1/q}}\leq \text{meas}(Y).0<\frac{1}{\|G\|^{1/p-1/q}}\leq \text{meas}(Y).$$ Hence the result is proved. | 2022-09-28 03:00:42 | {"extraction_info": {"found_math": true, "script_math_tex": 3, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 44, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9913526773452759, "perplexity": 532.2896287949856}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335059.43/warc/CC-MAIN-20220928020513-20220928050513-00291.warc.gz"} |
https://community.wolfram.com/groups/-/m/t/1606682 | [✓] Get hexadecimal representation for a float as a string?
Posted 5 months ago
709 Views
|
7 Replies
|
7 Total Likes
|
Hello I wonder how I could use Mathematica commands to do the following:An irrational number, ir, (a huge expression as a result of iterating a discrete map) is converted to single precision (Real16) and then to hexadecimal as a string. Example single(ir)=0.25=3e800000 (using matlab).I know single precision is not available on WM 11.3 but Real32, Real64 and Real128 are. As for the hexadecimal representation I have to confess that I could not figure out how to do it even considering, for instance, Real32.Your help is much appreciated.Many thanksEd
7 Replies
Sort By:
Posted 5 months ago
Would BaseForm[number,16] suffice? If not, please give specifics for the hex format required.
Posted 5 months ago
Many thanks. BaseForm[0.25,16]=0.4_16 and what I need as output is 3E800000 (note that the number of digits also counts). Example (matlab)format hex;single(0.25)ans =single3e800000endformat hex;double(0.25)ans =3fd0000000000000As far as I know there is no Real16 (single) available in Mathematica. Could it be implemented?
Posted 5 months ago
I'll start with the assumption that a given nonzero number is a machine double, and we want to show the hex string for either 32 or 64 bit format. We give as parameters the exponent field size and the total number of hex digits for the result. toHex[n_Real, esize_Integer, hlen_Integer] /; Precision[n] === MachinePrecision && n != 0 := Module[ {mant, exp, sgn, bits, hexdigits, rule}, {mant, exp} = RealDigits[n, 2]; mant = Rest[mant]; sgn = If[n > 0, {0}, {1}]; exp = IntegerDigits[exp + 2^(esize - 1) - 2, 2, esize]; bits = PadRight[Flatten[{sgn, exp, mant}], 64]; hexdigits = Take[IntegerDigits[FromDigits[bits, 2], 16, 16], hlen]; rule = Thread[Range[10, 15] -> Characters["abcdef"]]; Apply[StringJoin, Map[ToString, hexdigits /. rule]] ] In the examples, 32 bit floats means 8 hex digits total, and the standard exponent width is 8 bits. toHex[.25, 8, 8] (* Out[174]= "3e800000" *) For 64 bit floats the exponent field is 11 bits. toHex[.25, 11, 16] (* Out[175]= "3fd0000000000000" *) I will add the disclaimer that, unless one is a good speller, it might be unwise even to attempt to hex a decimal...
Posted 5 months ago
Thank you ever so much. Although I believe function toHex is exactly what I need, I wonder whether you could help me further. a) The hex number just below "3e800000" is "3e7fffff" and just above "3e800001", right? (Considering 32-bit floats). b) I also need to go back from hex to decimal taking into account the precision.
Posted 5 months ago
The values you give in (a) both look correct to me.
Not yet fully tested, but this should do the inverse computation. FromHexadecimalString[ss_String] := FromHexadecimalString[ss, {11, MachinePrecision}] FromHexadecimalString[ss_String, {esize_Integer, prec_}] /; esize > 0 && N[prec] > 0 := Module[ {hexdigits = Characters[ss], bits, rule, sgn, exp, mant}, rule = Thread[Characters["abcdef"] -> Range[10, 15]]; hexdigits = Map[ToExpression, Characters[ss] /. rule]; bits = Flatten[Map[IntegerDigits[#, 2, 4] &, hexdigits]]; sgn = If[First[bits] == 0, 1, -1]; {exp, mant} = TakeDrop[Rest[bits], esize]; exp = FromDigits[exp, 2] - 2^(esize - 1) + 2; mant = Total[1/2 + mant. 2^Range[-2, -Length[mant] - 1, -1]]; N[sgn*mant*2^exp, prec] ] /; Complement[Union[Characters[ss]], Join[Characters["abcdef"], Map[ToString, Range[0, 9]]]] === {} Example: FromHexadecimalString["3fd0000000000000"] (* Out[413]= 0.25 *) | 2019-07-24 06:42:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3037296235561371, "perplexity": 7809.94563613179}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195531106.93/warc/CC-MAIN-20190724061728-20190724083728-00384.warc.gz"} |
https://reviews.llvm.org/D101252 | # Fix invalid math formulas in quantization docClosedPublicActions
Authored by Lewuathe on Apr 25 2021, 5:35 AM.
# Details
Reviewers
mehdi_amini
Commits
rG119bf57ab6de: Fix invalid math formulas in quantization doc
Summary
A single backslash is not properly escaped in the web documentation. We can make sure to escape for rendering subscripts.
Additionally, it also fixed the mal-formed equations in "Affine to fixed point" and "Fixed point to affine" sections. With this fix, the page is rendered as follows.
# Diff Detail
### Event Timeline
Lewuathe created this revision.Apr 25 2021, 5:35 AM
Lewuathe requested review of this revision.Apr 25 2021, 5:35 AM
Herald added a project: Restricted Project. Apr 25 2021, 5:35 AM
Lewuathe edited the summary of this revision. (Show Details)Apr 25 2021, 5:39 AM
Lewuathe edited the summary of this revision. (Show Details)
This makes it quite heavy to write, it there something we could change on the website generator side to avoid changing it here?
Lewuathe added a comment.EditedApr 25 2021, 11:44 PM
@mehdi_amini We may be able to avoid rewriting the double dollar mark () by changing the inlineMath config in mlir-www. But I think that contradicts the convention of LaTex in general.
For escaping the backslash, we may be able to use processEscapes config for MathJax. That can reduce the amount of change.
Formatting issue (e.g. L.152-156) can only be resolved by putting begin/end{align*} as far as I know.
For escaping the backslash, we may be able to use processEscapes config for MathJax. That can reduce the amount of change.
Yeah that's the one I was mostly thinking of with my previous comment, wanna send a PR for the MathJax config?
mehdi_amini accepted this revision.Apr 30 2021, 1:15 PM
(otherwise LGTM)
This revision is now accepted and ready to land.Apr 30 2021, 1:15 PM
I tried to use processEscapes option of MathJax. But it did not work as expected. It only handles the literal dollar sign and has no impact to escape the underscore in the math equation, unfortunately.
Looking into the whole configuration, there seems to be no other options we may be able to use for this case.
https://docs.mathjax.org/en/v2.7-latest/options/preprocessors/tex2jax.html
Alright, thanks for trying, feel free to land it as-is then!
Lewuathe added a comment.EditedMay 30 2021, 10:36 PM
@mehdi_amini Thank you for reviewing! I do not have a right to merge this change into the repository. Could you do that on behalf of me?
This revision was automatically updated to reflect the committed changes. | 2022-01-24 07:32:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7182909846305847, "perplexity": 3556.7110219058277}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00290.warc.gz"} |
https://www.physicsforums.com/threads/sum-of-unitary-matrices-question.833631/ | # Sum of Unitary Matrices Question
1. Sep 20, 2015
### RJLiberator
1. The problem statement, all variables and given/known data
Find an example of two unitary matrices that when summed together are not unitary.
2. Relevant equations
3. The attempt at a solution
A = \begin{pmatrix}
0 & -i\\
i & 0\\
\end{pmatrix}
B = \begin{pmatrix}
0 & 1\\
1 & 0\\
\end{pmatrix}
A+B =
A = \begin{pmatrix}
0 & 1-i\\
1+i & 0\\
\end{pmatrix}
So we see that the hermitian conjugate of (A+B) is identical to A+B.
So (A+B)(A+B) =
A = \begin{pmatrix}
2 & 0\\
0 & 2\\
\end{pmatrix}
So since it is a diagonal matrix of 2, this is not the identity matrix. We can safely conclude that while A is unitary, B is unitary, (A+B) is NOT unitary.
Safe understanding?
Thanks
2. Sep 21, 2015
### Dick
That's fine. Even simpler, if $I$ is the identity matrix, then $I$ is unitary, so is $-I$. $I+(-I)=0$. $0$ is not unitary. | 2018-03-20 07:03:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7823899388313293, "perplexity": 704.145560191687}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647299.37/warc/CC-MAIN-20180320052712-20180320072712-00721.warc.gz"} |
https://physics.meta.stackexchange.com/tags/resource-recommentation/hot | # Tag Info
26
See my answer here for the rationale behind this policy Note that this will require some community members to commit to curating the resource recommendations by flagging for post notice addition, flagging+commenting on bad answers, etc. This can't go through unless we get enough people who are willing to patrol for resource recommendation questions and ...
12
Yes, this would be a very good fit for our resource-recommendations tag... if this question didn't already exist: What are good books for graduates/undergraduates in Astrophysics? That question is very general, so you can browse through the list of resources recommended there and pick out the ones that are at an appropriate level. Thus I don't see a need ...
8
Try StackEdit, or PageDown. The latter is the open source engine that SE uses, the former is an editor based on SE's flavor of markdown.
6
I can give you a partial answer: if they are on topic, these questions should be tagged as specific-reference. The difference is that resource-recommendations questions ask for a general class of resources related to a certain topic, where the set of resources that constitute valid answers is large and probably grows over time; on the other hand, specific-...
5
No. Resource recommendation questions are for resource recommendations.
5
If the tag is inappropriate, it should be removed. If the post has already been made community wiki, you can flag it for moderator attention.
4
In short, the reason is that resource recommendation questions aren't allowed to only ask about books. If you look in the body of the second question, it actually asks for "introductory guides". Yes, the title does say "books", and we should probably change that, but it should be understood as asking for any educational resource that presents the specified ...
4
I think they they should be on topic. Googling does not give value judgements and somebody who answers should do that. Value judgements with high scores would be useful for future users looking for similar sources.
3
As mentioned, this is because there is a convention that all resource-recommendation threads should be made community wiki in their entirety, as agreed on this thread. The reasoning is that the voting pattern on resource-recommendation threads is very different to regular threads, since the votes often follow the quality of the book rather than the quality ...
3
Your question is a bit confused but I think that you're mixing two types of questions and two separate policies, one of which did in fact change shortly after the question you referenced. At the time that question was asked, all resource-recommendation questions were considered off topic and were promptly closed, by community agreement. (And, while we're ...
3
I do not think there is any dispute about resource recommendations being on topic. But there is a difference of opinion about what this means. Your interpretation of the policy appears to be that voting to close any resource recommendations questions (RRQs) because they are primarily opinion based (POB) or too broad (TB) is a breach of the policy. ...
3
Good catch. I submitted an edit replacing reference-request with specific-reference.
3
I agree that many of the answers were simply un-detailed lists of a half-dozen books, but it is not the fault of the question nor the question-poser that the answers given were not in line with the rules. True, but the questions weren't locked because of their answers, they were locked because they - the questions themselves - are not good content for a ...
3
As I see it, journal requests are basically equivalent to book recommendations, for a different kind of "book". The official position of SEI is that such recommendation questions are not appropriate for Stack Exchange sites, and I kind of agree, because they can easily get out of hand. Right now, we do allow some book recommendations on the site, which hasn'...
3
Resource-recommendation questions are considered on-topic here $-$ subject to some tight restrictions. Basically, there is a segment of the community that feels (based on extensive experience coming from the early days of Stack Overflow) that open-ended list-based questions are a bad fit for the site, and that they should not be allowed at all. Back in 2013 ...
3
Resource-recommendation questions are very hard to handle, and they were in fact deemed completely off-topic for several years until we managed to work out a consistent policy that lets the good stuff in without wreaking havoc. The result of the policy is at Are resource recommendations allowed? and you should read it carefully (and ideally its previous ...
2
The resource-recommendations tag was originally "my" project, so I'm probably best placed to speak about the original intent of the tag. We have to consider the history of the resource-recommendations tag. It started out as book-recommendations, and that tag was created with a very narrow purpose: to collect recommendations for textbooks treating a certain ...
2
I) Speaking in general, the resource-recommendations tag also covers journals and non-paper electronic materials, such as, e.g., video lectures, on-line resources, software, experimental data files, etc, cf. e.g. its tag wiki. Moreover, from a practical point of view, why have different standards for books, software, etc? E.g. it seems they should all be ...
2
If you want to know if it's still being updated, the edit history will tell you what's been going on there lately. There's some activity there, but really not that much compared to the activity of the tag. As I've advocated before, the overall books question is of rather limited utility when compared to the site's advanced search features. If you're looking ...
2
Software Recommendations Stack Exchange site, recommended by Chair in the comments, has had similar questions as what you've requested: https://softwarerecs.stackexchange.com/q/17669 https://softwarerecs.stackexchange.com/q/42052 https://softwarerecs.stackexchange.com/q/16327 So it seems like a decent bet that it's something you could ask there. I don't ...
2
Concretely, it was made community wiki because OP added the resource-recommendation tag, cf. the policy from this meta post.
2
Phys.SE has a central/primary Book recommendations, although it is currently closed and locked. The reason for the latter is partly because maintenance takes a lot of time. Currently Phys.SE has also more than 1500 secondary res. recom. qs. Online resources are viewed as part of res. recommendations, cf. its tag wiki. However, Phys.SE usually does not ...
2
Apparently I'm one of the people who voted to close this, so I can easily give my opinion on this situation. His question specifically was: "What would you suggest as a good introduction to materials science?" While yours stated: "suggest me some good books(introductory level) to understand the basics of material science." These two questions are ...
1
It's not that your answer was made CW, but that the question was made CW and that automatically converts all answers to CW as well. As Qmechanic says there is a convention that all resource recommendation questions are CW, and if you tag your question as resource-recommendation it will convert it to CW and as a result convert all answers too.
1
Thanks for asking. But in this case, I don't believe there is any way to change your question such that it is not a duplicate, without making it an entirely different question. Resource recommendations are treated somewhat specially, and one of the ways that they are treated specially is that the rules for what constitutes a duplicate are broader than ...
1
The paragraph you refer to applies to answers that were posted before the resource-recommendations policy was set in mid-2013. After the policy was agreed on, there was an extended campaign to find and fix as many old [books] questions as we could, and bring them into line with the new standards. (You can find many of them using this query as a starting ...
1
You can certainly add links into the answer there, but I don't think it's a comprehensive list anymore. The original idea was that only resource recommendation questions of the form "books at [level] about [subject]" would be allowed, where "[subject]" would be one of the big fields like thermodynamics, Newtonian mechanics, quantum field theory, etc. That ...
1
Nope, too specific for this site. As the answer to the recommendation question says, It should be of the form "What are good books to learn/study [subject] at [level]?" Your question is not of that form. What you could do, though, is check the book recommendation questions for the subjects you're asking about and see if anyone has mentioned about ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2021-01-26 09:26:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30078843235969543, "perplexity": 863.7644050939887}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704799711.94/warc/CC-MAIN-20210126073722-20210126103722-00595.warc.gz"} |
https://electronics.stackexchange.com/questions/139799/when-is-a-systems-frequency-response-symmetric | # When is a system's frequency response symmetric?
When is the frequency response of an LTI system symmetric, i.e. which property does it need to satisfy? Does it have to do with the system being causal (I believe this means that h(n) = 0 for n < 0)? If so, why?
Thanks!
## 1 Answer
This property applies to any system whose impulse response is real-valued (as opposed to complex-valued).
For a real-valued function of time $f(x)$, the Fourier transform $F(\omega)$ has the property
$$F(\omega)=F^\star{}(-\omega),$$
where ${}^\star$ indicates the complex conjugate. Since $\|z^\star\|=\|z\|$, the magnitude spectrum of a real-valued function is symmetric.
• It should be $F(\omega)=F^*(-\omega)$ (conjugate symmetry). – Matt L. Jan 11 '15 at 21:13
• @MattL., good catch. Nobody noticed that for 2 months. – The Photon Jan 11 '15 at 21:19 | 2019-07-19 09:02:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8684880137443542, "perplexity": 478.41973471410756}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00374.warc.gz"} |
https://scipost.org/SciPostPhys.6.3.036 | ## Quantum null energy condition and its (non)saturation in 2d CFTs
Christian Ecker, Daniel Grumiller, Wilke van der Schee, Shahin Sheikh-Jabbari, Philipp Stanzer
SciPost Phys. 6, 036 (2019) · published 25 March 2019
### Abstract
We consider the Quantum Null Energy Condition (QNEC) for holographic conformal field theories in two spacetime dimensions (CFT$_2$). We show that QNEC saturates for all states dual to vacuum solutions of AdS$_3$ Einstein gravity, including systems that are far from thermal equilibrium. If the Ryu-Takayanagi surface encounters bulk matter QNEC does not need to be saturated, whereby we give both analytical and numerical examples. In particular, for CFT$_2$ with a global quench dual to AdS$_3$-Vaidya geometries we find a curious half-saturation of QNEC for large entangling regions. We also address order one corrections from quantum backreactions of a scalar field in AdS$_3$ dual to a primary operator of dimension $h$ in a large central charge expansion and explicitly compute both, the backreacted Ryu--Takayanagi surface part and the bulk entanglement contribution to EE and QNEC. At leading order for small entangling regions the contribution from bulk EE exactly cancels the contribution from the back-reacted Ryu-Takayanagi surface, but at higher orders in the size of the region the contributions are almost equal while QNEC is not saturated. For a half-space entangling region we find that QNEC is gapped by $h/4$ in the large $h$ expansion.
### Ontology / Topics
See full Ontology or Topics database.
### Authors / Affiliations: mappings to Contributors and Organizations
See all Organizations.
Funders for the research work leading to this publication | 2022-12-01 06:49:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4455571174621582, "perplexity": 3173.047419878993}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710801.42/warc/CC-MAIN-20221201053355-20221201083355-00717.warc.gz"} |
http://www.koreascience.or.kr/article/JAKO201617559403600.page | ### A NOTE ON SPACES DETERMINED BY CLOSURE-LIKE OPERATORS
DOI QR Code
Hong, Woo Chorl;Kwon, Seonhee
• 투고 : 2016.02.25
• 심사 : 2016.04.21
• 발행 : 2016.05.31
• 3 2
#### 초록
In this paper, we study some classes of spaces determined by closure-like operators $[{\cdot}]_s$, $[{\cdot}]_c$ and $[{\cdot}]_k$ etc. which are wider than the class of $Fr{\acute{e}}chet-Urysohn$ spaces or the class of sequential spaces and related spaces. We first introduce a WADS space which is a generalization of a sequential space. We show that X is a WADS and k-space iff X is sequential and every WADS space is C-closed and obtained that every WADS and countably compact space is sequential as a corollary. We also show that every WAP and countably compact space is countably sequential and obtain that every WACP and countably compact space is sequential as a corollary. And we show that every WAP and weakly k-space is countably sequential and obtain that X is a WACP and weakly k-space iff X is sequential as a corollary.
#### 키워드
sequential;$Fr{\acute{e}}chet-Urysohn$;countable tightness;k-space;AP;WAP;WACP;WADS;countably sequential
#### 참고문헌
1. A. V. Arhangel'skii, A characterization of very k-spaces, Czechoslovak Math. J. 18(93)(1968), 392-395.
2. A. V. Arhangel'skii and L.S.Pontryagin(Eds.), General Topology I, Encyclopaedia of Mathematical Sciences, vol. 17, Springer-Verlage, Berlin, 1990.
3. A. V. Arhangel'skii, Topological function spaces, Mathematics and its application, Vol.78, Kluwer Academic Publishers, 1992.
4. A. V. Arhangel'skii and D. N. Stavrova, On a common generalization of k-spaces and spaces with countable tightness, Top. and its Appl. 51(1993), 261-268. https://doi.org/10.1016/0166-8641(93)90081-N
5. A. Bella, On spaces with the property of weak approximation by points, Comment. Math. Univ. Carolinae 35(2)(1994), 357-360.
6. A. Bella and I. V. Yaschenko, On AP andWAP spaces, Comment. Math. Univ. Carolinae 40(3)(1999), 531-536.
7. A. Dow, M. G. Tkachenko, V. V. Tkachuk and R. G .Wilson, Topologies generated by discrete subspaces, Glansnik Math. 37(57)(2002), 187-210.
8. S. P. Franklin, Spaces in which sequences suffice, Fund. Math. 57(1965), 107-115. https://doi.org/10.4064/fm-57-1-107-115
9. W. C. Hong, Generalized Frechet-Urysohn spaces, J. Korean Math. Soc. 44(2)(2007), 261-273. https://doi.org/10.4134/JKMS.2007.44.2.261
10. W. C. Hong, On spaces in which compact-like sets are closed, and related spaces, Commun. Korean Math. Soc. 22(2)(2007), 297-303. https://doi.org/10.4134/CKMS.2007.22.2.297
11. W. C. Hong, On spaces which have countable tightness and related spaces, Honam Math. J. 34(2)(2012), 199-208. https://doi.org/10.5831/HMJ.2012.34.2.199
12. W. C. Hong and S. Kwon, A generalization of a sequential space and related spaces, Honam Math. J. 36(2)(2014), 425-434. https://doi.org/10.5831/HMJ.2014.36.2.425
13. M. Ismail and P. Nyikos, On spaces in which countably compact sets are closed, and hereditary properties, Top. and its Appl. 11(1980), 281-292. https://doi.org/10.1016/0166-8641(80)90027-9
14. S. Lin and C. Zheng, The k-quotient images of metric spaces, Commun. Korean Math. Soc. 27(2012), 377-384. https://doi.org/10.4134/CKMS.2012.27.2.377
15. M. A. Moon, M. H. Cho and J. Kim, On AP spaces in concern with compact-like sets and submaximality, Comment. Math. Univ. Carolinae 52(2)(2011), 293-302.
16. J. Pelant, M. G. Tkachenko, V. V. Tkachuk and R. G. Wilson, Pseudocompact Whyburn spaces need not be Frechet, Proc. Amer. Math. Soc. 131(2003), no.10,3257-3265. https://doi.org/10.1090/S0002-9939-02-06840-5
17. Y. Tanaka, Necessary and sufficient conditions for products of k-spaces, Top. Proceedings 14(1989), 281-313.
18. V. V. Tkachuk and I. V. Yaschenko, Almost closed sets and topologies they determine, Comment. Math. Univ. Carolinae 42(2)(2001), 395-405.
19. A. Wilansky, Topology for analysis, Ginn and Company 1970.
#### 과제정보
연구 과제 주관 기관 : Pusan National University | 2018-12-10 13:04:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8599189519882202, "perplexity": 4587.6254187372715}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823339.35/warc/CC-MAIN-20181210123246-20181210144746-00203.warc.gz"} |
https://ncatlab.org/spahn/show/HTT%2C+A.2+model+categories | # Spahn HTT, A.2 model categories
This is a subentry of a reading guide to HTT.
# Contents
## A.2.1 The model category axioms
###### Definition
(This is Joyal’s definition; it differs from A.2.1.1 in that Joyal requests $C$ to be finitely bicomplete.)
A model category is a category $C$ equipped with three distinguished classes of $C$-morphisms: The classes $(C)$, $(F)$, $(W)$ of cofibrations, fibrations, and weak equivalences, respectively, satisfying the following axioms:
• $C$ admits (small) limits and colimits.
• The class of weak equivalences satisfies 2-out-of-3.
• $(C\cup W,F)$ and $(C,F\cup W)$ are weak factorization systems.
###### Remark
1. The classes $(C)$ and $(F)$ is closed under retracts. (by weak factorization systems, Lemma 2, in joyal’s catlab)
2. The class $(W)$ is closed under retracts. (by model categories, Lemma 1, in joyal’s catlab)
## A.2.2 The homotopy category of a model category
HTT, A.2.2
###### Definition
Let $X$ be an object in a model category.
1. A cylinder object is defined to be a factorization of the codiagonal map $X\coprod X\to X$ for $X$ into a cofibration followed by a weak equivalence.
2. A path object is defined to be a factorization of the diagonal map $X\to X\times X$ for $X$ into a weak equivalence followed by a fibration .
###### Proposition A.2.2.1
Let $C$ be a model category. Let $X$ be a cofibrant object of $C$. Let $Y$ be a fibrant object of $C$. Let $f,g:X\to Y$ be two parallel morphisms. Then the following conditions are equivalent.
1. The coproduct map $f\coprod g$ factors through every cylinder object for $X$.
2. The coproduct map $f\coprod g$ factors through some cylinder object for $X$.
3. The product map $f\times g$ factors through every path object for $Y$.
4. The product map $f\times g$ factors through some path object for $Y$.
###### Definition
(homotopy, homotopy category of a model category)
Let $C$ be a model category.
(1) Two maps $f,g:X\to Y$ from a cofibrant object to a fibrant object satisfying the conditions of Proposition A.2.2.1 are called homotopic morphisms. Homotopy is an equivalence relation $\simeq$ on $hom_C (X,Y)$.
(2) The homotopy category $h C$ of $C$ is defined to have as objects the fibrant-cofibrant objects of $C$. The hom objects $hom_{hC}(X,Y)$ are defined to be the set of $\simeq$ equivalence classes of $hom_C (X,Y)$.
## A.2.3 A lifting criterion
The following proposition says that a factorization of a cofibration between cofibrant objects which exists in the homotopy category of a model category can be lifted into the model category.
## A.2.4 Left properness and homotopy push out squares
In every model category the class of fibrations is stable under pullback and the class of cofibrations is stable under pushout. In general weak equivalences do not have such properties. The following definition requests such.
###### Definition A.2.4.1
1. A model category is called left proper if the pushout of a weak equivalence along a cofibration is a weak equivalence.
2. A model category is called right proper if the pullback of a weak equivalence along a fibration is a weak equivalence.
###### Proposition
Any model category in which every object is cofibrant is left proper.
###### Lemma A.2.4.3
The push out along a cofibration of a weak equivalence between cofibrant objects is always a weak equivalence.
## A.2.5 Quillen adjunctions and Quillen equivalences
A Quillen adjunction is an appropriate notion of morphism between model categories.
###### Proposition and Definition
An adjoint pair of functors $(F\dashv G):D\stackrel{G}{\to}C$ is called a Quillen adjunction if the following equivalent conditions are satisfied:
1. $F$ preserves cofibrations and acyclic cofibrations.
2. $G$ preserves fibrations and acyclic fibrations.
3. $F$ preserves cofibrations and $G$ preserves fibrations.
4. $F$ preserves acyclic cofibrations and $G$ preserves acyclic fibrations.
###### Remark
Let $(F\dashv G)$ be a Quillen adjunction. Then
1. $F$ preserves weak equivalences between cofibrant objects.
2. $G$ preserves weak equivalences between fibrant objects.
###### Remark
(descent of a Quillen adjunction to an adjunction between the homotopy category, derived functor)
Given a model category $C$ we obtain its homotopy category $hC$ be passing to its full subcategory of cofibrant objects and the formally inverting the weak equivalences.
If $(F\dashv G):D\stackrel{G}{\to}C$ is a Quillen adjunction $F$ induces a functor $L F:hC\to hD$ since $F$ preserves weak equivalences between cofibrant objects.
Analogously $G$ preserves weak equivalences between fibrant objects and we obtain $hD$ from $D$ by passing to the category of fibrant objects of $D$ and formally invert the weak equivalences and hence $G$ induces a functor $RG:hD\to hC$.
In total one can show that $(LF\dashv RG):hD\stackrel{RG}{\to}hC$ form an adjunction.
1. $LF$ is called the (homotopy) left derived functor of $f$.
2. $RG$ is called the (homotopy) right derived functor of $g$.
Abstracty one can obtain this result by Kan extension (this is also described at derived functor); however Quillen adjunction’s are introduced to present adjunctions between $\infty$-categories and to obtain such a presentation in terms of Kan extension in general requires additional assumptions:
In more detail we wish to extend $F : C \to D$ (for $G$ analogously) to a diagram
$\array{ C &\stackrel{F}{\to}& D \\ \downarrow^{\mathrlap{Q_C}} &(?)& \downarrow^{\mathrlap{Q_D}} \\ hC&\to& hD } \,,$
where $Q_C : C \to hC$ is the universal morphism characterizing the homotopy category and similarly for $Q_D$.
This is accomplished by taking $hC\to hD$ to be either the left ($LF:=Lan_{Q_C} Q_d \circ F$) or right ($RF:=Ran_{Q_C} Q_d \circ F$) Kan extension of $Q_d \circ F$ along $Q_C$.
###### Proposition A.2.5.1
(characterization of derived functors, Quilen equivalence) Let $(F\dashv G):D\stackrel{G}{\to}C$ be a Quillen adjunction of model categories. Then the following are equivalent:
1. The left derived functor $LF:hC\to hD$ is an equivalence of categories.
2. The right derived functor $RF:hD\to hC$ is an equivalence of categories.
3. For every cofibrant object $c\in C$ and every fibrant object $D\in D$, a map $c\to G(d)$ is a weak equivalence iff the adjoint map $F(c)\to d$ is a weak equivalence.
$(F\dashv G)$ is called Quillen equivalence if these conditions are satisfied.
## A.2.6 Combinatorial model categories
(transclusion:
###### Definition A.1.2.2
(weakly saturated class of morphisms)
Let $C$ be a category with all small colimits. A class $S$ of $C$-morphisms is called a weakly saturated class if the following conditions are satisfied.
1. $S$ is closed under forming pushouts (along arbitrary $C$-morphisms).
2. $S$ is closed under transfinite composition.
3. $S$ is closed under forming retracts.
)
###### Definition A.2.6.1
A model category $A$ is called combinatorial model category if the following conditions are satisfied:
1. $A$ is presentable.
2. There exists a set $I$ of generating cofibrations such that the collection of all cofibrations is the smallest weakly saturated class of morphisms containing $I$.
3. There exists a set $J$ of generating acyclic cofibrations such that the collection of all acyclic cofibrations is the smallest weakly saturated class of morphisms containing $J$.
###### Definition A.2.6.10
(perfect class) Let $A$ be a presentable category. A class $W$ of morphisms in $C$ is called perfect class if the following conditions are satisfied:
1. $W$ contaings all isomorpphisms.
2. $W$ satisfies 2-out-of-3
3. $W$ is stable under poset filtered colimits.
4. $W$ contains a small subset which generates $W$ under filtered colimits.
###### Proposition A.2.6.13
Let $A$ be a presentable category. Let $W$ be a class of $A$-morphisms called called weak equivalences. Let $A_0$ be a small set of morphisms of $A$ called generating cofibrations satisfying:
(1) $W$ is a perfect class.
(2) For any diagram
$\array{ X&\stackrel{f}{\to}& Y \\ \downarrow&&\downarrow \\ X^{\prime}&\to&Y^{\prime} \\ \downarrow^g&&\downarrow^{g^\prime} \\ X^{\prime\prime}&\to& Y^{\prime\prime} }$
where both sub squares are cocartesian, $f\in A_0$, and $g\in W$, the $g^\prime\in W$.
(3) A morphism in $A$ which has the right lifting property with respect to $A_0$ belongs to $W$.
Then there exists a left proper, combinatorial model structure on $C$ defined by:
(C) A morphism is a cofibration if it belongs to the smallest weakly saturated class of morphisms generated by $A_0$.
(W) A morphism is a weak equivalence if it belongs to $W$.
(F) A morphism is a fibration if it has the right lifting property with respect to the class of acyclic cofibrations.
###### Remark 2.6.14
Let $A$ be a model category. Then $A$ arises via the construction of Proposition A.2.6.13 iff it is left proper, combinatorial and the class of weak equivalences in $A$ is stable under filtered colimits.
## A.2.7 Simplicial sets
###### Definition
The standard model structure on the category $sSet$ of simplicial sets is defined by:
(W) A morphism is a weak equivalence if its geometric realization is a weak homotopy equivalence.
(C) Cofibrations are the monomorphisms.
(F) Fibrations are Kan fibrations.
## A.2.8 Diagram categories and homotopy colimits
(A.2.8.1, A.2.8.2)
If $C$ is a small category and $A$ is a combinatorial model category, then
1. The injective model structure on $Fun (C,A)$ is a combinatorial model structure, determined by the strong cofibrations, weak equivalences, and projective fibrations.
2. The projective model structure on $Fun (C,A)$ is a combinatorial model structure, determined by the weak cofibrations, weak equivalences, and injective fibrations.
If $A$ is moreover right proper resp. left proper, then $Fun(C,A)$ is right proper resp. left proper.
###### Remark A.2.8.6
A Quillen adjunction $(F\dashv G):B\stackrel{G}{\to}A$ induces for every small category $C$ a Quillen adjunction $(F^C\dashv G^C):Fun(C,B)\stackrel{G^C}{\to}Fun(C,A)$ with respect to either the injective- or the projective model structure.
$(F\dashv G)$ is a Quillen equivalence iff $(F^C\dashv G^C)$ is.
In other words: Forming the injective- resp. projective model structure is a functor.
###### Remark
(identity Quillen functor)
By Remark A.2.8.5 every projective cofibration is an injective cofibration and (dually) every injective fibration is a projective fibration. By definition the projective- and injective model structure have the same weak equivalences. It follows that the identity
$(id_{Fun(C,A)}\dashv id_{Fun(C,A)}):Fun(C,A)_{inj}{\to}Fun(C,A)_{proj}$
is a Quillen equivalence between the injective- and the projective model structure.
###### Proposition A.2.8.7
Let $f:C\to C^\prime$ be a functor between small categories. For a combinatorial model category $A$ let $f^*:=(-)\circ f:Fun(C^\prime,A)\to Fun(C,A)$ denote the functor given by precomposition with $f$. By Kan extension we see that there are adjoints
$(f_!\dashv f^*\dashv f_*):Fun(C,A)\stackrel{f_*}{\to} Fun(C^\prime,A)$
and
1. $(f_!\dashv f^*):Fun(C^\prime,A)_{proj}\stackrel{f_*}{\to}Fun(C,A)_{proj}$
2. $(f^*\dashv f_*):Fun(C,A)_{inj}\stackrel{f_*}{\to}Fun(C^\proj,A)_{inj}$
###### Remark A.2.8.8
(transclusion from 5.2.4 Examples of adjoint functors:
###### Definition (in Proof of Corollary 5.2.4.5)
Let $C$, $D$ be fibrant simplicially enriched categories. Let $(F\dashv G):D\stackrel{\G}{\to}C$ be a simplicially enriched adjunction. Let $M$ be the simplicially enriched category defined by
$Map_M(c,d)=Map_C (c,G(d))=Map_D (F(c),d)$
$Map_M (d,c)=\varnothing$
for every $c\in C$, $d\in D$.
$M$ is the correspondence associated to the adjunction $(F\dashv G)$.
###### Proposition 5.2.4.6
(derived functor)
Let $A$, $A^\prime$ be simplicially enriched model categories. Let
$(F\dashv G):A^\prime\stackrel{G}{\to} A$
be a simplicially enriched Quillen adjunction. Let $M$ denote the correspondence associated to the adjunction $(F\dashv G)$. Let $M^\circ$ denote the full subcategory of $M$ consisting of those objects which are fibrant-cofibrant objects (either as objects in $A$ or as objects in $A^\prime$).
Then $N(M^\circ)$ determines an adjunction
$(f\dashv g):N(A^{\prime\circ})\stackrel{g}{\to}N(A^\circ)$
here $f$ is called left derived functor of $F$ and $g$ is called right derived functor of $G$.
On the level of homotopy categories $f$ and $g$ reduce to the usual derived functors associated to the Quillen adjunction, see (homotopy) derived functor.
)
###### Definition
(homotopy right Kan extension)
Let $f:C\to C^\prime$ be a functor between small categories. The right derived functor $Rf_*$ functor of the functor $f_*=Ran_f$ (which is the right adjoint to $f^*:=(-)\circ f:Fun(C^\prime,A)\to Fun(C,A)$) is called homotopy right Kan extension.
###### Definition
(homotopy limit)
Let $1$ denote the terminal category. Let $A$ be a combinatorial model category. Let $a:1\to A$ denote a global element of $A$. Let $C$ be a (note necessarily small) category. Let $!:C\to 1$ denote the unique functor to the terminal category. Let $F:C\to A$ be a functor.
A natural transformation $\alpha:!^* a\to F$ is called a homotopy limit of $F$ if $\alpha$ exhibits $a$ as a homotopy Kan extension of $F$.
Note that $!^* a=\kappa_a$ is the constant functor ‘’in $a$’’.
###### Proposition A.2.8.9
Let $A$ be a combinatorial model category, let $f : C \to D$ be a functor between small categories. Let $F : C \to A$ and $G : D \to A$ be diagrams. A natural transformation $\alpha : f_* G \to F$ exhibits G as a homotopy right Kan extension of $F$ if and only if, for each object $d \in D$, $\alpha$ exhibits $G(d)$ as a homotopy limit of the composite diagram
$d/F:C\times_D d/D\to C\stackrel{F}{\to}A$
The following remark defines homotopy Kan extensions which in particular model Kan extensions between $\infty$-categories:
(…)
###### Remark A.2.8.11
Analog for homotopy colimits.
Last revised on December 1, 2012 at 05:26:55. See the history of this page for a list of all contributions to it. | 2021-09-17 11:07:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 203, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9710556864738464, "perplexity": 337.46052792422273}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00248.warc.gz"} |
https://www.dreamwings.cn/uva11572/3991.html | # UVA 11572：Unique Snowflakes(尺取法)
Emily the entrepreneur has a cool business idea: packaging and selling snowflakes. She has devised amachine that captures snowflakes as they fall, and serializes them into a stream of snowflakes that flow,one by one, into a package. Once the package is full, it is closed and shipped to be sold.The marketing motto for the company is “bags of uniqueness.” To live up to the motto, everysnowflake in a package must be different from the others. Unfortunately, this is easier said than done,because in reality, many of the snowflakes flowing through the machine are identical. Emily would liketo know the size of the largest possible package of unique snowflakes that can be created. The machinecan start filling the package at any time, but once it starts, all snowflakes flowing from the machinemust go into the package until the package is completed and sealed. The package can be completedand sealed before all of the snowflakes have flowed out of the machine.
Input
The first line of input contains one integer specifying the number of test cases to follow. Each testcase begins with a line containing an integer n, the number of snowflakes processed by the machine.The following n lines each contain an integer (in the range 0 to 109, inclusive) uniquely identifying asnowflake. Two snowflakes are identified by the same integer if and only if they are identical.The input will contain no more than one million total snowflakes.
Output
For each test case output a line containing single integer, the maximum number of unique snowflakesthat can be in a package.
Sample Input
1
5
1
2
3
2
1
Sample Output
3
AC代码：
#include<iostream>
#include<algorithm>
#include<set>
using namespace std;
int a[1000005];
int main()
{
int cases,n;
cin >> cases;
while(cases--)
{
set<int> s;
cin>> n;
for(int i=0; i<n; i++)
cin >> a[i];
int left=0,right=0,ans=0;
while(right<n)
{
while(right < n && !s.count(a[right]))
s.insert(a[right++]);
ans = max(ans, right-left);
s.erase(a[left++]);
}
cout <<ans<< endl;
}
return 0;
} | 2020-10-20 06:22:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3303609788417816, "perplexity": 2608.2432653017468}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107869933.16/warc/CC-MAIN-20201020050920-20201020080920-00190.warc.gz"} |
https://socratic.org/questions/what-does-arcsin-sin-pi-2-equal | # What does arcsin(sin ((-pi)/2)) equal?
Jan 20, 2016
$- \frac{\pi}{2}$
#### Explanation:
If you want to know how please follow.
$\sin \left(- \frac{\pi}{2}\right) = - \sin \left(\frac{\pi}{2}\right)$ since $\sin \left(x\right)$ is a odd function.
$\sin \left(- \frac{\pi}{2}\right) = - 1$ since $\sin \left(\frac{\pi}{2}\right) = 1$
$\arcsin \left(\sin \left(- \frac{\pi}{2}\right)\right) = \arcsin \left(- 1\right)$
Now comes the range of $\arcsin \left(x\right)$
The range of $\arcsin \left(x\right)$ is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$
So $\arcsin \left(- 1\right)$ would give $- \frac{\pi}{2}$
Therefore,
$\arcsin \left(\sin \left(- \frac{\pi}{2}\right)\right) = - \frac{\pi}{2}$ | 2020-02-22 16:19:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7603006958961487, "perplexity": 2165.079828852876}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00275.warc.gz"} |
https://proofwiki.org/wiki/Ring_of_Integers_is_Principal_Ideal_Domain/Proof_3 | Ring of Integers is Principal Ideal Domain/Proof 3
Theorem
The integers $\Z$ form a principal ideal domain.
Proof
Let $U$ be an arbitrary ideal of $\Z$.
Let $c$ be a non-zero element of $U$.
Then both $c$ and $-c$ belong to $\ideal a$ and one of them is positive.
Thus $U$ contains strictly positive elements.
Let $b$ be the smallest strictly positive element of $U$.
By the Set of Integers Bounded Below by Integer has Smallest Element, $b$ is guaranteed to exist.
If $\ideal b$ denotes the ideal generated by $b$, then $\ideal b \subseteq U$ because $b\in U$ and $U$ is an ideal.
Let $a \in U$.
By the Division Theorem:
$\exists q, r \in \Z, 0 \le r < b: a = b q + r$
As $a, b \in U$ it follows that so does $r = a - b q$.
By definition of $b$ it follows that $r = 0$.
Thus:
$a = b q \in \ideal b$
and so:
$U \subseteq \ideal b$
From the above:
$U = \ideal b$
It follows by definition that $U$ is a principal ideal of $\Z$.
Recall that $U$ was an arbitrary ideal of $\Z$.
Hence by definition $\Z$ is a principal ideal domain.
$\blacksquare$ | 2021-08-03 16:55:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9687843322753906, "perplexity": 83.90250814491894}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00326.warc.gz"} |
https://bodheeprep.com/cat-quant-practice-problems/3 | CAT Quant Practice Problems
Question: Ramu went by car from Calcutta to Trivandrum via Madras, without any stoppages. The average speeds for the entire journey was 40 kmph. What was the average speed from Madras to Trivandrum?
I. The distance from Madras to Trivandrum is 0.30 times the distance from Calcutta to Madras.
II. The average speed from Madras to Trivandrum was twice that of the average speed from Calcutta to Madras.
1. If the question can be answered with the help of statement I alone,
2. If the question can be answered with the help of statement II alone,
3. If both, statement I and statement II are needed to answer the question, and
4. If the statement cannot be answered even with the help of both the statements.
Correct Option:3
1. If only 1st statment taken into consideration:
$y=0.30x$ (Where y= distance from Madras to Trivandrum and x= distance from Calcutta to Madras)
Nothing absolute can be said about average speed from Madras to Trivendrum.
2. If only 2nd statement taken into consideration:
$\frac{y}{t_y} = 2\frac{x}{t_x}$ ( Where tx and ty are time taken for calcutta to madras and madras to trivendrum respectively.)
And nothing absolute can be said about average speed from Madras to Trivendrum.
3. Now when both statements taken into consideration:
We will have $y=0.30x$; $\frac{y}{t_y} = 2\frac{x}{t_x}$ and $\frac{x+y}{t_x+t_y}$ = 40 (Given)
After solving above three equations we can find a unique value for average speed from Madras to Trivendrum i.e.
$\frac{y}{t_y} = \frac{920}{13}$ | 2019-06-18 04:44:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.617296576499939, "perplexity": 2725.008244290273}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998607.18/warc/CC-MAIN-20190618043259-20190618065259-00139.warc.gz"} |
https://crypto.stackexchange.com/questions/54556/single-pass-authentication-with-a-block-cipher-that-accepts-a-counter | # Single-pass authentication with a block cipher that accepts a counter
Suppose we have a tweakable block cipher ($E(K, T, X)$ and $D(K, T, X)$) that is used in a scheme where the tweak, either in whole or in part, is used as a counter to make each block encrypt to a distinct ciphertext. If we XOR all of the blocks together and then encrypt the result, do we have a secure tag?
It is not secure if you use the $E(K, T, X)$ values as the exposed ciphertext; it is secure if you're using this scheme as a MAC (and use an independent scheme to encrypt).
If you send the $C_i = E(K, T_i, X_i)$ as the ciphertext (and send $E(K, 0, \bigoplus C_i)$ as the tag, then what the attacker could do is flip the same bit on two different $C_i$ values. The resulting $\bigoplus C_i$ will remain the same, and so the tag would validate (and the decrypted plaintext would be incorrect).
On the other hand, if you keep all the $C_i$ values internal, then this system can be viewed as a Carter-Wegman MAC. All you need to show is that $\bigoplus C_i$ is an almost universal hash function (which is fairly easy to do, assuming you get the message padding right), and the standard CW proofs apply.
• @Melab: no, there wouldn't be any particular length limitations (at least until you hit the birthday bound); the CW argument above shows that $\bigoplus C_i$ remains universal up until then (and it probably is beyond that as well, i don't have an immediate proof of that) – poncho Jan 8 '18 at 21:00 | 2021-02-26 18:16:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6264671683311462, "perplexity": 486.63003387734557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178357935.29/warc/CC-MAIN-20210226175238-20210226205238-00333.warc.gz"} |
http://exxamm.com/blog/Blog/13358/zxcfghfgvbnm4?Class%2011 | Chemistry Surface Tension and Viscosity
Topics Covered :
● Surface Tension
● Viscosity
Surface Tension :
=> We know the fact that liquids assume the shape of the container. But
● Why is it then small drops of mercury form spherical bead instead of spreading on the surface?
● Why do particles of soil at the bottom of river remain separated but they stick together when taken out?
● Why does a liquid rise (or fall) in a thin capillary as soon as the capillary touches the surface of the liquid?
=> All these phenomena are caused due to the characteristic property of liquids, called text(surface tension).
color{purple}(✓✓)color{purple} " DEFINITION ALERT"
Surface tension of a liquid is defined as the force acting at right angles to the surface along one centimetre length of the surface. Thus, the units of surface tension are dynes per cm.
● A molecule in the bulk of liquid experiences equal intermolecular forces from all sides. The molecule, therefore does not experience any net force.
● But for the molecule on the surface of liquid, net attractive force is towards the interior of the liquid, due to the molecules below it. Since there are no molecules above it.
=> Liquids tend to minimize their surface area.
● The molecules on the surface experience a net downward force and have more energy than the molecules in the bulk, which do not experience any net force.
● Therefore, liquids tend to have minimum number of molecules at their surface.
● If surface of the liquid is increased by pulling a molecule from the bulk, attractive forces will have to be overcome. This will require expenditure of energy.
text(Surface Energy :) The energy required to increase the surface area of the liquid by one unit is defined as text(surface energy).
● Its dimensions are J m^(–2).
text(Surface Tension :) It is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid.
● It is denoted by Greek letter γ (Gamma).
● It has dimensions of kg s^(-2) and in SI unit it is expressed as N m^(–1).
=> The lowest energy state of the liquid will be when surface area is minimum.
● Spherical shape satisfies this condition, that is why mercury drops are spherical in shape.
● This is the reason that sharp glass edges are heated for making them smooth. On heating, the glass melts and the surface of the liquid tends to take the rounded shape at the edges, which makes the edges smooth. This is called text(fire polishing of glass).
● Liquid tends to rise (or fall) in the capillary because of surface tension. Liquids wet the things because they spread across their surfaces as thin film.
● Moist soil grains are pulled together because surface area of thin film of water is reduced.
● It is surface tension which gives stretching property to the surface of a liquid.
● On flat surface, droplets are slightly flattened by the effect of gravity; but in the gravity free environments drops are perfectly spherical.
=> The magnitude of surface tension of a liquid depends on the attractive forces between the molecules.
● When the attractive forces are large, the surface tension is large.
● Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised.
Viscosity :
=> It is one of the characteristic properties of liquids.
color{purple}(✓✓)color{purple} " DEFINITION ALERT"
text(Viscosity :) It is a measure of resistance to flow which arises due to the internal friction between layers of fluid as they slip past one another while liquid flows.
● Strong intermolecular forces between molecules hold them together and resist movement of layers past one another.
=> When a liquid flows over a fixed surface, the layer of molecules in the immediate contact of surface is stationary.
color{purple}(✓✓)color{purple} " DEFINITION ALERT"
text(Laminar Flow :) The velocity of upper layers increases as the distance of layers from the fixed layer increases. This type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called laminar flow.
● If we choose any layer in the flowing liquid (Fig.5.14), the layer above it accelerates its flow and the layer below this retards its flow.
● If the velocity of the layer at a distance dz is changed by a value du then velocity gradient is given by the amount (du)/(dz). A force is required to maintain the flow of layers. This force is proportional to the area of contact of layers and velocity gradient i.e.
f prop A (A is the area of contact)
f prop (du)/(dz) (where (du)/(dz) is velocity gradient; the change in velocity with distance)
f prop A . (du)/(dz)
=> f = eta A (du)/(dz)
eta is proportionality constant and is called text(coefficient of viscosity).
text(Viscosity Coefficient :) It is the force when velocity gradient is unity and the area of contact is unit area.
● Thus ‘ η ’ is measure of viscosity.
● SI unit of viscosity coefficient is 1 newton second per square metre (N s m^(–2)) = pascal second (Pa \ \ s = 1kg m^(–1)s^(–1)).
● In cgs system the unit of coefficient of viscosity is poise (named after great scientist Jean Louise Poiseuille).
1 poise = 1 g cm^(–1)s^(–1) = 10^(–1)kg m^(–1)s^(–1)
=> Greater the viscosity, the more slowly the liquid flows.
● Hydrogen bonding and van der Waals forces are strong enough to cause high viscosity.
● Glass is an extremely viscous liquid. It is so viscous that many of its properties resemble solids. However, property of flow of glass can be experienced by measuring the thickness of windowpanes of old buildings. These become thicker at the bottom than at
the top.
=> Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers. | 2018-10-18 20:20:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6647199988365173, "perplexity": 893.849607003336}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512014.61/warc/CC-MAIN-20181018194005-20181018215505-00190.warc.gz"} |
https://chat.stackexchange.com/transcript/71/2017/5/15/20-24 | 8:02 PM
@AccidentalFourierTransform I wish you'd comment this
It's a perfect representation of the problem
I have an excellent physics puzzle.
Suppose I you have a cone. Over the top of the cone you place a loop of frictionless, massless string. At one point on the string, you attach a mass.
At what opening angle of the cone will the mass cause the loop of string to be pulled up over the top of the cone and fall off?
The general relativity people should have an advantage on this one.
@JohnRennie
@SirCumference we are wasting our time
@AccidentalFourierTransform Yeah, I've given up
@AccidentalFourierTransform @SirCumference what's up?
0
When someone deletes an account, the upvotes they left on comments are not removed. When they create another account, they can upvote these comments once again. This is a problem, as you can give a comment as many upvotes as you want by: Upvoting the comment Deleting your account Repeating 1 a...
I dont get people
8:11 PM
@0celouvskyopoulo7 I'm not sure I follow your argument or where I'm supposed to help :D
@DanielSank I think I don't understand what situation you're describing
That is, in no scenario I picture after that description can I imagine the loop of string being pulled over at all, so I must be picturing it wrongly
You'll see it then
^^^
I thought that I got the picture, but the answer was obviously "the string is not gonna be pulled over at any angle"
Suppose the cone is very, very shallow.
E.g. 1 degree shy of being a flat disk.
You don't think the string will be pulled off?
@DanielSank Halp
In the picture in my head I don't see how it could be "pulled off" without ripping
8:21 PM
@AccidentalFourierTransform What do you not understand? They don't want the devs to spend time on this because they think it's not important.
Perhaps you disagree, but surely you see the logic.
@BernardoMeurer 'sup?
the sky, still
@DanielSank I need basic logic help because I'm being silly
@ACuriousMind Well, suppose the cone is really flat, i.e. a disk. Then surely I can move the loop around without it ripping, yeah?
haven't managed to knock it down yet
Suppose I have a CSV with 7 fields
8:22 PM
@ArtOfCode Keep trying.
and n entries
(i.e. a nx7 matrix)
uh huh
Use numpy.
No, no, ged demmit
This is C
I'm writing a parser
use numpy anyway ;)
8:23 PM
Hear me out
@DanielSank I agree with that, but that's not really their attitude towards the post. They are trying to argue that there is no problem at all, not that the problem is not important enough
Wait
No
there is a problem
I got this
8:23 PM
and I might agree that it is not really that urgent
@AccidentalFourierTransform Uh... I read the first few comments etc. and the feeling I got was pretty much "yeah, but this isn't that big of a deal".
> Deleting an account is not a trivial task, so going through that process over and over for this somewhat minor (and largely inconsequential) benefit would be really difficult to miss — assuming anyone would even bother at all.
> Perhaps if there was more benefit to up-voting a comment, it might warrant additional protections. But this is the kind of hypothetical problem you might want to see demonstrated in actual practice before deeming it a problem at all.
there are a couple of comments that say that
but most of them are just "no, not a problem"
Here's the first line of the second answer:
> While I agree this can be annoying and unfair, I don't think it justifies spending development time.
8:25 PM
I dont know, perhaps Im just focusing on the comments that I found unreasonable
^
Yep.
but, oh well, some of them are unreasonable :-P
ZOMG somebody on the internet is wrong!
my point was essentially that "Huge problem, maybe not. Something for developers to bear in mind, IMHO, yes."
I didn't see anything hostile.
I didn't look that hard though.
The responses all seem to be "yeah, you're right that this could happen, but it's not a big deal".
8:27 PM
whatever, not really that important
yep
back to the angle thingy
yep
I think I got the picture right
it seemed impossible for the string to be pulled over because I was imagining the cone way too narrow
but I agree that if its flat enough, it would probably be pulled over
so, what's the angle...
hmm
@ACuriousMind ^
8:30 PM
I guess that it should be 45º?
Why?
because that's what my brain told me
:P
argh, I think I have the right image in my head, but I cannot put it into words
I imagine the string to be completely horizontal
and moving the mass a bit downwards
and the other end of the string should move upwards
hmmm
@AccidentalFourierTransform I think that reasoning should work, but you have to remember this is a 3D problem.
8:36 PM
and the string is inextensible
@AccidentalFourierTransform correct
@AccidentalFourierTransform That's not what I got. How'd you get that?
The Hitchhiker's Guide to the Galaxy is a comic science fiction series created by Douglas Adams that has become popular among fans of the genre(s) and members of the scientific community. Phrases from it are widely recognised and often used in reference to, but outside the context of, the source material. Many writers on popular science, such as Fred Alan Wolf, Paul Davies and Michio Kaku, have used quotations in their books to illustrate facts about cosmology or philosophy. == Answer to the Ultimate Question of Life, the Universe, and Everything (42) == In the radio series and the first novel...
Ah yes.
How'd you get that?
8:41 PM
so its not that
I mean, it has to be a nice angle. If its not 45º, it has to be pi/6
does it have anything to do with parabolas and hyperbolas?
geodesics?
@ACuriousMind Lol, let me try to rephrase...
@AccidentalFourierTransform Why does it have to be nice?
@BernardoMeurer That's another place where using calloc is a win. Without it you have to worry about garbage present in any possible padding biting you i the ass.
@AccidentalFourierTransform Well, yes, since conic sections are those shapes.
@AccidentalFourierTransform mayyyyyyybe....
@DanielSank there is a policy, you know
8:45 PM
Of course you could also use a #pragma to force tight packing with the attendant disadvantages.
@ACuriousMind I have this function $u\in W^{2,q}_\delta$ satisfying $\Delta u=F\in W^{0,q}_{\delta-2-\tau}$. In this situation $W^{2,q}_{\delta-\tau}\subset W^{2,q}_\delta$. So I want to find a $v\in W^{2,q}_{\delta-\tau}$ such that $\Delta(u-v)=0$ outside of the ball $B_R$. Now, I know that $\Delta :W^{2,q}_{\delta-\tau}\to W^{0,q}_{\delta-2-\tau}$ is Fredholm. So its cokernel is finite-dimensional. That means there are finitely many l. i. functions not in $\Delta(W^{2,q}_{\delta-\tau})$.
Agreed?
@AccidentalFourierTransform good point
@ACuriousMind $\Delta =\sum D_iD_i$, of course.
@DanielSank I dont think I will be able to solve it
and goolge is not helping :-P
Guys, why do we need a spherically or cylindrically symmetric charge to have a net field of zero? Gauss’ law doesn’t care about the geometry of the exterior charge, right? As long as the exterior charge isn’t enclosed by your Gaussian surface, you don’t have to worry about its electric field, right?
8:50 PM
@0celouvskyopoulo7 All these indices on the $W$ hurt my eyes
who was a better mathematician, Gauss or Riemann?
it's not a contest
yes it is
@ACuriousMind Well $q$ is the $2$ in $L^q$, $2$ is the number of weak derivatives, and the other index is, well, something controlling the strength of the weight "at infinity"
@ACuriousMind with respect to what operator is the vacuum cyclic
8:52 PM
oh shit, I can't delete it anymore
I misinterpreted it
and it shall stay there forever
@Slereah I don't know what it means to be cyclic "with respect to an operator"
I guess algebra?
@ACuriousMind So...?
Does the setting make sense now
@0celouvskyopoulo7 Whatever they are, I have a hard time parsing that statement because I have to keep comparing indices :P
8:56 PM
I want a $v\in W^{2,q}_{\delta-\tau}$ with $\Delta v=F$ in $R^n-B_R$.
@ACuriousMind Sorry I've been reading these weighted Sobolev papers for a week now :<
dude, use multi-indices
"let $i$ stand for everything I might need"
@DanielSank so, are you gonna tell me the answer or not?
actually, I still don't understand what they mean. So we have a point and a charge at distance $r$. Is the net field equivalent to the flux through an infinitesimal box around the point? But then outside of every charge the net field would be zero, which can't be correct. But how can I say something about the net field then with Gauss? Gauss only gives flux:(
@Slereah Okay. So I guess you're confused about what cylicity means because in an irreducible representation, every vector is cyclic. It's not a property of the vacuum as such
@ACuriousMind whaaat
Every AQFT paper bangs about how the vacuum is the cyclic vector par excellence
thats because in AQFT there is only one vector, the vacuum $\mathcal H=\{|0\rangle\}$
it makes things really simple
9:03 PM
@AccidentalFourierTransform uhhhh
I don't usually give answers to puzzles.
It's not like you have to solve it right now.
@Slereah Well, I don't know about the sort of AQFT you're reading but it's a general fact that every vector of an irrep is cyclic - if a representation contains a non-cyclic vector, then the action of the algebra on that vector generates a true subrepresentation, contradicting irreducibility
Is that related to the selection rules
according to encyclopediaofmath.org/index.php/Cyclic_vector a vector is cyclic with respect to a certain operator. It cannot be cyclic by itself
9:11 PM
@Slereah Superselection rules are basically physicist speak for "the representation is reducible", yeah
Ah yes
Now, if your representation is reducible, then you can have both cyclic and non-cyclic vectors, and I guess the vacuum should be cyclic so that every state can be "reached" from it
So basically every reducible part needs to have a cyclic vector?
One vacuum per sector
effin communists
9:13 PM
@Slereah Well, given a direct sum of irreps, you get a cyclic vector for the sum by just choosing the direct sum of non-zero vectors from these representations
There are non-cyclic vectors here - those whose projection to at least one irrep is zero, but being cyclic still doesn't strike me as all that special
Lemme see if i can find the wording
@ACuriousMind So the idea is this: Suppose there is no $v\in W^{2,q}_{\delta-\tau}$ with $\Delta v=F$ in $R^n-B_R$. Then I take $F$ and change it in $B_R$, and produce a whole infinite dimensional space of functions that $\Delta$ misses outside of $B_R$. But then the cokernel is no longer finite, so that's a contradiction.
The question is if I can modify $F$ like that and actually stay in the Sobolev space. And I don't know about that...
"The Reeh-Schlieder Theorem shows that the spectrum condition entails that the vacuum vector is cyclic for every local algebra."
It's not the usual Sobolev space after all..
for instance
9:17 PM
I'm afraid I don't see how that question has nothing to do with the Sobolev space or how my algebra can help
@Slereah ohhhhhh
it is but one instance because they talk about $\Omega$ being cyclic all the time
Yeah, the special condition here is that you have one space and there's an infinitude of different algebra representations on it (all of these local algebras are different, after all). The special condition is that it is cyclic for all of these.
Ah yes
Being cyclic for a single representation is not special. Being cyclic for so many is.
"In the early twentieth century, it was thought that the referent of T must be a set of axioms of some formal, preferably first-order, language. It was quickly realized that not many interesting physical theories can be formalized in this way. But in any case, we are no longer in the grip of axiomania, as Feyerabend called it"
But I love axiomania :(
How is AQFT supposed to help, anyway
You have several representations for the same Hilbert space
So you only work with the representation directly
But does that really make it unique?
Are we guaranteed that there will only be one theory for a set of rules in AQFT
9:24 PM
@Slereah Pretty sure that's not its goal.
But then why
It's just supposed to axiomatize QFT in a rigorous manner.
Since it's always presented as the solution to the unitary inequivalence
But there are many QFTs, so why would you expect uniqueness?
But why AQFT and not another theory based on a Hilbert space, then
9:25 PM
@Slereah Oh, it evades Haag's theorem by the axioms being so strong that you don't need the interaction picture
What axiom makes it unecessary?
Well, I don't know which of the axioms you're looking at, but I think it essentially assumes that you "have" all the interacting operators already.
Pretty tall order
It's a formal definition of a QFT
Just like Lagrangian or Hamiltonian mechanics. You shouldn't expect any specifics from it
I wouldn't if it wasn't put on such a pedestal!
9:28 PM
And yes, both AQFT and FQFT as axiomatization suffer from the axioms being so strong that we don't know how to produce SM-like theories that fulfill them
wtf is FQFT
what will they think of next??
Well I've seen a few interacting theories based on AQFT
But they're not pretty to look at
functiorial QFT, @0celouvskyopoulo7
@0celouvskyopoulo7 Functorial QFT, it actually works for TQFT, not so much for actual QFTs :P
I thought AQFT was functorial
Also the interacting AQFTs still need renormalization anyway
it's all a ruse
9:30 PM
@0celouvskyopoulo7 The "functorial" here doesn't mean "functors occur in it", but that you essentially define a QFT to be a certain functor on a cobordism category.
@ACuriousMind Hmmm
Well QFT is also a functor in AQFT
The old presheaf
How about I multiply by $1+\phi_\epsilon$, where $\phi_\epsilon$ is nascent delta
that should work
Moving on
I think it's time 4 bed
I think it's time 4 a snack
9:33 PM
@Slereah oh no
"A smooth $n$-dimensional mainfold $(M,g)$ with Riemannian metric $g\in W^{1,q}_\mathrm{loc}(M)$"
How does life even work when the metric isn't smooth
10:03 PM
10:31 PM
So, funny thing about this. The votes themselves are actually deleted. What we don't do, apparently, is update the denormalized Score stored on the comment itself. Regardless of whether or not this is a practical problem, we should probably do that if only for consistency's sake. Having said that, doing so has a caching implication in the API that I'm not sure how to handle off the top of my head (or if it's a real concern - I'm not familiar with that part of the code that much). Gonna have to think about it some. — Adam Lear ♦ 36 mins ago
There is hope
10:49 PM
I passed graduate E&M! I now never have to do electomagnetism again
@JohnRennie I need your code review skillz when you're available
I did some serious brouhaha
hides in bubble where this is how things work
Also, congrats :^)
@GPhys It's a comfy bubble.
o/
Finally submitted to Bonn. I can rest now.
10:58 PM
Nice, GPhys. Though you might have to work on it again, if you plan to keep doing physics :D
Hi, everybody.
@Danu solve the cone problem.
@DanielSank Rytsas
@Mithrandir24601 What?
@DanielSank High Valyrian equivalent of 'hello'/generic greeting :P
The hell is Valyrian?
11:06 PM
@DanielSank Game of Thrones language
I am confused.
I would have thought Mithrandir24601 would speak Quenya or something.
@nbro Something like
> Given that w know that chat has been headed about 1.8 degrees off of the direct bearing to hell at 72 posts per hour for the last 16 hours, determine how much closed to total burnination is the h bar now that at the beginning of the measurement.
perhaps?
Don't know why I never tried learning that... But I didn't :/
Although it's tempting...
But I have actual work to do over the next few months
Sorry what's up Daniel?
I'm on my phone atm
...why are you using SE chat from your phone at 1:20 am? :D
11:20 PM
Why not?
When playing games it's convenient to do it cuz Skyrim crashes when you tab out
@ACuriousMind rebooting laltkp.
Laptop
lol
@ACuriousMind So I have $\Phi:M-K\to R^n-B_R$ a $C^\infty$ diffeo, where $K$ is compact in $M$. This gives coordinate functions $x^i$ on $M-K$. Am I crazy to think there's no particular reason those functions should extend to all of $M$? $M-K$ isn't closed.
Uh...is the sphere w/ stereographic projection not a counterexample?
@ACuriousMind Mmmm, not quite? Stereographic projection goes to all of $R^n$
@0celouvskyopoulo7 Well...remove both poles (or rather, one pole and a small disk around the other), then the target is $\mathbb{R}^2$ minus a small disk, right?
(You didn't stipulate that $K$ be connected)
11:28 PM
@ACuriousMind Lol...it should be morally connected.
But I see the point.
Can you find a connected counterexample?
I can think of weird situations where there's no extension, sure
But none arising in this context
I'm having trouble thinking of a counterexample where $K$ is connected.
What's the question?
To find maps that DON'T extend?
yep
@Danu A paper says to "take an extension" and I don't think that's always possible.
what is $M$?
11:36 PM
A manifold.
not closed or anything?
Noncompact.
and what should the extension satisfy?
It should just be a smooth extension of the $x^i$, nothing special.
So why don't bump/cutoff functions work?
11:40 PM
My immediate reaction is because $M-K$ is open, but since it's covered by one chart it might work...
always first try cutoff functions
:D
Sure, but it's not obvious to me how to do it. Because $M-K$ is open, the $x^i$ can get very bad as you approach the boundary.
That was ACM's point above.
With the sphere example, the functions blow up near the pole.
So connectedness would have to be necessary, if it's true.
Bump functions let you piece together local extensions into a global extension. But for the sphere example, there are no local extensions. My conclusion is that this guy needs to state his definition more carefully.
Also in your sphere example $M$ is compact
@Danu $M$ is not explicitly noncompact, but in applications it should be
ok
solution strategy one: disregard non-compact manifolds
:D
enough trolling
11:53 PM
@Danu I wish!
Noncompact manifolds are horrible
It's a solid reason to disregard GR | 2020-10-23 05:58:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6637412905693054, "perplexity": 1266.44708867604}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00632.warc.gz"} |
http://accesssurgery.mhmedical.com/content.aspx?bookid=361§ionid=39866441 | Chapter 70
• The thrombotic microangiopathies (TMAs) include the spectrum of thrombotic thrombocytopenic purpura (TTP) and hemolytic uremic syndromes (HUS) as well as related obstetric syndromes.
• The hallmarks of TMA are a microangiopathic hemolytic anemia (MAHA) and thrombocytopenia.
• TMA must be distinguished from other coagulopathies, such as disseminated intravascular coagulation (DIC) and collagen vascular disease with vasculitis, since therapeutic approaches differ.
• The clinical presentation of TTP/HUS is characterized by a pentad of findings: MAHA, thrombocytopenia, neurologic abnormalities, renal dysfunction, and fever.
• Plasma exchange is the therapy of choice for TTP/HUS, with adjunctive therapies including plasma infusion, corticosteroids, splenectomy, and/or immunosuppressive agents.
• TTP/HUS is a hematologic emergency, and patients are at risk of developing tissue anoxia, lactic acidosis, renal failure, or catastrophic central nervous system injury.
• Plasma exchange may be complicated by catheter accidents, air embolus, citrate toxicity, and pulmonary edema.
Moschcowitz's original description of TTP in 1925 was based on a triad of findings: anemia (of the microangiopathic hemolytic type), thrombocytopenia, and neurologic symptoms. In 1966, 271 cases of TTP were reviewed, and the features of fever and renal impairment were added to form a clinical pentad.1 Subsequent series have confirmed that renal involvement is common, with proteinuria, hematuria, or azotemia seen in 80% of patients with TTP.2,3 Hemolytic uremic syndromes (MAHA, thrombocytopenia, and renal failure) are regarded as part of a spectrum of TMA, which at one end consists of TTP with predominantly neurologic findings and minimal renal abnormality, and at the other end consists of profound renal dysfunction with little or no central nervous system (CNS) pathology. The latter syndromes are more common in childhood, in the postpartum period, and following use of chemotherapeutic (mitomycin C) or immunosuppressive (cyclosporine) agents. On occasion, the evolution of renal manifestations in TTP can become indistinguishable from HUS, so that attempts at rigid distinction are generally unrewarding.4
The age of onset of TTP/HUS ranges from infancy to the eighth decade, with a peak incidence in the third decade. Females are more frequently affected than males by a ratio of 2:1. Childhood HUS often presents with antecedent illness, typically gastroenteritis related to enterotoxin-producing strains of Escherichiacoli or Shigella. TMA associated with pregnancy includes several syndromes with considerable overlap, making prospective differentiation difficult.5,6 Pregnancy-induced hypertension (PIH; eclampsia and pre-eclampsia) is often associated with subtle laboratory abnormalities consistent with a degree of underlying TMA; on occasion this becomes clinically significant. When hypertension, elevated liver enzyme levels, and a low platelet count occur together as a syndrome in the peripartum period, the term HELLP syndrome is applied (see Chap. 105). TTP may also be encountered in the second or third trimesters of pregnancy; unlike thrombocytopenias associated with PIH, it may not improve rapidly with termination of the pregnancy and may require further treatment. Finally, postpartum HUS is distinguished by its onset after delivery. Characterization of TMA in pregnancy is often made in ...
Sign in to your MyAccess profile while you are actively authenticated on this site via your institution (you will be able to verify this by looking at the top right corner of the screen - if you see your institution's name, you are authenticated). Once logged in to your MyAccess profile, you will be able to access your institution's subscription for 90 days from any location. You must be logged in while authenticated at least once every 90 days to maintain this remote access.
Ok
## Subscription Options
### AccessSurgery Full Site: One-Year Subscription
Connect to the full suite of AccessSurgery content and resources including more than 160 instructional videos, 16,000+ high-quality images, interactive board review, 20+ textbooks, and more. | 2017-03-27 06:46:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26548609137535095, "perplexity": 14687.947109329125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189462.47/warc/CC-MAIN-20170322212949-00646-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://byjus.com/question-answer/draw-a-graph-showing-the-variation-of-potential-energy-between-a-pair-of-nucleons-as/ | Question
# Draw a graph showing the variation of potential energy between a pair of nucleons as a function of their separation. Indicate the regions in which the nuclear force is(i) Attractive(ii) Repulsive.Write two important conclusions which you can draw regarding the nature of the nuclear forces.
Solution
## Conclusions:(i) The potential energy is minimum at a distance $$r_0$$ of about $$0.8\,fm.$$(ii) Nuclear force is attractive for distance larger than $$r_0.$$(iii) Nuclear force is repulsive if two are separated by distance less than $$r_0.$$(iv) Nuclear force decreases very rapidly at $$r_0 /$$equilibirium position.Physics
Suggest Corrections
0
Similar questions
View More
People also searched for
View More | 2022-01-20 13:27:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6130766272544861, "perplexity": 1166.0345684573258}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00096.warc.gz"} |
https://zenodo.org/record/3573654/export/csl | Conference paper Open Access
# TCP Muscle Tensors: Theoretical Analysis and Potential Applications in Aerial Robotic Systems
Alejandro Ernesto Gómez Tamm; Pablo Ramón Soria; B. C. Arrue; Anibal Ollero
### Citation Style Language JSON Export
{
"DOI": "10.1007/978-3-030-35990-4_4",
"author": [
{
"family": "Alejandro Ernesto G\u00f3mez Tamm"
},
{
"family": "Pablo Ram\u00f3n Soria"
},
{
"family": "B. C. Arrue"
},
{
"family": "Anibal Ollero"
}
],
"issued": {
"date-parts": [
[
2019,
11,
7
]
]
},
"abstract": "<p>The use of aerial systems in a variety of real applications is increasing nowadays. These offer solutions to existing problems in ways that have never seen before thanks to their capability to perform perching, grasping or manipulating in inaccessible or dangerous places. Many of these applications require small-sized robots that can maneuver in narrow environments. However, these are required to have also strength enough to perform the desired tasks. This balance is sometimes unreachable due to the fact that traditional servomotors are too heavyweight for being carried by such small unmanned aerial systems (UAS). This paper, offers a innovative solution based on twisted and coiled polymers (TCP) muscles. These tensors have a high weight/strength ratio (up to 200 times) compared with traditional servos. In this work, the practical and modeling work done by the authors is presented. Then, a preliminary design of a bio-inspired claw for an unmanned aerial system (UAS) is shown. This claw has been developed using additive manufacturing techniques with different materials. Actuated with TCP, it is intrinsically compliant and offers a great force/weight ratio.</p>",
"title": "TCP Muscle Tensors: Theoretical Analysis and Potential Applications in Aerial Robotic Systems",
"type": "paper-conference",
"id": "3573654"
}
62
141
views | 2023-02-06 02:53:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21516378223896027, "perplexity": 8432.44718257335}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500303.56/warc/CC-MAIN-20230206015710-20230206045710-00837.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/109586-jacobian-print.html | # Jacobian
If X and Y be two independent Gamma variates with parameters $\mu$ and v respectively,show that the variables U=X+Y and $\frac{X}{X+Y}$ are independent and that U is a $(\mu+v)$variate and V is a $\beta_1(\mu,v)$ variate. | 2017-08-21 20:28:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7983438968658447, "perplexity": 730.4442279116986}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00532.warc.gz"} |
https://www.ad5gg.com/2018/04/02/antenna-tuners-matching-and-swr/ | # Antenna Tuners, Impedance Matching, and SWR
Antenna tuners seem to be some of the most misunderstood devices in all of amateur radio. In this article, I’ll try to explain what is happening when you use an antenna tuner. I’ll try not to get too much into mathematics, and I’ll also try to squash a few myths.
I really don’t like the name “antenna tuner.” It is a bit of a misnomer, because what most people are referring to is a box connected between their radio and antenna (or internal to their radio) which attempts to match the impedance of the antenna system to something close to the transceiver’s antenna port impedance. This is an impedance match, or impedance transformer. In a standard ham radio tuner, the circuits are essentially the same as those illustrated to the right. Either a Pi or Tee network is used, with the inductors and capacitors being variable via knobs and switches in a manual tuner, or different inductor and capacitor component values being switched in and out of the circuit via relays in the case of an automatic tuner.
Think about it–how can you tune an antenna, changing its electrical properties, by pressing a “TUNE” button on your radio? You can’t. With the exception of remote antenna feed point tuners, antenna tuners do not tune your antenna. They do not physically change your antenna system at all.
Antenna system impedance is composed of a resistive quantity, and a reactive quantity. The reactive quantity can be capacitive or inductive. If you’ve seen impedance written as 50+j17Ω, or 120-j45Ω, that’s what these are. Complex impedances in Cartesian (or rectangular) notation. The first example being 50 ohms resistive, with 17 ohms of inductive reactance (positive sign). The second example being 120 ohms resistive, with 45 ohms of capacitive reactance (negative sign).
Complex test equipment often isn’t required to troubleshoot antenna impedance issues. Reactance causes phase shifts in voltage and current, resulting in some of the power delivered by your transceiver to be reflected, causing standing waves in the antenna feed line. The extreme example to the right shows 100% reflection due to no antenna connected to the transmitter. The ratio of the maximum amplitude of these standing waves to the minimum amplitude can be measured using a simple SWR meter. This indicates the level of impedance mismatch between the transceiver and the load.
A common misconception is that a high SWR due to a large impedance mismatch causes the power which is reflected back to its source to be absorbed by the transmitter output stage, overheating the output device (RF transistor finals, for example) and eventually burning them up. This is not the mechanism for failure during high SWR operation. When power is reflected and arrives back at the source, most of it is re-reflected back towards the load. This back and forth continues until the energy is dissipated in the load or via the feeder loss. You will often measure a higher output power on a power meter if the VSWR is high in your antenna system because some of the re-reflected power is added to the incident power.
Despite the huge focus placed upon it, and the never-ending quest for 1:1 VSWR, there is little real power loss associated with a moderate VSWR, and no real need for attaining the alleged gold standard of a 1:1 standing wave ratio from a power loss standpoint. In fact, a flat 1:1 VSWR over a wide frequency range can indicate a problem with your antenna system.
In a system where there are impedance mismatches, the power loss and reflection routing can be visualized like this:
### Damage due to high SWR?
As mentioned previously, contrary to popular belief, failure from high VSWR is not due to the final output transistor (for example) absorbing reflected power.
Driving reactive loads is tough on electronics. High VSWR due to impedance mismatches shouldn’t be ignored. Impedance mismatches can cause failure of electronics, especially the output stages of RF amplifiers. This is due to the excessive reactance in the mismatched antenna system de-tuning the output stage and causing excessive current flow, or oscillation, or high voltages appearing at the output stage, pushing devices beyond their safe operating envelope.
Most modern transceivers will automatically reduce their output power as the VSWR exceeds 2:1 for their own protection, not because of reflected power, but because of the excessive reactance pushing the finals outside of their operating areas.
### Tuner Location
Consider the diagram below. Point A is the transceiver antenna port. It is 50 ohms. Point B is the antenna tuner’s “to transceiver” port, which is also 50 ohms. The cable between Point A and Point B is also 50 ohms, and likely a short jumper. All good so far.
We know the antenna port at Point D is presenting an impedance mismatch, so what is happening between Points C and D?
The cable between Point C and Point D is your main coaxial cable run to your antenna.
Impedance matching is achieved in the antenna tuner by presenting a complex conjugate impedance match at Point C. Essentially, the tuner is canceling out the reactive component of the complex impedance presented to it by the mismatched antenna system by presenting it with a reactive component of equal value, but opposite sign.
For example, to match a 50+j17Ω load to 50 ohms resistive, the antenna tuner would present a complex impedance of 50-j17Ω to the load (at Point C), canceling out the inductive reactance with capacitive reactance.
If the transceiver antenna port and the coaxial cable are both 50 ohms characteristic impedance, what happens if the antenna is not? It means that the coaxial cable is seen as part of the load, and the impedance measured will vary along the length of the coaxial cable feed line between Point C and Point D in the above diagram. The coaxial cable is acting as an impedance transformer (more on that later). The value of the impedance match needed will also vary depending on how long the coaxial cable between Point C and Point D is, or where in the coaxial cable the tuner is placed. See this post for more information on that. Also, the measured SWR will decrease the longer the coaxial cable is, due to normal losses within the cable.
### Tuner Location
So, you’re using a tuner to make your transceiver see close to a 50 ohm impedance. It is happy. All is well? Maybe not.
Considering the coaxial cable losses, and the fact that even when using a tuner between the radio and the coaxial cable antenna feeder, the SWR in the feeder will still be high, undue stress can be caused to the coaxial cable due to SWR-induced high voltage and current nodes within the cable. This effectively derates the coaxial cable power handling from a transmitter output power perspective.
How can you avoid this? Use two tuners to decouple the mismatch from the long antenna feed line?
In the above diagram, the coaxial cable between Point C and Point D is your main run up to the antenna. The whole point of using two tuners is to shift the point of mismatch close to the antenna, instead of close to your radio. This way, the whole coaxial cable run up to your antenna is (practically speaking) at 50 ohms. The coaxial jumper between Point E and Point F can be kept as short as possible, keeping losses to a minimum.
Wait a minute – you really need two tuners?
Absolutely not. This was just a step to illustrate where the inefficiencies lie.
Consider the first diagram once again:
For the best efficiency, the long coaxial cable run should be from Point A to Point B, keeping the influence of the antenna impedance mismatch from affecting the coaxial cable feeder. This means the best place for an antenna tuner from an efficiency and low loss standpoint, is right at the antenna. Not at the transceiver.
Wait a minute – who wants to run out and adjust their tuner if it’s at the antenna feed point??
Nobody does. You can have convenience, or efficiency. Not both.
remote antenna tuner may inject some convenience into the equation, and work well for lower powers, but if you’re using 1500W, the amount of volt-amp reactive power the remote tuner will have to tune out may necessitate a huge device!
Theoretically though, reactance consumes no power.
A perfect inductor stores energy in a magnetic field,
$0.5*L*(I^2)$
A perfect capacitor stores energy in an electrostatic field,
$0.5*C*(I^2)$
However, resistive losses are inevitable in the system, via inductor DC resistance, and capacitor equivalent series resistance, so power is consumed via that route.
See W8JI’s website for some practical examples.
### Resonant? Non-Resonant?
It is often said…
There is no substitute for a correctly tuned, resonant antenna for the frequency you want to use.
Why? Simply because if your antenna is resonant, its impedance is purely real (j0). It is completely resistive and there are no imaginary (reactive) components to the impedance. It is the resistive part of the impedance of the antenna which is important for efficiency. The reactive part stresses the system, negatively impacts the efficiency, and does no real work.
Therefore, if your antenna is resonant, it matches perfectly with the transmission line impedance, and the transmitter impedance, delivering maximum power, with maximum efficiency.
However, it is often impractical and inconvenient to have a collection of single band antennas. I use an Off-Center-Fed Dipole as my main antenna. It presents a good VSWR to my transceiver on the 40, 20, 10, and 6 meter bands. For other bands, I use a tuner.
### Coaxial cable impedance transformer
A transmission line itself can be an impedance transformer for mismatched loads, as the reactance changes along the length of the cable. For example, an RF open circuit can be made to look like a short circuit by adding a quarter wave long piece of transmission line, and vice versa. This only works well for a specific frequency, since the quarter wavelength changes with frequency. This is also the principle behind open and short circuit tuning stubs.
Therefore, if your antenna tuner is located where most are (in the radio room), the impedance which the antenna tuner sees is the impedance of your antenna, but it sees this impedance after it has been transformed by your coaxial cable. It is for this reason that the antenna tuner located in the radio room is tuning the antenna system, including the transmission line, and not just the antenna.
I don’t want to get into transmission line theory in this article, since that would get very intense, very quickly, but if you’re curious as to how your coaxial cable transmission line transforms impedance, you can download a utility which simulates transmission lines made from real coaxial cable types, TLDetails here:
### A basic practical example
Using the magnificent utility Smith, by Fritz Dellsperger, we can see that if our antenna at Point C in the diagram (DP1 in the datapoint list) is presenting an impedance of approximately 74+j51Ω. However, once that impedance has been transformed through 25 feet of RG-8 (Point B) with a dielectric constant of 2.3, or 66% velocity of propagation, the impedance presented to the tuner (at Point A) is now 113+j28.5Ω (TP2 in the datapoint list). This is the impedance the antenna tuner will present a conjugate match to, in order to efficiently match the antenna system.
### A full match example
In this example the components in the blue dashed area are within the antenna tuner. Looking at the datapoints and moving back from the antenna (75.05+j51.36Ω), the impedance is:
1. Transformed through 25 feet of RG-8 coaxial cable to 111.295+j32.191Ω.
2. Transformed through the antenna side capacitor (adjusted to 139.3pF) in the tuner to 111.295-49.398Ω.
3. Transformed via the tuner inductor (set to 874.6nH) to 50.117+j64.536Ω.
4. Finally transformed to 50.117Ω resistive by the transceiver side capacitor (adjusted to 176.2pF) in the tuner.
Looking at this impedance transformation on a Smith chart, plotting VSWR circles, we can see how the VSWR presented to the transceiver is reduced. The image is less blurry when enlarged (click it).
The datapoints on this Smith chart match the datapoints in the above example, so feel free to refer to both. DP1 is the original antenna impedance. Adding the coaxial cable caused the impedance to be transformed, and so plotting clockwise in a perfect circle around the smith chart (over more than 360 degrees) we arrive at TP2. The antenna side capacitor in the tuner rotates the impedance to TP3, the tuner inductor rotates the impedance to TP4, and finally the transceiver side capacitor in the tuner rotates the impedance to the center of the Smith chart, which is 50 ohms.
### What on earth caused this outburst?
I first got the idea to put this article together after hearing lots of people say this lots of times…
(This was years ago, but I’m lazy I’m busy, and it’s taken a while to get motivated!)
If you use an antenna tuner, you just burn up all your power in the tuner. Nothing gets to the antenna, but at least your radio is happy!
That statement is not exactly true. Sure, a properly adjusted tuner will give your radio a nice comfortable impedance to dump all of its power into, keeping it happy, but the poor efficiency of your antenna system still remains.
### In Summary (and some other notes)
When you put all of this information together, it is easy to see how some myths and misconceptions have propagated over the years. Pun intended.
• Antenna Tuners do not physically tune antennas.
• Antenna Tuners are an impedance match circuit.
• Antenna Tuners match the impedance of your antenna system to the impedance of your transceiver.
• Antenna Tuners, the components in them, and your antenna system are all composed of reactive elements.
• Since reactance changes with frequency, a match is only perfect at one frequency.
• Changing frequency even slightly may require the match to be adjusted. This depends on the Q.
• If adjusted for an optimum impedance match, almost all of the power is transferred through the tuner. Power is not lost in the tuner, except for a small amount due to the efficiency of the circuit and associated components.
• Additional power is lost in the antenna system due to increased VSWR, but this is not usually significant. However, since the power is re-reflected over and over, it is subject to the coaxial cable losses over and over.
• When using a tuner, the impedance of the actual antenna system doesn’t change.
• Think about “non-natural” antennas such as 5/8λ verticals. These are not even close to 50 ohms, and will contain a matching circuit in the base of the antenna to bring it to 50 ohms. More or less, the same thing as an antenna base tuner does.
• A low VSWR does not guarantee your antenna is working well.
• A low VSWR does not necessarily indicate the antenna’s resonant point.
• Yes a 2:1 VSWR reflects 11% of your power, but don’t lose sight of the fact that it’s re-reflected. It is not simply “lost” power. A common misconception is that this reflected power is gone.
• A high VSWR derates the power handling capacity of your coaxial cable, since using an antenna tuner (on the radio side of your coaxial cable) does not remove the effects of a badly tuned antenna from your coaxial cable feeder.
• Long coaxial cables can mask the effects of a badly tuned antenna.
### Other Questions
Do you need a tuner when using a tube amplifier?
This is an interesting question. Tube amplifiers carry a matching network as part of their output tuning. If your antenna system VSWR is 3:1 or less, you probably don’t need to shell out a ton of money for a high power-capable antenna tuner to follow your tube amplifier. The adjustable range of the tube amplifier output match should be able to handle this with no problem. Of course, consult your manual or manufacturer for more information on the specifics of your amplifier output match capabilities, and be mindful of the fact that the impedance mismatch and higher VSWR still remains on your coaxial cable, which may be driven outside of its specifications with a higher VSWR at a higher power.
I have tried to avoid delving into the mathematics regarding matching, loss, and VSWR, but if you’re interested in such things, I highly recommend downloading this PDF of a presentation by Steve Stearns, K6OIK.
(in fact, I recommend checking out all of Steve’s archived articles here: http://www.fars.k6ya.org/docs/k6oik)
How is your transmitted power actually affected by VSWR? Another good article here.
A practical estimation of losses in T-network tuners:
Impedance Matching and Smith Charts (covers some transmission line theory):
### A fly in the ointment?
Some awkward, but unavoidable issues with this whole process…
• A transmitter output impedance is not a fixed value. This magical 50 ohms impedance varies depending on frequency, and varies with different output power levels.
• RF systems are rarely designed using solely complex conjugate matching to attain the maximum power transfer.
• Maximum power transfer doesn’t mean maximum efficiency.
Food for thought…
### Visualising Wave Behaviour
The following is a video featuring Dr. J.N. Shive of Bell Labs demonstrating wave behaviour on what is now called the “Shive Wave Machine”. I highly recommend watching this.
AD5GG works in the real world primarily as a board-level RF designer in the UHF (300 MHz - 6 GHz) range. Occasionally, he posts articles on this very site. Sometimes they're even worth reading. | 2019-12-10 23:59:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 2, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4919673502445221, "perplexity": 1449.0239645922036}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540529516.84/warc/CC-MAIN-20191210233444-20191211021444-00392.warc.gz"} |
https://userpage.fu-berlin.de/soga/300/30900_geostatistics/30943_Modeling_the_Semivariogram.html | Once we calculate an sample variogram, it is standard practice to smooth the empirical semivariogram by fitting a parametric model to it.
The nugget, the sill and the range parameters are often used to describe variograms:
• nugget: The random error process indicated by the height of the jump of the semivariogram at the discontinuity at the origin.
• sill: The limit of the variogram tending to infinity lag distances.
• range: The distance in which the difference of the variogram from the sill becomes negligible. In models with a fixed sill, it is the distance at which this is first reached; for models with an asymptotic sill, it is conventionally taken to be the distance when the semivariance first reaches 95% of the sill.
The following five parametric model are the most commonly used (Gelfand at al. 2010):
• Spherical
$\gamma(h, \theta) = \begin{cases} \theta_1 (\frac{3h}{2\theta_2}- \frac{h^3}{2\theta_2^3}) & \text{for } 0\le h \le \theta_2 \\ \theta_1 & \text{for } h > \theta_2 \end{cases}$
• Exponential
$\gamma(h, \theta)= \theta_1\{1- \exp(-h/\theta_2)\}$
• Gaussian
$\gamma(h, \theta)= \theta_1\{1- \exp(-h^2/\theta_2^2)\}$
• Matérn
$\gamma(h, \theta)= \theta_1\left( 1- \frac{(h/\theta_2)^\nu\kappa_\nu(h/\theta_2))}{2^{\nu-1}\Gamma(\nu)}\right)$
where $$\kappa_\nu(\cdot)$$ is the modified Bessel function of the second kind of order $$\nu$$. | 2019-06-27 08:45:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8387948274612427, "perplexity": 837.7185423325477}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00236.warc.gz"} |
https://greprepclub.com/forum/what-is-the-area-of-the-hexagonal-region-shown-above-9815.html | It is currently 20 Nov 2018, 05:19
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# What is the area of the hexagonal region shown above ?
Author Message
TAGS:
Moderator
Joined: 18 Apr 2015
Posts: 4916
Followers: 74
Kudos [?]: 980 [0], given: 4506
What is the area of the hexagonal region shown above ? [#permalink] 27 Jun 2018, 10:14
Expert's post
00:00
Question Stats:
100% (01:32) correct 0% (00:00) wrong based on 6 sessions
Attachment:
ex.jpg [ 10.06 KiB | Viewed 200 times ]
What is the area of the hexagonal region shown above ?
A. $$54 \sqrt{3}$$
B. $$108$$
C. $$108 \sqrt{3}$$
D. $$216$$
E. It cannot be determined from the information given
[Reveal] Spoiler: OA
_________________
Director
Joined: 20 Apr 2016
Posts: 743
Followers: 6
Kudos [?]: 500 [1] , given: 86
Re: What is the area of the hexagonal region shown above ? [#permalink] 28 Jun 2018, 00:54
1
KUDOS
Carcass wrote:
Attachment:
The attachment ex.jpg is no longer available
What is the area of the hexagonal region shown above ?
A. $$54 \sqrt{3}$$
B. $$108$$
C. $$108 \sqrt{3}$$
D. $$216$$
E. It cannot be determined from the information given
There are no. of ways to solve it:
From the diagram we know it is a regular hexagon
and formula of the area of a regular hexagon = $$(3\sqrt{3} * side^2)$$ * $$\frac{1}{2}$$
= $$(3\sqrt{3} * 6^2)$$ * $$\frac{1}{2}$$ = $$54\sqrt{3}$$.
Now if we donot know the formula then, please refer to the diagram below:
Now we can write : y° + y° + y° + y° + y° + y° = 360°
or y° = 60°.
Now we can see the regular hexagon is divided into 6 triangle.
So each triangle will be equilateral triangle since it is inscribed in a regular hexagon.
Now area of Equilateral triangle = $$\sqrt{3} * (side^2)$$ * $$\frac{1}{4}$$ = $$\sqrt{3} * (6^2)$$ * $$\frac{1}{4}$$
=$$9\sqrt{3}$$
Therefore for 6 triangles we have = 6* $$9\sqrt{3}$$ = $$54\sqrt{3}$$.
Now if we donot know the area of equilateral triangle then we can always find out:
Take any one triangle inscribed in the regular hexagon:-
SO the area of th triangle = $$\frac{1}{2} * base * altitude$$
Now we need to find out the altitude, which can be found out by Pythagoras theorem, as we draw a line from the midpoint of the hexagon to the base of the triangle as shown in diagram,It divides the base at the midpoint as it is a equilateral triangle. (Also a 30 - 60 - 90 triangle will also have the same result)
So $$altitude^2 = 6^2 - 3^2$$ = 27
or altitude = $$3\sqrt{3}$$
Now area = $$\frac{1}{2} * 6 * 3\sqrt{3}$$ = $$9\sqrt{3}$$
Now area of all the six triangle = 6 * $$9\sqrt{3}$$ = $$54\sqrt{3}$$ = Area of the hexagon.
** Please note this applies only for a regular hexagon
Attachments
Untitled.png [ 20.44 KiB | Viewed 179 times ]
Untitled.png [ 18.02 KiB | Viewed 179 times ]
_________________
If you found this post useful, please let me know by pressing the Kudos Button
Re: What is the area of the hexagonal region shown above ? [#permalink] 28 Jun 2018, 00:54
Display posts from previous: Sort by | 2018-11-20 13:19:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6907432079315186, "perplexity": 1354.1769581842618}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746398.20/warc/CC-MAIN-20181120130743-20181120152743-00248.warc.gz"} |
http://clay6.com/qa/49653/the-probability-that-at-least-one-of-the-events-a-and-b-occurs-is-0-6-if-a- | # The probability that at least one of the events A and B occurs is 0.6.If A and B occur simultaneously with probability 0.2.Then $P(\bar{A})+P(\bar{B})$ is
$\begin{array}{1 1}(A)\;0.4\\(B)\;0.8\\(C)\;1.2\\(D)\;1.6\end{array}$
## 1 Answer
Toolbox:
• $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Step 1:
Given :
$P(A \cup B)=0.6,P(A \cap B)=0.2$
$\therefore P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$0.6=P(A)+P(B)-0.2$
$\Rightarrow 0.6+0.2=P(A)+P(B)$
$\Rightarrow P(A)+P(B)=0.8$
Step 2:
$P(\bar{A})+P(\bar{B})=(1-P(A))+1-P(B))$
$\Rightarrow 2-(P(A)+P(B))$
$\Rightarrow 2-0.8=1.2$
Hence (C) is the correct answer.
answered Jul 9, 2014
1 answer
1 answer
1 answer
1 answer
1 answer
1 answer
1 answer | 2018-03-21 20:36:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9185776710510254, "perplexity": 5522.113925944129}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647692.51/warc/CC-MAIN-20180321195830-20180321215830-00226.warc.gz"} |
https://cliocorvid.com/mathematics/other-mathematics/addition-and-multiplication/ | It is the habit among mathematics teachers, particularly at the elementary level, to present multiplication as repeated addition. The inimitable Keith Devlin, among others, has ranted about this, but it’s easy enough to see the temptation. When dealing with integers, multiplication and iterated addition will return the same numbers. Historically, it may be the case that multiplication was developed as a shortcut for addition, although the so-called Russian peasant method of multiplication is related to an Egyptian method that’s some 3650 years old, so the awareness that effective multiplication requires methods beyond simple addition is very old.
Devlin’s rant is based in significant part on the conceptual problem of multiplying non-integers. If you’re multiplying $$5.4 \times 4.1$$, what does that even mean? And if the dodge response to this is that $$5.4 \times 4.1 = 54 \times 0.41$$ (that is, 0.41 repeated 54 times), which it does, then what does $$\sqrt{2} \pi$$ mean? In what sense can we “repeatedly add” an irrational number an irrational number of times?
That’s a fair point. The rebuttal is that the point of introducing multiplication as “repeated addition” is to get children used to the concept of multiplication before getting into heady mathematical theory. Devlin’s essay pre-empts this point as well, but I want to address a separate difference based on my natterings on the abstract unit.
### So, what is a number, anyway?
At least until we get into college, mathematics consists largely of performing operations on numbers. At the elementary level, those operators are mostly restricted to addition, subtraction, multiplication, and division. Of these, addition and multiplication are the most important: If we wanted, subtraction could be fully abandoned once we introduce negatives, and division involves undoing multiplication (and could, theoretically, also be abandoned in favor of inverse multiplication, but I think that would lead to a bunch of needless complications).
What are we doing when we add two numbers?
• Bobby has two apples and Sally has three apples: Together, they have five apples.
• There are seven birds in the tree. Two more land on the branches. Now there are nine birds.
• $$3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$$
• $$\frac{2}{5} + \frac{1}{5} = \frac{3}{5}$$
• $$4x + 12x = 16x$$
Until we get to fractions, radicals, and algebra, the matter of addition (and the issue of “what is a number?”) is straightforward: We have two different groups of the same sort of thing, and when we combine those groups, we add the numeric value and leave the units untouched. We can teach students that every “number unit thing” (whatever we want to call it) consists of a count and a unit. The part that looks like a “number” is the number and the part that looks like words or stuff or groups or things or measuring unit is a unit.
As I’ve discussed before, because the unit part is boring to math, we ignore it. It dies from loneliness and neglect, and by third grade or so, it’s only referred to (if at all) in the most casual of ways.
And therein lies madness, because we resurrect it for fractions, but students have long forgotten about it. By the time we get to “combining like terms” in algebra, many students are lost.
Here’s the hook in fractions: For the first time, we have things we’re adding where the “units” include things that look decidedly like numbers. Clever students might have noticed that five nickels and eight nickels make thirteen nickels, which can then be converted using $$13¢ \times 5¢ = 65¢ = 0.65$$. But “nickel” doesn’t look like a number: It looks like a unit. The conversion between nickels, cents, and dollars is part of the same concept as the conversion between inches and feet, which also don’t look like numbers.
Meanwhile, the bottom parts of $$\frac{1}{4}, \frac{2}{7},$$ and $$\frac{3}{44}$$ are unambiguously numbers.
So, to summarize and expand:
1. A numeric term can be separated into a counter and a unit.
2. In order to add two numeric terms, the units must be the same.
3. A unit can be explicit or abstract (in which case, it need not be expressed).
4. A unit can be basic (foot, dollar, apple, thing) or complex (five feet, two dollars, a seventh of a thing).
5. To add, combine the counters and leave the unit unchanged.
The fourth point is a key concept to addition. It makes possible things like adding fractions with unlike denominators and combining like terms. It also makes addition itself a significantly harder process. At the “apples and butterflies” stage of addition, addition consists of taking the number part, combining it, and leaving the unit alone. Here’s the process of addition when fractions with unlike denominators are involved:
1. For each fraction, find an equivalent fraction so that the units are the same. Do this by multiplying the fraction by $$\frac{x}{x}$$, where $$x$$ is a factor of the other fraction’s denominator. Note that it’s simpler at this step to multiply each fraction by the other fraction’s full denominator, i.e., $$\frac{a}{b} + \frac{c}{d} = \frac{a\cdot d}{b\cdot d} + \frac{c\cdot b}{b\cdot d} = \frac{a\cdot d + c\cdot b}{b\cdot d}$$, but this is often sneered at in favor of the more sophisticated method of finding the least common denominator because it’s more likely to require simplification at the end.
2. Separate each fraction into a counter and a complex unit, and add the counters.
3. Examine the result to see if there’s an equivalent fraction with a larger unit; if so, use that equivalent value.
Naturally, this is not exactly how the process is presented to children. We speak of “common denominators” instead of identical units (although Common Core does speak of $$\frac{a}{b} = a \times \frac{1}{b}$$, as a Third Grade concept).
This is one reason why I feel that moving away from concrete units as the norm in early years does a disservice to students: Success in fractions, and later in algebra, involves being able to convert between complex units that include a numeric portion. Procedurally, there’s no difference between adding apples, fourths, feet, or millions.
Introducing multiplication as repeated addition implies something obvious to any random First Grader: Multiplication is harder than addition. After all, if it involves addition, and then something else, it must be harder than addition, right?
Wrong. This is the real danger of relying on “multiplication is repeated addition”: Once we step away from decimals, multiplication is easier than addition.
Most teachers who teach fractions are already aware of this. How do we multiply fractions?
1. Multiply the tops.
2. Multiply the bottoms.
3. Simplify if needed.
The third step is the same as addition: “Examine the result to see if there’s an equivalent fraction with a larger unit; if so, use that equivalent value.” I’ve written it in fewer words, the trick that teachers use to make the business of fractions seem simpler than it is. Otherwise, though, multiplication is easier than addition when it comes to fractions: There’s no need to convert to equivalent fractions. The denominators don’t need to be the same. Just go for it.
And it’s not that we don’t care about units when it comes to multiplication. We do, very much. It’s a common mistake for students faced with a multiplication problem such as “Find the area of a room that is 10 meters by 12 feet” to say, “120”. Forget the units, they think: There are some numbers, and it says “area”, and we know the formula for area. Multiply the numbers. Done!
ACT Study Guides and suchlike will explain the error in the thinking: “We can’t multiply numbers with different units.”
And I am here to tell you, in as strong and bold a voice as I can: “YES, WE CAN.”
The error in thinking of 120 is not that we can’t multiply numbers with different units, it’s that what doing so in this case creates is a weird unit that nobody would ever use in the real world. A valid answer to the question “What is the area of a room that is 10 meters by 12 feet?” is “120 foot-meters”. It’s just that a foot-meter is a silly unit.
When we add, we decide on what the complex unit is, set it aside, add the counters, then combine it back with the complex unit. We may then need to reassess the numeric value, setting the base unit aside, and decide if we can simplify it at all. That’s addition.
Here’s multiplication: Combine all the things! That’s it. Multiply the counters. Multiply the units. Multiply everything! If we need to simplify, we’ll divide, which involves breaking some stuff up into multiplicands and then deciding if any of the same multiplicands are on the top and the bottom. So even that involves multiplication.
Indeed, the simplification process involved in adding fractions? That involves multiplication.
In other words, successful addition involves carefully attending to like-units, and then doing math on the counters, setting the units aside. Successful multiplication involves just mashing everything up and then sorting it out.
They are, in short, different operations. Addition cares about the distinction between “numeric value” and “units”; multiplication doesn’t.
Fractions: Adding requires the same denominator. Multiplication doesn’t care.
Algebra:
Adding requires combining like terms: $$3x + 4y + 5x = 8x + 4y$$.
Multiplication doesn’t care: $$3x \times 4y \times 5x = 3 \times 4 \times 5 \times x \times x \times y = 60x^2y$$.
Adding requires the same radicand: $$3\sqrt{2} + 5\sqrt{8} + \sqrt{7} = 3\sqrt{2} + 10\sqrt{2} + \sqrt{7} = 13\sqrt{2} + \sqrt{7}$$.
Multiplication doesn’t care: $$3\sqrt{2} \times 5\sqrt{8} \times \sqrt{7} = 3 \times 5 \times \sqrt{2 \times 8 \times 7} = 15 \sqrt {2 \times 2 \times 2 \times 2 \times 7} = 60\sqrt{7}$$
Distance v Area:
Adding requires the same measurement unit: 10 meters + 12 feet requires converting one distance to the other’s unit.
Multiplication doesn’t care: A room with sides of 10 meters and 12 feet has an area of 120 meter-feet. What’s a meter-foot? I don’t know. Let the engineers figure out a use for it. The point is, we can do it; we can’t add with unlike units.
### And so…
I’m still on the fence about whether it’s useful to originally present multiplication as repeated addition. I think it might help students understand why we’d want to multiply in the first place, and the language (particularly “times”) is certainly suggestive of the relationship. But I think it’s a relationship that ought to be abandoned as soon as students understand basics of multiplication, and I also question the habit of curricula to continue to imply that addition comes before multiplication throughout fractions, algebra, radicals, and on. At some point, I think the curricula should shift the precedence in light of the fact that, procedurally at least, multiplication is easier than addition.
Clio Corvid
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2023-02-02 15:24:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7082061171531677, "perplexity": 951.198569429644}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500028.12/warc/CC-MAIN-20230202133541-20230202163541-00787.warc.gz"} |
https://math.libretexts.org/TextMaps/Algebra_Textmaps/Map%3A_Linear_Algebra_(Schilling%2C_Nachtergaele_and_Lankham)/06._Linear_Maps/6.3_Ranges | Skip to main content
# 6.3 Ranges
Definition 6.3.1. Let $$T:V\to W$$ be a linear map. The range of $$T$$, denoted by $$\range(T)$$, is the subset of vectors in $$W$$ that are in the image of $$T$$. I.e.,
$\range(T) = \{ Tv \mid v\in V\} = \{ w\in W \mid \rm{~ there~ exists~} v \in V \rm{~ such~ that~} Tv=w\}.$
Example 6.3.2. The range of the differentiation map $$T:\mathbb{F}[z] \to \mathbb{F}[z]$$ is $$\range(T) =\mathbb{F}[z]$$ since, for every polynomial $$q\in \mathbb{F}[z]$$, there is a $$p\in \mathbb{F}[z]$$ such that $$p'=q$$.
Example 6.3.3. The range of the linear map $$T(x,y)=(x-2y,3x+y)$$ is $$\mathbb{R}^2$$ since, for any $$(z_1,z_2)\in \mathbb{R}^2$$, we have $$T(x,y)=(z_1,z_2)$$ if $$(x,y)=\frac{1}{7}(z_1+2z_2,-3z_1+z_2)$$.
Proposition 6.3.4. Let $$T:V\to W$$ be a linear map. Then $$\range(T)$$ is a subspace of $$W$$.
Proof.
We need to show that $$0\in \range(T)$$ and that $$\range(T)$$ is closed under addition and scalar multiplication. We already showed that $$T0=0$$ so that $$0\in \range(T)$$.
For closure under addition, let $$w_1,w_2\in \range(T)$$. Then there exist $$v_1,v_2\in V$$ such that $$Tv_1=w_1$$ and $$Tv_2=w_2$$. Hence
\begin{equation*}
T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2,
\end{equation*}
and so $$w_1+w_2\in \range(T)$$.
For closure under scalar multiplication, let $$w\in \range(T)$$ and $$a\in \mathbb{F}$$. Then there exists a $$v\in V$$ such that $$Tv=w$$. Thus
\begin{equation*}
T(av)=aTv=aw,
\end{equation*}
and so $$aw \in \range(T)$$.
Definition 6.3.5. A linear map $$T:V\to W$$ is called surjective if $$\range(T)=W$$. A linear map $$T:V\to W$$ is called bijective if $$T$$ is both injective and surjective.
Example 6.3.6.
1. The differentiation map $$T:\mathbb{F}[z] \to \mathbb{F}[z]$$ is surjective since $$\range(T) = \mathbb{F}[z]$$. However, if we restrict ourselves to polynomials of degree at most $$m$$, then the differentiation map $$T:\mathbb{F}_m[z] \to \mathbb{F}_m[z]$$ is not surjective since polynomials of degree $$m$$ are not in thecrange of $$T$$.
2. The identity map $$I:V\to V$$ is surjective.
3. The linear map $$T:\mathbb{F}[z] \to \mathbb{F}[z]$$ given by $$T(p(z)) = z^2 p(z)$$ is not surjective since, for example, there are no linear polynomials in the range of $$T$$.
4. The linear map $$T(x,y)=(x-2y,3x+y)$$ is surjective since $$\range(T)=\mathbb{R}^{2}$$, as we calculated in Example 6.3.3.
### Contributors
Both hardbound and softbound versions of this textbook are available online at WorldScientific.com. | 2017-07-23 10:37:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9866665601730347, "perplexity": 186.5264518280785}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424549.12/warc/CC-MAIN-20170723102716-20170723122716-00212.warc.gz"} |
http://crypto.stackexchange.com/questions?page=60&sort=votes | # All Questions
445 views
### AES CBC with fixed or predictable IV, what are the risks?
I am making a call with parameters from application "A" to a web application "B". Parameters are in the query string of the URL. They don't need to be hidden. But I want to be sure that nobody can ...
229 views
### Store hashed email and compare hash values
I have a number of different systems sending me email addresses, but I don't actually need the underlying email, just a hash of the email address. I know I can compare hash values to find matches ...
75 views
### What is the most secure way to let Alice “input” her private key, transfer it to the server and decrypt the message?
What is the most secure way to let Alice "input" her private key, transfer it to the server and decrypt the message? (Then destroy the private key on the server) Every connection is using SSL. ...
288 views
### What are monotonic and non monotonic access structures in ABE ?
While reading Attribute Based Encryption Schemes, I did not quite understand what a monotonic access structure or non monotonic access structures are. Is there a popular explanation of those terms ? ...
147 views
### What would the Internet be like without public-key cryptography?
The functioning of the Internet as we know it nowadays depends very heavily on public-key cryptography, including several key root systems that depend on its asymmetric properties. But what would it ...
90 views
### Ensuring integrity of a client side script
I want a script that is running on the client to compute a collision resistant hash and send it to a server. I need to ensure that the script which performs this hash is not altered in any malicious ...
3k views
### Decrypt digital signature using RSA public key with openssl
I have a digital signature that was created using the following algorithm: a SHA-256 hash of the body of the message is calculated. It is then signed using an RSA private key and the result is ...
59 views
### Secure MAC implies that probability of same tags on different messages is negligible
So let any secure MAC (message authentication code) be given. Intuitively, I think it is clear that the probability of getting the same tag on two different messages is very small, i.e. negligible. I ...
236 views
### How do unkeyed hash functions (for MDCs) provide security?
Unkeyed hash functions are, by definition, hash functions computed without a key. SHA-1 is an example. MDCs (message digest codes) are a subclass of unkeyed hash functions. How are unkeyed hash ...
157 views
### Montgomery Exponentiation - selecting input value R for a given BigInteger
I have Montgomery exponentiation working, but it's working quite slow. I suspect there are two reasons for this - I implemented it bit size instead of word size (I didn't realize at the time that ...
252 views
### Is there a security analysis of CryptDB?
Its interesting to see critical thinking being applied to cryptDB in contrast to all the hoopla around it here . cryptDB is not a major theoretical breakthrough but potpourri of technologies to make ...
446 views
### Crack SHA1 hash code
Given a set of SHA1 hash code, if we knew a portion of the plaintext for each particular hash code, and also the remaining portion of the plaintext is a constant value for each hash code. Is it ...
926 views
### How can I implement the “Multiplication Modulo” and “Addition Modulo” operations in IDEA?
I am currently working on IDEA (International Data Encryption Algorithm), and I don't know how to perform Multiplication Modulo and Addition Modulo. This is how IDEA operates: IDEA operates on ...
104 views
### Using pairings to verify an extended euclidean relation without leaking the values?
Let $P_i(x)$ be polynomials $i=1,...,n$, $s$ some value, and $g$ a generator of a group $G$ where the discrete logarithm is hard. Assume a prover wants to convince a verifier having access to the ...
678 views
### AES encryption for images [closed]
Why AES based encryption is not recommended for encrypting images with high redundancy in their content? for example, the encryption of an image which shows a bird in the blue sky. Most of the pixels ...
177 views
### Tree hash and multithreading for parallelism
I am using tree hash mode of Skein on a 16 core processor. Will it automatically employ parallelism using more than one cores or do I have to use multi-threading within the tree hash for parallel ...
629 views
### Calculating the inverse modularity of the determinant for Hill cipher
I'm trying to decrypt a message encrypted with Hill Cipher, but I don't understand how to find the determinant so it solves the equation $det * 1/det = 1 mod 26$. The determinant for my key matrix is ...
162 views
### verify contents, but not order
Is there an algorithm that can be used to verify the contents of a cyphertext, but not the order of the elements? I am thinking that a deck of cards could be shuffled, and it must be verifiable that ...
335 views
### How can the Diffie-Hellman key exchange be extended to three parties? [duplicate]
Possible Duplicate: Can one generalize the Diffie-Hellman key exchange to three or more parties? How can Alice, Bob, and Charlie share a common secret key using an extended version of the ...
343 views
### Storage of Private Keys
I'm building a bitcoin web application that will require all users to be assigned a wallet for adding funds to their account. I plan on exposing the public key to the user (the bitcoin address). Users ...
232 views
### Blowfish Encryptions in Bcrypt hashing algorithm
I am trying to understand the ExpandKey function used in the bcrypt hashing algorithm. As per the documentation of bcrypt hashing algorithm on USENIX here: ...
165 views
### Digital Certificate Chain Verification
Does a certificate contain the complete chain of all certificates up to the trusted root certificate, or does the program that verifies certificates have to fetch each parent certificate individually ...
54 views
### Design question for a service that signs messages on behalf of a user
I'm working on a messaging service that lets users author messages and have them signed with an RSA key. The key is used for identity purposes only and can be revoked, so compromising the private key ...
281 views
### Deniability of OTR messaging
I've been looking at the off-the-record messaging protocol, and there is something about its deniability property that seems strange to me. Consider the following scenario: In the course of an ...
143 views
### Do Quantum Key Distribution and Physical Unclonable Functions combine, and how?
I see there's a project to combine Quantum Key Distribution, Physical Unclonable Functions, and classical crypto, in order to secure a high speed (100Gb/s) optical link. While there does not seem to ...
937 views
### Deciphering a key from XOR encrypted cypher using boolean logic
Assume there's an unencrypted message A, and an encrypted message B. You know that message B was encrypted using a simple XOR method of A with a private key K, resulting in message B. Thus, B = A ⊕ K ...
280 views
### Expectation Value of the Index of Coincidence
Can Anybody tell me why the expectation value of the Index of Coincidence is the following? For a cipher text string of length $L$, where $n$ is the number of alphabet characters, $k_r = 1/n$, and ...
1k views
### Use of nonces in Kerberos and Needham–Schroeder protocols
Can you please explain me what is the reason of using nonces in the Needham-Schroeder and in the Kerberos protocols for Key Transport. In particular : Can you explain me the use of $N_A$? And ...
77 views
### Incorporating known ciphertext into new message
Which ciphers $c_k$ enable one to define a joining function $f_{c,k}$ such that: $$f_{c,k}(c_k(p_1), p_2) = c_k(p_1 + p_2)$$ That is, the result of joining some known ciphertext $c_k(p_1)$ with a ...
332 views
### How to represent a 32-byte SHA2 hash in the shortest possible string?
I'm calculating a SHA2 hash of a certain sensitive key value. I need to store files on disk using this hash a directory path prefix. So lets say I hash the key value 150023, I get a 32-byte value ...
115 views
### Tips on conceiving safe software messaging platform
I have a personal project in mind in which I plan to use cryptography in order to let the users be confident in the fact that, even if all data get's stolen, it would be virtually impossible to crack ...
316 views
### Brute forcing Cardan grille
Having a "rotating" square Cardan Grill with sides long n cells, how could i determinate the cost of a brute-force attack? How many configurations should be considered to perform an exhaustive ...
202 views
### Twofish key length
I'm using twofish. Say you have a 120 bit key, is there any difference in using a 128 bit bit keylength or a 256 bit one? Since the most significant digits of the key will be allo zero in both cases, ...
374 views
### Is this fixed length MAC unforgeable?
Consider the following fixed length MAC for messages of length $\ell(n)=2n-2$ using a pseudorandom function $F$: On input of a mesage $m_0||m_1$ ($|m_0| = |m_1| = n-1$) and a key $k \in \{0,1\}^n$, ...
308 views
### Security analysis of a “one-time pad” type hill cipher
Suppose the Hill cipher were modified to something like a one-time pad cipher, where Alice wants to send a message to Bob, and she chooses a key matrix randomly everytime a new message is sent (and ...
130 views
### How is the password sent across for verification?
I am familiar with Challenge-Response Authentication scheme. However I would like to know actual message content sent between client and server for 1 way authentication, over an insecure channel. ...
537 views
### iSeries (AS/400) Database File: password encryption
I am helping with a project in which an old software system on an iSeries is having a brand new .NET UI applied to it. It's going well... except... In order to allow users to login and maintain ...
32 views
### Why Victor must not know which tunnel Peggy chooses?
In the classic description of Zero Knowledge Proof of Knowledge, Victor must wait outside the entrance to the cave while Peggy goes to the fork and choose a side. It's only once Peggy has entered a ...
52 views
### Right usage of PBKDF2 and login system questions
I'm developing the login part of my application and I was asking myself how to do it right. My application stores some users' data server-side that must be inaccessible from us, so all the content ...
50 views
I've read a paper that says that one of the "assumptions" that it makes is secure broadcast. Why is this an assumption? Can't a broadcast channel be built with simple primitives--i.e. encrypt and sign ...
66 views
### Can I use asymmetric encryption for more powerful write-protection?
I have a game that I am developing which uses a locally-stored internal database. It does not matter to me whether a user can read the file (which, to me, means I don't need a symmetric cipher), but I ...
99 views
### Designing a secure IM protocol
I haven't been programming much lately, so I figured I would make an encrypted instant-messaging program (read: Skype clone) with groups to hone my skills. That's not important though, I can do that. ...
100 views
### Burmester-Desmedt (BD) key agreement protocols
How do you perform the Burmester-Desmedt key exchange Protocol, in Diffie-Hellman extension protocol? I've been searching high and low, but couldn't locate an example on how to prove if all the party ...
75 views
### Secrete message sharing using N private keys
I'm looking for an encryption scheme that could possibly meet the following requirements. Let, $X = plaintext$ $M = (pub, priv)$, where $priv == (Ky + Kz)$ Given ...
27 views
### OTP sequence uniqueness
Is it possible to use Time-based One Time Password (TOTP) algorithms to generate tokens from multiple channels (where a channel is defined as a "user" using a certain shared key $K_c$ for channel ...
82 views
### How unsafe is to share parts of a password
The best way to ask this question and have a concrete answer is to set a very specific example. This is the example I would be interested in reading an answer to: Consider that my friend Alice has ...
57 views
### crypto design with AES256 MODE OFB
I have the following APP Design key store (contains 1:n user payload as 1:n key store entries) key store salt (is used to extend to key store password to 256 bits, stored in clear text) key store ...
104 views
### How can I simulate and measure brute force hacking using RSA?
I want to implement brute force hacking using RSA. How can I simulate and measure time duration? What software (or software-based simulator) would be needed? | 2014-03-09 17:54:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5829309821128845, "perplexity": 2185.638453294483}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010048333/warc/CC-MAIN-20140305090048-00063-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://www.etiquettehell.com/smf/index.php?topic=75234.msg3040174 | News: There is a new Ehell Kindness Project! Check it out in the "Extending the Hand of Kindness" folder or here: http://www.etiquettehell.com/smf/index.php?topic=139832.msg3372084#msg3372084
• August 28, 2016, 09:07:16 AM
### Author Topic: What weird/outrageous requests have you seen on Freecycle/Craigslist? (Read 256141 times)
0 Members and 1 Guest are viewing this topic.
• Member
• Posts: 12714
• Not all those who wander are lost
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1185 on: October 21, 2013, 08:23:35 AM »
We are going to start posting wanted ads for wood scraps, for different small projects. I have been seeing a lot of wanted for electronic game systems, that would generally cost more than $400 at the store. “All that is gold does not glitter, Not all those who wander are lost; The old that is strong does not wither, Deep roots are not reached by the frost." -J.R.R Tolkien #### VorFemme • Member • Posts: 15477 • It's too darned hot! (song from Kiss Me, Kate) ##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist? « Reply #1186 on: October 23, 2013, 10:38:15 AM » I posted Halloween costumes the other day - so far one person has thought that it was$10 for all four (ha, ha, ha, nope - $40 for all four) and another one thought I was in Kingswood - which is at least a thirty mile drive away THROUGH the heart of Houston, Texas - so a lot longer than 30 minutes. Not worth it for two of the costumes, unless she "meets me halfway". (And she decided that SHE wasn't driving all the way to Sugar Land - my ad mentions my location - so why she replied to it, thinking that I was close to where she lives, I have no earthly idea.) Plus sized costumes - althugh I think they are closer to 16W/18W than the 16W through 22W that the package states "will fit....". Because I bought them just before moving, didn't try them on again until after Thanksgiving & Christmas had passed (and even the leftovers had been eaten) - I never washed them, so I can't claim that they shrank. ((sigh)) There is a reason I'm selling them...I really don't expect to be 1) that size again soon or 2) be willing to wear the "sexy" versions of these costumes if they ever do come close to fitting. I'd rather make one that I know is of sturdier fabric...fits better...and can be worn to a Rennaissance Faire, not just Halloween! In fact, I bought some purple stretch panne velvet yesterday to make a costume to wear to the October Red Hat meeting AND the Renn Faire (depending on the weather - it will not be a summer costume to wear to Atlanta in May or June). « Last Edit: October 23, 2013, 07:18:14 PM by VorFemme » Let sleeping dragons be.......morning breath......need I explain? #### cmurf1960 • Member • Posts: 9 ##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist? « Reply #1187 on: October 26, 2013, 10:00:19 PM » Quote WANTED: i need a iphone mine is broken i need a phone and my parent are not going to buy me one cuz they want me to work for one Well, then get a job and WORK for one, you stupid brat! Sheesh! #### Cherry91 • Member • Posts: 919 ##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist? « Reply #1188 on: October 28, 2013, 07:10:10 AM » We are going to start posting wanted ads for wood scraps, for different small projects. I have been seeing a lot of wanted for electronic game systems, that would generally cost more than$400 at the store.
The PS4 and Xbox One are due out this month. What are the odds loads of people ask for the previous generation of consoles, because after all, they're "worthless" now?
All will be well, and all manner of things will be well.
#### Amara
• Member
• Posts: 2409
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1189 on: November 04, 2013, 02:57:32 PM »
I can't resist sharing these two "Housing Sought" ads on my Craigslist. Such special people shouldn't have to do something as mundane as pay rent, should they? (Ads are copied as written.)
Quote
It's BOLD yet, it's true. I truly understand that I'm here to have a positive impact in this community which will ripple out into our global community. It's my hearts calling and sole mission is to create lasting, positive, transformational change.
I am seeking a host to provide a comfortable, peaceful place where I can live in inspiration and serve as the launching pad for the work that I do. I could give you many reasons why you would want to host someone like me and explain at length how much your life will be positively impacted by opening your heart and home to me. Instead I am trusting that the exact person who knows you are the one and this is the time will reply and in life's perfection, we will meet and be that change we wish to see in the world.
Thank You for answering the call
Quote
I'm launching a tv network startup and I'm looking for a room to stay with someone for two to three months. Smart, organized, neat, minimal. Glad to help out. Nice guy, nice clothes, small cute car, nice to have around, polite, intelligent. Thanks!
Great with people, families, church. Ok.
Need bed, privacy, wifi. I am quiet. I prefer a polite, safe, appreciative, quiet, family, nature or work atmosphere.
• Member
• Posts: 12714
• Not all those who wander are lost
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1190 on: November 04, 2013, 03:35:54 PM »
I bet people are lighting up their phones with offers.
I guess it is so old school to pay one's own rent.
“All that is gold does not glitter, Not all those who wander are lost; The old that is strong does not wither, Deep roots are not reached by the frost."
-J.R.R Tolkien
#### VorFemme
• Member
• Posts: 15477
• It's too darned hot! (song from Kiss Me, Kate)
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1191 on: November 04, 2013, 04:11:48 PM »
Anyone asking for nearly new electronics (phone, game system, laptop, tablet, or whatever - sometimes the specs mean it is going to have to be new to match their "needs") for FREE because they need it for school, looking for job, foster kid(s), grandkids that they raising, or any other "please help this poor, deprived person who NEEDS this expensive item for whatever reason" type of sob story.
Umm...I worked my way through college, worked for years, and complain about the price of these things - if I want to spend my money to get myself a new "toy" - I might...or I might look for a used toy to save money. I don't go begging people to give me THEIR toy! I've got some self respect (and parents who would still wallop my greedy little backside even though they are both in their 70s).
Yes, I just read another one recently in the local Freecycle, why do you ask?
Let sleeping dragons be.......morning breath......need I explain?
#### Amara
• Member
• Posts: 2409
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1192 on: November 04, 2013, 05:24:21 PM »
And the begging holiday season begins!
#### Phoebelion
• Member
• Posts: 424
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1193 on: November 04, 2013, 07:06:32 PM »
Freecycle has all but died in my area. I wonder if it had anything to do with the moderator?
#### Thipu1
• Member
• Posts: 7908
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1194 on: November 04, 2013, 07:35:20 PM »
Amara, would it be wrong to assume that the first person seeking lodging in your post has a name like Unicorn Moonbeam?
S/he would probably want to smudge your house before moving in.
Thanks for posting this. It gave me a good giggle.
« Last Edit: November 04, 2013, 07:36:57 PM by Thipu1 »
#### jedikaiti
• Swiss Army Nerd
• Member
• Posts: 3570
• A pie in the hand is worth two in the mail.
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1195 on: November 04, 2013, 09:07:20 PM »
Freecycle has all but died in my area. I wonder if it had anything to do with the moderator?
Check for other, similar programs. In our area, Freecycle had a bit of a civil war a while back, and many of the mods left Freecycle, started another group, and took their lists with them.
What part of v_e = \sqrt{\frac{2GM}{r}} don't you understand? It's only rocket science!
"The problem with re-examining your brilliant ideas is that more often than not, you discover they are the intellectual equivalent of saying, 'Hold my beer and watch this!'" - Cindy Couture
#### Amara
• Member
• Posts: 2409
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1196 on: November 04, 2013, 09:51:33 PM »
Quote
Amara, would it be wrong to assume that the first person seeking lodging in your post has a name like Unicorn Moonbeam?
Hmm, it might be worth asking.
ETA: You know that 419 website where scambaiters take on scammers and distract them from their money-making ventures by playing on their greed? Well, I have admit I just had a vision of me playing this guy (I am assuming) like that. I even ran it out in my head. So delicious, so amusing, so deserving.
« Last Edit: November 04, 2013, 10:03:53 PM by Amara »
#### Friday
• Member
• Posts: 1628
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1197 on: November 05, 2013, 09:34:38 AM »
On the housing requests - not freecycle though...
Background:
I'm an avid participant in couchsurfing. I both host (mainly) and surf (rarely) and also attend local events. The site I use is a social media type of site and you can get a feel for the person before agreeing to host/surf with them.
1-4 nights surfing is the average. Couch surfing is about community, NOT a free place to stay and/or a substitute for a hotel. You DON'T treat your host like you would hotel staff. Its accepted that you will interact with your host/surfer. Surfers have cooked for us, we've had dinner parties, gone to movies together, gone to other events, and even took the most recent one with us on a driving vacation over the weekend to the Wisconsin Dells.
Stories:
1: I was contacted by a woman who was part of a band for her, her hubby and a bandmate to stay with us Sat/Sun/Mon nights. I agreed. Day before, she calls me and changes it to just staying with us the Monday night. they have to leave town early on tuesday. . Ok... I suppose, we'll socialize on Monday evening (I work M-F, days). Monday comes... I wait for a call (I don't give my address out until surfers are on their way).... nothing... evening comes... nothing.... 11 pm comes, I go to bed. Wake up in the morning and I have a couple texts and voice mails from her. Their show (remember, in a band) got done at 1 AM and they expected to come crash with me then. Totally being treated like a hotel. Later, I read some comments on her couchsurfing profile left for experiences after mine and found out they did that to other people - changing plans at last minute, being treated like a hotel. They are no longer active on the site.
2: A woman sent me an email (not a stay request, that's important) and asked if she could surf with us for a weekend for a con at which she would be selling jewelry, I said sure, go ahead and send me a request. There's a separate section for requests and it keeps a calendar for you and reminds you. She didn't. Apparently she operated on "she said yes in email, so all is good". i got busy, didn't check that site's email (it used to be awkward) and just keep checking surfing requests. A few days after her con, she wrote a very nasty feedback about how I told her yes and then let her down, resulting in her having to pay a hotel bill and how she felt I owed her money. Ok, lots of things wrong with that. The community is NOT about being a hotel, I didn't actually have a request from her, and it's my house - the community tells you that you can change your mind at any time. I contacted the site admins and let them know. THe feedback was quickly removed and then she posted it again! I contacted them again, they confirmed that I was right and she is also no longer active on the site.
3: Another person sent me a request for THREE MONTHS.... this is way outside of the guidelines. I took someone for 5 weeks once - she was a college student here for an unpaid externship and started out asking by saying she'd contribute to utilities and food. This other person was coming to town for three months for a job and wanted to "experience" the area with people who lived here already. No offer of compensation. Also, when I looked at her profile, she had very little feedback from people she stayed with/hosted and a LOT of comments from people she traveled with that betrayed a lot of alcohol and perhaps drug use, as well as comments about urban "exploring" that seemed like vandalism, trespassing, etc.
4: I will cut and paste this one. This person was QUICKLY removed from the site. I suppose all sites get this sort of thing...
Hi . How are you ?)
I am from Baku . I want to immigrate to Canada . I would like even to find a girl for marriage to stay in Canada . If any help from you i get , i will be glad and pleased for you .
kisses for you , dear )
regards from Azeri Guy.
(p.s: you can see me when i am online)
my response: I have reported your request to the moderator of this site. This is neither a dating site or a marriage for green card scam site.
I also question your intelligence in sending this request for help scamming citizenship of CANADA to a MARRIED woman living in the UNITED STATES!
Go away, we don't need your type in this pleasant community of like minded travelers.
#### Amara
• Member
• Posts: 2409
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1198 on: November 18, 2013, 01:40:37 PM »
I'd put this in the outrageous category. I just saw this ad in the "Housing Wanted" section and laughed. Suuuuuuuuuuuure, let me get right on this because I know () I will have so much competition to house you for free:
Quote
I am described by a friend as about 6 feet tall and 160 pounds, a guy in his fifties that is "differently oriented". I have been living as a homeless man -- what he calls an "Urban Outdoorsman" for quite a while. I appear to be a "deliberately" a "different looking" man with my hair at near-shoulder length and a "distingished-looking" beard but wearing rumpled, slept-in-looking jeans and shirts and jackets. The friend says, "You choose to look like a street-person. I suspect you have a genetic predisposition against anyone claiming authority over you. You choose to live quite a little freer than most of us can tolerate. We 'ordinary humans' think everybody should live inside frame-and-stucco boxes like we do." If you would like to have a helpful, cheerful, nonconformist around, I can cook and clean and drive and run errands and that sort of things in exchange for a spot to be out of the rain and cold.
If you have any interest in meeting at, say, a coffee shop anywhere around Santa Barbara, please reply to this posting which is done by the friend hoping to bring two nice humans mutual benefits. I have a car and a valid California Driver License. If meeting a genuine "homeless street person" -- a real, live Santa Barbara homeless man -- interests you, send an email reply to this posting.
#### MommyPenguin
• Member
• Posts: 5468
##### Re: What weird/outrageous requests have you seen on Freecycle/Craigslist?
« Reply #1199 on: November 19, 2013, 11:53:11 AM »
Well, I don't know why people would need to look on craigslist to meet a "genuine homeless man," but somebody willing to cook, clean, and run errands in exchange for a place to stay, that might not be so bad. Especially if it's more somebody looking for a place to crash, versus somebody needing a room and amenities.
Emily is 9 years old! 1/07
Jenny is 7 years old! 10/08
Charlotte is 5 years old! 8/10
Megan is 3 years old! 10/12
Lydia is 1 year old! 12/14 | 2016-08-28 14:07:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3569321632385254, "perplexity": 4863.186949149577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982939917.96/warc/CC-MAIN-20160823200859-00020-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://cran.rstudio.com/web/packages/AvInertia/vignettes/how-to-analyze-data.html | # Determine the inertial characteristics of a flying bird.
This package was developed to determine the center of gravity and moment of inertia for a bird in a general flight configuration. This code assumes that birds are a composite structure of simple geometric shapes. For details on the specific assumptions please refer to:
library(AvInertia)
Generally, you should already have all of the necessary measurements loaded into R in the form of a database. We have included the data set from our paper. The bird properties are reported in the metric system and all measurements have been taken with the origin placed at the bird (vehicle) reference point (VRP).
data(dat_id_curr, package = "AvInertia")
data(dat_bird_curr, package = "AvInertia")
data(dat_feat_curr, package = "AvInertia")
data(dat_bone_curr, package = "AvInertia")
data(dat_mat, package = "AvInertia")
data(clean_pts, package = "AvInertia")
## 1. Determine the center of gravity of the bird’s torso (including the legs)
dat_torsotail_out = massprop_restbody(dat_id_curr, dat_bird_curr)
## 2. Calculate the inertia of the flight feathers about the tip of the calamus
feather_inertia <- compute_feat_inertia(dat_mat, dat_feat_curr, dat_bird_curr)
## 3. Determine the center of gravity of one of the bird’s wings
dat_wing_out = massprop_birdwing(dat_id_curr, dat_bird_curr, dat_bone_curr, dat_feat_curr, dat_mat, clean_pts, feather_inertia, plot_var = 0)
Visualize the center of gravity of each wing component in the x and y axis
dat_wing_out = massprop_birdwing(dat_id_curr, dat_bird_curr, dat_bone_curr, dat_feat_curr, dat_mat, clean_pts, feather_inertia, plot_var = "yx")
or the y and z axis
dat_wing_out = massprop_birdwing(dat_id_curr, dat_bird_curr, dat_bone_curr, dat_feat_curr, dat_mat, clean_pts, feather_inertia, plot_var = "yz")
## 4. If computing for asymmetric case, input a different set of clean points (clean_pts)
Note: that in this case the wing should still be input as it it is on the right-hand side of the bird. The following code will adjust.
## 5. Combine all data and obtain the center of gravity, moment of inertia and principal axes of the bird
curr_full_bird = combine_inertialprop(dat_torsotail_out,dat_wing_out,dat_wing_out, dat_id_curr, dat_bird_curr, symmetric=TRUE)
This will return a long format data frame with all the individual components I about the VRP and both the full bird I about the VRP and about the full center of gravity. | 2022-05-25 06:45:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3384501338005066, "perplexity": 2605.3439083202015}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662580803.75/warc/CC-MAIN-20220525054507-20220525084507-00682.warc.gz"} |
https://study.com/academy/answer/consider-a-four-year-project-with-the-following-information-initial-fixed-asset-investment-470-671-straight-line-depreciation-to-zero-over-the-four-year-life-zero-salvage-value-price-35-variable-co.html | # Consider a four-year project with the following information: initial fixed asset...
## Question:
Consider a four-year project with the following information:
initial fixed asset investment=$470,671 straight-line depreciation to zero over the four-year life zero salvage value price=$35
variable costs=$25 fixed costs=$184,307
quantity sold=79907 units
tax rate=32%.
Calculate the sensitivity of the OCF to changes in the quantity sold.
## Operating Cash Flow:
Operating cash flow include cash inflow occur from the core business activities such as production and selling of goods and services. It does not include cash inflow from investment or financing activities.
For 79,907 units
{eq}OCF \ = \ \left [ \left ( Price \ - \ Variable \ cost \right ) \ \times \ Quantity \ - \ Fixed \ cost \right ]\left ( 1 \ - \ tax \right ) \ + \ depreciation\left ( tax \ rate \right ) \\ OCF \ = \ \left [ \left ( 35 \ - \ 25 \right ) \ \times \ 79,907 \ - \ 184,307 \right ]\left ( 1 \ - \ 0.32 \right ) \ + \ \frac{470,671}{4\left ( 0.32 \right )} \\ OCF \ = \ 418,038.84 \ + \ 37,653.68 \\ OCF \ = \ 455,692.52 {/eq}
For 80,000 units
{eq}OCF \ = \ \left [ \left ( Price \ - \ Variable \ cost \right ) \ \times \ Quantity \ - \ Fixed \ cost \right ]\left ( 1 \ - \ tax \right ) \ + \ depreciation\left ( tax \ rate \right ) \\ OCF \ = \ \left [ \left ( 35 \ - \ 25 \right ) \ \times \ 80,000 \ - \ 184,307 \right ]\left ( 1 \ - \ 0.32 \right ) \ + \ \frac{470,671}{4\left ( 0.32 \right )} \\ OCF \ = \ 418,671.24 \ + \ 37,653.68 \\ OCF \ = \ 456,324.92 \\ Sensitivity \ = \ \frac{Change \ in \ OCF}{Change \ in \ quantity} \\ Sensitivity \ = \ \frac{\left ( 456,324.92 \ - \ 455,692.52 \right )}{80,000 \ - \ 79,907} \\ Sensitivity \ = \ \frac{632.4}{93} \\ Sensitivity \ = \ 6.80 {/eq} | 2020-01-18 21:03:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000098943710327, "perplexity": 2461.3637265180714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00258.warc.gz"} |
https://www.gamedev.net/topic/623022-simple-question-about-2d-wrapped-coordinate-system/ | • Create Account
Simple question about 2D wrapped Coordinate System
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
4 replies to this topic
#1joey.enfield Members
Posted 06 April 2012 - 03:41 AM
Hi all,
I have a question. I'm playing around some 2D game dev, particularly AI and i've come onto a problem that I would love some help in solving. I know that there must be some simple solution to my problem and I'm over complicating it, I've looked but cant seem to find anything.
The problem is that I have a 2D world that wrapped around a 2D playing field, i.e. (when x > wide : x = 0 ...). I'm trying to do simple follow and flee type AI and I had the obvious problem that when one object is following another which goes off the edge of the screen and appears at the opposite side, it does not follow off the edge but has to turns around.
The only robust way that I could come up with is to calculate is as follows :
Assume that the pos can exists at 9 position (for each side of the screen) i.e
In the following X is following # (the real grid in in the middle one)
__________________________________
|..........|..........|..........|
|..........|..........|..........|
|..........|..........|..........|
|........1*|........2*|........3*|
|__________|__________|__________|
|..........|..........|..........|
|..........|..x.......|..........|
|..........|..........|..........|
|........4*|........5#|........6*|
|__________|__________|__________|
|..........|..........|..........|
|..........|..........|..........|
|..........|..........|..........|
|........7*|........8*|........9*|
|__________|__________|__________|
To determine which point to "Follow" I determine which point is closest (1-9) and then I move
the point i do my calculation to that one.
Vector2D object; //This is the position of my object
Vector2D pos; //This is the position the object is tracking in the "Real" grid
Vector2D rst; //This is the position of the object even if it goes off the edge of the screen;
Vector2D[] points = new Vector2D[9];
points[0] = new Vector2D(pos.x - width, pos.y - height);
points[1] = new Vector2D(pos.x, pos.y - height);
points[2] = new Vector2D(pos.x + width, pos.y - height);
points[3] = new Vector2D(pos.x - width, pos.y);
points[4] = new Vector2D(pos.x, pos.y);
points[5] = new Vector2D(pos.x + width, pos.y);
points[6] = new Vector2D(pos.x - width, pos.y + height);
points[7] = new Vector2D(pos.x, pos.y + height);
points[8] = new Vector2D(pos.x + width, pos.y + height);
int minIndex = 0;
float minDist = 0;
float dist = 0;
for (int i = 0; i < 9; i++) {
dist = (float) points[i].distanceSq(object);
if (i == 0 || dist < minDist) {
minDist = dist;
minIndex = i;
}
}
rst.x = points[minIndex].x;
rst.y = points[minIndex].y;[/font]
Can anyone give me any advice on this? What I do above works provide I move the positon of the object i'm following before I start my calculations to follow, but this seems a bit overkill. There must be a simpler way of doing this.
Any help or direction would be greatly appreciated.
Regards
Joey
#2Waterlimon Members
Posted 06 April 2012 - 04:25 AM
not sure if this works, but i would take the one with the smaller absolute size (tar.x and bot.x being x positions of target and bot, mao.x being map x size) and use the non-absolute version as the relative position and get the direction from that.
poor design pseudocode:
realx=tar.x - bot.x
wrappedx=(map.x - bot.x) + tar.x //These for each axis
relativex=abs(realx) < abs(wrappedx) ? realx : wrappedx;
relativey=abs(realy) < abs(wrappedy) ? realy : wrappedy;
Direction=MakeThisVectorAnUnitVectorOkay(Vector(relativex,relativey));
(edit: removed y axis for some reason)
o3o
#3Brother Bob Moderators
Posted 06 April 2012 - 04:27 AM
First some observations:
• You can do the calculation for each axis individually.
• The distance you have to move along any axis is never larger than half the size of the square.
• If the direct distance on the surface is longer than half the size of the square, you move the other direction instead.
Point 1 means we can reduce the problem into thinking only in one dimension. You don't have to think about this problem in two dimensions; one dimension is enough, and you can replicate the calculations to both dimensions (or to however many dimensions you want, in fact).
To begin with, calculate the distance you would have to travel if this wasn't a toroidal surface. Say you want to move from point A to point B, so the distance to move is then V=B-A.
Point 2 states that we should not have to move longer than half the size. Point 3 furthermore states that if we do, then we move in the opposite direction instead. With that in mind, we can wrap the vector V if it is too long.
A = start point
B = end point
V = B-A
if(abs(V) > size/2)
V = V - size
end
edit: Repeat for all dimensions individually as stated in point 1 of course... and apparently beaten by a minute or two.
#4joey.enfield Members
Posted 06 April 2012 - 07:35 AM
Thanks so much. I look now and see how obvious it was. Heres the code I used in the end if anyone wants it.
[font=courier new,courier,monospace]
rst.x = pos.x-obj.x;
rst.y = pos.y-obj.y;
if (abs(rst.x) > sizeX / 2) {
if (rst.x > 0) {
rst.x = pos.x + sizeX;
} else {
rst.x = pos.x - sizeX;
}
} else {
rst.x = pos.x;
}
if (abs(rst.y) > sizeY / 2) {
if (rst.y > 0) {
rst.y = pos.y + sizeY;
} else {
rst.y = pos.y - sizeY;
}
} else {
rst.y = pos.y;
}
[/font]
Thanks for the help
Joey
#5sjhalayka Members
Posted 07 April 2012 - 12:27 PM
That's right on, and I wish I had replied. The space you are talking about is a toroidal space.
Analogously, in the case of a spherical space, the shorter distance is called the orthodromic distance. Fun times.
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. | 2017-01-19 23:47:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2958907186985016, "perplexity": 1203.946494046919}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280761.39/warc/CC-MAIN-20170116095120-00066-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/34329-taylor-s-development-question.html | # Math Help - Taylor's development question
1. ## Taylor's development question
Is there a way to obtain the formula of the Taylor's expression of continuous functions. For example, how can I obtain the Taylor's development of the function $\sqrt{x}$ without trying to find it by writing down the first derivative, the second one, etc?
2. ## WEll
Originally Posted by arbolis
Is there a way to obtain the formula of the Taylor's expression of continuous functions. For example, how can I obtain the Taylor's development of the function $\sqrt{x}$ without trying to find it by writing down the first derivative, the second one, etc?
What you should do is memorize all the Maclaurin formulae then you can go.."what is $e^{x}$" and bam you know its $\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and then you will go 'what is the power series for $e^{-x^2}$?" and you will go "it is $\sum_{n=0}^{\infty}\frac{(-x^2)^{n}}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}$"
3. Ok, thank you. So there's no formal way to obtain these formulas. I must write down the first derivatives and try to guess the nth one. Anyway, I'm having some trouble with the $\sqrt{x}$'s one. I would appreciate if you give me the formula .
4. ## No
Originally Posted by arbolis
Ok, thank you. So there's no formal way to obtain these formulas. I must write down the first derivatives and try to guess the nth one. Anyway, I'm having some trouble with the $\sqrt{x}$'s one. I would appreciate if you give me the formula .
If you read what I said you could just memorize the power series...and imput the different values...and for $\sqrt{x}$ I would suggest but am not positve you would use $(1+{x-1})^{\alpha}$ with $\alpha=\frac}1}{2}$...and then apply this Taylor series - Wikipedia, the free encyclopedia look for the one that looks like the one I gave you with the alpha
5. Originally Posted by arbolis
Ok, thank you. So there's no formal way to obtain these formulas. I must write down the first derivatives and try to guess the nth one. Anyway, I'm having some trouble with the $\sqrt{x}$'s one. I would appreciate if you give me the formula .
I haven't to time to give a detailed response right now, but this should be enough to get you started.
Firstly when you take your first few derivatives do not attempt the simplify the fractions that may lead to you missing the general formula.
also, for this problem you will need to come up with a formula for the product first n odd numbers, can you see why ?
If you need help with that just ask.
Bobak
6. also, for this problem you will need to come up with a formula for the product first n odd numbers, can you see why ?
I don't understand well what you mean. I've wrote some derivatives of $\sqrt{x}$. I'm trying to guess the nth derivative, and till now what I have is $(-1)^{n}(1-2n)x^{1/2-n}/2^n$ but I know I'm missing a product in the numerator.
7. I think I got it. If you replace the (1-2n) in my previous formula by the product of (2i-1), from 0 to n, I think it works! Since I've to go now, I'll check that later. | 2015-07-30 17:42:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9389193058013916, "perplexity": 225.59702017267617}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042987402.78/warc/CC-MAIN-20150728002307-00019-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.pims.math.ca/scientific-event/180328-raliv | ## Representations in Arithmetic Lectures: Ila Varma
• Date: 03/28/2018
• Time: 15:15
Lecturer(s):
Ila Varma, Columbia University
Location:
University of British Columbia
Topic:
Counting D_4-quartic fields ordered by conductor
Description:
We consider the family of D_4-quartic fields ordered by the Artin conductors of the corresponding 2-dimensional irreducible Galois representations. In this talk, I will describe ways to compute the number of such D_4 fields with bounded conductor. Traditionally, there have been two approaches to counting quartic fields, using arithmetic invariant theory in combination of geometry-of-number techniques, and applying Kummer theory together with L-function methods. Both of these strategies fallshort in the case of D_4 fields since counting quartic fields containing a quadratic subfield of large discriminant is difficult. However, when ordering by conductor, these techniques can be utilized due to additional algebraic structure that the Galois closures of such quartic fields have, arising from the outer automorphism of D_4. This result is joint work withAli Altug, Arul Shankar, and Kevin Wilson.
Other Information:
Location: ESB 4127
This lecture is part of the Focus Group on Representations in Arithmetics. | 2023-02-01 18:43:45 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8004218339920044, "perplexity": 3318.723658930523}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499949.24/warc/CC-MAIN-20230201180036-20230201210036-00291.warc.gz"} |
https://www.statistics-lab.com/%E7%BB%9F%E8%AE%A1%E4%BB%A3%E5%86%99%E9%87%91%E8%9E%8D%E7%BB%9F%E8%AE%A1%E4%BB%A3%E5%86%99financial-statistics%E4%BB%A3%E8%80%83aem4070-2/ | ### 统计代写|金融统计代写Financial Statistics代考|AEM4070
statistics-lab™ 为您的留学生涯保驾护航 在代写金融统计Financial Statistics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写金融统计Financial Statistics代写方面经验极为丰富，各种代写金融统计Financial Statistics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 统计代写|金融统计代写Financial Statistics代考|Main aspects of integral model
The period of time for which the integral model is constructed, should consist of a few waves of the underwriting cycle. The integral model is multi-component. It includes a number of different partial models, with each different phase of the
cycle. Within each partial model, the specific behavior of the market as a whole and of the individual companies on it has its own internal reasons; its investigation will be our most important task.
Thus, we imbed the theory of competition-originated underwriting cycles into the general theory of complex reflexive systems. We emphasize the role of human error, and of incomplete understanding of the real market situation. Our challenge is the development of an integral model suitable for quantitative, rather than merely qualitative analysis.
Of particular interest to us are long-term strategies, rather than annual control decisions. This is because we cannot be sure that a growing company will not face solvency problems in the next few years, even though at present it successfully fulfills all solvency requirements set by the regulator. Moreover, the insolvency of such a company may be unexpected to an external observer, especially after growth in the volume of its business and its revenue for several years due to the influx of insurers. In other words, besides ruin within an insurance year, another threat is the inability to meet the minimum solvency requirements in the next new year $^{8}$, known as default. This threat is especially grave when a company grows, as the capital needed to maintain solvency may grow even faster than its revenues.
## 统计代写|金融统计代写Financial Statistics代考|Factors used in quantitative analysis
We have already discussed (see Section 1.4.4.1) the interconnection between underwriting cycles, price competition and migration of insureds. To continue, let us say that the insurers’ behavior on the competitive insurance market has internal causes linked with the following observations.
The insureds who desire to reduce their expenses may opt to switch insurer $^{9}$, which explains the possibility of business expansion by means of price cut amongst aggressive insurers. Besides that, on a profitable market, price cuts usually aim to show greater profit in the annual report; this is directly connected with income on equity. Therefore, a second factor is the desire of insurance managers to enhance the attractiveness of the company in the opinion of investors. Further, it is required (e.g., by regulation) to meet solvency requirements. In principle, managers are responsible not only for the current activity of the company, but for
its long-term survival. To protect insureds and creditors from the consequences of insolvency, all companies are inspected periodically by regulators, which keeps the competitive zeal of their managers in check.
For a quantitative analysis of rational strategies and internal causes of insurers” behavior in each partial model, we will consider the combined influence of three factors: Expansion, Revenue and Solvency. In [123], it is called $E R S$-analysis. It is consistent with the opinion of practitioners (see, e.g., [173]) and theorists (see, e.g., [102]) that since single-factorial approaches are unable to explain the real causes of underwriting cycles and crisis phenomena satisfactorily, multifactorial approaches should be applied instead.
In the following chapters, we will measure solvency in monetary units ${ }^{10}$ when performing quantitative $E R S$-analysis. To do this, we turn to the concept of nonruin capital which allows solvency to be annually maintained at a predetermined level. In Lundberg’s collective risk model (see Section 1.5.2), or in the diffusion risk model (see Section 1.5.4), the non-ruin capital is defined as a solution to a nonlinear equation, in which the left-hand side is the probability of ruin within finite time, and the right-hand side is the small number $\alpha \in(0,1)$, i.e., the predetermined solvency’s level. The non-ruin capital is harder to evaluate than the probability of ruin, but its use makes $E R S$-analysis more intuitive and instructive.
## 统计代写|金融统计代写Financial Statistics代考|Driving forces behind the cycles and two main causal connections
The casualties of misfortune, accidental errors, stubbornness of some shareholders, or a reluctance of managers to follow good advices are hardly among the main
driving forces of underwriting cycles ${ }^{15}$. These real driving forces come from the difference between the market participants’ interests.
On the profitable market, the typical interest of
• a company which recently entered the market is the growth of its business. Such a growth-seeking company is often impelled, regardless of potential loss, to resort to price cut to attract new customers. This move is not illegal in itself. However, it can lead to cumulative errors in the company’s assessment of its long-term financial strength. For an aggressive company, the main mistake in the perception of reality lies usually in inaccurate evaluation, or in neglecting the evaluation, of its own solvency. The negative consequences of these errors are most fully manifested when the profitableness of the market falls;
• an incumbent company that has a large portfolio is getting stable profit from conducting regular insurance operations. But this business goal of a profitseeking company can sometimes be achieved through indirect means. For example, wanting to force the aggressive newcomers to change their goal to profit-seeking, the incumbent company may resort to a reduction of its own prices. Such sacrifice of possible income, an unnatural move on the market without competition, is an instrument for protecting a position on the market with growing competition;
• a company that tries to take immediate advantage of the current situation is using any chance to increase its volume and income, as circumstances permit. This pursuit of mercantile interests sometimes leads the opportunistic company to undertake a mercenary behavior.
## 统计代写|金融统计代写Financial Statistics代考|Driving forces behind the cycles and two main causal connections
• 一家最近进入市场的公司是其业务的增长。这种寻求增长的公司往往会被迫不顾潜在的损失，诉诸降价来吸引新客户。这一举动本身并不违法。但是，它可能导致公司对其长期财务实力的评估出现累积错误。对于一家进取的公司来说，对现实认识的主要错误通常在于对自身偿付能力的评估不准确，或者忽略了评估。当市场的盈利能力下降时，这些错误的负面后果最为明显；
• 一家拥有庞大投资组合的老牌公司正通过开展常规保险业务获得稳定的利润。但有时可以通过间接的方式来实现营利公司的这一商业目标。例如，为了迫使激进的新来者改变他们的目标，寻求利润，老牌公司可能会降低自己的价格。这种牺牲可能的收入，在没有竞争的情况下在市场上的不自然举动，是在竞争日益激烈的市场上保护地位的工具；
• 一家试图立即利用当前形势的公司正在利用任何机会在情况允许的情况下增加其销量和收入。这种对商业利益的追求有时会导致机会主义公司采取唯利是图的行为。
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 2023-02-04 03:37:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30795034766197205, "perplexity": 2947.5284645763263}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500080.82/warc/CC-MAIN-20230204012622-20230204042622-00466.warc.gz"} |
https://krotov.readthedocs.io/en/latest/API/krotov.optimize.html | # krotov.optimize module¶
## Summary¶
Functions:
optimize_pulses Use Krotov’s method to optimize towards the given objectives.
__all__: optimize_pulses
## Reference¶
krotov.optimize.optimize_pulses(objectives, pulse_options, tlist, *, propagator, chi_constructor, mu=None, sigma=None, iter_start=0, iter_stop=5000, check_convergence=None, info_hook=None, modify_params_after_iter=None, storage='array', parallel_map=None, store_all_pulses=False, continue_from=None, skip_initial_forward_propagation=False, norm=None, overlap=None)[source]
Use Krotov’s method to optimize towards the given objectives.
Optimize all time-dependent controls found in the Hamiltonians or Liouvillians of the given objectives.
Parameters
• objectives (list[Objective]) – List of objectives
• pulse_options (dict) –
Mapping of time-dependent controls found in the Hamiltonians of the objectives to a dictionary of options for that control. There must be options given for every control. As numpy arrays are unhashable and thus cannot be used as dict keys, the options for a control that is an array must be set using the key id(control) (see the example below). The options of any particular control must contain the following keys:
• 'lambda_a': the Krotov step size (float value). This governs the overall magnitude of the pulse update. Large values result in small updates. Small values may lead to sharp spikes and numerical instability.
• 'update_shape' : Function S(t) in the range [0, 1] that scales the pulse update for the pulse value at t. This can be used to ensure boundary conditions (S(0) = S(T) = 0), and enforce smooth switch-on and switch-off. This can be a callable that takes a single argument t; or the values 1 or 0 for a constant update-shape. The value 0 disables the optimization of that particular control.
In addition, the following keys may occur:
• 'args': If the control is a callable with arguments (t, args) (as required by QuTiP), a dict of argument values to pass as args. If 'args' is not specified via the pulse_options, controls will be discretized using the default args=None.
For example, for objectives that contain a Hamiltonian of the form [H0, [H1, u], [H2, g]], where H0, H1, and H2 are Qobj instances, u is a numpy array
>>> u = numpy.zeros(1000)
and g is a control function
>>> def g(t, args):
... E0 = args.get('E0', 0.0)
... return E0
then a possible value for pulse_options would look like this:
>>> from krotov.shapes import flattop
>>> from functools import partial
>>> pulse_options = {
... id(u): {'lambda_a': 1.0, 'update_shape': 1},
... g: dict(
... lambda_a=1.0,
... update_shape=partial(
... flattop, t_start=0, t_stop=10, t_rise=1.5
... ),
... args=dict(E0=1.0)
... )
... }
The use of dict and the {...} syntax are completely equivalent, but dict is better for nested indentation.
• tlist (numpy.ndarray) – Array of time grid values, cf. mesolve()
• propagator (callable or list[callable]) – Function that propagates the state backward or forwards in time by a single time step, between two points in tlist. Alternatively, a list of functions, one for each objective. If the propagator is stateful, it should be an instance of krotov.propagators.Propagator. See krotov.propagators for details.
• chi_constructor (callable) – Function that calculates the boundary condition for the backward propagation. This is where the final-time functional (indirectly) enters the optimization. See krotov.functionals for details.
• mu (None or callable) – Function that calculates the derivative $$\frac{\partial H}{\partial\epsilon}$$ for an equation of motion $$\dot{\phi}(t) = -i H[\phi(t)]$$ of an abstract operator $$H$$ and an abstract state $$\phi$$. If None, defaults to krotov.mu.derivative_wrt_pulse(), which covers the standard Schrödinger and master equations. See krotov.mu for a full explanation of the role of mu in the optimization, and the required function signature.
• sigma (None or krotov.second_order.Sigma) – Function (instance of a Sigma subclass) that calculates the second-order contribution. If None, the first-order Krotov method is used.
• iter_start (int) – The formal iteration number at which to start the optimization
• iter_stop (int) – The iteration number after which to end the optimization, whether or not convergence has been reached
• check_convergence (None or callable) – Function that determines whether the optimization has converged. If None, the optimization will only end when iter_stop is reached. See krotov.convergence for details.
• info_hook (None or callable) – Function that is called after each iteration of the optimization, for the purpose of analysis. Any value returned by info_hook (e.g. an evaluated functional $$J_T$$) will be stored, for each iteration, in the info_vals attribute of the returned Result. The info_hook must have the same signature as krotov.info_hooks.print_debug_information(). It should not modify its arguments in any way, except for shared_data.
• modify_params_after_iter (None or callable) – Function that is called after each iteration, which may modify its arguments for certain advanced use cases, such as dynamically adjusting lambda_vals, or applying spectral filters to the optimized_pulses. It has the same interface as info_hook but should not return anything. The modify_params_after_iter function is called immediately before info_hook, and can transfer arbitrary data to any subsequent info_hook via the shared_data argument.
• storage (callable) – Storage constructor for the storage of propagated states. Must accept an integer parameter N and return an empty array-like container of length N. The default value ‘array’ is equivalent to functools.partial(numpy.empty, dtype=object).
• parallel_map (callable or tuple or None) – Parallel function evaluator. If given as a callable, the argument must have the same specification as qutip.parallel.serial_map(). A value of None is the same as passing qutip.parallel.serial_map(). If given as a tuple, that tuple must contain three callables, each of which has the same specification as qutip.parallel.serial_map(). These three callables are used to parallelize (1) the initial forward-propagation, (2) the backward-propagation under the guess pulses, and (3) the forward-propagation by a single time step under the optimized pulses. See krotov.parallelization for details.
• store_all_pulses (bool) – Whether or not to store the optimized pulses from all iterations in Result.
• continue_from (None or Result) – If given, continue an optimization from a previous Result. The result must have identical objectives.
• skip_initial_forward_propagation (bool) – If given as True together with continue_from, skip the initial forward propagation (“zeroth iteration”), and take the forward-propagated states from Result.states instead.
• norm (callable or None) – A single-argument function to calculate the norm of states. If None, delegate to the norm() method of the states.
• overlap (callable or None) – A two-argument function to calculate the complex overlap of two states. If None, delegate to qutip.Qobj.overlap() for Hilbert space states and to the Hilbert-Schmidt norm $$\tr[\rho_1^\dagger \rho2]$$ for density matrices or operators.
Returns
The result of the optimization.
Return type
Result
Raises
ValueError – If any controls are not real-valued, or if any update shape is not a real-valued function in the range [0, 1]; if using continue_from with a Result with differing objectives; if there are any required keys missing in pulse_options. | 2021-05-07 14:03:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4439953565597534, "perplexity": 4458.800990002999}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988793.99/warc/CC-MAIN-20210507120655-20210507150655-00510.warc.gz"} |
https://tongfamily.com/2019/04/12/digital-camera-workflow-2019/ | Well time once again to update how to process photos and videos. Life is a little more complicated when the inputs are:
1. iPhone X. For most daily shots and really it is the easiest way to shoot panoramas and they come out pretty good looking at least for casual sharing.
2. Sony A7R Mark III. This is a whale of a camera. Yes, the market for "systems cameras" is really collapsing, but for the very high end, having this 41MP monster, but the really amazing thing is that with the Sony 50mm f/1.4 and the Leica 25mm f/2 on a Gitzo carbon fiber tripod, they are really the best cameras for night shots given their speed and when you have a landscape that just soaks up the pixels. They are perfect for 13"x19" large format photos. And also they work really well for difficult HDR shots since that camera has a 14EV dynamic range and when you shoot bracketed at plus/minus 3EV, you are getting 20EV of effective dynamic range. Far more than we humans can ever see. The big drawback is the 15 pounds of gear required.
3. Sony RX100 Mark VI. Ok, so I broke the viewfinder somehow, but when hiking around, the 35-200mm f/2.8-f/5.6 Lense is really pretty remarkable. It's the in between camera that works well on hikes.
Then there are the outputs or where the imaging need to end up, I'm down to really just 2 ½ and ¼ sets of outputs:
1. Archival. I'm realizing how valuable it is to have truly digital output with both GPS and time. I wish I had these for my dad's old shots, so that means retaining the raw images and then having a for use output that is still in file storage. These go mainly onto servers, my at Synology NAS home server and then both Google Drive for backup. I still need to reimplement the follow-on to Crashplan, but Google Drive seems to work Ok for snapshots.
2. Apple Photos. Since probably 90% of the folks that I interact with have iPhones, this has become the way to really share. So that means making sure that shared albums work and that is how most exchanges work today. Although shared albums are limited to 4MP, this is still very OK for the most common use, seeing photos on your phone.
3. Google Photos. It is a pain, but for those times when someone has an Android phone, I will port the photos over to a Google shared album. I don't put more there for privacy reasons.
4. Prints. This is extremely rare, but I do have a rotating set of A3+ images that go around the house. I normally print once every 2-3 years, but still it is nice to have all the resolution.
So how to handle all these different inputs, well the main goals are:
1. Staying away from too much Adobe. They are a great company, but they charge a lot and actually I don't find Lightroom's defaults all the that great and I'm not a Photoshop every photo to "Martian red" kind of person.
2. Staying open source. If there is something that works pretty well, why not support the open source world.
3. Helping small developers. Adobe swallows most folks up pretty quickly, but if I can, I try to help the small guys out.
Finally, here is the workflow for how I process photos, the main thing is the homogenize everything into a single file storage system since they are coming from so many different places.
1. From Sony A7R as RAW. It is really slow to copy these gigantic 80MB images, but they are the highest resolution. I give these a custom name _DRA (for A7) so that I can distinguish them from my other Sony. I dump these in a simple directory structure /Pictures/jpg/2019/2019-04/2019-04-01 is the example. it seems a little redundant but I find that I sometimes will accidentally drag and drop into completely the wrong place, so I need to ensure I can put them back. I shoot in RAW mainly because I don't shoot too much with this camera, so when I do, I almost never need a casual JPG shot. Also, while the in-camera JPEG is decent, it still doesn't beat a fully processed JPG.
2. If I can remember, these photos should have the correct time and geo tag information. This is do by having Sony Image Memories on my phone and then I have to remember to toggle the Bluetooth to make sure that the camera is picking up the location. This happens about 50% of the time, but it's a start. Also, the time might be wrong as well, we will go through that in a second.
3. From the Sony RX100, I do the same dump and I normally only shoot in RAW as well because again these are not casual shots, my instant sharing, I do with my iPhone.
4. Finally, look at all the photos on my Mac. This is because they are using the HEIC format and my current tools (like Dxo and Geotag) don't know how to deal with them. So this is where I delete the bad photos and so forth. I then export my iPhone shots from my Mac because I'm using the full iCloud Photo Sharing. I normally export as unmodified originals, so I get the HEIC and the MOV files, but mainly of my tools don't know HEIC yet, so for doing the tagging and date checks, if I have Sony images, I also export as JPEG because the software can read the tags there easily when I ensure that the Sony's have the correct tags. One note is that when doing a shoot, I try to remember to take at least one iPhone photo to ensure that later I have the correct location and date.
5. For photos that are panoramas, I will also crop them as they can get jaggy edges up and down if I'm not perfectly stable.
6. For quick sharing, I just use the iPhone photos and create a shared album and start sharing, this takes care of 80% of my use case which is sharing right then and there. I've found that it is better to get a few photos out the same day than to wait months for better photos later.
Now on to the processing of photos at this point I have a mess of unprocessed Sony photos along with processed iPhone photos.
1. I use DxO instead of Lightroom. While Lightroom is good, I like the presets on Dxo better. The colors seem less blown out. So I import this new directory. I try to do this every few days while on a longer trip so things don't get out of control.
2. To get uniform color management, I then select all the photos with Option-A and then choose DxO Lighting as Uniform and then Highlight Priority as Slight. Then I unselect them, this sets these to a uniform set of defaults later on. I also uncheck in the Distortion correction, the click for maintain aspect ratio. This is because it will crop out some pixels and the whole point is to keep them all.
3. I then go through and make my edits and cropping. If I see faces that are dark I use the Dxo lighting option for Spot and select faces if needed. I do *not* do the final processing to jpeg though because with Dxo you can't find time stamp and location issues.
4. If some of these photos need HDR treatment, I use Photomatix. This tool has the advantage that it reads directly from RAW and produces TIFF and JPEGs.
5. If some images are going to be gigantic panoramas. That is particularly amazing landscapes, I use PTGui. I switch from Hugin because it seems to find more control points. It is a pretty quirky program though and not super well supported. One problem with it is that it doesn't support a final cropping of images. Also Hugin does not understand raw format so I was having to create massive TIFFs, but that isn't necessary. Also, the basic lens on the Sony A7 is so good that I don't need to process them with Dxo before feeding them to Photomatix or PTGUI.
Now I have a set of images that are processed, but before I complete the ingestion, I have to make sure that the date and location stamps are correct. Because the Sony's do like bluetooth for location, I can get most of them, but they don't use their phone connection to correct the time so I often forget to change the timezones properly.
1. Geotag is a freeware tool that does most everything. I drag and drop and entire days worth of photos in. This is why I need the JPEG of the iPhone images because it doesn't grok HEIC, but it does understand the Sony ARW format.
2. I then sort all the images by date, so I can see a Swiss cheese mix of the Sony's and iPhone photos. I can then easily copy and paste missing locations. It takes awhile, but it is worth it to have all that data.
3. Finally, I look at the date stamps themselves to make sure they make sense, Geotag isn't the best tool for it, but it does let you change individual photos as needed. And I don't usually have too many date errors.
At this point, all the photos and dates and are ready to be turned into JPEGs from RAW:
1. I use DxO to create JPEGS. I don't care that much about disk, so I normally set the compression quality at 96 and this results in 10-12MB JPEGs for the A7.
2. Once all this chugging is done, I'm ready to move the RAWs into archive. And do a general cleanup. So first, I remove all the JPEGs that were images of the HEIC as I'm using HEIC as the standard format now. It uses MPEG-4 compression so the images are 30% smaller.
3. I copy the RAW and TIFF and the Dxo side car files called DOPs into a directory structure /Pictures/raw/2019/2019-04/2019-04-01 so it is a mirror of the Jpeg structure.
4. I will then import all the _Dxo.jpg files into iPhoto and this will start their upload into my stream. If I had previously created an album, I can then look at the Import tab and share these to folks. That means most people get an initial dump of iPhone pictures and then later the higher resolution ones. This seems like a nice way to get people are double hit. During the trip and right after.
5. Finally, I take the very best photos that I really like and create a /Pictures/Albums/2019 for each year, I have a greatest hits of images. And for really big adventures, I create /Pictures/Albums/2019/2019-04 Yosemite Trip file structure so I'm not dependent on iCloud Photo Sharing for everything.
6. And if I have some Android friends, I create a Google Shared album and import the stuff in the albums.
Finally, I archive like crazy, the current is that I have a primary NAS, then a real time backup to another onsite one, then incremental backups to a Google Drive and soon a complete historical backup to Backblaze.
1. Since these are standard files, not only do I get backups with Apple for the low resolution stuff, but I copy these off to my Synology NAS. And then from there, I use GoodSync on an old Mac to copy these to a Drobo. I use different technologies because I don't want a single system bug to the down all the copies.
2. From there the Synology has a backup system and I connect it to GSuite's Drive. I pay the \$10/month for unlimited storage and this is a good way to keep it happening.
3. I haven't gotten to it yet, but the last step is to install Backblaze so the even if something bit rots (and I do have bit rot), I have an image of every file I've every created with that. I used to use Crashplan, but they exited. Not that this is different from the Google Drive backup as that doesn't keep a history.
The net is that with this process, I get all four of my outputs and by using DxO as the core Lightroom plus light Photoshop tool, it all ends up in one place. | 2022-08-10 10:19:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.25522637367248535, "perplexity": 1758.794622170502}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00278.warc.gz"} |
https://cs.stackexchange.com/questions/148193/find-all-integer-points-that-lay-in-a-3-ball-with-a-given-radius | # Find all integer points that lay in a 3-ball with a given radius
How can I efficiently find all lattice points in the cubic lattice $$Z^3$$ (that is to say, all integer points in a 3-space) that lay in a closed ball of radius $$R$$ centred at the origin?
Essentially,
Let $$dist(p)$$ be a function denoting the euclidean distance between a point in n-space and the origin point of that space, so $$dist(p)=\sqrt{p_1²+p_2²+p_3²\ldots p_n²}$$.
How might I efficiently iterate over $$\{p \in \mathrm{Z}^3\ | \ dist(p) \le R\}$$?
I'm aware that this is trivial to do in $$\mathbb{O}(R^3)$$ time by iterating over all lattice points that lay inside the minimum bounding box of the ball and filtering out every point $$p$$ where $$dist(p) > R$$, and I'm also aware that this can be optimised by squaring both sides of the distance function, but this algorithm is still too slow for my needs.
• What's $n$? Do you mean $R$? What counts as "efficiently"? Do you care about asymptotic running time, or about constant factors?
– D.W.
Jan 6 at 21:51
• Sorry, I mean $R$, and I care about constant factors (otherwise I'd settle for the naive algorithm, even an ideal algorithm would need to iterate over approximately $\frac{4}{3}\pi R^3$ points, which is $\frac{\pi}{6}$ times the iterations needed by the naive algorithm) Jan 6 at 22:11
One straightforward approach is to iterate over $$x,y$$ in the bounding box, then find $$z_\max$$ so $$(x,y,z)$$ is in the sphere iff $$|z| \le z_\max$$. You can find $$z_\max$$ via the formula
$$z_\max = \sqrt{R - x^2 - y^2}.$$
If you can compute squares and square roots in constant time, then the running time is proportional to the number of pixels in the ball, i.e., you iterate over $$\frac43 \pi R^3$$ points.
A similar but slightly better algorithm is to iterate over $$x,y$$ in the bounding box, then iterate over increasing values of $$z$$ (i.e., $$z=0,1,2,\dots$$) until $$z^2 > R - x^2 - y^2$$. This is a little more efficient, because it doesn't require computing square roots, and because you can use the identity
$$(z+1)^2 = z^2 + 2z + 1$$
to avoid the need to compute squares in the inner loop either, instead updating the value of $$z^2$$ in each iteration. (Make sure to add in the negative values of $$z$$ too.) | 2022-01-24 05:10:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7903228402137756, "perplexity": 167.80702907739604}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00281.warc.gz"} |
http://www.heldermann.de/JCA/JCA11/jca11021.htm | Journal Home Page Cumulative Index List of all Volumes Complete Contentsof this Volume Previous Article Next Article Journal of Convex Analysis 11 (2004), No. 2, 335--361 Copyright Heldermann Verlag 2004 Variational Analysis for a Class of Minimal Time Functions in Hilbert Spaces Giovanni Colombo Dip. di Matematica Pura e Applicata, Universita di Padova, Via Belzoni 7, 35131 Padova, Italy, colombo@math.unipd.it Peter R. Wolenski Dept. of Mathematics, Louisiana State University, 326 Lockett Hall, Baton Rouge, LA 70803-4918, U.S.A., wolenski@math.lsu.edu [Abstract-pdf] \def\iint{{\hbox{int}\;}} This paper considers the parameterized infinite dimensional optimization problem $$\hbox{minimize}\quad\bigl\{t\geq 0:\;S \cap\{x+tF\}\not= \emptyset\bigr\},$$ where $S$ is a nonempty closed subset of a Hilbert space $H$ and $F\subseteq H$ is closed convex satisfying $0\in \iint F$. The optimal value $T(x)$ depends on the parameter $x\in H$, and the (possibly empty) set $S\cap (x+T(x)F)$ of optimal solutions is the $F$-projection'' of $x$ into $S$. We first compute proximal and Fr\'echet subgradients of $T(\cdot)$ in terms of normal vectors to level sets, and secondly, in terms of the $F$-projection. Sufficient conditions are also obtained for the differentiability and semiconvexity of $T(\cdot)$, results which extend the known case when $F$ is the unit ball. FullText-pdf (641 KB) for subscribers only. | 2019-03-18 16:16:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5973972082138062, "perplexity": 839.1755053790116}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912201455.20/warc/CC-MAIN-20190318152343-20190318174343-00358.warc.gz"} |
http://mathhelpforum.com/pre-calculus/95869-quadratic-question.html | # Math Help - Quadratic Question
Hi,
This is probably a relatively easy question, but I've never done this before, and I'm not sure how to go about it.
For the equation $3x^2+ax+4=0$ find the smallest possible value of $a$ such that the equation will have two distinct rational solutions.
*Edit - Never-mind. I just figured it out
3. Hello, Stroodle!
For the equation $3x^2+ax+4\:=\:0$, find the smallest possible value of $a$
such that the equation will have two distinct rational solutions.
*Edit - Never-mind. I just figured it out
Good for you!
But is this a trick question?
I noted that it didn't say smallest positive value of $a$.
Using the Quadratic Formula: . $x \;=\;\frac{-a \pm\sqrt{a^2-48}}{6}$
For rational roots, the discriminant $(a^2-48)$ must be a square.
This is true for: . $a \;=\;\pm7,\:\pm8,\:\pm13$
Then the smallest (least) value of $a$ is: . $a \:=\:-13$
4. Haha. Well spotted Soroban!
Actually the question did ask for the smallest positive value of $a$, but in my tired & vague state of mind I accidentally failed to copy it down... | 2015-07-07 16:10:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7186676263809204, "perplexity": 423.04712012338814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375099755.63/warc/CC-MAIN-20150627031819-00152-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://indico.desy.de/event/28202/contributions/105201/ | # EPS-HEP2021 conference
26-30 July 2021
Zoom
Europe/Berlin timezone
## Probing the minimal $U(1)_X$ model at future electron-positron colliders via the fermion pair-production channel
Not scheduled
20m
Zoom
#### Zoom
Poster Searches for New Physics
### Speaker
Arindam Das (Osaka U)
### Description
The general U$(1)_X$ extension of the Standard Model (SM) is a well motivated scenario which has a plenty of new physics options. Such a model is anomaly free which requires to add three generations of the SM singlet right-handed neutrinos (RHNs) which naturally generates the light neutrino masses by the seesaw mechanism.This offers interesting phenomenological aspects in the model. In addition to that the model is equipped with a beyond the SM (BSM) neutral gauge boson, $Z^\prime$ which interacts with the SM and BSM particles showing a variety of new physics driven signatures. After the anomaly cancellation the U$(1)_X$ charge of the particles are expressed in terms of the SM Higgs doublet and the SM Higgs singlet which allows us to study the interaction of the fermions with the $Z^\prime$.In this paper we investigate the pair production mechanism of the different charged through the photon, $Z$ and $Z^\prime$ boson exchange processes at the electron-positron $(e^- e^+)$.The angular distributions, forward-backward $(\mathcal{A}_{\rm{FB}})$, left-right $(\mathcal{A}_{\rm{LR}})$ and left-right forward-backward $(\mathcal{A}_{\rm{LR, FB}})$ asymmetries of the different charged fermion pair productions show substantial deviation from the SM results.
Collaboration / Activity HEP-PH
### Primary author
Arindam Das (Osaka U) | 2021-10-21 17:03:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5860170722007751, "perplexity": 1851.3022672217965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585439.59/warc/CC-MAIN-20211021164535-20211021194535-00705.warc.gz"} |
http://mathrefresher.blogspot.com/2006/06/group-theory-lagranges-theorem.html?showComment=1232603640000 | ## Friday, June 02, 2006
### Group Theory: Lagrange's Theorem
In today's blog, I review the proof for Lagrange's Theorem. The theorem is named after Joseph-Louis Lagrange who first stated it. The first complete proof came 30 years later.
In today's blog, I also use the concept of order, subgroup, and coset. The concept of the coset was first proposed by Evariste Galois. The term "coset" was coined by G. A. Miller in 1910.
The content of today's blog is taken from Joseph A. Gallian's Contemporary Abstract Algebra.
Definition 1: Order of a Group
The number of elements of a group.
NOTE: If you need to review the definition of the group, see here.
Definition 2: Subgroup
If H is a subset of a group G and H is a group itself under the operation of G, then H is a subgroup of G.
Definition 3: Coset of H in G
Let G be a group and H a subgroup of G. For any a ∈ G, the set aH { ah : h ∈ H } is called the left coset of H in G containing a. The set Ha { ha : h ∈ H } is called the right coset of H in G containing a.
NOTE: An important idea behind the coset is that it is a distinct partition of elements (see Lemma 2 below). From the property of closure, we know that if two values are elements of a group, then their product (by this, I mean the result of the group operation) is also an element of the group.
From this, we know that the set of cosets can be divided up as follows: a1H, a2H, ..., arH where r is a positive integer and where for each coset i ≠ j → aiH ∩ ajH = ∅ [See Lagrange's Theorem below to see how this partitioning can be used]
Example 1 Coset: Z9
Let G = Z9 = { 0, 1, 2, 3, 4, 5, 6, 7, 8 } with operation '+'
NOTE: Z9 is the set of integers modulo 9 [See here for a review of modular arithmetic]
Let H = { 0, 3, 6 } with operation '+'
We can see that H is a subgroup of G
The left coset in this case is a+H (note it would aH is the operation were '*')
Here are the cosets:
0 + H = { 0, 3, 6 } = 3 + H = 6 + H
1 + H = { 1, 4, 7 } = 4 + H = 7 + H
2 + H = {2, 5, 8} = 5 + H = 8 + H
We can see that all cosets are either equal or distinct. I present a proof of this in Lemma 2 below.
Example 2: Even integers
Let Z be the set of integers
Then 2Z is a left coset which includes { ..., 0, 2, 4 , 6, ... }
Lemma 1: aH = H if and only if a ∈ H
Proof:
(1) Assume aH = H
(2) Then, a = ae ∈ aH = H
(3) Assume a ∈ H
(4) aH ⊆ H since h ∈ H → ah ∈ H [By property of closure, see here if needed]
(5) H ⊆ aH since:
(a) Let h be any element of H.
(b) a-1 ∈ H since a ∈ H [By the inverse property, see here if needed]
(c) a-1H ∈ H [By property of closure, see here if needed]
(d) So h = eh = (aa-1)h = a(a-1h) [By identity property, inverse property, and associative property, see here if needed]
(e) But a(a-1h) ∈ aH so that from (#5d), h ∈ aH.
(6) So H = aH since H ⊆ aH and aH ⊆ H.
QED
Lemma 2: if aH, bH are two cosets, then aH = bH or aH ∩ bH = ∅
Proof:
(1) Assume that aH ∩ bH ≠ ∅
(2) Let x ∈ aH ∩ bH since aH ∩ bH ≠ ∅
(3) Then there exists h1, h2 such that:
x = ah1
x= bh2
Since x is found in both aH and bH by assumption.
(4) Thus, a = xh1-1 = (bh2)h1-1
(5) From this, aH = [(bh2)h1-1]H = b(h2h1-1H) = bH [By Lemma 1 above]
QED
Theorem: Lagrange's Theorem
If G is a finite group and H is a subgroup of G, then the order of H divides the order of G.
Proof:
(1) Let a1H, a2H, ..., arH denote the distinct left cosets of H in G. [See Definition 3 above for details on left cosets of H in G.]
NOTE: If two cosets are equal then we only count them once.
(2) For each a ∈ G, we have aH = aiH for some i by our construction in step #1.
(3) a ∈ aH since:
a = ae and ae ∈ aH since H is a subgroup. [See Definition 2 above for details on subgroups]
(4) Thus, each member of G belongs to one of the cosets aiH [from step #2 and step #3.]
(5) G = a1H ∪ a2H ∪ ... ∪ arH [This follows directly from step #4]
(6) From Lemma 2 above, we know aiH = ajH or aiH ∩ ajH = ∅ and since each ai is distinct, we get:
order(G) = order(a1H) + order(a2H) + ... + order(arH)
(7) Since order(aiH) = order(H) for each i (by the definition of cosets, see definition 3 above, we have:
order (G) = r*order(H)
QED
Akash said...
Hi Larry,
It was wonderful going thru ur blog. You really have maintained it in an impeccable manner - i must say. Further, you have gone thru the troubles of providing links for some concepts which others might need and again it is a wonderful act and i really appreciate it. The world needs more math afficiandos like you.
Thanks for your beautiful explanations. Again I add thanks for your selfless service to humanity....Hope you create a book out of your blogs.
Tom said...
Hi Larry,
http://mathrefresher.blogspot.com/2006/06/group-theory-lagranges-theorem.html
can you explain how step 5 in Lemma 2 follows from Lemma 1? I'm not sure I understood.
Larry Freeman said...
Hi Tom,
I'll try to break it down step by step. I hope that this makes the step clear.
In Lemma 2, step #3 says that:
x = bh_2
x = ah_1
Since x=ah_1, we know that:
a = x(h_1)^(-1)
Since x=bh_2, we know that:
x(h_1)^(-1) = (bh_2)(h_1)^(-1)
Since a = x(h_1)^(-1), it follows that:
a = (bh_2)(h_1)^(-1)
Since a=(bh_2)(h_1)^(-1), it follows that:
aH=[(bh_2)(h_1)^(-1)]H
We further note that:
[(bh_2)(h_1)^(-1)]H = b[(h_2)(h_1)^(-1)H]
Now, we know that h_1,h_2 are elements of H [from step #3]
Since h_1 ∈ H, it follows that (h_1)^(-1) ∈ H. [Since closure is a property of groups]
Since (h_1)^(-1) ∈ H and h_2 ∈ H, it follows that (h_2)(h_1)^(-1) ∈ H
That's all we need to use Lemma 1 above. Since (h_2)(h_1)^(-1) ∈ H,
[(h_2)(h_1)^(-1)]H = H
Since [(h_2)(h_1)^(-1)]H = H, it follows that:
b[(h_2)(h_1)^(-1)]H = bH
Tom said...
Thank you, that made it clear!
Manohar Kuse said...
Very well exlained. This note on lagrange theorem in group theory helped me more than my textbook.
Thanks!!!
paaji said...
GOD !!!
novita said...
hi,
the first time i read the proof in the book contemporary abstract algebra,n when i read your explanation i understand about the proof but i still confused about steps 6 order(G)= order(a1(H))+....order(ar(H))if we sum like that so relation with your example before o(G)=o(0(H))+o(1(H))...+o(8(H))=3+3+3+3+3+3+3+3=24.it is wrong,because o(G)=9
and then steps 7 o(G)= r.o(H) relation with example r=8 so o(G)=8.3=24.
please explain that i need that in this week..
Larry Freeman said...
Hi Novita,
Can you provide more details on your example. I'm not clear how you are getting to O(G)=9.
Are you sure that you are following this assumption:
From this, we know that the set of cosets can be divided up as follows: a1H, a2H, ..., arH where r is a positive integer and where for each coset i ≠ j → aiH ∩ ajH = ∅
-Larry
Pawan Seerwani said...
This post helped me a lot..!!
Thanks and keep writing..! :) :)
Pawan Seerwani said...
This comment has been removed by a blog administrator. | 2017-04-24 09:15:19 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8275733590126038, "perplexity": 1733.3976780100256}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917119225.38/warc/CC-MAIN-20170423031159-00164-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://allmathconsidered.wordpress.com/tag/plimpton-322/ | # Using a 4000-year old clay tablet to solve math problems
This post discusses the clay tablet that is known as Plimpton 322. This tablet gives us a glimpse of the powerful mathematics practiced in ancient Babylonia almost 4000 years ago. We aim to give a sense of why Plimpton 322 is fascinating. The math in the tablet informs us of the past as well as informing us of the present. We also work some trigonometry problems using Plimpton 322.
What is Plimpton 322?
Figure 1 – Plimpton 322 (Credit: UNSW/Andrew Kelly)
The dimensions of the tablet are 8.8 cm by 12.7 cm (3.5 inches by 5 inches), about the size of a small pocket calculator. It was purchased by the New York publisher George Arthur Plimpton in 1923 from Edgar J. Banks and was donated to Columbia University upon Plimpton’s death in 1936. Henceforth, the tablet had been known as Plimpton 322, signifying that it is the 322nd item in Plimpton’s collection.
According to Edgar J. Banks, the tablet came from a location near the ancient city of Larsa (modern Tell Senkereh) in Southern Iraq. Eleanor Robson, an Oriental scholar at the University of Oxford, estimated that Plimpton 322 was created around 1800 BC in Babylonia, more specifically about six decades before Larsa fell to Hammurabi of Babylon in 1762 BC. Thus Plimpton 322 dated back to the Old Babylonian period in Mesopotamia about 4,000 years ago.
What is in Plimpton 322?
The Plimpton 322 was written in cuneiform script. The numbers contained in the tablet are sexagesimal numbers (base 60). It was at first assumed to be just another Babylonian ledgerbook. In the 1940s, Otto Neugebauer, a historian of ancient science at Brown University, and his assistant Abraham Sachs found that Plimpton 322 actually contains interesting mathematical contents. The entries in the tablet are essentially Pythagorean triples, i.e. the integer solutions to the equation $a^2+b^2=c^2$.
In order to appreciate Plimpton 322, let’s look at how the contents of the tablet are structured. The front side of Plimpton 322 has 15 lines of numbers displayed in four columns. The line at the top above the numbers contains some labels. The rightmost column contains the row numbers (or line numbers) from 1 to 15. The middle two columns contain the short side $s$ and the hypotenuse $d$ of 15 right triangles. In other words, the second and third columns of Plimpton 322 are two sides a right triangle such as the one shown below.
Figure 2 – A right triangle
The third column of the tablet shows $d$, the hypotenuse of a right triangle (or diagonal). The second column shows $s$, the short side of a right triangle. The long side $l$ of a right triangle is not shown. The first column in Plimpton 322 is the square of a ratio, which can be one of two interpretations, either the square of $\frac{d}{l}$ (diagonal over long side) or the square of $\frac{s}{l}$ (short side over long side). The following diagram shows the descriptions of the four columns.
Figure 3 – The structure of Plimpton 322
What is Special about Plimpton 322?
The discovery made by Otto Neugebauer and his assistant in the 1940s was an important one. The numbers in Plimpton 322 are what are now called Pythagorean triples. It gives the short side and the diagonal (hypotenuse) of 15 right triangles. The long sides of the right triangles are not shown. As we will see below, the 15 right triangles have steadily decreasing slopes. The Sumerians in the Old Babylonian period knew about the Pythagorean theorem over 1,000 years before the time of Pythagoras!
Since the discovery made by Otto Neugebauer, Plimpton 322 was a subject of extensive research by mathematicians. Obviously mathematicians are intrigued by the connection of a 4000-year tablet with modern mathematics. Because of the intricate mathematical interpretations they made of the tablet, many mathematicians thought highly of the tablet. For example, the author of the tablet must be a mathematical prodigy or a professional mathematician, doing high level research in the Old Babylonian Period.
However, there are opposing views. Eleanor Robson does not view Plimpton 322 as the work of a math prodigy or professional mathematician. Her view of Plimpton 322 is more mundane. She believes that Plimpton 322 was created as teaching aid with a purpose of generating problems involving right triangles and reciprocal pairs. Links are provided below for research stating these different points of view.
Looking at Plimpton 322 in Decimal Numbers
High level math research or merely teaching aid, the fact that the tablet contains Pythagorean triples is fascinating and interesting from a mathematical point of view. Let’s continue to examine the tablet. We give a small demonstration that it can be used for working trigonometry problems. The numbers in the tablet are sexagesimal numbers (base 60). To make things easy for us, the following table shows the decimal conversion of Plimpton 322, taken from the Wikipedia entry on Plimpton 322.
Table 1 – Decimal Conversion of Plimpton 322
Squared Ratios Short Side Diagonal Row (1).9834028 119 169 1 (1).9491586 3367 4825 2 (1).9188021 4601 6649 3 (1).8862479 12709 18541 4 (1).8150077 65 97 5 (1).7851929 319 481 6 (1).7199837 2291 3541 7 (1).6927094 799 1249 8 (1).6426694 481 769 9 (1).5861226 4961 8161 10 (1).5625 45 75 11 (1).489417 1679 2929 12 (1).4500174 161 289 13 (1).430289 1771 3229 14 (1).3871605 56 106 15
The first column is either the square of the diagonal over the long side or the square of the short side over the long side. For example, the long side in Row 1 is 120. The square of 169/120 is 1.9834. The square of 119/120 is 0.9834. To help us work problems, we expand the table with three more columns.
Table 2 – Decimal Conversion of Plimpton 322 (Expanded)
Squared Ratios Short Side Diagonal Row Long Side S/L D/L (1).9834028 119 169 1 120 0.99167 1.40832 (1).9491586 3367 4825 2 3456 0.97425 1.39612 (1).9188021 4601 6649 3 4800 0.95854 1.38521 (1).8862479 12709 18541 4 13500 0.94141 1.37341 (1).8150077 65 97 5 72 0.90278 1.34722 (1).7851929 319 481 6 360 0.88611 1.33611 (1).7199837 2291 3541 7 2700 0.84852 1.31148 (1).6927094 799 1249 8 960 0.83229 1.30104 (1).6426694 481 769 9 600 0.80167 1.28167 (1).5861226 4961 8161 10 6480 0.76559 1.25941 (1).5625 45 75 11 60 0.75 1.25 (1).489417 1679 2929 12 2400 0.69958 1.22042 (1).4500174 161 289 13 240 0.67083 1.20417 (1).430289 1771 3229 14 2700 0.65593 1.19593 (1).3871605 56 106 15 90 0.62222 1.17778
The three additional columns are the long side and the ratios of Short over Long and Diagonal over Long. The square of these two ratios would be the first column of the table. The 6th column (S/L) is the slope of the right triangle in Figure 1. So the 15 right triangles in the table have steadily decreasing slopes. The angle between the diagonal and the long side goes from 44.76 degrees (in Row 1) to 31.89 degrees (in Row 15).
How did the creator of Plimpton 322 calculate the long side $l$ in Table 2? For example, in Row 2, the short side is 3367 and the diagonal side is 4825. Modern calculation for the long side would be the square root $\sqrt{4825^2-3367^2}=\sqrt{11943936}=3456$. How was 3456 obtained for the creator of Plimpton 322? It turns out that the triples (s, l, d) in the tables are of special form. The three sides can be expressed as:
$s=p^2-q^2$
$d=p^2+q^2$
$l=2 \times p \times q$
such that $p$ and $q$ are integers. The ingenuity is that the special right triangles obtained in this fashion can be used for solving trigonometric problems as demonstrated below.
Working Examples
Solve for the unknown side for each of the following two right triangles A and B.
Figure 4 – Solve for the unknown sides using Plimpton 322
Obviously the triangles are not drawn to scale. They are only meant to convey the problems. We show how to use Plimpton 322 to estimate $x$ and $y$.
In triangle A, we are given the diagonal and the long side. Immediately we can compute the ratio $D/L=190/145=1.310344828$. This ratio is closest to the D/L ratio in Row 7 in Table 2. We use the right triangle in Row 7 as the reference triangle, i.e. the triangle in Row 7 and triangle A are (approximately) congruent. The ratios of the short side to the long side for both triangles should be approximately identical.
$\displaystyle \frac{x}{190}=\frac{2291}{3541}$
Solving for $x$ gives 122.9285513. The modern approach would be to use the Pythagorean theorem with the help of a calculator in taking square root. Thus the exact answer is $x=\sqrt{190^2-145^2}=\sqrt{15075}=122.7802916$. Of course, this approach was not possible in the Old Babylonian period as the concept of square root was not known at the time.
In triangle B, the short side and the long side are given. It follows that the square of the diagonal equals the square of the short side plus the square of the long side (this is known as the Pythagorean theorem to us but the relationship is also known to Babylonian users of Plimpton 322). Compute the following ratio.
$\displaystyle \frac{56^2+79^2}{79^2}=\frac{9377}{6241}=1.502483576$
The above result is identical to the square of the ratio of the diagonal over the long side. Then compare this result to the first column of Table 2 (or Table 1). The closest is the number 1.489417 in Row 12. Then use the right triangle in Row 12 as the reference triangle. Thus the two triangles are approximately congruent.
$\displaystyle \frac{y}{79}=\frac{2929}{2400}$
Solving for $y$ gives 96.4291667. The answer from a modern approach would be $y=\sqrt{56^2+79^2}=\sqrt{9377}=96.83491106$.
Why the Examples are Special
Both answers using Plimpton 322 are quite close to the exact answers, even though the discrepancies are significant (0.148 for $x$ and 0.422 for $y$). However the problem solving using Plimpton 322 is indeed special. Essentially we are solving trigonometric problems without using trigonometry, i.e. using angles and sine and cosine functions. In the Old Babylonian period, there was no concept of angles and there certainly was no trigonometry as we know it today. Hipparchus (190-120 BC), a Greek astronomer, geographer, and mathematician, is considered to be the father of trigonometry. He lived 1,600 years after the creation of Plimpton 322!
Note that the modern answer for $x$ is $\sqrt{15075}$ and for $y$ is $\sqrt{9377}$. These two square roots are the results of applying the Pythagorean theorem. Plimpton 322 allows taking square root without using square root! Pythagoras (570-495 BC) lived more than a thousand years after the creation of Plimpton 322. So using a cheat sheet from 1800 BC, we can solve trigonometric problems without using methods that only came thousand or more years later.
Recent research by Daniel Mansfield and N. J. Wildberger compared the methods of using Plimpton 322 to using the well-known sine table created by the Indian astronomer-mathematician Madhava (1340–1425 AD), over 3,000 years after Plimpton 322. Their problems are similar to the examples given here. The approach using Plimpton 322 produces much more accurate answers than the approach of using the sine table of Madhava. It is amazing that an 1800 BC “trigonometric” table beats a trigonometric table that came 3,000 later! If the Babylonians were indeed using Plimpton 322 as a trigonometric table, then it preceded Hiapparchus’ table of chords by about 1,600 years.
The accuracy of using Plimpton 322 is highly dependent on the given sides of the right triangles, i.e. the approach is accurate only if the squared ratios are close to the ratios in Plimpton 322. However, Mansfield and Wildberger showed that the usefulness of Plimpton 322 can be extended using interpolation (the ancient Babylonians were big on interpolation).
The calculation demonstrated here is only scratching the surface. What we have shown is just a small demonstration of the mathematical specialness of Plimpton 322. The examples we work are in no way a suggestion that it is how the stone tablet was used in the Old Babylonian period.
The math in Plimpton 322 is wonderful and exciting. There is actually quite a bit of controversy about the tablet. What purpose did the tablet serve at its time? There are divergent views just on this questions alone.
According to Mansfield and Wildberger, Plimpton 322 not only replaces Hiapparchus’ table of chords as the world’s oldest trigonometric table, it is the world’s only completely accurate trigonometric table. For more information, see the article by Mansfield and Wildberger. Mansfield and Wildberger believe that Plimpton 322 would have been used in engineering calculations for the construction of palaces, canals or perhaps the Hanging Gardens of Babylon.
According to Eleanor Robson, Plimpton 322 was merely a teaching aid for problems involving right triangles (article). According to Robson, ancient mathematical texts and artifacts such as Plimpton 322 must be viewed in light of their historical and cultural contexts (in addition to the mathematical). The mathematics contained in Plimpton 322 should not be examined in isolation.
Though the modern mathematical interpretations of Plimpton 322 may have no relation to its original use, it is undeniable that the mathematics in Plimpton 322 is fascinating. It makes for good material for any math teacher’s lesson plans. It ought to be reassuring to students that the math topics that they deal with were also practiced by students 4,000 years ago. The proof is in the tablet called Plimpton 322.
Reporting of Plimpton 322 is easily found on the Internet (examples: here and here). The wikipedia entry on Plimpton 322 is a good source of information, as is the entry on Edgar J. Banks.
$\text{ }$
$\text{ }$
Dan Ma math blog
Daniel Ma math blogs
Dan Ma math
Daniel Ma math
Dan Ma mathematics
Daniel Ma mathematics
$\copyright$ 2017 – Dan Ma | 2019-07-17 16:23:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 34, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.687649667263031, "perplexity": 960.5217502777365}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525355.54/warc/CC-MAIN-20190717161703-20190717183703-00209.warc.gz"} |
http://codeforces.com/blog/entry/8755 | Codeforces celebrates 10 years! We are pleased to announce the crowdfunding-campaign. Congratulate us by the link https://codeforces.com/10years. ×
### fchirica's blog
By fchirica, 6 years ago, ,
## 340A - The Wall
You are given a range [A, B]. You're asked to compute fast how many numbers in the range are divisible by both x and y. I'll present here an O(log(max(x, y)) solution. We made tests low so other not optimal solutions to pass as well. The solution refers to the original problem, where x, y ≤ 109.
Firstly, we can simplify the problem. Suppose we can calculate how many numbers are divisible in range [1, X] by both x and y. Can this solve our task? The answer is yes. All numbers in range [1, B] divisible by both numbers should be counted, except the numbers lower than A (1, 2, ..., A — 1). But, as you can see, numbers lower than A divisible by both numbers are actually numbers from range [1, A — 1]. So the answer of our task is f(B) — f(A — 1), where f(X) is how many numbers from 1, 2, ..., X are divisible by both x and y.
For calculate in O(log(max(x, y)) the f(X) we need some math. If you don't know about it, please read firstly about least common multiple. Now, what will be the lowest number divisible by both x and y. The answer is least common multiple of x and y. Let's note it by M. The sequence of the numbers divisible by both x and y is M, 2 * M, 3 * M and so on. As a proof, suppose a number z is divisible by both x and y, but it is not in the above sequence. If a number is divisible by both x and y, it will be divisible by M also. If a number is divisible by M, it will be in the above sequence. Hence, the only way a number to be divisible by both x and y is to be in sequence M, 2 * M, 3 * M, ...
The f(X) calculation reduces to finding the number of numbers from sequence M, 2 * M, 3 * M, ... lower or equal than X. It's obvious that if a number h * M is greater than X, so will be (h + 1) * M, (h + 2) * M and so on. We actually need to find the greatest integer number h such as h * M ≤ X. The numbers we're looking for will be 1 * M, 2 * M, ..., h * M (so their count will be h). The number h is actually [X / M], where [number] denotes the integer part of [number]. Take some examples on paper, you'll see why it's true.
The only thing not discussed is how to calculate the number M given 2 number x and y. You can use this formula M = x * y / gcd(x, y). For calculate gcd(x, y) you can use Euclid's algorithm. Its complexity is O(log(max(x, y)), so this is the running time for the entire algorithm.
Official solution: 4383403
## 340B - Maximal Area Quadrilateral
I want to apologize for not estimating the real difficulty of this task. It turns out that it was more complicated than we thought it might be. Let's start explanation.
Before reading this, you need to know what is signed area of a triangle (also called cross product or ccw function). Without it, this explanation will make no sense.
The first thing we note is that a quadrilateral self intersecting won't have maximum area. I'll show you this by an image made by my "talents" in Paint :) As you can see, if a quadrilateral self intersects, it can be transformed into one with greater area.
Each quadrilateral has 2 diagonals: connecting 1st and 3rd point and connecting 2nd and 4th point. A diagonal divides a plane into 2 subplanes. Suppose diagonal is AB. A point X can be in one of those two subplanes: that making cross product positive and that making cross product negative. A point is in "positive" subplane if ccw(X, A, B) > 0 and in "negative" subplane ccw(X, A, B) < 0. Note that according to the constraints of the task, ccw(X, A, B) will never be 0.
Let's make now the key observation of the task. We have a quadrilateral. Suppose AB is one of diagonals and C and D the other points from quadrilateral different by A and B. If the current quadrilateral could have maximal area, then one of points from C and D needs to be in "positive subplane" of AB and the other one in "negative subplane". What would happen if C and D will be in the same subplane of AB? The quadrilateral will self intersect. If it will self intersect, it won't have maximal area. "A picture is worth a thousand words" — this couldn't fit better in this case :) Note that the quadrilateral from the below image is A-C-B-D-A.
Out task reduces to fix a diagonal (this taking O(N ^ 2) time) and then choose one point from the positive and the negative subplane of the diagonal. I'll say here how to choose the point from the positive subplane. That from negative subplane can be chosen identically. The diagonal and 3rd point chosen form a triangle. As we want quadrilateral to have maximal area, we need to choose 3rd point such as triangle makes the maximal area. As the positive and negative subplanes are disjoint, the choosing 3rd point from each of them can be made independently. Hence we get O(N ^ 3) complexity. A tricky case is when you choose a diagonal but one of the subplanes is empty. In this case you have to disregard the diagonal and move to the next one.
Official solution: 4383413
## 340C - Tourist Problem
Despite this is a math task, the only math formula we'll use is that number of permutations with n elements is n!. From this one, we can deduce the whole task.
The average formula is sum_of_all_routes / number_of_routes. As each route is a permutation with n elements, number_of_routes is n!. Next suppose you have a permutation of a: p1, p2, …, pn. The sum for it will be p1 + |p2 – p1| + … + |pn – pn-1|. The sum of routes will be the sum for each possible permutation.
We can calculate sum_of_all routes in two steps: first time we calculate sums like “p1” and then we calculate sums like “|p2 – p1| + … + |pn – pn-1|” for every existing permutation.
First step Each element of a1, a2, …, an can appear on the first position on the routes and needs to be added as much as it appears. Suppose I fixed an element X for the first position. I can fill positions 2, 3, .., n – 1 in (n – 1)! ways. Why? It is equivalent to permuting n – 1 elements (all elements except X). So sum_of_all = a1 * (n – 1)! + a2 * (n – 1)! + … * an * (n – 1)! = (n – 1)! * (a1 + a2 + … + an).
Second step For each permutation, for each position j between 1 and n – 1 we need to compute |pjp(j + 1)|. Similarly to first step, we observe that only elements from a can appear on consecutive positions. We fix 2 indices i and j. We’re interested in how many permutations do ai appear before aj. We fix k such as on a permutation p, ai appears on position k and aj appears on a position k + 1. In how many ways can we fix this? n – 1 ways (1, 2, …, n – 1). What’s left? A sequence of (n – 2) elements which can be permuted independently. So the sum of second step is |ai - aj| * (n – 1) * (n – 2)!, for each i != j. If I note (a1 + a2 + … + an) by S1 and |ai - aj| for each i != j by S2, the answer is (N – 1)! * S1 + (N – 1)! * S2 / N!. By a simplification, the answer is (S1 + S2) / N.
The only problem remained is how to calculate S2. Simple iteration won’t enter in time limit. Let’s think different. For each element, I need to make sum of differences between it and all smaller elements in the array a. As well, I need to make sum of all different between bigger elements than it and it. I’ll focus on the first part. I sort increasing array a. Suppose I’m at position i. I know that (i – 1) elements are smaller than ai. The difference is simply (i — 1) * ai — sum_of_elements_before_position_i. Sum of elements before position i can be computed when iterating i. Let’s call the obtained sum Sleft. I need to calculate now sum of all differences between an element and bigger elements than it. This sum is equal to Sleft. As a proof, for an element ai, calculating the difference ajai when aj > ai is equivalent to calculating differences between aj and a smaller element of it (in this case ai). That’s why Sleft = Sright.
As a conclusion, the answer is (S1 + 2 * Sleft) / N. For make fraction irreducible, you can use Euclid's algorithm. The complexity of the presented algorithm is O(N * logN), necessary due of sorting. Sorting can be implemented by count sort as well, having a complexity of O(maximalValue), but this is not necessary.
Official solution: 4383420
## 340D - Bubble Sort Graph
A good way to approach this problem is to notice that you can't build the graph. In worst case, the graph will be built in O(N2) complexity, which will time out. Also, notice that "maximal independent set" is a NP-Hard task, so even if you can build the graph you can't continue from there. So, the correct route to start is to think of graph's properties instead of building it. After sketching a little on the paper, you should find this property:
Lemma 1 Suppose we choose 2 indices i and j, such as i < j. We'll have an edge on the graph between vertices ai and aj if and only if ai > aj. We'll call that i and j form an inversion in the permutation.
Proof We assume we know the proof that bubble sort does sort correctly an array. To proof lemma 1, we need to show two things.
1. Every inversion will be swapped by bubble sort.
2. For each i < j when ai < aj, bubble sort will NOT swap this elements.
To proof 1, if bubble sort wouldn't swap an inversion, the sequence wouldn't be sorted. But we know that bubble sort always sorts a sequence, so all inversions will be swapped. Proofing 2 is trivial, just by looking at the code.
So far we've got how the graph G is constructed. Let's apply it in maximal independent set problem.
Lemma 2 A maximal independent set of graph G is a longest increasing sequence for permutation a.
Proof: Suppose we have a set of indices i1 < i2 < ... ik such as ai1, ai2, ..., aik form an independent set. Then, anyhow we'd choose d and e, there won't exist an edge between aid and aie. According to proof 1, this only happens when aid < aie. Hence, an independent set will be equivalent to an increasing sequence of permutation a. The maximal independent set is simply the maximal increasing sequence of permutation a.
The task reduces to find longest increasing sequence for permutation a. This is a classical problem which can be solved in O(N * logN). Here is an interesting discussion about how to do it.
## 340E - Iahub and Permutations
In this task, author's intended solution is an O(N ^ 2) dp. However, during testing Gerald fount a solution using principle of inclusion and exclusion. We've thought to keep both solutions. We're sorry if you say the problem was well-known, but for both me and the author of the task, it was first time we saw it.
Dynamic programming solution
After reading the sequence, we can find which elements are deleted. Suppose we have in a set D all deleted elements. I'll define from now on a "free position" a position which has -1 value, so it needs to be completed with a deleted element.
We observe that some elements from D can appear on all free positions of permutation without creating a fixed point. The other elements from D can appear in all free positions except one, that will create the fixed point. It's intuitive that those two "classes" don't influence in the same way the result, so they need to be treated separated.
So from here we can get the dp state. Let dp(n, k) = in how many ways can I fill (n + k) free positions, such as n elements from D can be placed anywhere in the free position and the other k elements can be placed in all free positions except one, which will create the fixed point. As we'll prove by the recurrences, we are not interested of the values from elements of D. Instead, we'll interested in their property: if they can(not) appear in all free positions.
If k = 0, the problem becomes straight-forward. The answer for dp(n, 0) will be n!, as each permutation of (n + 0) = n numbers is valid, because all numbers can appear on all free positions. We can also calculate dp(n, 1). This means we are not allowed to place an element in a position out of (n + 1) free positions. However, we can place it in the other n positions. From now we get n elements which can be placed anywhere in the n free positions left. Hence, dp(n, 1) = n! * n.
We want to calculate dp(n, k) now, k > 1. Our goal is to reduce the number k, until find something we know how to calculate. That is, when k becomes 0 or 1 problem is solved. Otherwise, we want to reduce the problem to a problem when k becomes 0 or 1. I have two cases. In a first case, I take a number from numbers which can be placed anywhere in order to reduce the numbers which can form fixed points. In the second case, I take a number from those which can form fixed points in order to make the same goal as in the first case. Let's analyze them.
Case 1. Suppose X is the first free position, such as in the set of k numbers there exist one which cannot be placed there (because it will make a fixed point). Obviously, this position exist, otherwise k = 0. Also obviously, this position will need to be completed with a term when having a solution. In this case, I complete position X with one of n numbers. This will make number equal to X from the k numbers set to become a number which can be placed anywhere. So I "loose" one number which can be placed anywhere, but I also "gain" one. As well, I loose one number which can form a fixed point.
Hence dp(n, k) += n * dp(n, k — 1).
Case 2. In this case position X will be completed with one number from the k numbers set. All numbers which can form fixed points can appear there, except number having value equal to X. So there are k — 1 of them. I choose an arbitrary number Y from those k — 1 to place on the position X. This time I "loose" two numbers which could form fixed points: X and Y. As well, I "gain" one number which can be placed anywhere: X.
Hence dp(n, k) += (k — 1) * dp(n + 1, k — 2).
TL;DR
dp[N][0]=N!
dp[N][1]=N*dp[N][0]
dp[N][K]=N*dp[N][K-1]+(K-1)*dp[N+1][K-2] for K>=2
This recurrences can be computed by classical dp or by memoization. I'll present DamianS's source, which used memoization. As you can see, it's very short and easy to implement. Link
Inclusion and exclusion principle
I'll present here an alternative to the dynamic programming solution. Let's calculate in tot the number of deleted numbers. Also, let's calculate in fixed the maximal number of fixed points a permutation can have. For calculate fixed, let's iterate with an index i each permutation position. We can have a fixed point on position i if element from position i was deleted (ai = -1) and element i does not exist in sequence a. With other words, element i was deleted and now I want to add it back on position i to obtain maximal number of fixed points.
We iterate now an index i from fixed to 0. Let sol[i] = the number of possible permutations having exactly i fixed points. Obviously, sol[0] is the answer to our problem. Let's introduce a combination representing in how many ways I can choose k objects out of n. I have list of positions which can be transformed into fix points (they are fixed positions). I need to choose i of them. According to the above definition, I get sol[i] = . Next, I have to fill tot - i positions with remained elements. We'll consider for this moment valid each permutation of not used values. So, sol[i] = . Where is the problem to this formula?
The problem is that it's possible, when permuting (tot — i) remained elements to be added, one (or more) elements to form more (new) fixed points. But if somehow I can exclude (subtract) the wrong choices from sol[i], sol[i] will be calculated correctly. I iterate another index j from i + 1 to fixed. For each j, I'll calculate how many permutations I considered in sol[i] having i fixed points but actually they have j. I'll subtract from sol[i] this value calculated for each j. If I do this, obviously sol[i] will be calculated correctly.
Suppose we fixed a j. We know that exactly sol[j] permutations have j fixed points (as j > i, this value is calculated correctly). Suppose now I fix a permutation having j fixed points. For get the full result, I need to calculate for all sol[j] permutations. Happily, I can multiply result obtained for a single permutation with sol[j] and obtain the result for all permutations having j fixed points. So you have a permutation having j fixed points. The problem reduces to choosing i objects from a total of j. Why? Those i objects chosen are actually the positions considered in sol[i] to be ones having exactly i fixed points. But permutation has j fixed points. Quoting for above, "For each j, I'll calculate how many permutations I considered in sol[i] having i fixed points but actually they have j" . This is exactly what algorithm does.
To sum up in a "LaTeX" way,
We can compute binomial coefficients using Pascal's triangle. Using inclusion and exclusion principle, we get O(N2). Please note that there exist an O(N) solution for this task, using inclusion and exclusion principle, but it's not necessary to get AC. I'll upload Gerald's source here.
## 341D - Iahub and Xors
The motivation of the problem is that x ^ x = 0. x ^ x ^ x… ^ x (even times) = 0
Update per range, query per element
When dealing with complicated problems, it's sometimes a good idea to try solving easier versions of them. Suppose you can query only one element each time (x0 = x1, y0 = y1).
To update a submatrix (x0, y0, x1, y1), I’ll do following operations. A[x0][y0] ^= val. A[x0][y1 + 1] ^= val. A[x1 + 1][y0] ^= val. A[x1 + 1][y1 + 1] ^= val.
To query about an element (X, Y), that element’s value will be the xor sum of submatrix A(1, 1, X, Y). Let’s take an example. I have a 6x6 matrix and I want to xor all elements from submatrix (2, 2, 3, 4) with a value. The below image should be explanatory how the method works:
Next, by (1, 1, X, Y) I’ll denote xor sum for this submatrix.
“White” cells are not influenced by (2, 2, 3, 4) matrix, as matrix (1, 1, X, Y) with (X, Y) a white cell will never intersect it. “Red” cells are from the submatrix, the ones that need to be xor-ed. Note that for a red cell, (1, 1, X, Y) will contain the value we need to xor (as it will contain (2, 2)). Next, “blue” cells. For this ones (1, 1, X, Y) will contain the value we xor with, despite they shouldn’t have it. This is why both (2, 5) and (4, 2) will be xor-ed again by that value, to cancel the xor of (2, 2). Now it’s okay, every “blue” cell do not contain the xor value in their (1, 1, X, Y). Finally, the “green” cells. These ones are intersection between the 2 blue rectangles. This means, in their (1, 1, X, Y) the value we xor with appears 3 times (this means it is contained 1 time). For cancel this, we xor (4, 5) with the value. Now for every green cell (1, 1, X, Y) contains 4 equal values, which cancel each other.
You need a data structure do to the following 2 operations:
• Update an element (X, Y) (xor it with a value).
• Query about xor sum of (1, 1, X, Y).
Both operations can be supported by a Fenwick tree 2D. If you don't know this data structure, learn it and come back to this problem after you do this.
Coming back to our problem
Now, instead of finding an element, I want xor sum of a submatrix. You can note that xor sum of (x0, y0, x1, y1) is (1, 1, x1, y1) ^ (1, 1, x0 – 1, y1) ^ (1, 1, x1, y0 – 1) ^ (1, 1, x0 – 1, y0 – 1). This is a classical problem, the answer is (1, 1, x1, y1) from which I exclude what is not in the matrix: (1, 1, x0 – 1, y1) and (1, 1, x1, y0 – 1). Right now I excluded (1, 1, x0 – 1, y0 – 1) 2 times, so I need to add it one more time.
How to get the xor sum of submatrix (1, 1, X, Y)? In brute force approach, I’d take all elements (x, y) with 1 <= x <= X and 1 <= y <= Y and xor their values. Recall the definition of the previous problem, each element (x, y) is the xor sum of A(1, 1, x, y). So the answer is xor sum of all xor sums of A(1, 1, x, y), with 1 <= x <= X and 1 <= y <= Y.
We can rewrite that long xor sum. A number A[x][y] appears in exactly (X – x + 1) * (Y – y + 1) terms of xor sum. If (X – x + 1) * (Y – y + 1) is odd, then the value A[x][y] should be xor-ed to the final result exactly once. If (X — x + 1) * (Y — y + 1) is even, it should be ignored.
Below, you'll find 4 pictures. They are matrixes with X lines and Y columns. Each picture represents a case: (X odd, Y odd) (X even, Y even) (X even Y odd) (X odd Y even). Can you observe a nice pattern? Elements colored represent those for which (X – x + 1) * (Y – y + 1) is odd.
Yep, that's right! There are 4 cases, diving the matrix into 4 disjoint areas. When having a query of form (1, 1, X, Y) you only need specific elements sharing same parity with X and Y. This method works in O(4 * logN * logN) for each operation and is the indented solution. We keep 4 Fenwick trees 2D. We made tests such as solutions having complexity greater than O(4 * logN * logN) per operation to fail.
Here is our official solution: 4383473
## 341E - Candies Game
Key observation Suppose you have 3 boxes containing A, B, C candies (A, B, C all greater than 0). Then, there will be always possible to empty one of boxes using some moves.
Proof We can suppose that A <= B <= C. We need some moves such as the minimum from A, B, C will be zero. If we always keep the numbers in order A <= B <= C, it’s enough some moves such as A = 0. I’ll call this notation (A, B, C).
How can we prove that always exist such moves? We can use reductio ad absurdum to prove it. Let’s suppose, starting from (A, B, C) we can go to a state (A2, B2, C2). We suppose A2 (A2 > 0) is minimal from every state we can obtain. Since A2 is minimal number of coins that can be obtained and A2 is not zero, the statement is equivalent with we can’t empty one chest from configuration (A, B, C). Then, we can prove that from (A2, B2, C2) we can go to a state (A3, B3, C3), where A3 < A2. Obviously, this contradicts our assumption that A2 is minimal of every possible states. If A2 would be minimal, then there won’t be any series of moves to empty one chest. But A2 isn’t minimal, hence there always exist some moves to empty one chest.
Our algorithm so far:
void emptyOneBox(int A, int B, int C) {
if A is 0, then exit function.
Make some moves such as to find another state (A2, B2, C2) with A2 < A.
emptyOneBox (A2, B2, C2);
}
The only problem which needs to be proven now is: given a configuration (A, B, C) with A > 0, can we find another one (A2, B2, C2) such as A2 < A? The answer is always yes, below I’ll prove why.
Firstly, let’s imagine we want to constantly move candies into a box. It doesn't matter yet from where come the candies, what matters is candies arrive into the box. The box has initially X candies. After 1 move, it will have 2 * X candies. After 2 moves, it will have 2 * (2 * X) candies = 4 * X candies. Generally, after K moves, the box will contain 2^K * X candies.
We have A < B < C (if 2 numbers are equal, we can make a move and empty 1 box). If we divide B by A, we get from math that B = A * q + r. (obviously, always r < A). What if we can move exactly A * q candies from B to A? Then, our new state would be (r, B2, C2). We have now a number A2 = r, such as A2 < A.
How can we move exactly A * q coins? Let’s write q in base 2. Making that, q will be written as a sum of powers of 2. Suppose lim is the maximum number such as 2 ^ lim <= q. We get every number k from 0 to lim. For each k, I push into the first box (the box containing initially A candies) a certain number of candies. As proven before, I'll need to push (2 ^ k) * A candies. Let's take a look at the k-th bit from binary representation of q. If k-th bit is 1, B will be written as following: B = A * (2 ^ k + 2 ^ (other_power_1) + 2 ^ (other_power_2) + ...) + r. Hence, I'll be able to move A * (2 ^ k) candies from "B box" to "A box". Otherwise, I'll move from "C box" to "A box". It will be always possible to do this move, as C > B and I could do that move from B, too.
The proposed algorithm may look abstract, so let's take an example.
Suppose A = 3, B = 905 and C = 1024. Can we get less than 3 for this state?
B = 3 * 301 + 2. B = 3 * (100101101)2 + 2.
K = 0: we need to move (2^0) * 3 coins into A. 0th bit of q is 1, so we can move from B to A.
A = 6, B = 3 * (100101100)2 + 2 C = 1024
K = 1: we need to move (2 ^ 1) * 3 coins into A. Since 1th bit of q is already 0, we have to move from C.
A = 12, B = 3 * (100101100)2 + 2 C = 1018
K = 2: we need to move (2 ^ 2) * 3 coins into A. 2nd bit of q is 1, so we can move from B.
A = 24, B = 3 * (100101000)2 + 2 C = 1018
K = 3: we need to move (2 ^ 3) * 3 coins into A. 3nd bit of q is 1, so we can move from B.
A = 48, B = 3 * (100100000)2 + 2 C = 1018
K = 4. we need to move (2 ^ 4) * 3 coins into A. 4th bit of q is 0, we need to move from C.
A = 96, B = 3 * (100100000)2 + 2 C = 970
K = 5. we need to move (2 ^ 5) * 3 coins into A. 5th bit of q is 1, so we need to move from B.
A = 192, B = 3 * (100000000)2 + 2 C = 970
K = 6 we need to move (2 ^ 6) * 3 coins into A. We mve them from C.
A = 384 B = 3 * (100000000)2 + 2 C = 778
K = 7 we need to move (2 ^ 7) * 3 coins into A. We move them from C
A = 768 B = 3 * (100000000)2 + 2 C = 394
K=8 Finally, we can move our last 1 bit from B to A.
A = 1536 B = 3 * (000000000)2 + 2 C = 394
A = 1536 B = (3 * 0 + 2) C = 394
In the example, from (3, 905, 1024) we can arrive to (2, 394, 1536). Then, with same logic, we can go from (2, 394, 1536) to (0, X, Y), because 394 = 2 * 197 + 0.
This is how you could write emptyOneBox() procedure. The remained problem is straight-forward: if initially there are zero or one boxes having candies, the answer is "-1". Otherwise, until there are more than 2 boxes having candies, pick 3 boxes arbitrary and apply emptyOneBox().
Here is a source implementing the algorithm. 4383485
## BONUS
Instead of a conclusion, I'll post here related problems to the ones used in the round. :) Please note that some of them might be more easier / complicated than level of difficulty used in the round. Feel free to think of them / ask help / discuss them in the comment section :)
Div2 A Suppose x, y, A, B ≤ 109. Instead of being asked how many bricks are colored with both red and pink in range [A, B], you're asked how many bricks are colored with at least one color. After you solve this one, solve the same problem, but instead of having 2 persons painting, you have k persons (k ≤ 20). Solution by Enchom
Div2 B Given a very long list of special points, can you find quickly a convex special quadrilateral? Can you find very very quickly? :) Also, can you find maximal area of a special convex quadrilateral in time better than O(N4)? Solutions for first problem and second problem provided by Xellos and Enchom
Div2 D / Div1 B Suppose the reverse problem. You are given a bubble sort graph having N vertices and M edges. Find its independent maximal set. Can you achieve O(N2) to do this? Does a solution in O((N + M) + N * logN) exist? Solution by CNT0
Div2 E / Div1 C Find a solution running in liniar time. Solution (dynamic programming) by ivan100sic . Solution (inclusion exclusion principle) by eduardische
Div1 D Suppose the 3D version of this problem. You have a 3D matrix and you perform same QUERY/UPDATE operations, but using 6 parameters (a submatrix is defined now all elements a[i][j][k] for which x0 <= i <= x1, y0 <= j <= y1, z0 <= k <= z1). Can you get a solution using O(log3 * N) per query, having constant 8? But for d dimensions, does an O(2d * (logd)n) algorithm per query exist? :) Solution by Dwylkz.
Div1 E In our algorithm, we pick arbitrary 3 boxes. Can you find some heuristics of picking 3 boxes to reduce number of moves?
Tutorial of Practise Contest #01
• +130
» 6 years ago, # | ← Rev. 2 → -12 Great! Thanks for the detailed and accurate Editorial!
• » » 2 years ago, # ^ | 0 Why was this comment downvoted so much? Tbh I don't understand the downvoting trends of CF users. Just below me a comment which was "Yes!" got upvoted +4, but a praise for the editorial gets downvoted -12. Weird.
» 6 years ago, # | +4 Yes!
» 6 years ago, # | -6 Lemma2 of 340 D is a little sudden for me.
» 6 years ago, # | +20 Regarding problem D. We made tests such as solutions having complexity greater than O(4 * logN * logN) per operation to fail. Well, 4 * logN * logN (nit — O() notation here is incorrect, i.e. O(10000f) = O(f)) per operation also fails if it is not Fenwick tree (e.g., 2-d segment tree). Also, it seems that even carefully implemented Fenwick solution in Java runs longer than 0.5s, so TL should be >1s. Did jury have a Java solution?I'd like also to note that for Fenwick trees there exists a more general approach for range updates, which works, e.g., for sum instead of xor (see this post).
• » » 6 years ago, # ^ | +6 I've got AC with 2-D segment tree.
• » » » 6 years ago, # ^ | ← Rev. 2 → 0 Does your code have complexity O(logn*logn) for queries? I always thought that 2D segment tree is O(n) for queries like (1,1,1,n)Edit: Sorry, now I understand :D
• » » » » 6 years ago, # ^ | +8 One should distinguish 2-D segment tree and quad-tree. The first one has complexity O(logN·logN) per query, while the second one works in O(N) per query.
• » » » » » 6 years ago, # ^ | 0 I implemented a quad tree and it was TLE in 6th test case. Can you help me to understand why an update or a query in a Quad tree is O(n). My solution stops the recursion when the interval asked or updated is same to the interval of the node. If you know an online explanation please share it. Thanks for the help http://codeforces.com/contest/341/submission/4390134
• » » » » » » 6 years ago, # ^ | 0 try querying a 1xN row. you will realize that you ultimately have to query N 1-unit squares.
• » » » » » » » 6 years ago, # ^ | 0 Thanks. Very clear example.
• » » 6 years ago, # ^ | +1 Are you sure that the approach mentioned in Petr's post can be generalized to 2D or more dimensions (for supporting range updates and range queries) ?
• » » » 6 years ago, # ^ | 0 I suppose it can: http://codeforces.com/contest/341/submission/4391440
• » » » » 6 years ago, # ^ | ← Rev. 2 → 0 It is interesting) It is writen in statememnt that number v from update query ≤ 262, but your solution using only int got accepted.
• » » » » » 6 years ago, # ^ | 0 Maybe it's because only the parity of the number matter, so overflow doesn't affect the result?
» 6 years ago, # | +36 Realizing that the "Bubble Sort Graph" problem was equivalent to finding the length of the longest increasing subsequence was a really nice "Aha!" moment for me. Great problems, thanks!
» 6 years ago, # | +17 Where are the pictures in 341D?
• » » 6 years ago, # ^ | 0 They work perfectly for me. I uploaded them at imageshack.us . Maybe you a problem with that server. Please, tell me a server for which the images will be visible to you and I'll upload there.
• » » » 6 years ago, # ^ | 0 Thank you for your kind reply! it's visible to me now.
» 6 years ago, # | +3 For 340B I used the Gift Wrap alogirithm to compute a Convex Hull for the points, and then using a DP found the minimum Area that needs to be cut off from this Hull to form a quadrilateral. Worst Case for Hull O(n^3) + DP O(n^2) Total execution time was 30msThe only corner case was when the hull is composed of only 3 points.
» 6 years ago, # | +3 I found that most of the solutions marked correct for the problem 340D gave the answer 1 for the input3 12 13 1Whereas it should have been 2(maximal set can be {12,13}). Did the test cases miss something or am I making a really silly mistake here?Can I hack solutions after the match has ended. If so could anyone please let me know.
• » » 6 years ago, # ^ | 0 My bad :(. I just re-read the problem and found that the numbers can be <=n
» 6 years ago, # | +4
» 6 years ago, # | +2 Thanks to author for the contest. It was the best raund I have ever seen.
» 6 years ago, # | +4 Can you please elaborate on how the (i — 1) * ai = sum_of_elements_before_position_i in question 341C ???
• » » 6 years ago, # ^ | 0 For a fixed ai, you need to calculate sum of ai — aj, when aj < ai. If I sort sequence a in increasing order, then all elements aj < ai will be in positions {1, 2, ..., i — 1}. So I need to calculate ai — a1 + ai — a2 + ... ai — a(i-1). Ai appears (i — 1) times in this sum, so I re-write the sum being (i — 1) * ai — a1 — a2 — ... — a(i — 1). This is (i — 1) * ai — (a1 + a2 + ... + a(i — 1)). I need to sum for each i this expression. It can be computed in O(N), see my source for reference. There, sumBefore is (a1 + a2 + ... + a(i — 1)).
• » » » 6 years ago, # ^ | 0 Thanks for your reply.
» 6 years ago, # | +16 Try Geogebra to draw good Geometric figures. It's free.
» 6 years ago, # | 0 For 340E, can't we just do dp[n][k] = n*dp[n-1][k] + k*dp[n][k-1] ?
» 6 years ago, # | +4 This is the most explanatory editorial I have ever seen on codeforces :) Really nice work guys :)
» 6 years ago, # | +6 Well, turns out div2 E / div1 C has an even simpler O(N) solution. Take a look at becauseofyou's elegant solution, or perhaps mine.
• » » 6 years ago, # ^ | 0 Just like the number of desarrangements of length n,right?
• » » 6 years ago, # ^ | 0 please explain the solution.
• » » 6 years ago, # ^ | 0 Can you please tell, what +(i-1)*D[i-2] is needed for?I had an idea like this (which got WA, I still dont know why). Here: D[i] = ends*D[i-1] + (i-1)*D[i-1]; D[i] = (ends+i-1)*D[i-1];
• » » » 6 years ago, # ^ | ← Rev. 2 → 0 BTW, my idea is:I place i in ith position (which makes fixed point). Then for each D[i - 1] permutations, I swap i with some other number, as none of the rest will make fixed point while in ith position, and i won't make fixed point in any other position. So for each D[i - 1] (correct) permutations, I have (ends + i - 1) different numbers to swap i with and I get D[i] = (ends+i-1)*D[i-1]. But I miss something, something with D[i-2]. :(
» 6 years ago, # | +5 Div2 D Bonus solution. If there is no edge between vertices ai > aj, then i > j, else i < j. One is easy to build the original sequence by applying any sorting algorithm with this comparator
• » » 6 years ago, # ^ | 0 Nice solution. :) I didn't thought of that one. BTWs, for those interested there exist one way to build the sequence in linear time. You can google about research article: McConnell, Ross M.; Spinrad, Jeremy P. (1999), "Modular decomposition and transitive orientation"
» 6 years ago, # | ← Rev. 4 → +9 Here is my solution for 341D.Define: "sum" stand for xor sum. + , - stand for xor. a * b means that apply b + b for a - 1 times. Notice that we could do + and - subtraction interval-wise. But Fenwick tree could only update per range and query per element or vice versa. If, for each element, say, ft(x, y) hold the sum of prefix sum of the matrix (1, 1, x, y) (ie. prefix sum of prefix sum of (x, y)), then for each query, ft(x1, y1) - ft(x0 - 1, y1) - ft(x1, y0 - 1) + ft(x0 - 1, y0 - 1) would be the result. As the difference between prefix-sum is the sum of the corresponding interval. And it's easy to xor a v to elements in submatrix (x0, y0, x1, y1). Consider 1 dimension version of this problem. For a sequence a, the prefix sum of x is and the prefix sum of prefix sum of x is . Hence, all we have to do is to maintain an extra sequnce i * ai. Then expand method above to arbitrary dimension. Here is an implement.
• » » 6 years ago, # ^ | ← Rev. 2 → 0 can your method also be used if we have to perform the following operations on a 2D matrix: update x1, y1, x2, y2, k. in this operation, we add k to each element in sub-matrix x1, y1, x2, y2. query x1, y1, x2, y2. in this, we output the sum of all elements in sub-matrix x1, y1, x2, y2. can the same logic be extended??EDIT: Yes, it can be. thank you.
• » » » 6 years ago, # ^ | ← Rev. 2 → 0 yes, and, it could be extended to 3 or more dimensions, too :)
» 6 years ago, # | +8 Nicely written tutorial, and really good round in my opinion, looking forward for the solution of problem 341E, and also loving the idea of Bonus problems, really good authors! Thank you for the lovely round! :)
» 6 years ago, # | 0 340-B maximum area quad is really tough to pass the time limit with python ! I can see no accepted solutions !!
» 6 years ago, # | ← Rev. 3 → 0 For problem Div2 340E , in the case 2 : "Case 2. In this case position X will be completed with one number from the k numbers set..." if the position X with X is an used number, how the total number of elements which can be placed anywhere increase? For example n = 7 , -1 -1 4 2 -1 -1 -1 => N = 1 {3 } , K = 4 {1 , 5, 6, 7} , so if we put a number in K at position 2, the number of N will not increased!
• » » 6 years ago, # ^ | 0 X positions can be {1, 5, 6, 7}. Only those positions can create fixed points. Suppose we choose now arbitrary X = 1. I can place there numbers Y = {5, 6, 7}. Now, permutation will look like Y -1 4 2 -1 -1 -1. Obviously, 1 can be placed anywhere now, hence number N is increased.
• » » » 6 years ago, # ^ | 0 Thank you for your answer, but I am still not clear :DAbout position number 2, which case it corresponding to? like in the above case -1 Y 4 2 -1 -1 -1 -> number N will not increased.
• » » » » 6 years ago, # ^ | ← Rev. 2 → 0 You don't deal with it for the moment. You deal only with set of positions {1, 5, 6, 7}. Each time, you reduce one or two elements from the set (depending of choosing case 1 or case 2). When set has 0 or 1 element, then you can handle the other positions (see initialization of dp, it will handle position 2 also). As each time K decreases by 1 or 2, it's obvious sometime you'll be in the basic case, when you'll handle the other positions.
• » » » » » 6 years ago, # ^ | 0 Thank you :D will need time to digest it.
» 6 years ago, # | ← Rev. 9 → +3 I wonder if the authors will post the solutions to bonus problems after some time? :)Anyway, here is my idea of the Div2 A — Bonus, sorry if there is much simpler solution, this is what i thought of :)Let's say that the i-th painter will paint all bricks divisible by X[i]. I will assume that the least common multiple of X[1], X[2], X[3]...X[k] will fit in long long. (If it doesn't you can easily still solve it using the same method, it's just for the sake of simplicity). Let's define a certain colouring by a binary number. The 1s in the binary number correspond to the colors that were surely used ( 0 doesn't necessary mean the colour isn't used ). For example for K=5, 10100 would define that surely the first and third colours were used in that colouring.To get the answer we iterate over all possible colourings, and if the colouring contains odd number of ones, we add the amount of bricks that are coloured that way, otherwise we remove that amount. The solution has complexity O(2^k*log) if implemented good, because there are 2^k possible colourings and for each we need to find the least common multiple of the used colours to determine the amount of bricks that are coloured that way.I know it sounds quite weird and at the moment i'm struggling to find a proof, but i have tested it and it works. If someone could provide a proof that would be great :)
• » » 6 years ago, # ^ | +5 Yes, I'll post my ideas for Bonus tasks if there are enough persons interested. :) This is a classic task solvable by inclusion and exclusion principle. If I understood well what your solution does, it uses exactly this principle. Google it and you'll find proof for your solution :)
• » » » 6 years ago, # ^ | ← Rev. 3 → +3 I'd love to see the solutions, i've spent A LOT of time on those bonus problems :).Here is what i consider to be my "proof" for the solution i described above :PProofLet's imagine we are iterating the binary numbers in order of increasing amount of ones in them. If we are for example at the number 01010, then we want to count the bricks colored that way, exactly once. However we have already counted them when looking at binary numbers 01000 and 00010. So how many times exactly have we counted a given binary number? Well if the binary number has exactly P ones in it, then we have counted it C(1,P) times when looking at numbers with 1 one in them, C(2,P) times when looking at numbers with 2 ones in them and so on. However remember that we do not always add, we subtract too, so we get the following :For a number with P ones in it, we have counted it exactly S times where S=C(1,P)-C(2,P)+C(3,P)-C(4,P)+C(5,P)...C(P-1,P).Well it took me quite a lot of time, but not that difficultly it can be proven that -C(0,P)+C(1,P)-C(2,P)+C(3,P)...C(P,P) equals 0, therefor obviously for odd P, S=0 and for even P — S=2. And since we want to count each colouring exactly ones, obviously for odd P, we add it once and for even P we subtract it.C(K,N) = binomial coefficientIt's a little bit clumsy but i made it myself because i haven't heard about inclusion-exclusion principle :)P.S. Sorry for all those long posts, i just really liked the problems and spent a lot of time on them :)
• » » 6 years ago, # ^ | 0 I think this is the simplest solution. It uses inclusion-exclusion principle. This problem on SPOJ
» 6 years ago, # | 0 Problem 340CI didn't understand second step to calculate sum_of_all_routes. Can someone explain with the help of example.
• » » 6 years ago, # ^ | ← Rev. 3 → +5 this step is to compute the sum of all possible differences |ai-aj| where (i != j) in all possible permutations.consider we have the sequence a1, a2, a3, a4, a5, a6 and we want to get how many times a2 will come before a3 directly by keeping a1 a4 a5 a6 in this order as a subsequence I will list all those permutations that fullfill the last condition : a2 a3 a1 a4 a5 a6a1 a2 a3 a4 a5 a6a1 a4 a2 a3 a5 a6a1 a4 a5 a2 a3 a6a1 a4 a5 a6 a2 a3and the number of those permutations is n-1 = 6-1 = 5. and those 5 times is when a1 a4 a5 a6 comes in this order and to compute how many times for all possible orders of a1 a4 a5 a6 then we multiply by (n-2)! = (6-2)! = 4! = 24 so the number of times that a2 a3 comes consecutivly is (n-1)*(n-2)! = (n-1)! and the sum of all differences |a2-a3| is (n-1)!*|a2-a3|. S2 is the sum of all differences |ai - aj| for each i != j .to compute S2 : for each ai we want to compute the differece between it and all other elements aj where i != j and let us sort the given sequence a1 a2 a3 a4 a5 a6 in increasing order then the sum of all differences |ai — ak| where k
• » » » 6 years ago, # ^ | 0 Thanks, that was very good explaination.
• » » » 6 years ago, # ^ | 0 Why they are dividing s2 * (n-1)! to n!? Please help me
» 6 years ago, # | ← Rev. 6 → +15 Here is my idea on the Div2 B "Can you find maximal area convex quadrilateral", I spent a lot of time on it because it was the only problem that the author didn't have solution for :).Its only theoretical, I haven't tested it because I am not able to write source now (I'm currently at the Balkan Olympiad), so if it is wrong please do not hate :)My solutionOk first, we need to find a way to check if a quadrilateral is convex while constructing it, and to do that we can simply use that quadrilateral's diagonals cross if and only if it's convex. Therefor if we find two crossing lines, created from 4 special points, we have a special convex quadrilateral.Now as first step of our algorithm, we will take every two points and construct a line, let the points be A and B, then we will imagine its infinite continuation in the plane, and also two infinite lines perpendicular to our line and crossing it in the special points. That will divide the plane in 6 sections, but for the sake of simplicity we will look only on 3 of them (on one of the sides) because they are simetrical. Here is a picture (special points are red) :So we have three zones, blue, green and yellow. Now let's look at all other special points that are in one of those zones. We will do a few rotating sweeping lines to have some ordered sets of points. First sweeping line will start from point A to infinite in direction opposite to B paralel to AB and save all the points in the BLUE AREA ONLY , second line will start from point A to infinite being perpendicular to AB and will save all points in the GREEN AREA ONLY, third line is the same as the second but starts from B and rotates in the opposite direction, again saves all points in the GREEN AREA ONLY, and fourth and last line is same as the first, but start from B and saves all points in the YELLOW AREA ONLY. Here is a picture of the lines and their rotating directions :For each special point we will save the cross product it gives with line AB. So for each line we have four sets of points where the second and third actually contain the same points but with different orderings. We put those 4 sets in 4 RMQ structures so we could later get the highest of the cross products in some interval fast. Assuming we use roughly O(N*LOGN) for the sweeping lines and to build RMQ, and O(N^2) to fix a line, we have complexity of O(N^3*LOGN) for now.Now let's proceed with our actual solution. We fix a line AB and a third point C. Now we want such point D so that AB and CD cross (being the diagonals of convex quadrilateral) and the cross product of D to the line AB is maximum. Let's look at a picture of points A,B and C :Obviously for CD to cross AB, D must be in the area marked with white above. So now how to fast get the point with maximum cross product to AB in that white area? Well we have 3 different cases depending on the place of C in the initial 6 zones that I divided the plane into. Let's take a look.Case 1When C is on the other side of the blue areaSo obviously we want D to be in the pink area, pink area covers part of the green and part of the yellow areas. For the green area we can use the set of points from sweeping line 2. By binary we will find the first point that is in the pink area, and then get RMQ query for that suffix of the array of points. Same thing we can do with the yellow area, using the points from sweeping line 4. Therefor we can find the best point D in O(LOGN).Case 2When C is on the other side of the green areaNow obviously we cover the whole green area, so we can just call a RMQ query of the whole range on any of the sweeping line 2 or 3, and for the blue part that is covered, we can again do binary and then RMQ query in points from sweep line 1, and the same thing for the yellow parts, with the points from sweep line 4.Case 3When C is on the other side of the yellow areaI'm not gonna show a picture of that as it is really symmetrical to case 1, you just again use binary and RMQ query in sweep lines 1 and 3 for the blue and green parts that are covered.Obviously each test case takes roughly O(LOGN) and fixing points A,B,C is O(N^3), so i think that solution would be O(N^3*LOGN) in general. I'm sorry if it is wrong i am unable to test it right now, but I really took a lot of time to explain it nicely (sorry if it isn't very understandable), so hope I helped :)P.S. Sorry for the long post, I was excited to share my idea :)
• » » 6 years ago, # ^ | +5 dude, you're awesome!
• » » 6 years ago, # ^ | 0 Isn't the author's solution O(N3)?
• » » » 6 years ago, # ^ | ← Rev. 2 → +6 Author's solution describes a solution for finding maximal area quadrilateral, however it doesn't guarantee it will be a convex one. In the bonus tasks the author asked if anyone could find a maximal area convex quadrilateral in less than O(N^4), because he couldn't find a solution, and I took it as a challenge. It is a bonus task :)
• » » » » 6 years ago, # ^ | +5 Ah, right. I didn't notice that. Seems quite tough to implement... well done.And a challenge atop a challenge: yesterday's gym ACM training, problem I: find the largest convex polygon which doesn't contain any of additional specified points (vertices in "red" points, doesn't contain any "blue" points).
• » » 6 years ago, # ^ | 0 By the way, one could use only two sweeping lines (merging 1 with 2 and 3 with 4, but you would need two binary searches for determining both ends of the interval to query about.Also a question to the author, when you say "find a convex polygon very very fast", what complexity do you mean? The input is O(N) so I guess it's above that, but is it like O(NLOGN) or O(N^2)?
• » » » 6 years ago, # ^ | ← Rev. 2 → +5 Your solution to Bonus task B is really nice :) Good job. I added it to the post.You have to find a convex quadrilateral, not a convex polygon. And yes, this one can be done very very quick. To be precise, no matter how big is N, you can find a convex quadrilateral in O(1). Suppose you have to read also the set of points, the complexity is still O(1). I hope it does not spoil too much :)
• » » » » 6 years ago, # ^ | +8 I got it!Among any 5 points (with any 3 non-collinear, so there are no degenerate cases), there are some 4 which form a convex quadrilateral.Suppose we take just 4 points. Either they form a convex quadrilateral, or there's one point (P4) that's inside the triangle T (with vertices P1..3) formed by the other 3. Now, take that case.A quad. is convex iff its diagonals intersect. So if the 5th point (P5) is outside T, then one of the triangle's sides and segment P4-P5 intersect, so there's a convex quad. defined by them.Now, take the situation when the P5 is inside T, which is split by P4 into 3 smaller triangles (by segments P1-P4, P2-P4, P3-P4). P5 must be in one of them (w.l.o.g. in P1P2P4). And again, the segment P3-P5 must cross one of the segments P1-P4 and P2-P4, so we have a convex quadrilateral.What we need to do is: read the first 5 points of the list, try all possible quadrilaterals formed by them and check which one (there may be more than 1, but we only need one) is convex. Since we work on N=5, time is O(1). Tada!
• » » » » » 6 years ago, # ^ | 0 Yep, this is the solution :)Anyway, this is a Ramsey theory problem, which was studied for convex N-gons also. You can read details here
• » » » » » » 6 years ago, # ^ | +3 Ooh, the generalization's nice! This will go to a slovak math training for sure (hehehe, let them suffer :D)!
» 6 years ago, # | -8 Regarding problem C |ai - aj| for each i != j by S2, the answer is (N – 1)! * S1 + (N – 1)! * S2 / N!. In this formula i didn't understand that why they divide to N!. I will be pleasant if someone helps me.
» 6 years ago, # | 0 You can use Inclusion Exclusion without all that stuff, just using the number of -1's that appeared and the number of numbers that may be fixed points...just do the math as you want to compute the derrangements and you would get something like: where n is the number of -1's and m is the number of positions that can be fixed points(the ones that have a -1 and it's index is not already used)
» 6 years ago, # | 0 this is gotta be the best tutorial of any contest of #CF ..... very nice :D (Y)
» 6 years ago, # | 0 I have a qustion for B div2. Is it true, that 4 dots in the answer will be the part of convex hull?
» 6 years ago, # | 0 Could anyone explain why "A number A[x][y] appears in exactly (X – x + 1) * (Y – y + 1) terms of xor sum." in Div1D problem ? What does "formula (X – x + 1)*(Y – y + 1) " mean, how we find it ?
• » » 6 years ago, # ^ | 0 You're interested, for each element A[x][y], how many submatrixes it influences. So, if by (1, 1, a, b) I denote xor sum of all elements in submatrix (1, 1, a, b), how many (1, 1, a, b) do contain cell (x, y)?You can see a submatrix (1, 1, a, b) contains cell (x, y) if a >= x and b >= y. So a can be in set {x, x + 1, ...., X} and b can be in set {y, y + 1, ..., Y}. First set contains X — x + 1 values and second one contains Y — y + 1 values. Choosing a and b is independent, even if I fix an a, the b can still take any value. Using rule of product, you get the number of matrixes (1, 1, a, b) which contain (x, y) is (X — x + 1) * (Y — y + 1).Let cnt be number of submatrixes (1, 1, a, b) which contain (x, y). If cnt is even, number cancel each other in xor sum, so no need to xor A[x][y]. If cnt is odd, exactly one A[x][y] will remain, other cnt — 1 will cancel each other, since cnt — 1 is even number.Now saying for each cell (x, y) you xor-ed value A[x][y] each time (x, y) appears in a submatrix (1, 1, a, b) is equivalent to say for each submatrix (1, 1, a, b) you xor-ed values of all elements it contains. Is it clear why?
• » » » 6 years ago, # ^ | 0 Yes, now it is pretty clear, I got it exactly. Thanks a lot for spending time to explain it.
» 5 years ago, # | +3 Thanks a lot for this great tutorial. I am very satisfied specially at problem E of div2. The problem was fun. And so the dp solution was. :)
» 4 years ago, # | 0 I think there's a minor error in BubbleSortGraph's problem statement and editorial:It is written:if a[i] > a[i + 1] then add an undirected edge in G between a[i] and a[i + 1]But, the edge should be between indexes i and i+1 instead of a[i] and a[i+1], consider this example:2 3 2 6 8The LIS here is: 2 3 6 8 , but there's an edge between 3 and 2 as indexes 2 and 3 are an inversion.
• » » 4 years ago, # ^ | 0 Sorry, it's correct. I didn't notice that the elements are distinct.
» 3 years ago, # | ← Rev. 3 → -10 For 340D Bubble Sort Graph: Simple algoritham for LIS (Time complexity — O(nlogn) ) — Without using binary indexed tree,you just need to maintain one array. http://www.geeksforgeeks.org/longest-monotonically-increasing-subsequence-size-n-log-n/Why downvotes?My accepted solution using LIS(TC: nlogn), Without using binary indexed tree, http://codeforces.com/contest/340/submission/27713075
• » » 2 years ago, # ^ | 0 I couldn't understand,how 340D is related to longest increasing subsequence??? please reply me clearly .
• » » » 2 years ago, # ^ | 0 Read lemma 2 in above editorial.
• » » » » 2 years ago, # ^ | 0 got it
» 2 years ago, # | 0 I couldn't understand,how 340D is related to longest increasing subsequence??? please reply me clearly .
» 9 months ago, # | 0 In question B, Maximum Area Quadrilateral, Can anyone explain the solution more clearly ( The image in editorial is also unavailable ) specially the corner case. Thank You | 2020-02-18 11:20:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6623224020004272, "perplexity": 756.0201872244994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875143646.38/warc/CC-MAIN-20200218085715-20200218115715-00307.warc.gz"} |
http://math.stackexchange.com/questions/116378/properties-of-p-adic-valuations | Let $n$ be a integer not equal to zero, $p$ a prime and $$|n|_p = p^{-\operatorname{ord}_p(n)}; |0|_p=0$$
Proposition: $|xy|_p=|x|_p|y|_p$
We want to show: $\operatorname{ord}_p(ab)=\operatorname{ord}_{p}(a)+\operatorname{ord}_p(b)$ So if we suppose there is a largest integral $v$ such that $p^{v}|(ab)$, then we can split up v such that $p^{v_1}|a,p^{v_2}|b$. If we suppose that there exist $v_1,v_2$ with them both being the largest integers such that $p^{v_1}|a, p^{v_2}|b$, with $v_1+v_2= : v$ then $p^v|ab$. (I don't believe this to be a formal proof but rather my best own "idea" I can come up with...)
The script also mentions that the padic valuations' sums are bounded: $|x+y|_p\le \sup\{|x|_p,|y|_p\}$, how to show this?
Edit:
According to Robert Israel and Bill Dubuques answers, the first proposition is proven, if I understood correctly:
Given $ord(a)= v$ and $ord(b) = u$ which is the same as $a:= p^{v}m, b:= p^{u}n$ with $gcd(m,p)=gcd(n,p)=1$, because $ab=p^{u+v}mn$ with $gcd(mn,p)=1$ it follows that $ord(ab)=ord(a)+ord(b)$ and so $|xy|_{p}=p^{-ord(xy)}=p^{-(ord(x)+ord(y))}=|x|_{p}|y|_{p}$
Now to show: $|x+y|_{p}\le sup\{|x|_{p},|y|_{p} \}$
and the second can be tackled like this: set $a:=p^{v}m. b:=p^{u}n$, then checking two cases $1:(u=v)$ and $2:(v<u)$: $(a+b)=p^{v}m+p^{u}n= p^{v}(m+p^{u-v}n)$ so in case of (u=v) it follows that $ord(a+b)= u = sup\{ord(a),ord(b)\}$
and in case of $(v<u)$ because the biggest v is $v:= u-1$: $ord(a+b)\le p^{u-1}(m+pn) \le p^{u}(m+n)= sup\{ord(a),ord(b)\}$
It seems that I did something wrong in both cases...
-
Notes: $\newcommand{\ord}{\operatorname{ord}}$Your second part is equivalent to proving that \$\ord_p(x + y) \geq \min(\ord_p(x), \ord_p(y)). \$ In fact, equality holds if $\ord_p(x)$ and $\ord_p(y)$ are not equal. See Prof Israel's answer. – Dylan Moreland Mar 4 '12 at 18:52
Thank you Dylan Moreland. However, I don't get to this conclusion. – VVV Mar 5 '12 at 12:24
I don't think you can conclude, in the case of $u = v$, that the order is $u$. It could be the case that $m + n$ is divisible by $p$. For example, with $p = 2$ and $u = v = 0$ one could have $x = y = 1$, whereupon $\operatorname{ord}_p(x + y) = 1$. – Dylan Moreland Mar 5 '12 at 16:37
It's very simple. If $a = p^u m$ and $b = p^v n$ with $(m,p) = (n,p) = 1$, then $ab = p^{u+v} m n$ with $(mn,p) = 1$.
For the other question, if $u \le v$ then $a + b = p^u (m + p^{v-u} n)$ so $\ldots$
-
Thank you Robert Israel. – VVV Mar 5 '12 at 12:27
Hint $\rm\ (p^i m)\:(p^j n)\ =\ p^{i+j}\:mn\$ and $\rm\ p\nmid m,n\ \Rightarrow\ p\nmid mn\:$ by $\rm\:p\:$ prime.
This is equivalent to uniqueness of prime factorizations.
-
Thank you Bill Dubuque. – VVV Mar 5 '12 at 12:24 | 2014-11-28 14:41:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9625096917152405, "perplexity": 199.04813706384488}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931010469.50/warc/CC-MAIN-20141125155650-00204-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2187797/inertia-group-in-composite-galois-extension-of-linearly-disjoint-fields | Inertia group in composite galois extension of linearly disjoint fields
I am struggling to understand the behavior of inertia groups and ramification:
Let $L/K$ and $N/K$ be finite galois extensions of number fields and $L\cap N = K$. And let $\mathfrak{P}_{LN}$ be a prime ideal in $\mathcal{O}_{LN}$ (the integral closure of a dedekind domain $o_K \subset K$ in $LN$) and $\mathfrak{P}_L := \mathfrak{P}_{LN} \cap L, \mathfrak{P}_N := \mathfrak{P}_{LN} \cap N$ and $\mathfrak{p} := \mathfrak{P}_{LN} \cap K$ be the corresponding prime ideals below. For the grades of ramifications we get: $e(\mathfrak{P}_{LN} \mid \mathfrak{p}) = e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L})\cdot e(\mathfrak{P}_{L} \mid \mathfrak{p})$. Now my question is, does $e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L}) = e(\mathfrak{P}_{N} \mid \mathfrak{p})$ hold?
I think it does, by the following argument: $\mathfrak{P}_{N}^{e(\mathfrak{P}_{N} \mid \mathfrak{p})}=\mathfrak{p}\mathcal{O}_N = \mathfrak{P}_L \mathcal{O}_{LN} \cap N = \mathfrak{P}_{LN}^{e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L})} \cap N = \mathfrak{P}_N^{e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L})}$
Yet I end up with the following contradiction:
Suppose $K = \mathbb{Q}$ and $[L:K] = [N:K] = l$ where $l$ is prime. Let $\mathfrak{p} \neq (l)$ be ramified in both $L/K$ and $N/K$ (i.e. $\mathfrak{p}\mathcal{O}_L = \mathfrak{P}_L^l$ and $\mathfrak{p}\mathcal{O}_N = \mathfrak{P}_N^l$). If the above holds, then $I_{\mathfrak{P}_{LN}}(LN/K) \cong C_l \times C_l$. However, as $(\mathfrak{p}, (l) ) = 1$, $\mathfrak{p}$ is only tamely ramified in $LN/K$. Therefore $I_{\mathfrak{P}_{LN}}(LN/K)$ is cyclic, which is obviously not true for $C_l \times C_l$.
Where am I messing up? Thank you for your help!
No, it doesn't necessarily hold. For example, take $L = \mathbf Q(\sqrt{3})$, $N = \mathbf Q(\sqrt{6})$ and $K = \mathbf Q$, and look at the behavior of the prime $\mathfrak p = 3$. (This is exactly the setting you outline in your second argument.)
I am not very sure what you are doing in your first argument; if you try to justify your manipulations explicitly, you will see that they don't work. Taking powers of ideals doesn't behave so regularly under taking intersections, in other words, you don't generally have $\mathfrak q^n \cap \mathcal O_K = (\mathfrak q \cap \mathcal O_K)^n$; which invalidates the final equality in your argument, for example.
• Thank you! May I ask a follow up question: Is there something like an inequality, e.g. $e(\mathfrak{P}_{LN} \mid \mathfrak{p}) = e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L})\cdot e(\mathfrak{P}_{L} \mid \mathfrak{p}) \leq e(\mathfrak{P}_{N} \mid \mathfrak{p})\cdot e(\mathfrak{P}_{L} \mid \mathfrak{p})$. I.e. if a prime is neither ramified in $L/K$ nor in $N/K$, does this insure that it is unramified in $LN/K$? Mar 15, 2017 at 16:43 | 2023-03-26 21:52:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.951884388923645, "perplexity": 106.82678454675316}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00605.warc.gz"} |
http://physicspages.com/2011/09/26/gausss-law/ | ## Gauss’s Law
Required math: vectors, calculus
Required physics: electrostatics
Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Sec 2.2.
Gauss’s law in electrostatics is a relation between the charge contained by a closed surface and the electric field that crosses that surface. The easiest way to see how it works is to begin with a point charge at the origin and a spherical surface centred at the origin. By symmetry the electric field due to the point charge is
$\displaystyle \mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}\hat{\mathbf{r}}$
The flux of this field through the sphere is defined as the surface integral of the component of the field that is normal to the surface. That is the flux ${\Phi}$ is defined as
$\displaystyle \Phi=\int\mathbf{E}\cdot d\mathbf{a}$
where the integral extends over the surface, and ${d\mathbf{a}}$ is a differential vector whose magnitude is a differential area element and whose direction is normal to the surface at each point.
In the case of a sphere, it is not surprisingly easiest to use spherical coordinates, and in that case
$\displaystyle d\mathbf{a}=r^{2}\sin\theta d\theta d\phi\hat{\mathbf{r}}$
That is, the area element points radially outwards at each point on the sphere.
Combining these results, we see that for a point charge
$\displaystyle \Phi$ $\displaystyle =$ $\displaystyle \int\mathbf{E}\cdot d\mathbf{a}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\left(\frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}\hat{\mathbf{r}}\right)\cdot\left(r^{2}\sin\theta d\theta d\phi\hat{\mathbf{r}}\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\int_{0}^{2\pi}\int_{0}^{\pi}\sin\theta d\theta d\phi$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{\epsilon_{0}}$
That is, the flux due to a point charge depends only on the magnitude of the charge and not on the radius of the sphere that contains it. This makes intuitive sense, since if we imagine a point charge ‘emitting’ the electric field, then as long as we provide a surface that wraps up the charge completely, the flux through that surface will be the same regardless of the size of the surface. In fact, it shouldn’t depend on the shape of the surface either, so long as that surface completely encloses the charge. A more general derivation using an arbitrary closed surface is possible (using the solid angle subtended by the elemental area), but it adds nothing to the general idea.
From here, we can generalize the idea to a collection of point charges using the principle of superposition, and get, for a collection of ${n}$ point charges, all enclosed by the surface:
$\displaystyle \int\mathbf{E}\cdot d\mathbf{a}=\frac{1}{\epsilon_{0}}\sum_{i=1}^{n}q_{i}$
For a continuous charge distribution, where the charge density is ${\rho(\mathbf{r})}$, we get
$\displaystyle \int\mathbf{E}\cdot d\mathbf{a}=\frac{1}{\epsilon_{0}}\int\rho(\mathbf{r})d^{3}\mathbf{r}$
where it is important to note that the integral on the left is over the enclosing surface, while that on the right is over the volume enclosed by that surface. This is the integral form of Gauss’s law for electrostatics.
Using the divergence theorem, we can equate the charge density with the divergence of the electric field:
$\displaystyle \nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_{0}}$
This is the differential form of Gauss’s law. Both these forms are very powerful in solving various types of problems since they allow electric fields to be calculated, often without requiring complicated integrals. | 2013-12-13 16:17:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9508875012397766, "perplexity": 111.50236673089432}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164960531/warc/CC-MAIN-20131204134920-00026-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://geologus.eu/two-way-oowry/word-equation-shortcuts-6de6bd | Although you can also click on “Equations” under the “Insert” Tab to get it. Request. Like matrix, @ is used as a row separator. On the File tab, choose Options. However, when parenthesis is required to be displayed, it must be doubled (one for grouping which will vanish in final formatted expression and other for displaying). In both the cases, desired piecewise functions are entered inside the parenthesis. But if you change to Print Layout View and click on the equation, there will be a new purple "Equation Tools" tab on the Ribbon. The shortcuts are very intuitive and easy to remember. This shortcut works in Microsoft Word and PowerPoint to quickly create (or remove) subscripts. Microsoft Word Mathematics Equation Shortcuts (For Math Equation in Chemistry Lab Report) - Duration: 6:39. For e.g., to get Greek letter , you can type \alpha instead of going to Symbols in Insert Tab and searching for . While adding pre-subscript or pre-superscript, use \zwsp along with _ and ^ sign as shown below. In Word 2016 equation many symbols can be inserted using a \+name of the symbol : \alpha. If you have to press one key immediately after another, the keys are separated by a comma (,). \prod - Product Sign. Some of the commonly used symbols: \infty - Infinity. These shortcuts can then be used in lieu of the buttons on MathType's tab in Word, or on the MathType menu or toolbar (Word 2011). Alt + = List of Equation editor shortcut for typing square root, cube root, 4th root & nth root and text inside root sign. BHH�zxXX@����?�IDUWeTWUwvͱ�ݴ%;�;�Ό/"22���Jo$��O�]~����?MYN?y���V7QD�����1H�6�6Hx�g����V ��o��0� ��lqSi�\?� �xlB�VM$��X��l :ZG. skewed fraction) is obtained using \ldiv (resp. This method of typing Greek letters is as easy as typing its spelling after \ (backslash). Equation editor shortcut for scientific and mathematical symbols like infinity, different arrows, operators (like partial, del and nabla), conditional symbols, dot, cross, mapsto, perpendicular, set symbols, for all, equivalent, congruent, angle, proportional etc are given in the following table. Safari Thumbnails on a MacBook - … The standard form of linear equations in two variables is Ax+By=C , where A, B and C are constants. Equation editor shortcut for subscript and superscript is _ and ^. \sum - Summa. �z٣�X����y��#"J2�FKT�6�Q:��$�DB�=��|ڌ6D�}YB��� �{ �ة�v�����yG��ـ�����-�����s��)0u���u������H]�����=��Y�Z�Pd���b�q��SEE#J�� n����]@zH$TJ��k��P i����E������r_��7"�]M lLS]��G��^j_�G�e��1N��Q��8ݷ��'݉ڭ�8�K�Lo��(ɫY��j�5�_�����7͡PE�c�Q5Ϛl\��F�獻���� &��'�:bm��D_.�L#2'l�����7��!����,����"�TOKJ��G1�$��g����zY���a*K7ЬCO3���1y�X���Uۗ�U4��/��z��c�hӁ�y�PJ���a5u�MA��,��� ]Ȣ|h�[a���ȠS �a ���ȮID���,�}� �3ɐ�> f� _���Բ�!��c�ZѮ(���@���x{8�=V�-��=I����ey�s>g�u� from Darryl Yong Plus . Escape sequence (\ followed by desired bracket is used to prevent bracket from being reformatted. How to insert equation numbering in MS Word Utpal Kumar 2 minute read It is of essential to insert equation number if you are working on your thesis and/or any scientific paper that consists of a lot of equations. Equation editor shortcut for Sub script and Super script In the above table, stands for space, pressing space bar is important and is required to … The easiest and probably the best way of doing this is to write your manuscript in latex. Using this shortcut for equation in Word, you can type equation in Word as fast as LaTeX. Templates, style changes, and spaces are inserted by simply pressing Ctrl followed by another … Using them you can quickly type variety of arrows, set symbols, partial, nabla, degree C, angle, etc. stream %�쏢 \Gamma for ). Use the first of the "Script" structures. \leq - Less than or equal. Microsoft Word Equation Editor Keyboard Shortcuts. Although you can also click on “Equations” under the “Insert” Tab to get it. We can easily write it in Ms Word using \frakturXX where XX is an uppercase letter. 1 \sdiv 2) and pressing space. There are multiple ways to display a fraction. Notes: • If a shortcut requires two or more keys at the same time, the keys are separated by a plus sign (+). Go to the Customize Ribbon category. These grouping parenthesis don’t appear after Math AutoCorrect. You can use the decimal values of the Unicode points to use with the alt keys on Windows based documents. This short video illustrates how to create mathematical documents in Microsoft Word very quickly by using keyboard shortcuts. How to quickly insert Fraktur symbols in Word equations. Word 2016. To enter the Equation Editor mode, either key-in [Alt=]or select the [Insert] [Equation]menu item: a) If [Alt=]is implemented inside a document paragraph, then the equation is known as “in-line” e… Hope that above shortcut helps you to type mathematical equation in Ms Word faster. Unicode has a code point from 2200 to 22FF for mathematical operators. Keep on doing your good work. Use a keyboard for regular Latin letters. \alpha for ) and for upper case Greek Alphabet type name of Greek letter after \ in Title Case (e.g. use Ms Word shortcut “\frakturB” for ““ and “\frakturG” for ““. Ctrl + f creates the fraction object. x��V�n#7}����j��V���E/A���}(��:q�v�4�t�|R���43�xm�E��5"y�sH�So4o�^�ݛ���}쌎�oW���Wr���s�7{��9�X�@O��b�9t?����hoХ��a��6. 365 2013 2010 2007 2003. endobj First one uses \cases() method while the second one uses \matrix(). \partial - Partial differential. For drawn one, you can use \doubleNto create , \frakturNto create , \scriptNto create . 8 years ago. How to insert an equation with a limit. Shortcut for integral sign, sum and product signs are \int, \sum and \prod. Thank you is not enough for the knowledge you have shared. How to Customize Keyboard Shortcuts in Word 2016. Equation editor shortcut provides a quick way to write Scientific and Mathematical symbols in Ms Word. It exists in Word 2016 (Ctrl+=) but this shortcut creates a superscript in Powerpoint. \lambda. The trick here is to use Equation Editor for typing fraction in Word. \sdiv) and pressing space (twice) or by typing 1 \ldiv 2 (resp. They are often the fastest and efficient way of typing equation in Ms Word. These shortcut work only inside Equation Editor. (ƨ���9�Lm�I��/m�u�����^7�AG��:�!�#�-=q�ī��7&7����&d�ٗ˃v�r�ƔLR�F[��j7����>�{������>G �0xm��^8xo����٨��Z�����l�t۶� �u�=v5��j x.�O˦ �Z�Cb%R��oZ��a?�dtJ�D��d\��T57}�F��� Equation Editor Shortcut for Mac Powerpoint 2016 I have been unable to locate a shortcut that inserts a new equation in Powerpoint 2016 for Mac. 15 0 obj Choose symbols and templates from MathType 's palettes. Other than that, building an equation is easy. 3 Minutes Tutorial 4,746 views. \subset - Contained in. We use cookies to ensure that we give you the best experience on our website. Tip A) Use Alt-= to start an equation. Alternatively, use the corresponding keyboard shortcuts. After enabling the Math AutoCorrect, you can use those text replacement shortcuts on your document. Unicode math. It may sound like quite a task, but believe me if you know correct shortcut then it becomes a piece of cake. However, typing it in Ms Word is cumbersome and tiring task. Ms Word and Power Point shortcut for equation editor is “Alt + =” (i.e. 3. 1074 You can also create matrix filled with elements using above shortcut. Shortcut keys for inserting Greek symbols into the equation. Equation editor causes brackets (such as [], {} and ()) to grow to fit the size of expression within them. ���o�!���zki9t��*���t��K�q\��v��������k�֕ŕ�4�T�W�_6[�ْ ��q���0�mh�p�v"6q��Km� Л�R��vN��ց���l{>eD��o;(���we$��F�� �~�568)�X��HhzM�gO('���B+*���5ZѠ.������zH�;��j\���^'^��� \> �"~�k�_}��ZbU���S�489p�O�f 6��&�'� 6N��ā��b���_�;���.4R Grouping is also important as it distinguish between and . For more details and example, visit our blog on different methods and equation editor shortcut for typing matrix in Ms Word. To ease it, Microsoft Word has LaTeX type equation editor shortcut feature for typing equation. Use & to move to the next column and @ for the next row. 6:39. Equation Editor Shortcut. Parenthesis can be used for grouping text with spaces. Complete Reference on Ms Word Equation Editor Shortcut, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to share on Reddit (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on WhatsApp (Opens in new window). This will insert an … ���'� �5� ������?������� K��V}b�e��� �IVV�'(Y5�T�T�d*1�Q��G�oά�Q������az���A,���n����qw>endstream Although you can get equation editor by going to Insert Tab and clicking on Equation, it is an unproductive way of getting it if you need it often. use \doubleA for and \doubleR for . 1. Letters. For various reasons you need to have an accent like bar, grave, tilde, dot (for denoting derivative) above symbol. Shortcut keys for inserting Greek symbols into the equation. These equation editor shortcut as termed as Math AutoCorrect and are available in versions of Microsoft Word 2007 and above. To include space in subscript or superscript, group them in () or parenthesis. Number of @ and & symbol is one less than number of rows and columns, respectively. Equation editor shortcut for subscript and superscript, Shortcut for Square root, and higher order roots, how to insert an arrow in word using keyboard, How to type multiplication & division symbol in Word, How to set default font in Word – Smart Word User, Automated interpolation formula for Excel: Define excel interpolate function & use it forever, Pi symbol in Word: Type π or Π faster with this shortcut, [] bracket automatically expands to adjust fraction, Parentheses displayed as they not used for grouping, Parentheses used for grouping (denominator here) are not displayed, Parentheses used for grouping (denominator here) is not displayed, Again parentheses used for grouping are not displayed, @ is used as row separator and \close is required to ensure opening { expands vertically to cover all cases, Similar to above, & is used as column separator, Without \close, opening '{' doesn't expands, \oint for cyclic integral, similarly use \oiint for cyclic double Sum, integral, \sum_(i=1)^nA_i\sum for sum symbol and _ & ^ sign for getting text below and above sum. \beta. Use \doubleXX, where XX is the required uppercase letter for e.g. H 2 O) is Ctrl + = The keyboard shortcut for superscript (e.g. While you can also do this by right-clicking on the equation and clicking Linear, this affects the whole equation and not just the fraction. I cannot thank you enough for the article. The equation editor in ExamView Test Generator includes numerous shortcut keys to make inserting symbols, inserting templates, adding accents, inserting spaces, and changing font styles as easy as possible. For MS Word 2007/2010/2013: use the equation feature, designed for math, but works okay for chemistry. We can easily achieve these using following word shortcut. All of the keyboard shortcuts involve the Ctrl key. Matrix size decided by number of @ (for rows) and & (for columns) respectively. You can get equation editor by navigating to Insert Menu and clicking on Equation or by using a keyboard shortcut i.e. v / p {\displaystyle {v}/{p}} For e.g. Hi, this video tutorial will show shortcuts on inserting equation from equation editor in Ms Word. However, if you want to use it outside Equation Editor, then check “Use Math Autocorrect Rules outside of math regions“. The Subscript shortcut is Ctrl + = on a PC and Ctrl + Cmd + + on a Mac. Equation Tools Design tab contains dozens of equation templates. You can use _ and ^ for inserting text below and above signs, respectively. To get lower case Greek Alphabet, type name of Greek letter after \ in lower case (for e.g. Ribbon keyboard shortcuts. Anything after _ or ^ will get converted into subscript or superscript respectively, after hitting space. How to quickly insert Script symbols in Word equations. How to quickly insert Double strike or Blackboard bold symbols in Word equations. 2. Ms Word and Power Point shortcut for equation editor is “Alt + =” (i.e. Most version of Microsoft Word, Math AutoCorrect is enabled by default. To get only the opening curly braces ‘{‘ which automatically extends the height of piecewise function, use \close in place of closing ‘}’. x��]Y�$G��ݵ,s��isV�;�G�� Fraktur is a calligraphic hand of the Latin alphabet. Obtain this by typing the fraction and pressing space: 1/2 1 2 {\displaystyle {\frac {1}{2}}} Linear fraction (resp. There are two ways to insert piece wise function in using Equation Editor shortcut in Ms Word. +5_m�AZ��t��BJ�\%'ɐ�M��J�����/j�ex�W�V9��տ��zFL���7�����Z��u�����oj#�� m��/��-��?j)�uN��I�����;�d_=����k��Ok+��ֲJY��jT�Jś.�+5�����vQ���=N$+�z��f~�kh[�W�U It is typically used for footnotes, endnotes, and mathematical or scientific formulas.. Insert a subscript. The default is vertically aligned as illustrated below. It's equivalent to equation editor in LaTeX. If you continue to use this site we will assume that you are happy with it. Microsoft Word Equation Editor 1. Dozens of equation into mathematical Symbols/Operators create, \scriptNto create space are shown word equation shortcuts! Less than number of @ ( for rows ) and pressing space ( twice ) or parenthesis entered the... For fraction, Ctrl+R for square root, and higher order roots of arrows, set symbols,,. Of these, Math AutoCorrect shortcut has LaTeX type equation editor shortcut in Ms Word will that! Editor is “ Alt + = on a Mac “ \frakturB ” for “ “ as it distinguish and... Latex type equation editor shortcut as termed as Math AutoCorrect Rules outside of Math regions “ thesis and research.! The decimal values of the commonly used symbols: \infty - Infinity as fast as LaTeX we can easily it... Use \doubleNto create, \scriptNto create pre-superscript, use \zwsp along with _ and ^ the contained! Next column and @ for the web, access keys all start with Alt+Windows logo key, add... Code shortcuts for typing most of the Latin Alphabet ” under the “ Insert > equation ” option under symbols. It may sound like quite a task, but believe me if you continue to use this site will. Shortcuts the equation in documents the ribbon Math equation in Chemistry Lab Report ) - Duration: 6:39 shown... After enabling the option “ Replace text as you type ” common things row separator reasons you to! Be inserted using a \+name of the mathematical expression superscript, group them in ( method! Lower case ( e.g letters is as easy as typing its spelling after \ in lower case ( e.g... Title case ( for e.g use equation editor word equation shortcuts “ Alt + = ” ( i.e into Symbols/Operators... Typing ‘ = ’ ) several tools available and ^ sign as shown below for integral sign visit. Editor for word equation shortcuts Greek letters escape sequence ( \ followed by desired bracket used. Instead of going to symbols in Ms Word using \frakturXX where XX is an important of! You enough for the ribbon save a lot of time and effort is a representation of Math shortcut!, \scriptNto create Word using \frakturXX where XX is the easiest and probably the best way of this... Point when it is typically used for grouping text with spaces it is to! Inserted using a \+name of the keyboard shortcut for square root, and higher order roots 2 )! Site we will assume that you don ’ t have to Press one immediately. Are available in versions of Microsoft Word equation editor by enabling the Math objects.! Xx is the complete list of Alt code shortcuts for typing matrix in Ms Word faster bar, grave tilde! Footnotes, endnotes, and mathematical or Scientific formulas.. Insert a subscript Insert > equation option! Insert Double strike or Blackboard bold symbols in Word equations process installs several keyboard shortcuts involve the key! Of @ and & symbol is missing displayed in the final formatted expression has a code Point from to. Potential to save a lot of time and effort \zwsp along with _ and ^ / { }... Ctrl+= ) but this shortcut creates a super Script shortcuts on your document and ^ are entered inside the.. In Title case ( for rows ) and & symbol is one less than number of @ ( columns... ) respectively AutoCorrect Rules outside of Math AutoCorrect shortcut start with Alt+Windows logo key, then check use! Feature for typing fraction in Word as fast as LaTeX when it is to! Very quickly by using keyboard shortcuts … a Word on shortcut keys for inserting Greek symbols into the.! 1 \ldiv 2 ( resp, Ctrl+R for square root, and higher order roots offers access all... Our blog on shortcut for equation in Ms Word shortcut “ \frakturB for! Of arrows, set symbols, partial, nabla, degree C, angle, etc add letter! Are very intuitive and easy to remember is that this method of Greek! This is to write your manuscript in LaTeX easiest and probably the best experience on our website a. Often the fastest and efficient way of doing this is to use this site we will assume that you happy! \Alpha instead of going to symbols in Word is cumbersome and tiring.... To use the decimal values of the matrix row wise, starting with the top row with it these editor! Or remove ) subscripts Alphabet type name of Greek letter, you can use text. List of Alt code shortcuts for typing most of the matrix row wise starting. “ symbols ” group using equation editor shortcut for equation in Ms Word is “ Alt + = a! You type ” by typing 1 \ldiv 2 ( resp Press Alt and.. - … Microsoft Word, Math AutoCorrect and are available in versions of Microsoft very... In ( ) or parenthesis, \sum and \prod fraction in Word as fast LaTeX... Means that you are right and for some reason, Word wo n't let me use the shortcut for in. \Infty - Infinity start an equation is easy this short video illustrates to. But this shortcut creates a superscript in Powerpoint is enabled by default equations! For Math equation in Ms Word a super Script @ for the.... Xx is the complete list of Alt code shortcuts for mathematics symbols Ax+By=C... Shortcut then it becomes a piece of cake are right and for upper Greek... No more a tedious job & to move to the next row it in Ms Word an accent like,! Shortcut has a potential to save a lot of time and effort achieve these using following shortcut! Works in Microsoft Word a lot of time and effort click on MacBook! Button on the toolbar, there are also shortcut commands to render most common things its. One thing to remember and the fastest of all four Blackboard bold symbols in Word 2016 equation many can! When you install MathType, the keys are separated by a comma (, ) complete list of Alt shortcuts! In Math AutoCorrect shortcut is also important as it distinguish between and use the shortcut for square root, mathematical... ) is Ctrl + Cmd + + on a word equation shortcuts to see tools. Spelling after \ in lower case ( for rows ) and for some,... That means that you don ’ t appear after Math AutoCorrect shortcut provides a quick way write. Word supports: and searching for editor by navigating to Insert piece wise in. Typing ‘ = ’ ) replacement shortcuts on your document are very intuitive and easy to remember on... ) is obtained using \ldiv ( resp of Math on one line in documents or remove ).... A quick way to write your manuscript in LaTeX \frakturXX where XX is uppercase. Ax+By=C, where XX is the required uppercase letter for e.g, parenthesis for! Scientific and mathematical or Scientific formulas.. Insert a subscript and a ^! Clarity whenever necessary, space are shown as < sp > in AutoCorrect! Version of Microsoft Word very quickly word equation shortcuts using keyboard shortcuts … a Word on shortcut for (. Insert Menu and clicking on equation or by typing 1 \ldiv 2 resp... Web, access keys all start with Alt+Windows logo key, then check “ use Math replacements in equation for. Within each button on the toolbar, there are two linear formats Math! Denoting derivative ) above symbol Word wo n't let me use the shortcut for equation editor typing! Less than number of rows and columns, respectively for clarity whenever necessary, space are as! & ( for Math that Word supports: Title case ( for columns ) respectively root. Quickly create ( or remove ) subscripts + = on a button to the... Of doing this is to use equation editor shortcut feature for typing most of Unicode. Above signs, respectively mathematical equation in Ms Word time and effort can use first. Piece of cake to include space in subscript or superscript, group them (. Inside the parenthesis keys are separated by a comma (, ) in. The web offers access keys all start with Alt+Windows logo key, then check “ Math... Duration: 6:39 tools available linear formats for Math that Word supports: of the ''... Shortcut and example, underscore _ creates a superscript in Powerpoint one, you can use Math replacements in editor... Editor shortcut for superscript ( e.g after another, the keys are separated by comma... \Cases ( ) a MacBook - … Microsoft Word very quickly by using a keyboard i.e. Than that, building an equation is easy a lot of time and effort required uppercase letter a to... Word as fast as LaTeX ensure that we give you the best way of doing this to. \Frakturg ” for “ “ ( \ followed by desired bracket is used as a row separator the are! P { \displaystyle { v } / { p } } some of the Script structures... Desired bracket is used as a row separator contains dozens of equation into Symbols/Operators... They are often the fastest and efficient way of typing equation in Word as fast as LaTeX shortcut feature typing. Xx is an important part of Math regions “ letter after \ in case... To write Scientific and mathematical symbols part of Unicode system a button to the. \Int, \sum and \prod columns, respectively dozens of equation into mathematical Symbols/Operators for various you... Is time to translate a part of equation templates enabled by default replacement! To navigate the ribbon Tab common things Insert equation feature Power Point when it is time to translate part! | 2021-04-16 02:59:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5426521897315979, "perplexity": 3521.3466208688696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00193.warc.gz"} |
https://zbmath.org/?q=ut%3Apath | ## Found 21,490 Documents (Results 1–100)
100
MathJax
Newest 1st
### The 3-path-connectivity of the $$k$$-ary $$n$$-cube. (English)Zbl 07589675
MSC: 05Cxx 68Mxx 68Rxx
Full Text:
### Universal quantum semigroupoids. (English)Zbl 07584342
MSC: 16S10 16T99 16W50
Full Text:
Full Text:
### Closure operators associated with relations. (English)Zbl 07568615
MSC: 54A05 54H30 08A02
Full Text:
MSC: 70E55
Full Text:
### Adapting paths against zero-determinant strategies in repeated prisoner’s dilemma games. (English)Zbl 07593000
MSC: 91A20 91A05
Full Text:
### Hamiltonian and long paths in bipartite graphs with connectivity. (English)Zbl 07592409
MSC: 05C45 05C12 05C40
Full Text:
### Fifth-order A-WENO schemes based on the path-conservative central-upwind method. (English)Zbl 07592126
MSC: 65Mxx 35Lxx 76Mxx
Full Text:
Full Text:
### An asymptotic radius of convergence for the Loewner equation and simulation of $$SLE_{\kappa}$$ traces via splitting. (English)Zbl 07591717
MSC: 60Hxx 60Jxx 30Cxx
Full Text:
### Realization utility with path-dependent reference points. (English)Zbl 07591710
MSC: 91G80 35Q91 49J40
Full Text:
### Unidirectional Littelmann paths for crystals of type $$A$$ and rank 2. (English)Zbl 07590583
MSC: 17B10 17B35
Full Text:
### Limit laws for two distance-based indices in random recursive tree models. (English)Zbl 07589927
MSC: 05C05 60F05
Full Text:
### Strength-based concurrent shape and fiber path optimization of continuous fiber composites. (English)Zbl 07589848
MSC: 74-XX 90-XX
Full Text:
### Concurrent optimization of topological configuration and continuous fiber path for composite structures – a unified level set approach. (English)Zbl 07589827
MSC: 74-XX 90-XX
Full Text:
### The overfullness of graphs with small minimum degree and large maximum degree. (English)Zbl 07589623
MSC: 05C07 05C15
Full Text:
### Path connectedness of Volterra type integral operators on Bergman and Dirichlet type spaces. (English)Zbl 07589100
MSC: 30B40 30D10
Full Text:
Full Text:
MSC: 90Cxx
Full Text:
MSC: 90Cxx
Full Text:
### Monotone path-connectedness of strict suns. (English)Zbl 07587447
MSC: 41Axx 54Cxx 52Axx
Full Text:
Full Text:
MSC: 83-XX
Full Text:
### Leavitt path algebras for power graphs of finite groups. (English)Zbl 07584874
MSC: 16S88 05C25
Full Text:
### Lie solvable Leavitt path algebras. (English)Zbl 07584710
MSC: 16S88 17B30
Full Text:
Full Text:
### A translation of an analogue of Wiener space with its applications on their product spaces. (English)Zbl 07584431
MSC: 28C20 46G12 46T12
Full Text:
### Efficiency analysis of the forward error correction at the transport protocol level. (English. Russian original)Zbl 07583570
Autom. Remote Control 83, No. 7, 1059-1077 (2022); translation from Avtom. Telemekh. 2022, No. 7, 59-81 (2022).
MSC: 93C55 93E03
Full Text:
### Two optimal path planning problems for a moving object in the case of degeneration of necessary extremum conditions. (English. Russian original)Zbl 07583567
Autom. Remote Control 83, No. 7, 1011-1035 (2022); translation from Avtom. Telemekh. 2022, No. 7, 3-32 (2022).
MSC: 93C20 93C10
Full Text:
### Safe interval path planning and flatness-based control for navigation of a mobile robot among static and dynamic obstacles. (English. Russian original)Zbl 07583560
Autom. Remote Control 83, No. 6, 903-918 (2022); translation from Avtom. Telemekh. 2022, No. 6, 96-117 (2022).
MSC: 93C85 93B45
Full Text:
MSC: 74-XX
Full Text:
### An algorithm for updating betweenness centrality scores of all vertices in a graph upon deletion of a single edge. (English)Zbl 07583048
MSC: 05C38 05C12 05C35
Full Text:
MSC: 05C50
Full Text:
MSC: 82-XX
Full Text:
### Alexandroff homology and path homology. (English. Russian original)Zbl 07581879
Math. Notes 112, No. 1, 159-162 (2022); translation from Mat. Zametki 112, No. 1, 148-152 (2022).
MSC: 55N35 57M15 05C20
Full Text:
### Gas path parameter prediction of aero-engine based on an autoregressive discrete convolution sum process neural network. (English)Zbl 07581684
MSC: 68-XX 82-XX
Full Text:
### On size multipartite Ramsey numbers of large paths versus wheel on five vertices. (English)Zbl 07581635
MSC: 05C55 05D10
Full Text:
### Algorithm to check the existence of $$H$$ for a given $$G$$ such that $$A(G)A(H)$$ is graphical. (English)Zbl 07581625
MSC: 05C85 05C50
Full Text:
### On $$H$$-irregularity strength of cycles and diamond graphs. (English)Zbl 07581623
MSC: 05C70 05C78
Full Text:
### Decomposition of hypercube graphs into paths and cycles of length four. (English)Zbl 07579612
MSC: 05C70 05C38 05C12
Full Text:
MSC: 74-XX
Full Text:
### An efficient routing heuristic for a drone-assisted delivery problem. (English)Zbl 07578501
MSC: 90-XX 91-XX
Full Text:
### Fast and optimal sequence-to-graph alignment guided by seeds. (English)Zbl 07577831
Pe’er, Itsik (ed.), Research in computational molecular biology. 26th annual international conference, RECOMB 2022, San Diego, CA, USA, May 22–25, 2022. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 13278, 306-325 (2022).
MSC: 92D20 05C90
Full Text:
### A shift Gray code for fixed-content Łukasiewicz words. (English)Zbl 07577713
Bazgan, Cristina (ed.) et al., Combinatorial algorithms. 33rd international workshop, IWOCA 2022, Trier, Germany, June 7–9, 2022. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 13270, 383-397 (2022).
MSC: 68Rxx 68Wxx
Full Text:
### Chevalley formula for anti-dominant minuscule fundamental weights in the equivariant quantum $$K$$-group of partial flag manifolds. (English)Zbl 07577510
MSC: 17Bxx 20Fxx 05Exx
Full Text:
### Connectedness in structures on the real numbers: o-minimality and undecidability. (English)Zbl 07576906
MSC: 03C64 03D35 54D99
Full Text:
Full Text:
### A sufficient condition for mobile sampling in terms of surface density. (English)Zbl 07576441
MSC: 68-XX 62-XX
Full Text:
### A near-optimal algorithm for shortest paths among curved obstacles in the plane. (English)Zbl 07575637
MSC: 68Q25 68Q20 68P05
Full Text:
### Stress-driven nonlinear behavior of curved nanobeams. (English)Zbl 07575006
MSC: 74-XX 76-XX
Full Text:
### Gallai-Ramsey numbers for rainbow $$P_5$$ and monochromatic fans or wheels. (English)Zbl 07574776
MSC: 05C55 05D10 05C15
Full Text:
Full Text:
### Extended path partition conjecture for semicomplete and acyclic compositions. (English)Zbl 07574749
MSC: 05C70 05C38 05C20
Full Text:
### Comparison theorem for path dependent SDEs driven by $$G$$-Brownian motion. (English)Zbl 07574577
MSC: 60J65 47G20 60G52
Full Text:
### Path independence of the additive functionals for stochastic differential equations driven by $$G$$-Lévy processes. (English)Zbl 07574061
MSC: 60H10 60G51
Full Text:
Full Text:
Full Text:
### Decorated Dyck paths, polyominoes, and the delta conjecture. (English)Zbl 07573750
Memoirs of the American Mathematical Society 1370. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-7157-6/pbk; 978-1-4704-7170-5/ebook). iii, 119 p. (2022).
MSC: 05-02 05E05 05B50
Full Text:
### Decomposing $$2k$$-regular graphs into paths of length $$k$$. (English)Zbl 07572745
MSC: 05C70 05C38 05C12
Full Text:
### The two-stripe symmetric circulant TSP is in P. (English)Zbl 07572298
Aardal, Karen (ed.) et al., Integer programming and combinatorial optimization. 23rd international conference, IPCO 2022, Eindhoven, The Netherlands, June 27–29, 2022. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 13265, 319-332 (2022).
MSC: 90C27
Full Text:
### Faster goal-oriented shortest path search for bulk and incremental detailed routing. (English)Zbl 07572276
Aardal, Karen (ed.) et al., Integer programming and combinatorial optimization. 23rd international conference, IPCO 2022, Eindhoven, The Netherlands, June 27–29, 2022. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 13265, 15-28 (2022).
MSC: 90C35 90C27 90C10
Full Text:
### Phase space Feynman path integrals of parabolic type on the torus as analysis on path space. (English)Zbl 07571553
MSC: 81S40 35S05 35K25
Full Text:
Full Text:
MSC: 93-XX
Full Text:
Full Text:
Full Text:
### The median of Sierpiński graphs. (English)Zbl 07567749
MSC: 05C12 05C35 05C38
Full Text:
Full Text:
### The microlocal irregularity of Gaussian noise. (English)Zbl 07566092
MSC: 60G15 60G60
Full Text:
Full Text:
Full Text:
### Coexistence of two kinds of superfluidity at finite temperatures in optical lattices. (English)Zbl 07565269
MSC: 82-XX 81-XX
Full Text:
MSC: 62-XX
Full Text:
Full Text:
### Heavy traffic limits for queues with non-stationary path-dependent arrival processes. (English)Zbl 1491.60159
MSC: 60K25 60F17 90B22
Full Text:
### Group inverses of adjacency matrices of cycles, wheels and brooms. (English)Zbl 07562939
MSC: 05C50 15A09
Full Text:
### Decomposing 10-regular graphs into paths of length 5. (English)Zbl 07562657
MSC: 05C38 05C70
Full Text:
Full Text:
MSC: 90-XX
Full Text:
MSC: 90-XX
Full Text:
MSC: 60H10
Full Text:
### Construction of reduced chemical mechanisms orientated toward specific applications: a case study of primary reference fuel. (English)Zbl 07561985
MSC: 80Axx 76-XX
Full Text:
Full Text:
Full Text:
### On paths with three blocks $$P(k,1,l)$$. (English)Zbl 07560305
MSC: 05C38 05C45
Full Text:
### Primal-dual path-following methods and the trust-region updating strategy for linear programming with noisy data. (English)Zbl 07560295
MSC: 65L20 65K05 65L05
Full Text:
### Strongly graded Leavitt path algebras. (English)Zbl 07559651
MSC: 16S88 16W50
Full Text:
Full Text:
### Distance constrained vehicle routing problem to minimize the total cost: algorithms and complexity. (English)Zbl 07558393
MSC: 90C35 90C27 90B06
Full Text:
MSC: 90C31
Full Text:
MSC: 90C35
Full Text:
MSC: 90C27
Full Text:
### Stochastic control of a class of dynamical systems via path limits. (English)Zbl 07557898
MSC: 60F10 90C40 93E03
Full Text:
### 3-D path planning system for autonomous vehicle considering the rollover and path length. (English)Zbl 07557780
MSC: 93C85 93C42
Full Text:
### The vectorial kernel method for walks with longer steps. (English)Zbl 07557724
MSC: 05A15 05A16
Full Text:
Full Text:
### The crossing numbers of join products of paths with three graphs of order five. (English)Zbl 07557566
MSC: 05C76 05C10 05C38
Full Text:
Full Text:
### KMS states and continuous orbit equivalence for ultragraph shift spaces with sinks. (English)Zbl 07556781
MSC: 16S88 37B10 46L55
Full Text:
### Invariant ideals in Leavitt path algebras. (English)Zbl 07556776
MSC: 16D25 16S88
Full Text:
all top 5
all top 5
all top 5
all top 3
all top 3 | 2022-09-28 22:55:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7356767654418945, "perplexity": 14429.141161430802}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335286.15/warc/CC-MAIN-20220928212030-20220929002030-00455.warc.gz"} |
https://www.ncatlab.org/nlab/show/symmetric+smash+product+of+spectra | # nLab symmetric smash product of spectra
Contents
### Context
#### Stable Homotopy theory
stable homotopy theory
Introduction
# Contents
#### Higher algebra
higher algebra
universal algebra
# Contents
## Idea
A symmetric smash product of spectra is a realization of the smash product of spectra such as to make a symmetric monoidal model category presentation of the symmetric monoidal (infinity,1)-category of spectra.
In higher algebra and stable homotopy theory one is interested in monoid objects in the stable (∞,1)-category of spectra – called $A_\infty$-rings – and commutative monoid objects – called $E_\infty$-rings. These monoid objects satisfy associativity, uniticity and, in the $E_\infty$-case, commutativity up to coherent higher homotopies.
For concretely working with these objects, it is often useful to have concrete 1-categorical algebraic models for these intricate higher categorical/homotopical entities. The symmetric monoidal smash product of spectra is a structure that allows to model A-infinity rings as ordinary monoids and E-infinity rings as ordinary commutative monoids in a suitable ordinary category – one speaks of highly structured ring spectra.
Historically, this had been desired but out of reach for a long time, due to the initial focus on the model by plain sequential spectra. By this remark at smash product of spectra, plain sequential spectra do not reflect the graded-commutativity implicit in the braiding of the smash product of n-spheres and thus do not admit a symmetric smash product of spectra.
When the relevant highly structured ring spectra were finally found that do admit symmetric smash products, the relief was substantial and led to terminology such as “brave new algebra”. More recently maybe the term higher algebra is becoming more popular.
Then, model structures were found which also admit symmetric monoidal smash products, but which are not of the form “highly structured spectra”: model structure for excisive functors.
As a first step one wants a model category of spectra $\mathcal{S}$ that presents the full (infinity,1)-category of spectra. This allows to model the notion of equivalence of spectra and of homotopies between maps of spectra. Several such model categories have been known for a long time; all are Quillen equivalent and their common homotopy category is called “the” stable homotopy category $Ho \mathcal{S}$.
Now, for some of the model categories $\mathcal{S}$ of spectra, the smash product on $Ho \mathcal{S}$ can be lifted to a functor
$\wedge\colon \mathcal{S} \times \mathcal{S} \to \mathcal{S} \,,$
but for the most part these functors were neither associative nor unital nor commutative at the level of the 1-category $\mathcal{S}$. In fact (Lewis 91) proved a theorem that there could be no symmetric monoidal category $\mathcal{S}$ modeling the stable homotopy category and satisfying a couple of other natural requirements.
However, in the 1990s it was realized that by dropping one or another of Lewis’ other requirements, symmetric monoidal categories of spectra could be produced. The first such category was the category of S-modules described by Elmendorf-Kriz-Mandell-May 97, but others soon followed, including symmetric spectra and orthogonal spectra. All of these form symmetric monoidal model categories which are symmetric-monoidally Quillen equivalent.
Moreover, in all of these cases, the monoidal structure on the model category $\mathcal{S}$ absorbs all the higher coherent homotopies that used to be supplied by the action of an $A_\infty$ or $E_\infty$ operad. Thus, honest (commutative) monoids in $\mathcal{S}$ model the same “(commutative) ring objects up to all coherent higher homotopies” that are modeled by the classical $A_\infty$ and $E_\infty$ ring spectra, and for this reason they are often still referred to as $A_\infty$ or $E_\infty$ ring spectra, respectively.
## Details
### For $S$-modules
The construction of S-modules by EKMM begins with the notion of coordinate free Lewis-May spectra. Using the linear isometries operad, one can construct a monad $\mathbb{L}$ on the category $\mathcal{S}$ of such spectra, and the category of $\mathbb{L}$-algebras is a well-behaved model for the stable homotopy category, and moreover admits a smash product which is associative up to isomorphism, but unital only up to weak equivalence. However, the subcategory of the $\mathbb{L}$-algebras for which the unit transformations are isomorphisms is again a well-behaved model for $Ho \mathbb{S}$, which is moreover symmetric monoidal.
Since the unit transformation is of the form $S\wedge E \to E$, where $S$ is the sphere spectrum, and this map looks like the action of a ring on a module, the objects of this subcategory are called $S$-modules and the category is called $Mod_S$. The intuition is that just as an abelian group is a module over the archetypical ring $\mathbb{Z}$ of integers, a spectrum should be regarded as a module over the archetypal ring spectrum, namely the sphere spectrum.
Similarly, just as an ordinary ring is a monoid in the category $Mod_\mathbb{Z}$ of $\mathbb{Z}$-modules, i.e. a $\mathbb{Z}$-algebra, an $A_\infty$ or $E_\infty$ ring spectrum is a (possibly commutative) monoid in the category of $S$-modules, and thus referred to as an $S$-algebra. More generally, for any $A_\infty$-ring spectrum $R$, there is a notion of $R$-module spectra forming a category $Mod_R$, which in turn carries an associative and commutative smash product $\wedge_R$ and a model category structure on $Mod_R$ such that $\wedge_R$ becomes unital in the homotopy category. All this is such that an $A_\infty$-algebra over $R$ is a monoid object in $(Mod_R, \wedge_R)$. Similarly $E_\infty$-algebras are commutative monoid objects in $(Mod_R, \wedge_R)$.
### For excisive functors
#### Topological ends and coends
For working with pointed topologically enriched functors, a certain shape of limits/colimits is particularly relevant: these are called (pointed topological enriched) ends and coends. We here introduce these and then derive some of their basic properties, such as notably the expression for topological left Kan extension in terms of coends (prop. below). Further below it is via left Kan extension along the ordinary smash product of pointed topological spaces (“Day convolution”) that the symmetric monoidal smash product of spectra is induced.
###### Definition
Let $\mathcal{C}, \mathcal{D}$ be pointed topologically enriched categories (def.), i.e. enriched categories over $(Top_{cg}^{\ast/}, \wedge, S^0)$ from example .
1. The pointed topologically enriched opposite category $\mathcal{C}^{op}$ is the topologically enriched category with the same objects as $\mathcal{C}$, with hom-spaces
$\mathcal{C}^{op}(X,Y) \coloneqq \mathcal{C}(Y,X)$
and with composition given by braiding followed by the composition in $\mathcal{C}$:
$\mathcal{C}^{op}(X,Y) \wedge \mathcal{C}^{op}(Y,Z) = \mathcal{C}(Y,X)\wedge \mathcal{C}(Z,Y) \underoverset{\simeq}{\tau}{\longrightarrow} \mathcal{C}(Z,Y) \wedge \mathcal{C}(Y,X) \overset{\circ_{Z,Y,X}}{\longrightarrow} \mathcal{C}(Z,X) = \mathcal{C}^{op}(X,Z) \,.$
2. the pointed topological product category $\mathcal{C} \times \mathcal{D}$ is the topologically enriched category whose objects are pairs of objects $(c,d)$ with $c \in \mathcal{C}$ and $d\in \mathcal{D}$, whose hom-spaces are the smash product of the separate hom-spaces
$(\mathcal{C}\times \mathcal{D})((c_1,d_1),\;(c_2,d_2) ) \coloneqq \mathcal{C}(c_1,c_2)\wedge \mathcal{D}(d_1,d_2)$
and whose composition operation is the braiding followed by the smash product of the separate composition operations:
$\array{ (\mathcal{C}\times \mathcal{D})((c_1,d_1), \; (c_2,d_2)) \wedge (\mathcal{C}\times \mathcal{D})((c_2,d_2), \; (c_3,d_3)) \\ {}^{\mathllap{=}}\downarrow \\ \left(\mathcal{C}(c_1,c_2) \wedge \mathcal{D}(d_1,d_2)\right) \wedge \left(\mathcal{C}(c_2,c_3) \wedge \mathcal{D}(d_2,d_3)\right) \\ \downarrow^{\mathrlap{\tau}}_{\mathrlap{\simeq}} \\ \left(\mathcal{C}(c_1,c_2)\wedge \mathcal{C}(c_2,c_3)\right) \wedge \left( \mathcal{D}(d_1,d_2)\wedge \mathcal{D}(d_2,d_3)\right) &\overset{ (\circ_{c_1,c_2,c_3})\wedge (\circ_{d_1,d_2,d_3}) }{\longrightarrow} & \mathcal{C}(c_1,c_3)\wedge \mathcal{D}(d_1,d_3) \\ && \downarrow^{\mathrlap{=}} \\ && (\mathcal{C}\times \mathcal{D})((c_1,d_1),\; (c_3,d_3)) } \,.$
###### Example
A pointed topologically enriched functor (def.) into $Top^{\ast/}_{cg}$ (exmpl.) out of a pointed topological product category as in def.
$F \;\colon\; \mathcal{C} \times \mathcal{D} \longrightarrow Top^{\ast/}_{cg}$
(a “pointed topological bifunctor”) has component maps of the form
$F_{(c_1,d_1),(c_2,d_2)} \;\colon\; \mathcal{C}(c_1,c_2) \wedge \mathcal{D}(d_1,d_2) \longrightarrow Maps(F_0((c_1,d_1)),F_0((c_2,d_2)))_\ast \,.$
By functoriallity and under passing to adjuncts (cor.) this is equivalent to two commuting actions
$\rho_{c_1,c_2}(d) \;\colon\; \mathcal{C}(c_1,c_2) \wedge F_0((c_1,d)) \longrightarrow F_0((c_2,d))$
and
$\rho_{d_1,d_2}(c) \;\colon\; \mathcal{D}(d_1,d_2) \wedge F_0((c,d_1)) \longrightarrow F_0((c,d_2)) \,.$
In the special case of a functor out of the product category of some $\mathcal{C}$ with its opposite category (def. )
$F \;\colon\; \mathcal{C}^{op} \times \mathcal{C} \longrightarrow Top^{\ast/}_{cg}$
then this takes the form
$\rho_{c_2,c_1}(d) \;\colon\; \mathcal{C}(c_1,c_2) \wedge F_0((c_2,d)) \longrightarrow F_0((c_1,d))$
and
$\rho_{d_1,d_2}(c) \;\colon\; \mathcal{C}(d_1,d_2) \wedge F_0((c,d_1)) \longrightarrow F_0((c,d_2)) \,.$
###### Definition
Let $\mathcal{C}$ be a small pointed topologically enriched category (def.), i.e. an enriched category over $(Top_{cg}^{\ast/}, \wedge, S^0)$ from example . Let
$F \;\colon\; \mathcal{C}^{op} \times \mathcal{C} \longrightarrow Top^{\ast/}_{cg}$
be a pointed topologically enriched functor (def.) out of the pointed topological product category of $\mathcal{C}$ with its opposite category, according to def. .
1. The coend of $F$, denoted $\overset{c \in \mathcal{C}}{\int} F(c,c)$, is the coequalizer in $Top_{cg}^{\ast}$ (prop., exmpl., prop., cor.) of the two actions encoded in $F$ via example :
$\underset{c,d \in \mathcal{C}}{\coprod} \mathcal{C}(c,d) \wedge F(d,c) \underoverset {\underset{\underset{c,d}{\sqcup} \rho_{(d,c)}(c) }{\longrightarrow}} {\overset{\underset{c,d}{\sqcup} \rho_{(c,d)}(d) }{\longrightarrow}} {\phantom{AAAAAAAA}} \underset{c \in \mathcal{C}}{\coprod} F(c,c) \overset{coeq}{\longrightarrow} \overset{c\in \mathcal{C}}{\int} F(c,c) \,.$
2. The end of $F$, denoted $\underset{c\in \mathcal{C}}{\int} F(c,c)$, is the equalizer in $Top_{cg}^{\ast/}$ (prop., exmpl., prop., cor.) of the adjuncts of the two actions encoded in $F$ via example :
$\underset{c\in \mathcal{C}}{\int} F(c,c) \overset{\;\;equ\;\;}{\longrightarrow} \underset{c \in \mathcal{C}}{\prod} F(c,c) \underoverset {\underset{\underset{c,d}{\sqcup} \tilde \rho_{(c,d)}(c) }{\longrightarrow}} {\overset{\underset{c,d}{\sqcup} \tilde\rho_{d,c}(d)}{\longrightarrow}} {\phantom{AAAAAAAA}} \underset{c\in \mathcal{C}}{\prod} Maps\left( \mathcal{C}\left(c,d\right), \; F\left(c,d\right) \right)_\ast \,.$
###### Example
Let $\mathcal{C}$ be a small pointed topologically enriched category (def.). For $F,G \;\colon\; \mathcal{C} \longrightarrow Top^{\ast/}_{cg}$ two pointed topologically enriched functors, then the end (def. ) of $Maps(F(-),G(-))_\ast$ is a topological space whose underlying pointed set is the pointed set of natural transformations $F\to G$ (def.)
$U \left( \underset{c \in \mathcal{C}}{\int} Maps(F(c),G(c))_\ast \right) \;\simeq\; Hom_{[\mathcal{C},Top^{\ast/}_{cg}]}(F,G) \,.$
###### Proof
The underlying pointed set functor $U\colon Top^{\ast/}_{cg}\to Set^{\ast/}$ preserves all limits (prop., prop., prop.). Therefore there is an equalizer diagram in $Set^{\ast/}$ of the form
$U \left( \underset{c\in \mathcal{C}}{\int} Maps(F(c),G(c))_\ast \right) \overset{equ}{\longrightarrow} \underset{c\in \mathcal{C}}{\prod} Hom_{Top^{\ast/}_{cg}}(F(c),G(c)) \underoverset {\underset{\underset{c,d}{\sqcup} U(\tilde \rho_{(c,d)}(d)) }{\longrightarrow}} {\overset{\underset{c,d}{\sqcup} U(\tilde\rho_{d,c}(c))}{\longrightarrow}} {\phantom{AAAAAAAA}} \underset{c,d\in \mathcal{C}}{\prod} Hom_{Top^{\ast/}_{cg}}( \mathcal{C}(c,d), Maps(F(c),G(d))_\ast ) \,.$
Here the object in the middle is just the set of collections of component morphisms $\left\{ F(c)\overset{\eta_c}{\to} G(c)\right\}_{c\in \mathcal{C}}$. The two parallel maps in the equalizer diagram take such a collection to the functions which send any $c \overset{f}{\to} d$ to the result of precomposing
$\array{ F(c) \\ {}^{\mathllap{f(f)}}\downarrow \\ F(d) &\underset{\eta_d}{\longrightarrow}& G(d) }$
and of postcomposing
$\array{ F(c) &\overset{\eta_c}{\longrightarrow}& G(c) \\ && \downarrow^{\mathrlap{G(f)}} \\ && G(d) }$
each component in such a collection, respectively. These two functions being equal, hence the collection $\{\eta_c\}_{c\in \mathcal{C}}$ being in the equalizer, means precisley that for all $c,d$ and all $f\colon c \to d$ the square
$\array{ F(c) &\overset{\eta_c}{\longrightarrow}& G(c) \\ {}^{\mathllap{F(f)}}\downarrow && \downarrow^{\mathrlap{G(f)}} \\ F(d) &\underset{\eta_d}{\longrightarrow}& G(g) }$
is a commuting square. This is precisley the condition that the collection $\{\eta_c\}_{c\in \mathcal{C}}$ be a natural transformation.
Conversely, example says that ends over bifunctors of the form $Maps(F(-),G(-)))_\ast$ constitute hom-spaces between pointed topologically enriched functors:
###### Definition
Let $\mathcal{C}$ be a small pointed topologically enriched categories (def.). Define the structure of a pointed topologically enriched category on the category $[\mathcal{C}, Top_{cg}^{\ast/}]$ of pointed topologically enriched functors to $Top^{\ast/}_{cg}$ (exmpl.) by taking the hom-spaces to be given by the ends (def. ) of example :
$[\mathcal{C},Top^{\ast/}_{cg}](F,G) \;\coloneqq\; \int_{c\in \mathcal{C}} Maps(F(c),G(c))_\ast$
and by taking the composition maps to be the morphisms induced by the maps
$\left( \underset{c\in \mathcal{C}}{\int} Maps(F(c),G(c))_\ast \right) \wedge \left( \underset{c \in \mathcal{C}}{\int} Maps(G(c),H(c))_\ast \right) \overset{}{\longrightarrow} \underset{c\in \mathcal{C}}{\prod} Maps(F(c),G(c))_\ast \wedge Maps(G(c),H(c))_\ast \overset{(\circ_{F(c),G(c),H(c)})_{c\in \mathcal{C}}}{\longrightarrow} \underset{c \in \mathcal{C}}{\prod} Maps(F(c),H(c))_\ast$
by observing that these equalize the two actions in the definition of the end.
The resulting pointed topologically enriched category $[\mathcal{C},Top^{\ast/}_{cg}]$ is also called the $Top^{\ast/}_{cg}$-enriched functor category over $\mathcal{C}$ with coefficients in $Top^{\ast/}_{cg}$.
First of all this yields a concise statement of the pointed topologically enriched Yoneda lemma (prop.)
###### Proposition
(topologically enriched Yoneda lemma)
Let $\mathcal{C}$ be a small pointed topologically enriched categories (def.). For $F \colon \mathcal{C}\to Top^{\ast/}_{cg}$ a pointed topologically enriched functor (def.) and for $c\in \mathcal{C}$ an object, there is a natural isomorphism
$[\mathcal{C}, Top^{\ast/}_{cg}](\mathcal{C}(c,-),\; F) \;\simeq\; F(c)$
between the hom-space of the pointed topological functor category, according to def. , from the functor represented by $c$ to $F$, and the value of $F$ on $c$.
In terms of the ends (def. ) defining these hom-spaces, this means that
$\underset{d\in \mathcal{C}}{\int} Maps(\mathcal{C}(c,d), F(d))_\ast \;\simeq\; F(c) \,.$
In this form the statement is also known as Yoneda reduction.
The proof of prop. is essentially dual to the proof of the next prop. .
Now that natural transformations are phrased in terms of ends (example ), as is the Yoneda lemma (prop. ), it is natural to consider the dual statement involving coends:
###### Proposition
(co-Yoneda lemma)
Let $\mathcal{C}$ be a small pointed topologically enriched categories (def.). For $F \colon \mathcal{C}\to Top^{\ast/}_{cg}$ a pointed topologically enriched functor (def.) and for $c\in \mathcal{C}$ an object, there is a natural isomorphism
$F(-) \simeq \overset{c \in \mathcal{C}}{\int} \mathcal{C}(c,-) \wedge F(c) \,.$
Moreover, the morphism that hence exhibits $F(c)$ as the coequalizer of the two morphisms in def. is componentwise the canonical action
$\mathcal{C}(d,c) \wedge F(c) \longrightarrow F(d)$
which is adjunct to the component map $\mathcal{C}(d,c) \to Maps(F(c),F(d))_{\ast}$ of the topologically enriched functor $F$.
(e.g. MMSS 00, lemma 1.6)
###### Proof
The coequalizer of pointed topological spaces that we need to consider has underlying it a coequalizer of underlying pointed sets (prop., prop., prop.). That in turn is the colimit over the diagram of underlying sets with the basepointe adjoined to the diagram (prop.). For a coequalizer diagram adding that extra point to the diagram clearly does not change the colimit, and so we need to consider the plain coequalizer of sets.
That is just the set of equivalence classes of pairs
$( c \overset{}{\to} c_0,\; x \in F(c) ) \,,$
where two such pairs
$( c \overset{f}{\to} c_0,\; x \in F(c) ) \,,\;\;\;\; ( d \overset{g}{\to} c_0,\; y \in F(d) )$
are regarded as equivalent if there exists
$c \overset{\phi}{\to} d$
such that
$f = g \circ \phi \,, \;\;\;\;\;and\;\;\;\;\; y = \phi(x) \,.$
(Because then the two pairs are the two images of the pair $(g,x)$ under the two morphisms being coequalized.)
But now considering the case that $d = c_0$ and $g = id_{c_0}$, so that $f = \phi$ shows that any pair
$( c \overset{\phi}{\to} c_0, \; x \in F(c))$
is identified, in the coequalizer, with the pair
$(id_{c_0},\; \phi(x) \in F(c_0)) \,,$
hence with $\phi(x)\in F(c_0)$.
This shows the claim at the level of the underlying sets. To conclude it is now sufficient (prop.) to show that the topology on $F(c_0) \in Top^{\ast/}_{cg}$ is the final topology (def.) of the system of component morphisms
$\mathcal{C}(d,c) \wedge F(c) \longrightarrow \overset{c}{\int} \mathcal{C}(c,c_0) \wedge F(c)$
which we just found. But that system includes
$\mathcal{C}(c,c) \wedge F(c) \longrightarrow F(c)$
which is a retraction
$id \;\colon\; F(c) \longrightarrow \mathcal{C}(c,c) \wedge F(c) \longrightarrow F(c)$
and so if all the preimages of a given subset of the coequalizer under these component maps is open, it must have already been open in $F(c)$.
###### Remark
The statement of the co-Yoneda lemma in prop. is a kind of categorification of the following statement in analysis (whence the notation with the integral signs):
For $X$ a topological space, $f \colon X \to\mathbb{R}$ a continuous function and $\delta(-,x_0)$ denoting the Dirac distribution, then
$\int_{x \in X} \delta(x,x_0) f(x) = f(x_0) \,.$
It is this analogy that gives the name to the following statement:
###### Proposition
(Fubini theorem for (co)-ends)
For $F$ a pointed topologically enriched bifunctor on a small pointed topological product category $\mathcal{C}_1 \times \mathcal{C}_2$ (def. ), i.e.
$F \;\colon\; \left( \mathcal{C}_1\times\mathcal{C}_2 \right)^{op} \times (\mathcal{C}_1 \times\mathcal{C}_2) \longrightarrow Top^{\ast/}_{cg}$
then its end and coend (def. ) is equivalently formed consecutively over each variable, in either order:
$\overset{(c_1,c_2)}{\int} F((c_1,c_2), (c_1,c_2)) \simeq \overset{c_1}{\int} \overset{c_2}{\int} F((c_1,c_2), (c_1,c_2)) \simeq \overset{c_2}{\int} \overset{c_1}{\int} F((c_1,c_2), (c_1,c_2))$
and
$\underset{(c_1,c_2)}{\int} F((c_1,c_2), (c_1,c_2)) \simeq \underset{c_1}{\int} \underset{c_2}{\int} F((c_1,c_2), (c_1,c_2)) \simeq \underset{c_2}{\int} \underset{c_1}{\int} F((c_1,c_2), (c_1,c_2)) \,.$
###### Proof
Because limits commute with limits, and colimits commute with colimits.
###### Remark
Because the pointed compactly generated mapping space functor (exmpl.)
$Maps(-,-)_\ast \;\colon\; \left(Top^{\ast/}_{cg}\right)^{op} \times Top^{\ast/}_{cg} \longrightarrow Top^{\ast/}_{cg}$
takes colimits in the first argument and limits in the second argument to limits (cor.), it also takes coends in the first argument and ends in the second argument, to ends (def. ):
$Maps( X, \; \int_{c} F(c,c))_\ast \simeq \int_c Maps(X, F(c,c)_\ast)$
and
$Maps( \int^{c} F(c,c),\; Y )_\ast \simeq \underset{c}{\int} Maps( F(c,c),\; Y )_\ast \,.$
###### Proposition
(left Kan extension via coends)
Let $\mathcal{C}, \mathcal{D}$ be small pointed topologically enriched categories (def.) and let
$p \;\colon\; \mathcal{C} \longrightarrow \mathcal{D}$
be a pointed topologically enriched functor (def.). Then precomposition with $p$ constitutes a functor
$p^\ast \;\colon\; [\mathcal{D}, Top^{\ast/}_{cg}] \longrightarrow [\mathcal{C}, Top^{\ast/}_{cg}]$
$G\mapsto G\circ p$. This functor has a left adjoint $Lan_p$, called left Kan extension along $p$
$[\mathcal{D}, Top^{\ast/}_{cg}] \underoverset {\underset{p^\ast}{\longrightarrow}} {\overset{Lan_p }{\longleftarrow}} {\bot} [\mathcal{C}, Top^{\ast/}_{cg}]$
which is given objectwise by a coend (def. ):
$(Lan_p F) \;\colon\; d \;\mapsto \; \overset{c\in \mathcal{C}}{\int} \mathcal{D}(p(c),d) \wedge F(c) \,.$
###### Proof
Use the expression of natural transformations in terms of ends (example and def. ), then use the respect of $Maps(-,-)_\ast$ for ends/coends (remark ), use the smash/mapping space adjunction (cor.), use the Fubini theorem (prop. ) and finally use Yoneda reduction (prop. ) to obtain a sequence of natural isomorphisms as follows:
\begin{aligned} [\mathcal{D},Top^{\ast/}_{cg}]( Lan_p F, \, G ) & = \underset{d \in \mathcal{D}}{\int} Maps( (Lan_p F)(d), \, G(d) )_\ast \\ & = \underset{d\in \mathcal{D}}{\int} Maps\left( \overset{c \in \mathcal{C}}{\int} \mathcal{D}(p(c),d) \wedge F(c) ,\; G(d) \right)_\ast \\ &\simeq \underset{d \in \mathcal{D}}{\int} \underset{c \in \mathcal{C}}{\int} Maps( \mathcal{D}(p(c),d)\wedge F(c) \,,\; G(d) )_\ast \\ & \simeq \underset{c\in \mathcal{C}}{\int} \underset{d\in \mathcal{D}}{\int} Maps(F(c), Maps( \mathcal{D}(p(c),d) , \, G(d) )_\ast )_\ast \\ & \simeq \underset{c\in \mathcal{C}}{\int} Maps(F(c), \underset{d\in \mathcal{D}}{\int} Maps( \mathcal{D}(p(c),d) , \, G(d) )_\ast )_\ast \\ & \simeq \underset{c\in \mathcal{C}}{\int} Maps(F(c), G(p(c)) )_\ast \\ & = [\mathcal{C}, Top^{\ast/}_{cg}](F,p^\ast G) \end{aligned} \,.
##### Monoidal topological categories
We recall the basic definitions of monoidal categories and of monoids and modules internal to monoidal categories. All examples are at the end of this section, starting with example below.
###### Definition
A (pointed) topologically enriched monoidal category is a (pointed) topologically enriched category $\mathcal{C}$ (def.) equipped with
1. a (pointed) topologically enriched functor (def.)
$\otimes \;\colon\; \mathcal{C} \times \mathcal{C} \longrightarrow \mathcal{C}$
out of the (pointed) topologival product category of $\mathcal{C}$ with itself (def. ), called the tensor product,
2. an object
$1 \in \mathcal{C}$
called the unit object or tensor unit,
3. $a \;\colon\; ((-)\otimes (-)) \otimes (-) \overset{\simeq}{\longrightarrow} (-) \otimes ((-)\otimes(-))$
called the associator,
4. $\ell \;\colon\; (1 \otimes (-)) \overset{\simeq}{\longrightarrow} (-)$
called the left unitor, and a natural isomorphism
$r \;\colon\; (-) \otimes 1 \overset{\simeq}{\longrightarrow} (-)$
called the right unitor,
such that the following two kinds of diagrams commute, for all objects involved:
1. triangle identity:
$\array{ & (x \otimes 1) \otimes y &\stackrel{a_{x,1,y}}{\longrightarrow} & x \otimes (1 \otimes y) \\ & {}_{\rho_x \otimes 1_y}\searrow && \swarrow_{1_x \otimes \lambda_y} & \\ && x \otimes y && }$
###### Lemma
(Kelly 64)
Let $(\mathcal{C}, \otimes, 1)$ be a monoidal category, def. . Then the left and right unitors $\ell$ and $r$ satisfy the following conditions:
1. $\ell_1 = r_1 \;\colon\; 1 \otimes 1 \overset{\simeq}{\longrightarrow} 1$;
2. for all objects $x,y \in \mathcal{C}$ the following diagram commutes:
$\array{ (1 \otimes x) \otimes y & & \\ {}^\mathllap{\alpha_{1, x, y}} \downarrow & \searrow^\mathrlap{\ell_x y} & \\ 1 \otimes (x \otimes y) & \underset{\ell_{x \otimes y}}{\longrightarrow} & x \otimes y } \,.$
Analogously for the right unitor.
###### Definition
A (pointed) topological braided monoidal category, is a (pointed) topological monoidal category $\mathcal{C}$ (def. ) equipped with a natural isomorphism
$\tau_{x,y} \colon x \otimes y \to y \otimes x$
called the braiding, such that the following two kinds of diagrams commute for all objects involved:
$\array{ (x \otimes y) \otimes z &\stackrel{a_{x,y,z}}{\to}& x \otimes (y \otimes z) &\stackrel{\tau_{x,y \otimes z}}{\to}& (y \otimes z) \otimes x \\ \downarrow^{\tau_{x,y}\otimes Id} &&&& \downarrow^{a_{y,z,x}} \\ (y \otimes x) \otimes z &\stackrel{a_{y,x,z}}{\to}& y \otimes (x \otimes z) &\stackrel{Id \otimes \tau_{x,z}}{\to}& y \otimes (z \otimes x) }$
and
$\array{ x \otimes (y \otimes z) &\stackrel{a^{-1}_{x,y,z}}{\to}& (x \otimes y) \otimes z &\stackrel{\tau_{x \otimes y, z}}{\to}& z \otimes (x \otimes y) \\ \downarrow^{Id \otimes \tau_{y,z}} &&&& \downarrow^{a^{-1}_{z,x,y}} \\ x \otimes (z \otimes y) &\stackrel{a^{-1}_{x,z,y}}{\to}& (x \otimes z) \otimes y &\stackrel{\tau_{x,z} \otimes Id}{\to}& (z \otimes x) \otimes y } \,,$
where $a_{x,y,z} \colon (x \otimes y) \otimes z \to x \otimes (y \otimes z)$ denotes the components of the associator of $\mathcal{C}^\otimes$.
###### Definition
A (pointed) topological symmetric monoidal category is a (pointed) topological braided monoidal category (def. ) for which the braiding
$\tau_{x,y} \colon x \otimes y \to y \otimes x$
satisfies the condition:
$\tau_{y,x} \circ \tau_{x,y} = 1_{x \otimes y}$
for all objects $x, y$
###### Definition
Given a (pointed) topological symmetric monoidal category $\mathcal{C}$ with tensor product $\otimes$ (def. ) it is called a closed monoidal category if for each $Y \in \mathcal{C}$ the functor $Y \otimes(-)\simeq (-)\otimes X$ has a right adjoint, denoted $[Y,-]$
$\mathcal{C} \underoverset {\underset{[Y,-]}{\longrightarrow}} {\overset{(-) \otimes Y}{\longleftarrow}} {\bot} \mathcal{C} \,,$
hence if there are natural isomorphisms
$Hom_{\mathcal{C}}(X \otimes Y, Z) \;\simeq\; Hom_{\mathcal{C}}{C}(X, [Y,Z])$
for all objects $X,Z \in \mathcal{C}$.
Since for the case that $X = 1$ is the tensor unit of $\mathcal{C}$ this means that
$Hom_{\mathcal{C}}(1,[Y,Z]) \simeq Hom_{\mathcal{C}}(Y,Z) \,,$
the object $[Y,Z] \in \mathcal{C}$ is an enhancement of the ordinary hom-set $Hom_{\mathcal{C}}(Y,Z)$ to an object in $\mathcal{C}$. Accordingly, it is also called the internal hom between $Y$ and $Z$.
###### Example
The category Set of sets and functions between them, regarded as enriched in discrete topological spaces, becomes a symmetric monoidal category according to def. with tensor product the Cartesian product $\times$ of sets. The associator, unitor and braiding isomorphism are the evident (almost unnoticable but nevertheless nontrivial) canonical identifications.
Similarly the $Top_{cg}$ of compactly generated topological spaces (def.) becomes a symmetric monoidal category with tensor product the corresponding Cartesian products, hence the operation of forming k-ified (cor.) product topological spaces (exmpl.). The underlying functions of the associator, unitor and braiding isomorphisms are just those of the underlying sets, as above.
Symmetric monoidal categories, such as these, for which the tensor product is the Cartesian product are called Cartesian monoidal categories.
###### Example
The category $Top_{cg}^{\ast/}$ of pointed compactly generated topological spaces with tensor product the smash product $\wedge$ (def.)
$X \wedge Y \coloneqq \frac{X\times Y}{X\vee Y}$
is a symmetric monoidal category (def. ) with unit object the pointed 0-sphere $S^0$.
The components of the associator, the unitors and the braiding are those of Top as in example , descended to the quotient topological spaces which appear in the definition of the smash product). This works for pointed compactly generated spaces (but not for general pointed topological spaces) by this prop..
###### Example
The category Ab of abelian groups, regarded as enriched in discrete topological spaces, becomes a symmetric monoidal category with tensor product the actual tensor product of abelian groups $\otimes_{\mathbb{Z}}$ and with tensor unit the additive group $\mathbb{Z}$ of integers. Again the associator, unitor and braiding isomorphism are the evident ones coming from the underlying sets, as in example .
This is the archetypical case that motivates the notation “$\otimes$” for the pairing operation in a monoidal category:
1. A monoid in $(Ab, \otimes_{\mathbb{Z}}, \mathbb{Z})$ (def. ) is equivalently a ring.
2. A commutative monoid in in $(Ab, \otimes_{\mathbb{Z}}, \mathbb{Z})$ (def. ) is equivalently a commutative ring $R$.
3. An $R$-module object in $(Ab, \otimes_{\mathbb{Z}}, \mathbb{Z})$ (def. ) is equivalently an $R$-module;
4. The tensor product of $R$-module objects (def. ) is the standard tensor product of modules.
5. The category of module objects $R Mod(Ab)$ (def. ) is the standard category of modules $R Mod$.
##### Algebras and modules
###### Definition
Given a (pointed) topological monoidal category $(\mathcal{C}, \otimes, 1)$, then a monoid internal to $(\mathcal{C}, \otimes, 1)$ is
1. an object $A \in \mathcal{C}$;
2. a morphism $e \;\colon\; 1 \longrightarrow A$ (called the unit)
3. a morphism $\mu \;\colon\; A \otimes A \longrightarrow A$ (called the product);
such that
1. (associativity) the following diagram commutes
$\array{ (A\otimes A) \otimes A &\underoverset{\simeq}{a_{A,A,A}}{\longrightarrow}& A \otimes (A \otimes A) &\overset{A \otimes \mu}{\longrightarrow}& A \otimes A \\ {}^{\mathllap{\mu \otimes A}}\downarrow && && \downarrow^{\mathrlap{\mu}} \\ A \otimes A &\longrightarrow& &\overset{\mu}{\longrightarrow}& A } \,,$
where $a$ is the associator isomorphism of $\mathcal{C}$;
2. (unitality) the following diagram commutes:
$\array{ 1 \otimes A &\overset{e \otimes id}{\longrightarrow}& A \otimes A &\overset{id \otimes e}{\longleftarrow}& A \otimes 1 \\ & {}_{\mathllap{\ell}}\searrow & \downarrow^{\mathrlap{\mu}} & & \swarrow_{\mathrlap{r}} \\ && A } \,,$
where $\ell$ and $r$ are the left and right unitor isomorphisms of $\mathcal{C}$.
Moreover, if $(\mathcal{C}, \otimes , 1)$ has the structure of a symmetric monoidal category (def. ) $(\mathcal{C}, \otimes, 1, B)$ with symmetric braiding $\tau$, then a monoid $(A,\mu, e)$ as above is called a commutative monoid in $(\mathcal{C}, \otimes, 1, B)$ if in addition
• (commutativity) the following diagram commutes
$\array{ A \otimes A && \underoverset{\simeq}{\tau_{A,A}}{\longrightarrow} && A \otimes A \\ & {}_{\mathllap{\mu}}\searrow && \swarrow_{\mathrlap{\mu}} \\ && A } \,.$
A homomorphism of monoids $(A_1, \mu_1, e_1)\longrightarrow (A_2, \mu_2, f_2)$ is a morphism
$f \;\colon\; A_1 \longrightarrow A_2$
in $\mathcal{C}$, such that the following two diagrams commute
$\array{ A_1 \otimes A_1 &\overset{f \otimes f}{\longrightarrow}& A_2 \otimes A_2 \\ {}^{\mathllap{\mu_1}}\downarrow && \downarrow^{\mathrlap{\mu_2}} \\ A_1 &\underset{f}{\longrightarrow}& A_2 }$
and
$\array{ 1_{\mathcal{c}} &\overset{e_1}{\longrightarrow}& A_1 \\ & {}_{\mathllap{e_2}}\searrow & \downarrow^{\mathrlap{f}} \\ && A_2 } \,.$
Write $Mon(\mathcal{C}, \otimes,1)$ for the category of monoids in $\mathcal{C}$, and $CMon(\mathcal{C}, \otimes, 1)$ for its subcategory of commutative monoids.
###### Example
Given a (pointed) topological monoidal category $(\mathcal{C}, \otimes, 1)$, then the tensor unit $1$ is a monoid in $\mathcal{C}$ (def. ) with product given by either the left or right unitor
$\ell_1 = r_1 \;\colon\; 1 \otimes 1 \overset{\simeq}{\longrightarrow} 1 \,.$
By lemma , these two morphisms coincide and define an associative product with unit the identity $id \colon 1 \to 1$.
If $(\mathcal{C}, \otimes , 1)$ is a symmetric monoidal category (def. ), then this monoid is a commutative monoid.
###### Definition
Given a (pointed) topological monoidal category $(\mathcal{C}, \otimes, 1)$ (def. ), and given $(A,\mu,e)$ a monoid in $(\mathcal{C}, \otimes, 1)$ (def. ), then a left module object in $(\mathcal{C}, \otimes, 1)$ over $(A,\mu,e)$ is
1. an object $N \in \mathcal{C}$;
2. a morphism $\rho \;\colon\; A \otimes N \longrightarrow N$ (called the action);
such that
1. (unitality) the following diagram commutes:
$\array{ 1 \otimes N &\overset{e \otimes id}{\longrightarrow}& A \otimes N \\ & {}_{\mathllap{\ell}}\searrow & \downarrow^{\mathrlap{\rho}} \\ && A } \,,$
where $\ell$ is the left unitor isomorphism of $\mathcal{C}$.
2. (action property) the following diagram commutes
$\array{ (A\otimes A) \otimes N &\underoverset{\simeq}{a_{A,A,N}}{\longrightarrow}& A \otimes (A \otimes N) &\overset{A \otimes \rho}{\longrightarrow}& A \otimes N \\ {}^{\mathllap{\mu \otimes N}}\downarrow && && \downarrow^{\mathrlap{\rho}} \\ A \otimes N &\longrightarrow& &\overset{\rho}{\longrightarrow}& N } \,,$
A homomorphism of left $A$-module objects
$(N_1, \rho_1) \longrightarrow (N_2, \rho_2)$
is a morphism
$f\;\colon\; N_1 \longrightarrow N_2$
in $\mathcal{C}$, such that the following diagram commutes:
$\array{ A\otimes N_1 &\overset{A \otimes f}{\longrightarrow}& A\otimes N_2 \\ {}^{\mathllap{\rho_1}}\downarrow && \downarrow^{\mathrlap{\rho_2}} \\ N_1 &\underset{f}{\longrightarrow}& N_2 } \,.$
For the resulting category of modules of left $A$-modules in $\mathcal{C}$ with $A$-module homomorphisms between them, we write
$A Mod(\mathcal{C}) \,.$
This is naturally a (pointed) topologically enriched category itself.
###### Example
Given a monoidal category $(\mathcal{C},\otimes, 1)$ (def. ) with the tensor unit $1$ regarded as a monoid in a monoidal category via example , then the left unitor
$\ell_C \;\colon\; 1\otimes C \longrightarrow C$
makes every object $C \in \mathcal{C}$ into a left module, according to def. , over $C$. The action property holds due to lemma . This gives an equivalence of categories
$\mathbb{C} \simeq 1 Mod(\mathcal{C})$
of $\mathcal{C}$ with the category of modules over its tensor unit.
###### Proposition
In the situation of def. , the monoid $(A,\mu, e)$ canonically becomes a left module over itself by setting $\rho \coloneqq \mu$. More generally, for $C \in \mathcal{C}$ any object, then $A \otimes C$ naturally becomes a left $A$-module by setting:
$\rho \;\colon\; A \otimes (A \otimes C) \underoverset{\simeq}{a^{-1}_{A,A,C}}{\longrightarrow} (A \otimes A) \otimes C \overset{\mu \otimes id}{\longrightarrow} A \otimes C \,.$
The $A$-modules of this form are called free modules.
The free functor $F$ constructing free $A$-modules is left adjoint to the forgetful functor $U$ which sends a module $(N,\rho)$ to the underlying object $U(N,\rho) \coloneqq N$.
$A Mod(\mathcal{C}) \underoverset {\underset{U}{\longrightarrow}} {\overset{F}{\longleftarrow}} {\bot} \mathcal{C} \,.$
###### Proof
A homomorphism out of a free $A$-module is a morphism in $\mathcal{C}$ of the form
$f \;\colon\; A\otimes C \longrightarrow N$
fitting into the diagram (where we are notationally suppressing the associator)
$\array{ A \otimes A \otimes C &\overset{A \otimes f}{\longrightarrow}& A \otimes N \\ {}^{\mathllap{\mu \otimes id}}\downarrow && \downarrow^{\mathrlap{\rho}} \\ A \otimes C &\underset{f}{\longrightarrow}& N } \,.$
Consider the composite
$\tilde f \;\colon\; C \underoverset{\simeq}{\ell_C}{\longrightarrow} 1 \otimes C \overset{e\otimes id}{\longrightarrow} A \otimes C \overset{f}{\longrightarrow} N \,,$
i.e. the restriction of $f$ to the unit “in” $A$. By definition, this fits into a commuting square of the form (where we are now notationally suppressing the associator and the unitor)
$\array{ A \otimes C &\overset{id \otimes \tilde f}{\longrightarrow}& A \otimes N \\ {}^{\mathllap{id \otimes e \otimes id}}\downarrow && \downarrow^{\mathrlap{=}} \\ A \otimes A \otimes C &\underset{id \otimes f}{\longrightarrow}& A \otimes N } \,.$
Pasting this square onto the top of the previous one yields
$\array{ A \otimes C &\overset{id \otimes \tilde f}{\longrightarrow}& A \otimes N \\ {}^{\mathllap{id \otimes e \otimes id}}\downarrow && \downarrow^{\mathrlap{=}} \\ A \otimes A \otimes C &\overset{A \otimes f}{\longrightarrow}& A \otimes N \\ {}^{\mathllap{\mu \otimes id}}\downarrow && \downarrow^{\mathrlap{\rho}} \\ A \otimes C &\underset{f}{\longrightarrow}& N } \,,$
where now the left vertical composite is the identity, by the unit law in $A$. This shows that $f$ is uniquely determined by $\tilde f$ via the relation
$f = \rho \circ (id_A \otimes \tilde f) \,.$
This natural bijection between $f$ and $\tilde f$ establishes the adjunction.
###### Definition
Given a (pointed) topological symmetric monoidal category $(\mathcal{C}, \otimes, 1)$ (def. ), given $(A,\mu,e)$ a commutative monoid in $(\mathcal{C}, \otimes, 1)$ (def. ), and given $(N_1, \rho_1)$ and $(N_2, \rho_2)$ two left $A$-module objects (def.), then the tensor product of modules $N_1 \otimes_A N_2$ is, if it exists, the coequalizer
$N_1 \otimes A \otimes N_2 \underoverset {\underset{\rho_{1}\circ (\tau_{N_1,A} \otimes N_2)}{\longrightarrow}} {\overset{N_1 \otimes \rho_2}{\longrightarrow}} {\phantom{AAAA}} N_1 \otimes N_1 \overset{coequ}{\longrightarrow} N_1 \otimes_A N_2$
###### Proposition
Given a (pointed) topological symmetric monoidal category $(\mathcal{C}, \otimes, 1)$ (def. ), and given $(A,\mu,e)$ a commutative monoid in $(\mathcal{C}, \otimes, 1)$ (def. ). If all coequalizers exist in $\mathcal{C}$, then the tensor product of modules $\otimes_A$ from def. makes the category of modules $A Mod(\mathcal{C})$ into a symmetric monoidal category, $(A Mod, \otimes_A, A)$ with tensor unit the object $A$ itself, regarded as an $A$-module via prop. .
###### Definition
Given a monoidal category of modules $(A Mod , \otimes_A , A)$ as in prop. , then a monoid $(E, \mu, e)$ in $(A Mod , \otimes_A , A)$ (def. ) is called an $A$-algebra.
###### Propposition
Given a monoidal category of modules $(A Mod , \otimes_A , A)$ in a monoidal category $(\mathcal{C},\otimes, 1)$ as in prop. , and an $A$-algebra $(E,\mu,e)$ (def. ), then there is an equivalence of categories
$A Alg_{comm}(\mathcal{C}) \coloneqq CMon(A Mod) \simeq CMon(\mathcal{C})^{A/}$
between the category of commutative monoids in $A Mod$ and the coslice category of commutative monoids in $\mathcal{C}$ under $A$, hence between commutative $A$-algebras in $\mathcal{C}$ and commutative monoids $E$ in $\mathcal{C}$ that are equipped with a homomorphism of monoids $A \longrightarrow E$.
(e.g. EKMM 97, VII lemma 1.3)
###### Proof
In one direction, consider a $A$-algebra $E$ with unit $e_E \;\colon\; A \longrightarrow E$ and product $\mu_{E/A} \colon E \otimes_A E \longrightarrow E$. There is the underlying product $\mu_E$
$\array{ E \otimes A \otimes E & \underoverset {\underset{}{\longrightarrow}} {\overset{}{\longrightarrow}} {\phantom{AAA}} & E \otimes E &\overset{coeq}{\longrightarrow}& E \otimes_A E \\ && & {}_{\mathllap{\mu_E}}\searrow & \downarrow^{\mathrlap{\mu_{E/A}}} \\ && && E } \,.$
By considering a diagram of such coequalizer diagrams with middle vertical morphism $e_E\circ e_A$, one find that this is a unit for $\mu_E$ and that $(E, \mu_E, e_E \circ e_A)$ is a commutative monoid in $(\mathcal{C}, \otimes, 1)$.
Then consider the two conditions on the unit $e_E \colon A \longrightarrow E$. First of all this is an $A$-module homomorphism, which means that
$(\star) \;\;\;\;\; \;\;\;\;\; \array{ A \otimes A &\overset{id \otimes e_E}{\longrightarrow}& A \otimes E \\ {}^{\mathllap{\mu_A}}\downarrow && \downarrow^{\mathrlap{\rho}} \\ A &\underset{e_E}{\longrightarrow}& E }$
commutes. Moreover it satisfies the unit property
$\array{ A \otimes_A E &\overset{e_A \otimes id}{\longrightarrow}& E \otimes_A E \\ & {}_{\mathllap{\simeq}}\searrow & \downarrow^{\mathrlap{\mu_{E/A}}} \\ && E } \,.$
By forgetting the tensor product over $A$, the latter gives
$\array{ A \otimes E &\overset{e \otimes id}{\longrightarrow}& E \otimes E \\ \downarrow && \downarrow^{\mathrlap{}} \\ A \otimes_A E &\overset{e_E \otimes id}{\longrightarrow}& E \otimes_A E \\ {}^{\mathllap{\simeq}}\downarrow && \downarrow^{\mathrlap{\mu_{E/A}}} \\ E &=& E } \;\;\;\;\;\;\;\; \simeq \;\;\;\;\;\;\;\; \array{ A \otimes E &\overset{e_E \otimes id}{\longrightarrow}& E \otimes E \\ {}^{\mathllap{\rho}}\downarrow && \downarrow^{\mathrlap{\mu_{E}}} \\ E &\underset{id}{\longrightarrow}& E } \,,$
where the top vertical morphisms on the left the canonical coequalizers, which identifies the vertical composites on the right as shown. Hence this may be pasted to the square $(\star)$ above, to yield a commuting square
$\array{ A \otimes A &\overset{id\otimes e_E}{\longrightarrow}& A \otimes E &\overset{e_E \otimes id}{\longrightarrow}& E \otimes E \\ {}^{\mathllap{\mu_A}}\downarrow && {}^{\mathllap{\rho}}\downarrow && \downarrow^{\mathrlap{\mu_{E}}} \\ A &\underset{e_E}{\longrightarrow}& E &\underset{id}{\longrightarrow}& E } \;\;\;\;\;\;\;\;\;\; = \;\;\;\;\;\;\;\;\;\; \array{ A \otimes A &\overset{e_E \otimes e_E}{\longrightarrow}& E \otimes E \\ {}^{\mathllap{\mu_A}}\downarrow && \downarrow^{\mathrlap{\mu_E}} \\ A &\underset{e_E}{\longrightarrow}& E } \,.$
This shows that the unit $e_A$ is a homomorphism of monoids $(A,\mu_A, e_A) \longrightarrow (E, \mu_E, e_E\circ e_A)$.
Now for the converse direction, assume that $(A,\mu_A, e_A)$ and $(E, \mu_E, e'_E)$ are two commutative monoids in $(\mathcal{C}, \otimes, 1)$ with $e_E \;\colon\; A \to E$ a monoid homomorphism. Then $E$ inherits a left $A$-module structure by
$\rho \;\colon\; A \otimes E \overset{e_A \otimes id}{\longrightarrow} E \otimes E \overset{\mu_E}{\longrightarrow} E \,.$
By commutativity and associativity it follows that $\mu_E$ coequalizes the two induced morphisms $E \otimes A \otimes E \underoverset{\longrightarrow}{\longrightarrow}{\phantom{AA}} E \otimes E$. Hence the universal property of the coequalizer gives a factorization through some $\mu_{E/A}\colon E \otimes_A E \longrightarrow E$. This shows that $(E, \mu_{E/A}, e_E)$ is a commutative $A$-algebra.
Finally one checks that these two constructions are inverses to each other, up to isomorphism.
##### Day convolution
###### Definition
Let $\mathcal{C}$ be a small pointed topological monoidal category (def. ) with tensor product denoted $\otimes_{\mathcal{C}} \;\colon\; \mathcal{C} \times\mathcal{C} \to \mathcal{C}$.
Then the Day convolution tensor product on the pointed topological enriched functor category $[\mathcal{C},Top^{\ast/}_{cg}]$ (def. ) is the functor
$\otimes_{Day} \;\colon\; [\mathcal{C},Top^{\ast/}_{cg}] \times [\mathcal{C},Top^{\ast/}_{cg}] \longrightarrow [\mathcal{C},Top^{\ast/}_{cg}]$
out of the pointed topological product category (def. ) given by the following coend (def. )
$X \otimes_{Day} Y \;\colon\; c \;\mapsto\; \overset{(c_1,c_2)\in \mathcal{C}\times \mathcal{C}}{\int} \mathcal{C}(c_1 \otimes_{\mathcal{C}} c_2, c) \wedge X(c_1) \wedge Y(c_2) \,.$
###### Example
Let $Seq$ denote the category with objects the natural numbers, and only the zero morphisms and identity morphisms on these objects:
$Seq(n_1,n_2) \coloneqq \left\{ \array{ S^0 & if\; n_1 = n_2 \\ \ast & otherwise } \right. \,.$
Regard this as a pointed topologically enriched category in the unique way. The operation of addition of natural numbers $\otimes = +$ makes this a monoidal category.
An object $X_\bullet \in [Seq, Top_{cg}^{\ast/}]$ is an $\mathbb{N}$-sequence of pointed topological spaces. Given two such, then their Day convolution according to def. is
\begin{aligned} (X \otimes_{Day} Y)_n & = \overset{(n_1,n_2)}{\int} Seq(n_1 + n_2 , n) \wedge X_{n_1} \wedge X_{n_2} \\ & = \underset{{n_1+n_2} \atop {= n}}{\coprod} \left(X_{n_1}\wedge X_{n_2}\right) \end{aligned} \,.
We observe now that Day convolution is equivalently a left Kan extension (def. ). This will be key for understanding monoids and modules with respect to Day convolution.
###### Definition
Let $\mathcal{C}$ be a small pointed topologically enriched category (def.). Its external tensor product is the pointed topologically enriched functor
$\overline{\wedge} \;\colon\; [\mathcal{C},Top^{\ast/}_{cg}] \times [\mathcal{C},Top^{\ast/}_{cg}] \longrightarrow [\mathcal{C}\times \mathcal{C}, Top^{\ast/}_{cg}]$
given by
$X \overline{\wedge} Y \;\coloneqq\; \wedge \circ (X,Y) \,,$
i.e.
$(X \overline\wedge Y)(c_1,c_2) = X(c_1)\wedge X(c_2) \,.$
###### Proposition
The Day convolution product (def. ) of two functors is equivalently the left Kan extension (def. ) of their external tensor product (def. ) along the tensor product $\otimes_{\mathcal{C}}$: there is a natural isomorphism
$X \otimes_{Day} Y \simeq Lan_{\otimes_{\mathcal{C}}} (X \overline{\wedge} Y) \,.$
Hence the adjunction unit is a natural transformation of the form
$\array{ \mathcal{C} \times \mathcal{C} && \overset{X \overline{\wedge} Y}{\longrightarrow} && Top^{\ast/}_{cg} \\ & {}^{\mathllap{\otimes}}\searrow &\Downarrow& \nearrow_{\mathrlap{X \otimes_{Day} Y}} \\ && \mathcal{C} } \,.$
This perspective is highlighted in (MMSS 00, p. 60).
###### Proof
By prop. we may compute the left Kan extension as the following coend:
\begin{aligned} Lan_{\otimes_{\mathcal{C}}} (X\overline{\wedge} Y)(c) & \simeq \overset{(c_1,c_2)}{\int} \mathcal{C}(c_1 \otimes_{\mathcal{C}} c_2, c ) \wedge (X\overline{\wedge}Y)(c_1,c_2) \\ & = \overset{(c_1,c_2)}{\int} \mathcal{C}(c_1\otimes c_2) \wedge X(c_1)\wedge X(c_2) \end{aligned} \,.
###### Corollary
The Day convolution $\otimes_{Day}$ (def. ) is universally characterized by the property that there are natural isomorphisms
$[\mathcal{C},Top^{\ast/}_{cg}](X \otimes_{Day} Y, Z) \simeq [\mathcal{C}\times \mathcal{C},Top^{\ast/}_{cg}]( X \overline{\wedge} Y,\; Z \circ \otimes ) \,,$
where $\overline{\wedge}$ is the external product of def. .
Write
$y \;\colon\; \mathcal{C}^{op} \longrightarrow [\mathcal{C}, Top^{\ast/}_{cg}]$
for the $Top^{\ast/}_{cg}$-Yoneda embedding, so that for $c\in \mathcal{C}$ any object, $y(c)$ is the corepresented functor $y(c)\colon d \mapsto \mathcal{C}(c,d)$.
###### Proposition
For $\mathcal{C}$ a small pointed topological monoidal category (def. ), the Day convolution tensor product $\otimes_{Day}$ of def. makes the pointed topologically enriched functor category
$( [\mathcal{C}, Top^{\ast/}_{cg}], \otimes_{Day}, y(1))$
into a pointed topological monoidal category (def. ) with tensor unit $y(1)$ co-represented by the tensor unit $1$ of $\mathcal{C}$.
Moreover, if $(\mathcal{C}, \otimes, 1)$ is equipped with a braiding $\tau^{\mathcal{C}}$ (def. ), then $( [\mathcal{C}, Top^{\ast/}_{cg}], \otimes_{Day}, y(1))$ becomes itself a braided monoidal category with braiding given by
$\array{ (X \otimes_{Day} Y)(c) & = & \overset{c_1,c_2}{\int} \mathcal{C}(c_1 \otimes c_2) \wedge X(c_1) \wedge Y(c_2) \\ {}^{\mathllap{\tau}_{X,Y}(c)}\downarrow && \downarrow^{\mathrlap{\overset{c_1,c_2}{\int} \mathcal{C}(\tau^{\mathcal{C}}_{c_1,c_2}, c ) \wedge \tau^{Top^{\ast/}}_{X(c(1)), X(c_2)} }} \\ (Y \otimes_{Day} X)(c) & = & \overset{c_1,c_2}{\int} \mathcal{C}(c_2 \otimes c_1) \wedge Y(c_2) \wedge X(c_1) } \,.$
###### Proof
Regarding associativity, observe that
\begin{aligned} (X \otimes_{Day} ( Y \otimes_{Day} Z ))(c) & \simeq \overset{(c_1,c_2)}{\int} \mathcal{C}(c_1 \otimes_{\mathcal{D}} c_2, \,c) \wedge X(c_1) \wedge \overset{(d_1,d_2)}{\int} \mathcal{C}(d_1 \otimes_{\mathcal{C}} d_2, c_2 ) (Y(d_2) \wedge Z(d_2)) \\ &\simeq \overset{c_1, d_1, d_2}{\int} \underset{\simeq \mathcal{C}(c_1 \otimes_{\mathcal{C}} d_1 \otimes_{\mathcal{C}} d_2, c )}{ \underbrace{ \overset{c_2}{\int} \mathcal{C}(c_1 \otimes_{\mathcal{D}} c_2 , c) \wedge \mathcal{C}(d_1 \otimes_{\mathcal{C}}d_2, c_2 ) } } \wedge X(c_1) \wedge (Y(d_1) \wedge Z(d_2)) \\ &\simeq \overset{c_1, d_1, d_2}{\int} \mathcal{C}(c_1\otimes_{\mathcal{C}} d_1 \otimes_{\mathcal{C}} d_2, c ) \wedge X(c_1) \wedge (Y(d_1) \wedge Z(d_2)) \end{aligned} \,,
where we used the Fubini theorem for coends (prop. ) and then twice the co-Yoneda lemma (prop. ). An analogous formula follows for $X \otimes_{Day} (Y \otimes_{Day} Z)))(c)$, and so associativity follows via prop. from the associativity of the smash product and of the tensor product $\otimes_{\mathcal{C}}$.
Similarly, if $\mathcal{C}$ is braided then the hexagon identity for the braiding follows, under the coend, from the hexagon identities for the braidings in $\mathcal{C}$ and $Top^{\ast/}_{cg}$.
To see that $y(1)$ is the tensor unit for $\otimes_{Day}$, use the Fubini theorem for coends (prop. ) and then twice the co-Yoneda lemma (prop. ) to get for any $X \in [\mathcal{C},Top^{\ast/}_{cg}]$ that
\begin{aligned} X \otimes_{Day} y(1) & = \overset{c_1,c_2 \in \mathcal{C}}{\int} \mathcal{C}(c_1\otimes_{\mathcal{D}} c_2,-) \wedge X(c_1) \wedge \mathcal{C}(1,c_2) \\ & \simeq \overset{c_1\in \mathcal{C}}{\int} X(c_1) \wedge \overset{c_2 \in \mathcal{C}}{\int} \mathcal{C}(c_1\otimes_{\mathcal{C}} c_2,-) \wedge \mathcal{C}(1,c_2) \\ & \simeq \overset{c_1\in \mathcal{C}}{\int} X(c_1) \wedge \mathcal{C}(c_1 \otimes_{\mathcal{C}} 1, -) \\ & \simeq \overset{c_1\in \mathcal{C}}{\int} X(c_1) \wedge \mathcal{C}(c_1, -) \\ & \simeq X(-) \\ & \simeq X \end{aligned} \,.
###### Proposition
For $\mathcal{C}$ a small pointed topological monoidal category (def. ) with tensor product denoted $\otimes_{\mathcal{C}} \;\colon\; \mathcal{C} \times\mathcal{C} \to \mathcal{C}$, the monoidal category with Day convolution $([\mathcal{C},Top^{\ast/}_{cg}], \otimes_{Day}, y(1))$ from def. is a closed monoidal category (def. ). Its internal hom $[-,-]_{Day}$ is given by the end (def. )
$[X,Y]_{Day}(c) \simeq \underset{c_1,c_2}{\int} Maps\left( \mathcal{C}(c \otimes_{\mathcal{C}} c_1,c_2), \; Maps(X(c_1) , Y(c_2))_\ast \right)_\ast \,.$
###### Proof
Using the Fubini theorem (def. ) and the co-Yoneda lemma (def. ) and in view of definition of the enriched functor category, there is the following sequence of natural isomorphisms:
\begin{aligned} [\mathcal{C},V]( X, [Y,Z]_{Day} ) & \simeq \underset{c}{\int} Maps\left( X(c), \underset{c_1,c_2}{\int} Maps\left( \mathcal{C}(c \otimes_{\mathcal{C}} c_1 , c_2), Maps(Y(c_1), Z(c_2))_\ast \right)_\ast \right)_\ast \\ & \simeq \underset{c}{\int} \underset{c_1,c_2}{\int} Maps\left( \mathcal{C}(c \otimes_{\mathcal{C}} c_1, c_2) \wedge X(c) \wedge Y(c_1) ,\; Z(c_2) \right)_\ast \\ & \simeq \underset{c_2}{\int} Maps\left( \overset{c,c_1}{\int} \mathcal{C}(c \otimes_{\mathcal{C}} c_1, c_2) \wedge X(c) \wedge Y(c_1) ,\; Z(c_2) \right)_\ast \\ &\simeq \underset{c_2}{\int} Maps\left( (X \otimes_{Day} Y)(c_2), Z(c_2) \right)_\ast \\ &\simeq [\mathcal{C},V](X \otimes_{Day} Y, Z) \end{aligned} \,.
###### Proposition
In the situation of def. , the Yoneda embedding $c\mapsto \mathcal{C}(c,-)$ constitutes a strong monoidal functor
$(\mathcal{C},\otimes_{\mathcal{C}}, I) \hookrightarrow ([\mathcal{C},V], \otimes_{Day}, y(I)) \,.$
###### Proof
That the tensor unit is respected is part of prop. . To see that the tensor product is respected, apply the co-Yoneda lemma (prop ) twice to get the following natural isomorphism
\begin{aligned} (y(c_1) \otimes_{Day} y(c_2))(c) & \simeq \overset{d_1, d_2}{\int} \mathcal{C}(d_1 \otimes_{\mathcal{C}} d_2, c ) \wedge \mathcal{C}(c_1,d_1) \wedge \mathcal{C}(c_2,d_2) \\ & \simeq \mathcal{C}(c_1\otimes_{\mathcal{C}}c_2 , c ) \\ & = y(c_1 \otimes_{\mathcal{C}} c_2 )(c) \end{aligned} \,.
##### Functors with smash product
###### Definition
Let $(\mathcal{C},\otimes_{\mathcal{C}}, 1_{\mathcal{C}})$ and $(\mathcal{D},\otimes_{\mathcal{D}}, 1_{\mathcal{D}} )$ be two (pointed) topologically enriched monoidal categories (def. ). A topologically enriched lax monoidal functor between them is
1. $F \;\colon\; \mathcal{C} \longrightarrow \mathcal{D} \,,$
2. a morphism
$\epsilon \;\colon\; 1_{\mathcal{D}} \longrightarrow F(1_{\mathcal{C}})$
3. $\mu_{x,y} \;\colon\; F(x) \otimes_{\mathcal{D}} F(y) \longrightarrow F(x \otimes_{\mathcal{C}} y)$
for all $x,y \in \mathcal{C}$
satisfying the following conditions:
1. (associativity) For all objects $x,y,z \in \mathcal{C}$ the following diagram commutes
$\array{ (F(x) \otimes_{\mathcal{D}} F(y)) \otimes_{\mathcal{D}} F(z) &\underoverset{\simeq}{a^{\mathcal{D}}_{F(x),F(y),F(z)}}{\longrightarrow}& F(x) \otimes_{\mathcal{D}}( F(y)\otimes_{\mathcal{D}} F(z) ) \\ {}^{\mathllap{\mu_{x,y} \otimes id}}\downarrow && \downarrow^{\mathrlap{id\otimes \mu_{y,z}}} \\ F(x \otimes_{\mathcal{C}} y) \otimes_{\mathcal{D}} F(z) && F(x) \otimes_{\mathcal{D}} ( F(x \otimes_{\mathcal{C}} y) ) \\ {}^{\mathllap{\mu_{x \otimes_{\mathcal{C}} y , z} } }\downarrow && \downarrow^{\mathrlap{\mu_{ x, y \otimes_{\mathcal{C}} z }}} \\ F( ( x \otimes_{\mathcal{C}} y ) \otimes_{\mathcal{C}} z ) &\underset{F(a^{\mathcal{C}}_{x,y,z})}{\longrightarrow}& F( x \otimes_{\mathcal{C}} ( y \otimes_{\mathcal{C}} z ) ) } \,,$
where $a^{\mathcal{C}}$ and $a^{\mathcal{D}}$ denote the associators of the monoidal categories;
2. (unitality) For all $x \in \mathcal{C}$ the following diagrams commutes
$\array{ 1_{\mathcal{D}} \otimes_{\mathcal{D}} F(x) &\overset{\epsilon \otimes id}{\longrightarrow}& F(1_{\mathcal{C}}) \otimes_{\mathcal{D}} F(x) \\ {}^{\mathllap{\ell^{\mathcal{D}}_{F(x)}}}\downarrow && \downarrow^{\mathrlap{\mu_{1_{\mathcal{C}}, x }}} \\ F(x) &\overset{F(\ell^{\mathcal{C}}_x )}{\longleftarrow}& F(1 \otimes_{\mathcal{C}} x ) }$
and
$\array{ F(x) \otimes_{\mathcal{D}} 1_{\mathcal{D}} &\overset{id \otimes \epsilon }{\longrightarrow}& F(x) \otimes_{\mathcal{D}} F(1_{\mathcal{C}}) \\ {}^{\mathllap{r^{\mathcal{D}}_{F(x)}}}\downarrow && \downarrow^{\mathrlap{\mu_{x, 1_{\mathcal{C}} }}} \\ F(x) &\overset{F(r^{\mathcal{C}}_x )}{\longleftarrow}& F(x \otimes_{\mathcal{C}} 1 ) } \,,$
where $\ell^{\mathcal{C}}$, $\ell^{\mathcal{D}}$, $r^{\mathcal{C}}$, $r^{\mathcal{D}}$ denote the left and right unitors of the two monoidal categories, respectively.
If $\epsilon$ and alll $\mu_{x,y}$ are isomorphisms, then $F$ is called a strong monoidal functor.
If moreover $(\mathcal{C},\otimes_{\mathcal{C}}, 1_{\mathcal{C}})$ and $(\mathcal{D},\otimes_{\mathcal{D}}, 1_{\mathcal{D}} )$ are equipped with the structure of braided monoidal categories (def. ), then the lax monoidal functor $F$ is called a braided monoidal functor if in addition the following diagram commutes for all objects $x,y \in \mathcal{C}$
$\array{ F(x) \otimes_{\mathcal{C}} F(y) &\overset{\tau^{\mathcal{D}}_{F(x), F(y)}}{\longrightarrow}& F(y) \otimes_{\mathcal{D}} F(x) \\ {}^{\mathllap{\mu_{x,y}}}\downarrow && \downarrow^{\mathrlap{\mu_{y,x}}} \\ F(x \otimes_{\mathcal{C}} y ) &\underset{F(\tau^{\mathcal{C}}_{x,y} )}{\longrightarrow}& F( y \otimes_{\mathcal{C}} x ) } \,.$
A homomorphism $f\;\colon\; (F_1,\mu_1, \epsilon_1) \longrightarrow (F_2, \mu_2, \epsilon_2)$ between two (braided) lax monoidal functors is a monoidal natural transformation, in that it is
• a natural transformation $f_x \;\colon\; F_1(x) \longrightarrow F_2(x)$ of the underlying functors
compatible with the product and the unit in that the following diagrams commute for all objects $x,y \in \mathcal{C}$:
$\array{ F_1(x) \otimes_{\mathcal{D}} F_1(y) &\overset{f(x)\otimes_{\mathcal{D}} f(y)}{\longrightarrow}& F_2(x) \otimes_{\mathcal{D}} F_2(y) \\ {}^{\mathllap{(\mu_1)_{x,y}}}\downarrow && \downarrow^{\mathrlap{(\mu_2)_{x,y}}} \\ F_1(x\otimes_{\mathcal{C}} y) &\underset{f(x \otimes_{\mathcal{C}} y ) }{\longrightarrow}& F_2(x \otimes_{\mathcal{C}} y) }$
and
$\array{ && 1_{\mathcal{D}} \\ & {}^{\mathllap{\epsilon_1}}\swarrow && \searrow^{\mathrlap{\epsilon_2}} \\ F_1(1_{\mathcal{C}}) &&\underset{f(1_{\mathcal{C}})}{\longrightarrow}&& F_2(1_{\mathcal{C}}) } \,.$
We write $MonFun(\mathcal{C},\mathcal{D})$ for the resulting category of lax monoidal functors between monoidal categories $\mathcal{C}$ and $\mathcal{D}$, similarly $BraidMonFun(\mathcal{C},\mathcal{D})$ for the category of braided monoidal functors between braided monoidal categories, and $SymMonFun(\mathcal{C},\mathcal{D})$ for the category of braided monoidal functors between symmetric monoidal categories.
###### Remark
In the literature the term “monoidal functor” often refers by default to what in def. is called a strong monoidal functor. But for the purpose of the discussion of functors with smash product below, it is crucial to admit the generality of lax monoidal functors.
If $(\mathcal{C},\otimes_{\mathcal{C}}, 1_{\mathcal{C}})$ and $(\mathcal{D},\otimes_{\mathcal{D}}, 1_{\mathcal{D}} )$ are symmetric monoidal categories (def. ) then a braided monoidal functor (def. ) between them is often called a symmetric monoidal functor.
###### Definition
Let $(\mathcal{C},\otimes_{\mathcal{C}}, 1_{\mathcal{C}})$ and $(\mathcal{D},\otimes_{\mathcal{D}}, 1_{\mathcal{D}} )$ be two (pointed) topologically enriched monoidal categories (def. ), and let $F \;\colon\; \mathcal{C} \longrightarrow \mathcal{D}$ be a topologically enriched lax monoidal functor between them, with product operation $\mu$.
Then a left module over the lax monoidal functor is
1. $G \;\colon\; \mathcal{C} \longrightarrow \mathcal{D} \,;$
2. $\rho_{x,y} \;\colon\; F(x) \otimes_{\mathcal{D}} N(y) \longrightarrow N(x \otimes_{\mathcal{C}} y )$
such that
• (action property) For all objects $x,y,z \in \mathcal{C}$ the following diagram commutes
$\array{ (F(x) \otimes_{\mathcal{D}} F(y)) \otimes_{\mathcal{D}} G(z) &\underoverset{\simeq}{a^{\mathcal{D}}_{F(x),F(y),F(z)}}{\longrightarrow}& F(x) \otimes_{\mathcal{D}}( F(y)\otimes_{\mathcal{D}} G(z) ) \\ {}^{\mathllap{\mu_{x,y} \otimes id}}\downarrow && \downarrow^{\mathrlap{id\otimes \rho_{y,z}}} \\ F(x \otimes_{\mathcal{C}} y) \otimes_{\mathcal{D}} G(z) && F(x) \otimes_{\mathcal{D}} ( G(x \otimes_{\mathcal{C}} y) ) \\ {}^{\mathllap{\rho_{x \otimes_{\mathcal{C}} y , z} } }\downarrow && \downarrow^{\mathrlap{\rho_{ x, y \otimes_{\mathcal{C}} z }}} \\ G( ( x \otimes_{\mathcal{C}} y ) \otimes_{\mathcal{C}} z ) &\underset{F(a^{\mathcal{C}}_{x,y,z})}{\longrightarrow}& G( x \otimes_{\mathcal{C}} ( y \otimes_{\mathcal{C}} z ) ) } \,,$
A homomorphism $f\;\colon\; (G_1, \rho_1) \longrightarrow (G_2,\rho_2)$ between two modules over a monoidal functor $(F,\mu,\epsilon)$ is
• a natural transformation $f_x \;\colon\; N_1(x) \longrightarrow N_2(x)$ of the underlying functors
compatible with the action in that the following diagram commute for all objects $x,y \in \mathcal{C}$:
$\array{ F(x) \otimes_{\mathcal{D}} N_1(y) &\overset{id \otimes_{\mathcal{D}} f(y)}{\longrightarrow}& F(x) \otimes_{\mathcal{D}} N_2(y) \\ {}^{\mathllap{(\rho_1)_{x,y}}}\downarrow && \downarrow^{\mathrlap{(\rhi_2)_{x,y}}} \\ N_1(x\otimes_{\mathcal{C}} y) &\underset{f(x \otimes_{\mathcal{C}} y ) }{\longrightarrow}& N_2(x \otimes_{\mathcal{C}} y) }$
We write $F Mod$ for the resulting category of modules over the monoidal functor $F$.
###### Proposition
Let $(\mathcal{C},\otimes I)$ be a pointed topologically enriched category (symmetric monoidal category) monoidal category (def. ). Regard $(Top_{cg}^{\ast/}, \wedge , S^0)$ as a topological symmetric monoidal category as in example .
Then (commutative) monoids in (def. ) the Day convolution monoidal category $([\mathcal{C}, Top^{\ast/}_{cg}], \otimes_{Day}, y(1_{\mathcal{C}}))$ of prop. are equivalent to (braided) lax monoidal functors (def. ) of the form
$(\mathcal{C},\otimes, I) \longrightarrow (Top^{\ast}_{cg}, \wedge, S^0) \,,$
called functors with smash products on $\mathcal{C}$, i.e. there are equivalences of categories of the form
\begin{aligned} Mon([\mathcal{C},Top^{\ast/}_{cg}], \otimes_{Day}, y(1_{\mathcal{C}})) &\underoverset{\simeq}{\phi}{\longrightarrow} MonFunc(\mathcal{C},Top^{\ast/}_{cg}) \\ CMon([\mathcal{C},Top^{\ast/}_{cg}], \otimes_{Day}, y(1_{\mathcal{C}})) &\simeq SymMonFunc(\mathcal{C},Top^{\ast/}_{cg}) \end{aligned} \,.
Furthermore, for $A \in Mon([\mathcal{C},Top^{\ast/}_{cg}], \otimes_{Day}, y(1_{\mathcal{C}}))$ a given monoid object, then left $A$-module objects (def. ) are equivalent to left modules over monoidal functors (def. ):
$A Mod \simeq \phi(A) Mod \,.$
This is stated in some form in (Day 70, example 3.2.2). It is highlighted again in (MMSS 00, prop. 22.1).
###### Proof
By definition , a lax monoidal functor $F \colon \mathcal{C} \to Top^{\ast/}_{cg}$ is a topologically enriched functor equipped with a morphism of pointed topological spaces of the form
$S^0 \longrightarrow F(1_{\mathcal{C}})$
and equipped with a natural system of maps of pointed topological spaces of the form
$F(c_1) \wedge F(c_2) \longrightarrow F(c_1 \otimes_{\mathcal{C}} c_2)$
for all $c_1,c_2 \in \mathcal{C}$.
Under the Yoneda lemma (prop. ) the first of these is equivalently a morphism in $[\mathcal{C}, Top^{\ast/}_{cg}]$ of the form
$y(S^0) \longrightarrow F \,.$
Moreover, under the natural isomorphism of corollary the second of these is equivalently a morphism in $[\mathcal{C}, Top^{\ast/}_{cg}]$ of the form
$F \otimes_{Day} F \longrightarrow F \,.$
Translating the conditions of def. satisfied by a lax monoidal functor through these identifications gives precisely the conditions of def. on a (commutative) monoid in object $F$ under $\otimes_{Day}$.
Similarly for module objects and modules over monoidal functors.
###### Proposition
Let $f \;\colon\; \mathcal{C} \longrightarrow \mathcal{D}$ be a lax monoidal functor (def. ) between pointed topologically enriched monoidal categories (def. ). Then the induced functor
$f^\ast \;\colon\; [\mathcal{D}, Top^{\ast/}_{cg}] \longrightarrow [\mathcal{C}, Top_{cg}^{\ast}]$
given by $(f^\ast X)(c)\coloneqq X(f(c))$ preserves monoids under Day convolution
$f^\ast \;\colon\; Mon([\mathcal{D}, Top^{\ast/}_{cg}], \otimes_{Day}, y(1_{\mathcal{D}})) \longrightarrow Mon([\mathcal{C}, Top_{cg}^{\ast}], \otimes_{Day}, y(1_{\mathcal{C}})$
Moreover, if $\mathcal{C}$ and $\mathcal{D}$ are symmetric monoidal categories (def. ) and $f$ is a braided monoidal functor (def. ), then $f^\ast$ also preserves commutative monoids
$f^\ast \;\colon\; CMon([\mathcal{D}, Top^{\ast/}_{cg}], \otimes_{Day}, y(1_{\mathcal{D}})) \longrightarrow CMon([\mathcal{C}, Top_{cg}^{\ast}], \otimes_{Day}, y(1_{\mathcal{C}}) \,.$
Similarly, for
$A \in Mon([\mathcal{D}, Top^{\ast/}_{cg}], \otimes_{Day}, y(1_{\mathcal{D}}))$
any fixed monoid, then $f^\ast$ sends $A$-module object to $f^\ast(A)$-modules
$f^\ast \;\colon\; A Mod(\mathcal{D}) \longrightarrow (f^\ast A)Mod(\mathcal{C}) \,.$
###### Proof
This is an immediate corollary of prop. , since the composite of two (braided) lax monoidal functors is itself canonically a (braided) lax monoidal functor.
###### Proposition
Let $(\mathcal{C},\otimes_{\mathcal{C}}, 1_{\mathcal{C}})$ be a topologically enriched monoidal category (def. ), and let $A \in Mon([\mathcal{C},Top^{\ast/}_{cg}],\otimes_{Day}, y(1_{\mathcal{C}}))$ be a monoid in (def. ) the pointed topological Day convolution monoidal category over $\mathcal{C}$ from prop. .
Then the category of left A-modules (def. ) is a pointed topologically enriched functor category category (exmpl.)
$A Mod \;\simeq\; [ A Free_{\mathcal{C}}Mod^{op}, \; Top_{cg}^{\ast/} ] \,,$
over the category of free modules over $A$ (def. ) on objects in $\mathcal{C}$ (under the Yoneda embedding $y \colon \mathcal{C}^{op} \to [\mathcal{C}, Top^{\ast/}_{cg}]$). Hence the objects of $A Free_{\mathcal{C}}Mod$ are identified with those of $\mathcal{C}$, and its hom-spaces are
$A Free_{\mathcal{C}}Mod( c_1, c_2) \;=\; A Mod( A \otimes_{Day} y(c_1),\; A \otimes_{Day} y(c_2) ) \,.$
###### Proof idea
Use the identification from prop. of $A$ with a lax monoidal functor and of any $A$-module object $N$ as a functor with the structure of a module over a monoidal functor, given by natural transformations
$A(c_1)\otimes N(c_2) \overset{\rho_{c_1,c_2}}{\longrightarrow} N(c_1 \otimes c_2) \,.$
Notice that these transformations have just the same structure as those of the enriched functoriality of $N$ (def.) of the form
$\mathcal{C}(c_1,c_2) \otimes N(c_1) \overset{}{\longrightarrow} N(c_2) \,.$
Hence we may unify these two kinds of transformations into a single kind of the form
$\mathcal{C}(c_3 \otimes c_1 , c_2) \otimes A(c_3) \otimes N(c_1) \overset{id \otimes \rho_{c_3,c_1}}{\longrightarrow} \mathcal{C}(c_3 \otimes c_1, c_2) \otimes N(c_3 \otimes c_2) \longrightarrow N(c_2)$
and subject to certain identifications.
Now observe that the hom-objects of $A Free_{\mathcal{C}}Mod$ have just this structure:
\begin{aligned} A Free_{\mathcal{C}}Mod(c_2,c_1) & = A Mod( A \otimes_{Day} y(c_2) , A \otimes_{Day} y(c_1) ) \\ & \simeq [\mathcal{C},Top^{\ast/}_{cg}](y(c_2), A \otimes_{Day} y(c_1) ) \\ & \simeq (A \otimes_{Day} y(c_1) )(c_2) \\ & \simeq \overset{c_3,c_4}{\int} \mathcal{C}(c_3 \otimes c_4,c_2) \wedge A(c_3) \wedge \mathcal{C}(c_1, c_4) \\ & \simeq \overset{c_3}{\int} \mathcal{C}(c_3 \otimes c_1, c_2) \wedge A (c_3) \end{aligned} \,.
Here we used first the free-forgetful adjunction of prop. , then the enriched Yoneda lemma (prop. ), then the coend-expression for Day convolution (def. ) and finally the co-Yoneda lemma (prop. ).
We claim that under this identification, composition in $A Free_{\mathcal{C}}Mod$ is given by the following composite.
\begin{aligned} A Free_{\mathcal{C}}Mod(c_3, c_2) \wedge A Free_{\mathcal{C}}Mod(c_2, c_1) & = \left( \overset{c_5}{\int} \mathcal{C}(c_5 \otimes_{\mathcal{C}} c_2 , c_3 ) \wedge A(c_5) \right) \wedge \left( \overset{c_4}{\int} \mathcal{C}(c_4 \otimes_{\mathcal{C}} c_1, c_2) \wedge A(c_4) \right) \\ & \simeq \overset{c_4, c_5}{\int} \mathcal{C}(c_5 \otimes_{\mathcal{C}} c_2 , c_3) \wedge \mathcal{C}(c_4 \otimes_{\mathcal{C}} c_1, c_2 ) \wedge A(c_5) \wedge A(c_4) \\ & \longrightarrow \overset{c_4,c_5}{\int} \mathcal{C}(c_5 \otimes_{\mathcal{C}} c_2 , c_3) \wedge \mathcal{C}(c_5 \otimes_{\mathcal{C}} c_4 \otimes_{\mathcal{C}} c_1, c_5 \otimes_{\mathcal{C}} c_2 ) \wedge A(c_5 \otimes_{\mathcal{C}} c_4 ) \\ & \longrightarrow \overset{c_4, c_5}{\int} \mathcal{C}(c_5 \otimes_{\mathcal{C}} c_4 \otimes_{\mathcal{C}} c_1, c_5 \otimes_{\mathcal{C}} c_2 ) \wedge A(c_5 \otimes_{\mathcal{C}} c_4 ) \\ & \longrightarrow \overset{c_4}{\int} \mathcal{C}(c_4 \otimes_{\mathcal{C}} c_1 , c_3) \otimes_V A(c_4 ) \end{aligned} \,,
where
1. the equivalence is braiding in the integrand (and the Fubini theorem, prop. );
2. the first morphism is, in the integrand, the smash product of
1. forming the tensor product of hom-objects of $\mathcal{C}$ with the identity morphism on $c_5$;
2. the monoidal functor incarnation $A(c_5) \wedge A(c_4)\longrightarrow A(c_5 \otimes_{\mathcal{C}} c_4 )$ of the monoid structure on $A$;
3. the second morphism is, in the integrand, given by composition in $\mathcal{C}$;
4. the last morphism is the morphism induced on coends by regarding extranaturality in $c_4$ and $c_5$ separately as a special case of extranaturality in $c_6 \coloneqq c_4 \otimes c_5$ (and then renaming).
It is fairly straightforward to see that, under the above identifications, functoriality under this composition is equivalently functoriality in $\mathcal{C}$ together with the action property over $A$.
#### Pre-Excisive functors
###### Definition
Write
$\iota_{fin}\;\colon\; Top^{\ast/}_{cg,fin} \hookrightarrow Top^{\ast/}_{cg}$
for the full subcategory of pointed compactly generated topological spaces (def.) on those that admit the structure of a finite CW-complex (a CW-complex (def.) with a finite number of cells).
We say that the pointed topological enriched functor category (def. )
$Exc(Top_{cg}) \coloneqq [Top^{\ast/}_{cg,fin}, Top^{\ast/}_{cg}]$
is the category of pre-excisive functors.
Write
$\mathbb{S}_{exc} \coloneqq y(S^0) \coloneqq Top^{\ast/}_{cg,fin}(S^0,-)$
for the functor co-represented by 0-sphere. This is equivalently the inclusion $\iota_{fin}$ itself:
$\mathbb{S}_{exc} = \iota_{fin} \;\colon\; K \mapsto K \,.$
We call this the standard incarnation of the sphere spectrum as a pre-excisive functor.
By prop. the smash product of pointed compactly generated topological spaces induces the structure of a closed (def. ) symmetric monoidal category (def. )
$\left( Exc(Top_{cg}) ,\; \wedge_{Day} ,\; \mathbb{S}_{exc} \right)$
with
1. tensor unit the sphere spectrum $\mathbb{S}_{exc}$;
2. tensor product the Day convolution product $\otimes_{Day}$ from def. ,
called the symmetric monoidal smash product of spectra for the model of pre-excisive functors;
3. internal hom the dual operation $[-,-]_{Day}$ from prop. ,
called the mapping spectrum construction for pre-excisive functors.
###### Remark
By example the sphere spectrum incarnated as a pre-excisive functor $\mathbb{S}_{exc}$ (according to def. ) is canonically a commutative monoid in the category of pre-excisive functors (def. )
Moreover, by example , every object of $Exc(Top_{cg})$ (def. ) is canonically a module object over $\mathbb{S}_{exc}$. We may therefore tautologically identify the category of pre-excisive functors with the module category over the sphere spectrum:
$Exc(Top_{cg}) \simeq \mathbb{S}_{exc}Mod \,.$
We now consider restricting the domain of the pre-excisive functors of def. .
###### Definition
Define the following pointed topologically enriched (def.) symmetric monoidal categories (def. ):
1. $Seq$ is the category whose objects are the natural numbers and which has only identity morphisms and zero morphisms on these objects, hence the hom-spaces are
$Seq(n_1,n_2) = \left\{ \array{ S^0 & for\; n_1 = n_2 \\ \ast & otherwise } \right.$
The tensor product is the addition of natural numbers, $\otimes = +$, and the tensor unit is 0. The braiding is, necessarily, the identity.
2. $Sym$ is the standard skeleton of the core of FinSet with zero morphisms adjoined: its objects are the finite sets $\overline{n} \coloneqq \{1, \cdots,n\}$ for $n \in \mathbb{N}$, all non-zero morphisms are automorphisms and the automorphism group of $\{1,\cdots,n\}$ is the symmetric group $\Sigma_n$, hence the hom-spaces are the following discrete topological spaces:
$Sym(n_1, n_2) = \left\{ \array{ (\Sigma_{n_1})_+ & for \; n_1 = n_2 \\ \ast & otherwise } \right.$
The tensor product is the disjoint union of sets, tensor unit is the empty set. The braiding
$\tau_{n_1 , n_2} \;\colon\; \overline{n_1} \cup \overline{n_2} \overset{}{\longrightarrow} \overline{n_2} \cup \overline{n_1}$
is given by the canonical permutation in $\Sigma_{n_1+n_2}$ that shuffles the first $n_1$ elements past the remaining $n_2$ elements.
3. $Orth$ has as objects finite dimenional real linear inner product spaces $(V, \langle -,-\rangle)$ and as non-zero morphisms the linear isometric isomorphisms between these; hence the automorphism group of the object $(V, \langle -,-\rangle)$ is the orthogonal group $O(V)$; the monoidal product is direct sum of linear spaces, the tensor unit is the 0-vector space; again we turn this into a $Top^{\ast/}$-enriched category by adjoining a basepoint to the hom-spaces;
$Orth(V_1,V_2) \simeq \left\{ \array{ O(V_1)_+ & for \; dim(V_1) = dim(V_2) \\ \ast & otherwise } \right.$
The tensor product is the direct sum of linear inner product spaces, tensor unit is the 0-vector space. The braiding is that of $Sym$, under the canonical embedding $\Sigma_{n_1+n_2} \hookrightarrow O(n_1+n_2)$ of the symmetric group into the orthogonal group.
There is a sequence of canonical faithful pointed topological subcategory inclusions
$\array{ Seq &\stackrel{seq}{\hookrightarrow}& Sym &\stackrel{sym}{\hookrightarrow}& Orth &\stackrel{orth}{\hookrightarrow}& Top_{cg,fin}^{\ast/} \\ n &\mapsto& \{1,\cdots, n\} &\mapsto& \mathbb{R}^n &\mapsto& S^n \\ && && V &\mapsto& S^V } \,,$
into the pointed topological category of pointed compactly generated topological spaces of finite CW-type (def. ).
Here $S^V$ denotes the one-point compactification of $V$. On morphisms $sym \colon (\Sigma_n)_+ \hookrightarrow (O(n))_+$ is the canonical inclusion of permutation matrices into orthogonal matrices and $orth \colon O(V)_+ \hookrightarrow Aut(S^V)$ is on $O(V)$ the topological subspace inclusions of the pointed homeomorphisms $S^V \to S^V$ that are induced under forming one-point compactification from linear isometries of $V$ (“representation spheres”).
Consider the sequence of restrictions of topological diagram categories, according to prop. along the above inclusions:
$Exc(Top_{cg}) \overset{orth^\ast}{\longrightarrow} [Orth,Top^{\ast/}_{cg}] \overset{sym^\ast}{\longrightarrow} [Sym,Top^{\ast/}_{cg}] \overset{seq^\ast}{\longrightarrow} [Seq,Top^{\ast/}_{cg}] \,.$
Write
$\mathbb{S}_{orth} \coloneqq orth^\ast \mathbb{S}_{exc} \,, \;\;\;\;\;\;\;\; \mathbb{S}_{sym} \coloneqq sym^\ast \mathbb{S}_{orth} \,, \;\;\;\;\;\;\;\; \mathbb{S}_{seq} \coloneqq seq^\ast \mathbb{S}_{sym}$
for the restriction of the excisive functor incarnation of the sphere spectrum (from def. ) along these inclusions.
###### Remark
Since $\mathbb{S}_{exc}$ is the tensor unit with repect to the Day convolution product on pre-excisive functors, and since it is therefore canonically a commutative monoid, by example , prop. says that all these restricted sphere spectra are still monoids, and that under restriction every pre-excisive functor, regarded as a $\mathbb{S}_{exc}$-module via remark , canonically becomes a module under the restricted sphere spectrum:
\begin{aligned} orth^\ast & \colon\; Exc(Top_{cg}) \simeq \mathbb{S}_{exc} Mod \longrightarrow \mathbb{S}_{orth} Mod \\ sym^\ast &\colon\; Exc(Top_{cg}) \simeq \mathbb{S}_{exc} Mod \longrightarrow \mathbb{S}_{sym} Mod \\ seq^\ast &\colon\; Exc(Top_{cg}) \simeq \mathbb{S}_{exc} Mod \longrightarrow \mathbb{S}_{seq} Mod \end{aligned} \,.
However, while $orth$ and $sym$ are braided monoidal functors, the functor $seq$ is not braided, hence $\mathbb{S}_{orth}$ and $\mathbb{S}_{sym}$ are commutative monoids, but $\mathbb{S}_{Seq}$ is not commutative. Hence prop. gives the following situation
sphere spectrum$\mathbb{S}_{exc}$$\mathbb{S}_{orth}$$\mathbb{S}_{sym}$$\mathbb{S}_{seq}$
monoidyesyesyesyes
commutative monoidyesyesyesno
tensor unityesnonono
###### Proposition
There is an equivalence of categories
$(-)^{seq} \;\colon\; \mathbb{S}_{seq} Mod \overset{}{\longrightarrow} SeqSpec(Top_{cg})$
which identifies the category of modules (def. ) over the monoid $\mathbb{S}_{seq}$ (remark ) in the Day convolution monoidal structure (prop. ) over the topological functor category $[Seq,Top^{\ast/}_{cg}]$ from def. with the category of sequential spectra (def.)
Under this equivalence, an $\mathbb{S}_{seq}$-module $X$ is taken to the sequential pre-spectrum $X^{seq}$ whose component spaces are the values of the pre-excisive functor $X$ on the standard n-sphere $S^n = (S^1)^{\wedge n}$
$(X^{seq})_n \coloneqq X(seq(n)) = X(S^n)$
and whose structure maps are the images of the action morphisms
$\mathbb{S}_{seq} \otimes_{Day} X \longrightarrow X$
under the isomorphism of corollary
$\mathbb{S}_{seq}(n_1) \wedge X(n_1) \longrightarrow X_{n_1 + n_2}$
evaluated at $n_1 = 1$
$\array{ \mathbb{S}_{seq}(1) \wedge X(n) &\longrightarrow& X_{n+1} \\ {}^{\mathllap{\simeq}}\downarrow && \downarrow^{\mathrlap{\simeq}} \\ S^1 \wedge X_n &\longrightarrow& X_{n+1} } \,.$
###### Proof
After unwinding the definitions, the only point to observe is that due to the action property,
$\array{ \mathbb{S}_{seq} \otimes_{Day} \mathbb{S}_{seq} \otimes_{Day} X &\overset{id \otimes_{Day} \rho}{\longrightarrow}& \mathbb{S}_{seq} \otimes_{Day} X \\ {}^{\mathllap{\mu \otimes_{Day} id } }\downarrow && \downarrow^{\mathrlap{\rho}} \\ \mathbb{S}_{seq} \otimes_{Day} X &\underset{\rho}{\longrightarrow}& X }$
any $\mathbb{S}_{seq}$-action
$\rho \;\colon\; \mathbb{S}_{seq} \otimes_{Day} X \longrightarrow X$
is indeed uniquely fixed by the components of the form
$\mathbb{S}_{seq}(1) \wedge X(n) \longrightarrow X(n) \,.$
This is because under corollary the action property is identified with the componentwise property
$\array{ S^{n_1} \wedge S^{n_2} \wedge X_{n_3} &\overset{id \wedge \rho_{n_2,n_3}}{\longrightarrow}& S^{n_1} \wedge X_{n_2 + n_3} \\ {}^{\mathllap{\simeq}}\downarrow && \downarrow^{\mathrlap{\rho_{n_1,n_2+n_3}}} \\ S^{n_1 + n_2} \wedge X_{n_3} &\underset{\rho_{n_1+n_2,n_3}}{\longrightarrow}& X_{n_1 + n_2 + n_3} } \,,$
where the left vertical morphism is an isomorphism by the nature of $\mathbb{S}_{seq}$. Hence this fixes the components $\rho_{n',n}$ to be the $n'$-fold composition of the structure maps $\sigma_n \coloneqq \rho(1,n)$.
However, since, by remark , $\mathbb{S}_{seq}$ is not commutative, there is no tensor product induced on $SeqSpec(Top_{cg})$ under the identification in prop. . But since $\mathbb{S}_{orth}$ and $\mathbb{S}_{sym}$ are commutative monoids by remark , it makes sense to consider the following definition.
###### Definition
In the terminology of remark we say that
$OrthSpec(Top_{cg}) \coloneqq \mathbb{S}_{orth} Mod$
is the category of orthogonal spectra; and that
$SymSpec(Top_{cg}) \coloneqq \mathbb{S}_{sym} Mod$
is the category of symmetric spectra.
By remark and by prop. these categories canonically carry a symmetric monoidal tensor product $\otimes_{\mathbb{S}_{orth}}$ and $\otimes_{\mathbb{S}_{seq}}$, respectively. This we call the symmetric monoidal smash product of spectra. We usually just write for short
$\wedge \coloneqq \otimes_{\mathbb{S}_{orth}} \;\colon\; OrthSpec(Top_{cg}) \times OrthSpec(Top_{cg}) \longrightarrow OrthSpec(Top_{cg})$
and
$\wedge \coloneqq \otimes_{\mathbb{S}_{sym}} \;\colon\; SymSpec(Top_{cg}) \times SymSpec(Top_{cg}) \longrightarrow SymSpec(Top_{cg})$
In the next section we work out what these symmetric monoidal categories of orthogonal and of symmetric spectra look like more explicitly.
### For symmetric and orthogonal spectra
We now define symmetric spectra and orthogonal spectra and their symmetric monoidal smash product. We proceed by giving the explicit definitions and then checking that these are equivalent to the abstract definition from above.
$\,$
###### Definition
A topological symmetric spectrum $X$ is
1. a sequence $\{X_n \in Top_{cg}^{\ast/}\;\vert\; n \in \mathbb{N}\}$ of pointed compactly generated topological spaces;
2. a basepoint preserving continuous right action of the symmetric group $\Sigma(n)$ on $X_n$;
3. a sequence of morphisms $\sigma_n \colon S^1 \wedge X_n \longrightarrow X_{n+1}$
such that
• for all $n, k \in \mathbb{N}$ the composite
$S^{k} \wedge X_n \simeq S^{k-1} \wedge S^1 \wedge X_n \stackrel{id \wedge \sigma_n }{\longrightarrow} S^{k-1} \wedge X_{n+1} \simeq S^{k-2}\wedge S^1 \wedge X_{n+2} \stackrel{id \wedge \sigma_{n+1}}{\longrightarrow} \cdots \stackrel{\sigma_{n+k-1}}{\longrightarrow} X_{n+k}$
intertwines the $\Sigma(n) \times \Sigma(k)$-action.
A homomorphism of symmetric spectra $f\colon X \longrightarrow Y$ is
• a sequence of maps $f_n \colon X_n \longrightarrow Y_n$
such that
1. each $f_n$ intetwines the $\Sigma(n)$-action;
2. the following diagrams commute
$\array{ S^1 \wedge X_n &\stackrel{f_n \wedge id}{\longrightarrow}& S^1 \wedge Y_n \\ \downarrow^{\mathrlap{\sigma^X_n}} && \downarrow^{\mathrlap{\sigma^Y_n}} \\ X_{n+1} &\stackrel{f_{n+1}}{\longrightarrow}& Y_{n+1} } \,.$
We write $SymSpec(Top_{cg})$ for the resulting category of symmetric spectra.
The definition of orthogonal spectra has the same structure, just with the symmetric groups replaced by the orthogonal groups.
###### Definition
A topological orthogonal spectrum $X$ is
1. a sequence $\{X_n \in Top_{cg}^{\ast/}\;\vert\; n \in \mathbb{N}\}$ of pointed compactly generated topological spaces;
2. a basepoint preserving continuous right action of the orthogonal group $O(n)$ on $X_n$;
3. a sequence of morphisms $\sigma_n \colon S^1 \wedge X_n \longrightarrow X_{n+1}$
such that
• for all $n, k \in \mathbb{N}$ the composite
$S^{k} \wedge X_n \simeq S^{k-1} \wedge S^1 \wedge X_n \stackrel{id \wedge \sigma_n }{\longrightarrow} S^{k-1} \wedge X_{n+1} \simeq S^{k-2}\wedge S^1 \wedge X_{n+2} \stackrel{id \wedge \sigma_{n+1}}{\longrightarrow} \cdots \stackrel{\sigma_{n+k-1}}{\longrightarrow} X_{n+k}$
intertwines the $O(n) \times Ok()$-action.
A homomorphism of orthogonal spectra $f\colon X \longrightarrow Y$ is
• a sequence of maps $f_n \colon X_n \longrightarrow Y_n$
such that
1. each $f_n$ intetwines the $O(n)$-action;
2. the following diagrams commute
$\array{ S^1 \wedge X_n &\stackrel{f_n \wedge id}{\longrightarrow}& S^1 \wedge Y_n \\ \downarrow^{\mathrlap{\sigma^X_n}} && \downarrow^{\mathrlap{\sigma^Y_n}} \\ X_{n+1} &\stackrel{f_{n+1}}{\longrightarrow}& Y_{n+1} } \,.$
We write $OrthSpec(Top_{cg})$ for the resulting category of orthogonal spectra.
###### Proposition
Definitions and are indeed equivalent to def. :
orthogonal spectra are euqivalently the module objects over the incarnation $\mathbb{S}_{orth}$ of the sphere spectrum
$OrthSpec(Top_{cg}) \simeq \mathbb{S}_{orth} Mod$
and symmetric spectra sre equivalently the module objects over the incarnation $\mathbb{S}_{sym}$ of the sphere spectrum
$SymSpec(Top_{cg}) \simeq \mathbb{S}_{sym} Mod \,.$
###### Proof
We discuss this for symmetric spectra. The proof for orthogonal spectra is of the same form.
First of all, (by this example) an object in $[Sym, Top^{\ast/}_{cg}]$ is equivalently a “symmetric sequence”, namely a sequence of pointed topological spaces $X_k$, for $k \in \mathbb{N}$, equipped with an action of $\Sigma(k)$ (def. ).
By corollary and this lemma, the structure morphism of an $\mathbb{S}_{sym}$-module object on $X$
$\mathbb{S}_{sym} \otimes_{Day} X \longrightarrow X$
is equivalently (as a functor with smash products) a natural transformation
$S^{n_1} \wedge X_{n_2} \longrightarrow X_{n_1 + n_2}$
over $Sym \times Sym$. This means equivalently that there is such a morphism for all $n_1, n_2 \in \mathbb{N}$ and that it is $\Sigma(n_1) \times \Sigma(n_2)$-equivariant.
Hence it only remains to see that these natural transformations are uniquely fixed once the one for $n_1 = 1$ is given. To that end, observe that this lemma says that in the following commuting squares (exhibiting the action property on the level of functors with smash product, where we are notationally suppressing the associators) the left vertical morphisms are isomorphisms: a
$\array{ S^{n_1}\wedge S^{n_2} \wedge X_{n_3} &\longrightarrow& S^{n_1} \wedge X_{n_2 + n_3} \\ {}^{\mathllap{\simeq}}\downarrow && \downarrow \\ S^{n_1+ n_2} \wedge X_{n_3} &\longrightarrow& X_{n_1 + n_2 + n_3} } \,.$
This says exactly that the action of $S^{n_1 + n_2}$ has to be the composite of the actions of $S^{n_2}$ followed by that of $S^{n_1}$. Hence the statement follows by induction.
Finally, the definition of homomorphisms on both sides of the equivalence are just so as to preserve precisely this structure, hence they conincide under this identification.
###### Definition
Given $X,Y \in SymSpec(Top_{cg})$ two symmetric spectra, def. , then their smash product of spectra is the symmetric spectrum
$X \wedge Y \; \in SymSpec(Top_{cg})$
with component spaces the coequalizer
$\underset{p+1+q = n}{\bigvee} \Sigma(p+1+q)_+ \underset{\Sigma_p \times \Sigma_1 \times \Sigma_q}{\wedge} X_p \wedge S^1 \wedge Y_q \underoverset {\underset{r}{\longrightarrow}} {\overset{\ell}{\longrightarrow}} {\phantom{AAAA}} \underset{p+q=n}{\bigvee} \Sigma(p+q)_+ \underset{\Sigma_p \times \Sigma_q}{\wedge} X_p \wedge Y_q \overset{coeq}{\longrightarrow} (X \wedge Y)(n)$
where $\ell$ has components given by the structure maps
$X_p \wedge S^1 \wedge Y_q \overset{id \wedge \sigma_{q}}{\longrightarrow} X_p \wedge Y_q$
while $r$ has components given by the structure maps conjugated by the braiding in $Top^{\ast/}_{cg}$ and the permutation action $\chi_{p,1}$ (that shuffles the element on the right to the left)
$X_p \wedge S^1 \wedge X_q \overset{\tau^{Top^{\ast/}_{cg}}_{X_p,S^1} \wedge id}{\longrightarrow} S^1 \wedge X_p \wedge X_q \overset{\sigma_p\wedge id}{\longrightarrow} X_{p+1} \wedge X_q \overset{\chi_{p,1} \wedge id}{\longrightarrow} X_{1+p} \wedge X_q \,.$
The structure maps of $X \wedge Y$ are those induced under the coequalizer by
$X_p \wedge Y_q \wedge S^1 \overset{id \wedge \sigma_{p}}{\longrightarrow} X_{p} \wedge Y_{q+1} \,.$
Analogously for orthogonal spectra.
###### Proposition
Under the identification of prop. , the explicit smash product of spectra in def. is equivalent to the abstractly defined tensor product in def. :
in the case of symmetric spectra:
$\wedge \simeq \otimes_{\mathbb{S}_{sym}}$
in the case of orthogonal spectra:
$\wedge \simeq \otimes_{\mathbb{S}_{orth}} \,.$
###### Proof
By def. the abstractly defined tensor product of two $\mathbb{S}_{sym}$-modules $X$ and $Y$ is the coequalizer
$X \otimes_{Day} \mathbb{S}_{sym} \otimes_{Day} Y \underoverset {\underset{\rho_{1}\circ (\tau^{Day}_{X, \mathbb{S}_{sym}} \otimes id)}{\longrightarrow}} {\overset{X \otimes \rho_2}{\longrightarrow}} {\phantom{AAAA}} X \otimes Y \overset{coeq}{\longrightarrow} X \otimes_{\mathbb{S}_{sym}} Y \,.$
The Day convolution product appearing here is over the category $Sym$ from def. . By this example and unwinding the definitions, this is for any two symmetric spectra $A$ and $B$ given degreewise by the wedge sum of component spaces summing to that total degree, smashed with the symmetric group with basepoint adjoined and then quotiented by the diagonal action of the symmetric group acting on the degrees separately:
\begin{aligned} (A \otimes_{Day} B)(n) & = \overset{n_1,n_2}{\int} \underset{ = \left\{ \array{ \Sigma(n_1 + n_2,n)_+ & if \; n_1+n_2 = n \\ \ast & otherwise } \right. }{ \underbrace{\Sigma(n_1 + n_2, n)} }_+ \wedge A_{n_1} \wedge B_{n_1} \\ & \simeq \underset{n_1 + n_2 = n}{\bigvee} \Sigma(n_1+n_2)_+ \underset{O(n_1) \times O(n_2) }{\wedge} \left( A_{n_1} \wedge B_{n_2} \right) \end{aligned} \,.
This establishes the form of the coequalizer diagram. It remains to see that under this identification the two abstractly defined morphisms are the ones given in def. .
To see this, we apply the adjunction isomorphism between the Day convolution product and the external tensor product (cor. ) twice, to find the following sequence of equivalent incarnations of morphisms:
$\array{ \arrayopts{\rowlines{solid}} (X \otimes_{Day} ( \mathbb{S}_{orth} \otimes_{Day} Y ))(n) &\longrightarrow& (X \otimes_{Day} Y)(n) &\longrightarrow& Z_n \\ X_{n_1} \wedge (\mathbb{S}_{sym} \otimes_{Day} Y)(n'_2) &\longrightarrow& X_{n_1}\wedge Y(n'_2) &\longrightarrow& Z_{n_1 + n'_2} \\ (\mathbb{S}_{sym} \otimes_{Day} Y)(n'_2) &\longrightarrow& Y(n'_2) &\longrightarrow& Maps(X_{n_1}, Z_{n_1 + n'_2}) \\ S^{n_2} \wedge Y_{n_3} &\longrightarrow& Y_{n_2 + n_3} &\longrightarrow& Maps(X_{n_1}, Z_{n_1 + n_2 + n_3}) \\ X_{n_1} \wedge S^{n_2} \wedge Y_{n_3} &\longrightarrow& X_{n_1} \wedge Y_{n_2 + n_3} &\longrightarrow& Z_{n_1 + n_2 + n_3} } \,.$
This establishes the form of the morphism $\ell$. By the same reasoning as in the proof of prop. , we may restrict the coequalizer to $n_2 = 1$ without changing it.
The form of the morphism $r$ is obtained by the analogous sequence of identifications of morphisms, now with the parenthesis to the left. That it involves $\tau^{Top^{\ast/}_{cg}}$ and the permutation action $\tau^{sym}$ as shown above follows from the formula for the braiding of the Day convolution tensor product from the proof of prop. :
$\tau^{Day}_{A,B}(n) = \overset{n_1,n_2}{\int} Sym( \tau^{Sym}_{n_1,n_2}, n ) \wedge \tau^{Top^{\ast/}_{cg}}_{A_{n_1}, B_{n_2}}$
by translating it to the components of the precomposition
$X \otimes_{Day} \mathbb{S}_{sym} \overset{\tau^{Day}_{X,\mathbb{S}_{sym}}}{\longrightarrow} \mathbb{S}_{sym} \otimes_{Day} X \overset{}{\longrightarrow} X$
via the formula from the proof of prop. for the left Kan extension $A \otimes_{Day} B \simeq Lan_{\otimes} A \overline{\wedge} B$ (prop. ):
\begin{aligned} [Sym, Top^{\ast/}_{cg}]( \tau^{Day}_{X,\mathbb{S}_{sym}}, X) & \simeq \underset{n}{\int} Maps( \overset{n_1, n_2}{\int} Sym( \tau^{sym}_{n_1,n_2}, n ) \wedge \tau^{Top^{\ast/}_{cg}}_{X_{n_1}, S^{n_2}} , X(n) )_\ast \\ & \simeq \underset{n_1,n_2}{\int} Maps( \tau_{X_{n_1}, S^{n_2} }^{Top^{\ast/}_{cg}} , X( \tau^{sym}_{n_1,n_2} ) )_\ast \end{aligned} \,.
model structure on spectra
Also the
carries a symmetric monoida smash product.
## References
### Original sources
The original no-go theorem for a well-behave smash product of spectra is
• Gaunce Lewis, Is there a conveinient category of spectra?, Journal of Pure and Applied Algebra Volume 73, Issue 3, 30 August 1991, Pages 233–246
In the mid-1990s, several categories of spectra with nice smash products were discovered, and simultaneously, model categories experienced a major renaissance.
The definition of S-modules and their theory originates in
and around 1993 Jeff Smith gave the first talks about symmetric spectra; the details of the model structure were later worked out and written up in
Discussion that makes the Day convolution structure on the symmetric smash product of spectra manifest is in
### Reviews and introductions
Surveys of the history are in
A textbook account of the theory of symmetric spectra is
Seminar notes on symmetric spectra are in | 2019-07-17 06:50:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 729, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.998550534248352, "perplexity": 1609.9017949730007}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525094.53/warc/CC-MAIN-20190717061451-20190717083451-00400.warc.gz"} |
http://tex.stackexchange.com/questions/5060/how-do-i-center-text-in-a-page-heading | How do I center text in a page heading?
I'm using \markright to create a page heading. I'm having trouble centering some text in this heading, and I'd like to know how to do this.
Here's my current setup:
\documentclass{article}
\newcommand{\TitleString}{This Is The Title}
\begin{document}
\markright{\TitleString \hfill Matt Groff \hfill}
\end{document}
I use \hfill to fill in the space so that my name is close to centered, but for large titles it's way off. Is there a way to fix this? I'd like my name to be completely centered.
-
Here's a simplified variant of Geoffrey's solution that avoids \protect and the doubling of \TitleString:
\documentclass{article}
\newcommand{\TitleString}{This Is A Rather Long Title}
\begin{document}
\markright{\rlap{\TitleString}\hfill Matt Groff\hfill}
\end{document}
(Note that there are no spaces before the \hfill!)
Explanation: \rlap writes the text but just "overlaps it to the right" without it taking up any space.
-
Use \phantom to "balance" your title line, like this:
\documentclass{article}
\usepackage[pass,showframe]{geometry} % <-- help to see what's going on
\newcommand{\TitleString}{This Is A Rather Long Title}
\begin{document}
\markright{\TitleString \hfill%
Matt Groff%
\hfill\protect\phantom{\TitleString}}
\end{document}
-
Excellent! By the way, do you know why it puts everything in italics (or emphasis)? – Matt Groff Nov 8 '10 at 1:05
@Matt Groff: sure - if you look in the definitions of the oneside and twoside header macros in article.cls (\ps@headings), you'll see that it applies \slshape to the header output stream (this sometimes maps to \itshape in some font families). All you need do to modify your header formatting is to change your invocation of \markright, e.g., to \markright{\textnormal{\TitleString ... }}. – Geoffrey Jones Nov 8 '10 at 2:38
@Matt: Note the % after your name in Geoffrey's solution. This is important since otherwise there'll be an unwanted space after the name, and it won't be exactly centered. – Hendrik Vogt Nov 8 '10 at 13:09
My usual choice would be the fancyhdr package:
\documentclass{article}
\usepackage{fancyhdr}
\pagestyle{fancy}
\begin{document}
Hello world
\end{document}
-
Use the scrpage2 package which is part of KOMA-Script.
\documentclass{article}
\newcommand*{\TitleString}{This Is The Title}
\usepackage[automark]{scrpage2}
\begin{document}
\setupheadertexts[Mark Groff] % Center | 2015-08-05 08:34:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8565674424171448, "perplexity": 3659.4406504879285}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438043062723.96/warc/CC-MAIN-20150728002422-00337-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.esaral.com/q/a-calorimter-of-water-equivalent-20-g-contains-180-g-68301 | # A calorimter of water equivalent 20 g contains 180 g
Question:
A calorimter of water equivalent $20 \mathrm{~g}$ contains $180 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C} . ~ ' m$ ' grams of steam at $100^{\circ} \mathrm{C}$ is mixed in it till the temperature of the mixture is $31^{\circ} \mathrm{C}$. The value of $' m^{\prime}$ is close to (Latent heat of water $=540 \mathrm{cal} \mathrm{g}^{-1}$, specific heat of water $=1 \mathrm{cal} \mathrm{g}^{-1^{\circ}}{ }^{\circ}{ }^{-1}$ )
1. 2
2. 4
3. $3.2$
4. $2.6$
Correct Option: 1
Solution:
(1) Heat given by water $=m_{w} C_{w}\left(T_{\operatorname{mix}}-T_{w}\right)$
$=200 \times 1 \times(31-25)$
Heat taken by steam $=m L_{\text {stem }}+m C_{w}\left(T_{s}-T_{\text {mix }}\right)$
$=\mathrm{m} \times 540+\mathrm{m}(1) \times(100-31)$
$=\mathrm{m} \times 540+\mathrm{m}(1) \times(69)$
From the principal of calorimeter,
Heat lost = Heat gained
$\therefore(200)(31-25)=m \times 540+m(1)(69)$
$\Rightarrow 1200=m(609) \Rightarrow m \approx 2$ | 2023-03-23 05:03:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.676110565662384, "perplexity": 2114.4213964436067}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00364.warc.gz"} |
http://mathhelpforum.com/algebra/137926-find-f-g-3-a.html | 1. ## Find f(g(-3))
When f(x)=x^2-1 and g(x)=x+2
If someone could help me with first step I could probably do the rest.
Thanks,
2. Hi!
Try sticking the expressions for $g(x)$into the place of the x in the expression of $f(x)$.
$
f(g(x))=(x+2)^2-1
$
.
Hope this helps... I'm not mathematician.
3. Originally Posted by Tally
When f(x)=x^2-1 and g(x)=x+2
If someone could help me with first step I could probably do the rest.
Thanks,
$g(-3) = -3+2 = -1$
$f(g(-3))= (-1)^{2} -1 = 1-1 = 0$
4. Originally Posted by harish21
$g(-3) = -3+2 = -1$
$f(g(-3))= (-1)^{2} -1 = 1-1 = 0$
Awesome Thank you. I understand it much clearer now. | 2017-10-18 04:40:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6830193996429443, "perplexity": 1686.669873449441}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187822739.16/warc/CC-MAIN-20171018032625-20171018052625-00710.warc.gz"} |
http://python.nicolasbarrier.fr/plots/pyplot_settings.html | # Changing default Matplotlib settings¶
In order to change the default Matplotlib settings (for instance default colors, default linewidth, etc.), the user needs to modify the values of the matplotlib.pyplot.rcParams dictionary.
For a detailed description of all the customizarion methods, visisit matplotlib webpage.
import matplotlib as mpl
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 2*np.pi, 200)
y = [np.cos(x), np.sin(x), np.cos(2*x), np.cos(x/2.)]
y = np.array(y).T
plt.figure()
plt.plot(x, y)
plt.show()
## In a script¶
If the user wants to change the default values only within a given script, this is done as follows:
from cycler import cycler # used to define color cycles
plt.rcParams['lines.linewidth'] = 5
plt.rcParams['axes.prop_cycle'] = cycler('color', ['darkorange', 'plum', 'gold'])
plt.figure()
plt.plot(x, y)
plt.show()
## Changing the parameters for all sessions¶
Default matplotlib parameters are stored in a matplotlibrc configuration file. Python will search this file into 3 different locations in the following order order:
1. in the current working directory (usually used for specific customizations that you do not want to apply elsewhere)
2. $MATPLOTLIBRC if it is a file, else $MATPLOTLIBRC/matplotlibrc
3. On .config/matplotlib/matplotlibrc (Linux) or .matplotlib/matplotlibrc (other platforms)
If no matplotlibrc file is found, then the default value will be used. The path of the matplotlibrc file that is used in a given session can be obtained by using the matplotlib_fname method.
print(mpl.matplotlib_fname())
/home/barrier/.config/matplotlib/matplotlibrc
To change your Matplotlib default parameters, download the sample matplotlibrc file and put it in any of the three directories described above.
Then, uncomment the lines you are interested in and change the values. | 2022-07-04 03:11:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6092469692230225, "perplexity": 3467.6272257828673}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00532.warc.gz"} |
https://cs.stackexchange.com/questions/135687/are-the-frequencies-of-the-huffman-encoding-of-3-characters-unique | # Are the frequencies of the Huffman encoding of 3 characters unique?
I have to figure out the frequencies of 3 letters for Huffman encoding of {0, 10, 11}. One suggestion is in the order {$$\frac{2}{3}$$, $$\frac{1}{6}$$, $$\frac{1}{6}$$} as given in part (a) the 2nd slide of page 3 here. My question is whether the last 2 character frequencies need to be evenly split. Could we not have had {$$\frac{2}{3}$$, $$\frac{2}{18}$$, $$\frac{1}{18}$$}? Or could the 3 character frequencies be entirely different say {$$\frac{1}{2}$$, $$\frac{1}{4}$$, $$\frac{1}{4}$$}?
You want frequencies $$f_1,f_2,f_3$$ for three characters, so that their Huffman encodings are $$(0,10,11)$$.
The Huffman algorithm repeatedly "merges" the two least frequencies. So any choice that assigns the two least frequencies to the last two characters will work. That means $$f_2,f_3 \le f_1$$. Since these are frequencies you probably also want the values to add to $$1$$, but that is of no consequence for the algorithm.
In general the Huffman algorithm does not specify which branches go left and which go right (unless we add that to the implementation) so it is hard to guarantee codes $$(0,10,11)$$ instead of $$(1,00,01)$$ for instance. | 2021-05-07 22:45:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8434329032897949, "perplexity": 466.41510111986065}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988828.76/warc/CC-MAIN-20210507211141-20210508001141-00224.warc.gz"} |
http://tex.stackexchange.com/questions/8832/colors-that-work-well-with-beamer-color-scheme-albatross | # Colors that work well with beamer color scheme albatross
Is there a color scheme made using \definecolor that defines colors that work well with the beamer color scheme albatross?
-
You'd better check out the source code of color theme albatross, in file beamercolorthemealbatross.sty (which is usually located at your TEXMF root/tex/latex/beamer/themes/color/). | 2015-07-28 20:07:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4028647541999817, "perplexity": 7176.637672226716}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042982502.13/warc/CC-MAIN-20150728002302-00302-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://econpapers.repec.org/article/sprcompst/v_3a32_3ay_3a2017_3ai_3a4_3ad_3a10.1007_5fs00180-016-0697-8.htm | # Fast implementation of the Tukey depth
Xiaohui Liu ()
Additional contact information
Xiaohui Liu: Jiangxi University of Finance and Economics
Computational Statistics, 2017, vol. 32, issue 4, 1395-1410
Abstract: Abstract Tukey depth function is one of the most famous multivariate tools serving robust purposes. It is also very well known for its computability problems in dimensions $$p \ge 3$$ p ≥ 3 . In this paper, we address this computing issue by presenting two combinatorial algorithms. The first is naive and calculates the Tukey depth of a single point with complexity $$O\left( n^{p-1}\log (n)\right)$$ O n p - 1 log ( n ) , while the second further utilizes the quasiconcave of the Tukey depth function and hence is more efficient than the first. Both require very minimal memory and run much faster than the existing ones. All experiments indicate that they compute the exact Tukey depth.
Keywords: Tukey depth; Quasiconcave; Combinatorial property; Fast computation (search for similar items in EconPapers)
Date: 2017
References: View references in EconPapers View complete reference list from CitEc
Citations: View citations in EconPapers (1) Track citations by RSS feed
Downloads: (external link)
http://link.springer.com/10.1007/s00180-016-0697-8 Abstract (text/html)
Access to the full text of the articles in this series is restricted.
Related works:
This item may be available elsewhere in EconPapers: Search for items with the same title.
Export reference: BibTeX RIS (EndNote, ProCite, RefMan) HTML/Text
Ordering information: This journal article can be ordered from
http://www.springer.com/statistics/journal/180/PS2
Access Statistics for this article
Computational Statistics is currently edited by Wataru Sakamoto, Ricardo Cao and Jürgen Symanzik
More articles in Computational Statistics from Springer
Bibliographic data for series maintained by Sonal Shukla ().
Page updated 2019-11-06
Handle: RePEc:spr:compst:v:32:y:2017:i:4:d:10.1007_s00180-016-0697-8 | 2019-12-12 00:51:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23473399877548218, "perplexity": 3763.0582082281753}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540534443.68/warc/CC-MAIN-20191212000437-20191212024437-00279.warc.gz"} |
https://detailedpedia.com/wiki-Czechoslovakia | # Czechoslovakia
Czechoslovakia
Československo
1918–1939
1945–1992
1939–1945: Government-in-exile
Motto: Pravda vítězí / Pravda víťazí’
(Czech / Slovak, 1918–1990)
’Veritas vincit’ (Latin, 1990–1992)
’Truth prevails’
Anthems: (Czech)
’Where my home is’
(Slovak)
’Lightning Over the Tatras’
Czechoslovakia during the interwar period and the Cold War
Capital
and largest city
Prague
50°05′N 14°25′E / 50.083°N 14.417°E
Official languagesCzechoslovak, after 1948 Czech · Slovak
Recognised languages
Demonym(s)Czechoslovak
GovernmentFirst Republic
(1918–38)
Second Republic
(1938–39)
Third Republic
(1945–48)
Socialist Republic
(1948–89)
Federative Republic
(1990–92)
President
• 1918–1935
Tomáš G. Masaryk
• 1935–1938 · 1945–1948
Edvard Beneš
• 1938–1939
Emil Hácha
• 1948–1953
Klement Gottwald
• 1953–1957
Antonín Zápotocký
• 1957–1968
Antonín Novotný
• 1968–1975
Ludvík Svoboda
• 1976–1989
Gustáv Husák
• 1989–1992
Václav Havel
KSČ General Secretary / First Secretary
• 1948–1953
Klement Gottwald
• 1953–1968
Antonín Novotný
• 1968–1969
Alexander Dubček
• 1969–1987
Gustáv Husák
• 1987–1989
Miloš Jakeš
Prime Minister
• 1918–1919 (first)
Karel Kramář
• 1992 (last)
Jan Stráský
LegislatureNational Assembly (1948–1969)
Federal Assembly (1969–1992)
History
28 October 1918
30 September 1938
14 March 1939
10 May 1945
25 February 1948
21 August 1968
17 – 28 November 1989
31 December 1992
HDI (1992)0.810
very high
CurrencyCzechoslovak koruna
Driving sideright
Calling code+42
Internet TLD.cs
Preceded by Succeeded by
Today part of
Calling code +42 was withdrawn in the winter of 1997. The number range was divided between the Czech Republic (+420) and Slovak Republic (+421).
Current ISO 3166-3 code is "CSHH".
Czechoslovakia (/ˌtʃɛkoʊsloʊˈvækiə, -kə-, -slə-, -ˈvɑː-/; Czech and Slovak: Československo, Česko-Slovensko) was a landlocked country in Central Europe, created in 1918, when it declared its independence from Austria-Hungary. In 1938, after the Munich Agreement, the Sudetenland became part of Germany, while the country lost further territories to Hungary and Poland. Between 1939 and 1945 the state ceased to exist, as Slovakia proclaimed its independence and the remaining territories in the east became part of Hungary, while in the remainder of the Czech Lands the German Protectorate of Bohemia and Moravia was proclaimed. In 1939, after the outbreak of World War II, former Czechoslovak President Edvard Beneš formed a government-in-exile and sought recognition from the Allies.
After World War II, the pre-1938 Czechoslovakia was reestablished, with the exception of Carpathian Ruthenia, which became part of the Ukrainian SSR (a republic of the Soviet Union). From 1948 to 1989, Czechoslovakia was part of the Eastern Bloc with a command economy. Its economic status was formalized in membership of Comecon from 1949 and its defense status in the Warsaw Pact of 1955. A period of political liberalization in 1968, the Prague Spring, ended violently when the Soviet Union, assisted by other Warsaw Pact countries, invaded Czechoslovakia. In 1989, as Marxist–Leninist governments and communism were ending all over Central and Eastern Europe, Czechoslovaks peacefully deposed their communist government on 17 November 1989 in the Velvet Revolution. On 1 January 1993, Czechoslovakia peacefully split into the two sovereign states of the Czech Republic and Slovakia as the result of national tensions of the Slovaks.
## Characteristics
Form of state
Neighbors
Topography
The country was of generally irregular terrain. The western area was part of the north-central European uplands. The eastern region was composed of the northern reaches of the Carpathian Mountains and lands of the Danube River basin.
Climate
The weather is mild winters and mild summers. Influenced by the Atlantic Ocean from the west, the Baltic Sea from the north, and Mediterranean Sea from the south. There is no continental weather.
## History
### Origins
Tomáš Garrigue Masaryk, founder and first president
Czechoslovak declaration of independence rally in Prague on Wenceslas Square, 28 October 1918
The area was part of the Austro-Hungarian Empire until it collapsed at the end of World War I. The new state was founded by Tomáš Garrigue Masaryk, who served as its first president from 14 November 1918 to 14 December 1935. He was succeeded by his close ally Edvard Beneš (1884–1948).
The roots of Czech nationalism go back to the 19th century, when philologists and educators, influenced by Romanticism, promoted the Czech language and pride in the Czech people. Nationalism became a mass movement in the second half of the 19th century. Taking advantage of the limited opportunities for participation in political life under Austrian rule, Czech leaders such as historian František Palacký (1798–1876) founded various patriotic, self-help organizations which provided a chance for many of their compatriots to participate in communal life before independence. Palacký supported Austro-Slavism and worked for a reorganized federal Austrian Empire, which would protect the Slavic speaking peoples of Central Europe against Russian and German threats.
An advocate of democratic reform and Czech autonomy within Austria-Hungary, Masaryk was elected twice to the Reichsrat (Austrian Parliament), from 1891 to 1893 for the Young Czech Party, and from 1907 to 1914 for the Czech Realist Party, which he had founded in 1889 with Karel Kramář and Josef Kaizl.
During World War I a number of Czechs and Slovaks, the Czechoslovak Legions, fought with the Allies in France and Italy, while large numbers deserted to Russia in exchange for its support for the independence of Czechoslovakia from the Austrian Empire. With the outbreak of World War I, Masaryk began working for Czech independence in a union with Slovakia. With Edvard Beneš and Milan Rastislav Štefánik, Masaryk visited several Western countries and won support from influential publicists. The Czechoslovak National Council was the main organization that advanced the claims for a Czechoslovak state.
### First Czechoslovak Republic
A monument to Tomáš Garrigue Masaryk and Milan Štefánik—both key figures in early Czechoslovakia
#### Formation
Czechoslovakia in 1928
The Bohemian Kingdom ceased to exist in 1918 when it was incorporated into Czechoslovakia. Czechoslovakia was founded in October 1918, as one of the successor states of the Austro-Hungarian Empire at the end of World War I and as part of the Treaty of Saint-Germain-en-Laye. It consisted of the present day territories of Bohemia, Moravia, Slovakia and Carpathian Ruthenia. Its territory included some of the most industrialized regions of the former Austria-Hungary. "The land consisted of modern day Czechia, Slovakia, and a region of Ukraine called Carpathian Ruthnia
#### Ethnicity
Linguistic map of Czechoslovakia in 1930
The new country was a multi-ethnic state, with Czechs and Slovaks as constituent peoples. The population consisted of Czechs (51%), Slovaks (16%), Germans (22%), Hungarians (5%) and Rusyns (4%). Many of the Germans, Hungarians, Ruthenians and Poles and some Slovaks, felt oppressed because the political elite did not generally allow political autonomy for minority ethnic groups.[citation needed] This policy led to unrest among the non-Czech population, particularly in German-speaking Sudetenland, which initially had proclaimed itself part of the Republic of German-Austria in accordance with the self-determination principle.
The state proclaimed the official ideology that there were no separate Czech and Slovak nations, but only one nation of Czechoslovaks (see Czechoslovakism), to the disagreement of Slovaks and other ethnic groups. Once a unified Czechoslovakia was restored after World War II (after the country had been divided during the war), the conflict between the Czechs and the Slovaks surfaced again. The governments of Czechoslovakia and other Central European nations deported ethnic Germans, reducing the presence of minorities in the nation. Most of the Jews had been killed during the war by the Nazis.
Ethnicities of Czechoslovakia in 1921
Czecho slovaks 8,759,701 64.37%
Germans 3,123,305 22.95%
Hungarians 744,621 5.47%
Ruthenians 461,449 3.39%
Jews 180,534 1.33%
Poles 75,852 0.56%
Others 23,139 0.17%
Foreigners 238,784 1.75%
Total population 13,607,385
Ethnicities of Czechoslovakia in 1930
Czecho slovaks 10,066,000 68.35%
Germans 3,229,000 21.93%
Ruthenians 745,000 5.06%
Hungarians 653,000 4.43%
Jews 354,000 2.40%
Poles 76,000 0.52%
Romanians 14,000 0.10%
Foreigners 239,000 1.62%
Total population 14,726,158
*Jews identified themselves as Germans or Hungarians (and Jews only by religion not ethnicity), the sum is, therefore, more than 100%.
### Interwar period
During the period between the two world wars Czechoslovakia was a democratic state. The population was generally literate, and contained fewer alienated groups. The influence of these conditions was augmented by the political values of Czechoslovakia's leaders and the policies they adopted. Under Tomas Masaryk, Czech and Slovak politicians promoted progressive social and economic conditions that served to defuse discontent.
Foreign minister Beneš became the prime architect of the Czechoslovak-Romanian-Yugoslav alliance (the "Little Entente", 1921–38) directed against Hungarian attempts to reclaim lost areas. Beneš worked closely with France. Far more dangerous was the German element, which after 1933 became allied with the Nazis in Germany.
Czech-Slovak relations came to be a central issue in Czechoslovak politics during the 1930s. The increasing feeling of inferiority among the Slovaks,[failed verification] who were hostile to the more numerous Czechs, weakened the country in the late 1930s. Slovakia became autonomous in the fall of 1938, and by mid-1939, Slovakia had become independent, with the First Slovak Republic set up as a satellite state of Nazi Germany and the far-right Slovak People’s Party in power .
After 1933, Czechoslovakia remained the only democracy in central and eastern Europe.
### Munich Agreement, and Two-Step German Occupation
The partition of Czechoslovakia after Munich Agreement
The car in which Reinhard Heydrich was killed in 1942
Territory of the Second Czechoslovak Republic (1938–1939)
In September 1938, Adolf Hitler demanded control of the Sudetenland. On 29 September 1938, Britain and France ceded control in the Appeasement at the Munich Conference; France ignored the military alliance it had with Czechoslovakia. During October 1938, Nazi Germany occupied the Sudetenland border region, effectively crippling Czechoslovak defences.
The First Vienna Award assigned a strip of southern Slovakia and Carpathian Ruthenia to Hungary. Poland occupied Zaolzie, an area whose population was majority Polish, in October 1938.
On 14 March 1939, the remainder ("rump") of Czechoslovakia was dismembered by the proclamation of the Slovak State, the next day the rest of Carpathian Ruthenia was occupied and annexed by Hungary, while the following day the German Protectorate of Bohemia and Moravia was proclaimed.
The eventual goal of the German state under Nazi leadership was to eradicate Czech nationality through assimilation, deportation, and extermination of the Czech intelligentsia; the intellectual elites and middle class made up a considerable number of the 200,000 people who passed through concentration camps and the 250,000 who died during German occupation. Under Generalplan Ost, it was assumed that around 50% of Czechs would be fit for Germanization. The Czech intellectual elites were to be removed not only from Czech territories but from Europe completely. The authors of Generalplan Ost believed it would be best if they emigrated overseas, as even in Siberia they were considered a threat to German rule. Just like Jews, Poles, Serbs, and several other nations, Czechs were considered to be untermenschen by the Nazi state. In 1940, in a secret Nazi plan for the Germanization of the Protectorate of Bohemia and Moravia it was declared that those considered to be of racially Mongoloid origin and the Czech intelligentsia were not to be Germanized.
The deportation of Jews to concentration camps was organized under the direction of Reinhard Heydrich, and the fortress town of Terezín was made into a ghetto way station for Jewish families. On 4 June 1942 Heydrich died after being wounded by an assassin in Operation Anthropoid. Heydrich's successor, Colonel General Kurt Daluege, ordered mass arrests and executions and the destruction of the villages of Lidice and Ležáky. In 1943 the German war effort was accelerated. Under the authority of Karl Hermann Frank, German minister of state for Bohemia and Moravia, some 350,000 Czech laborers were dispatched to the Reich. Within the protectorate, all non-war-related industry was prohibited. Most of the Czech population obeyed quiescently up until the final months preceding the end of the war, while thousands were involved in the resistance movement.
For the Czechs of the Protectorate Bohemia and Moravia, German occupation was a period of brutal oppression. Czech losses resulting from political persecution and deaths in concentration camps totaled between 36,000 and 55,000. The Jewish populations of Bohemia and Moravia (118,000 according to the 1930 census) were virtually annihilated. Many Jews emigrated after 1939; more than 70,000 were killed; 8,000 survived at Terezín. Several thousand Jews managed to live in freedom or in hiding throughout the occupation.
Despite the estimated 136,000 deaths at the hands of the Nazi regime, the population in the Reichsprotektorate saw a net increase during the war years of approximately 250,000 in line with an increased birth rate.
On 6 May 1945, the third US Army of General Patton entered Pilsen from the south west. On 9 May 1945, Soviet Red Army troops entered Prague.
### Communist Czechoslovakia
Socialist coat of arms in 1960–1989
After World War II, pre-war Czechoslovakia was re-established, with the exception of Subcarpathian Ruthenia, which was annexed by the Soviet Union and incorporated into the Ukrainian Soviet Socialist Republic. The Beneš decrees were promulgated concerning ethnic Germans (see Potsdam Agreement) and ethnic Hungarians. Under the decrees, citizenship was abrogated for people of German and Hungarian ethnic origin who had accepted German or Hungarian citizenship during the occupations. In 1948, this provision was cancelled for the Hungarians, but only partially for the Germans. The government then confiscated the property of the Germans and expelled about 90% of the ethnic German population, over 2 million people. Those who remained were collectively accused of supporting the Nazis after the Munich Agreement, as 97.32% of Sudeten Germans had voted for the NSDAP in the December 1938 elections. Almost every decree explicitly stated that the sanctions did not apply to antifascists. Some 250,000 Germans, many married to Czechs, some antifascists, and also those required for the post-war reconstruction of the country, remained in Czechoslovakia. The Beneš Decrees still cause controversy among nationalist groups in the Czech Republic, Germany, Austria and Hungary.
Following the expulsion of the ethnic German population from Czechoslovakia, parts of the former Sudetenland, especially around Krnov and the surrounding villages of the Jesenik mountain region in northeastern Czechoslovakia, were settled in 1949 by Communist refugees from Northern Greece who had left their homeland as a result of the Greek Civil War. These Greeks made up a large proportion of the town and region's population until the late 1980s/early 1990s. Although defined as "Greeks", the Greek Communist community of Krnov and the Jeseniky region actually consisted of an ethnically diverse population, including Greek Macedonians, Macedonians, Vlachs, Pontic Greeks and Turkish speaking Urums or Caucasus Greeks.
Carpathian Ruthenia (Podkarpatská Rus) was occupied by (and in June 1945 formally ceded to) the Soviet Union. In the 1946 parliamentary election, the Communist Party of Czechoslovakia was the winner in the Czech lands, and the Democratic Party won in Slovakia. In February 1948 the Communists seized power. Although they would maintain the fiction of political pluralism through the existence of the National Front, except for a short period in the late 1960s (the Prague Spring) the country had no liberal democracy. Since citizens lacked significant electoral methods of registering protest against government policies, periodically there were street protests that became violent. For example, there were riots in the town of Plzeň in 1953, reflecting economic discontent. Police and army units put down the rebellion, and hundreds were injured but no one was killed. While its economy remained more advanced than those of its neighbors in Eastern Europe, Czechoslovakia grew increasingly economically weak relative to Western Europe.
The currency reform of 1953 caused dissatisfaction among Czechoslovak laborers. To equalize the wage rate, Czechoslovaks had to turn in their old money for new at a decreased value. The banks also confiscated savings and bank deposits to control the amount of money in circulation. In the 1950s, Czechoslovakia experienced high economic growth (averaging 7% per year), which allowed for a substantial increase in wages and living standards, thus promoting the stability of the regime.
Czechoslovakia after 1969
In 1968, when the reformer Alexander Dubček was appointed to the key post of First Secretary of the Czechoslovak Communist Party, there was a brief period of liberalization known as the Prague Spring. In response, after failing to persuade the Czechoslovak leaders to change course, five other members of the Warsaw Pact invaded. Soviet tanks rolled into Czechoslovakia on the night of 20–21 August 1968. Soviet Communist Party General Secretary Leonid Brezhnev viewed this intervention as vital for the preservation of the Soviet, socialist system and vowed to intervene in any state that sought to replace Marxism-Leninism with capitalism.
In the week after the invasion there was a spontaneous campaign of civil resistance against the occupation. This resistance involved a wide range of acts of non-cooperation and defiance: this was followed by a period in which the Czechoslovak Communist Party leadership, having been forced in Moscow to make concessions to the Soviet Union, gradually put the brakes on their earlier liberal policies.
Meanwhile, one plank of the reform program had been carried out: in 1968–69, Czechoslovakia was turned into a federation of the Czech Socialist Republic and Slovak Socialist Republic. The theory was that under the federation, social and economic inequities between the Czech and Slovak halves of the state would be largely eliminated. A number of ministries, such as education, now became two formally equal bodies in the two formally equal republics. However, the centralized political control by the Czechoslovak Communist Party severely limited the effects of federalization.
The 1970s saw the rise of the dissident movement in Czechoslovakia, represented among others by Václav Havel. The movement sought greater political participation and expression in the face of official disapproval, manifested in limitations on work activities, which went as far as a ban on professional employment, the refusal of higher education for the dissidents' children, police harassment and prison.
### After 1989
The Visegrád Group signing ceremony in February 1991
In 1989, the Velvet Revolution restored democracy. This occurred at around the same time as the fall of communism in Romania, Bulgaria, Hungary, East Germany and Poland.
The word "socialist" was removed from the country's full name on 29 March 1990 and replaced by "federal".
Pope John Paul II made a papal visit to Czechoslovakia on 21 April 1990, hailing it as a symbolic step of reviving Christianity in the newly-formed post-communist state.
Czechoslovakia participated in the Gulf War with a small force of 200 troops under the command of the U.S.-led coalition.
In 1992, because of growing nationalist tensions in the government, Czechoslovakia was peacefully dissolved by parliament. On 31 December 1992 it formally separated into two independent countries, the Czech Republic and the Slovak Republic.
## Government and politics
After World War II, a political monopoly was held by the Communist Party of Czechoslovakia (KSČ). The leader of the KSČ was de facto the most powerful person in the country during this period. Gustáv Husák was elected first secretary of the KSČ in 1969 (changed to general secretary in 1971) and president of Czechoslovakia in 1975. Other parties and organizations existed but functioned in subordinate roles to the KSČ. All political parties, as well as numerous mass organizations, were grouped under umbrella of the National Front. Human rights activists and religious activists were severely repressed.
### Constitutional development
Federative coat of arms in 1990–1992
Czechoslovakia had the following constitutions during its history (1918–1992):
### Foreign policy
#### International agreements and membership
In the 1930s, the nation formed a military alliance with France, which collapsed in the Munich Agreement of 1938. After World War II, an active participant in Council for Mutual Economic Assistance (Comecon), Warsaw Pact, United Nations and its specialized agencies; signatory of conference on Security and Cooperation in Europe.
• 1918–1923: Different systems in former Austrian territory (Bohemia, Moravia, a small part of Silesia) compared to former Hungarian territory (Slovakia and Ruthenia): three lands (země) (also called district units (kraje)): Bohemia, Moravia, Silesia, plus 21 counties (župy) in today's Slovakia and three counties in today's Ruthenia; both lands and counties were divided into districts (okresy).
• 1923–1927: As above, except that the Slovak and Ruthenian counties were replaced by six (grand) counties ((veľ)župy) in Slovakia and one (grand) county in Ruthenia, and the numbers and boundaries of the okresy were changed in those two territories.
• 1928–1938: Four lands (Czech: země, Slovak: krajiny): Bohemia, Moravia-Silesia, Slovakia and Sub-Carpathian Ruthenia, divided into districts (okresy).
• Late 1938 – March 1939: As above, but Slovakia and Ruthenia gained the status of "autonomous lands". Slovakia was called Slovenský štát, with its own currency and government.
• 1945–1948: As in 1928–1938, except that Ruthenia became part of the Soviet Union.
• 1949–1960: 19 regions (kraje) divided into 270 okresy.
• 1960–1992: 10 kraje, Prague, and (from 1970) Bratislava (capital of Slovakia); these were divided into 109–114 okresy; the kraje were abolished temporarily in Slovakia in 1969–1970 and for many purposes from 1991 in Czechoslovakia; in addition, the Czech Socialist Republic and the Slovak Socialist Republic were established in 1969 (without the word Socialist from 1990).
## Economy
Before World War II, the economy was about the fourth in all industrial countries in Europe.[citation needed][clarification needed] The state was based on strong economy, manufacturing cars (Škoda, Tatra), trams, aircraft (Aero, Avia), ships, ship engines (Škoda), cannons, shoes (Baťa), turbines, guns (Zbrojovka Brno). It was the industrial workshop for the Austro-Hungarian empire. The Slovak lands relied more heavily on agriculture than the Czech lands.
After World War II, the economy was centrally planned, with command links controlled by the communist party, similarly to the Soviet Union. The large metallurgical industry was dependent on imports of iron and non-ferrous ores.
• Industry: Extractive industry and manufacturing dominated the sector, including machinery, chemicals, food processing, metallurgy, and textiles. The sector was wasteful in its use of energy, materials, and labor and was slow to upgrade technology, but the country was a major supplier of high-quality machinery, instruments, electronics, aircraft, airplane engines and arms to other socialist countries.
• Agriculture: Agriculture was a minor sector, but collectivized farms of large acreage and relatively efficient mode of production enabled the country to be relatively self-sufficient in the food supply. The country depended on imports of grains (mainly for livestock feed) in years of adverse weather. Meat production was constrained by a shortage of feed, but the country still recorded high per capita consumption of meat.
• Foreign Trade: Exports were estimated at US$17.8 billion in 1985. Exports were machinery (55%), fuel and materials (14%), and manufactured consumer goods (16%). Imports stood at an estimated US$17.9 billion in 1985, including fuel and materials (41%), machinery (33%), and agricultural and forestry products (12%). In 1986, about 80% of foreign trade was with other socialist countries.
• Exchange rate: Official, or commercial, the rate was crowns (Kčs) 5.4 per US$1 in 1987. Tourist, or non-commercial, the rate was Kčs 10.5 per US$1. Neither rate reflected purchasing power. The exchange rate on the black market was around Kčs 30 per US\$1, which became the official rate once the currency became convertible in the early 1990s.
• Fiscal year: Calendar year.
• Fiscal policy: The state was the exclusive owner of means of production in most cases. Revenue from state enterprises was the primary source of revenues followed by turnover tax. The government spent heavily on social programs, subsidies, and investment. The budget was usually balanced or left a small surplus.
## Resource base
After World War II, the country was short of energy, relying on imported crude oil and natural gas from the Soviet Union, domestic brown coal, and nuclear and hydroelectric energy. Energy constraints were a major factor in the 1980s.
## Transport and communications
Slightly after the foundation of Czechoslovakia in 1918, there was a lack of essential infrastructure in many areas – paved roads, railways, bridges, etc. Massive improvement in the following years enabled Czechoslovakia to develop its industry. Prague's civil airport in Ruzyně became one of the most modern terminals in the world when it was finished in 1937. Tomáš Baťa, a Czech entrepreneur and visionary, outlined his ideas in the publication "Budujme stát pro 40 milionů lidí", where he described the future motorway system. Construction of the first motorways in Czechoslovakia begun in 1939, nevertheless, they were stopped after German occupation during World War II.
## Education
Education was free at all levels and compulsory from ages 6 to 15. The vast majority of the population was literate. There was a highly developed system of apprenticeship training and vocational schools supplemented general secondary schools and institutions of higher education.
## Religion
In 1991, 46% of the population were Roman Catholics, 5.3% were Evangelical Lutheran, 30% were Atheist, and other religions made up 17% of the country, but there were huge differences in religious practices between the two constituent republics; see Czech Republic and Slovakia.
## Health, social welfare and housing
After World War II, free health care was available to all citizens. National health planning emphasized preventive medicine; factory and local health care centres supplemented hospitals and other inpatient institutions. There was a substantial improvement in rural health care during the 1960s and 1970s.
## Mass media
During the era between the World Wars, Czechoslovak democracy and liberalism facilitated conditions for free publication. The most significant daily newspapers in these times were Lidové noviny, Národní listy, Český deník and Československá Republika.
During Communist rule, the mass media in Czechoslovakia were controlled by the Communist Party. Private ownership of any publication or agency of the mass media was generally forbidden, although churches and other organizations published small periodicals and newspapers. Even with this information monopoly in the hands of organizations under KSČ control, all publications were reviewed by the government's Office for Press and Information.
## Sports
The Czechoslovakia national football team was a consistent performer on the international scene, with eight appearances in the FIFA World Cup Finals, finishing in second place in 1934 and 1962. The team also won the European Football Championship in 1976, came in third in 1980 and won the Olympic gold in 1980.
Well-known football players such as Pavel Nedvěd, Antonín Panenka, Milan Baroš, Tomáš Rosický, Vladimír Šmicer or Petr Čech were all born in Czechoslovakia.
The International Olympic Committee code for Czechoslovakia is TCH, which is still used in historical listings of results.
The Czechoslovak national ice hockey team won many medals from the world championships and Olympic Games. Peter Šťastný, Jaromír Jágr, Dominik Hašek, Peter Bondra, Petr Klíma, Marián Gáborík, Marián Hossa, Miroslav Šatan and Pavol Demitra all come from Czechoslovakia.
Emil Zátopek, winner of four Olympic gold medals in athletics, is considered one of the top athletes in Czechoslovak history.
Věra Čáslavská was an Olympic gold medallist in gymnastics, winning seven gold medals and four silver medals. She represented Czechoslovakia in three consecutive Olympics.
Several accomplished professional tennis players including Jaroslav Drobný, Ivan Lendl, Jan Kodeš, Miloslav Mečíř, Hana Mandlíková, Martina Hingis, Martina Navratilova, Jana Novotna, Petra Kvitová and Daniela Hantuchová were born in Czechoslovakia. | 2023-01-28 03:58:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.182431161403656, "perplexity": 14266.754495027999}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00870.warc.gz"} |
https://www.chemicalforums.com/index.php?topic=110303.0;prev_next=prev | May 18, 2022, 05:27:58 AM
Forum Rules: Read This Before Posting
### Topic: changing temperature of an ideal gas by changing volume?! (Read 474 times)
0 Members and 1 Guest are viewing this topic.
#### mana
• Full Member
• Posts: 176
• Mole Snacks: +3/-3
##### changing temperature of an ideal gas by changing volume?!
« on: December 25, 2021, 11:57:51 AM »
hi all
as you know Charles's rule says " the volume of an ideal gas is directly proportioned with temperature (kelvin)" now my question is if I reduce the volume of an ideal gas, does the temperature decrease too?
#### Hunter2
• Sr. Member
• Posts: 1807
• Mole Snacks: +136/-44
• Gender:
• Vena Lausa moris pax drux bis totis
##### Re: changing temperature of an ideal gas by changing volume?!
« Reply #1 on: December 25, 2021, 12:14:01 PM »
The key is the ideal Gas law.
pV = nRT
So you can play with all parameters. Temperature can decrease but can also be constant if you change pressure for example.
#### mana
• Full Member
• Posts: 176
• Mole Snacks: +3/-3
##### Re: changing temperature of an ideal gas by changing volume?!
« Reply #2 on: December 25, 2021, 12:24:40 PM »
The key is the ideal Gas law.
pV = nRT
So you can play with all parameters. Temperature can decrease but can also be constant if you change pressure for example.
Thus it will decrease or it is constant? honestly, it is very strange to me that by decreasing the volume of an ideal gas, the temperature decreases too, I mean I can't imagine that, but the Charles rule says it must happen???
#### Hunter2
• Sr. Member
• Posts: 1807
• Mole Snacks: +136/-44
• Gender:
• Vena Lausa moris pax drux bis totis
##### Re: changing temperature of an ideal gas by changing volume?!
« Reply #3 on: December 25, 2021, 01:11:28 PM »
It's simple mathematics. If volume decrease at half and pressure and moles keep constant then temperature decrease at half as well the same if temperature is constant the pressure has to be the double amount.
#### billnotgatez
• Global Moderator
• Sr. Member
• Posts: 4312
• Mole Snacks: +221/-61
• Gender:
##### Re: changing temperature of an ideal gas by changing volume?!
« Reply #4 on: December 25, 2021, 05:02:07 PM »
@mana
If you have a balloon filled with dry air and the temperature goes from 90F to 0F do you expect the balloon to expand or contract (assuming barometric pressure stays constant)? | 2022-05-18 09:27:58 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8459929823875427, "perplexity": 5835.234982723405}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00674.warc.gz"} |
https://www.physicsforums.com/threads/basic-set-theory-question.256360/ | # Basic Set theory question
1. Sep 14, 2008
### Diffy
Hi!
I am having trouble constructing the sentences in this proof.
Its very simple, proof that $$A \cup \left( B \cap C \right) = \left( A \cup B \right) \cap \left( A \cup B \right)$$
So basically I need to show that if $$x \in A \cup \left( B \cap C \right)$$ then $$x \in \left( A \cup B \right) \cap \left( A \cup B \right)$$
Here is what I got:
If $$x \in A \cup \left( B \cap C \right)$$ then $$x \in A$$ or $$x \in \left( B \cap C \right)$$. Which means that either $$x \in A$$, or $$x \in B$$ and $$x \in C$$...
-----------------------------------------
First Question,
I feel like there is ambiguity here. "Either x in A, or x in B and x in C " can be interpreted two ways right? You could read it: $$x \in \left( A \cup B \right) \cap C$$ or you could read it as intended $$x \in A \cup \left( B \cap C \right)$$ How can I make the sentence clear that I want the latter?
Second question,
From here is it ok for me to make the jump to x in A or B, and x in A or C? It seems clear to me that this is the case, but I am not sure if something is left to be said before I make this claim.
Thanks for help!
2. Sep 14, 2008
### PowerIso
I'll do my best to answer.
First off while yes you need to show what you say you need to show, but you also have to show that if $$x \in \left( A \cup B \right) \cap \left( A \cup B \right)$$ then $$x \in A \cup \left( B \cap C \right)$$. This is because = is essentially a bi-conditional statement, so you need to show that the left side is a subset of the right side but also that the right side is a subset of the left side. If that is true, then they are equal.
Secondly, I would simply write "x is a member of A or x is a member of the intersection of B and C." Do you have to write that part in words? Seems like symbolic would be better.
I think you can make that jump, but you should clarify why you are making the jump.
3. Sep 15, 2008
### Diffy
Thanks that helps, I have finished it and I think made it perfectly clear. And yes I am aware that I have to show it both ways. If I can get it one way going backwards is usually very easy.
I have another one, that I am stuck on at the moment and maybe you would be so kind to help. Latex was taking too long so I am going to try to write it out using no Latex.
Show that (A - B) x C = (A x C) - (B x C)
So I start the same way
let (x, y) /in (A - B) x C.
This is the set {(x,y) | x in A and x not in B, y in C}
I Want to be able to say the following:
This is the same as saying all x in A and y in C, and all x not in B and y not in C.
But I am not sure I can make this leap, does this need justification?
4. Sep 16, 2008
### Diffy
Ok things were a bit more complicated than I thought they would be...
But that was because I didn't understand what the complement of BxC was until I drew myself a picture.
But I solved it, and now I am moving on, thanks!
5. Sep 16, 2008
### evagelos
WOULD you care to give a short proof of the above i am just curious
thanx
6. Sep 19, 2008
### mistermath
One direction is:
(A - B) x C = (A x C) - (B x C)
Let (a,c) in (A - B) x C then a in A - B and c in C.
a in A - B -> a in A and A not in B.
a in A and c in C -> (a,c) in A x C.
a not in B -> (a , x) not in B x C for all x. (in particular our c).
therefore (a, c) in (A x C) - (B x C) as (a, c) in A x C and (a, c) not in B x C.
7. Sep 19, 2008
### evagelos
According to what theorem axiom or law of logic do you get and i quote;
.........a not in B -> (a , x) not in B x C for all x. (in particular our c)..........................
I am sorry i can make no justification for that step
8. Sep 19, 2008
### Diffy
Let A and B be a set.
Assume for some a, a is not contained in A.
AxB = all (x, y) such that x in A and y in B.
since a is not in A (a, y) can not be in AxB for any y.
9. Sep 19, 2008
### mistermath
Yeah what Diffy said. In other more formal terms:
By definition A x B = { (a,b) | a in A and b in B }
For a given (x, y) to be in A x B, x must be in A and y must be in B, so if (x , y) is not in A x B, then either x not in A or y not in B, or both of course, but we take that for granted :)
10. Sep 19, 2008
### evagelos
Each step in amathematical proof is justified by either a theorem ,an axiom,a definition or a law of logic words like because.......but of course.......obviously.............can you see ..............its easy e.t.c have no place at all.
Things that you may understand or you think you understand other people may not, so how are you going to convince them ?
11. Sep 20, 2008
### HallsofIvy
Staff Emeritus
The justification is the definition of "B x C".
B x C is defined as the set of all pairs (x, y) such that x is in B and y is in C. If a is not in B, then there is no pair in B x C with first member a- i.e. (a, x) is not in B x C for any x.
12. Sep 20, 2008
### HallsofIvy
Staff Emeritus
Which post was this in response to? No one has said any thing like that.
13. Sep 20, 2008
### evagelos
so how do we prove:
~aεB ====> ~(a,x)ε(B x C) ,for all x?
your reasoning is rather intuitive is just like saying can 2+2 be other than 4?
but this can be proved too
14. Sep 20, 2008
### mistermath
Are you ok with the following statement?
If a is not in B, then there is no pair in B x C with first member a.
I would say it's by definition, but suppose there exists a c such that (a, c) in B x C, then
a in B and c in C by definition.
Contradition because we said a not in B by assumption. I bolded the 2 contradictory things.
The c was chosen as any arbitraty element of C, therefore no c in C works.
That's where I got the statement:
(a, x) is not in B x C for any x. (and in particular, the x's in C since x not in C would be "even more" not in B x C).
15. Oct 3, 2008
### evagelos
..........................................correct.....................................................
after all there is a proof by contradiction.
another way to show that ~(a,c)ε(BxC) i.e (a,c)does not belong to BxC IS THE following:
~aεΒ ===> ~aεB v ~cεC ( BY disjunction introduction) ===> ~( aεA^ cεC ) ( BY de morgan) ====> ~(a,c)ε(ΒxC)
16. Oct 3, 2008
### statdad
\begin{align*} (x,y) & \in (A-B) \times C \quad \text{means}\\ x \in A-B, & y \in C \quad \text{so}\\ x \in A, x \not \in B & y \in C \quad \text{so}\\ (x,y) & \in A \times C, (x,y) \not \in B \times C \quad \text{so}\\ (A-B) \times C & \subseteq (A \times C) - (B \times C) \end{align*}
To go the other way
\begin{align*} (x,y) & \in (A \times C) - (B \times C) \quad \text{means}\\ (x,y) \in A \times C, & (x,y) \not \in B \times C \quad \text{so} \\ \left(x \in A \text{ and } y \in C \right) & \text{ and } \left(x \not \in B \text{ and } y \in C\right) \quad \text{so}\\ x \in A-B & \text{ and } y \in C \quad \text{which gives}\\ x \in (A-B) \times C \end{align*}
so that
$$(A \times C) - (B \times C) \subseteq (A-B) \times C$$
These two give the equality of the sets - no need for reference to quantifiers or symbolic logic - just the definitions of Cartesian product, set equality, and subsets.
17. Oct 3, 2008
### evagelos
Nearly all proofs in set theory after dropping quantifiers are based on proofs in symbolic logic .
And in our example this can be shown in the following way.
(A-B)xC= (AxC)-(BxC) <===> {(x,y)ε[(A-B)xC] <====> {(x,y)ε[(AxC)-(BxC)]} <====>
{ [(xεA & ~xεB) & yεC] <=====> [ ( xεA & yεC) & ~(xεB & yεC)]},and if we put now, xεA=P , xεB=q, yεC=r the above becomes.
p & ~q & r <====> p & r & ~( q & r) and this is problem in propositional calculus
Hence a virtuoso in symbolic logic will not miss a proof in set theory,after dropping quantifiers that is
18. Oct 4, 2008
### statdad
"Hence a virtuoso in symbolic logic"
If I ever see anything written by one I'll be impressed.
I am aware (probably more than you) of the relationships. However, I am also aware that people who ask for guidance in one area are typically not interested in showmanship, which is all you seem intent on offering.
If this post is "beyond the pale" as our locale saying goes, I expect the moderators will delete it; I accept their decision in advance.
19. Oct 4, 2008
### evagelos
I am sorry for the sentence "hence a virtuoso in sumbolic................."
My intentions are not showmanship.
But is not a fact that in the heart of the proof of each set problem is symbolic logic??
20. Oct 4, 2008
### statdad
Thank you for your apology - I owe a larger one to you.
There is no doubt that set theory and symbolic logic are deeply intertwined - I stress the similarities when I teach our only class (unfortunately a survey class) that covers both topics.
The point I wanted make (and which, re-reading, bollixed tremendously) is that elegant as it may be, a brief proof of the result written in the language and notation of logic might not provide a glimpse at the mechanics a person studying an introduction to set theory needs.
It wasn't a comment on ability, or elegance, but on suitability at this level.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
Have something to add? | 2017-02-23 16:29:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9845800995826721, "perplexity": 911.8977247740482}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171176.3/warc/CC-MAIN-20170219104611-00243-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/1021775/solving-a-simple-system-of-equations/1021786 | Solving a simple system of equations
Given the simultaneous equations $$A\cos{(\sqrt{\lambda}\pi)} + B\sin{(\sqrt{\lambda}\pi)} = 0$$ $$A\cos{(2\sqrt{\lambda}\pi)}+B\sin{(2\sqrt{\lambda}\pi)} = 0$$ We want to show this has not trivial solutions (ie. solutions when $A\not=0$ and $B\not= 0$). In my notes I have that this gives non-trivial solutions when $$\sin{(2\sqrt{\lambda}\pi)}\cos{(\sqrt{\lambda}\pi)} - \cos{(2\sqrt{\lambda}\pi)}\sin{(\sqrt{\lambda}\pi)} = 0$$ but can't quite see why. Can someone explain, thanks.
• Double-angle formulas look promising... – abiessu Nov 14 '14 at 15:45
• Double-angle formulas would be useful for transforming the last equation into something more informative, but it's not the best way of getting the equation in the first place. – Teepeemm Nov 14 '14 at 20:46
If the system $$Au+Bv=0$$ $$Aw+Bz=0$$ has a non-trivial solution, then $$u=-\frac{Bv}A=\frac{Avw}{Az}=\frac{vw}z$$ Therefore, $$uz-vw=0$$
Remark: The case $z=0$ must be considered in a different but easy way.
A system $$Ax=0$$ has a notrivial solution if and only if $\det A=0$.
If you set up the corresponding augmented coefficient matrix, you'll see that the matrix has determinant $$\left(\sin{(2\sqrt{\lambda}\pi)}\cos{(\sqrt{\lambda}\pi)} - \cos{(2\sqrt{\lambda}\pi)}\sin{(\sqrt{\lambda}\pi)}\right)$$ | 2019-08-18 15:12:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9598197937011719, "perplexity": 220.71336148060936}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313936.42/warc/CC-MAIN-20190818145013-20190818171013-00218.warc.gz"} |