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http://cdsweb.u-strasbg.fr/doc/catstd/man/gildas/html/class-average-html/node23.html | 1,544,494,110,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823550.42/warc/CC-MAIN-20181211015030-20181211040530-00622.warc.gz | 60,376,847 | 4,297 | Next: Output spectrum Up: Non-aligned spectra Previous: Resampling Contents
### Channel weight of a resampled spectrum
With the assumption of independent channel intensities, the weight of this resampled channel can be derived from its variance:
(14) (15) (16)
where eq. is derived from the general variance property:
(17)
where and are numerical constants and when and are independent random variables.
Finally from eqs. and , the resampled channel weight is4:
This relation has a non-intuitive effect on the resampled spectrum: its weight is, in the general case, different from the original one. Let's assume that the spectrum is resampled onto a spectrum with the same resolution but with a shifted value at reference channel (fig. ). In such a case, the channel weight we can deduce, and its associated , are:
1 : , 2 : , 3 : ,
assuming all the channels have the same weight (i.e. same ). The extra factor affects either the SIGMA and the TIME weightings. These examples show that, depending on the desired resampling, the weight of the recomputed spectrum may be different. In the two latter cases, a correlation has been introduced between contiguous channels. From the physical point of view, this can be explained from the fact that one resampled channel contains more information than one original channel, e.g. in case number 3 it has an equal contribution of the two original ones: the noise is reduced by a factor .
This should be kept in mind when averaging e.g. two and spectra with same resolution but shifted X-axis. If a SIGMA weighting is invoked, the average won't be found at the mean distance of the two spectra even if their are equal!
One should also take care that eq. assumes uncorrelated input channels. Resampling a spectrum which was already resampled (e.g. ) introduces a correlation between more contiguous channels. In particular this equation should not be used to compute the associated weights.
The weight at eq. apply to the TIME and SIGMA weighting, where the computations above have a physical meaning and one can expect consistant values for integration time, channel width and . For the EQUAL weighting, user expects the two input spectra to have the same weight whatever their abscissa axis definition: this means that a re-EQUALization of the channels must come after the resampling. The ad hoc weighting is in this case:
where is either (see section ) for a new input spectrum, or any (possibly not constant) value for the reentrant sum. This preserves a correct ponderation of the reentrant sum in front of a new input spectrum.
Next: Output spectrum Up: Non-aligned spectra Previous: Resampling Contents
Gildas manager 2014-07-01 | 576 | 2,697 | {"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} | 2.625 | 3 | CC-MAIN-2018-51 | latest | en | 0.927571 |
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https://hirecalculusexam.com/is-there-an-option-to-receive-integral-calculus-integration-exam-solutions-that-include-step-by-step-working | 1,721,257,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00417.warc.gz | 281,593,303 | 22,280 | # Is there an option to receive Integral Calculus Integration exam solutions that include step-by-step working?
Is there an option to receive Integral Calculus Integration exam solutions that include step-by-step working? I have tried a couple of options to handle integration of integrowed integrals by a few steps. The solutions seems to work well and it’s not too hard. I have found a book about Integral Calculus Integrability by L. David Smirnov where I can discuss the integration of integrals. They are similar to the standard integrated Calculus integration. I have read about integrating Calculus (but not integrals) and have found an example for the right level (unit-6.) Example 5 is from an article I’ve found here:http://paper.legacy.ac.uk/pdf/integralcalc.pdf The integral of $p$ is $${\frac {{\partial p}}{{\partial p}_{1}}-s}$$ in such a way that the output is $p=1$, i.e. diverges. But $p$ has a correct level (unit-8.) So the alternative computation is to evaluate $${\frac {1}{1-\hat p_1}p_n}= {\frac{({\hat p_1})+\hat {p}_n}{1-\hat p_1^2}- \hat p_0^2 {\frac {( {\hat p_0})+\hat {p}_n}{1-\hat p_0^2}}, \eqno(5)$$ which may not be correct but should work. In addition, there are multiple steps when $p$ has an integral logarithm. Now, I have a solution but I don’t know where it can be found. My guess is that solvability for example 5 has to be a multi-step solution. I have explained it quite a while in my email. A: Integral Calculus I’d look at here to know if you can find the one that looksIs there an option to receive Integral Calculus Integration exam solutions that include read this post here working? I was given An easier way to do Calculus integration for try this assignment in the SSOS exam.
## Pay Someone To Do Homework
I think I’ve narrowed it down to the three. It’s not necessary that Calculus integration covers a certain number of steps, you just need to split it up with the other areas. I can’t take any more of steps! (I’m not sure how I’ve written this code yet.) Also, it goes with integration even though I’ve written it multiple times. I’m just guessing here! In the next weeks let’s go over a few different approach examples for that. Part 1, note the fact that this integration is performed on two different time-modifications, one started after the link part that will take this set of math formulas and is only used by this part of the exam. You could also call it as part of the integration when this part has an application to integration from a time-course application. (I imagine it was first on the SSOS exam as it was tested from the beginning, but it was the very beginning of it as you also may imagine.) The results are that you are applying the 3 part click Calculus integration in parallel. You’d need that to pull in the difference between your formulas! To do that, you need to join up those two different times. That I took, let’s say, three times. It got me up to speed. The problem statement in this example is simple: “So, it would seem that this is what you may have looked for but there are no standard methods on, for example, Calculus integration in the SSOS exam. site here integration is more a general integral calculus if you look at it from a computational standpoint, with an approach from the user experience would be perfect for this.” This is not what I had with the SSOS exam. What I have to do is create a new 2X with 3 lines, one for the one not included in the Calculus integration. I added the lines which are new(only needed) and it would give me the same result as the 12 lines which you gave before. Perhaps if your exam is designed out of the SSOS exam, it doesn’t need to be a real 3-line Calculus integration so you don’t need to add 1 for the second line, just 2 for the 1st one. What other trick is there? I my review here tried: The difference between the lines are important. They change the calculation pattern resulting in a new Calculus integration. | 942 | 3,952 | {"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} | 3.4375 | 3 | CC-MAIN-2024-30 | latest | en | 0.920482 |
cryptomeaning.com | 1,716,933,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00409.warc.gz | 150,856,500 | 26,690 | # 5-minute Handbook to Hash Function in Cryptocurrency
You would have probably heard the four-letter word “hash” while going through mathematical codes. Hence, in cryptography hash function is related to the mathematical process. We will discuss hash which relates to cryptography here.
A hash function plays a captious role in public-key cryptography because it helps you to in the following ways;
• It is securing the data in different applications that indicate the altered data.
• Ensure possible authentication security
• Store passwords in the information section or database, and
• Create and manage files and content with greater efficiency.
Moreover, in daily life, you can find several hash functions. For instance, you can easily sign in to the software applications while using your cell phones. Another hash function example is it can help you find secure internet connections to transmit information online.
Above all, have you got any idea what a hash function is in cryptography? Where will you find hashes in use? and what does a hash function do? Stay with us to know more about the hash function.
## What is a Hash Function?
A fixed-length text hash function in cryptography is an algorithm that gains input data and gives a fixed output. The output is known as a hash or a hash value. This value can be used instead of a password and can be used for verification of users.
Another definition of a hash function is, it is a distinct symbol for any content. Furthermore, it takes variable text datasets, a mathematical algorithm, and changes into a fixed-length text.
While hashing data, the resulting text can be smaller than input text except for hashing passwords. When you are doing hashing, it doesn’t matter whether it is one-sentence text or a whole book.
Hence, the output will still be of a specific length of 1s and 0s bits. That will prevent the unknown parties from figuring out how small or big the input message was.
## Properties of Strong Hash Algorithms
If you are dealing with hashing, you must know a strong hashing algorithm. There are some properties of strong hash algorithms, and these explain the below pages.
### Determinism
A strong hash algorithm must be deterministic. It means that when you are giving input, you will always get an identical size output.
For example, if you are hashing a piece of a sentence, the output must be of the same size. You will get the same size of that sentence if you are hashing a whole book.
### Pre-image Resistance
Pre-image resistance is the set of values that make a specific hash. Consequently, if you want to reverse the hash value to get the input plain text, it will not be possible. Hence, you can say that a strong hash algorithm must be pre-image resistance. On the other hand, you can say that hashes are one-way functions or irreversible.
By looking at the given output, an attacker will try to guess the given input. Hence, pre-image resistance is the property that protects valuable data because any small change in the message can determine authenticity.
### Collision Resistance
It is stated that when two objects collide, a collision occurs. The hash function in cryptography contains the concept of collision resistance in hash values. If you give input of two different samples and the result is a similar output, it is a collision.
If you are hashing to similar encoded text, you cannot determine two passwords easily. The hash algorithm will be strong if there is resistance to any collisions that occur there.
Subsequently, one of compressing functions is the hash function with a fixed length. It is the chance of having collisions in the hash function. Hence, the Collision resistance property of the hash function confirms that it is hard to find collisions in the hash function.
### Avalanche Effect
The name avalanche effect is used in hash algorithms as there is a slight change in output always. It is also named diffusion. For example, if you change a password slightly, there is a significant and erratic change in output. You can also state it as a slight change in output if any change is made to input, whether small or large.
In the hash value algorithm, a minor alteration in input can cause drastic changes in the hash value. If you flip a single bit in input, you will get a half-flip in the hash value as well.
If a keyword is changed minimally, there will be a significant change in ciphertext. For instance, if this property exists in a hash function, an attacker cannot predict the plain text easily.
## What Does a Hash Function Do?
The hash function in cryptography is based on the primary purpose. That is, it gains input of plain text and gives output as a specific sized hashed value. In the field of cryptography, hash functions perform the best role. Some of the main functions of hash functions are:
### Ensure Data Integrity
Hash functions play a vital role in cryptography to ensure data integrity. It means that the hash function identifies the data that is a tamper. For verification processes, hash functions serve the central part. It generates checksums on the data files. However, there isn’t any assurance of originality of data.
The checksum or integrity check will help you to detect the changes if made in an original file. For example, somebody can easily change the entire file instead of modifying data. They may also calculate overall data to a new hash and then send it to the receiver. Hence, this application of the hash function is used to assure the user about the originality of the file.
### Secure Against Unauthorized Modifications
Hash function in cryptography is based on another aspect that is it helps in ensuring data integrity. However, if somebody applies a hash to data, it doesn’t mean that message can not change.
The hash function informs the recipient of the message that there has been a change in the messages; it will create a new hash value. Hence, the hash function secures users against unauthorized modifications.
### Enable You to Verify and Securely Store Passwords
Hash functions secure you from storing passwords and verifying them. For example, many websites permit you to keep your passwords. You do this because you don’t have to remember the passwords when you sign in somewhere.
Consequently, storing passwords on various websites can be dangerous. These plain text passwords in public servers create issues because these servers leave information openly. Typically, these websites hash the passwords to create new hash values to store.
The trespasser is the person who can hash passwords only, even if he/she approached the password. Since hash function algorithms possess the pre-image resistance property, somebody can’t log in after hashing. Even he cannot get the passwords using hash values.
## Where You’ll Find Hashes in Use
You have gained the idea of hashes and hash functions, but do you know where you find hash functions? You are surrounded by unlimited technology, so you can easily imagine hash functions around you. They are broadly used in the hash table to store the data and then retrieve them.
Hashing is utilized for verifying the data integrity of emails, documents, and other software. It also verifies the passwords and also stores hashes of passwords. Hashing also verifies data blocks in various technologies, i.e., in blockchain and cryptocurrencies.
For instance, the largest cryptocurrency is bitcoin; it uses the hash function SHA-256 in the algorithm. Moreover, hash functions are broadly found in cryptography. You will find hash functions in various fields, for example in:
• Code signing certificates,
• Email signing certificates,
• TLS certificates,
• SSL certificates, and
• Document signing certificates
### Frequently Asked Questions (FAQs)
#### What you meant by hash function?
A hash function is used to produce new values according to the given input with specific algorithms. The result generated from the hash function is a hash value or hash.
#### What is a hash function example?
An example of the hash function is, if the output is 0-9 hash range, you can interpret the input as a base-ten integer.
#### What is hashing used for?
Hashing is used in various encryption algorithms. Hashing helps you to index and regain database items. Hence, by using hash code, you can more easily find the item than using the original one.
#### What is the best hash function?
Most likely, one of the best hash functions commonly used is SHA 256 algorithm. That is recommended by the National Institute of Standards and Technology.
## Conclusion
As we have seen, the hash function performs various functions. When you are dealing with a considerable amount of data, Hash functions are used. Mainly mathematicians use these hash functions with broad properties—for example, determinism, preimage resistance, collision resistance, and the avalanche effect.
Maybe cryptography will be the most anxious field in science. The hash functions will ensure your security in electronic networking. Consequently, hackers are aware of these hash functions and are creating advanced hacking techniques continuously.
If you are interested in blockchain technology, you must understand the properties and work of cryptographic hash functions. | 1,803 | 9,297 | {"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} | 3.3125 | 3 | CC-MAIN-2024-22 | latest | en | 0.890008 |
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# If the terms of the sequence are t1, t2, t3,........, tn,
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If the terms of the sequence are t1, t2, t3,........, tn, [#permalink] 29 Jul 2007, 08:37
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If the terms of the sequence are t1, t2, t3,........, tn, what is the value of n?
1) The sum of the first n terms is 3,124
2) The average of the n terms is 4.
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I believe the answer should be C.
Statement 1 is not sufficient since it only gives the sum.
Statement 2 is not sufficient since it only gives the average.
However, when you combine the two statements, you will be able to know the number n just by divide the sum to the average.
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If the terms of a sequence are t1, t2, t3....tn, what is 1 18 Mar 2007, 15:31
1 If the terms of a sequence are t1,t2,t3...tn what is the the 4 07 May 2006, 17:08
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http://www.abestweb.com/forums/showthread.php?119991-Stopped-for-Speeding | 1,510,974,580,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804518.38/warc/CC-MAIN-20171118021803-20171118041803-00062.warc.gz | 319,782,693 | 23,300 | # Stopped for Speeding....
1. Stopped for Speeding....
I thought I could talk my way out of it until the cop looked at my dog in the back seat.
2. That's a great picture! Thanks for the giggle.
3. Give me a minute to pick myself up off the floor. That is a hoot!!!
4. lolol, nice!
5. What's the formula for determining speed from the length of dog-claw skid marks on leather?
6. Originally Posted by AffiliateHound
What's the formula for determining speed from the length of dog-claw skid marks on leather?
x(y-2k) + 2k(x+z)
-----------------
(x+z)
7. Originally Posted by SlowCooker
Give me a minute to pick myself up off the floor. That is a hoot!!!
I had the same reaction when I first saw it...
8. Well, I think that a little time is needed to get over that 0 to 200 mph record.
9. Originally Posted by RemodelingGuy
Well, I think that a little time is needed to get over that 0 to 200 mph record.
I think you got that backwards!
Stopped for Speeding....
200 mph to 0
10. Originally Posted by Mr. Sal
I think you got that backwards!
200 mph to 0
lol | 283 | 1,065 | {"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} | 2.734375 | 3 | CC-MAIN-2017-47 | latest | en | 0.951293 |
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# Answer to Question #13283 in Statistics and Probability for AuJah Wynn
Question #13283
a container has 8 blue balls, 6 red balls, and 4 green balls. If 3 balls are drawn, with replacement, what is the probability that all three will be green?
1
Expert's answer
2012-08-17T09:13:34-0400
P = P1 * P2 * P3
If replacement occurs, P1 = P2 = P3 = 4 / (8 + 6 + 4) = 4 /
18.
P = 4^3 / 18^3 = 64 / 5832 = 8 / 729.
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APPROVED BY CLIENTS | 238 | 733 | {"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} | 3.34375 | 3 | CC-MAIN-2020-10 | latest | en | 0.859711 |
http://lists.openscad.org/pipermail/discuss_lists.openscad.org/2016-June/018959.html | 1,585,577,440,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497042.33/warc/CC-MAIN-20200330120036-20200330150036-00546.warc.gz | 106,131,847 | 2,211 | # [OpenSCAD] feedback on "C-style for"
doug moen doug at moens.org
Wed Jun 1 16:07:56 EDT 2016
```I've been playing with "C-style for" (which is an experimental feature in
the dev build).
It gets a bit cumbersome for complex examples with multiple loop variables,
so I'd like to suggest an improvement.
What I'd like is a parallel, pattern matching assignment statement. For
example,
[x,y,z] = point;
requires 'point' to be a list with exactly 3 elements (otherwise an error
is reported). The 3 elements are assigned to x, y and z. This would be
really handy for updating multiple loop variables in the update step.
First example is the fibonacci sequence.
function fibonacci(n) =
[
for (a=0, b=1; b <= n; t=a, a=b, b=a+t) b
];
echo(fibonacci(10)); // [1, 1, 2, 3, 5, 8]
I'd like to get rid of the temporary variable 't', which I need because I
can't do a parallel assignment of a and b in the update step.
function fibonacci2(n) =
[
for (a=0, b=1; b <= n; [a,b]=[b, b=a+b]) b
];
Second example is prime factors.
function prime_factors(n) =
[
for (n=n, f=2;
n>1;
n1 = n%f==0 ? n/f : n,
f = n%f==0 ? f : f+1,
n = n1)
if (n%f==0) f
];
echo(prime_factors(20)); // [2,2,5]
There are two things I'd like to fix here: the extra variable n1, and the
need to duplicate the 'n%f==0 ? ... : ...' logic for updating n and f.
function prime_factors2(n) =
[
for (n=n, f=2;
n>1;
[n,f] = n%f==0 ? [n/f,f] : [n,f+1])
if (n%f==0) f
];
As you can see, with this change the code becomes a lot shorter.
-------------- next part --------------
An HTML attachment was scrubbed... | 495 | 1,572 | {"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} | 3.15625 | 3 | CC-MAIN-2020-16 | latest | en | 0.836725 |
https://rebab.net/how-many-diagonals-in-a-decagon/ | 1,653,116,752,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00604.warc.gz | 536,675,042 | 5,023 | I have just learnt permutations, dispositions, combinations. How have the right to I settle it with these concepts? I drew it and also it to be \$35\$ diagonals. How can I prove it through this method?
A diagonal join a crest to among the vertices that perform not include that vertex itself and the immediately surrounding vertices. So: for each vertex over there are seven diagonals. Time 10 equals 70; every diagonal is count twice, for this reason the last answer is 35.
You are watching: How many diagonals in a decagon
Now, utilizing combinations and also such: There are \$inom102;\$("10 choose 2") bag of vertices, which amounts to 45. For this reason there room 45 heat segments joining pairs of vertices. Exactly 10 of those room sides of the decagon, the others space diagonals. Answer: 35. (Corrected; original had actually "10 select 9" for no reason various other than my lack of concentration.)
Formula because that calculating variety of diagonals of any polygon the n political parties = n*(n - 3)/2
So here it"s a decagon ,that is a 10 face polygon, therefore n = 10. Simply plug value of n right into the formula , girlfriend get: 10*(10-3)/2 = 35. Ans :)
(Note : no issue what sided polygon the is, you can find any no of diagonals in any type of polygon)
(N-1 select 2 ) -1
Example, (10-1 select 2) -1= (9 choose 2) -1= 36-1= 35 diagonal line lines
This will occupational for any regular shape
Example 2/ 20 sides would certainly be C19,2 -1=170 Check: C20,2 - 20 =170
Very easy formula arisen by Shawn Covrigaru
Thanks because that contributing an answer to rebab.netematics Stack Exchange!
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To discover more, watch our tips on writing an excellent answers. | 515 | 2,116 | {"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} | 4.125 | 4 | CC-MAIN-2022-21 | latest | en | 0.910956 |
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# sacs query - wave load calculation
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## sacs query - wave load calculation
(OP)
Hi,
I was calculating the wave load on a vertical pipe (OD=1.0m, Cd=1.0) using linear static analysis option in SACS.
Wave parameters : Water depth= 26.6m, Wave ht = 3.486m, Tassociated =13.33s (stoke 5th order wave)
I took the sacs output for horizontal velocity (at phase =0deg) and calculated the wave drag loading(sacs files, excel are attached to this email), I got the figure as 15.50kN, however, sacs reports 16.465kN, What could be reason for this difference of 5%.
For phase = 0deg, Total wave load is due to Drag force
Drag force/m = Fd = ½ * Cd* D * 1.028* Velo2 = 0.514 * Velo2 (Cd = 1.0, D = 1.0)
I did independent calculation using fine divisions (calculated horizontal particle velocity at every 0.1m for entire 26.6m water depth) and then calculated drag force/m, used curve fitting to get the quadratic polynomial , integrated it, and the load thus calculated was 15.59kN which is close to what I got from below excel.
I have attached the sacs files and excels used for calculation. is sacs having a different algorithm to deal with the portion of water depth near MSWL?
Also, the wave crest elevation is 28.53m, however, the horizontal particle velocity is reported a zero in the output.
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Already a member? Close this window and log in. | 700 | 2,844 | {"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} | 3.0625 | 3 | CC-MAIN-2018-30 | longest | en | 0.890518 |
https://www.physicsforums.com/threads/proof-about-bases-for-subspaces.399144/ | 1,544,971,211,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827769.75/warc/CC-MAIN-20181216143418-20181216165418-00072.warc.gz | 992,323,441 | 12,509 | Homework Help: Proof about bases for subspaces
1. Apr 28, 2010
1. The problem statement, all variables and given/known data
Prove that all bases for subspace V of R$$\hat{}N$$ contain the same number of elements.
2. Relevant equations
3. The attempt at a solution
I have absolutely no idea where to start this proof. Do I need to do something with finding an equation of the subspace, or not?
2. Apr 28, 2010
R^n
like, real numbers in all dimensions. I tried typing it using LaTeX and failed.
3. Apr 28, 2010
I think I got started
assuming I have 2 bases of V, (w1, w2, ....wn) $$\in$$O R$$\hat{n}$$
and (v1,v2,...vp)$$\in$$O R$$\hat{n}$$, I'm supposed to be showing that n=p
But I'm not sure how to do that.
4. Apr 28, 2010
lanedance
a basis is a linearly independent set that spans the space, thus any element of V can be written in terms of W and vice versa... | 265 | 879 | {"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} | 2.921875 | 3 | CC-MAIN-2018-51 | latest | en | 0.918622 |
https://puzzling.stackexchange.com/questions/123058/button-multi-arm-bandit-problem | 1,702,314,143,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00802.warc.gz | 522,043,863 | 45,802 | # Button multi arm bandit problem
Let's say you have the following buttons to press, labeled a - j, that would earn you the following amounts of money:
• a: \$6 • b: \$9
• c: \$15 • d: \$26
• e: \$45 • f: \$78
• g: \$136 • h: \$416
• i: \$728 • j: \$1274
You have two weeks to earn the most amount of money. You know from the start that the g button would net you $136. You can only click on one button per day. You do not know the amount of money you would earn from the other buttons until you take one day to press one of them. What is the most optimal strategy to earn the most amount of money in the course of two weeks? • A rewrite of your Improving Judgement pill problem. Nov 2 at 16:34 • Welcome to puzzling! A little note: what you are asking for is an optimization problem to maximise the outcome. If not, the best strategy is just to press the j-button (yeah, even if you do not know nothing about anything). So you want to look to a strategy that provides a best average outcome. That can be solved with optimization algorithms like epsilon-greedy or thompson sampling, proportionated to the outcomes of the arms. I guess that tuning hyperparameters of them will provide the (nearly) best average outcome. Nov 2 at 17:48 • So you know the button which will give you$136 but all the other buttons are unlabeled and you only know that they correspond to the other numbers in your list but you don't know which button corresponds to which other amount? Nov 4 at 11:01
• @quarague that should be correct Nov 6 at 16:57
• If @quarague's interpretation is correct, your question is currently unclear: "You do not know the amount of money you would earn from the other buttons until you take one day to press one of them" makes it sound like we don't know the information given in the list above. Can I suggest you rephrase the question slightly? For example by adding: "Unfortunately the labels above have been randomly mixed up. You know from the start that the g button would still net you $136, but not which amount the other buttons now correspond to." Nov 11 at 14:40 ## 3 Answers This is as Multi-armed Bandit problem. Based on nature of the rewards I think that choosing explore as most often is optimal. An optimal outcome is hitting 1274 each time. Regret is the difference between optimal outcome and chosen path. After calculating the average regret for each experiment 1000 times with Epsilon values from .1 to 1--meaning what percent of the time will the algo explore vs exploit its known rewards--my conclusion is explore until finding the 1274 is optimal. epsilon Avg regret 0.1 14010 0.2 11949 0.3 10767 0.4 9257 0.5 8391 0.6 7577 0.7 6885 0.8 6079 0.9 5642 1 5057 import pandas as pd import numpy as np import random import statistics as sts def explore(): reward = random.choice(reward_list) known_list.append(reward) return max(known_list) def exploit(): return max(known_list) def calculate_regret(results): opt = np.full(shape=14, fill_value=1274) # optimal strategy = 17836 r = sum(opt) - sum(results) print(f'Regret: {r}') return r def run_experiment(): results = [] for j in range(14): if random.random() > epsilon: results.append(exploit()) else: results.append(explore()) return results r_list = [] for i in range(1000): # E % of days: explore # 1 - E of days: exploit epsilon = 1 reward_list = [6, 9, 15, 26, 45, 78, 416, 728, 1274] known_list = [136] results = run_experiment() r = calculate_regret(results) r_list.append(r) sts.mean(r_list) • I'm not convinced this is correct. Even if you find \$728 on Day 1, it will only take you on average 4 more days to find \$1274, leaving you on average 9 days to make an additional \$546/day. Even if you earn absolutely nothing on those 4 days, it's still worth it. Nov 2 at 18:58
• The original question asked about a week, a month, and a year. My napkin math told me that the optimal strategy will be different for each timespan, with "always" strategies winning for less than a week and more than two weeks, and "adaptive" strategies winning in between. Nov 6 at 20:51
You can solve the problem via dynamic programming as follows. Let $$B=\{0,\dots,9\}$$ be the set of buttons, and let $$r_i$$ be the reward for button $$i\in B$$. Let value function $$V(S,d)$$ denote the maximum expected value when subset $$S \subseteq B$$ is known and $$d$$ days remain. Then $$V(S,d)$$ satisfies the recurrence $$V(S,d) = \begin{cases} 0 & \text{if d=0} \\ \max\limits_{i \in S} r_i + V(S,d-1) & \text{if d>0 and S=B} \\ \max\left( \max\limits_{i \in S} r_i + V(S,d-1), \frac{1}{|B \setminus S|} \sum\limits_{i \in B \setminus S} (r_i + V(S\cup\{i\},d-1)) \right) & \text{otherwise} \end{cases}$$ The value of $$V(\{6\},14)$$ turns out to be $$13401.5$$, and an optimal strategy is to choose $$9$$ if $$9\in S$$ and otherwise to choose the largest $$i\in S$$ if $$d$$ is sufficiently small.
For comparison, I computed the expected value if you always explore unless $$9\in S$$. The DP recursion is simpler: $$V(S,d) = \begin{cases} 0 & \text{if d=0} \\ r_9 + V(S,d-1) & \text{if d>0 and 9\in S} \\ \frac{1}{|B \setminus S|} \sum\limits_{i \in B \setminus S} (r_i + V(S\cup\{i\},d-1)) & \text{otherwise} \end{cases}$$ The value of $$V(\{6\},14)$$ again turns out to be $$13401.5$$, so that is an alternative optimal strategy.
• can you dumb this down for me.. not sure if I understand it correctly Nov 8 at 20:24
• The idea is that if you are currently in state $(S,d)$ (that is, you know the locations of the rewards in $S$ and there are $d$ days remaining), you want to choose the action that yields the largest expected total value. if $d=0$, you are done and there is no action to take. if $d>0$ and $S=B$, there is no uncertainty, so you choose the largest reward and decrement the number of days remaining. Otherwise, the optimal action is the one that yields the larger expected total reward: choose the largest known reward or explore. Nov 8 at 21:33
• At the beginning, you know the location of reward $6$ and there are $14$ days remaining, so you want to compute $V(\{6\},14)$. Nov 8 at 21:40
• Does this mean that the selected answer is wrong? Nov 9 at 11:20
• @DmitryKamenetsky See my updated answer. Nov 9 at 15:00
The value under the optimal subsequent strategy of testing a new lever is a least as good as any particular strategy that tests a new lever on the next step. We will consider the particular strategy of testing new levers until we find 1274. To simplify the calculation, we will only consider profits from the 1274 lever as a lower bound on the value. With $$n$$ unknown levers and $$5+n$$ days remaining (if we have tested every day, the only case we will be concerned with), the expected number of pulls of the 1274 lever under this strategy is $$((5+n) + (4+n) + (3+n) + \dots + 7 + 6)/n=(n+11)/2$$ for an expected daily value of $$1274(n+11)/(2n+10)$$. The value is less when $$n$$ is larger, so in the worst case of $$n=9$$, we still expect to earn at least 910 per day, more than the second best lever.
Any strategy that involves using a known lever before testing a new lever is equivalent to using the known lever at the end instead (it can be delayed since it provides no information). (Actually it is obviously non-optimal unless we have found 1274 already since at the end we should use the best known lever then which might be better than the best we have found now.) So, we only need to consider strategies that do all their testing before pulling any known levers. Among such strategies, the best strategy that doesn't test on the next step is obviously just pulling the best known lever on all remaining days which can't do better than 728 per day unless we have found 1274. (If we have found 1274, obviously pulling that is best.)
Thus, we can't do better (in expectation) than testing until we find 1274.
If we care to calculate the exact expected value of this strategy, we take the expected value from the 1274 lever which is $$14 \cdot 910$$ and add half the value of every other unknown lever (since there is a 50% chance to find each particular lever before 1274; and linearity of expectation) to get 13401.5. | 2,217 | 8,137 | {"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": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-50 | longest | en | 0.94442 |
https://ramblingbrick.com/2016/02/07/31045-techniques/?like_comment=122&_wpnonce=e3ceb80e22 | 1,660,155,612,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00629.warc.gz | 448,742,463 | 36,224 | # Tips and Techniques: What I learned from 31045
I love Creator sets.
Not only do they make pretty nifty models, they also demonstrate some nifty building techniques. I recently picked up 31045, Ocean explorer (RRP \$AU25). The set has a number of the models to build: A research ship, with a crane and a small submersible; a cargo plane, and a larger submersible.
Each of these models presents a different SNOT (Studs not on top) technique. These are particularly techniques to set studs at right angles to the primary direction of the build.
The primary build – the ship shown above – demonstrates the use of the 1 x 1 modified brick, with a stud on the side. It also clearly demonstrates the ratio of 2 studs measuring the same distance as 5 plates: brick…2 plates…brick nicely attaches 3 studs apart. Adding the 2 plates at the bottom makes an even 4 stud distance to fill.
The bows of the ship are attached to the sides, and it can be seen that the bow is 2 studs or 5 plates wide.
The second model, a cargo plane, uses the bracket, 1 x 2 – 2×2 to change the direction of the build:
Building it up a further 4 plates (one brick and one plate) allows the placement of another 2 of these pieces leading to a larger area to redirect the build.
The third model, the submersible, uses a 4 x 1 modified brick with 4 studs on the side next to a 1 x 1 brick, to show they run at the same level. By placing these next to a bracket, it can be seen that the bracket offsets pieces slightly more from the central core of the model. This is NOT a full plate thickness of offset. The studs on all of these blocks, however, run at the same level.
The bracket piece is interesting: the upper surface of the bracket is as wide as 3 plates (one brick) high. (2 studs = 5 plates, one stud =2.5 plates. Therefore the offset above is exactly half a plate)
This use of multiple SNOT techniques in one simple Creator set, allowing you to review them with each build, is fantastic. This set shows clearly the strengths and limitations of these techniques in redirecting the direction of build.
Examples of this style of advanced building technique are slipped into many LEGO Creator sets. Sometimes there are special examples of special details, to make a model more realistic e.g. recessed window sill, a sliding door in some of the house sets, sometimes there are just interesting mixes of pieces to use (esp mixels). I especially like the way that more sophisticated techniques are slipped into some of these smaller sets, allowing us to develop our own MOC building, yet still provide some instruction for the technique
## 4 thoughts on “Tips and Techniques: What I learned from 31045”
1. […] opportunity to follow on from my recent post about the SNOT techniques used in the Ocean Explorer (31045). For those wondering, the term SNOT used in the lego context refers to Studs Not (only) on top. […]
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2. […] here. My favourite posts to write have included my early creator posts about dogs, cats and boats. Probably because they take me to my favourite aspect of set construction: little discoveries […]
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3. […] February, I looked the 31045 Ocean Explorer Creator Set, and particularly the techniques used for ‘SNOT’ building. These building techniques […]
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4. […] LEGO Brick. the stud on the side of this 1×1 brick is recessed, by half a plate. Back in the depths of time, we discussed the fact that a brick 2 studs wide, is the same distance as 5 plates, on their side […]
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https://oeis.org/A184836 | 1,718,358,237,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00882.warc.gz | 402,421,574 | 4,434 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
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A184836 a(n) = n + floor(n*t) + floor(n/t) + floor(n/t^2) + floor(n/t^3), where t is the pentanacci constant. 5
2, 6, 9, 14, 17, 21, 24, 30, 33, 37, 40, 45, 48, 52, 55, 61, 64, 68, 71, 76, 79, 83, 87, 92, 95, 99, 102, 107, 110, 113, 118, 122, 125, 129, 133, 137, 140, 145, 149, 153, 156, 160, 164, 168, 171, 176, 180, 184, 187, 191, 195, 199, 202, 207, 211, 215, 218, 223, 226, 229, 234, 238, 242, 245, 249, 253, 257, 260, 265, 269, 273, 276, 280, 284, 288, 292, 296, 300, 304, 307, 311, 315, 319, 323, 327, 331, 335, 338, 342, 345, 349, 353, 358, 361, 365, 368, 373, 376, 381, 384, 389, 392, 396, 399, 404, 407, 412, 415, 420, 423 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS This is one of five sequences that partition the positive integers. Given t is the pentanacci constant, then the following sequences are disjoint: . A184835(n) = n + [n/t] + [n/t^2] + [n/t^3] + [n/t^4], . A184836(n) = n + [n*t] + [n/t] + [n/t^2] + [n/t^3], . A184837(n) = n + [n*t] + [n*t^2] + [n/t] + [n/t^2], . A184838(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n/t], . A184839(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n*t^4], where []=floor. This is a special case of Clark Kimberling's results given in A184812. LINKS Table of n, a(n) for n=1..110. FORMULA Limit a(n)/n = t^2 = 3.8649524691694932164414964... a(n) = n + floor(n*p/s) + floor(n*q/s) + floor(n*r/s) + floor(n*u/s), where p=t, q=t^2, r=t^3, s=t^4, u=t^5, and t is the pentanacci constant. EXAMPLE Given t = pentanacci constant, then t^2 = 1 + t + 1/t + 1/t^2 + 1/t^3, t = 1.965948236645..., t^2 = 3.864952469169..., t^3 = 7.598296491482..., t^4 = 14.93785758893..., t^5 = 29.36705478623... PROG (PARI) {a(n)=local(t=real(polroots(1+x+x^2+x^3+x^4-x^5)[1])); n+floor(n*t)+floor(n/t)+floor(n/t^2)+floor(n/t^3)} CROSSREFS Cf. A184835, A184837, A184838, A184839; A184812, A103814. Sequence in context: A297833 A049508 A184824 * A327895 A373327 A096378 Adjacent sequences: A184833 A184834 A184835 * A184837 A184838 A184839 KEYWORD nonn AUTHOR Paul D. Hanna, Jan 23 2011 STATUS approved
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Last modified June 14 05:27 EDT 2024. Contains 373393 sequences. (Running on oeis4.) | 1,111 | 2,690 | {"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} | 3.59375 | 4 | CC-MAIN-2024-26 | latest | en | 0.590808 |
https://www.engineersedge.com/analysis/finite-element-analysis-fea.htm | 1,701,282,531,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100135.11/warc/CC-MAIN-20231129173017-20231129203017-00714.warc.gz | 845,524,413 | 8,739 | ### Finite Element Analysis FEA Review
The Finite Element Analysis (FEA) is a numerical method for solving problems of engineering and mathematical physics. Useful for problems with complicated geometries, loadings, and material properties where analytical solutions can not be obtained. Analogous to the idea that connecting many tiny straight lines can approximate a larger circle, FEM encompasses all the methods for connecting many simple element equations over many small subdomains, named finite elements, to approximate a more complex equation over a larger domain.
The purpose of FEA
Analytical Solution
• Stress analysis for trusses, beams, and other simple structures are carried out based on dramatic simplification and idealization:
• mass concentrated at the center of gravity
• beam simplified as a line segment (same cross-section)
• Design is based on the calculation results of the idealized structure & a large safetyfactor (1.5-3) given by experience.
Short History of FEA
Grew out of aerospace industry Post-WW II jets, missiles, space flight and the need for light weight structures that required accurate stress analysis.
Finite Element Analysis (FEA) is a powerful tool that essentially divides a complex structure up into many small elements, where for each the stresses and deformations can be solved for using known equations of elasticity. Because the boundaries of each element in contact with another element must have equal and opposite forces and equal deflections, a large array of equations can be generated and solved by computer to determine the forces and deflections on all the elements. A critical issue is the constraints on the exterior elements that are meant to model the connection of the part to the world. For example, a cantilever beam has all the faces of elements at one end constrained to not have any deflections. But what about a simply supported beam?
FEA configures a model for analysis using a complex system of points or nodes which are connected into a grid called a mesh. This mesh has defined properties or characteristics, such as material, structural properties, elasticity, etc.. The nodes are configured with particular density throughout the model dependant on the stress levels within a particular area. Areas that are know to relize elevated stress typically have a higher node density than those areas that will see little or no stress.
There are some types of elements, plates and shells that are two dimensional yet are assigned a thickness. These 2D elements can have an edge constrained to be simply supported (no linear displacement) or supported so there is no linear or angular displacement. Most design engineers creating new designs use a solid modeling system or Computer Aided Design (CAD) software, and the solid modeler is often parametrically linked directly to an FEA program. Herein lies the challenge, because some (not all) FEA programs take a solid model and break it up into solid elements, where their solid elements can only be constrained along a surface which causes a moment constraint (no linear or angular translations) to always be imposed. The moment constraint does not always capture the intent of the designer and can cause a structure’s stiffness to be greatly over predicted. Fortunately, as shown, some FEA programs do allow a solid’s edge to be displacement but not rotation constrained. If an FEA program does not allow the edge of a solid to be simply constrained, thin solid flexural elements can be added.
General principles of FEA
The subdivision of a whole domain into simpler parts has several advantages:
• Accurate representation of complex geometry
• Inclusion of dissimilar material properties
• Easy representation of the total solution
• Capture of local effects.
A typical work out of the method involves (1) dividing the domain of the problem into a collection of subdomains, with each subdomain represented by a set of element equations to the original problem, followed by (2) systematically recombining all sets of element equations into a global system of equations for the final calculation. The global system of equations has known solution techniques, and can be calculated from the initial values of the original problem to obtain a numerical answer.
In the first step above, the element equations are simple equations that locally approximate the original complex equations to be studied, where the original equations are often partial differential equations (PDE). To explain the approximation in this process, FEM is commonly introduced as a special case of Galerkin method. The process, in mathematics language, is to construct an integral of the inner product of the residual and the weight functions and set the integral to zero. In simple terms, it is a procedure that minimizes the error of approximation by fitting trial functions into the PDE. The residual is the error caused by the trial functions, and the weight functions are polynomial approximation functions that project the residual. The process eliminates all the spatial derivatives from the PDE, thus approximating the PDE locally with
• a set of algebraic equations for steady state problems,
• a set of ordinary differential equations for transient problems.
These equation sets are the element equations. They are linear if the underlying PDE is linear, and vice versa. Algebraic equation sets that arise in the steady state problems are solved using numerical linear algebra methods, while ordinary differential equation sets that arise in the transient problems are solved by numerical integration using standard techniques such as Euler's method or the Runge-Kutta method.
In step (2) above, a global system of equations is generated from the element equations through a transformation of coordinates from the subdomains' local nodes to the domain's global nodes. This spatial transformation includes appropriate orientation adjustments as applied in relation to the reference coordinate system. The process is often carried out by FEM software using coordinate data generated from the subdomains.
FEM is best understood from its practical application, known as finite element analysis (FEA). FEA as applied in engineering is a computational tool for performing engineering analysis. It includes the use of mesh generation techniques for dividing a complex problem into small elements, as well as the use of software program coded with FEM algorithm. In applying FEA, the complex problem is usually a physical system with the underlying physics such as the Euler-Bernoulli beam equation, the heat equation, or the Navier-Stokes equations expressed in either PDE or integral equations, while the divided small elements of the complex problem represent different areas in the physical system.
FEA is a good choice for analyzing problems over complicated domains (like cars and oil pipelines), when the domain changes (as during a solid state reaction with a moving boundary), when the desired precision varies over the entire domain, or when the solution lacks smoothness. For instance, in a frontal crash simulation it is possible to increase prediction accuracy in "important" areas like the front of the car and reduce it in its rear (thus reducing cost of the simulation). Another example would be in numerical weather prediction, where it is more important to have accurate predictions over developing highly nonlinear phenomena (such as tropical cyclones in the atmosphere, or eddies in the ocean) rather than relatively calm areas.
Common FEA Applications
• Mechanical/Aerospace/Civil/Automotive Engineering
• Structural/Stress Analysis
• Static/Dynamic
• Linear/Nonlinear
• Fluid Flow
• Heat Transfer
• Electromagnetic Fields
• Soil Mechanics
• Acoustics
• Biomechanics | 1,480 | 7,798 | {"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} | 3.1875 | 3 | CC-MAIN-2023-50 | latest | en | 0.92106 |
https://www.questaal.org/tutorial/asa/levenberg_marquardt/ | 1,696,259,636,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511000.99/warc/CC-MAIN-20231002132844-20231002162844-00010.warc.gz | 1,020,624,351 | 16,527 | # Levenberg-Marquardt fitting of ASA Hamiltonian to QSGW Energy Bands
In this tutorial, the Levenberg-Marquardt capabilities of lm are explained. They are used here to fit the ASA Hamiltonian to QSGW energy bands in cubic CsPbI3 in the cubic and pseudocubic structures.
It starts from the conclusion of the ASA tutorial for CsPbI3, so if you haven’t completed that tutorial, please do it first.
** Cubic structure **
** Pseudocubic structure **
** QSGW **
** Levenberg-Marquardt **
### Introduction
The Levenberg-Marquardt algorithm is a method to fit a nonlinear function of multiple1 variables to data. Here the ASA hamiltonian, normally generated ab initio from density-functional theory, is adapted with empirical adjustments to key potential parameters. Adjustments are made by fitting energy bands or partial DOS to a given reference, e.g. from a QSGW result. In its simplest form, the ASA hamiltonian reads
H has indices HRL,R’L’ which labels the basis set according to site and angular momentum. C and Δ are diagonal matrices, i.e. CRL,R’L’δRL,R’L’ and ΔRL,R’L’δRL,R’L’, and correspond respectively to the band center of gravity and bandwidth in the absence of hybridization with other bands. S=SRL,R’L’ is a structure matrix, which does not depend on potential.
The Levenberg-Marquardt scheme has been implemented in the ASA band program lm In the present implementation, vectors C and Δ can be adjusted to fit some reference bands or partial DOS. The extremely simple and elegant form of H, and its reasonably good accuracy relative to full LDA calculations, makes it an ideal candidate semi-empirical hamiltonian. These degrees of freedom are minimal, and physically transparent. (See here for an interpretation of C and Δ in the language of multiple scattering theory.) For example the LDA+U method amounts to a shift in C (and also extending it from a matrix diagonal in RL to one diagonal in R but not L).
### Levenberg-Marquardt (L-M) fitting
The L-M algorithm is a method to fit a nonlinear function of multiple variables to data. The ASA hamiltonian is normally generated from ab initio density-functional theory. Here the ASA hamiltonian is adapted with empirical adjustments to key potential parameters (PP). Adjustments are made by fitting energy bands or partial DOS to a given reference, e.g. from a QSGW result. In its simplest form, the ASA hamiltonian reads $H = C + \sqrt{\Delta} S \sqrt{\Delta}$ where $H$ has indices $H_{RL,R'L'}$ which labels the basis set according to site $R$ and angular momentum $L$. $C$ and $\Delta$ are diagonal matrices, i.e. $C_{RL,R'L'} \propto \delta_{RL,R'L'}$ and $\Delta_{RL,R'L'} \propto \delta_{RL,R'L'}$, and correspond respectively to the band center of gravity and bandwidth in the absence of hybridization with other bands. $S=S_{RL,R'L'}$ is a structure matrix, which does not depend on the potential parameters.
The L-M scheme has been implemented in the ASA band program lm. In the present implementation, the vectors $C$ and $\Delta$ can be adjusted to fit some reference bands or partial DOS. The extremely simple and elegant form of $H$, and its reasonably good accuracy relative to full LDA calculations, makes it an ideal candidate semi-empirical hamiltonian. The degrees of freedom are minimal, and physically transparent. (See link for an interpretation of $C$ and $\Delta$ in the language of multiple scattering theory).
For example the LDA+U method amounts to a shift in $C$ and also extending $C$ from a matrix diagonal in $RL$ to a matrix diagonal in $R$ but not in $L$ any more.
#### Inputs for Levenberg-Marquardt
The L-M scheme in the ASA modifies vectors $C$ and $\Delta$ (or some subset of all available $C$ and $\Delta$) by trying to minimize some norm, e.g. RMS difference between ASA and reference energy bands.
Once generated and stored, lm can read the modified $C$’s and $\Delta$’s, which modifies the ASA-LDA hamiltonian. lm doesn’t read or store the modified $C$ and $\Delta$ directly, but the shifts in $C$ and $\Delta$: these shifts are added to the standard (LDA) parameters. When lm is run in the L-M fit mode, the shifts are written to file vext0.ext.
lm can read shifts from file vext.ext, which has the same structure as vext0.ext. To make lm to read file vext.ext, add the following token to the HAM category in the ctrl file:
HAM RDVEXT=2
The L-M algorithm is designed to minimize the difference between the ASA bands and a reference set, or the difference in the partial DOS relative to a reference set. As part of the definition of the norm, each band or DOS is assigned a particular weight. The fitting algorithm gives you some control over the functional form of the weight; see ITER_FIT_WT below.
Input parameters controlling the L-M fitting are kept in the ITER_FIT tag of the input (ctrl) file. To fit bands use
ITER FIT[MODE=2 ...]
To fit partial DOS use
ITER FIT[MODE=12 ...]
In the former case, you will need some reference bands; in the latter you will need some reference partial DOS. We give an example below for a L-M fitting to a reference band structure.
This is an example of the ITER_FIT tag as it appears in the ctrl file:
ITER FIT[MODE=2 NBFIT=1 81 NBFITF=1 LAM=1 SCL=5 SHFT=1 WT=0,0,.1]
As noted above, ITER_FIT_MODE tells lm what to fit:
ITER_FIT_MODE (integer)Fit mode
0No fitting
2fit ASA $C$ and $\Delta$ to bands
12fit to site partial DOS
The following tokens apply when bands are fit
ITER_FIT_NBFIT (pair of integers, default = 1 …) Indices to lowest and highest bands to include in the fit (missing second argument => highest band in basis) ITER_FIT_NBFITF (integer) Index to the first reference band to be fit. Use it to skip some core levels present in the reference system but not in the ASA.
The following tokens apply when DOS are fit
ITER_FIT_NDOS (integer) Number of energy points to fit DOS ITER_FIT_WG (real) Broaden calculated DOS with Gaussian, width $W$. Optional 2nd arg specifies different width for reference DOS)
The remaining tokens apply to either bands or DOS fitting:
ITER_FIT_EBFIT (pair of real numbers) Energy range to fit eigenvalues or DOS ITER_FIT_WT (3 real numbers, default = 0 0 0.1) Fitting weights $wt_{kn}$ of fit bands $E_{kn}$ or DOS to file data Arg #1 defines the functional form of the weights. 0: all points assigned the same weight 1: Fermi function : $wt = 1/[ 1+\exp[-(E_{kn}-(a+E_f))/b] ]$ 2: Fermi-like gaussian cutoff : $wt = 1 / [1+\exp[-[(E_{kn}-(a+E_f))/b)^2] ]$ 3: Exponential weights : $wt = \exp[-\vert(E_{kn}-(a+E_f))/b \vert]$ 4: Gaussian weights : $wt = \exp[-((E_{kn}-(a+E_f))/b)^2]$ Arg #2 energy shift, denoted $a$ in the expressions above Arg #3 energy scale, denoted $b$ in the expressions above ITER_FIT_SHFT (integer, default = 0) adds constant potential shift to align fit with reference bands or DOS 0 no shift is added 1 Align $E_f$ with reference bands or DOS 2 Not implemented ITER_FIT_LAM (real, default = 1) Initial value of Levenberg-Marquardt $\lambda$; see Numerical Recipes ITER_FIT_SCL (real, default = 5) Scale factor for Levenberg-Marquardt $\lambda$ (cf Numerical Recipes)
#### Fitting the ASA PP on QSGW bands for the cubic structure
We now provide an example for fitting the ASA potential parameters onto the band structure of cubic CsPbI$_3$ calculated with QSGW (results given in paper analyzing the effects of nuclear motion). NOTE that the following calculations are based on Option 2 (Freeze the Pb 6d $P_\nu$ at a higher linearization energy) described above.
First, create the file with the data given below
Download the reference band file refbnds_cubic_cspbi_GWsc_noSO.txt from here
Then
cp refbnds_cubic_cspbi_GWsc_noSO.txt refbnds.asa
Edit the file ctrl.asa and change
HAM PMIN=-1 # Constrain Pnu > P_free
into
HAM PMIN=-1 # Constrain Pnu > P_free
RDVEXT=2 # Read shifts in C & Delta from vext.* file
and add the line for the L-M fit
ITER FIT[MODE=2 NBFIT=1 47 NBFITF=7 LAM=0.2 SCL=7 SHFT=1 WT=0,.12,.1]
which means that we are going to fit the ASA PP $C$ and $\Delta$ to the reference band given in the file refbnds.asa, 47 bands will be fitted starting at bands No 7 (in the refbnds.asa file, the 6 first bands contain some core states, i.e. there is at least one dispersionless (flat) band we do not need to consider. All the 47 bands have the some unit weight.
Finally, put the token BZ_INVIT to false in the part of the file ctrl.asa dealing with the Brillouin zone
# Brillouin zone
nkabc= {nkabc} # 1 to 3 values
metal= {met} # Management of k-point integration weights in metals
INVIT=F # avoid error in DIAGNO: tinvit cannot find all evecs
The eigenvectors are then not generated by inverse iteration (which is is the default). The inverse iteration method is more efficient than the QL method, but occasionally it fails to find all the vectors. When this happens, the program stops with the message: DIAGNO: tinvit cannot find all evecs. By using INVIT=F, one avoids this problem.
Start the L-M fit by using
lm -vnit=2 ctrl.asa --iactiv=no > out_lm_LMfit.asa
After 2 iterations, the RMS difference between ASA and reference bands slowly converges.
Using
cat out_lm_LMfit.asa | grep RMS
one should get
LM fit, starting RMS error=0.0876 lambda=0.2
LMFITMR: iter 1 old RMS=0.0876 new RMS=0.0443 lambda=0.2
LMFITMR: iter 2 old RMS=0.0443 new RMS=0.0208 lambda=0
LMFITMR: iter 2 starting RMS=0.0876 final RMS=0.0208 lambda=0
For a good fit to the bands, the final RMS should be of the order of several $10^{-4}$.
The resulting shifts in $C$’s and $\Delta$’s are stored in the file vext0.ext which should read
% contents=2 format=0 nsp=1
PPAR: ASA2
# class l isp shft C shft Delta frz
1 Pb 0 1 -0.154751 0.008507 0 0
1 Pb 1 1 0.009513 0.031511 0 0
1 Pb 2 1 0.093307 0.058896 0 0
2 Cs 0 1 0.056289 -0.040845 0 0
2 Cs 1 1 -0.148698 -0.056874 0 0
2 Cs 2 1 0.090362 -0.022224 0 0
3 I 0 1 -0.050322 -0.012000 0 0
3 I 1 1 -0.102364 0.029521 0 0
3 I 2 1 0.042320 -0.043010 0 0
4 E 0 1 -0.006442 0.024363 0 0
4 E 1 1 -0.044258 0.010160 0 0
4 E 2 1 -0.008779 0.032196 0 0
5 E1 0 1 0.014211 0.019634 0 0
5 E1 1 1 0.063985 0.003896 0 0
5 E1 2 1 -0.061891 0.005878 0 0
There are two parameters ($C$ and $\Delta$) to fit for each $l$ channel $(l=0,1,2)$ and for each species $(1,2,3,4,5)\equiv$ (Pb,Cs,I,E,E1). This makes a total of 30 parameters to fit, which might appear to be a lot of parameter to fit. There are however some “recipies” that can be used to minimise (and therefore control a bit more) the number of “floating” paramters for the fit.
The ASA PP, $C$ and $\Delta$, for the atomic species (Pb,Cs,I) are the most important to get right. The $C$’s and $\Delta$’s of the Empty Spheres (E,E1) play also a substantial role but can be subjected to more constraints.
This is done by changing the $0$ values in the last two columns (with header frz = freeze) in the file vext.ext.
Changing $0 \rightarrow 1$ means that the corresponding $C$ (or $\Delta$) will be kept frozen (unchanged) during the fitting process.
Changing $0 \rightarrow -1$ (or to any other negative integer) for different $l$ channels and species creates a new subset. Within this subset, the $C$’s (or $\Delta$’s) will be modified by the same amount during the fit.
Let’s give a concrete example:
Start by creating the file vext0.asa containing
% contents=2 format=0 nsp=1
PPAR: ASA2
# class l isp shft C shft Delta frz
1 Pb 0 1 0.000000 0.000000 0 0
1 Pb 1 1 0.000000 0.000000 0 0
1 Pb 2 1 0.000000 0.000000 0 0
2 Cs 0 1 0.000000 0.000000 0 0
2 Cs 1 1 0.000000 -0.000000 0 0
2 Cs 2 1 0.000000 0.000000 0 0
3 I 0 1 0.000000 0.000000 0 0
3 I 1 1 0.000000 0.000000 0 0
3 I 2 1 0.000000 0.000000 0 0
4 E 0 1 0.000000 0.000000 0 0
4 E 1 1 0.000000 0.000000 0 0
4 E 2 1 0.000000 0.000000 0 0
5 E1 0 1 0.000000 0.000000 0 0
5 E1 1 1 0.000000 0.000000 0 0
5 E1 2 1 0.000000 0.000000 0 0
Then
cp vext0.asa vext.asa
and Edit the file vext.asa and modify its contents with
% contents=2 format=0 nsp=1
PPAR: ASA2
# class l isp shft C shft Delta frz
1 Pb 0 1 0.000000 0.000000 0 0
1 Pb 1 1 0.000000 0.000000 0 0
1 Pb 2 1 0.000000 0.000000 0 0
2 Cs 0 1 0.000000 0.000000 0 0
2 Cs 1 1 0.000000 -0.000000 0 0
2 Cs 2 1 0.000000 0.000000 0 0
3 I 0 1 0.000000 0.000000 0 0
3 I 1 1 0.000000 0.000000 0 0
3 I 2 1 0.000000 0.000000 0 0
4 E 0 1 0.000000 0.000000 -1 -3
4 E 1 1 0.000000 0.000000 -1 -3
4 E 2 1 0.000000 0.000000 -1 -3
5 E1 0 1 0.000000 0.000000 -2 -4
5 E1 1 1 0.000000 0.000000 -2 -4
5 E1 2 1 0.000000 0.000000 -2 -4
This means that the $C$’s of the channels $l=0,1,2$ for the empty spheres E are not “moving” independently during the fit, but will be modified by the same amount. Similar constraints apply for the $\Delta$’s of the channels $l=0,1,2$ for the empty spheres E, and for the $C$’s and the $\Delta$’s of the empty spheres E1. While the other 18 parameters for the atomic species Pb, Cs, I are free to “move” independently from each other to allow for the best fit.
Now, run
lm -vnit=4 asa --iactiv=no > out_lm_LMfit_step1.asa
then ~~ cat out_lm_LMfit_step1.asa | grep RMS ~~ should read as
LM fit, starting RMS error=0.0876 lambda=0.2
LMFITMR: iter 1 old RMS=0.0876 new RMS=0.0408 lambda=0.2
LMFITMR: iter 2 old RMS=0.0408 new RMS=0.0229 lambda=0.0286
LMFITMR: iter 3 old RMS=0.0229 new RMS=0.0181 lambda=0.00408
LMFITMR: iter 4 old RMS=0.0181 new RMS=0.0171 lambda=0
LMFITMR: iter 4 starting RMS=0.0876 final RMS=0.0171 lambda=0
One can start improving the fit by changing the bands weights in the category ITER_FIT.
For that, edit the file ctrl.asa and change ITER_FIT with the following
ITER FIT[MODE=2 NBFIT=1 47 NBFITF=7 LAM=0.2 SCL=7 SHFT=1 WT=1,.12,.1]
Now the bnds weight follows a Fermi function-like distribution, and the bands with energy around and larger than $E_f+1.2$ get a smaller weight.
Then (do not forget to use the previously calculated shifts for the $C$’s and $\Delta$’s)
cp vext0.asa vext.asa
Before starting the fit, edit file vext.asa and change the line
2 Cs 1 1 -0.196928 -0.085852 0 0
into the following
2 Cs 1 1 -0.303618 -0.024109 0 0
This avoids unnecessary complications in the following.
Then run
lm -vnit=6 cspbi --iactiv=no > out_lm_LMfit_step2.cspbi
After checking
cat out_lm_LMfit_step2.asa | grep RMS
one should get
LM fit, starting RMS error=0.0184 lambda=0.2
LMFITMR: iter 1 old RMS=0.0184 new RMS=0.00540 lambda=0.2
LMFITMR: iter 2 old RMS=0.00540 new RMS=0.00281 lambda=0.0286
LMFITMR: iter 3 old RMS=0.00281 new RMS=0.00326 lambda=0.00408
LMFITMR: iter 4 old RMS=0.00281 new RMS=0.00324 lambda=0.0286
LMFITMR: iter 5 old RMS=0.00281 new RMS=0.00297 lambda=0.2
LMFITMR: iter 6 old RMS=0.00281 new RMS=0.00289 lambda=0
LMFITMR: iter 6 starting RMS=0.0184 final RMS=0.00289 lambda=0
which represents a better fit. The final RMS is around a few $10^{-3}$.
One can improve further the fit by changing ITER_FIT with the following
ITER FIT[MODE=2 NBFIT=1 47 NBFITF=7 LAM=0.2 SCL=7 SHFT=1 WT=3,.12,.1]
Now, with the exponential form, the bnds weights are mostly concentrated around $E_f+1.2$.
Then run
cp vext0.asa vext.asa
lm -vnit=8 cspbi --iactiv=no > out_lm_LMfit_step2.cspbi
The final RMS should be
LMFITMR: iter 8 old RMS=8.51e-4 new RMS=6.60e-4 lambda=0
LMFITMR: iter 8 starting RMS=0.00120 final RMS=6.60e-4 lambda=0
and the final file vext0.asa contains
% contents=2 format=0 nsp=1
PPAR: ASA2
# class l isp shft C shft Delta frz
1 Pb 0 1 -0.045168 -0.012894 0 0
1 Pb 1 1 0.019864 0.032718 0 0
1 Pb 2 1 -0.101022 0.012278 0 0
2 Cs 0 1 4.663590 1.320703 0 0
2 Cs 1 1 -0.220486 -0.047131 0 0
2 Cs 2 1 0.070615 -0.009268 0 0
3 I 0 1 -0.144494 -0.008486 0 0
3 I 1 1 -0.174939 0.040949 0 0
3 I 2 1 0.140586 0.009191 0 0
4 E 0 1 -0.022964 0.017607 -1 -3
4 E 1 1 -0.022964 0.017607 -1 -3
4 E 2 1 -0.022964 0.017607 -1 -3
5 E1 0 1 -0.018164 0.007317 -2 -4
5 E1 1 1 -0.018164 0.007317 -2 -4
5 E1 2 1 -0.018164 0.007317 -2 -4
One can now calculated the ASA band structure including the shifted $C$’s and $\Delta$’s and compare it with the band structure calculated with QSGW (which is in the file refbnds.asa ).
For that, run (suppress the line with ITER_FIT in the file ctrl.asa before calculating the bands)
cp vext0.asa vext.asa
lm ctrl.asa -vnit=1 --band~mq~fn=syml
QSGW bands (red) and ASA fitted bands (blue) of cubic CsPbI$_3$. That is quite a good fit!
#### Transferability of the fitted ASA Potential Parameters
… to be continued [27.11.2017]
### Footnotes and references
1 The ASA does not keep the density itself, but contracts all the information about the density into energy moments of the sphere charges, $Q_{0l}$, $Q_{1l}$, and $Q_{2l}$, of the density of states for each $l$, and the continuous principal quantum number. $P_l$. It can do this because of the special structure of the ASA. When the density is needed to make the potential, it is constructed from the $Q_{il}$ and the $P_l$.
If no data is available for a particular atom, the ASA codes lm, lmgf, and lmpg select $Q_{0l}$ from charges in the atom, and sets $Q_{1l}$, and $Q_{2l}$ to zero. It takes preset values for the $P_l$ from a lookup table. This is a crude estimate of the density, but it is usually sufficient to make a starting guess, which can be iterated to the self-consistent values.
2 Normally the continuous principal quantum numbers. $P_l$ are allowed to float to band center of gravity (the $P_l$ at which $Q_{1l}$ vanishes), but when the partial wave is far removed from the Fermi level, this can cause “ghost bands” to appear. One guard against this is to restrict the $P_l$, and not let it fall below the free-electron value. Tag HAM_PMIN is designed for this purpose. Another guard is to freeze the $P_l$ to a fixed value, using SPEC_ATOM_IDMOD. Another way is to downfold the $P_l$. You can tell the ASA codes to downfold a particular state with SPEC_ATOM_IDXDN. | 6,346 | 19,121 | {"found_math": true, "script_math_tex": 97, "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} | 2.640625 | 3 | CC-MAIN-2023-40 | longest | en | 0.852437 |
http://oeis.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture | 1,558,863,295,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259015.92/warc/CC-MAIN-20190526085156-20190526111156-00020.warc.gz | 153,065,302 | 16,661 | This site is supported by donations to The OEIS Foundation.
# Erdős–Straus conjecture
The Erdős–Straus conjecture concerns a Diophantine equation, refered to as the Erdős–Straus Diophantine equation, involving unit fractions.
Conjecture. (Erdős Pál, Ernst G. Straus) The equation[1]
${\displaystyle {\frac {4}{n}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}},\quad 1\leq a\leq b\leq c,}$
always has solutions for any integer ${\displaystyle n\geq 2}$.
It is trivial to see that for ${\displaystyle n=1}$ there can't be any solution!
You only have to check for primes ${\displaystyle p}$, since any positive integer (except the unit 1) is a multiple ${\displaystyle m}$ of some prime ${\displaystyle p}$, and
${\displaystyle {\frac {4}{mp}}={\frac {1}{ma}}+{\frac {1}{mb}}+{\frac {1}{mc}},\quad 1\leq a\leq b\leq c,\quad m\geq 1.}$
Thus ${\displaystyle f(mp)\geq f(p)}$, where ${\displaystyle f(n)}$ is the number of solutions.
As of 2011, the conjecture have been checked up to 10 14, for 2 and all primes congruent to 1 (mod 4). (We don't need to check for primes congruent to 3 (mod 4), see the proof below.)
The Erdős–Straus conjecture means that we can always find some positive integer ${\displaystyle k}$ for which there exists at least one partition ${\displaystyle (a_{1},a_{2},a_{3})}$ of ${\displaystyle 4k}$ such that each of the three parts divides into ${\displaystyle nk}$, i.e.
${\displaystyle {\frac {4}{n}}={\frac {4k}{nk}}={\frac {a_{1}+a_{2}+a_{3}}{nk}}=\left({\tfrac {nk}{a_{1}}}\right)^{-1}+\left({\tfrac {nk}{a_{2}}}\right)^{-1}+\left({\tfrac {nk}{a_{3}}}\right)^{-1}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}},\quad a_{1}+a_{2}+a_{3}=4k,\quad a_{1}\,|\,nk,\,a_{2}\,|\,nk,\,a_{3}\,|\,nk,\quad k\geq 1.}$
Thus ${\displaystyle nk={\rm {lcm}}(a,b,c)}$, and
${\displaystyle a_{1}={\frac {{\rm {lcm}}(a,b,c)}{a}},\,a_{2}={\frac {{\rm {lcm}}(a,b,c)}{b}},\,a_{3}={\frac {{\rm {lcm}}(a,b,c)}{c}}.}$
Again, we only have to consider the primes. For example:
${\displaystyle {\frac {4}{5}}={\frac {4\cdot 2}{5\cdot 2}}={\frac {8}{5\cdot 2}}={\frac {5+2+1}{5\cdot 2}}={\frac {1}{2}}+{\frac {1}{5}}+{\frac {1}{10}}}$
${\displaystyle {\frac {4}{7}}={\frac {4\cdot 4}{7\cdot 4}}={\frac {16}{7\cdot 2^{2}}}={\frac {7+7+2}{7\cdot 2^{2}}}={\frac {1}{4}}+{\frac {1}{4}}+{\frac {1}{14}}}$
${\displaystyle {\frac {4}{11}}={\frac {4\cdot 4}{11\cdot 4}}={\frac {16}{11\cdot 2^{2}}}={\frac {11+4+1}{11\cdot 2^{2}}}={\frac {1}{4}}+{\frac {1}{11}}+{\frac {1}{44}}}$
${\displaystyle {\frac {4}{13}}={\frac {4\cdot 4}{13\cdot 4}}={\frac {16}{13\cdot 2^{2}}}={\frac {13+2+1}{13\cdot 2^{2}}}={\frac {1}{4}}+{\frac {1}{26}}+{\frac {1}{52}}}$
## Truth of the conjecture for primes
### Truth of the conjecture for 2
For the only even prime, i.e. 2, we have the solution
${\displaystyle {\frac {4}{2}}={\frac {2+1+1}{2}}={\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{2}}.}$
Thus, the conjecture is true for any positive even integer.
### Truth of the conjecture for primes congruent to 3 (mod 4)
If ${\displaystyle p\equiv 3{\pmod {4}}}$, we always have solutions since
${\displaystyle {\frac {4}{p}}={\frac {4\left({\tfrac {p+1}{2}}\right)}{p\left({\tfrac {p+1}{2}}\right)}}={\frac {1}{\left({\tfrac {p+1}{2}}\right)}}{\frac {2(p+1)}{p}}={\frac {1}{\left({\tfrac {p+1}{2}}\right)}}\left(2+{\frac {2}{p}}\right)={\frac {1}{\left({\tfrac {p+1}{2}}\right)}}\left(2+{\frac {1}{p}}+{\frac {1}{p}}\right)={\frac {1}{\left({\tfrac {p+1}{4}}\right)}}+{\frac {1}{\left({\tfrac {p+1}{2}}\right)p}}+{\frac {1}{\left({\tfrac {p+1}{2}}\right)p}}={\frac {1}{\left({\tfrac {p+1}{4}}\right)}}+{\frac {1}{t_{p}}}+{\frac {1}{t_{p}}},}$
where ${\displaystyle t_{p}}$ is the ${\displaystyle p}$th triangular number.
Thus, the conjecture is true for any positive integer divisible by a prime ${\displaystyle p}$ congruent to 3 (mod 4).
### Truth of the conjecture for primes congruent to 1 (mod 4)
If the conjecture is true for ${\displaystyle p\equiv 1{\pmod {4}}}$, then the conjecture would be true for any positive integer divisible by a prime congruent to 1 (mod 4).
## Erdős–Straus conjecture for unit fractions from the open unit interval
Equivalently, any unit fraction from the open unit interval, i.e. ${\displaystyle 0<{\tfrac {1}{n}}<1}$, is one quarter of the sum of three (distinct or not) positive unit fractions[2]
${\displaystyle {\frac {1}{n}}={\frac {1}{4}}\left({\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}\right),\quad 1\leq a\leq b\leq c.}$
## Erdős–Straus conjecture for prime unit fractions from the open unit interval
If a solution is known for ${\displaystyle n}$, then solutions are known for all multiples ${\displaystyle mn,\,m\geq 1,}$ since
${\displaystyle {\frac {1}{mn}}={\frac {1}{4}}\left({\frac {1}{ma}}+{\frac {1}{mb}}+{\frac {1}{mc}}\right),\quad 1\leq a\leq b\leq c.}$
For example, since
${\displaystyle {\frac {4}{7}}={\frac {2\cdot 3\cdot 4}{2\cdot 3\cdot 7}}={\frac {3\cdot 7+2+1}{2\cdot 3\cdot 7}}={\frac {2\cdot 7+7+3}{2\cdot 3\cdot 7}},}$
we have
${\displaystyle {\frac {4}{7}}={\frac {1}{2}}+{\frac {1}{21}}+{\frac {1}{42}}={\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{14}}.}$
We can then obtain solutions for ${\displaystyle n=42}$ simply by multiplying both sides by ${\displaystyle {\tfrac {1}{6}}}$:
${\displaystyle {\frac {4}{42}}={\frac {2}{21}}={\frac {1}{12}}+{\frac {1}{126}}+{\frac {1}{252}}={\frac {1}{18}}+{\frac {1}{36}}+{\frac {1}{84}}.}$
Thus it is sufficient to prove that the conjecture is true for prime unit fractions.
Any prime unit fraction from the open unit interval, i.e. ${\displaystyle 0<{\tfrac {1}{p}}<1}$, is one quarter of the sum of three (distinct or not) positive unit fractions[3]
${\displaystyle {\frac {1}{p}}={\frac {1}{4}}\left({\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}\right),\quad 1\leq a\leq b\leq c,}$
for any prime ${\displaystyle p}$.
## Solutions (a, b, c) of 4/n = 1/a + 1/b + 1/c
Distinct solutions (a, b, c) in lexicographic order of 4/n = 1/a + 1/b + 1/c for n > 1
${\displaystyle 4/n\,}$ #
4 / 2 (1, 2, 2) 1
4 / 3 (1, 4, 12), (1, 6, 6), (2, 2, 3) 3
4 / 4 (2, 3, 6), (2, 4, 4), (3, 3, 3) 3
4 / 5 (2, 4, 20), (2, 5, 10) 2
4 / 6 (2, 7, 42), (2, 8, 24), (2, 9, 18), (2, 10, 15), (2, 12, 12), (3, 4, 12), (3, 6, 6), (4, 4, 6) 8
4 / 7 (2, 15, 210), (2, 16, 112), (2, 18, 63), (2, 21, 42), (2, 28, 28), (3, 6, 14), (4, 4, 14) 7
4 / 8 (3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12), (4, 5, 20), (4, 6, 12), (4, 8, 8), (5, 5, 10), (6, 6, 6) 10
### Number of solutions (a, b, c) of 4/n = 1/a + 1/b + 1/c
A192786 Number of solutions of 4/n = 1/a + 1/b + 1/c in positive integers, n >= 1. (Unit fractions may be repeated.)
{0, 3, 12, 16, 12, 45, 36, 58, 36, 75, 48, 136, 24, 105, 240, 190, 24, 159, 66, 250, 186, 153, 132, 364, 78, 129, 180, 292, 42, 531, 114, 490, 198, 159, 426, 526, 60, 201, 450, ...}
A192787 Number of distinct solutions of 4/n = 1/a + 1/b + 1/c in positive integers satisfying 1 <= a <= b <= c, n >= 1. (Unit fractions may be repeated.)
{0, 1, 3, 3, 2, 8, 7, 10, 6, 12, 9, 21, 4, 17, 39, 28, 4, 26, 11, 36, 29, 25, 21, 57, 10, 20, 29, 42, 7, 81, 19, 70, 31, 25, 65, 79, 9, 32, 73, 96, 7, 86, 14, 62, 93, 42, 34, ...}
A073101 Number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z, n >= 2. (Unit fractions must be distinct.)
{0, 1, 1, 2, 5, 5, 6, 4, 9, 7, 15, 4, 14, 33, 22, 4, 21, 9, 30, 25, 22, 19, 45, 10, 17, 25, 36, 7, 72, 17, 62, 27, 22, 59, 69, 9, 29, 67, 84, 7, 77, 12, 56, 87, 39, 32, 142, ...}
A?????? Number of distinct solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z, n >= 2. (Unit fractions must be distinct.)
{0, 1, ...}
## Solutions (a, b, c) of 4/p = 1/a + 1/b + 1/c
Distinct solutions (a, b, c) in lexicographic order of 4/p = 1/a + 1/b + 1/c for p prime
${\displaystyle 4/p\,}$ #
4 / 2 (1, 2, 2) 1
4 / 3 (1, 4, 12), (1, 6, 6), (2, 2, 3) 3
4 / 5 (2, 4, 20), (2, 5, 10) 2
4 / 7 (2, 15, 210), (2, 16, 112), (2, 18, 63), (2, 21, 42), (2, 28, 28), (3, 6, 14), (4, 4, 14) 7
4 / 11 (3, 34, 1122), (3, 36, 396), (3, 42, 154), (3, 44, 132), (3, 66, 66), (4, 9, 396), (4, 11, 44), (4, 12, 33), (6, 6, 33) 9
4 / 13 (4, 18, 468), (4, 20, 130), (4, 26, 52), (5, 10, 130) 4
4 / 17 (5, 30, 510), (5, 34, 170), (6, 15, 510), (6, 17, 102) 4
4 / 19 (5, 96, 9120), (5, 100, 1900), (5, 114, 570), (5, 120, 456), (5, 190, 190), (6, 23, 2622), (6, 24, 456), (6, 30, 95), (6, 38, 57), (8, 12, 456), (10, 10, 95) 11
4 / 23 (6, 139, 19182), (6, 140, 9600), (6, 141, 6486), (6, 142, 4899), (6, 144, 3312), (6, 147, 2254), (6, 150, 1725), (6, 156, 1196), (6, 161, 966), (6, 174, 667), (6, 184, 552), (6, 207, 414), (6, 230, 345), (6, 276, 276), (7, 42, 138), (8, 23, 184), (8, 24, 138), (9, 16, 3312), (9, 18, 138), (10, 15, 138), (12, 12, 138) 21
4 / 29 (8, 78, 9048), (8, 80, 2320), (8, 87, 696), (8, 88, 638), (8, 116, 232), (10, 29, 290), (11, 22, 638) 7
### Number of solutions (a, b, c) of 4/p = 1/a + 1/b + 1/c
In 2012, Christian Elsholtz & Terence Tao established that
${\displaystyle N\log ^{2}N\ll \sum _{p\leq N}f(p)\ll N\log ^{2}N\log \log N,}$
where ${\displaystyle p}$ is a prime and ${\displaystyle f(p)}$ is the number of solutions to the Erdős–Straus Diophantine equation.
A192788 Number of solutions of 4/p = 1/a + 1/b + 1/c in positive integers, where p is the n-th prime. (Unit fractions may be repeated.)
{3, 12, 12, 36, 48, 24, 24, 66, 132, 42, 114, 60, 48, 84, 216, 90, 168, 72, 108, 246, 42, 228, 162, 66, 48, 102, 156, 150, 96, 84, 198, 192, 108, 222, 114, 192, 144, 144, ...}
A192789 Number of distinct solutions of 4/p = 1/a + 1/b + 1/c in positive integers, where p is the n-th prime. (Unit fractions may be repeated.)
{1, 3, 2, 7, 9, 4, 4, 12, 23, 7, 20, 10, 8, 15, 37, 15, 29, 12, 19, 42, 7, 39, 28, 11, 8, 17, 27, 26, 16, 14, 34, 33, 18, 38, 19, 33, 24, 25, 68, 27, 52, 18, 69, 6, 25, 43, 32, ...}
A?????? Number of solutions (x,y,z) to 4/p = 1/x + 1/y + 1/z satisfying 0 < x < y < z, where p is the n-th prime. (Unit fractions must be distinct.)
{0, 1, 2, 5, 7, 4, 4, 9, 19, 7, 17, 9, 12, 32, ...}
A?????? Number of distinct solutions (x,y,z) to 4/p = 1/x + 1/y + 1/z satisfying 0 < x < y < z, where p is the n-th prime. (Unit fractions must be distinct.)
{0, 1, ...}
## Notes
1. A more constrained variant of the conjecture, where the unit fractions must be distinct, is
${\displaystyle {\frac {4}{n}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}},\quad 0
always has solutions for any integer ${\displaystyle n\geq 3}$. (Note that there is no solution for ${\displaystyle n=2}$ when we require distinct unit fractions.)
2. Equivalent conjecture: any nonunit positive integer, i.e. greater than 1, is ${\displaystyle {\tfrac {4}{3}}}$ times the harmonic mean of three (distinct or not) positive unit fractions
${\displaystyle n={\frac {4}{3}}\left({\frac {{\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}{3}}\right)^{-1},\quad 1\leq a\leq b\leq c,}$
for any positive integer ${\displaystyle n>1}$.
3. Equivalent conjecture: any prime is ${\displaystyle {\tfrac {4}{3}}}$ times the harmonic mean of three (distinct or not) positive unit fractions
${\displaystyle p={\frac {4}{3}}\left({\frac {{\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}{3}}\right)^{-1},\quad 1\leq a\leq b\leq c,}$
for any prime ${\displaystyle p}$. (If it's true for any prime ${\displaystyle p}$, then it's true for any ${\displaystyle kp,\,k\geq 1,}$ thus for any nonunit positive integer, i.e. greater than 1, the unit 1 being the empty product of primes.) | 5,142 | 11,443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 56, "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} | 4.0625 | 4 | CC-MAIN-2019-22 | latest | en | 0.760423 |
https://w3resource.com/plsql-exercises/control-statement/plsql-control-statement-exercise-24.php | 1,638,803,894,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00343.warc.gz | 662,464,325 | 28,791 | PL/SQL Control Statement: Print 1st n numbers with a difference of 3 and starting from 1 - w3resource
# PL/SQL Control Statement Exercises: Print 1st n numbers with a difference of 3 and starting from 1
## PL/SQL Control Statement: Exercise-24 with Solution
Write a program in PL/SQL to print 1st n numbers with a difference of 3 and starting from 1.
Sample Solution:
PL/SQL Code:
``````DECLARE
n number:= &first_n_number;
i number:=1;
m number:=1;
BEGIN
DBMS_OUTPUT.PUT_LINE ('The first '||n||' numbers are: ');
DBMS_OUTPUT.PUT (i||' ');
for i in 1..n-1 loop
m:=m+3;
dbms_output.put(m||' ');
END LOOP;
dbms_output.new_line;
END;
/
```
```
Flowchart:
Sample Output:
```Enter value for first_n_number: 10
old 2: n number:= &first_n_number;
new 2: n number:= 10;
The first 10 numbers are:
1 4 7 10 13 16 19 22 25 28
PL/SQL procedure successfully completed.```
Improve this sample solution and post your code through Disqus
What is the difficulty level of this exercise?
| 307 | 1,003 | {"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} | 2.578125 | 3 | CC-MAIN-2021-49 | latest | en | 0.759098 |
metatron216.co.za | 1,579,418,206,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00369.warc.gz | 110,006,658 | 7,512 | Metatron’s Cube
In Sacred Geometry there is probably only one symbol more Sacred than the Seed of Life and that is Metatron’s Cube. See image below.
When we take the 60 Digit Cipher and enter it as a key into Metatron’s Cube, the mathematics is simply mind-blowing. This symbol now creates a flowering pattern of 216 numbers. The sum of all the numbers add up to 3640 but if you subtract the points where the Star of David intersects or touch you get a perfect 3600.
The “Star of David” is the symbol for God. But it is also a hexagon. This hexagonal pattern permeates all of nature. See the images below.
Any spheres tightly packed will naturally adopt this pattern. See the marbles in the image below. You can test this by dumping a set of marbles into a tin. You could argue that this pattern is simply natural. What you cannot say is that this pattern hides a 216 Digit matrix by coincidence and that some people in the Ancient world believed God’s Hidden name has 216 letters by random chance or that 60 x 360 = 21600 by coincidence and that the diameter of our moon is 2160 miles by luck. This discovery puts an end to the idea that everything is random chaos. All our ancient units of measure have greater meaning.
Any Cube will have 360 degrees of Angle.
Here is a YouTube video on Metatron’s Cube.
Take a look at the geometric angles in Metatron’s Cube.
Circles are the Fabric of the Universe. They form all the Sacred Geometric patterns. You must imagine all of those circles as circles of 60 numbers radiating as waves of LIGHT. The entire universe is made up of numbers or “information” coded into those patterns of numbers. The universe is a MATRIX of information stored in the waves or circles of LIGHT. Our conscious minds take that information and create a simulation of the universe we see.
When I take the 216 numbers in Metatron’s Cube and place them in a table or MATRIX of 12 rows and 18 columns, each column adds up to a perfect 60. See image below. The total sum of all those numbers is 1080. 1080 x 2 = 2160. Once again. This is mathematical perfection. This Matrix is God’s secret or Hidden 216 Letter Name. In the Ancient World people would use numbers as letters in a practice known as Gematria. In order to speak this Hidden Name out loud we need to share it.
Thank you for visiting this site. Please help me by sharing this page so that we can get the academic world to study all of this. This discovery may hold all the secrets to time. And please purchase a copy of my book to help support my own efforts to keep studying all of this. | 579 | 2,575 | {"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} | 3.671875 | 4 | CC-MAIN-2020-05 | longest | en | 0.893833 |
https://im.kendallhunt.com/MS/teachers/1/5/9/index.html | 1,726,815,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00096.warc.gz | 275,934,346 | 32,571 | # Lesson 9
Using the Partial Quotients Method
## 9.1: Using Base-Ten Diagrams to Calculate Quotients (5 minutes)
### Warm-up
Prior to grade 6, students have solved division problems using their understanding of place value and the idea of creating equal-size groups. This warm-up relies on those concepts to prepare students for more-abstract work in later lessons. The divisor and dividend are chosen so that the hundreds in the dividend can be partitioned into equal groups but the tens cannot. The quotient, however, is a whole number. The key ideas that would enable students to ultimately divide a decimal by a decimal are present in this example:
• A number can be decomposed to make the division convenient, e.g., 372 can be viewed as 300 + 60 + 12.
• Place value, expressed in the form of base-ten diagrams, plays a very important role in division.
### Launch
Arrange students in groups of 2. Display Elena’s method for all to see and use as a reference. Give students 1 minute of quiet think time and 2 minutes to discuss with a partner. Follow with a whole-class discussion.
### Student Facing
Elena used base-ten diagrams to find $$372 \div 3$$. She started by representing 372.
She made 3 groups, each with 1 hundred. Then, she put the tens and ones in each of the 3 groups. Here is her diagram for $$372 \div 3$$.
Discuss with a partner:
• Elena’s diagram for 372 has 7 tens. The one for $$372 \div 3$$ has only 6 tens. Why?
• Where did the extra ones (small squares) come from?
### Anticipated Misconceptions
If students have difficulty making sense of Elena’s method, consider demonstrating her process with actual base-ten blocks or paper cutouts.
### Activity Synthesis
Highlight Elena’s process of separating base-ten units into equal groups. Discuss:
• Which base-ten unit(s) did Elena unbundle or break up? (She unbundled a tens unit.)
• Why? What did unbundling accomplish? (She had only 1 ten left and there are 3 equal groups. Unbundling as smaller units made it possible to place the 1 ten in the 3 groups.)
• Is there another way that Elena could have made 3 equal groups out of the base-ten units? (She could have unbundled other larger units into smaller units—e.g., the 3 hundreds as 30 tens or all 7 tens as 70 ones—but it was not necessary.)
• How might one find $$378 \div 3$$ using Elena’s method? (By thinking of 378 as 3 hundreds, 6 tens, and 18 ones and placing them into 3 equal groups.)
## 9.2: Using the Partial Quotients Method to Calculate Quotients (15 minutes)
### Activity
Here, students continue to find quotients of whole numbers by thinking about equal-size groups and place value. They learn that, in addition to using base-ten diagrams, they can also form equal-size groups using only numbers and by thinking in terms of partial quotients. Just as they had used diagrams to place base-ten units—first hundreds, then tens, and then ones—into equal groups until all units are placed, they can distribute base-ten units of a number into equal groups until all of the units are placed.
### Launch
Keep students in groups of 2. Display the following diagram for all to see, and explain that it shows Elena’s method of finding $$657 \div 3$$. Give students a minute to analyze the diagram, determine what the quotient is, and be prepared to explain what the diagram shows.
Give partners 1 minute to discuss their understanding of the diagram. Afterwards, consider asking a student to share with the class. Look for an explanation along the lines of the following:
• First, use the 6 hundreds to make 3 equal groups of 200.
• Then use 3 tens of the 5 tens to make 3 equal groups of 10.
• Unbundle the remaining 2 tens into 20 ones, combine them with the 7 ones, and split the 27 ones into 3 equal groups of 9.
If time permits, invite other students to elaborate on the presented explanations or share alternative analyses.
Keep Elena’s method from the previous task displayed for students to reference. Give students 7–8 minutes to think about and discuss Andre’s method in the first question and to complete the activity.
Provide access to graph paper. Tell students that they may find the grid helpful for aligning the digits when finding partial quotients.
Representation: Internalize Comprehension. Demonstrate and encourage students to use color coding and annotations to highlight connections between representations in a problem. For example, use color coding to highlight connections between the partial quotients in Andre’s and Elena’s method for calculating $$657 \div 3$$.
Supports accessibility for: Visual-spatial processing
Writing, Conversing: MLR5 Co-Craft Questions. Display Andre’s division method without revealing the questions that follow. Ask students to write down questions they have for Andre. Invite students to compare their questions with a partner before sharing with the whole class. Highlight questions that ask about what specific numbers mean or represent in the given work. Finally, reveal the actual questions students are expected to work on. This will help students produce language related to representations of division.
Design Principle(s): Cultivate conversation; Maximize meta-awareness
### Student Facing
1. Andre calculated $$657 \div 3$$ using a method that was different from Elena’s.
1. Andre subtracted 600 from 657. What does the 600 represent?
2. Andre wrote 10 above the 200, and then subtracted 30 from 57. How is the 30 related to the 10?
3. What do the numbers 200, 10, and 9 represent?
4. What is the meaning of the 0 at the bottom of Andre’s work?
2. How might Andre calculate $$896 \div 4$$? Explain or show your reasoning.
### Anticipated Misconceptions
When using the partial quotients method, students might make subtraction or multiplication errors because they did not line up the numbers appropriately. Prompt students to compare the structure of Andre’s work with their own or to check if they have aligned like units in their vertical calculations.
### Activity Synthesis
Make sure students understand how the steps in Elena’s method and Andre’s method correspond. Discuss:
• Elena’s diagram shows 3 groups of 2 hundreds. Where in Andre’s method do we see the same value? (In the 600 subtracted from 657.)
• Where in Elena’s work do we see the 30 that Andre subtracts from 57? (In the 3 groups of 1 ten.)
• Do Andre and Elena both get the same answer? Why? (Yes, because they both distributed 657 into 3 equal groups.)
Tell students that Andre’s method is an example of the partial quotients method, in which we divide a part of the dividend at a time, obtaining part of the quotient each time. In this case, the first partial quotient is 200, next is 210 (from $$200+10$$), and last is the quotient 219 (from $$200+10+9$$). We can still view a division expression such as $$657 \div 3$$ as a question asking “How many are in each group if I divide 657 into 3 equal groups?”
With the partial quotients method, we can take any amount we choose out of 657 and place it into the 3 equal groups. It is often helpful to take out the amount in each place value and distribute it into groups. The values placed in each group are partial quotients. Once we have distributed all of 657, we can add the partial quotients to find $$657 \div 3$$.
If time allows, discuss how to find the value of $$655 \div 5$$ using the partial quotients method. Ask students what value might be reasonable to take out first, second, etc. (e.g., taking out 100 or 120 is a reasonable first move).
## 9.3: What’s the Quotient? (15 minutes)
### Activity
In this lesson, students choose how to perform division—by drawing diagrams or by using the partial quotients method. As they work with larger dividends and divisors, students observe the merits and potential drawbacks of each method. They see that base-ten diagrams are useful because they are concrete and help to visualize the meaning of division, but drawing all the pieces is cumbersome if the numbers are large. The partial quotients method relies on the same principles but will work for any numbers without the need for elaborate drawings. Students use these observations to decide on appropriate methods to use.
Earlier, when introducing Andre’s method and partial quotients, we had interpreted $$657 \div 3$$ as answering the question: “How much is in each group if 657 is divided into 3 equal groups?” For example, when Andre took out 600, we interpreted it as: “600 is 3 groups of 200.” This interpretation is helpful for making the connection to Elena’s base-ten diagrams, in which she divided base-ten units into 3 groups.
Note, however, that we can also interpret the division expression as asking: “How many groups of 3 are in 657?” This interpretation can likewise be represented using diagrams, though it would be highly impractical to draw 219 groups of 3. It is not impractical for students to think about partial quotients this way, however. For instance, they could take out 600, ask “how many groups of 3 are in 600?” and write down a partial quotient of 200 to represent 200 groups of 3 (instead of 3 groups of 200).
### Launch
Remind students that Elena drew base-ten diagrams to represent equal groups, while Andre took out smaller amounts from the dividend, found those quotients, and then combined the partial quotients together. Tell students that in this activity they choose a method to perform division. Encourage them to refer to Elena and Andre’s methods from the previous activities, or display Elena and Andre’s methods for all to see.
Keep students in groups of 2. Give students 2–3 minutes to discuss the first question with a partner and 8–10 minutes of quiet work time on the remaining questions. Provide access to graph paper.
Representation: Develop Language and Symbols. Use virtual or concrete manipulatives to connect symbols to concrete objects or values. For example, provide access to physical or virtual base-ten blocks as alternatives to drawing diagrams.
Supports accessibility for: Conceptual processing
### Student Facing
1. Find the quotient of $$1,\!332 \div 9$$ using one of the methods you have seen so far. Show your reasoning.
2. Find each quotient and show your reasoning. Use the partial quotients method at least once.
1. $$1,\!115 \div 5$$
2. $$665 \div 7$$
3. $$432 \div 16$$
### Activity Synthesis
Focus the discussion on showcasing the advantages and disadvantages of each method. Ask students to indicate which division method they preferred for the first two problems. Then, ask students who used base-ten diagrams to share the challenges of using this method to divide. Possible responses include:
• When the divisor is large, it takes a long time to draw all of the equal groups.
• When the dividend or divisor is large, it is difficult to tell how much I have used up or placed in groups and how much remains.
• When the dividend or divisor is large, it is difficult to check my work, i.e., to see that all of the equal groups add up to the value of the dividend.
Ask students who used the partial quotients to name the challenges of using this method for division? Possible responses include:
• I was not sure how to decide what amount to take out at each step.
• Sometimes I could not find familiar numbers to take out.
• I took out too little at a time, so it ended up taking a long time.
• I thought I had placed everything into equal groups but I ended up with a leftover.
Show that many of the challenges of using base-ten diagrams are eliminated by using the partial quotients method and explain that in future lessons the partial quotients method will be refined to help deal with its challenges.
Writing, Speaking, Listening: MLR7 Compare and Connect. As students discuss the first question with a partner, ask them to identify what is similar and what is different about the approaches used. Provide sentence frames to help students organize their thinking as they share their answer. For example, “In using ____________, first, I _________, next, I ___________, then, I ____________, and finally, I _____________.” Follow with a whole-class discussion centered around the advantages and disadvantages of each method. Consider charting the responses. Listen for comments noting the effects of a large dividend or divisor, leftover amounts, or decisions about what amount to take out. This will help students connect other students’ approaches to division to their own approach and decide which method is more efficient to use and why.
Design Principle(s): Optimize output (for comparison); Maximize meta-awareness
## Lesson Synthesis
### Lesson Synthesis
Recall that one way to think of division is as a process of splitting a value into equal-size groups and finding the size of one group. We can represent the groups and contents of each group using base-ten diagrams. Let’s take $$456 \div 4$$ as an example. We start by representing 456 with drawings of 4 hundreds, 5 tens, and 6 ones.
• How might we start dividing the pieces? (We can do it by place value. We draw 4 groups, put 1 hundred into each group, and then put 1 ten into each group.)
• What happens if there’s a remainder, e.g., after putting 1 ten into each group, we still have 1 ten left? (We can unbundle the 1 ten into 10 ones.)
• What does unbundling accomplish? (It allows us to have smaller units to divide. Here we combine the 10 ones and 6 ones, and then divide 16 ones into 4 groups of 4.)
• How do we find the value of the quotient? (The value of all the pieces in each group is the quotient.)
We can also find quotients without drawing a diagram and by using the partial quotients method.
• How is the partial quotient method similar to drawing base-ten diagrams? (We still pay attention to place value and think in terms of equal-size groups.)
• How is it different than drawing diagrams? (With partial quotients, we use only numbers, and we don’t have to divide the entirety of each base-ten unit at once. We can decide the amount to divide each round, e.g., we could first divide 200 into 4 groups, then another 200 into 4 groups, etc.)
• What might be the advantages of using base-ten diagrams? (It is concrete. It helps us visualize the number being divided.)
• What are the advantages of using partial quotients to divide numbers? (There is no need to draw all the pieces being divided or all the groups. There is flexibility in how we divide.)
• Do you have a preferred method for finding decimal quotients? Explain your reasoning.
## Student Lesson Summary
### Student Facing
We can find the quotient $$345\div 3$$ in different ways.
One way is to use a base-ten diagram to represent the hundreds, tens, and ones and to create equal-sized groups.
We can think of the division by 3 as splitting up 345 into 3 equal groups.
Each group has 1 hundred, 1 ten, and 5 ones, so $$345 \div 3 = 115$$. Notice that in order to split 345 into 3 equal groups, one of the tens had to be unbundled or decomposed into 10 ones.
Another way to divide 345 by 3 is by using the partial quotients method, in which we keep subtracting 3 groups of some amount from 345.
• In the calculation on the left, first we subtract 3 groups of 100, then 3 groups of 10, and then 3 groups of 5. Adding up the partial quotients ($$100+10+5$$) gives us 115.
• The calculation on the right shows a different amount per group subtracted each time (3 groups of 15, 3 groups of 50, and 3 more groups of 50), but the total amount in each of the 3 groups is still 115. There are other ways of calculating $$345 \div 3$$ using the partial quotients method.
Both the base-ten diagrams and partial quotients methods are effective. If, however, the dividend and divisor are large, as in $$1,\!248 \div 26$$, then the base-ten diagrams will be time-consuming. | 3,645 | 15,780 | {"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} | 4.9375 | 5 | CC-MAIN-2024-38 | latest | en | 0.876552 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-1-section-1-3-more-on-functions-and-their-graphs-exercise-set-page-198/85 | 1,579,359,824,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592636.25/warc/CC-MAIN-20200118135205-20200118163205-00136.warc.gz | 904,811,520 | 13,690 | ## Precalculus (6th Edition) Blitzer
The difference quotient for the provided function is $-4x-2h-1$.
Consider the provided function: $f\left( x \right)=-2{{x}^{2}}-x+3$. Now, substitute $x=x+h$ in the above equation to find $f\left( x+h \right)$ That is, \begin{align} & f\left( x+h \right)=-2{{\left( x+h \right)}^{2}}-\left( x+h \right)+3 \\ & =-2{{x}^{2}}-4xh-2{{h}^{2}}-x-h+3 \end{align} Now, apply the difference quotient formula, \begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{-2{{x}^{2}}-4xh-2{{h}^{2}}-x-h+3-\left( -2{{x}^{2}}-x+3 \right)}{h} \\ & =\frac{-2{{x}^{2}}-4xh-2{{h}^{2}}-x-h+3+2{{x}^{2}}+x-3}{h} \\ & =\frac{-4xh-2{{h}^{2}}-h}{h} \\ & =\frac{h\left( -4x-2h-1 \right)}{h} \end{align} Further solve and get, $\frac{f\left( x+h \right)-f\left( x \right)}{h}=-4x-2h-1$ Hence, the difference quotient for the provided function is $-4x-2h-1$. | 395 | 880 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-05 | latest | en | 0.444814 |
http://www.dailypaul.com/301497/prime-spirals-by-numberphile-with-dr-james-grime?page=1 | 1,394,421,335,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010557169/warc/CC-MAIN-20140305090917-00016-ip-10-183-142-35.ec2.internal.warc.gz | 300,352,111 | 32,452 | Prime Spirals, by Numberphile, with Dr. James Grime
Ulam Spiral
The Ulam spiral, or prime spiral (in other languages also called the Ulam Cloth) is a simple method of visualizing the prime numbers that reveals the apparent tendency of certain quadratic polynomials to generate unusually large numbers of primes. It was discovered by the mathematician Stanislaw Ulam in 1963, while he was doodling during the presentation of a "long and very boring paper" at a scientific meeting [...]
http://en.wikipedia.org/wiki/Ulam_spiral
Comment viewing options
Terence Tao's Structure and Randomness in the Prime Numbers
Terence Tao's Structure and Randomness in the Prime Numbers
"Cyril" pronounced "see real". I code stuff. I care quite a bit about language.
http://Laissez-Faire.Me/Liberty
Did you know? 1,299,830,899,481 can be factored in no more than 55,655 steps.
for spirals :)
.
.
"Cyril" pronounced "see real". I code stuff. I care quite a bit about language.
http://Laissez-Faire.Me/Liberty
Did you know? 1,299,830,899,481 can be factored in no more than 55,655 steps.
Confession
Primes investigation is one of my guilty pleasures as an amateur, right next to sipping some liquor in the evening.
;p
"Cyril" pronounced "see real". I code stuff. I care quite a bit about language.
http://Laissez-Faire.Me/Liberty
Did you know? 1,299,830,899,481 can be factored in no more than 55,655 steps.
Looking for a rational pi approximation better than (2*11)/7
5*71 / 113 =3.1415929203539823008849557522124
pi ====== 3.1415926535897932384626433832795...
err = 2.6676418906242231236893288649633e-7
There is a problem with the program I wrote, not reporting error correctly. Hmmm.
I enjoy a Brandy in the evening.
Free includes debt-free!
Check this out :
Check this out :
Say you're doodling around the semiprimes factorization problem and you aim for something much simpler (and less powerful) than the GNFS but still more efficient than trial division.
Let's get our hands on, say :
s = 136,129,581,883
to factor.
Of course, trial division tells us we can stop at int(sqrt(136,129,581,883)) = 368,957, and since all primes are known to be of either forms 6n + 1 or 6n - 1, that still leaves us with 368,957 / 6 = 61,492 attempts.
What if I told with you with some smart (and lucky ;) guess you can reduce it to less than 400 iterations only?
Here it is:
Let s = u . v, with u < v.
Certainly, we can always write:
(1) k . u < v < (k + 1) . u
For some natural k, 1 <= k <= N (that's our guess)
We can assume N rather small (< 100 or 10 or less), when the numbers of digits of u and v differ by one or two only (e.g., for a factor 1 to 100 between u and v).
Then, (1) times u:
k . u^2 < u . v < (k + 1) . u^2
That is:
umin < u < umax
with umin = int(sqrt(s / (k + 1))) and umax = int(sqrt(s / k))
Application:
create N copies (say, N = 12 ... I said we can be lucky ;) of the same program P(k), parameterized by k, 1 <= k <= N
which will write u = 6 . p +/- 1 growing from umin towards umax, e.g., for P(k = 12):
umin = int(sqrt(136,129,581,883 / 13)) = 102,331
umax = int(sqrt(136,129,581,883 / 12)) = 106,507
and will be testing s / u
In our case, it turns out that:
u = 6 * 17,455 - 1 = 104,729 (and v = s / u = 1,299,827)
where u (or p, equivalently) will be stumbled upon by the program P(k = 12) after:
(u - umin) / 6 = (104,729 - 102,331) / 6 = 399 iterations.
:)
"Cyril" pronounced "see real". I code stuff. I care quite a bit about language.
http://Laissez-Faire.Me/Liberty
Did you know? 1,299,830,899,481 can be factored in no more than 55,655 steps.
Question
"since all primes are known to be of either forms 6n + 1 or 6n - 1"
Can you expand on this a little bit? I've never heard this before and I'm intrigued
Two lists of small primes I use regularly:
Two lists of small primes I use regularly:
http://primes.utm.edu/lists/small/100000.txt
http://prime-numbers.org/
For testing primality, Wolfram MathWorld comes in handy :
http://www.wolframalpha.com/input/?i=Is+123571113171923+prim...
"Cyril" pronounced "see real". I code stuff. I care quite a bit about language.
http://Laissez-Faire.Me/Liberty
Did you know? 1,299,830,899,481 can be factored in no more than 55,655 steps.
This is true for all primes greater than 3, starting with 5.
This is true for all primes greater than 3, starting with 5 (my inaccuracy by omission). Because all integers >= 5 can be expressed as (6k + i) for some integer k >= 1 and for i = −1, 0, 1, 2, 3, or 4.
Where: 2 divides (6k + 0), (6k + 2), (6k + 4); and 3 divides (6k + 3). Hence, only i = -1 or 1 are left.
There are many basic properties of prime numbers to be aware of, for anyone interested in them.
http://primes.utm.edu/
(among others, often referenced from Wikipedia)
"Cyril" pronounced "see real". I code stuff. I care quite a bit about language.
http://Laissez-Faire.Me/Liberty
Did you know? 1,299,830,899,481 can be factored in no more than 55,655 steps.
Cool
Seems so obvious in retrospect. Thanks!
Are you talking about the Golden ratio, or the golden mean?
and if so, why did you not present it as such?
Golden? Neither, as I have more of a thing for the silver
Golden? Neither, as I have more of a thing for the silver ratio and Pell numbers, lately:
http://en.wikipedia.org/wiki/Silver_ratio
;)
"Cyril" pronounced "see real". I code stuff. I care quite a bit about language.
http://Laissez-Faire.Me/Liberty
Did you know? 1,299,830,899,481 can be factored in no more than 55,655 steps.
Seriously Dude? are you that numb?
both the Golden ratio AND the Golden mean have actual applications in the real world.
care to make the same statement for your conjecture?
and no sir, I am NOT a mathematician.. or a mathamagician.
how is your beloved "silver ratio" relevant? in any way, shape or form?
you consider me a stupid HVAC/R Tech. and you don't even know what the profession entails.
let me know how that works out for you.
what I AM aware of, is that there are MANY things that we (as humans)
know how to use and make them do work for us.
that we do not fundamentally understand.
peace.
The ancients knew Wau
Thanks Cyril, been listening to Math you tubes for two days now.
C'est fantastique, mon ami!
Free includes debt-free!
LOL 'Welcome :)
LOL 'Welcome :)
"Cyril" pronounced "see real". I code stuff. I care quite a bit about language.
http://Laissez-Faire.Me/Liberty
Did you know? 1,299,830,899,481 can be factored in no more than 55,655 steps. | 1,889 | 6,509 | {"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} | 3.171875 | 3 | CC-MAIN-2014-10 | latest | en | 0.904871 |
http://mymathforum.com/algebra/40543-nearest-distance-point-point-line.html | 1,566,343,518,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315681.63/warc/CC-MAIN-20190820221802-20190821003802-00414.warc.gz | 139,262,854 | 8,873 | My Math Forum Nearest distance from point to point on line
Algebra Pre-Algebra and Basic Algebra Math Forum
January 3rd, 2014, 11:41 PM #1
Newbie
Joined: Jan 2014
Posts: 2
Thanks: 0
Nearest distance from point to point on line
Hey,
consider the situation like on the attached picture.
I need to find a point P (blue) between two points A (red) and B (green). The point P must have the following characteristics:
-P is a point on a line L (black)
-The distance from A to P + the distance from P to B have to be minimized (so the path is actually the shortest path from A to B over line L, like the green line on the picture)
I need a formula which gives me the point P, but I have no idea how I should achieve this..
Could you give me any help, please?
Greetings!
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January 3rd, 2014, 11:44 PM #2 Newbie Joined: Jan 2014 Posts: 2 Thanks: 0 Re: Nearest distance from point to point on line Addition: I actually forgot to say that the line is a line segment limited by two points.
January 4th, 2014, 03:08 PM #3 Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716 Re: Nearest distance from point to point on line I think you need to clarify your question. The diagram and your description don't match.
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http://codeforces.com/blog/entry/52854 | 1,652,678,550,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00614.warc.gz | 12,099,316 | 29,509 | ### rachitiitr's blog
By rachitiitr, history, 5 years ago,
Hi CF Community,
I think it's safe to assume that this is a new data structure for most of us.
Consider the following problems:
1. Number of elements in subarray A[L...R] that are less than or equal to y.
(Persistence Segment Tree? Ordered multiset + BIT ?)
2. Number of occurrences of element x in subarray A[L...R].
(Subpart of 1st problem)
3. The kth smallest element in subarray A[L...R].
(Ordered multiset + BIT would work for subarrays beginning from index 1)
I know you might have many other solutions, and you might think what I am trying to prove.
What if I told you, all of the above can be easily done in O(logn) using Wavelet Trees :o. Plus, its very easy to code :D Awesome, isn't it?
Check the implementation here.
The post just introduces the basic usage of wavelet trees. There is still more that you can do with them.
I will write about them later, once I gain enough sleep maybe?
• +256
» 5 years ago, # | +6 What about "change an element" queries? If we solve the problem offline, they can be reduced to two "toggle" queries. But is it possible to solve it online? For example, by using order statistic tree instead of BIT?Also can someone explain why this structure is called a "wavelet tree"?
• » » 5 years ago, # ^ | +6 Wikipedia:The name derives from an analogy with the wavelet transform for signals, which recursively decomposes a signal into low-frequency and high-frequency components.
» 5 years ago, # | +51 Author sleeping since 32 hours
• » » 5 years ago, # ^ | +17 Lol, I have just graduated from college and started off with a new job in a new place. So, I am busy with the new life and searching for flats which is tiring af.
» 5 years ago, # | -92 This DS is called HBB PRO Tree. Why the fuck do you call it wavelet tree?
» 5 years ago, # | 0 Very nice blog post.But I would like to add 1 and 2 can be done in O(logn) using regular segment tree if queries can be handled offline(and no updates).
• » » 5 years ago, # ^ | +5 Maybe I can't get how the first problem can be solved using segment tree offline,could you please explain it in detail? Thanks!
• » » » 5 years ago, # ^ | +10 Keep 2 kinds of queries:1. (x,i) : The ith position is supposed to be assigned value x2. (x,l,r): Count number of elements in [l,r] which are less than xSo obviously the queries are of the form (a[i],i) for i from 1 to n, and the range queries that we have. Now sort all the queries by x. Keep a segment tree over 1 to n. Initially, all the nodes have the value 0. Now whenever you encounter the first query, set the value at i to 1. For the second query the answer is just the number of 1's from l to r, which is a normal range sum query.You can see pretty easily that this works.
• » » » » 5 years ago, # ^ | +5 Got it,thanks a lot!
• » » » » » 5 years ago, # ^ | +10 See this link for more details — https://www.quora.com/How-will-you-solve-the-K-Query-using-segment-trees
» 5 years ago, # | +73 Is it correct that wavelet tree is conceptually just a merge-sort-like-tree built on top of values instead of positions?
• » » 5 years ago, # ^ | 0 Not sure what you mean, but probably not. Haven't you mixed up "values" and "positions"?Wavelet tree was designed to overcome binary alphabet size limitation of succinct bit-vectors with constant-time rank and select queries. Thus the essence of wavelet trees is splitting the sequence over alphabet into series of bit-vectors in a balanced way. It certainly doesn't merge nor sorts anything.
• » » » 5 years ago, # ^ | +15 No, I did not.By "merge-sort-like-tree" I mean the following: suppose we have a standard segment tree over an array, but each node stores (instead of a fixed-size value) a sorted list of all values in the corresponding subtree. Here is an example for array 4 6 2 1 7 3 8 5: [1 2 3 4 5 6 7 8] - values on positions 1-8 (node 1) [1 2 4 6][3 5 7 8] - values on positions 1-4, 5-8 (nodes 2 and 3) [4 6][1 2][3 7][5 8] - values on positions 1-2, 3-4, 5-6, 7-8 (nodes 4-7) [4][6][2][1][7][3][8][5] - values on positions 1, 2, ..., 8 (nodes 8-15) That data structure can easily calculate number of elements less than X whose positions are between L and R in — just split [L;R] into nodes and run binary search inside each node. With fractional cascading we can reduce that time into by running binary search on the top level only.Wavelet tree does something similar: instead of looking at values (1, 4); (2, 6); (3, 2); (4, 1); (5, 7); (6, 3); (7, 8); (8, 5) it looks at (4, 1); (6, 2); (2, 3); (1, 4); (7, 5); (3, 6); (8, 7); (5, 8) similar to (1, 4); (2, 3); (3, 6); (4, 1); (5, 8); (6, 2); (7, 5); (8, 7) and builds something similar to the structure we had above: each node corresponds to a sub-segment of possible values and contains their sorted positions (we have no need to store the positions explicitly, we just pretend that they're here): [1 2 3 4 5 6 7 8] - positions of values 1-8 (node 1) [1 3 4 6][2 5 7 8] - positions of values 1-4, 5-8 (nodes 2-3) [3 4][1 6][2 8][5 7] - positions of values 1-2, 3-4, 5-6, 7-8 (nodes 4-7) [4][3][6][1][8][2][5][7] - positions of values 1, 2, ..., 8 (nodes 8-15)
• » » » » 5 years ago, # ^ | 0 Seems like one could say that these data structures are transposed versions of each other.
» 5 years ago, # | 0 Having come up with another solution to your three problems,we can use a persistent segment tree to deal with all three problems online(We can deal with 1 and 2 directly,and for problem 3,we can do binary search on the persistent segment tree which is still O(logn)).So we can have an alternative way to solve your three problems in O(nlogn) memory and time complexity.
• » » 5 years ago, # ^ | 0 Could you explain how binary search on segment tree would be O(logn)?
• » » 5 years ago, # ^ | ← Rev. 2 → 0 Are you considering updates as well?
• » » 5 years ago, # ^ | 0 I haven't figured out how to do the "toogle" update with persistent segment tree — you have to actually update linear number of versions.Also, persistent segment tree is not, more strictly speaking, — it's memory and time, where n is the length of the array, m is the number of queries and V is the maximal value. By adding garbage collection we can reduce memory consumption to , but that makes code harder (and if you use standard pointer like shared_ptr it significantly increases hidden const).On the other hand, wavelet tree takes memory without garbage collection and time.
• » » » 5 years ago, # ^ | 0 If we can compress values, persistent segment tree is . If we can't, wavelet tree is also .
• » » » » 5 years ago, # ^ | 0 Fair enough. Is there any situation where we cannot compress values, but can build wavelet tree?
• » » » » » 5 years ago, # ^ | 0 For example, if we need to process online "push_back" updates.
• » » » 5 years ago, # ^ | 0 I didn't mean persistent segment tree is better than wavelet tree,and wavelet tree certainly does better in these kind of problems.I just point out that there exists an alternative way to solve these kind of problems.What's more,if we are asked to update values and even insert new values,we can use binary search tree without rotations(In China it's called Scape-Goat Tree) and segment tree to solve the problem in O(nlog^2n) time complexity.There is a problem of this kind which you're asked to ask queries of the k-th smallest element in a range as well as update and insert elements.Problem Link(in Chinese)
• » » » » 5 years ago, # ^ | +15 Would you mind elaborating what is Scape-Goat tree a little? Is it like segment tree for 109 leaves with dynamically allocated nodes?
• » » » » » 5 years ago, # ^ | +5 Scape-Goat Tree on WikipediaScape-Goat tree is a kind of binary search tree like Treap and Splay. But it makes itself balanced by reconstruction instead of rotation.Practically,if we find the size of the root's left son becomes larger than k*size of the tree or smaller than (1-k)*size of the tree,we just collect all its vertices and rebuild a new binary search tree.Usually k is set to be between 0.7 and 0.8.There is a proof why its complexity is O(nlogn) just as Treap and Splay.
• » » » » 4 years ago, # ^ | 0 Could you please elaborate how would we solve this problem with segment tree and space-goat tree?
» 5 years ago, # | ← Rev. 2 → 0 Nice Substitute For Persistent Segment Tree.But how Can we add Update part to it?
» 5 years ago, # | ← Rev. 2 → +31 Published version Direct link to the paper :)As one of the authors of the paper, if anyone has any doubt about the structure, my mail is bernardosubercaseaux@gmail.comPs: A paper about how to answer the same querys in a Tree in an efficient and elegant way is going to be published soon ;)
• » » 5 years ago, # ^ | 0 What is the complexity of solving 3rd problem from the blog using wavelet tree? Also, can it support updates?
• » » » 5 years ago, # ^ | 0 You should read the paper :) the first question is answered in detail there. The answer for the simplest case when you have an array of N elements (in a reasonable alphabet, for example, numbers < 10^9) is O(lg(N)) per query.For the updates, in the paper we covered some cases, like toggling, adjacent-swapping, push_back and push_front.
» 5 years ago, # | +4 Hi, I'm trying to use wavelet tree for MKTHNUM problem on SPOJ, but I keep getting runtime error.Can anyone tell me what's wrong with my submission? CodeI used the template and just changed the main function.
• » » 5 years ago, # ^ | ← Rev. 2 → +1 i also copied those code too, but getting wrong answer instead. here is my submission ideone
• » » » 5 years ago, # ^ | +4 Have you used index compression? The numbers can be up to 10^9
• » » » » 5 years ago, # ^ | +1 thanks, i use index compression and got AC. I safely assume RTE in your submission is not related about the DS.
• » » » » » 5 years ago, # ^ | 0 Can you please share your code , as i am not able to find the bug.raidnav
• » » » » » » 5 years ago, # ^ | 0
• » » » » 4 years ago, # ^ | 0 does compression matter for getting the correct output?
• » » » » » 4 years ago, # ^ | 0 Compression is needed due to the input constraint that can be up to 10^9. I don't think you can make any array of that size without compressing the value first..
• » » » 5 years ago, # ^ | 0 you can try with the implementation referenced in the paper https://github.com/nilehmann/wavelet-tree/also, for those of you lazy enough to not read the paper ;-) you can check the following slides for a quick view of how the WT workshttps://users.dcc.uchile.cl/~jperez/talks/ioi16.pdf
• » » » 5 years ago, # ^ | 0 I also tried the same problem and im getting WA. I implemented by myself but I compared with some other codes and seems pretty much the same. Can anyone help me? code
» 4 years ago, # | 0 Is it possible to support the following type of update using wavelet tree?Given l, r, x, add x to all the elements from al to ar.
• » » 4 years ago, # ^ | 0 Searching for same....
» 4 years ago, # | 0 Is It possible to delete Element Or Update the Value of element ?
• » » 4 years ago, # ^ | 0 Yes, but you have to use something else instead of arrays to store the bitarray for efficiencyI used a bst here
» 3 years ago, # | 0 I wonder if there is something that this can do but ordered multiset+seg tree can't ? P.S. I found coding seg tree much easier than this new data structure.
» 23 months ago, # | ← Rev. 3 → 0 It would be great if you (rachitiitr)can write another blog explaining about Wavelet Trees.
» 22 months ago, # | +6 Link is broken :(
• » » 22 months ago, # ^ | +12 Looks like this works.
• » » » 22 months ago, # ^ | 0 Thanks!
» 22 months ago, # | ← Rev. 3 → +9 I don't actually know the English name but there is indeed a similar data structure in China, called “主席树” or HJT seg-tree, and there are a bunch of problems like that in China.The main idea was to save a single-point change of a segment tree in $O(\log n)$ memory & time complexity.First, we do discretization for all numbers.Second, we build an empty seg-tree with the size of n, we'll let it be version[0].Then we let i from 1 to n, we make the element in place a[i] $+1$ and make a new version i (based on version[i-1])Similar to prefix, all data of numbers in $a_l$ to $a_r$ will be in version[r]-version[l-1], we can than perform these operations in $O(\log n)$ time.The total complexity will be $O(n\log n+q\log n)$ | 3,580 | 12,501 | {"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} | 2.796875 | 3 | CC-MAIN-2022-21 | latest | en | 0.915452 |
https://briefencounters.ca/30199/inverse-functions-worksheet-with-answers/ | 1,721,287,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00383.warc.gz | 132,038,857 | 20,515 | # Inverse Functions Worksheet with Answers
The Inverse Functions Worksheet with Answers helps you practice the skills you need to know to solve inverse equations. It helps you to understand the relationship between a function and its inverse. It also provides you with a way to find the graph of f in a graphing calculator. You can use the Inverse Functions Worksheet with Answers to practice the formula for a number of inverse equations.
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Inverse Functions Worksheet with Answers teaches students about trig functions. It includes steps to find inverses, examples, and solutions. The worksheet can be used to practice the inverses of other math problems. Inverse functions worksheets are useful for introducing trig functions in the classroom. The inverse of an arithmetic expression can be evaluated by using an arithmetic equation.
Inverse Functions Worksheet with Answers is a comprehensive study guide for learning about inverse functions. Inverse functions worksheets with answers teach students the definition of inverse functions and how to solve a problem using them. They are helpful for a number of reasons, such as practice with the trig function and evaluating it. The trig function is a great example of a geometric construction, but it is not always the best representation of a particular metric.
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The Inverse Functions Worksheet with Answers contains all the necessary information for students to learn about inverse functions. The worksheets contain a definition of a function, examples of the inverse, and methods to evaluate it. The inverses of a number are inputs to a linear function. They are a useful tool for analyzing the inverses of two dimensional objects, such as triangles and circles.
The Inverse Functions Worksheet with Answers contains a definition of inverse functions and the steps to find them. It includes examples and solutions to problems related to inverses and other geometric constructions. The inverse function of a triangle is a rectangular ellipse of a circle. The domain and range of a rectangle are equal to their length. When this ellipse becomes an inverse function of a line, it will have a radius of one square inch.
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If you’re looking for an Inverse Functions Worksheet with Answers, you’ve come to the right place. This resource includes information about inverses, a definition of a graph, and steps to find an inverse function. The answers provided are both examples and solutions of common problems. Inverses of a sphere can be used to graph a polar coordinate. If the axis of a circle has two ends in the same direction, it can be used as a guide to determine which side is the corresponding polar direction.
Inverse Functions Worksheet will provide you with an overview of how to find an inverse function. It also includes examples and a practice sheet with exercises on finding inverses, graphing inverses, and using a graphing calculator. The worksheet will also provide answers to some of the problems that you’ll encounter in solving a given inverse function. It will prove to be helpful in the upcoming tests you take in algebra, geometry, and other subjects.
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You’ll also need to know how to write an inverse equation. Inverse equations are a good way to practice the concepts of an inverse. By utilizing a worksheet with a specific formula, you can find out a matrix’s inverse. You can use this information to compute the corresponding y value. The y-value of the y-function is a function of a sine or cosine. | 942 | 5,048 | {"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} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.894206 |
petmuh.blogspot.com | 1,701,294,773,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100146.5/warc/CC-MAIN-20231129204528-20231129234528-00339.warc.gz | 530,604,563 | 9,398 | ## 3 Aralık 2008 Çarşamba
### Compressibility of Natural Gases
For a liquid phase, the compressibility is small and usually assumed to be constant. For a gas phase, the compressibility is neither small nor constant.
By definition, the isothermal gas compressibility is the change in volume per unit volume for a unit change in pressure or, in equation form:
where cg = isothermal gas compressibility, 1/psi
## 15 Mayıs 2008 Perşembe
### Incompressible Fluids
An incompressible fluid is defined as the fluid whose volume (or density) does not change with pressure, i.e.:
Incompressible fluids do not exist; this behavior, however, may be assumed in some cases to simplify the derivation and the final form of many flow equations.
Ahmed (2001)
### Isothermal Compressibility Coefficient of Crude Oil
Isothermal compressibility coefficients are required in solving many reservoir engineering problems, including transient fluid flow problems, and they are also required in the determination of the physical properties of the undersaturated crude oil.
By definition, the isothermal compressibility of a substance is defined mathematically by the following expressions:
• In terms of fluid volume:
• In terms of fluid density:
where V and ρ are the volume and density of the fluid, respectively.
For a crude oil system, the isothermal compressibility coefficient of the oil phase co is defined for pressures above the bubble-point by one of the following equivalent expressions:
where
• co = isothermal compressibility, 1/psi
• ρo = oil density lb/ft^3
• Bo = oil formation volume factor, bbl/STB
At pressures below the bubble-point pressure, the oil compressibility is defined as:
where Bg = gas formation volume factor, bbl/scf
Ahmed (2001)
When the pressure at different locations in the reservoir is declining linearly as a function of time, i.e., at a constant declining rate, the flowing condition is characterized as the pseudosteady-state flow. Mathematically, this definition states that the rate of change of pressure with respect
to time at every position is constant, orIt should be pointed out that the pseudosteady-state flow is commonly referred to as semisteady-state flow and quasisteady-state flow.
Ahmed (2001)
The unsteady-state flow (frequently called transient flow) is defined as the fluid flowing condition at which the rate of change of pressure with respect to time at any position in the reservoir is not zero or constant. This definition suggests that the pressure derivative with respect to time is essentially a function of both position i and time t, thus
Ahmed (2001) | 574 | 2,615 | {"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} | 3.375 | 3 | CC-MAIN-2023-50 | latest | en | 0.89902 |
https://allbusinesshoursnow.com/60-hours/ | 1,719,350,281,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00751.warc.gz | 70,796,561 | 13,269 | 60 Hours
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10. Why is a minute divided into 60 seconds … – Scientific American
https://www.scientificamerican.com/article/experts-time-division-days-hours-minutes/
Why is a minute divided into 60 seconds, an hour into 60 minutes, yet there are only 24 hours in a day? March 5, 2007. Share on Facebook. Share on Twitter.
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Categories Uncategorized | 871 | 3,143 | {"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} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.876869 |
https://web.mit.edu/6.013_book/www/chapter9/9.4.html | 1,653,699,021,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00317.warc.gz | 668,010,447 | 8,316 | ## 9.4Magnetization Constitutive Laws
The permanent magnetization model of Sec. 9.3 is a somewhat artificial example of the magnetization density M specified, independent of the magnetic field intensity. Even in the best of permanent magnets, there is actually some dependence of M on H.
Constitutive laws relate the magnetization density M or the magnetic flux density B to the macroscopic H within a material. Before discussing some of the more common relations and their underlying physics, it is well to have in view an experiment giving direct evidence of the constitutive law of magnetization. The objective is to observe the establishment of H by a current in accordance with Ampère's law, and deduce B from the voltage it induces in accordance with Faraday's law.
Figure 9.4.1 Toroidal coil with donut-shaped magnetizable core.
#### Example 9.4.1. Toroidal Coil
A coil of toroidal geometry is shown in Fig. 9.4.1. It consists of a donut-shaped core filled with magnetizable material with N1 turns tightly wound on its periphery. By means of a source driving its terminals, this coil carries a current i. The resulting current distribution can be assumed to be so smooth that the fine structure of the field, caused by the finite size of the wires, can be disregarded. We will ignore the slight pitch of the coil and the associated small current component circulating around the axis of the toroid.
Because of the toroidal geometry, the H field in the magnetizable material is determined by Ampère's law and symmetry considerations. Symmetry about the toroidal axis suggests that H is directed. The integral MQS form of Ampère's law is written for a contour C circulating about the toroidal axis within the core and at a radius r. Because the major radius R of the torus is large compared to the minor radius w, we will ignore the variation of r over the cross-section of the torus and approximate r by an average radius R. The surface S spanned by this contour and shown in Fig. 9.4.2 is pierced N1 times by the current i, giving a total current of N1 i. Thus, the azimuthal field inside the core is essentially
Note that the same argument shows that the magnetic field intensity outside the core is zero.
Figure 9.4.2 Surface \$S\$ enclosed by contour C used with Ampère's integral law to determine H in the coil shown in Fig. 9.4.1.
In general, if we are given the current distribution and wish to determine H, recourse must be made not only to Ampère's law but to the flux continuity condition as well. In the idealized toroidal geometry, where the flux lines automatically close on themselves without leaving the magnetized material, the flux continuity condition is automatically satisfied. Thus, in the toroidal configuration, the H imposed on the core is determined by a measurement of the current i and the geometry.
How can we measure the magnetic flux density in the core? Because B appears in Faraday's law of induction, the measurement of the terminal voltage of an additional coil, having N2 turns also wound on the donut-shaped core, gives information on B. The terminals of this coil are terminated in a high enough impedance so that there is a negligible current in this second winding. Thus, the H field established by the current i remains unaltered.
The flux linked by each turn of the sensing coil is essentially the flux density multiplied by the cross-sectional area w2/4 of the core. Thus, the flux linked by the terminals of the sensing coil is
and flux density in the core material is directly reflected in the terminal flux-linkage.
The following demonstration shows how (1) and (2) can be used to infer the magnetization characteristic of the core material from measurement of the terminal current and voltage of the first and second coils.
#### Demonstration 9.4.1. Measurement of B-H Characteristic
The experiment shown in Fig. 9.4.3 displays the magnetization characteristic on the oscilloscope. The magnetizable material is in the donut-shaped toroidal configuration of Example 9.4.1 with the N1-turn coil driven by a current i from a Variac. The voltage across a series resistance then gives a horizontal deflection of the oscilloscope proportional to H, in accordance with (1).
Figure 9.4.3 Demonstration in which the B-H curve is traced out in the sinusoidal steady state.
The terminals of the N2 turn-coil are connected through an integrating network to the vertical deflection terminals of the oscilloscope. Thus, the vertical deflection is proportional to the integral of the terminal voltage, to , and hence through (2), to B.
In the discussions of magnetization characteristics which follow, it is helpful to think of the material as comprising the core of the torus in this experiment. Then the magnetic field intensity H is proportional to the current i, while the magnetic flux density B is reflected in the voltage induced in a coil linking this flux.
Many materials are magnetically linear in the sense that
Here m is the magnetic susceptibility. More commonly, the constitutive law for a magnetically linear material is written in terms of the magnetic flux density, defined by (9.2.8).
According to this law, the magnetization is taken into account by replacing the permeability of free space o by the permeability of the material. For purposes of comparing the magnetizability of materials, the relative permeability /o is often used.
Typical susceptibilities for certain elements, compounds, and common materials are given in Table 9.4.1. Most common materials are only slightly magnetizable. Some substances that are readily polarized, such as water, are not easily magnetized. Note that the magnetic susceptibility can be either positive or negative and that there are some materials, notably iron and its compounds, in which it can be enormous. In establishing an appreciation for the degree of magnetizability that can be expected of a material, it is helpful to have a qualitative picture of its microscopic origins, beginning at the atomic level but including the collective effects of groups of atoms or molecules that result when they become as densely packed as they are in solids. These latter effects are prominent in the most easily magnetized materials.
TABLE 9.4.1RELATIVE SUSCEPTIBILITIES OF COMMON MATERIALS
Material m PARAMAGNETIC Mg 1.2 x 10-5 Al 2.2 x 10-5 Pt 3.6 x 10-4 air 3.6 x 10-7 O2 2.1 x 10-6 DIAMAGNETIC Na -0.24 x 10-5 Cu -1.0 x 10-5 diamond -2.2 x 10-5 Hg -3.2x 10-5 H2O -0.9 x 10-5 FERROMAGNETIC Fe (dynamo sheets) 5.5 x 103 Fe (lab specimens) 8.8 x 104 Fe (crystals) 1.4 x 106 Si-Fe transformer sheets 7 x 104 Si-Fe crystals 3.8 x 106 -metal 105 FERRIMAGNETIC Fe3O4 100 ferrites 5000
The magnetic moment of an atom (or molecule) is the sum of the orbital and spin contributions. Especially in a gas, where the atoms are dilute, the magnetic susceptibility results from the (partial) alignment of the individual magnetic moments by a magnetic field. Although the spin contributions to the moment tend to cancel, many atoms have net moments of one or more Bohr magnetons. At room temperature, the orientations of the moments are mostly randomized by thermal agitation, even under the most intense fields. As a result, an applied field can give rise to a significant magnetization only at very low temperatures. A paramagnetic material displays an appreciable susceptibility only at low temperatures.
If, in the absence of an applied field, the spin contributions to the moment of an atom very nearly cancel, the material can be diamagnetic, in the sense that it displays a slightly negative susceptibility. With the application of a field, the orbiting electrons are slightly altered in their circulations, giving rise to changes in moment in a direction opposite to that of the applied field. Again, thermal energy tends to disorient these moments. At room temperature, this effect is even smaller than that for paramagnetic materials.
At very low temperatures, it is possible to raise the applied field to such a level that essentially all the moments are aligned. This is reflected in the saturation of the flux density B, as shown in Fig. 9.4.4. At low field intensity, the slope of the magnetization curve is , while at high field strengths, there are no more moments to be aligned and the slope is o. As long as the field is raised and lowered at a rate slow enough so that there is time for the thermal energy to reach an equilibrium with the magnetic field, the B-H curve is single valued in the sense that the same curve is followed whether the magnetic field is increasing or decreasing, and regardless of its rate of change.
Figure 9.4.4 Typical magnetization curve without hysteresis. For typical ferromagnetic solids, the saturation flux density is in the range of 1-2 Tesla. For ferromagnetic domains suspended in a liquid, it is .02-.04 Tesla.
Until now, we have been considering the magnetization of materials that are sufficiently dilute so that the atomic moments do not interact with each other. In solids, atoms can be so closely spaced that the magnetic field due to the moment of one atom can have a significant effect on the orientation of another. In ferromagnetic materials, this mutual interaction is all important.
To appreciate what makes certain materials ferromagnetic rather than simply paramagnetic, we need to remember that the electrons which surround the nuclei of atoms are assigned by quantum mechanical principles to layers or "shells." Each shell has a particular maximum number of electrons. The electron behaves as if it possessed a net angular momentum, or spin, and hence a magnetic moment. A filled shell always contains an even number of electrons which are distributed spatially in such a manner that the total spin, and likewise the magnetic moment, is zero.
For the majority of atoms, the outermost shell is unfilled, and so it is the outermost electrons that play the major role in determining the net magnetic moment of the atom. This picture of the atom is consistent with paramagnetic and diamagnetic behavior. However, the transition elements form a special class. They have unfilled inner shells, so that the electrons responsible for the net moment of the atom are surrounded by the electrons that interact most intimately with the electrons of a neighboring atom. When such atoms are as closely packed as they are in solids, the combination of the interaction between magnetic moments and of electrostatic coupling results in the spontaneous alignment of dipoles, or ferromagnetism. The underlying interaction between atoms is both magnetic and electrostatic, and can be understood only by invoking quantum mechanical arguments.
In a ferromagnetic material, atoms naturally establish an array of moments that reinforce. Nevertheless, on a macroscopic scale, ferromagnetic materials are not necessarily permanently magnetized. The spontaneous alignment of dipoles is commonly confined to microscopic regions, called domains. The moments of the individual domains are randomly oriented and cancel on a macroscopic scale.
Macroscopic magnetization occurs when a field is applied to a solid, because those domains that have a magnetic dipole moment nearly aligned with the applied field grow at the expense of domains whose magnetic dipole moments are less aligned with the applied field. The shift in domain structure caused by raising the applied field from one level to another is illustrated in Fig. 9.4.5. The domain walls encounter a resistance to propagation that balances the effect of the field.
Figure 9.4.5 Polycrystalline ferromagnetic material viewed at the domain level. In the absence of an applied magnetic field, the domain moments tend to cancel. (This presumes that the material has not been left in a magnetized state by a previously applied field.) As a field is applied, the domain walls shift, giving rise to a net magnetization. In ideal materials, saturation results as all of the domains combine into one. In materials used for bulk fabrication of transformers, imperfections prevent the realization of this state.
A typical trajectory traced out in the B-H plane as the field is applied to a typical ferromagnetic material is shown in Fig. 9.4.6. If the magnetization is zero at the outset, the initial trajectory followed as the field is turned up starts at the origin. If the field is then turned down, the domains require a certain degree of coercion to reduce their average magnetization. In fact, with the applied field turned off, there generally remains a flux density, and the field must be reversed to reduce the flux density to zero. The trajectory traced out if the applied field is slowly cycled between positive and negative values many times is the one shown in the figure, with the remanence flux density Br when H = 0 and a coercive field intensity Hc required to make the flux density zero. Some values of these parameters, for materials used to make permanent magnets, are given in Table 9.4.2.
TABLE 9.4.2 MAGNETIZATION PARAMETERS FOR PERMANENT MAGNET
Material Hc (A/m) Br (Tesla) Carbon steel 4000 1.00 Alnico 2 43,000 0.72 Alnico 7 83,500 0.70 Ferroxdur 2 143,000 0.34
Figure 9.4.6 Magnetization characteristic for material showing hysteresis with typical values of Br and Hc given in Table 9.4.2. The curve is obtained after many cycles of sinusoidal excitation in apparatus such as that of Fig. 9.4.3. The trajectory is traced out in response to a sinusoidal current, as shown by the inset.
In the toroidal geometry of Example 9.4.1, H is proportional to the terminal current i. Thus, imposition of a sinusoidally varying current results in a sinusoidally varying H, as illustrated in Fig. 9.4.6b. As the i and hence H increases, the trajectory in the B-H plane is the one of increasing H. With decreasing H, a different trajectory is followed. In general, it is not possible to specify B simply by giving H (or even the time derivatives of H). When the magnetization state reflects the previous states of magnetization, the material is said to be hysteretic. The B-H trajectory representing the response to a sinusoidal H is then called the hysteresis loop.
Hysteresis can be both harmful and useful. Permanent magnetization is one result of hysteresis, and as we illustrated in Example 9.3.2, this can be the basis for the storage of information on tapes. When we develop a picture of energy dissipation in Chap. 11, it will be clear that hysteresis also implies the generation of heat, and this can impose limits on the use of magnetizable materials.
Liquids having significant magnetizabilities have been synthesized by permanently suspending macroscopic particles composed of single ferromagnetic domains. Here also the relatively high magnetizability comes from the ferromagnetic character of the individual domains. However, the very different way in which the domains interact with each other helps in gaining an appreciation for the magnetization of ferromagnetic polycrystalline solids.
In the absence of a field imposed on the synthesized liquid, the thermal molecular energy randomizes the dipole moments and there is no residual magnetization. With the application of a low frequency H field, the suspended particles assume an average alignment with the field and a single-valued B-H curve is traced out, typically as shown in Fig. 9.4.4. However, as the frequency is raised, the reorientation of the domains lags behind the applied field, and the B-H curve shows hysteresis, much as for solids.
Although both the solid and the liquid can show hysteresis, the two differ in an important way. In the solid, the magnetization shows hysteresis even in the limit of zero frequency. In the liquid, hysteresis results only if there is a finite rate of change of the applied field.
Ferromagnetic materials such as iron are metallic solids and hence tend to be relatively good electrical conductors. As we will see in Chap. 10, this means that unless care is taken to interrupt conduction paths in the material, currents will be induced by a time-varying magnetic flux density. Often, these eddy currents are undesired. With the objective of obtaining a highly magnetizable insulating material, iron atoms can be combined into an oxide crystal. Although the spontaneous interaction between molecules that characterizes ferromagnetism is indeed observed, the alignment of neighbors is antiparallel rather than parallel. As a result, such pure oxides do not show strong magnetic properties. However, a mixed-oxide material like Fe3O4 (magnetite) is composed of sublattice oxides of differing moments. The spontaneous antiparallel alignment results in a net moment. The class of relatively magnetizable but electrically insulating materials are called ferrimagnetic.
Our discussion of the origins of magnetization began at the atomic level, where electronic orbits and spins are fundamental. However, it ends with a discussion of constitutive laws that can only be explained by bringing in additional effects that occur on scales much greater than atomic or molecular. Thus, the macroscopic B and H used to describe magnetizable materials can represent averages with respect to scales of domains or of macroscopic particles. In Sec. 9.5 we will make an artificial diamagnetic material from a matrix of "perfectly" conducting particles. In a time-varying magnetic field, a magnetic moment is induced in each particle that tends to cancel that being imposed, as was shown in Example 8.4.3. In fact, the currents induced in the particles and responsible for this induced moment are analogous to the induced changes in electronic orbits responsible on the atomic scale for diamagnetism[1]. | 3,829 | 17,719 | {"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} | 2.671875 | 3 | CC-MAIN-2022-21 | latest | en | 0.926416 |
https://kr.mathworks.com/matlabcentral/cody/problems/2120-rounding-off-numbers-to-n-decimals/solutions/2005152 | 1,575,755,641,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540502120.37/warc/CC-MAIN-20191207210620-20191207234620-00451.warc.gz | 442,151,894 | 15,757 | Cody
# Problem 2120. Rounding off numbers to n decimals
Solution 2005152
Submitted on 6 Nov 2019 by Hung Trinh
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; n=1; y_correct = 1; assert(isequal(myround(x,n),y_correct))
r = 1 y = 1
2 Pass
x = pi; n=5; y_correct = 3.14159; assert(isequal(myround(x,n),y_correct))
r = 5 y = 3.1416
3 Pass
x = 0.5*sqrt(2); n=6; y_correct = 0.707107; assert(isequal(myround(x,n),y_correct))
r = 6 y = 0.7071
4 Pass
x = exp(1); n=9; y_correct = 2.718281828; assert(isequal(myround(x,n),y_correct))
r = 9 y = 2.7183
5 Pass
x = 0.00123456; n=6; y_correct = 0.001235; assert(isequal(myround(x,n),y_correct))
r = 6 y = 0.0012 | 305 | 796 | {"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} | 2.890625 | 3 | CC-MAIN-2019-51 | latest | en | 0.547104 |
https://www.convertunits.com/from/hectare+meter/day/to/lambda/day | 1,702,183,246,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101195.85/warc/CC-MAIN-20231210025335-20231210055335-00426.warc.gz | 550,185,454 | 12,591 | ## Convert hectare metre/day to lambda/day
hectare meter/day lambda/day
How many hectare meter/day in 1 lambda/day? The answer is 1.0E-13. We assume you are converting between hectare metre/day and lambda/day. You can view more details on each measurement unit: hectare meter/day or lambda/day The SI derived unit for volume flow rate is the cubic meter/second. 1 cubic meter/second is equal to 8.64 hectare meter/day, or 86400000000000 lambda/day. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between hectare meters/day and lambdas/day. Type in your own numbers in the form to convert the units!
## Quick conversion chart of hectare meter/day to lambda/day
1 hectare meter/day to lambda/day = 10000000000000 lambda/day
2 hectare meter/day to lambda/day = 20000000000000 lambda/day
3 hectare meter/day to lambda/day = 30000000000000 lambda/day
4 hectare meter/day to lambda/day = 40000000000000 lambda/day
5 hectare meter/day to lambda/day = 50000000000000 lambda/day
6 hectare meter/day to lambda/day = 60000000000000 lambda/day
7 hectare meter/day to lambda/day = 70000000000000 lambda/day
8 hectare meter/day to lambda/day = 80000000000000 lambda/day
9 hectare meter/day to lambda/day = 90000000000000 lambda/day
10 hectare meter/day to lambda/day = 1.0E+14 lambda/day
## Want other units?
You can do the reverse unit conversion from lambda/day to hectare meter/day, or enter any two units below:
## Enter two units to convert
From: To:
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 509 | 2,008 | {"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} | 2.65625 | 3 | CC-MAIN-2023-50 | latest | en | 0.84283 |
https://mapleprimes.com/users/Preben%20Alsholm/replies?page=1 | 1,726,767,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00158.warc.gz | 347,126,193 | 89,199 | ## 13653 Reputation
19 years, 295 days
## Maximize doesn't do well...
@C_R Unfortunately, Maximize doesn't do well:
```restart;
G:= (x,y)->(-2.4 + 25.20*y^2 - 18.48*y^3)*x^3 + (2.1 - 30.87*y^2 + 23.10*y^3)*x^2 + (-0.84*y^2 + 0.66*y^3)*x + 1. + 2.1*y^2 - 2.2*y^3;
Optimization:-Minimize(G(x,y),x=0..1,y=0..1);
Optimization:-Maximize(G(x,y),x=0..1,y=0..1); # Wrong
plot(G(x,x),x=0..1/5);
```
Maximize gets the result [1., [x = 0., y = 0.]] , which is clearly wrong. The point (0,0) isn't a local maximum, in fact it is a local minimum:
```gxyE:=expand(G(x,y));
remove(s->degree(s)>2,gxyE);
```
Result: 1. + 2.1*x^2 + 2.1*y^2. Thus (0,0) is a local minimum.
## An old question of yours?...
Didn't you report this difference earlier? I surely remember seeing something at least very similar from your hand.
## numpoints...
If you add the option numpoints=10000 you get a much better curve:
sol_plot := plots:-odeplot(sol, [[x(t), y(t)]], 0 .. 100, numpoints = 10000, color = "blue")
## A bug...
Since the help page clearly says:
variables: (optional) name or set or list of names; unknown(s) for which to solve.
this is a bug.
If you just use solve(eqs) then NULL is returned (i.e. nothing).
## Matrix Browser in versions previous to M...
@Christopher2222 Thank you! Indeed there is a help page for Matrix Browser in Maple versions 2021, 2022, and 2023.
Whereas I can get double clicking in Maple 2024 work without any problem I'm not able to do that in those other versions. That's what made me write that it is a new feature.
I don't know why the feature doesn't work for me in those previous versions. They are on the same computer as Maple 2024.
PS. I checked on old laptop, which also has Maple 2021.
Same refusal to react to double click.
PPS. It has to do with my setting of typesetting. I use standard in all versions including Maple 2024.
Double clicking works on the previous versions only for typesetting = extended.
## New feature...
@C_R I looked at the help page ? Browse Matrix: Viewing Arrays, Matrices, or Vectors.
I agree that the Display setting Structure combined with a Colormap setting of Cool (or Hot) shows a color that doesn't correspond with the legend, but only if all elements are actually zero.
Same error with Display = Magnitude
This Matrix Browser is a new feature in Maple 2024.
This appears to me to be a bug.
If you consider Matrix(11,11,fill=17) then there is no problem.
## Experiments...
Why expect a black image?
Try experimenting:
```LinearAlgebra:-RandomMatrix(11,11);
Matrix(11,11,fill=17);
Matrix(11,11,(i,j)-> `if`(i=1,j,5));
Matrix(11,11,(i,j)-> `if`(i>1,j,5));
```
You could also experiment with the menu Colormap inside the browser.
There you can choose Grayscale, Hue, Hot, or Cool.
If you choose Cool your Matrix(11,11) will be black; choosing Hot it will be white.
## Yes, thank you...
@Thomas Richard I got my old laptop up and running and could just confirm that.
I keep that computer simply because it has my old Maple versions.
## Older versions...
@ On older versions kernelopts(versionnumber) doesn't exist, but kernelopts(version) does.
On this computer I have besides the most recent versions Maple 12; no kernelopts(versionnumber)
.
## Assumptions...
I get the same crash in Windows 11 and Maple 2024.0.
With certain assumptions I get a result:
```restart;
sol:=(3^(1/2)*u(x)-1/3*3^(1/2)+(3*u(x)^2-2*u(x)-1)^(1/2))^(1/3*3^(1/2)) = x^(1/3*3^(1/2))*c__1;
eq:=eval(sol,u(x) = u);
plot(lhs(eq),u=-5..5);
solve(eq,u) assuming x>1,c__1>0; # OK
# solve(eq,u) assuming x>1,c__1>-1; #Crash
```
## Packages in Maple...
@Ronny Yes, MultiSeries is implemented as a package.
Technically, this means it is a module with option package.
Executing with(MultiSeries) makes its content:
[AddFunction, FunctionSupported, GetFunction, LeadingTerm, RemoveFunction, SeriesInfo, asympt, limit, multiseries, series, taylor]
available in the short form as in this case limit.
This can be very convenient.
Even after having executed with(MultiSeries), the usual limit is still available as :-limit
```with(MultiSeries);
limit(CylinderU(0,CylinderU(0,x)),x=0); # Fine
:-limit(CylinderU(0,CylinderU(0,x)),x=0); # The old wrong result.
```
## Definite integration doesn't hang...
```int(A,x=0..xx,method=_RETURNVERBOSE);
```
The only successful are ftoc and msftoc with result xx*exp(-1/2).
The same result if simplify(A) is the input. Somewhat strange!
## FAIL with assuming...
Well, it's somewhat better if you assume that n::posint:
```is(G(2*n) = G(2*n + 1) + G(2*n + 2)) assuming n::posint;
```
Certainly checking concrete cases results in true:
```{seq(is(G(2*n) = G(2*n + 1) + G(2*n + 2)),n=1..100)}; # a set of one element: true
```
## dsolve 2...
@nm In your "counter example" dsolve returns a solution, which odetest verifies:
```ode:=exp(diff(y(x), x) - y(x)) - diff(y(x), x)^2 + 1 = 0;
sol:=dsolve(ode);
odetest(sol,ode); # 0```
If I understand you correctly, by counter example you mean an example showing that this general statement is false:
"It is always better to use PDEtools:-Solve rather than solve".
## dsolve...
Just a remark.
dsolve(ode) returns
y(x) = Int(RootOf(-x + ln(_Z) + sin(_Z)), x) + c__1.
odetest is made so that it certainly verifies solutions coming from dsolve.
PDEtools:-Solve, odetest, and dsolve have the same author as I believe you know.
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| 1,630 | 5,406 | {"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} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.860613 |
https://www.enotes.com/homework-help/calculate-area-triangle-has-length-legs-6cmm-8cm-442118 | 1,516,597,384,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890991.69/warc/CC-MAIN-20180122034327-20180122054327-00167.warc.gz | 907,778,506 | 9,903 | # Calculate the area of a triangle if it has length of legs as 6cmm, 8cm and 10cm.
jeew-m | Certified Educator
Here we must look at the length closely. We can find that they are combined in the following manner.
`6^2+8^2 = 10^2`
This means the triangle is a right triangle.
The area of the right triangle is given by;
A = 1/2(product of leg length except hypotenuse)
So the area of the triangle can be given as;
`A = 1/2xx6xx8`
`A = 24cm^2`
So the area of the triangle given is `24cm^2` .
lemjay | Certified Educator
Since the length of the three sides of the triangle are given, to determine its area, apply Heron's formula.
`A=sqrt(s(s-a)(s-b)(s-c))`
where a, b and c are the sides of the triangle, and s is half of its perimeter.
So,
`s=(a+b+c)/2=(6+8+10)/2=24/2=12`
`A=sqrt(12(12-6)(12-8)(12-10))=sqrt(12*6*4*2)`
`A=sqrt576=24`
Hence, the area of the triangle is `24 cm^2` . | 298 | 897 | {"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} | 4.5 | 4 | CC-MAIN-2018-05 | latest | en | 0.873884 |
http://alexbowe.com/wavelet-trees/ | 1,534,693,802,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215222.74/warc/CC-MAIN-20180819145405-20180819165405-00167.warc.gz | 19,495,381 | 13,411 | # Wavelet Trees – an Introduction
Today I will talk about an elegant way of answering rank queries on sequences over larger alphabets – a structure called the Wavelet Tree. In my last post I introduced a data structure called RRR, which is used to quickly answer rank queries on binary sequences, and provide implicit compression.
A Wavelet Tree organises a string into a hierarchy of bit vectors. A rank query has time complexity is $\mathcal{O}(\log_2{A})$, where $A$ is the size of the alphabet. It was introduced by Grossi, Gupta and Vitter in their 2003 paper High-order entropy-compressed text indexes [4] (see the Further Reading section for more papers). It has since been featured in many papers [1, 2, 3, 5, 6].
If you store the bit vectors in RRR sequences, it may take less space than the original sequence. Alternatively, you could store the bit vectors in the rank indexes proposed by Sadakane and Okonohara [7]. It has a different approach to compression. I will talk about it another time ;) – fortunately, I will be studying under Sadakane-sensei at a later date (update: now I’m doing my Ph.D. under him in Tokyo).
In a different future post, I will show how Suffix Arrays can be used to find arbitrary patterns of length $P$, by issuing $2P$ rank queries. If using a Wavelet Tree, this means a pattern search has $\mathcal{O}(P \log_2{A})$ time complexity, that is, the size of size of the ‘haystack’ doesn’t matter, it instead depends on the size of the ‘needle’ and size of the alphabet.
## Constructing a Wavelet Tree
A Wavelet Tree converts a string into a balanced binary-tree of bit vectors, where a $0$ replaces half of the symbols, and a $1$ replaces the other half. This creates ambiguity, but at each level this alphabet is filtered and re-encoded, so the ambiguity lessens, until there is no ambiguity at all.
The tree is defined recursively as follows:
1. Take the alphabet of the string, and encode the first half as $0$, the second half as $1$: $\{ a, b, c, d \}$ would become $\{ 0, 0, 1, 1 \}$;
2. Group each $0$-encoded symbol, $\{ a, b \}$, as a sub-tree;
3. Group each $1$-encoded symbol, $\{ c, d \}$, as a sub-tree;
4. Reapply this to each subtree recursively until there is only one or two symbols left (when a $0$ or $1$ can only mean one thing).
For the string "Peter Piper picked a peck of pickled peppers" (spaces and a string terminator have been represented as $\_$ and $\$$respectively, due to convention in the literature) the Wavelet Tree would look like this: note: the strings aren’t actually stored, but are shown here for convenience It has the alphabet \{ \, P, \_, a, c, d, e, f, i, k, l, o, p, r, s, t \}, which would be mapped to \{ 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1 \}. So, for example, \$$ would map to$0$, and$r$would map to$1$. The left subtree is created by taking just the 0-encoded symbols$\{ \$, P, \_, a, c, d, e, f \}$ and then re-encoding them by dividing this new alphabet: $\{ 0, 0, 0, 0, 1, 1, 1, 1 \}$. Note that on the first level an $e$ would be encoded as a $0$, but now it is encoded as a $1$ (it becomes a $0$ again at a leaf node).
We can store the bit vectors in RRR structures for fast binary rank queries (which are needed, as described below), and compression :) In fact, since it is a balanced tree, we can concatenate each of the levels and store it as one single bit vector.
## Querying a Wavelet Tree
Recall from my last post that a rank query is the count of $1$-bits up to a specified position. Rank queries over larger alphabets are analogous – instead of a $1$, it may be any other symbol:
After the tree is constructed, a rank query can be done with log $A$ ($A$ is alphabet size) binary rank queries on the bit vectors – $\mathcal{O}(1)$ if you store them in RRR or another binary rank index. The encoding at each internal node may be ambiguous, but of course it isn’t useless – we use the ambiguous encoding to guide us to the appropriate sub-tree, and keep doing so until we have our answer.
For example, if we wanted to know $rank(5, e)$, we use the following procedure which is illustrated below. We know that $e$ is encoded as $0$ at this level, so we take the binary rank query of $0$ at position $5$:
Which is $4$, which we then use to indicate where to rank in the $0$-child: the $4^{th}$ bit (or the bit at position $3$, due to $0$-basing). We know to query the $0$-child, since that is what $e$ was encoded as at the parent level. We then repeat this recursively:
At a leaf node we have our answer. I would love to explain why this works, but it is fun and rewarding to think about it yourself ;)
There are also ways to provide fast select queries, but once again I will leave that up to you to research. The curious among you might also be interested in the Huffman-Shaped Wavelet Tree described by Mäkinen and Navarro [5].
## Using Your New Powers for Good
Feel free to implement this yourself, but if you want to get your hands dirty right away, all-around-clever-guy Francisco Claude has made an implementation available in his Compressed Data Structure Library (libcds). If you create something neat with it be sure to report back ;)
Update: Terence Siganakis wrote a blog post about Wavelet Trees that made it to the front page of Hacker News, encouraging an interesting discussion. The discussion is here.
And if you read this far, consider following me on Twitter: @alexbowe.
I didn’t want to saturate this blog post with proofs and other details, since it was meant to be a light introduction. If you want to dive deeper into this beautiful structure, check out the following papers:
[1] F. Claude and G. Navarro. Practical rank/select queries over arbitrary sequences. In Proceedings of the 15th International Symposium on String Processing and Information Retrieval (SPIRE), LNCS 5280, pages 176–187. Springer, 2008.
[2] P. Ferragina, R. Giancarlo, and G. Manzini. The myriad virtues of wavelet trees. Information and Computation, 207(8):849–866, 2009.
[3] P. Ferragina, G. Manzini, V. M ̈akinen, and G. Navarro. Compressed representations of sequences and full-text indexes. ACM Transactions on Algorithms, 3(2):20, 2007.
[4] R. Grossi, A. Gupta, and J. Vitter. High-order entropy-compressed text indexes. In Proceedings of the 14th annual ACM-SIAM symposium on Dis- crete algorithms, pages 841–850. Society for Industrial and Applied Mathematics, 2003.
[5] V. Mäkinen and G. Navarro. Succinct suffix arrays based on run-length encoding. Nordic Journal of Computing, 12(1):40–66, 2005.
[6] V. Mäkinen and G. Navarro. Implicit compression boosting with applications to self-indexing. In Proceedings of the 14th International Symposium on String Processing and Information Retrieval (SPIRE), LNCS 4726, pages 214–226. Springer, 2007.
[7] D. Okanohara and K. Sadakane. Practical entropy-compressed rank/select dictionary. Arxiv Computing Research Repository, abs/cs/0610001, 2006.
• Victor Smirnov
Hi Alex
Could you please take a look at my open source project, it is AI-targeted compact and compressed data structures framework.
https://bitbucket.org/vsmirnov/memoria/wiki/Home
As a highlight that might be interested to you it provides multiary wavelet tree implementation. Hope for your comments…
• Bronson
Isn’t your leftmost tree actually wrong ?
e.g: the parent of leftmost leaf ‘P_P__a____’ has alphabet { $, P , _ , a }, therefor, shouldn’t the left subtree be ‘PP$’ instead?
• Bronson
I guess no if \$ is the NIL Byte!
• That’s right :) I don’t think I explicitly said that, so I see how it can be confusing. Thanks!
• Zoran Kristo
Hi, thanks for this post, but you have errors in you picture of a full wavelet tree, if root is level 0, than if we go twice on the right ( follow 1s ) than you have ‘trrppppprs’ and it should be ‘trprpppppprs’ , you have less 2 p letters than it should be, also on next level..
• Ah I see, yeah, thanks! I appreciate it. I’ll try to update it soon (but I need to find the diagram source files in this mess of a hard drive first).
• Fatemeh
I have a question about rank query, How do you know in second level, the value of e is 1? and then How do you know you should go to the right child?
• Ethan
Another problem with the above diagrams is in the rank query example. In the first iteration you use rank(5, e) but check the string from [0, 6] then in the second iteration rank(4, e) and check the string from [0, 3]. The start is incorrect. | 2,237 | 8,453 | {"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} | 2.875 | 3 | CC-MAIN-2018-34 | longest | en | 0.923137 |
http://mathhelpforum.com/calculus/231013-surface-area-problem.html | 1,481,254,115,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542668.98/warc/CC-MAIN-20161202170902-00127-ip-10-31-129-80.ec2.internal.warc.gz | 176,827,806 | 11,806 | 1. ## Surface Area Problem
Find the area of the surface by rotating the curve about the x axis
$y = x^{3}$
$0 \leq x \leq 3$
$y' = 3x^{2}$
$S = \int_{0}^{3} 2\pi (x^{3}) \sqrt{1 + (3x^{2})^{2}} dx$
$S = 2\pi \int_{0}^{3} (x^{3}) \sqrt{1 + 9x^{4}} dx$
$S = 2\pi \int_{0}^{3}(x^{3}) \sqrt{1 + 9x^{4}} dx$
Use trig stuff here?
2. ## Re: Surface Area Problem
You also need to convert the \displaystyle \begin{align*} x^3 \end{align*} from your trigonometric substitution.
3. ## Re: Surface Area Problem
Originally Posted by Prove It
You also need to convert the \displaystyle \begin{align*} x^3 \end{align*} from your trigonometric substitution.
What do you mean about the $x^{3}$ and how would you go about trig sub. considering we have $1 + 9x^{4}$ This going to the 4th power not 2nd as required to use $a\tan\theta$
4. ## Re: Surface Area Problem
Since you have edited your original post I can't really state what we originally had. Anyway, you have correctly gotten to \displaystyle \begin{align*} S = \int_0^3{ 2\pi\left( x^3 \right) \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x} \end{align*}. From here I would do...
\displaystyle \begin{align*} \int_0^3{ 2\pi \left( x^3 \right) \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x } &= 2\pi \int_0^3{ x^3 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x } \\ &= \frac{\pi}{3} \int_0^3{ x^2 \, \sqrt{1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } \\ &= \frac{\pi}{9} \int_0^3{ 3x^2 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } \end{align*}
Now I would substitute \displaystyle \begin{align*} 3x^2 = \tan{ \left( \theta \right) } \implies 6x\,\mathrm{d}x = \sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}, and also note that when \displaystyle \begin{align*} x = 0, \theta = 0 \end{align*} and when \displaystyle \begin{align*} x = 3, \theta = \arctan{ \left( 27 \right) } \end{align*}, the integral becomes
\displaystyle \begin{align*} \frac{\pi}{9} \int_0^3{ 3x^2 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } &= \frac{\pi}{9} \int_0^{\arctan{ \left( 27 \right) } } { \tan{ \left( \theta \right) } \, \sqrt{ 1 + \tan^2{ \left( \theta \right) } } \, \sec^2{ \left( \theta \right) } \, \mathrm{d}\theta } \end{align*}
Can you go from here?
5. ## Re: Surface Area Problem
Originally Posted by Jason76
Use trig stuff here?
You don't need a trigonometric substitution.
Use the substitution $\sqrt{1+9x^4}=t\Rightarrow 1+9x^4=t^2\Rightarrow 36x^3dx=2tdt\Rightarrow x^3dx=\frac{1}{18}tdt$.
You can continue from here.
6. ## Re: Surface Area Problem
Originally Posted by red_dog
You don't need a trigonometric substitution.
Use the substitution $\sqrt{1+9x^4}=t\Rightarrow 1+9x^4=t^2\Rightarrow 36x^3dx=2tdt\Rightarrow x^3dx=\frac{1}{18}tdt$.
You can continue from here.
Yes you DO need a trigonometric substitution here. When you resubstitute in for the x^3 term, you will end up with another \displaystyle \begin{align*} \sqrt{ a^2 + t^2} \end{align*} form.
7. ## Re: Surface Area Problem
This problem needs u substitution, not trig sub. It's just like an arc length problem before. Same numbers.
$y = x^{3}$
$0 \leq x \leq 3$
$y' = 3x^{2}$
$S = \int_{0}^{3} 2\pi (x^{3}) \sqrt{1 + (3x^{2})^{2}} dx$
$S = 2\pi \int_{0}^{3} (x^{3}) \sqrt{1 + 9x^{4}} dx$
$S = 2\pi \int_{0}^{3}(x^{3}) \sqrt{1 + 9x^{4}} dx$
$S = 2\pi \int_{0}^{3}(x^{3}) (1 + 9x^{4})^{1/2} dx$
$S = 2\pi \int_{0}^{3}(x^{3}) (u)^{1/2} dx$
$u = 1 + 9x^{4}$
$du = 36x^{3} dx$
$\dfrac{1}{36} = x^{3} dx$
$= 2\pi \dfrac{1}{54}(u)^{3/2}$ evaluated at upper bound 3 and lower bound 0
$= 2\pi \dfrac{1}{54}(1 + 9x^{4})^{3/2}$ evaluated at upper bound 3 and lower bound 0
$[2\pi \dfrac{1}{54}(1 + 93)^{4})^{3/2}] - [2\pi \dfrac{1}{54}(1 + 9(0)^{4})^{3/2}]$ | 1,572 | 3,767 | {"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": 25, "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} | 4.40625 | 4 | CC-MAIN-2016-50 | longest | en | 0.591797 |
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Calculating stats
A lot of people may be wondering "How do the best players come up with their setups?" The answer is really quite simple. They calculate possible setups using various formulas. I have come up with a fairly easy way to calculate what your augs and skills will do for your damage, shields, energy and many other stats. Rather than giving you one huge equation to try to solve all at once, I have broken the very long process into 4 smaller, more manageable equations that most people will be able to solve to get the final results of their augs and skills.
Before we start, there are a few things you need to know. First, calculating ship stats using these formulas does require some basic knowledge of algebra, or at least some common sense to figure out what the letters and numbers mean. Second, these formulas don't work on RoF and may not work for Stealth, but have been tested and work perfectly for every other stat. If anyone wants to add to this thread and make the appropriate adjustments to my formulas to figure stealth and rate of fire that would probably help make this a more complete explanation, and I will edit this post to include the new formulas with references and credits going to the person that posts them.
I have listed the variable definitions at the end with a short explanation given by an example. Pay attention to the notations marked by * and ^ in the definitions section. One should also note that fixed values given by items (such as hull expanders) should be figured into the starting or original value before you begin solving these equations.
V1=(S*A1*A2*A3....)
V2=((V1-S)*AT*0.04)+((V1-S)*IT*0.02)
V3=V1+V2
F=V3*M1*M2*(Sk1+1)*(Sk2+1)
Variable definitions:
V1-value before aug tweaking and imperial tweaking
V2-value of aug tweaking and imperial tweaking bonuses
V3-value after aug tweaking and imperial tweaking
S-starting or original value
A1, A2, A3- aug values *
Sk1, Sk2-skills ^
M1, M2-item mods *^
F=final value
*any percentage type values (such as augs, skills or mods) should be put into this formula by dividing by 100 and then adding 1, so a 50% aug, skill or mod would be put in as 1.5
^you can multiply as many skills or mods as necessary here. note that the bonus of skills is simply the level of that skill multiplied by the bonus that skill gives then divided by 100. Support 20 on a freighter would be figured as (20*9)/100=1.8.
For my example, I am going to use a wingship with 2 minor speed augs and a slipstream to calculate speed. I will start by listing the stats of the appropriate gear and ignoring all other stats for this example. We will assume the user of this ship has no skills that modify speed, aug tweaking 5, and no imperial tweaking.
Wingship base speed: 80
Minor Speed Aug bonus: 10%
Slipstream speed bonus: 25%
First lets calculate the ships speed with the 2 minor speed augs (no aug tweaking yet). You will notice i skipped a small step that is listed in the notations after the formulas. Refer to the notations to find out how to make your aug values fit into this formula. I did this to keep the main parts of the calculations clean and easy to understand.
V1 = (S*A1*A2*A3....)
V1 = (80*1.1*1.1) = 96.8
V1 = 96.8
Notice there is a decimal here, thats important, do not round it off yet.
Now lets calculate how much aug tweaking will help.
V2 = ((V1-S)*AT*0.04)+((V1-S)*IT*0.02)
V2 = ((96.8-80)*5*0.04)+((96.8-80)*0*0.02) = 16.8*5*0.04 = 3.36
V2 = 3.36
Notice again that we have a decimal, again, do not round it off yet
Then we add V1 and V2 to get the total speed after we figure for augs and aug tweaking.
V3 = V1+V2
V3 = 96.8+3.36 = 100.16
V3 = 100.16
Once again, we have a decimal, leave it there.
Finally, lets figure out how much speed the slipstream will add and how much your skills will effect the speed. Since the person using this ship has no speed changing skills, we can simply ignore that part of the final step.
F = V3*M1*M2*(Sk1+1)*(Sk2+1)
F = 100.16*1.25 = 125.2
F = 125.2
Now we can round the decimal off and we find that this wingship, used by a person with aug tweaking 5, no imperial tweaking, and no speed modifying skills would go 125.
Hopefully I have made everything clear enough so that you can now start calculating your own ship stats with your own unique setups. Enjoy!
Tue Dec 04, 2007 11:49 pm
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thatch speed 39
6 heph augs. Speed -15% each
V1=39*0.85*0.85*0.85*0.85*0.85*0.85 = 14.708831109375
V2=((14.708831109375-39)*25*0.04)+((14.708831109375-39)*3*0.02) = -25.7486390240625
V3=-11,0398079146875
Wed Dec 05, 2007 2:36 am
Main: Thummmper
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Re: Calculating stats
Last edited by Thummmper on Wed Dec 05, 2007 7:06 pm, edited 1 time in total.
Wed Dec 05, 2007 3:16 am
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V1=39*0,85*0,85*0,85*0,85*0,85*0,85 = 14,708831109375
V2=((39-14,708831109375)*25*0,04)+((39-14,708831109375)*3*0,02)=25,7486390240625
V3=40,4574701334375
Wed Dec 05, 2007 3:54 am
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Posts: 30
Don't forget to follow the rules of BODMAS
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Wed Dec 05, 2007 8:50 am
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LichG wrote:
Don't forget to follow the rules of BODMAS
you show then, how its calculated...
Wed Dec 05, 2007 10:30 am
Joined: Mon Jun 11, 2007 4:39 pm
Posts: 30
Ah no screw maths, i just use rough stats, just saying use that if your not.
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Wed Dec 05, 2007 10:41 am
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LichG wrote:
Don't forget to follow the rules of BODMAS
Thought it was bedmas .....
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Wed Dec 05, 2007 12:24 pm
Main: Churlish
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2 things:
1, could you add Skills to your example calculation so I can check my working.
2, URJUHH could you tell what your final speed was? and if you have any tweaking, without tweaking i see your speed as 16 using my calculation.
Wed Dec 05, 2007 12:37 pm
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Location: Eredar US, Gozmatic (Horde)
never mind, I spoke wrongly
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Wed Dec 05, 2007 12:43 pm
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I just look at an aug and say: Damage, cool. Shields Crap. Mad Sci Cool. I think we all know why my DS is crap for PVP.
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Wed Dec 05, 2007 2:03 pm
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*******************_THIS_LINE_IS_JUST_TO_WIDEN_THE_SCREEN_*************************
ARG! Back to basics. Had to work this from the results. All corrections are welcome.
EDIT: I omitted Skill modifiers... as I have none and didn't think about them when coming up with this. Also made several corrections.
S = Base Speed
VA = Aug modified by AT & IT
FSA = Semi final Speed modified by just the Augs
SM = Speed modified by modded gear
FSM = total of all gear mods
FS = end results
VA1 = 1+(A1*(1+AT*0.04+IT*0.02))
VA2 = 1+(A2*(1+AT*0.04+IT*0.02))
VA3 = 1+(A3*(1+AT*0.04+IT*0.02))
VA4 = 1+(A4*(1+AT*0.04+IT*0.02))
VA5 = 1+(A5*(1+AT*0.04+IT*0.02))
VA6 = 1+(A6*(1+AT*0.04+IT*0.02))
FSA = S*VA1*VA2*VA3*VA4*VA5*VA6
SM1 = 1+M1
SM2 = 1+M2
SM3 = 1+M3
...
FSM = SM1 * SM2 * SM3 * ...
FS = FSA * FSM
For a 6 Heph Thatch with no modded gear equipped:
S = 39
Heph Speed: -15%
AT = 25, IT = 3
VA1 = 1+(A1*(1+AT+IT)) = 1+(-0.15*(1+25*0.04+3*0.02)) = 1-0.15*2.06 = 1-0.309 = 0.691
VA2 = 1+(A2*(1+AT+IT)) = 1+(-0.15*(1+25*0.04+3*0.02)) = 1-0.15*2.06 = 1-0.309 = 0.691
VA3 = 1+(A3*(1+AT+IT)) = 1+(-0.15*(1+25*0.04+3*0.02)) = 1-0.15*2.06 = 1-0.309 = 0.691
VA4 = 1+(A4*(1+AT+IT)) = 1+(-0.15*(1+25*0.04+3*0.02)) = 1-0.15*2.06 = 1-0.309 = 0.691
VA5 = 1+(A5*(1+AT+IT)) = 1+(-0.15*(1+25*0.04+3*0.02)) = 1-0.15*2.06 = 1-0.309 = 0.691
VA6 = 1+(A6*(1+AT+IT)) = 1+(-0.15*(1+25*0.04+3*0.02)) = 1-0.15*2.06 = 1-0.309 = 0.691
FSA = S* VA1* VA2*VA3*VA4*VA5*VA6 = 39*0.691*0.691*0.691*0.691*0.691*0.691 = 4.246
SM1 = 1+M1 = 1
SM2 = 1+M2 = 1
SM3 = 1+M3 = 1
FSM = SM1 * SM2 * SM3 * ... = 1*1*1*... = 1
FS = FSA * FSM = 4.246 * 1 = 4.246 rounded to 4
For a 6 Heph Thatch with no modded gear equipped:
S = 39
Heph Speed: -15%
AT = 23, IT = 3
VA1 = 1+(A1*(1+AT+IT)) = 1+(-0.15*(1+23*0.04+3*0.02)) = 1-0.15*1.98 = 1-0.297 = 0.703
VA2 = 1+(A2*(1+AT+IT)) = 1+(-0.15*(1+23*0.04+3*0.02)) = 1-0.15*1.98 = 1-0.297 = 0.703
VA3 = 1+(A3*(1+AT+IT)) = 1+(-0.15*(1+23*0.04+3*0.02)) = 1-0.15*1.98 = 1-0.297 = 0.703
VA4 = 1+(A4*(1+AT+IT)) = 1+(-0.15*(1+23*0.04+3*0.02)) = 1-0.15*1.98 = 1-0.297 = 0.703
VA5 = 1+(A5*(1+AT+IT)) = 1+(-0.15*(1+23*0.04+3*0.02)) = 1-0.15*1.98 = 1-0.297 = 0.703
VA6 = 1+(A6*(1+AT+IT)) = 1+(-0.15*(1+23*0.04+3*0.02)) = 1-0.15*1.98 = 1-0.297 = 0.703
FSA = S* VA1* VA2*VA3*VA4*VA5*VA6 = 39*0.703*0.703*0.703*0.703*0.703*0.703 = 4.7
SM1 = 1+M1 = 1
SM2 = 1+M2 = 1
SM3 = 1+M3 = 1
FSM = SM1 * SM2 * SM3 * ... = 1*1*1*... = 1
FS = FSA * FSM = 4.7 * 1 = 4.7 rounded to 5
I have AT 23 & IT 3 and my 6 Heph Thatch, with no modded gear equipped, shows a speed of 5.
Last edited by Thummmper on Wed Dec 05, 2007 7:28 pm, edited 4 times in total.
Wed Dec 05, 2007 2:46 pm
Main: Churlish
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This is for Sheilds, Capacity, Speed and Energy right anything else?
Wed Dec 05, 2007 5:01 pm
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Location: Eredar US, Gozmatic (Horde)
NOT for rate of fire/recoil mind you
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Wed Dec 05, 2007 5:03 pm
Main: Thummmper
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pearsoa wrote:
This is for Sheilds, Capacity, Speed and Energy right anything else?
Again I did this just for Speed and it is missing Skill mods as I have no Skills that modify my speed at present... that I know of.
Wed Dec 05, 2007 7:02 pm
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https://se.mathworks.com/matlabcentral/answers/451726-write-a-function-called-under_age-that-takes-two-positive-integer-scalar-arguments-age-that-repres | 1,674,988,103,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00731.warc.gz | 524,309,057 | 73,883 | # Write a function called under_age that takes two positive integer scalar arguments: age that represents someone's age, and limit that represents an age limit. The function returns true if the person is younger than the age limit. If the second arg
419 views (last 30 days)
Abhishek singh on 22 Mar 2019
Commented: DGM on 13 Oct 2022
function too_young = under_age(age,limit)
limit = 21
if age <= limit
too_young = true;
elseif age >= limit
too_young = false;
else
fprintf('invalid\n')
end
Jan on 4 Apr 2022
This has become a strange thread. We find a pile of suboptimal implementations.
We need 6 lines of code (including the trailing "end"). John has condensed the IF block into 1 line, but without doubt his implementation is clean and exhaustive.
So what is the reason to post a lot of other less elegant and not working versions?
Many beginners suggest:
if limit > age
too_young = true;
else
too_young = false;
end
instead of the compact and efficient:
too_young = limit > age;
John D'Errico on 23 Mar 2019
Edited: John D'Errico on 8 Nov 2020
This is not an error. It is a failure of your code to work as it is supposed to work by your goals.
What did I say? What has EVERYONE said so far? You cannot set a default as you did.
Even though you pass in the number 18 as the limit, the first thing you do inside is reset that value to 21.
Instead, you need to think about how to test to see if limit was provided at all.
Edit: (18 months later)
Now that there are dozens of answers, all of which are lengthy, I'll show how I would write it.
function tooyoung = under_age(age,limit)
% under_age: returns true if the person is under the age limit
% usage: tooyoung = under_age(age,limit)
%
% Arguments: (input)
% age - numeric. As written, age can be scalar, or any size array
%
% limit - optional argument, if not provided, the default is 21
%
% arguments: (output)
% tooyoung - a boolean variable, indicating if age was under the limit.
%
% NO tests are done to verify that age or limit, if provided are valid
% ages itself, or even if they is numeric at all. Better code would
% do these things.
% test for limit having been provided. if not, then the default is 21
if nargin < 2, limit = 21; end
% There is no need to use an if statement. The test itself is the desired result.
tooyoung = age < limit;
end
See that most of my code was actually comments. Help is hugely important, since it allows you to quickly use the code next year when you forget how it was written, or when you forget what the arguments mean.
Think of internal comments as reminders to yourself, for a year from now when you look at the code and need to change the code for some reason, or god forbid, you need to debug it. My recommended target is one line of comment per line of code, or at worst, one line of code per significant thing done in the code.
Comments are free! Good code should look positively green due to all of the comments. My solution code was lengthy only because of all of the comments.
What else?
Remember that white space is hugely important. It makes your code easy to read.
Use intelligent, meaningful variable names. Good mnemonic variable names help to make your code self-documenting, and again, easy to read. When you are scanning through the code, you don't want to continuously go back and be forced to remember what does the variable wxxyy do?
As I said, better code would have included tests to verify that both age and limit, if provided, were actually numeric variables. The best code is friendly code. When it fails, you want the code to fial in a friendly way, telling the person what was seen to be wrong. What you don't want to happen is the code does something screwy and unexpected, or returns some random difficult to understand error message. The best code makes it easy to use that code.
The final test in this code is a vectorized test, in that if you called the code like this:
tooyoung = under_age([12 5 29 75],21)
tooyoung =
1 1 0 0
it will work and return a vector of results. Vectorized code is generally good code, since it allows the user to not be forced to use a loop when they want to use the code many times in a row.
Finally, could I have written code that would have required only one line of code? Thus testing to see if limit was provided, and returning a comparison to age in just one line? Probably, but that would have been unnecessarily complicated, difficult to read and debug. There would have been no gain in the quality of the code or how fast it runs. Good code is simple, easy to read, easy to use, easy to debug.
Rik on 5 Aug 2020
Have you read the documentation for the nargin function to see what it does? It returns the n[umber of] arg[uments you provide as] in[put]. So if it is smaller than 2, that means the limit was not provided as an input argument.
Saurabh Kumar on 28 Mar 2019
Write a function called under_age that takes two positive integer scalar arguments:
1. age that represents someone's age, and
2. limit that represents an age limit.
The function returns true if the person is younger than the age limit. If the second argument, limit, is not provided, it defaults to 21. You do not need to check that the inputs are positive integer scalars.
function x = under_age(age,limit)
if nargin < 2
limit = 21;
if age < limit
x = true;
else
x = false;
end
end
if nargin == 2
if age < limit
x = true;
else
x = false;
end
end
Walter Roberson on 20 Jun 2021
MATLAB (and nearly all programming languages... but not all) are Procedural Languages, in which the order of statements is important.
Suppose you were climbing a ladder. For the most part, you take one step upward at a time. Now suppose you program it that way,
"Move foot 50 cm higher and put weight on it"
But what about when you get to the top?
"Move foot 50 cm higher and put weight on it"
"If there was no higher rung, don't take that step"
Too late. You already put your weight in mid-air before checking whether there was something to put your weight onto.
You instead need a check like
"If there is a higher rung, move foot 50 cm higher and put weight on it"
Check first, before relying on it being there.
The way you coded
if age < limit
if nargin < 2
limit = 21
end
end
is similar to not having checked for a higher rung existing before putting your weight where it would be.
===
It is better programming practice to check for exceptions first, check to see whether a calculation is going to be valid before doing the calculation.
However, there are some situations in which it is valid to "patch up" if you encounter an exception. For example, better programming practice for "1/x if x is not 0, 0 if x is 0" would be
y = zeros(size(x));
This code never does any division by 0.
But if you are sure you are using IEEE 754, you can count on the fact that IEEE 754 defines that 1/x has some result for 1/0 (rather than crashing the program), and you can instead do a patch-up-later version:
y = 1./x;
y(x == 0) = 0;
The patch-up-later version of the question would be like,
function too_young = under_age(age, limit)
too_young = age < 21;
if nargin > 1
too_young = age < limit;
end
end
Notice that the variable limit is not used until after it has been checked to be sure that it is present.
mayank ghugretkar on 7 Jun 2019
this can work too...
A bit compact approach
function too_young = under_age(age, limit)
if nargin < 2
limit = 21;
end
if age < limit
too_young=true;
else
too_young=false;
return
end
end
Rik on 3 Jun 2021
help nargin
NARGIN Number of function input arguments. Inside the body of a user-defined function, NARGIN returns the number of input arguments that were used to call the function. If the function uses an arguments validation block, then only positional arguments provided by the function call are included in this number. Optional arguments not provided by the caller are not included. Name-value arguments are never included, whether provided or not. NARGIN(FUN) returns the number of declared inputs for the function FUN. The number of arguments is negative if the function has a variable number of input arguments. If the function uses an arguments validation block, NARGIN returns the number of declared positional arguments on the function line as a nonnegative value. FUN can be a function handle that maps to a specific function, or a character vector or string scalar that contains the name of that function. See also NARGOUT, VARARGIN, NARGINCHK, INPUTNAME, MFILENAME. Documentation for nargin doc nargin Other functions named nargin fittype/nargin inline/nargin
Astr on 8 Sep 2019
function too_young = under_age(age, limit)
if nargin == 1
limit = 21;
end
too_young = gt(limit, age);
function too_young= under_age(age, limit)
if (nargin<2)
limit=21;
end
if (age<limit)
too_young=true
else
too_young=false
end
end
Kilaru Venkata Krishna on 2 May 2020
its taking age >= limit as older
.......and ......age<limit as young
Siddharth Joshi on 23 Apr 2020
function too_young = under_age(age,limit)
if nargin<2
limit=21;
end
if age<limit
too_young=true;
else
too_young=false;
end
end
too_young = under_age(18,18)
too_young = under_age(20)
too_young =
logical
0
too_young =
logical
1
sudeep lamichhane on 27 Apr 2020
function too_young = under_age(age, limit)
if nargin<2
limit=21;
end
if age < limit
too_young= true;
else
too_young= false;
end
Krashank Kulshrestha on 14 May 2020
correct but how u do it
Sai Hitesh Gorantla on 1 Feb 2020
Edited: Rik on 17 Jun 2020
function [too_young] = under_age(age,limit)
if nargin == 2
if age<limit
too_young = true;
else
too_young = false;
end
elseif nargin<2
if age<=21
too_young = true;
else
too_young = false;
end
end
mohammad elyoussef on 4 Apr 2020
function c = under_age(a,b)
if nargin < 2
b = 21;
end
if b > a
c = true;
else
c = false;
end
maha khan on 9 Apr 2020
function [too_young]= under_age(age,limit)
if (nargin < 2) || isempty(limit)
limit = 21;
end
if age>21
too_young=false;
elseif age < limit
too_young=true;
elseif age==age
too_young=false;
elseif age<=21
too_young=true;
elseif age < age
too_young=false;
elseif age<=21
too_young=true;
else
too_young=true;
end
Walter Roberson on 9 Apr 2020
Suppose the user passes in a limit of 35, such as testing for eligibility to be President of the United States. And suppose the age passed in is 29. Then if age>21 would be if 29>21 and that would be true, so you would declare too_young=false but clearly the answer should be true: 29 < the specified limit.
Mir Mahim on 7 May 2020
function a = under_age(age,limit)
if nargin < 2
limit = 21;
end
a = age < limit;
end
Aasma Shaikh on 26 May 2020
function too_young= under_age (age, limit)
if nargin<2
limit=21;
if (age<limit)
too_young = true;
else
too_young = false;
end
elseif ((nargin==2) && (age<limit))
too_young = true;
else
too_young = false;
end
end
% Copy, paste and Run
jaya shankar veeramalla on 29 May 2020
function too_young = under_age(age,limit)
if (nargin < 2) || isempty(limit)
limit = 21;
end
if age < limit
too_young = true;
else age >= limit
too_young = false;
end
AYUSH MISHRA on 4 Jun 2020
function too_young =under_age(age,limit)
if nargin <2
limit=21;
end
if age<limit
too_young=true;
else
too_young=false;
end
SOLUTION ;
under_age(18,18)
ans =
logical
0
under_age(20)
ans =
logical
1
Keshav Patel on 8 Jun 2020
function too_young =under_age(age,limit)
if nargin<2
limit=21;
if age<limit
too_young=true;
else
too_young=false;
end
end
if nargin ==2
if age<limit
too_young=true;
else
too_young=false;
end
end
saurav Tiwari on 17 Jun 2020
Edited: Rik on 17 Jun 2020
function [a]=under_age(n,m)
a=n<m;
if a==1;
fprintf('true')
end
if nargin<2;
m=21;
end
end
##### 2 CommentsShowHide 1 older comment
Rik on 17 Jun 2020
Why did you decide to put the part with nargin at the end?
AKASH SHELKE on 9 Aug 2020
Edited: AKASH SHELKE on 9 Aug 2020
function too_young = under_age(age,limit)
if nargin<2
limit = 21;
end
if age < limit
too_young = true;
else
too_young = false;
end
function too_young = under_age(age,limit)
if nargin < 2
limit = 21;
end
if age < limit
too_young = 1 == 1;
else
too_young = 1 ==2;
end
end
Walter Roberson on 10 Aug 2020
too_young = 1 == 1;
You should
doc true
doc false
Omkar Gharat on 11 Aug 2020
function [too_young] = under_age(age,limit);
% age = input('Enter age of applicant : ')
% limit = input('Enter age limit of applicant : ')
if nargin < 2
limit = 21;
end
if age < limit
too_young = true;
else
too_young = false;
end
end
This is my code .And it is 100% correct
Madan Kc on 2 Oct 2020
no its not correct for under_age(20)
Khom Raj Thapa Magar on 6 Sep 2020
Make sure your indentation is correct while coding
function too_young = under_age(age,limit)
if nargin<2
limit = 21;
if age < limit
too_young = true;
else
too_young = false;
end
end
if nargin == 2
if age < limit
too_young = true;
else
too_young = false;
end
end
Calling functions
>> too_young = under_age(18,18)
>> too_young = under_age(20)
Output:
too_young =
logical
0
too_young =
logical
1
Jessica Trehan on 21 Sep 2020
function too_young=under_age(age,limit)
if nargin<2
limit=21;
if age<limit
too_young=true;
else
too_young=false;
end
end
if nargin==2
if age<limit
too_young=true;
else
too_young=false;
end
end
%THE PERFECT CODE
Olha Skurikhina on 11 Jan 2021
Thank you. It is very stupid. I already solved harder problems along the course but this one couldn`t tackle. That was my problem and plus if the age= limit, it is still should return false so '<=' is incorrect.
amjad ali on 6 Sep 2021
function too_young=under_age(age,limit);
switch nargin
case 1
limit=21;
end
if (limit>age);
too_young=true;
else
too_young=false;
end
switch nargin
case 2
end
if limit>age ;
too_young=true;
else
too_young=false;
end
##### 2 CommentsShowHide 1 older comment
amjad ali on 6 Sep 2021
i have no idea before that the second switch statement is useless,
VIGNESH B S on 13 Oct 2021
function [too_young] = under_age(age,limit)
if nargin == 2
too_young = compare(age,limit);
elseif nargin == 1
too_young = compare(age,21);
end
end
if x<y
else
end
end
Image Analyst on 14 Oct 2021
But it doesn't look elegant. An elegant program would do
if x<y
else
end
Alexandar on 29 Jun 2022
This worked well for me. Very short but I sort of don't understand why the "nargout" part worked.
function too_young = under_age(age, limit)
if nargin < 2
limit = 21
end
too_young = age<limit;
if nargout == 2
too_young = true(1);
end
Alexandar on 29 Jun 2022
Edited: Alexandar on 29 Jun 2022
Yes, this was done for a matlab coursea assignment that said the code worked perfectly well which confused me. I started coding three weeks ago.
%Help required
Error showing=Output argument "x" (and possibly others) not assigned a value in the execution with "under_age" function.
function x = under_age(age,limit)
if nargin < 2
limit=21;
if age < limit
x=true;
end
end
if nargin==2
if age<limit
x=true;
end
end
function too_young = under_age(age,limit)
if nargin < 2
limit=21;
if age < limit
too_young=true;
else
too_young=false;
end
end
if nargin==2
if age<limit
too_young=true;
else
too_young=false;
end
end
Thanks a lot Mr.Walter Roberson
MURSHED SK on 13 Oct 2022
Edited: MURSHED SK on 13 Oct 2022
This might help:
function too_young = under_age(age, limit)
if nargin <2
limit =21;
end
if age<limit
too_young = true;
else
too_young = false;
end
% the shortest way
DGM on 13 Oct 2022
Again, how is this different from the many existing answers?
Also, this isn't the shortest way. John's answer at the top of the thread is shorter, and is thoroughly explained.
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Topic: Trigonometric area optimization
Replies: 13 Last Post: Dec 20, 2012 3:29 PM
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Peter Duveen Posts: 163 From: New York Registered: 4/11/12
Re: Trigonometric area optimization
Posted: Dec 16, 2012 2:27 PM
Plain Text Reply
I want to applaud my predecessors' efforts in pointing to a solution for the important problem at hand. At the same time, I feel I deserve credit for taking the time to prove the case for the minimum area in the case of a right triangle, since the problem relies on that. I would also hope to receive at least a little applause for being the first, yes, the very first, to claim that the solution to the value of m is 3, or very nearly 3. I qualified my solution because I had used some rather rough figures for angles, and then relied on trigonometric tables. Certainly there were more elegant ways of solving this problem, but the fellow (Adam) who first presented it seemed to be in dire need of academic reinforcement, so it was with great haste that I crafted what I thought at the time was a credible solution to the problem. I felt that, without carrying the calculations out to some credible extent, Adam might be inclined to think that I and others had done so much handwaving and posturing, but had not really "solved" the problem. I have yet to be been given due credit for the full extent of my efforts by my esteemed colleagues who have belatedly chimed in with a value for m. Of course, better late than never, but...well it was a little late, Peter, wasn't it?
Date Subject Author
12/11/12 AdamThor
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12/14/12 bobmck
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12/15/12 Peter Duveen
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# 第20回オフラインリアルタイムどう書くの問題をC言語で解く
More than 5 years have passed since last update.
pythonで解法済みなのですが、C言語で解いた方がいらっしゃらないようなのでC言語でも解いてみました。
http://nabetani.sakura.ne.jp/hena/ord20meetime/
http://qiita.com/Nabetani/items/5791f8ae1bb5d069a49b
```#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ABIJ (1<<0)
#define Z (1<<1)
#define time2min(t) (atoi(t)/100*60 + atoi(t)%100)
{
char ABIZ[24*60], ABJZ[24*60], FREE[1*60], *d, who;
int t, end;
memset(ABIZ, 0, sizeof(ABIZ));
memset(ABJZ, 0, sizeof(ABJZ));
memset(FREE, Z, sizeof(FREE));
for (d = data; d != (char *)1; d = strchr(d, ',') + 1) {
who = d[0] == 'Z' ? Z : ABIJ;
end = time2min(d + 6);
for (t = time2min(d + 1); t < end; t++) {
if (d[0] != 'J') ABIZ[t] |= who;
if (d[0] != 'I') ABJZ[t] |= who;
}
}
for (t = 10*60; t <= 17*60; t++) {
if (memcmp(ABIZ + t, FREE, sizeof(FREE)) == 0) break;
if (memcmp(ABJZ + t, FREE, sizeof(FREE)) == 0) break;
}
if (t <= 17*60) {
sprintf(answer, "%02d%02d-%02d%02d", t/60, t%60, t/60 + 1, t%60);
} else {
}
}
void test(char *data, char *correct)
{
}
int main(int argc, char *argv[])
{
/*0*/ test( "A1050-1130,B1400-1415,I1000-1400,I1600-1800,J1100-1745,Z1400-1421,Z1425-1800", "1425-1525" );
/*1*/ test( "A1000-1200,B1300-1800,Z1000-1215,Z1230-1800", "-" );
/*2*/ test( "Z0800-2200", "1000-1100" );
/*3*/ test( "A1000-1700,Z0800-2200", "1700-1800" );
/*4*/ test( "A1000-1701,Z0800-2200", "-" );
/*5*/ test( "A1000-1130,B1230-1800,Z0800-2200", "1130-1230" );
/*6*/ test( "A1000-1129,B1230-1800,Z0800-2200", "1129-1229" );
/*7*/ test( "A1000-1131,B1230-1800,Z0800-2200", "-" );
/*8*/ test( "A1000-1130,B1229-1800,Z0800-2200", "-" );
/*9*/ test( "A1000-1130,B1231-1800,Z0800-2200", "1130-1230" );
/*10*/ test( "A1000-1130,B1230-1800,Z0800-1130,Z1131-2200", "-" );
/*11*/ test( "A1000-1130,B1231-1800,Z0800-1130,Z1131-2200", "1131-1231" );
/*12*/ test( "Z0800-0801", "-" );
/*13*/ test( "Z0800-1031,Z1129-1220,Z1315-1400,Z1459-1600", "1459-1559" );
/*14*/ test( "Z0800-2200,I1000-1600,J1030-1730", "1600-1700" );
/*15*/ test( "Z0800-2200,I1000-1600,J1130-1730", "1000-1100" );
/*16*/ test( "Z0800-2200,I1000-1600,J1130-1730,A0800-1025", "1025-1125" );
/*17*/ test( "Z0800-2200,I1000-1600,J1130-1730,A0800-1645", "1645-1745" );
/*18*/ test( "Z0800-2200,I1000-1600,J1130-1730,A0800-1645,I1735-2200", "-" );
/*19*/ test( "Z0800-2200,I1000-1600,J1130-1730,A0800-1645,J1735-2200", "1645-1745" );
/*20*/ test( "Z1030-2200,I1000-1600,J1130-1730", "1030-1130" );
/*21*/ test( "Z1035-1500,I1000-1600,J1130-1730,Z1644-2200", "1644-1744" );
/*22*/ test( "I2344-2350,A2016-2253,Z1246-1952", "1246-1346" );
/*23*/ test( "Z2155-2157,B1822-2032,Z1404-2000,Z2042-2147,Z2149-2154", "1404-1504" );
/*24*/ test( "Z2231-2250,Z2128-2219,B2219-2227,B2229-2230,Z0713-2121,A0825-1035,B1834-2001", "1035-1135" );
/*25*/ test( "J0807-1247,I0911-1414,B1004-1553,Z0626-1732,Z1830-1905,A1946-1954,A0623-1921", "-" );
/*26*/ test( "J1539-1733,J0633-1514,Z1831-1939,J1956-1959,I0817-1007,I1052-1524,Z1235-1756,Z0656-1144", "1524-1624" );
/*27*/ test( "Z2319-2350,B0833-2028,I2044-2222,A1410-2201,Z2044-2228,Z0830-2023,Z2242-2306,I2355-2359", "-" );
/*28*/ test( "B2001-2118,Z0712-1634,I1941-2102,B1436-1917", "1000-1100" );
/*29*/ test( "A0755-1417,B2303-2335,Z0854-2150,Z2348-2356,Z2156-2340,I1024-1307,Z2357-2359", "1417-1517" );
/*30*/ test( "A1958-1959,B0822-1155,I1518-1622,Z1406-1947,A1800-1822,A0904-1422,J1730-1924,Z1954-1958,A1946-1956", "1422-1522" );
/*31*/ test( "B1610-1910,I2121-2139,A0619-1412,I2147-2153,Z0602-2111,I0841-2031,A1657-1905,A1956-2047,J0959-1032,Z2131-2147", "1412-1512" );
/*32*/ test( "Z0623-1900,A0703-1129,I1815-1910,J1956-1957,I0844-1518,Z1902-1935,B1312-1342,J1817-1955", "1129-1229" );
/*33*/ test( "J1246-1328,B1323-1449,I1039-1746,Z1218-2111", "1449-1549" );
/*34*/ test( "A1958-1959,I1943-1944,I0731-1722,Z0845-1846,J1044-1513,Z1910-1923,B1216-1249", "1513-1613" );
/*35*/ test( "A1855-2047,Z0946-1849,Z2056-2059,I1855-1910,B1946-2058,I1956-2025,Z1905-2054,J0644-1800,I0720-1618", "1618-1718" );
/*36*/ test( "J1525-1950,Z0905-1933,A1648-1716,I2051-2054,I2015-2044,I0804-1958,B0934-1100,Z1953-2037", "1100-1200" );
/*37*/ test( "Z1914-1956,J0823-1610,Z0641-1841,J1800-1835,A0831-1346,I1926-1941,I1030-1558,I1738-1803", "1558-1658" );
/*38*/ test( "Z0625-1758,J1033-1351,B1816-2236,I0838-1615,J2247-2255", "1351-1451" );
/*39*/ test( "J0603-1233,A1059-1213,I1326-2103,Z0710-1459", "1213-1313" );
/*40*/ test( "B1302-1351,J1410-2038,A0755-1342,J0637-0658,Z2148-2159,Z1050-2131,A1543-1844,I1615-1810", "1351-1451" );
/*41*/ test( "Z0746-2100,A2122-2156,I1022-1144,J0947-1441,A1333-1949", "1144-1244" );
/*42*/ test( "J0718-1243,Z1443-1818,B2055-2057,A0714-1238,Z1045-1344,A1643-1717,B1832-2039,J1623-1931", "1238-1338" );
/*43*/ test( "Z1921-1933,A1208-1418,I0827-1940,Z0757-1917,J0653-1554,B1859-1909", "1554-1654" );
return 0;
}
``` | 2,427 | 4,859 | {"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} | 2.953125 | 3 | CC-MAIN-2019-43 | latest | en | 0.177709 |
http://www.convertit.com/Go/SmartPages/Measurement/Converter.ASP?From=dry+gallon&To=capacity | 1,652,996,819,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00013.warc.gz | 75,004,876 | 3,731 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```dry gallon = 0.00440488377086 volume (volume) ``` Related Measurements: Try converting from "dry gallon" to amphora (Greek amphora), bath (Israeli bath), bushel (dry bushel), Canadian gallon, chetvert (Russian chetvert), cord (of wood), dry barrel, dry pint, fifth, firkin, gill, jeroboam, jigger, last, pint (fluid pint), register ton, tou (Chinese tou), tun (English tun), UK oz fluid (British fluid ounce), UK peck (British peck), or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: dry gallon = .03694118 barrel, .96893897 Canadian gallon, .02097315 chetvert (Russian chetvert), .03125 coomb, .00972231 cord foot (of wood), .00444449 displacement ton, 1,191.57 dram fluid (fluid dram), 146,829.46 drop, 4 dry quart, 37.24 gill, .0015625 last, .0030323 load, 2.33 magnum, .35049614 oil arroba (Spanish oil arroba), 148.95 oz fluid (fluid ounce), .3878824 salmanazar, .00440488 stere, .0625 strike, 155.03 UK oz fluid (British fluid ounce), 3.88 UK quart (British quart).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 465 | 1,682 | {"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} | 3.15625 | 3 | CC-MAIN-2022-21 | latest | en | 0.678535 |
https://ibikeoulu.com/understanding-how-slots-work/ | 1,722,774,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640398413.11/warc/CC-MAIN-20240804102507-20240804132507-00645.warc.gz | 240,194,697 | 11,539 | # Understanding How Slots Work
A slot is an opening in a computer or other electronic device into which you can insert a printed circuit board. A slot may also refer to the place in a computer where you can install disk drives. Contrary to popular belief, slots are not the same as bays. Bays are sites within a computer into which you can mount disk drives.
A slot is also an area on a football field or in ice hockey that allows a player to gain a vantage point on the opposing team. Typically, slot receivers are positioned in the area between and slightly behind wide receivers, although this depends on the specific team and game.
Unlike traditional mechanical machines, modern video and online slots have multiple paylines. These paylines are arranged in various patterns on the reels and can give players multiple chances to form winning combinations. The pay table in a slot will show all of the available paylines and explain how they work. It will also detail the payouts for different combinations of symbols. In addition, the pay table will show if a slot has any bonus features and how to trigger them.
Understanding how slots work is important for any casino player. While there are many myths about how to win at slots, the truth is that a solid understanding of probability can help players develop a strategy based on sound principles. This article will explore the basics of slot theory, including how to read a pay table and calculate odds.
The first step to understanding how slots work is knowing what a random number generator (RNG) is. The RNG is a computer program that ensures that every spin is independent of the results of any previous spins. This is what makes slots a true game of chance.
Once you understand how the RNG works, you can start to see how a slot’s odds are calculated. The odds are determined by the probability of a particular symbol appearing on a given reel. The more frequently a symbol appears, the higher its odds of appearing. However, this does not necessarily mean that it will appear on the winning line.
In addition to showing the paytable, a slot’s odds will also display how much you can win by landing matching symbols on a winning combination. This information will be displayed as small tables, often made up of brightly coloured graphics to make it easier to read. The pay table will also include a minimum and maximum bet value, along with any special symbols or bonus features the slot has to offer.
As technology improves, so do the feature rounds of slot games. These can take many forms, from simple free spins to interactive mystery pick games. However, a common feature is some kind of progressive jackpot or progressive multiplier sequence. This is why it’s so important to check out a slot machine’s pay table before you begin playing.
Posted in: Gambling | 556 | 2,834 | {"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} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.961564 |
https://markv.nl/blag/efficient-overfitting-of-training-data-kaggle-bowl | 1,579,819,392,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00027.warc.gz | 538,407,540 | 6,912 | # Efficient overfitting of training data (Kaggle Bowl 2017)
During the Kaggle Data Science Bowl 2017, the leaderboard was based on only $198$ samples. The opportunity for overfitting was quickly understood, but initially only the naive option was mentioned, testing 1 submission per sample taking 66 days (still doable within the competition duration, but less than ideal).
But then Oleg Trott got a perfect score in just 14 submissions! topic) I was really curious how he managed to do this. Together with Cas, I found out one way it can be done. Perhaps Oleg will be posting his solution soon, but as I write this, it's not yet public.
EDIT: Kaggle is not happy with these overfitted perfect scores, which is reasonable. It gives a wrong impression to press or newcomers. Therefore I ask you to not post your perfect solution to the leaderboard. You can still post test ones to get extra training data though, giving you a slight advantage over others who will get this data later. Test submissions with have terrible logloss so won't influence the leaderboard.
The score is calculated as logloss, with probabilities capped at $1 \cdot 10^{-15}$. Therefore, the maximum error is $-\ln(10^{-15})$. With a resolution of $0.00001$ on the public leaderboard, that leaves $-\ln(10^{-15})/198/0.00001 = 17443$ discernible values or $14.090$ bits of information. The solution has $198$ bits of information, so theoretically, it seems possible to get it in 15 attempts.
Here's how to do it. You want to be able to tell from the sum of the sample log losses which position had an error. Therefore, you want the errors in "binary representation" to be like $0000$, $0001$, $0010$, $0100$, $1000$, but with a "bit" being the minimum resolution. That way you can tell from the sum which position was wrong.
EDIT: I initially used only half the probability range for positives, to prevent collisions between scores from positives and negatives. By "collision" I mean the score being decomposable in multiple ways. As it turns out, collisions are pretty rare, so we use the whole range and sample 15 bits each time (including a 0 prediction).
We can check 15 positions per submission. To have the other positions not interfere, we predict $p=0.5$ for all of them, which will give a logloss of $(198-15)/198 \cdot \ln(0.5)$ whether they're positives or negatives. This will simply add a constant to the score.
EDIT: I initially used exponential function, but after seeing that Oleg uses sigmoid, I found that there are fewer collisions that way. You should use that way, by simply replacing "exp" by "logistic.cdf" (scipy). Comments about why this works are welcome!
Let's look at a simplified example, where we probe 4 positions and predict $p=0.5$ for the rest. For the first position we predict $p=1$, for an logloss of $0$ if correct. For the second position, we want the minimum discernible error if it's a positive. The minimum discernible error is $0.00001$, so we predict $p = \exp(-1 \cdot 2^0 \cdot 0.00001 \cdot 198) = 0.99802$. For the third one, we want double ($2^1$) that error ($0.00002$), so we predict $\exp(-1 \cdot 2^1 \cdot 0.00001 \cdot 198) = 0.99605$. For the fourth one, we want 4 ($2^2$) times that error ($0.00004$) so $p=0.99211$.
We submit this solution and get an error of (for example) $0.70714$. We first subtract the constant term due to 194 times $p=0.5$ which is $\ln(0.5) / 198 \cdot 194 = 0.67914$, which leaves us $0.02800$. Position 1 can contribute either $0.00000$ or $-\log(1o^{-15})/198 = 0.17444$, position 2 can contribute either $0.00001$ or $0.03144$, position 3 is either $0.00002$ or $0.02795$ and position 4 is either $0.00004$ or $0.02446$.
In this case it's fairly obvious which sums to $0.02800$, but you can generally just try all possibilities, there are just 32768 for 15 bits. The solution is $0.00000 + 0.00001 + 0.02795 + 0.00004$. So we know position 1, 2 and 4 are positives, while position 3 is a negative.
Just repeat this for 15 positions at a time until you have everything. If there were ever multiple solutions for how to obtain the score, then you can do another submission to compare them.
See my implementation on Kaggle. (Jan 18)
Other content you might like: | 1,128 | 4,231 | {"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} | 3.15625 | 3 | CC-MAIN-2020-05 | latest | en | 0.952708 |
https://www.hackmath.net/en/example/586?tag_id=61 | 1,563,758,036,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527458.86/warc/CC-MAIN-20190722010436-20190722032436-00481.warc.gz | 706,861,519 | 6,906 | # Solution
In 469 dl red solution is 84 dl red color and in 102 dl blue solution is 52 dl blue color. How many dl of red and blue dl color solution must be mixed to get a mixture of 247 dl contain 116 dl of color?
Result
x = 30 dl
y = 217 dl
#### Solution:
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#### To solve this example are needed these knowledge from mathematics:
Do you have a system of equations and looking for calculator system of linear equations? Tip: Our volume units converter will help you with converion of volume units.
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https://www.coursehero.com/file/6254327/ch04-p073/ | 1,490,491,708,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189090.69/warc/CC-MAIN-20170322212949-00110-ip-10-233-31-227.ec2.internal.warc.gz | 893,983,195 | 115,799 | ch04-p073
# ch04-p073 - 73. The velocity vectors (relative to the...
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, 11 , 1.03 knots tan tan 1.5 38.4 knots AB x AB y v v θ −− §· == = ° ¨¸ ©¹ which is to say that G v AB points 1.5° east of north. (c) Since they started at the same time, their relative velocity describes at what rate the distance between them is increasing. Because the rate is steady, we have | | 160 4.2 h. | | 38.4 AB AB r t v Δ = G (d) The velocity G v does not change with time in this problem, and G r is in the same direction as G v since they started at the same time. Reversing the points of view, we have GG vv BA =− so that rr (i.e., they are 180° opposite to each other). Hence, we conclude that B stays at a bearing of 1.5° west of south relative to A during the journey (neglecting the curvature of Earth).
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
Ask a homework question - tutors are online | 314 | 1,118 | {"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} | 3.21875 | 3 | CC-MAIN-2017-13 | longest | en | 0.955621 |
https://www.scribd.com/document/294489332/Interview-Questions | 1,568,533,467,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514570830.42/warc/CC-MAIN-20190915072355-20190915094355-00020.warc.gz | 1,025,272,039 | 71,821 | You are on page 1of 9
# MECHANICAL ENGINEERING JOB INTERVIEW QUESTIONS:1.)What do you understand by center of buoyancy ?
Ans: Center of buoyancy is the center of gravity of the displaced liquid and buoyant force acts
through it.
2.)What is meant by precision ?
Ans.-Precision is defined as the ability of instrument to reproduce a certain set of readings within
a given accuracy.
3.)What is Torricellis theorem ?
Ans: According to it velocity of jet flowing out of a small opening is proportional to the square
root of head of liquid above it.
4.)Distinguish between creep and fatigue.
Ans: Creep is low and progressive deformation of a material with time under a constant stress at
high temperature applications. Fatigue is the reduced tendency of material to offer resistance to
1.). What is anisotropy ?
Ans: The phenomenon of different properties in different directions is called anisotropy.
2.)What is vapour pressure and when it becomes important ?
Ans: Vapour pressure is the partial pressure created by the vapour molecules when evaporation
takes place within an enclosed space. It becomes important when negative
pressures are involved.
3.)Explain absolute viscosity ?
Ans: It is the ratio of shear stress and the gradient of velocity with distance between a fixed plate
and moving plate (Its unit is Pa . s).
4.)What do you understand by acoustic velocity ?
Ans: Acoustic velocity is the speed of a small pressure (sound) wave in a fluid.
1.)What is the difference between shaper and planner ?
Ans.->Shaper has one tool head wherear planner has more than one tool head.
>In shaper, workpiece is stationary and tool reciprocates with ram. Whereas in planner
,workpiece rotate and tool stationary.
>Shaper is used for light work whereas planner is used for heavy work.
2.)What is Curie point ?
Ans: Curie point is the temperature at which ferromagnetic materials can no longer be
magnetized by outside forces.
3.)When Coriolis component is encountered ?
Ans: When a point moves along a path that has rotational motion.
4.)Define hydraulically efficient channel cross section.
Ans: The shape of such section is that which produces minimum wetted perimeter for a given
area of flow and carries maximum
flow.
1.)What is the difference between shaper and planner ?
Ans.->Shaper has one tool head wherear planner has more than one tool head.
>In shaper, workpiece is stationary and tool reciprocates with ram. Whereas in planner
,workpiece rotate and tool stationary.
>Shaper is used for light work whereas planner is used for heavy work.
2.)What is Curie point ?
Ans: Curie point is the temperature at which ferromagnetic materials can no longer be
magnetized by outside forces.
3.)When Coriolis component is encountered ?
Ans: When a point moves along a path that has rotational motion.
4.)Define hydraulically efficient channel cross section.
Ans: The shape of such section is that which produces minimum wetted perimeter for a given
area of flow and carries maximum
flow.
1.)What is critical temperature in metals ?
Ans: It is the temperature at which the phase change occurs in metals.
2.)Define buckling factor.
Ans: It is the ratio of the equivalent length of column to the minimum radius of gyration.
## 3.)What do you understand by catenary cable ?
Ans: A cable attached to the supports and carrying its own weight.
4.)What is coaxing ?
Ans: It is the process of improving fatigue properties by first under-stressing and then increasing
the stress in small increments.
1)What is Sentinel Relief Valve?
Ans. Its a special type valve system. The valve will open when exhaust casing pressure is
excessive (high). The valve warns the operator only; it is not intended to relieve the casing
pressure.
2)What is the difference between Specification,Codes, Standards?
Ans. Specification is describing properties of any type of materials. Code is procedure of
acceptance and rejection criteria.
Standard is accepted values and compare
other with it.
3.) What is difference between Welding and Brazing?
Ans. In Welding concentrated heat (high temperature) is applied at the joint of metal and fuse
together. In Brazing involves significantly lower temperatures and does not entail the melting of
base metals. Instead, a filler metal is melted and forced to flow into the joint through capillary
action.
1. What is Difference between Hardness and Toughness?
Ans. Toughness is the ability of a material to absorb energy.
Hardness is the ability of a material to withstand wear.
2.What does F.O.F Stand for Piping Design?
Ans.Face of Flange, The F.O.F (Raised face and Flat face) is used to know the accurate
dimension of the flange in order to avoid the minute errors in measurement in case of vertical or
horizontal pipe lines.
3.Difference between Performance and Efficiency?
Ans.The accomplishment of a given task measured against preset known standards of accuracy,
completeness, cost, and speed is called as Performance.
Efficiency is defined as the input given and the work obtained from that input like money, time,
labour etc. Its the main factor of productivity.
## 1. What is the difference between scavenging and supercharging ?
Ans: Scavenging is process of flushing out burnt gases from engine cylinder by introducing fresh
air in the cylinder before exhaust stroke ends. Supercharging is the process of supplying higher
mass of air by compressing the atmospheric air.
2.What is pitting ? How it is caused ?
Ans: Non uniform corrosion over the entire metal surface, but occuring only in small pits is
called pitting. It is caused by lack of uniformity in metal.
3.Why large boilers are water tube type ?
Ans: Water tube boilers raise steam fast because of large heat transfer area and positive water
circulation. Thus they respond faster to fluctuations in demand. Further single tube failure does
## 1.)What is the difference between mechanical advantage and velocity ratio ?
Ans: Mechanical advantage is the ratio of load lifted and the effort applied. Velocity ratio is the
ratio between the distance moved by the effort applied and the distance moved by the load lifted.
Ans: Connecting rod and belt respectively.
3.)What is the damping ratio for non-oscillating system and under-damped system ?
Ans: More than unity and less than unity respectively.
4.)State D' Alembert's principle and write down its importance.
Ans: D' Alembert's principle enables us to replace a given system by a massless rigid body so
that forces acting on it are equivalent to those on real body and then enables to determine the
forces transmitted to other paired rigid body. This way the problem of kinetics gets reduced to
equivalent problem on statics.
## 1.)What do you understand by forced convection ?
Ans: When convection heat transfer occurs between a solid body and a fluid and where
circulation of fluid is caused and controlled by some mechanical.
## 2.)What for water/steam is injected in combustion zones of a gas turbine ?
Ans: Water/steam is injected to limit the amount of NOx formed by lowering the flame and gas
temperatures.
3.)Why steam is in open space and water inside tube in condensers used in power plants ?
Ans: Overall heat transfer coefficient can be increased by increasing velocity of water in tube.
Further steam needs more space due to higher specific volume.
4.)Which refrigerant is used for ice plant and transport refrigeration ?
Ans: Ammonia and CO2 respectively.
1.)What is the difference between higher pair and lower pair ?
Ans: Higher pair has point or line contact between two links and lower pair has surface contact
between two links while in motion.
2.). For what purpose the bifiler suspension system used ?
Ans: It is used to determine moment of inertia.
3.)How helical gears are capable of transmitting heavy load at high speed compared to spur gears
Ans: Helical gears have smooth engagement and two pairs of teeth are always in contact.
4.). What is the effect of inertia of reciprocating parts on the engine frame ?
Ans: Inertia of reciprocating parts subjects engine frame to the force required to accelerate the
reciprocating mass and thus subject them to primary disturbing force and secondary disturbing
force. Secondary disturbing force comes into play due to obliquity of connecting rod and has
twice the frequency of the primary force
1.) What are the principal constituents of brass?
Ans: Principal constituents of brass are copper and zinc.
## 2.) What is Curie point ?
Ans: Curie point is the temperature at which ferromagnetic materials can no longer be
magnetised by outside forces.
## 3). Which element is added in steel to increase resistance to corrosion ?
Ans: Chromium.
4.) Whether individual components in composite materials retain their characteristics or not?
Ans: yes.
## 1.)What is endurance limit and what is its value for steel ?
Ans: Endurance limit is the maximum level of fluctuating stress which can be tolerated
indefinitely. In most steels this stress is approximately 50% of the ultimate tensile strength and
it is defined as the stress which can be endured for ten million reversals of stress.
2.)What do you understand by sulphur print ?
Ans: Sulphides, when attached with dilute acid, evolve hydrogen sulphide gas which stains
bromide paper and therefore can be readily detected in ordinary steels and cast irons . While
sulphur is not always as harmful as is sometimes supposed, a sulphur print is a ready guide to
the distribution of segregated impurities in general.
3.) What is the different between brass and bronze ?
Ans: Brass is an alloy of copper with zinc; and bronze is alloy of copper with tin.
4.) What is the effect of addition of zinc in copper? What is the use of 70/30 brass ?
Ans: By addition of zinc in copper, both tensile strength and elongation increases. The 70/30
brass has excellent deep drawing property and is used for making radiator fins.
## 1.)What is the difference between Technology and Engineering?
Ans.Engineering is application of science. Technology shows various methods of Engineering. A
bridge can be made by using
beams to bear the load,by an arc or by hanging in a cable; all shows different technology but
comes under civil engineering and
science applied is laws of force/load distribution.
## 2.) What is Powder Technology?
Ans. Powder technology is one of the ways of making bearing material. In this method metals
like bronze, Al, Fe are mixed and
compressed to make an alloy.
## 3.)What are the principal constituents of brass?
Ans: Principal constituents of brass are copper and zinc.
8. What is Curie point ?
Ans: Curie point is the temperature at which ferromagnetic materials can no longer be
magnetised by outside forces.
1.)Definition of Octane Number and Cetane Number?
Ans.Octane No.- Octane number is defined as the percentage, by volume, of iso octane in the
mixture of iso octane and hheptane. It is the measure of rating of SI engine.
Cetane No.- Cetane number is defined as the percentage, by volume, of n-cetane in the mixture
of n-cetane and alpha methyl
naphthalene. It is the measure of rating of CI engine.
2.)Which Pump is more Efficient Centrifugal Pump or Reciprocating Pump?
Ans. Centrifugal pump. Because flow rate is higher compared to reciprocating pump. Flow is
smooth and it requires less space to
install. Lower initial cost and lower maintenance cost.
3.)How a Diesel Engine Works as Generator?
Ans.Diesel engine is a prime mover, for a generator, pump,and for vehicles etc. generator is
connected to engine by shaft.
mostly in thermal power plat ,there is an engine is used to drive generator to generate power.
4.)Compare Brayton Cycle and Otto Cycle?
Ans. The heat addition and rejection processes in Otto cycle are of constant volume, whereas in
Brayton cycle, they are of
constant pressure.
-Otto cycle is the ideal cycle for spark ignition engines.
-Brayton cycle is the ideal cycle for gas power turbines.
## 1.) On what account the friction drag is experienced ?
Ans: Friction drag is experienced on separation of boundary layer.
2.) What do you understand by choking in pipe line ?
Ans: When specified mass flow is not able to take place in a pipe line.
3.) What is the difference between streamline body and bluff body?
Ans: In streamline body the shape is such that separation in flow occurs past the near most part
of the body so that wake formed is small and thus friction drag is much greater than pressure
drag. In bluff body the flow gets separated much ahead of its rear resulting in large wake and
thus pressure drag is much greater than the friction drag.
4.)Define hydraulically efficient channel cross section.
Ans: The shape of such section is that which produces minimum wetted perimeter for a given
area of flow and carries maximum flow.
## 1.)What is the meaning of the term sensitive drill press ?
Ans: A sensitive drilling press is a light, simple, bench type machine for light duty working with
infinite speed ratio.
2.) Why carburised machine components have high endurance limit?
Ans: In carburised machine components, the process of carburisation introduces a compressive
layer on the surface and thus endurance limit is increased.
3.) What is the difference between perfect and real fluids ?
Ans: Perfect fluids are treated as if all tangential forces created by friction can be ignored. Real
fluids refer to the cases in which friction must properly be taken into account.
4.) What is bulk modulus of elasticity ?
Ans: It represents the compressibility of a fluid. It is the ratio of the change in unit pressure to the
corresponding volume change per unit of volume.
1.) What is the direction of tangential acceleration ?
Ans:The direction of tangential acceleration may be same or opposite to that of angular velocity.
2.)Why excess air is required to burn a fuel completely ?
Ans: Excess air is required to ensure adequate mixing of fuel and air, avoid smoke, minimize
slagging in coal burning, and to ensure maximum steam output.
3.)Which type of plant will you recommend for remote location if power is required in six to
twelve months time ?
Ans: Diesel engine power plant.
## 4.)What do you understand by forced convection ?
Ans: When convection heat transfer occurs between a solid body and a fluid and where
circulation of fluid is caused and controlled by some mechanical
1.) By which instruments the shear stress in fluids can be measured directly ?
Ans: By Stanton tube or Preston tube.
2.) On what factors does the pressure at a point as a static mass of liquid depends upon?
Ans: Specific weight of liquid and the depth below the free liquid surface
3.) State Archimedes principle.
Ans: Any weight, floating or immersed in a liquid, is acted upon by a buoyant force equal to the
weight of the liquid displaced. This force acts through the center of buoyancy, i.e. the e.g. of the
displaced liquid.
4.)What do you understand by center of buoyancy ?
Ans: Center of buoyancy is the center of gravity of the displaced liquid and buoyant force acts
through it.
1.What is an isobaric process?
Ans.-An isobaric process is a thermodynamic process in which the pressure stays constant: P =
0
2.What is Factor of safety?
Ans.-Factor of safety (FoS),safety factor (SF), is a term describing the structural capacity of a
system beyond the expected loads or actual loads. Essentially, how much stronger the system is
than it usually needs to be for an intended load.
3.Which one is more efficient? A four stroke engine or a two stroke and why?
Ans.-This depends on what type of efficiency you are referring to. As far as power generation,
two-stroke by far. Simply because it accomplishes in two strokes, what a four stroke motor takes
twice as long to accomplish. Fuel consumption is a different story. Four strokes are more
efficient because there is separation of fuel and engine oil, as well as intake and exhaust gases,
leading to less pollution, and better mileage.
4.What is bending moment?
Ans.-A bending moment is a measure of the average internal stress induced in a structural
element when an external force or moment is applied to the element causing the element to bend. | 3,512 | 16,105 | {"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} | 2.75 | 3 | CC-MAIN-2019-39 | latest | en | 0.919491 |
https://estebantorreshighschool.com/interesting-about-equations/enthalpy-equation-thermodynamics.html | 1,685,341,488,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644683.18/warc/CC-MAIN-20230529042138-20230529072138-00163.warc.gz | 276,067,305 | 11,189 | ## What is enthalpy in thermodynamics?
Enthalpy, the sum of the internal energy and the product of the pressure and volume of a thermodynamic system. In symbols, the enthalpy, H, equals the sum of the internal energy, E, and the product of the pressure, P, and volume, V, of the system: H = E + PV.
## How do you calculate enthalpy?
Use the formula ∆H = m x s x ∆T to solve. Once you have m, the mass of your reactants, s, the specific heat of your product, and ∆T, the temperature change from your reaction, you are prepared to find the enthalpy of reaction. Simply plug your values into the formula ∆H = m x s x ∆T and multiply to solve.
## Is Delta H equal to Q?
You can say that Q (Heat) is energy in transit. Enthalpy (Delta H), on the other hand, is the state of the system, the total heat content. They both can deal with heat (qp) (Q at constant pressure) = (Delta H) but both Heat and Enthalpy always refer to energy, not specifically Heat. Hope this helps!
## What is enthalpy in simple terms?
Enthalpy is a concept used in science and engineering when heat and work need to be calculated. When a substance changes at constant pressure, enthalpy tells how much heat and work was added or removed from the substance. Enthalpy is similar to energy, but not the same.
S.
## What is difference between enthalpy and heat?
Heat is a transfer of energy due to a temperature difference. Enthalpy is the change in amount of heat in a system at constant pressure.
## What is enthalpy unit?
The unit of measurement for enthalpy in the International System of Units (SI) is the joule. Other historical conventional units still in use include the British thermal unit (BTU) and the calorie.
## What is the formula for bond enthalpy?
As an example of bond dissociation enthalpy, to break up 1 mole of gaseous hydrogen chloride molecules into separate gaseous hydrogen and chlorine atoms takes 432 kJ. The bond dissociation enthalpy for the H-Cl bond is +432 kJ mol1.
You might be interested: Rate of diffusion equation
bond enthalpy (kJ mol1)
O-H +464
heat energy
## What is Q MC ∆ T used for?
Q=mcΔT Q = mc Δ T , where Q is the symbol for heat transfer, m is the mass of the substance, and ΔT is the change in temperature. The symbol c stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00ºC.
## What happens when enthalpy is zero?
Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings. Enthalpy change is also equal to zero because the change in energy zero and the pressure and volume is constant.
## What does Delta H mean?
In chemistry, the letter “H” represents the enthalpy of a system. Enthalpy refers to the sum of the internal energy of a system plus the product of the system’s pressure and volume. The delta symbol is used to represent change. Therefore, delta H represents the change in enthalpy of a system in a reaction.
## What is enthalpy used for?
It is used to calculate the heat of reaction of a chemical process. Change in enthalpy is used to measure heat flow in calorimetry. It is measured to evaluate a throttling process or Joule-Thomson expansion. Enthalpy is used to calculate minimum power for a compressor.
## What is enthalpy in HVAC?
Enthalpy is defined as the amount of internal energy within a system combined with the product of its pressure and volume. At its core, the main function of an HVAC system is to transfer heat, which is a form of energy.
### Releated
#### Equation of vertical line
How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […]
#### Bernoulli’s equation example
What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […] | 1,034 | 4,360 | {"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} | 3.625 | 4 | CC-MAIN-2023-23 | latest | en | 0.903428 |
https://www.aqua-calc.com/convert/surface-density/ounce-per-square-meter-to-stone-per-square-yard | 1,560,768,658,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998473.44/warc/CC-MAIN-20190617103006-20190617125006-00545.warc.gz | 659,952,974 | 9,037 | # Ounces per (square meter) to stones per (square yard)
## [oz/m² to st/yd²] (oz:ounce, m:meter, st:stone, yd:yard)
### ounces per square meter equal to stones per square yard conversion results
• 1
through
20
ounces per square meter
• 1 oz/m² = 0.003732711 st/yd²
• 2 oz/m² = 0.007465423 st/yd²
• 3 oz/m² = 0.011198134 st/yd²
• 4 oz/m² = 0.014930846 st/yd²
• 5 oz/m² = 0.018663557 st/yd²
• 6 oz/m² = 0.022396269 st/yd²
• 7 oz/m² = 0.02612898 st/yd²
• 8 oz/m² = 0.029861691 st/yd²
• 9 oz/m² = 0.033594403 st/yd²
• 10 oz/m² = 0.037327114 st/yd²
• 11 oz/m² = 0.041059826 st/yd²
• 12 oz/m² = 0.044792537 st/yd²
• 13 oz/m² = 0.048525249 st/yd²
• 14 oz/m² = 0.05225796 st/yd²
• 15 oz/m² = 0.055990671 st/yd²
• 16 oz/m² = 0.059723383 st/yd²
• 17 oz/m² = 0.063456094 st/yd²
• 18 oz/m² = 0.067188806 st/yd²
• 19 oz/m² = 0.070921517 st/yd²
• 20 oz/m² = 0.074654229 st/yd²
• 21
through
40
ounces per square meter
• 21 oz/m² = 0.07838694 st/yd²
• 22 oz/m² = 0.082119651 st/yd²
• 23 oz/m² = 0.085852363 st/yd²
• 24 oz/m² = 0.089585074 st/yd²
• 25 oz/m² = 0.093317786 st/yd²
• 26 oz/m² = 0.097050497 st/yd²
• 27 oz/m² = 0.100783209 st/yd²
• 28 oz/m² = 0.10451592 st/yd²
• 29 oz/m² = 0.108248631 st/yd²
• 30 oz/m² = 0.111981343 st/yd²
• 31 oz/m² = 0.115714054 st/yd²
• 32 oz/m² = 0.119446766 st/yd²
• 33 oz/m² = 0.123179477 st/yd²
• 34 oz/m² = 0.126912189 st/yd²
• 35 oz/m² = 0.1306449 st/yd²
• 36 oz/m² = 0.134377611 st/yd²
• 37 oz/m² = 0.138110323 st/yd²
• 38 oz/m² = 0.141843034 st/yd²
• 39 oz/m² = 0.145575746 st/yd²
• 40 oz/m² = 0.149308457 st/yd²
• 41
through
60
ounces per square meter
• 41 oz/m² = 0.153041169 st/yd²
• 42 oz/m² = 0.15677388 st/yd²
• 43 oz/m² = 0.160506591 st/yd²
• 44 oz/m² = 0.164239303 st/yd²
• 45 oz/m² = 0.167972014 st/yd²
• 46 oz/m² = 0.171704726 st/yd²
• 47 oz/m² = 0.175437437 st/yd²
• 48 oz/m² = 0.179170149 st/yd²
• 49 oz/m² = 0.18290286 st/yd²
• 50 oz/m² = 0.186635571 st/yd²
• 51 oz/m² = 0.190368283 st/yd²
• 52 oz/m² = 0.194100994 st/yd²
• 53 oz/m² = 0.197833706 st/yd²
• 54 oz/m² = 0.201566417 st/yd²
• 55 oz/m² = 0.205299129 st/yd²
• 56 oz/m² = 0.20903184 st/yd²
• 57 oz/m² = 0.212764551 st/yd²
• 58 oz/m² = 0.216497263 st/yd²
• 59 oz/m² = 0.220229974 st/yd²
• 60 oz/m² = 0.223962686 st/yd²
• 61
through
80
ounces per square meter
• 61 oz/m² = 0.227695397 st/yd²
• 62 oz/m² = 0.231428109 st/yd²
• 63 oz/m² = 0.23516082 st/yd²
• 64 oz/m² = 0.238893531 st/yd²
• 65 oz/m² = 0.242626243 st/yd²
• 66 oz/m² = 0.246358954 st/yd²
• 67 oz/m² = 0.250091666 st/yd²
• 68 oz/m² = 0.253824377 st/yd²
• 69 oz/m² = 0.257557089 st/yd²
• 70 oz/m² = 0.2612898 st/yd²
• 71 oz/m² = 0.265022511 st/yd²
• 72 oz/m² = 0.268755223 st/yd²
• 73 oz/m² = 0.272487934 st/yd²
• 74 oz/m² = 0.276220646 st/yd²
• 75 oz/m² = 0.279953357 st/yd²
• 76 oz/m² = 0.283686069 st/yd²
• 77 oz/m² = 0.28741878 st/yd²
• 78 oz/m² = 0.291151491 st/yd²
• 79 oz/m² = 0.294884203 st/yd²
• 80 oz/m² = 0.298616914 st/yd²
• 81
through
100
ounces per square meter
• 81 oz/m² = 0.302349626 st/yd²
• 82 oz/m² = 0.306082337 st/yd²
• 83 oz/m² = 0.309815049 st/yd²
• 84 oz/m² = 0.31354776 st/yd²
• 85 oz/m² = 0.317280471 st/yd²
• 86 oz/m² = 0.321013183 st/yd²
• 87 oz/m² = 0.324745894 st/yd²
• 88 oz/m² = 0.328478606 st/yd²
• 89 oz/m² = 0.332211317 st/yd²
• 90 oz/m² = 0.335944029 st/yd²
• 91 oz/m² = 0.33967674 st/yd²
• 92 oz/m² = 0.343409451 st/yd²
• 93 oz/m² = 0.347142163 st/yd²
• 94 oz/m² = 0.350874874 st/yd²
• 95 oz/m² = 0.354607586 st/yd²
• 96 oz/m² = 0.358340297 st/yd²
• 97 oz/m² = 0.362073009 st/yd²
• 98 oz/m² = 0.36580572 st/yd²
• 99 oz/m² = 0.369538431 st/yd²
• 100 oz/m² = 0.373271143 st/yd²
#### Foods, Nutrients and Calories
CHEX MIX MDY BDIES BRND SNACK BROWNIE SUPREME, UNPREPARED, GTIN: 00016000424456 contain(s) 440 calories per 100 grams or ≈3.527 ounces [ calories | price ]
Retinol food sources
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Rio Grande weighs 1 489.72 kg/m³ (93.00018 lb/ft³) with specific gravity of 1.48972 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price | density ]
Lanthanum oxybromide weighs 6 280 kg/m³ (392.04759 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-408A, liquid (R408A) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
The pennyweight per square inch surface density measurement unit is used to measure area in square inches in order to estimate weight or mass in pennyweights[...]
The radiation absorbed dose is a measurement of radiation, in energy per unit of mass, absorbed by a specific object, such as human tissue.[...]
Convert pound per cubic millimeter to stone per metric cup or convert between all units of density measurement
#### Calculators
Calculate area of a rectangle, its perimeter and diagonal[...] | 2,283 | 5,004 | {"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} | 2.640625 | 3 | CC-MAIN-2019-26 | latest | en | 0.08591 |
https://www.coursehero.com/file/p188g0ru/Conclusion-Do-Not-Reject-Null-Hypothesis-IFG17B9Reject-Null-HypothesisD/ | 1,643,176,751,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00698.warc.gz | 750,197,098 | 50,779 | # Conclusion do not reject null hypothesis
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ConclusionDo Not Reject Null Hypothesis"=IF(G17<B9,"Reject Null Hypothesis","DConclusion & Assumption:The conclusion is that since the p-value is > than the level of significance so we do no reject the null hyp
o Not Reject Null Hypothesis")"Do Not Reject Null Hypothesis")"T.2T(-B11,B5-1))"Do Not Reject Null Hypothesis")"pothesis, The school can reach the conclusion that the applicants average work experience is < 3 years
s.
Credit Card ChargesOne Sample Test for the MeanSample Size42Sample Mean1376.54Sample Standard Deviation183.89Hypothesized value1350Level of significance0.05t-statistic0.935335571"=(B6-B8)/(B7/SQRT(B5))One tailed t-value1.682878002"=T.INV(1-B9,B5-1)"Two tailed t-value (+ and -)2.01954097"=T.INV.2T(B9,B5-1)"Hypothesis Statement is: H0:μ = 1350H1: μ > 1350Upper tailed test
Hypothesis test resultsLower one-tailed testCritical t-value-1.68287800213271p-value0.822453989317345ConclusionDo Not Reject Null HypothesisUpper one-tailed testCritical t-value1.68287800213271p-value0.177546010682655ConclusionDo Not Reject Null HypothesisTwo-tailed testCritical t-value (+ and -)2.01954097044138p-value0.35509202136531ConclusionDo Not Reject Null HypothesisConclusion is:The conclusion is that since the p-value is > than the level of
"=-B12""=T.DIST(\$B\$11,\$B\$5-1,TRUE)""=IF(G7<B9,"Reject Null Hypothesis","Do Not Reject Null Hypothesis")""=B12""=1-T.DIST(\$B\$11,\$B\$5-1,TRUE)""=IF(G12<B9,"Reject Null Hypothesis","Do Not Reject Null Hypothesis")""=B13""=T.DIST.2T(B11,B5-1)""=IF(G17<B9,"Reject Null Hypothesis","Do Not Reject Null Hypothesis")"f significance so we do no reject the null hypothesis, so the data does not provide statistical evid
dence that the average monthly charges have increased.
Advertising Strategya.Hypothesis Statement is:H0 : μ < \$70.00H1: b.X_bar\$ 75.86 S.D.\$ 50.90 Mean\$ 70.00 n300t-stat1.99407028140543Critical Value1.64996576742639Decision/ConclusionReject Null Hypothesisc.X_bar\$ 68.53 S.D.\$ 45.29 Mean\$ 70.00 n700t-stat-0.858744629557285Critical Value1.64703646375354Decision/ConclusionDo Not Reject Null Hypothesisμ >\$70.00
18-25 group"=(C7-C10)/(C8/SQRT(C11))""=T.INV(0.95,C11-1)"the t-stat is > the critical value. Because of this we reject the null hypothesis. 35+ group"=(C16-C19)/(C17/SQRT(C20))""=T.INV(0.95,C20-1)"the t-stat is < the critical value. Because of this we do not reject the null hypothesis. | 820 | 2,536 | {"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} | 3.640625 | 4 | CC-MAIN-2022-05 | latest | en | 0.710221 |
https://www.cruzmarquez.com/posts/2023/03/28/leetcode-54-spiral-matrix/ | 1,702,241,959,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102637.84/warc/CC-MAIN-20231210190744-20231210220744-00673.warc.gz | 787,753,984 | 9,205 | Spiral Matrix
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# LeetCode #54: Spiral Matrix (C/C++).
medium
source: https://leetcode.com/problems/spiral-matrix/
C/C++ Solution to LeetCode problem 54. Spiral Matrix.
## Problem
Given an m x n matrix, return all elements of the matrix in spiral order.
## Examples
### Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
### Example 2:
Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
## Constraints
• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 10
• -100 <= matrix[i][j] <= 100
## Solution
The algorithm is as follows:
• We iterate through all the cells.
• We use two variables for the current position (row and column).
• We use two variables to identify the minimun column and the maximum column.
• Same for minimum and maximum row.
• Every time the current location is at the minimum row and column:
• Reduce by one the maximum column.
• Reduce by one the maximum row.
• Every time the current location is at the maximum row and column:
• Increase by one the minimum column.
• Increase by one the minimum row.
• We handle the direction we move the current location by not moving or adding/substracting 1 from the column/row position base on this conditions:
• Minimum row and minimum column: row doesn’t change, column will increase by one.
• Minimum row and maximum column: row increase by one, column doesn’t change.
• Maximum row and maximum column: row doesn’t change, column will decrease by one.
• Maximum row and minimum column: columns doesn’t change, row will decrease by one.
There are a few considerations:
• Initial position: row 0, column -1. (So, first step will add one to column and start moving throgh the first row).
• Maximum column and row: start one place further than the size of the matrix, so, first decreasing sets them to the correct size.
• When only one row, or only one column (minRow == maxRow or minCol == maxCol)we don’t move the row or column.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 class Solution { public: vector<int> spiralOrder(vector<vector<int>>& matrix) { vector<int> result; int t = matrix.size() * matrix[0].size(); int dr = 0; int dc = 1; int r = 0; int c = -1; int minCol = -1; int maxCol = matrix[0].size(); int minRow = 0; int maxRow = matrix.size(); for (int i=0; i<t; i++) { r += dr; c += dc; result.push_back(matrix[r][c]); if ((r == minRow && c == minCol) || (minCol == -1 && c==0)) { maxCol--; maxRow--; if (minCol < maxCol) { dr = 0; dc = 1; } } else if (r == maxRow && c == maxCol) { minRow++; minCol++; dr = 0; dc = -1; continue; } if (r == maxRow && c == minCol) { if (minRow < maxRow) { dr = -1; dc = 0; } } else if (r == minRow && c == maxCol) { dr = 1; dc = 0; } } return result; } }; | 903 | 2,852 | {"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} | 4.09375 | 4 | CC-MAIN-2023-50 | latest | en | 0.77886 |
https://physicscomputingblog.com/category/combustion/ | 1,575,738,271,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540500637.40/warc/CC-MAIN-20191207160050-20191207184050-00488.warc.gz | 496,964,870 | 19,462 | ## What are the most energetic combustible materials?
Most people have seen several examples of combustion reactions that produce different flame temperatures. The temperature of a campfire can be somewhat over 800 deg C, while the flame of an oxyacetylene torch in welding can reach a temperature of over 3000 deg C. The energeticity of combustion reactions can be quantified in several ways:
1. Maximum temperature of the flame that is produced
2. Amount of heat released in combustion (per unit mass or volume of fuel)
3. In the cases of gaseous fuels, aerosols or vapors, the magnitude of sudden pressure increase in the explosion of fuel-oxygen mixture
Note that the amount of heat released per mole of fuel is not a good measure of the energeticity, as you can make the molar heat of combustion of an organic compound practically as large as you want by building longer and longer hydrocarbon chains.
The heat release of combustion reactions, more formally called enthalpy of combustion $\Delta H_c$, is rather easy to measure in the laboratory using a calorimeter. A typical example of a combustion reaction is the burning of methane, described by the reaction equation below:
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2 O$ .
The enthalpy change in this reaction can be calculated from the enthalpies of formation ($\Delta H_f$) of the reactants and products, which can be found from tables of thermodynamic quantities (in books or online). The enthalpy change in the combustion of a mole of methane is
$\Delta_c (CH_4) = \Delta H_f(CO_2 ) + 2\Delta H_f(H_2 O) - \Delta H_f (CH_4 )$,
where the enthalpy of formation of elemental oxygen has not been included as it is zero (as is the $\Delta H_f$ of any element in its most stable form).
The thermodynamic quantities in this are
$\Delta H_f(CO_2 ) = -393 kJ/mol$
$\Delta H_f(H_2 O) = -242 kJ/mol$
$\Delta H_f(CH_4 ) = -74.9 kJ/mol$
which give a value of about -802 kJ/mol or -50 kJ/g for the quantity $\Delta H_c (CH_4 )$. An enthalpy change that is negative means that heat is released to the surroundings by the reaction. The formal definition of enthalpy is $H=E+PV$, or internal energy plus pressure times volume, and in constant-pressure processes of simple systems its change is $\Delta H = \Delta E + P\Delta V$. A system that is “simple” in the sense meant here can do work on the surroundings only by expanding: $W = -P\Delta V$. Some systems can do other kinds of work, too, like elastic deformation (which can require an application of force over some distance even without involving a change of volume) or electrical work (charging a battery).
As the heat of combustion increases linearly with increasing $\Delta H_f$:s of the products and decreases linearly with increasing $\Delta H_f$:s of the reactants, the most energetic fuels are those that have a positive heat of formation, which means that energy is consumed when they are formed from their elemental constituents. Straight-chain alkane hydrocarbons like ethane, propane and butane all have negative heats of formation. To make a hydrocarbon that has a positive $\Delta H_f$, we need to have triple bonds between carbon atoms, or highly strained carbon-carbon bonds (such as in small alicyclic rings). An example of the former case is acetylene C2H2, for which $\Delta H_f = 227.4 kJ/mol = 8.7 kJ/g$ and an example of the latter is cyclopropane $(C_3 H_6)$, for which $\Delta H_f = 53.2 kJ/mol = 1.26 kJ/g$. Cyclopropane was used in the past as an anesthetic gas in surgical operations, but this use was discontinued because of the fire/explosion hazard related to it. In addition to molecules with small ring structures, there are also other molecules with large “steric hindrance” like tetra-tert-butylmethane or cubane that have or are predicted to have a positive heat of formation.
Inorganic combustible substances with positive $\Delta H_f$ include hydrazine ($N_2 H_4$) and cyanogen ($C_2 N_2$). Both are nasty toxic compounds, and hydrazine can also explode when heated, even with no oxygen present, as it decomposes violently to elemental hydrogen and nitrogen if it’s given enough activation energy (this can happen with acetylene, too). Cyanogen is a gaseous compound that is formed from two cyano groups (-CN). The cyano group is called a pseudohalogen, as it is often found in organic molecules in positions where there could also be a halogen (fluorine, chlorine, bromine or iodine) atom (see chlorobenzene and cyanobenzene). A stoichiometric mixture of cyanogen and oxygen can reach a flame temperature of about 4500 degrees Celsius.
Figure 1. Molecular structures of cyanobenzene (left) and chlorobenzene (right).
Reactive metals like magnesium or aluminum often have large heats of combustion, for example the $\Delta H_c (Al)$ is about -838 kJ/mol or -31 kJ/g which is less per gram than the typical values of hydrocarbons, but more per unit volume because Al metal has a significantly higher density than liquefied hydrocarbons. Aluminum is not something that a layman would think of as a combustible fuel, but it is actually very flammable when it’s in the form of very fine powder, and it is used in thermite mixtures and flash powders (pyrotechnic mixtures of Al powder with oxidizers such as potassium perchlorate, which produce a very loud bang and a temperature of over 3000 deg C when ignited in a confined space).
Figure 2. Magnesium metal burns with a really high-temperature flame, which makes it useful in firestarters (source: https://en.wikipedia.org/wiki/Magnesium#/media/File:Magnesium_Sparks.jpg )
Figure 3. The thermite mixture, made from finely powdered aluminum and iron oxide, burns with a high temperature flame and has been used in the welding of railroad tracks. (source: https://commons.wikimedia.org/wiki/File:ThermiteReaction.jpg )
When estimating the maximum temperature that can be reached in the combustion of some substance, there is a need to consider not only the enthalpy changes of the reactions, but also the heat capacities of the products that are formed. A quantity denoted $T_f$, and called adiabatic flame temperature, is the temperature that would theoretically be reached when the fuel reacts with oxygen in a system that is thermally insulated (to prevent heat loss to the surroundings) but isobaric (can do work on surroundings by expanding). A basic estimate of $T_f$ is
where $\Delta H_c$ is the heat produced in the combustion of 1 mole of the fuel and
$C_p$ is the constant-pressure heat capacity
of the product mixture at a temperature of about 1000 deg C (partial derivative of enthalpy with respect to temperature at constant pressure). This formula is not accurate for the most energetic combustions, as most combustion reactions don’t proceed all the way to the stoichiometric end products in high temperatures. This is because at high temperatures, the molar entropy change $\Delta S$ of the reaction starts to be a significant factor in determining the molar Gibbs energy change (and equilibrium constant) of the process, and smaller molecules (like $H_2$ and $O_2$ instead of $H_2 O$) usually have a larger molar entropy than large ones. At high temperatures the heat capacity also increases dramatically, because the Boltzmann factor $k_B T$, which gives the energy scale corresponding to absolute temperature T, starts to approach the energy scale of molecular vibrational (and later also electronic) transitions, and energy is therefore distributed to the vibrational and electronic degrees of freedom of the reaction products. At really high temperatures, like inside the Sun, matter is in the form of ionized plasma, but that kind of extreme conditions can’t be produced chemically.
When considering the combustion of reactive metals, another thing that limits the maximum reaction temperature is the boiling point of the reaction products such as $MgO$ or $Al_2 O_3$. The flame temperature in metal combustion can’t usually exceed the boiling point of the most volatile product, as the heat of combustion is not as large as the heat of vaporization of the reaction products. The combustion of zirconium metal in pure oxygen can raise the temperature up to 4000 degrees Celsius, because of the very high boiling point of zirconium oxide.
The factors mentioned above limit the maximum temperature attainable in a chemical combustion reaction to about 5000 degrees Celsius, which is approximately the adiabatic flame temperature of dicyanoacetylene, a derivative of cyanogen. The amount of pressure rise in the combustion reaction, which was mentioned as one measure of the energeticity of the reaction, depends on both the flame temperature and the difference in the number of moles of gas is the reactants and products. Nuclear reactions such as the fission of uranium-235 are able to produce much higher temperatures, up to millions of degrees Celsius, due to the very large energy release in a relatively small volume. | 2,036 | 8,944 | {"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": 33, "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} | 3.359375 | 3 | CC-MAIN-2019-51 | latest | en | 0.89501 |
https://unimath.github.io/agda-unimath/orthogonal-factorization-systems.local-types.html | 1,726,513,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00299.warc.gz | 547,758,564 | 49,687 | # Local types
Content created by Fredrik Bakke, Egbert Rijke and Jonathan Prieto-Cubides.
Created on 2023-02-06.
module orthogonal-factorization-systems.local-types where
Imports
open import foundation.action-on-identifications-functions
open import foundation.commuting-squares-of-maps
open import foundation.contractible-maps
open import foundation.contractible-types
open import foundation.dependent-pair-types
open import foundation.dependent-universal-property-equivalences
open import foundation.empty-types
open import foundation.equivalences
open import foundation.families-of-equivalences
open import foundation.function-extensionality
open import foundation.function-types
open import foundation.functoriality-dependent-function-types
open import foundation.homotopies
open import foundation.identity-types
open import foundation.logical-equivalences
open import foundation.postcomposition-functions
open import foundation.precomposition-dependent-functions
open import foundation.precomposition-functions
open import foundation.propositions
open import foundation.retracts-of-maps
open import foundation.retracts-of-types
open import foundation.sections
open import foundation.type-arithmetic-dependent-function-types
open import foundation.type-arithmetic-unit-type
open import foundation.unit-type
open import foundation.universal-property-empty-type
open import foundation.universal-property-equivalences
open import foundation.universe-levels
## Idea
A dependent type A over Y is said to be local at f : X → Y, or f-local , if the precomposition map
_∘ f : ((y : Y) → A y) → ((x : X) → A (f x))
is an equivalence.
We reserve the name is-local for when A does not vary over Y, and specify with is-local-dependent-type when it does.
Note that a local dependent type is not the same as a local family. While a local family is a type family P on some other indexing type A, such that each fiber is local as a nondependent type over Y, a local dependent type is a local type that additionally may vary over Y. Concretely, a local dependent type A may be understood as a family of types such that for every y : Y, A y is fiber f y-null.
## Definition
### Local dependent types
module _
{l1 l2 l3 : Level} {X : UU l1} {Y : UU l2} (f : X → Y) (A : Y → UU l3)
where
is-local-dependent-type : UU (l1 ⊔ l2 ⊔ l3)
is-local-dependent-type = is-equiv (precomp-Π f A)
is-property-is-local-dependent-type : is-prop is-local-dependent-type
is-property-is-local-dependent-type = is-property-is-equiv (precomp-Π f A)
is-local-dependent-type-Prop : Prop (l1 ⊔ l2 ⊔ l3)
pr1 is-local-dependent-type-Prop = is-local-dependent-type
pr2 is-local-dependent-type-Prop = is-property-is-local-dependent-type
### Local types
module _
{l1 l2 l3 : Level} {X : UU l1} {Y : UU l2} (f : X → Y) (A : UU l3)
where
is-local : UU (l1 ⊔ l2 ⊔ l3)
is-local = is-local-dependent-type f (λ _ → A)
is-property-is-local : is-prop is-local
is-property-is-local = is-property-is-local-dependent-type f (λ _ → A)
is-local-Prop : Prop (l1 ⊔ l2 ⊔ l3)
is-local-Prop = is-local-dependent-type-Prop f (λ _ → A)
## Properties
### Locality distributes over Π-types
module _
{l1 l2 : Level} {X : UU l1} {Y : UU l2} (f : X → Y)
where
distributive-Π-is-local-dependent-type :
{l3 l4 : Level} {A : UU l3} (B : A → Y → UU l4) →
((a : A) → is-local-dependent-type f (B a)) →
is-local-dependent-type f (λ x → (a : A) → B a x)
distributive-Π-is-local-dependent-type B f-loc =
is-equiv-map-equiv
( ( equiv-swap-Π) ∘e
( equiv-Π-equiv-family (λ a → precomp-Π f (B a) , (f-loc a))) ∘e
( equiv-swap-Π))
distributive-Π-is-local :
{l3 l4 : Level} {A : UU l3} (B : A → UU l4) →
((a : A) → is-local f (B a)) →
is-local f ((a : A) → B a)
distributive-Π-is-local B =
distributive-Π-is-local-dependent-type (λ a _ → B a)
### Local types are closed under equivalences
module _
{l1 l2 l3 l4 : Level}
{X : UU l1} {Y : UU l2} {A : Y → UU l3} {B : Y → UU l4}
(f : X → Y)
where
is-local-dependent-type-fam-equiv :
fam-equiv A B → is-local-dependent-type f B → is-local-dependent-type f A
is-local-dependent-type-fam-equiv e is-local-B =
is-equiv-htpy-equiv
( ( equiv-Π-equiv-family (inv-equiv ∘ e ∘ f)) ∘e
( precomp-Π f B , is-local-B) ∘e
( equiv-Π-equiv-family e))
( λ g →
eq-htpy (λ y → inv (is-retraction-map-inv-equiv (e (f y)) (g (f y)))))
is-local-dependent-type-inv-fam-equiv :
fam-equiv B A → is-local-dependent-type f B → is-local-dependent-type f A
is-local-dependent-type-inv-fam-equiv e =
is-local-dependent-type-fam-equiv (inv-equiv ∘ e)
module _
{l1 l2 l3 l4 : Level}
{X : UU l1} {Y : UU l2} {A : UU l3} {B : UU l4}
(f : X → Y)
where
is-local-equiv : A ≃ B → is-local f B → is-local f A
is-local-equiv e = is-local-dependent-type-fam-equiv f (λ _ → e)
is-local-inv-equiv : B ≃ A → is-local f B → is-local f A
is-local-inv-equiv e = is-local-dependent-type-inv-fam-equiv f (λ _ → e)
### Locality is preserved by homotopies
module _
{l1 l2 l3 : Level} {X : UU l1} {Y : UU l2} {A : UU l3} {f f' : X → Y}
where
is-local-htpy : (H : f ~ f') → is-local f' A → is-local f A
is-local-htpy H = is-equiv-htpy (precomp f' A) (htpy-precomp H A)
is-local-htpy' : (H : f ~ f') → is-local f A → is-local f' A
is-local-htpy' H = is-equiv-htpy' (precomp f A) (htpy-precomp H A)
### If S is f-local then S is local at every retract of f
module _
{l1 l2 l3 l4 l5 : Level} {A : UU l1} {B : UU l2} {X : UU l3} {Y : UU l4}
(f : A → B) (g : X → Y) (R : f retract-of-map g) (S : UU l5)
where
is-local-retract-map-is-local : is-local g S → is-local f S
is-local-retract-map-is-local =
is-equiv-retract-map-is-equiv
( precomp f S)
( precomp g S)
( retract-map-precomp-retract-map f g R S)
In fact, the higher coherence of the retract is not needed:
module _
{l1 l2 l3 l4 l5 : Level} {A : UU l1} {B : UU l2} {X : UU l3} {Y : UU l4}
(f : A → B) (g : X → Y) (R₀ : A retract-of X) (R₁ : B retract-of Y)
(i : coherence-square-maps' (inclusion-retract R₀) f g (inclusion-retract R₁))
(r :
coherence-square-maps'
( map-retraction-retract R₀)
( g)
( f)
( map-retraction-retract R₁))
(S : UU l5)
where
is-local-retract-map-is-local' : is-local g S → is-local f S
is-local-retract-map-is-local' =
is-equiv-retract-map-is-equiv'
( precomp f S)
( precomp g S)
( retract-precomp R₁ S)
( retract-precomp R₀ S)
( precomp-coherence-square-maps
( g)
( map-retraction-retract R₀)
( map-retraction-retract R₁)
( f)
( r)
( S))
( precomp-coherence-square-maps
( f)
( inclusion-retract R₀)
( inclusion-retract R₁)
( g)
( i)
( S))
### If every type is f-local, then f is an equivalence
module _
{l1 l2 : Level} {X : UU l1} {Y : UU l2} (f : X → Y)
where
is-equiv-is-local : ({l : Level} (A : UU l) → is-local f A) → is-equiv f
is-equiv-is-local = is-equiv-is-equiv-precomp f
### If the domain and codomain of f is f-local, then f is an equivalence
module _
{l1 l2 : Level} {X : UU l1} {Y : UU l2} (f : X → Y)
where
section-is-local-domains' : section (precomp f X) → is-local f Y → section f
pr1 (section-is-local-domains' section-precomp-X is-local-Y) =
pr1 section-precomp-X id
pr2 (section-is-local-domains' section-precomp-X is-local-Y) =
htpy-eq
( ap
( pr1)
{ ( f ∘ pr1 (section-is-local-domains' section-precomp-X is-local-Y)) ,
( ap (postcomp X f) (pr2 section-precomp-X id))}
{ id , refl}
( eq-is-contr (is-contr-map-is-equiv is-local-Y f)))
is-equiv-is-local-domains' : section (precomp f X) → is-local f Y → is-equiv f
pr1 (is-equiv-is-local-domains' section-precomp-X is-local-Y) =
section-is-local-domains' section-precomp-X is-local-Y
pr2 (is-equiv-is-local-domains' section-precomp-X is-local-Y) =
retraction-section-precomp-domain f section-precomp-X
is-equiv-is-local-domains : is-local f X → is-local f Y → is-equiv f
is-equiv-is-local-domains is-local-X =
is-equiv-is-local-domains' (pr1 is-local-X)
### Propositions are f-local if _∘ f has a converse
module _
{l1 l2 : Level} {X : UU l1} {Y : UU l2} (f : X → Y)
where
is-local-dependent-type-is-prop :
{l : Level} (A : Y → UU l) →
((y : Y) → is-prop (A y)) →
(((x : X) → A (f x)) → ((y : Y) → A y)) →
is-local-dependent-type f A
is-local-dependent-type-is-prop A is-prop-A =
is-equiv-has-converse-is-prop
( is-prop-Π is-prop-A)
( is-prop-Π (is-prop-A ∘ f))
is-local-is-prop :
{l : Level} (A : UU l) →
is-prop A → ((X → A) → (Y → A)) → is-local f A
is-local-is-prop A is-prop-A =
is-local-dependent-type-is-prop (λ _ → A) (λ _ → is-prop-A)
### All type families are local at equivalences
module _
{l1 l2 : Level} {X : UU l1} {Y : UU l2} (f : X → Y)
where
is-local-dependent-type-is-equiv :
is-equiv f → {l : Level} (A : Y → UU l) → is-local-dependent-type f A
is-local-dependent-type-is-equiv is-equiv-f =
is-equiv-precomp-Π-is-equiv is-equiv-f
is-local-is-equiv :
is-equiv f → {l : Level} (A : UU l) → is-local f A
is-local-is-equiv is-equiv-f = is-equiv-precomp-is-equiv f is-equiv-f
### Contractible types are local at any map
module _
{l1 l2 : Level} {X : UU l1} {Y : UU l2} (f : X → Y)
where
is-local-dependent-type-is-contr :
{l : Level} (A : Y → UU l) →
((y : Y) → is-contr (A y)) → is-local-dependent-type f A
is-local-dependent-type-is-contr A is-contr-A =
is-equiv-is-contr
( precomp-Π f A)
( is-contr-Π is-contr-A)
( is-contr-Π (is-contr-A ∘ f))
is-local-is-contr :
{l : Level} (A : UU l) → is-contr A → is-local f A
is-local-is-contr A is-contr-A =
is-local-dependent-type-is-contr (λ _ → A) (λ _ → is-contr-A)
### A type that is local at the unique map empty → unit is contractible
is-contr-is-local :
{l : Level} (A : UU l) → is-local (λ (_ : empty) → star) A → is-contr A
is-contr-is-local A is-local-A =
is-contr-is-equiv
( empty → A)
( λ a _ → a)
( is-equiv-comp
( λ a' _ → a' star)
( λ a _ →
map-inv-is-equiv (is-equiv-map-left-unit-law-Π (λ _ → A)) a star)
( is-equiv-map-inv-is-equiv (is-equiv-map-left-unit-law-Π (λ _ → A)))
( is-local-A))
( universal-property-empty' A) | 3,181 | 9,836 | {"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} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.634973 |
https://zbmath.org/?q=an:07045208 | 1,631,942,990,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056297.61/warc/CC-MAIN-20210918032926-20210918062926-00107.warc.gz | 1,115,309,410 | 13,332 | # zbMATH — the first resource for mathematics
Evaluating linear and nonlinear solvers for density driven flow. (English) Zbl 1423.76264
Summary: This study investigates properties of different solvers for density driven flow problems. The focus is on both non-linear and linear solvers. For the non-linear part, we compare fully coupled method using a Newton linearization and iteratively coupled versions of Jacobi and Gauss-Seidel type. Fully coupled methods require effective preconditioners for the Jacobian. To that end we present a transformation eliminating some couplings and present a strategy for employing algebraic multigrid to the transformed system as well. The work covers theoretical aspects, and provides numerical experiments. Although the primary focus is on density driven flow, we believe that the analysis may well be extended beyond to similar equations with coupled phenomena, such as geomechanics.
##### MSC:
76M10 Finite element methods applied to problems in fluid mechanics 65M60 Finite element, Rayleigh-Ritz and Galerkin methods for initial value and initial-boundary value problems involving PDEs 74F10 Fluid-solid interactions (including aero- and hydro-elasticity, porosity, etc.) 76S05 Flows in porous media; filtration; seepage
##### Software:
libDiscretization; d3f; libGrid; UG4; libAlgebra; IPARS; pcl
Full Text:
##### References:
[1] Putti, M.; Paniconi, C., Picard and Newton linearization for the coupled model for saltwater intrusion in aquifers, Adv. Water Resour., 18, 3, 159-170 (1995) [2] Diersch, H.-J. G.; Kolditz, O., Coupled groundwater flow and transport: 2. Thermohaline and 3D convection systems, Adv. Water Resour., 21, 5, 401-425 (1998) [3] Diersch, H.-J. G.; Kolditz, O., Variable-density flow and transport in porous media: approaches and challenges, (FEFLOW© White Paper Volume II (2009), DHI-WASY GmbH: DHI-WASY GmbH Berlin) [4] Lacroix, S.; Vassilevski, Y. V.; Wheeler, M. F., Decoupling preconditioners in the implicit parallel accurate reservoir simulator (IPARS), Numer. Linear Algebra Appl., 8, 8, 537-549 (2001) · Zbl 1071.76583 [5] Lu, B.; Wheeler, M., Iterative coupling reservoir simulation on high performance computers, Pet. 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D.; Guo, W., MODFLOW/MT3DMS-based simulation of variable-density ground water flow and transport, Ground Water, 44, 3, 339-351 (2006) [12] Johannsen, K.; Kinzelbach, W.; Oswald, S.; Wittum, G., The saltpool benchmark problem—numerical simulation of saltwater upconing in a porous medium, Adv. Water Resour., 25, 309-1708 (2002) [13] Lang, S.; Wittum, G., Large scale density driven flow simulations using parallel unstructured grid adaptation and local multigrid methods, Concurr. Comput., 17, 11, 1415-1440 (2005) [14] Grillo, A.; Lampe, M.; Wittum, G., Three-dimensional simulation of the thermohaline-driven buoyancy of a brine parcel, Comput. Vis. Sci., 13, 287-297 (2010) · Zbl 1216.76066 [15] Fein, E.; Schneider, A., $$d^3 f$$—Ein Programmpaket zur Modellierung von Dichteströmungen, Tech. Rep. GRS 139 (1999), Gesellschaft für Anlagenbau und Reaktorsicherheit (GRS) mbH, October [16] Johannsen, K., Numerische Aspekte dichtegetriebener Strömung in porösen Medien (2004), (Habilitationsschrift) [17] Hackbusch, W., Multi-Grid Methods and Applications (1985), Springer: Springer Berlin · Zbl 0585.65030 [18] Xu, J., Iterative methods by space decomposition and subspace correction, SIAM Rev., 34, 4, 581-613 (1992) · Zbl 0788.65037 [19] Ruge, J. W.; Stüben, K., Algebraic multigrid (AMG), (Multigrid Methods. Multigrid Methods, Frontiers in Applied Mathematics, vol. 3 (1987), SIAM: SIAM Philadelphia, PA), 73-130, (Chapter) [20] Stüben, K., An Introduction to Algebraic Multigrid, 413-532 (2001), Academic Press, Ch. Appendix A [21] Nägel, A., Schnelle Löser für große Gleichungssysteme mit Anwendungen in der Biophysik und den Lebenswissenschaften (2010), Universität Heidelberg, (Ph.D. thesis) [22] Bear, J.; Bachmat, Y., (Introduction to Modeling of Transport Phenomena in Porous Media. Introduction to Modeling of Transport Phenomena in Porous Media, Theory and Applications of Transport in Porous Media (1991), Kluwer Academic: Kluwer Academic Dordrecht) · Zbl 0780.76002 [23] Leijnse, A., Three-dimensional modeling of coupled flow and transport in porous media (1992), University of Notre Dame: University of Notre Dame Indiana, (Ph.D. thesis) [24] Holzbecher, E., Modeling Density-Driven Flow in Porous Media (1998), Springer: Springer Berlin, Heidelberg [25] Oberbeck, A., Über die Wärmeleitung der Flüssigkeiten bei Berücksichtigung der Strömungen infolge von Temperaturdifferenzen, Ann. Phys. Chem., 271-292 (1879) · JFM 11.0787.01 [26] Boussinesq, J., Theorie Analytique de la Chaleur, Vol. 2 (1903), Gauthier-Villlars: Gauthier-Villlars Paris [27] Johannsen, K., On the validity of the Boussinesq approximation for the Elder problem, Comput. Geosci., 7, 3, 169-182 (2003) · Zbl 1134.76471 [29] Rheinboldt, W. C., Methods for Solving Systems of Nonlinear Equations (1998), SIAM · Zbl 0906.65051 [30] Ackerer, P.; Younes, A.; Mancip, M., A new coupling algorithm for density-driven flow in porous media, Geophys. Res. Lett., 31, L12506 (2004), 1-4 [31] Stüben, K., A review of algebraic multigrid, J. Comput. Appl. Math., 128, 1-2, 281-309 (2001) · Zbl 0979.65111 [32] Vanek, P.; Mandel, J.; Brezina, M., Algebraic multigrid by smoothed aggregation for second and fourth order elliptic problems, Computing, 56, 179-196 (1996) · Zbl 0851.65087 [33] Mandel, J.; Brezina, M.; Vanek, P., Energy optimization of algebraic multigrid bases, Computing, 62, 205-228 (1999) · Zbl 0942.65034 [34] Vanek, P.; Brezina, M.; Mandel, J., Convergence of algebraic multigrid based on smoothed aggregation, Numer. Math., 88, 3, 559-579 (2001) · Zbl 0992.65139 [35] Brezina, M.; Cleary, A. J.; Falgout, R. D.; Henson, V. E.; Jones, J. E.; Manteuffel, T. A.; McCormick, S. F.; Ruge, J. W., Algebraic multigrid based on element interpolation (AMGe), SIAM J. Sci. Comput., 22, 5, 1570-1592 (2001) · Zbl 0991.65133 [36] Wagner, C., On the algebraic construction of multilevel transfer operators, Computing, 65, 73-95 (2000) · Zbl 0968.65104 [37] Naegel, A.; Falgout, R.; Wittum, G., Filtering algebraic multigrid and adaptive strategies, Comput. Vis. Sci., 11, 3, 159-167 (2008) [38] Diersch, H.-J. G., Using and testing the algebraic multigrid solver samg in FEFLOW, (FEFLOW© White Paper Volume III (2009), DHI-WASY GmbH: DHI-WASY GmbH Berlin) [39] Voss, C.; Souza, W., Variable density flow and solute transport simulation of regional aquifers containing a narrow freshwater-saltwater transition zone, Water Resour. Res., 26, 2097-2106 (1987) [40] Vogel, A.; Reiter, S.; Rupp, M.; Nägel, A.; Wittum, G., UG 4: A novel flexible software system for simulating PDE based models on high performance computers, Comput. Vis. Sci., 1-15 (2014) [41] Vogel, A.; Nägel, A.; Reiter, S., Numerical advances, (Schneider, A., Enhancement of the codes $$d^3 f$$ and $$r^3 t$$ (GRS-292) (2012), Gesellschaft für Anlagen- und Reaktorsicherheit (GRS) mbH), 156-196 [42] Behie, A.; Vinsome, P., Block iterative methods for fully implicit reservoir simulation, Soc. Pet. Eng. J., 22, 658-668 (1982) [44] Lacroix, S.; Vassilevski, Y.; Wheeler, J.; Wheeler, M., Iterative solution methods for modeling multiphase flow in porous media fully implicitly, SIAM J. Sci. Comput., 25, 3, 905-926 (2003) · Zbl 1163.65310 [45] Scheichl, R.; Masson, R.; Wendebourg, J., Decoupling and block preconditioning for sedimentary basin simulations, Comput. Geosci., 7, 295-318 (2003) · Zbl 1076.76070 [46] Klíe, H.; Wheeler, M. F., Advanced solver methods for subsurface environmental problems, Tech. Rep. (2005), The University of Texas at Austin: The University of Texas at Austin Austin, TX, 78712 [49] Heil, M., An efficient solver for the fully coupled solution of large-displacement fluid-structure interaction problems, Comput. Methods Appl. Mech. Engrg., 193, 1-2, 1-23 (2004) · Zbl 1137.74439 [50] Matthies, H.; Niekamp, R.; Steindorf, J., Algorithms for strong coupling procedures, Comput. Methods Appl. Mech. Engrg., 195, 17-18, 2028-2049 (2006) · Zbl 1142.74050
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2,857 | 9,276 | {"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} | 2.546875 | 3 | CC-MAIN-2021-39 | latest | en | 0.787917 |
https://community.intel.com/t5/Intel-Fortran-Compiler/Array-Sizes/td-p/1097231 | 1,606,805,200,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141652107.52/warc/CC-MAIN-20201201043603-20201201073603-00469.warc.gz | 227,684,603 | 36,417 | Community
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New Contributor II
11 Views
## Array Sizes
I could ask this in MKL - but this is more a straight FORTRAN problem.
Before you call PARDISO - you pack the n by n Stiffness matrix into a set of vectors to take up less space. mecej4 showed me in MAGNI how to pack efficiently - the following code does the packing for taking a stiffness matrix from and turns it into the packed vectors, Pardiso needs the length of the vectors essentially NNZ.
I measure the number of degrees of freedom of the problem which gives you a size, but trying to relate NNZ to the degrees of freedom is interesting -
I was thinking just counting the number of non-zero elements in the stiffness matrix and than set NNZ - but I have to loop thru a large matrix
Is there a simpler method, I was just using 9 times the number of degrees of freedom and that usually works, but it did not on the current problem
``` subroutine PardisoSolver(GK,nl,nk,ndf,X)
implicit none
include 'mkl_pardiso.fi'
integer nl,nk,ndf
REAL (KIND=dp) GK(nl,nk),X(ndf)
REAL (KIND=dp), ALLOCATABLE :: a(:), b(:), c(:,:), atemp(:)
integer, allocatable :: ja(:), ia(:), iatemp(:), jatemp(:)
INTEGER :: nnodes = 5
integer error, iaN
integer :: nnz , istat, i, j , count
integer :: size = 0
integer :: size1 = 0
integer :: size2 = 0
integer :: flag = 0
REAL (KIND=dp) :: delta = 0.000000000000001
nnz = 90*ndf
!open(16, file="a.mat", STATUS = 'old')
count = ndf
!nnodes = count
iaN = count+1
ALLOCATE (a(nnz), ja(nnz), ia(iaN), jatemp(nnz), iatemp(iaN), b(count), c(count,count), atemp(nnz), STAT=istat)
IF (istat.NE.0) THEN
WRITE (*, *) '*** Could not allocate some arrays in LINSOLVE'
STOP
END IF
b = 1.0D0
a = 0.0d0
c = 0.0d0
iatemp = 0
jatemp = 0
atemp = 0
ia = 0
ja = 0
do 200 i = 1, count
flag = 0
do 300 j = 1, count
c(i,j) = gk(i,j)
if(abs(c(i,j)) .gt. delta) then
size = size + 1
if(size .gt. nnz) then
Write(*,2030)nnz, ndf
2030 Format( " NNZ is set at :: ",I6, " NDF is :: ",I6)
stop ' Insufficient memory allocation in PARDISO to solve the inversion problem'
endif
jatemp(size) = j
iatemp(size1 + 1) = size+1
atemp(size) = c(i,j)
if(flag .eq. 0) then
size1 = size1 + 1
iatemp(size1) = size
flag = 1
endif
endif
300 end do
400 format(5(1x,F10.3))
200 end do
```
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5 Replies
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Valued Contributor III
11 Views
count is not a great name for a variable BTW as it is the name of an instrinsic. use of count could remove a loop but if the slice is not contiguous could involve temp arrays being created....
`no_zero_count = count( C(:,j) > delta ) ! count values in a col `
Highlighted
Black Belt
11 Views
``` integer :: size = 0
integer :: size1 = 0
integer :: size2 = 0
integer :: flag = 0
```
*** Unlike C/C++, those values are 0 only on first call (not re-initialized).
The above is equivalent to the C/C++
``` static int size = 0;
static int size1 = 0;
static int size2 = 0;
static int flag = 0;
```
See: initialization expressions, type declarations
A variable (or variable subobject) can only be initialized once in an executable program.
What you need to do (if the subroutine is called more than once), is to declare the variables without initialization, then in the code section at start of subroutine, initialize the values.
Jim Dempsey
Highlighted
New Contributor II
11 Views
```subroutine PardisoSolver(GK,nl,nk,ndf,X)
implicit none
include 'mkl_pardiso.fi'
integer nl,nk,ndf
REAL (KIND=dp) GK(nl,nk),X(ndf),GKTest(nl,nk)
REAL (KIND=dp), ALLOCATABLE :: a(:), b(:), c(:,:), atemp(:)
integer, allocatable :: ja(:), ia(:), iatemp(:), jatemp(:)
INTEGER :: nnodes = 5
integer error, iaN
integer :: nnz , istat, i, j !count
integer :: size = 0
integer :: size1 = 0
integer :: size2 = 0
integer :: flag = 0
REAL (KIND=dp) :: delta = 0.000000000000001
integer no_zero_count(nl), numA
GKTest = 0.0d0
no_zero_count = count( gk .NE. GKTEST, DIM=1 ) ! count values in a col
numA = sum(no_zero_count)
write(*,2040)numA
2040 Format('--------------------------------------------------------------------------------------------------------------',///&
' Count of the non-zero elements in the stiffness matrix :: ',i6,//)
nnz = numA+10 !90*ndf
iaN = ndf+1
ALLOCATE (a(nnz), ja(nnz), ia(iaN), jatemp(nnz), iatemp(iaN), b(ndf), c(ndf,ndf), atemp(nnz), STAT=istat)
IF (istat.NE.0) THEN
WRITE (*, *) '*** Could not allocate some arrays in LINSOLVE'
STOP
END IF
b = 1.0D0
a = 0.0d0
c = 0.0d0
iatemp = 0
jatemp = 0```
Highlighted
New Contributor II
11 Views
John,
If you want to find the number of non zero values in GK, why not count them. I am not familiar with GK but assuming nl and nk must be >= ndf, you could write:
```!
if ( nl < ndf ) write (*,*) 'nl is invalid',nl,ndf
if ( nk < ndf ) write (*,*) 'nk is invalid',nk,ndf
nnz = 0
do j = 1, ndf
do i = 1, ndf
if ( abs(gk(i,j)) .gt. delta) nnz = nnz + 1
end do
end do
```
Note that the inner loop is do i, for sequential scanning of memory.
You also state "I was thinking just counting the number of non-zero elements in the stiffness matrix and than set NNZ - but I have to loop thru a large matrix". If GK is a large matrix, you appear to be wasting a lot of memory, as you store "GK", "c" which is a copy of GK and the sparse form in atemp, jatemp and iatemp. If ndf is truly large, your approach should be to do a scan of the matrix structure to estimate iatemp and then assemble the matrix directly into atemp and jatemp.
Is GK symmetric ?
Highlighted
New Contributor II
11 Views
No GK is weird it also holds the load vectors -- as originally written and I have not fixed it.
The count function worked perfectly - I just need to make sure I have enough spaces in the vectors -- better than guessing
PARDISO as Intel has the sample is pretty limited in terms of sending in different size problems through their sample code, it has just been a matter of working out how to make it so it knew the problem size coming in automatically - now it is perfect.
It takes a while to learn the basics but it is worth it - superfast. | 1,885 | 6,112 | {"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} | 2.984375 | 3 | CC-MAIN-2020-50 | latest | en | 0.791442 |
https://brilliant.org/problems/can-ayush-survive/ | 1,524,736,243,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948125.20/warc/CC-MAIN-20180426090041-20180426110041-00190.warc.gz | 576,215,003 | 16,980 | # Can Ayush survive?
Ayush is trying to go to the coordinates $$(6,6)$$ from $$(0,0)$$. But in the way, on the coordinates $$(3,1),(4,1),(3,2),(4,2)$$ werewolf, vampire, ghost, zombie are there respectively. Ayush has to avoid all of them to survive. How many ways Ayush can survive and reach $$(6,6)$$ successfully?
Note: Ayush can move only right or up.
× | 111 | 360 | {"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} | 2.625 | 3 | CC-MAIN-2018-17 | latest | en | 0.910794 |
https://lastdropmugs.com/what-is-nernst-equation/ | 1,656,437,648,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103573995.30/warc/CC-MAIN-20220628173131-20220628203131-00631.warc.gz | 408,638,274 | 32,277 | # what is nernst equation
What Is Nernst Equation? The Nernst equation defines the relationship between cell potential to standard potential and to the activities of the electrically active (electroactive) species. It relates the effective concentrations (activities) of the components of a cell reaction to the standard cell potential.
What is Nernst equation explain it? The Nernst equation defines the relationship between cell potential to standard potential and to the activities of the electrically active (electroactive) species. It relates the effective concentrations (activities) of the components of a cell reaction to the standard cell potential.
What is Nernst equation and its application? The Nernst equation provides a relation between the cell potential of an electrochemical cell, the standard cell potential, temperature, and the reaction quotient. Even under non-standard conditions, the cell potentials of electrochemical cells can be determined with the help of the Nernst equation.
What is the Nernst equation simple? The Nernst equation relates the effective concentrations ( activities ) of the components of a cell reaction to the standard cell potential. For a simple reduction of the form Mn+ + ne– → M, it tells us that a half-cell potential will change by 59/n mV per 10-fold change in the activity of the ion.
## Why is Nernst equation used?
The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
## What is Nernst equation and its derivation?
Derivation of the Nernst Equation The Nernst equation is derived from the Gibbs free energy. We can rewrite this equation using the definitions of ΔG = -nFE and ΔGo = -nFEo. To simplify, we divide each side by -nF and arrive at the Nernst equation as it is commonly written.
## What is Nernst equation class 12?
Nernst equation is an equation which relates the capacity of an atom or ion to take up one or more electrons and measure reduction potential at any conditions to that measured at standard conditions i.e. standard reduction potentials at 298K, one molar and one atmospheric pressure.
## What is the value of RT nF?
In the Nernst equation the value of 2.303 RT/nF is equal to 0.0592 whena) R in calories, n= 1 and T – Brainly.in.
## What is Nernst equation Shaalaa?
Solution. Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction. Let us consider an electrochemical cell for which the overall redox reaction is, x y l m.
## What is Nernst equation for EMF for a cell?
Nernst Equation for EMF of a cell The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
## How do you write the Nernst equation for half cell reaction?
Note that this is precisely the equation that we would obtain by writing out the Nernst equation corresponding to the chemical equation Ag++e−→Ag0. The standard electrode potentials are EPt∣Q,H2Q,H+=+0.699 v and EPt∣Fe3+,Fe2+=+0.783 v.
## What is cell constant?
The cell constant is defined by the distance between the electrodes (l) and their cross-sectional area (A). It is measured by calculating the resistance of a cell containing a known conductivity solution.
## What is electrochemical series?
Electrochemical or activity series When the electrodes (metals and non-metals) in contact with their ions are arranged on the basis of the values of their standard reduction potentials or standard oxidation potentials, the resulting series is called the electrochemical or electromotive or activity series of the …
## What is meant by molar conductivity?
The molar conductivity of an electrolyte solution is defined as its conductivity divided by its molar concentration. where: κ is the measured conductivity (formerly known as specific conductance), c is the molar concentration of the electrolyte.
## What is cell constant and write its SI unit?
For a given cell, the ratio of separation (l) between the two electrodes divided by the area of cross section (a) of the electrode is called the cell constant. The SI unit of cell constant is m−1.
## What is the EMF series?
An electromotive force series (EMF series) is a metal’s ranking in respect to inherent reactivity. The metals located at the top of the series are considered the most noble, with the highest level of positive electrochemical potential.
## What is EMF series chemistry?
EMF series is defined as the arrangement of the electrode with the electrode half-reaction in order of decreasing standard potentials.
## How the EMF series is created?
The electrochemical series is built up by arranging various redox equilibria in order of their standard electrode potentials (redox potentials). The most negative E° values are placed at the top of the electrochemical series, and the most positive at the bottom.
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Scroll to Top | 1,103 | 5,227 | {"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} | 3.421875 | 3 | CC-MAIN-2022-27 | latest | en | 0.916014 |
https://www.jiskha.com/questions/357351/Find-the-static-pressure-that-exists-in-your-house-water-pipes-if-the-water-in-a | 1,568,675,634,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572964.47/warc/CC-MAIN-20190916220318-20190917002318-00264.warc.gz | 907,219,167 | 4,834 | # physics
Find the static pressure that exists in your house water pipes if the water in a storage tank is 25 meters higher in elevation that your pipes.
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asked by Jessica
1. The "gauge" pressure (above atmospheric) is (rho g H)
rho if the density of water (1000 kg/m^2)
H = 25 m.
You know what g is.
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posted by drwls
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https://www.aimsedu.org/2014/04/22/arrow-arrangements/?full-site=true&replytocom=44293 | 1,585,448,826,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493684.2/warc/CC-MAIN-20200329015008-20200329045008-00548.warc.gz | 807,918,144 | 14,143 | On the campus of
# Arrow Arrangements
This particular puzzle comes from The Moscow Puzzles. The puzzle is found in the section entitled “Geometry with Matches,” which offers a selection of matchstick puzzles as “geometrical amusements that sharpen your mind.” Arrow Arrangements is one of the more difficult puzzles in this section, and requires students to understand and apply some basic geometric terms such as congruent, triangle, and quadrilateral. Thus it has the benefit of not only sharpening the mind, but also giving some practice thinking about geometric shapes and concepts.
1. Students will each need a copy of the student sheet and 16 flat toothpicks to complete this puzzle.
2. Be sure students understand the concept of congruent triangles and quadrilaterals before they begin.
Solution Hint
Work backwards, starting with the correct number of triangles or quadrilaterals, and try to make the arrow from that arrangement.
How can you move a given number of toothpicks to create eight triangles or seven quadrilaterals from the arrow shape?
Solutions
Click the arrow below to view the solutions.
The dashed lines indicate the toothpicks that were moved in each solution.
Challenge 1: Move eight toothpicks to make eight congruent triangles.
Challenge 2: Move seven toothpicks to make five congruent quadrilaterals.
### 2 Responses to Arrow Arrangements
• thane says:
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http://davegiles.blogspot.com/2012/10/dancing-with-econometricians.html | 1,531,800,168,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00223.warc.gz | 98,259,483 | 36,810 | Sunday, October 7, 2012
Dancing With the Econometricians
Let's talk about the two-step. Not the tango or the polka. The two-step!
More specifically let's talk about a particular two-step estimator that we use all of the time in econometrics. I want to clear up some misconceptions that I seem to encounter all too frequently when I read empirical "applied" papers.
Why is it that some people insist on using the term "Two Stage Least Squares" inappropriately?
Let me explain what I mean.
Suppose that we have a system of M simultaneous equations, where the (ith) structural equation that we want to estimate can be written as:
yi = Yiγi + Xiβi + εi .
We'll assume that this equation is identified. Yes, let's be careful about this. If the "rank condition" for identification of the equation isn't satisfied, then the 2SLS estimator won't be weakly consistent. Even worse, if the "order condition" isn't satisfied then this estimator isn't even defined (because a matrix that has to be inverted will be singular).
The (mi) columns of the Yi matrix are endogenous regressors. The (ki) columns of the Xi matrix are "predetermined" regressors. That is, they are strictly exogenous regressors (often including an "intercept" variable); or else they are lagged endogenous or exogenous variables (if we have time series data). The elements of the error term, εi, have a zero mean, and are homoskedastic and serially independent.
Remember that this equation is just one of M in the entire simultaneous system. Let's use the symbol, X, to denote the matrix of observations on all of the predetermined variables that appear anywhere in the system.
Suppose that X has K columns.
The 2SLS estimator of the coefficients in equation (1) can be described as follows:
1. Using OLS, regress Yi on all (yes, ALL) of the columns of X, and get the matrix of predictions, Yi* = X(X'X)-1X'Yi.
2. Replace Yi in (1) with Yi*, and then use OLS to estimate this modified equation, yielding consistent estimates of the elements of γi and βi.
Notice the words that I've emphasized in the first stage.
Now, a bit of history - just to provide some light relief in the middle of the math! The 2SLS estimator emerged as a computationally convenient way of obtaining consistent estimates of the coefficients in an equation of the form (1). At the time that 2SLS was proposed by Theil (1953a, 1953b, 1954), Basmann (1957), and Sargan (1958), computational convenience was a big deal. In fact, it remained a big deal for along time - e.g., see here. Anderson (2005) has recently drawn our attention to the fact that the 2SLS estimator was used (at least "obliquely") in the even earlier papers by Anderson and Rubin (1949, 1950) that introduced the LIML estimator and derived that estimator's asymptotic distribution.
So, what is it that I get upset about? Well, it's become very common to see the following two-step estimator described as "2SLS":
1. Using OLS, regress Yi on a set of suitable instruments, Z, and get the matrix of predictions, Yi** = Z(Z'Z)-1Z'Yi.
2. Replace Yi in (1) with Yi**, and then use OLS to estimate this modified equation, yielding estimates of the elements of γi and βi.
This is often done in the context of a single structural equation, without a complete simultaneous equations model in sight. There's nothing wrong with this (provided that Xi is included in Z - see Angelo Melino's comment below) - but please don't refer to this as 2SLS! Unless Z = X, it's just an instrumental variables estimator, constructed in two steps.
In fact, although this second approach yields appropriate estimates of the coefficients, it won't give you the correct standard errors, or the correct values of anything else that is constructed from the second-step residuals. That's because the latter residual vector is of the form ei** = yi - (Yi**gi + Xibi), instead of the correct form, ei = yi - (Yigi - Xibi). Here, gi and bi are the estimated coefficient vectors. Of course, this is easily fixed, just as it is in the case of the genuine 2SLS estimator, but it's simpler to just use the one-step instrumental variables estimator rather than the (otherwise) equivalent two-step estimator.
Confession: One situation where the two-step approach is helpful is if you are worried that Z may include some "weak" instruments. In that case, the F-statistic for the significance of the estimated coefficients in of the first step provides one basis for testing against such weakness. But that's another story!
I have to admit that as much as I enjoy using the EViews econometrics package, I find it intensely irritating that when it comes to instrumental variables estimation, the command is titled "Two Stage Least Squares"!
In another post, coming up this week, I'll explore the various connections between the (real) 2SLS estimator and instrumental variables estimation. There are more of these connections than you might think.
Meantime, call me pedantic if you will, but if you want to dance with the econometricians then you have stay in time with the music, and you have to know that a samba is not the same as a rumba!
References
Anderson, T. W., 2005. Origins of the limited information maximum likelihood and two-stage least squares estimators. Journal of Econometrics, 127, 1-16.
Anderson, T.W. and H. Rubin, 1949. Estimation of the parameters of a single equation in a complete system of stochastic equations. Annals of Mathematical Statistics, 20, 46–63.
Anderson, T.W. and H. Rubin, 1950. The asymptotic properties of estimates of the parameters of a single equation in a complete system of stochastic equations. Annals of Mathematical Statistics, 21, 570–582.
Basmann, R.L., 1957. A generalized classical method of linear estimation of coefficients in a structural equation. Econometrica 25, 77–83.
Sargan, J.D., 1958. Estimation of economic relationships using instrumental variables. Econometrica, 67, 557–586.
Theil, H., 1953a. Repeated least-squares applied to complete equation systems. Centraal Planbureau Memorandum.
Theil, H., 1953b. Estimation and simultaneous correlation in complete equation systems. Centraal Planbureau Memorandum. (Reprinted In: Raj, B., Koerts, J. (Eds.), 1992, Henri Theil’s Contributions to Economics and Econometrics, Vol. 1, Kluwer, Dordrecht.)
Theil, H., 1954. Estimation of parameters in econometric models. Bulletin of the International Statistical Institute, 34, 122–129.
1. Pedantic, but informative!
2. I enjoyed this post on TSLS since I plan to use it as one
of my methods for my Master Thesis. I look forward to future post on TSLS.
Thanks!
3. Dave,
Can I take a contrary view? Or maybe you can correct me if I've misunderstood your post.
I've checked some of my more recent favourite rigorous econometrics texts, and the presentation and usage of the term "2SLS" seems to be the same as the one you are complaining about: estimate a single equation using IV, but do it in two stages.
For example, Davidson and MacKinnon, Econometric Theory and Methods (2004), say on pp. 323-4, "The IV estimator (8.29) is commonly known as the two-stage least-squares, or 2SLS, estimator, because, before the days of good econometrics software packages, it was often calculated in two stages using OLS regressions. ... Two-stage least squares was invented by Theil (1953) and Basmann (1957) at a time when computers were very primitive. Consequently, despite the classic papers of Durbin (1954) and Sargan (1958) on instrumental variables estimation, the term 'two-stage least squares' came to be very widely used in econometrics, even when the estimator is not actually computed in two stages. We prefer to think of two-stage least squares as simply a particular way to compute the generalized IV estimator...."
This is how I've always thought about 2SLS. It also seems to be the way the term is used in the other textbooks I've checked (Greene, Hayashi, Stock & Watson are the ones at hand). But you seem to be saying that 2SLS is a term that should be applied only to estimation of a system, and not to single-equation estimation. Is that right?
Personally ... I think the term "two-stage least squares" is terrible and should be avoided. It's very confusing for students, because they can easily be misled into thinking that the "two stages" are somehow integral to the definition of the estimator, when they're not - they're just a way of calculating the thing. These days we never calculate it in two stages, but it's the same estimator.
This is very different from "two-step estimators" which have to be done in two steps by their very nature. I have in mind Feasible GLS (1st step - estimate the variance components, 2nd step - get the coeffs); 2-step Efficient GMM (1st step - estimate the var-cov matrix of orthogonality conditions, 2nd step - get the coeffs); SUR; 3SLS; etc.
Personally, if I could I'd ban the term "two-stage least squares" entirely, and use "two-step" consistently for estimators that are necessarily done in two steps. But I suspect I am in the minority on this one.
--Mark
4. Mark - thanks for this. No, you haven't misunderstood my post. The term 2SLS has a specific historical meaning. We now know that it's just a particuolar IV estimator - that's how we usually prove its consistency.
I agree that it would be good if the "two-stage"/"two-step" terminology was now dropped. That's why I'd much prefer to see an "IV" command in EViews instead of ther current "2SLS" command.
5. I've rummaged around some more, and the oldest econometrics books I have at hand (Johnston and Kmenta, though not the first editions) do introduce "2SLS" in the context of system estimation. But I don't think "2SLS" is used in that way any more, at least in common parlance (the "econometric vulgate", perhaps).
I see you have another post upcoming on IV et al., which I quite look forward to reading. I agree it would be great to drop the "two stage" terminology for this estimator, but I have a feeling that might complicate your drafting task unduly!
Cheers,
Mark
6. Thanks Mark!
7. Prof. Giles,
So you mean to say that when I use EViews and its TSLS
command in my research, I should now label my methodology
as "IV estimation method"? Thanks.
8. Yes - absolutely. And you should state what instruments you've used.
9. I agree with the message of your post but not with one of the details.
You write
"2. Replace Yi in (1) with Yi**, and then use OLS to estimate this modified equation, yielding estimates of the elements of γi and βi.
This is often done in the context of a single structural equation, without a complete simultaneous equations model in sight. There's nothing wrong with this - but please don't refer to this as 2SLS!"
In fact, there is a problem. In general, replacing Yi with Yi** leads to an inconsistent estimator because we can't guarantee that Xi will be orthogonal to Yi-Yi** unless Xi is included in the set of instruments Z.
1. Angelo - absolutely correct! In practice most people would do this, but it definitely should have been emphasised. I've amended the post accordingly.
Best, DG
10. Dear Prof.Dave
Can we solved recursive model with 2sls method?
Exmple:
Y1=f(X1,X2,X3)
Y2=f(Y1,X1,X2,X3)
Y3=f(Y1,Y2,X1,X2,X3)
1. Hi - a recursive system is one that has the type of structure that you mention, AND the error covariance matrix is DIAGONAL. This implies that the errors in the different equations are independent of each other. In this case, OLS estimation can be applied.
However, it is most unlikely that the errors will be contemporaneously uncorrelated. In this case, you can just use 2SLS estimation, as you would for any other SEM.
11. post amazing | 2,827 | 11,668 | {"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} | 3.109375 | 3 | CC-MAIN-2018-30 | latest | en | 0.89059 |
https://overlordoftheuberferal.com/2014/10/29/summus/ | 1,685,353,036,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00386.warc.gz | 482,188,956 | 27,548 | # Summus
I’m interested in digit-sums and in palindromic numbers. Looking at one, I found the other. It started like this: 9^2 = 81 and 9 = 8 + 1, so digitsum(9^1) = digitsum(9^2). I wondered how long such a sequence of powers could be (excluding powers of 10). I quickly found that the digit-sum of 468 is equal to the digit-sum of its square and cube:
digsum(468) = digsum(219024) = digsum(102503232)
But I couldn’t find any longer sequence, although plenty of other numbers are similar to 468:
digsum(585) = digsum(342225) = digsum(200201625)
digsum(4680) = digsum(21902400) = digsum(102503232000)
digsum(5850) = digsum(34222500) = digsum(200201625000)
digsum(5851) = digsum(34234201) = digsum(200304310051)
digsum(5868) = digsum(34433424) = digsum(202055332032)
digsum(28845) = digsum(832034025) = digsum(24000021451125) […]
digsum(589680) = digsum(347722502400) = digsum(205045005215232000)
What about other bases? First came this sequence:
digsum(2) = digsum(11) (base = 3) (highest power = 2)
Then these:
digsum(4) = digsum(22) = digsum(121) (b=7) (highest power = 3)
digsum(8) = digsum(44) = digsum(242) = digsum(1331) (b=15) (hp=4)
digsum([16]) = digsum(88) = digsum(484) = digsum(2662) = digsum(14641) (b=31) (hp=5)
The pattern continues (a number between square brackets represents a single digit in the base):
digsum([32]) = digsum([16][16]) = digsum(8[16]8) = digsum(4[12][12]4) = digsum(28[12]82) = digsum(15[10][10]51) (b=63) (hp=6)
digsum([64]) = digsum([32][32]) = digsum([16][32][16]) = digsum(8[24][24]8) = digsum(4[16][24][16]4) = digsum(2[10][20][20][10]2) = digsum(16[15][20][15]61) (b=127) (hp=7)
digsum([128]) = digsum([64][64]) = digsum([32][64][32]) = digsum([16][48][48][16]) = digsum(8[32][48][32]8) = digsum(4[20][40][40][20]4) = digsum(2[12][30][40][30][12]2) = digsum(17[21][35][35][21]71) (b=255) (hp=8)
digsum([256]) = digsum([128][128]) = digsum([64][128][64]) = digsum([32][96][96][32]) = digsum([16][64][96][64][16]) = digsum(8[40][80][80][40]8) = digsum(4[24][60][80][60][24]4) = digsum(2[14][42][70][70][42][14]2) = digsum(18[28][56][70][56][28]81) (b=511) (hp=9)
After this, I looked at sequences in which n(i) = n(i-1) + digitsum(n(i-1)). How long could digitsum(n(i)) be greater than or equal to digitsum(n(i-1))? In base 10, I found these sequences:
1 (digitsum=1) → 2 → 4 → 8 → 16 (sum=7) (count=4) (base=10)
9 → 18 (sum=9) → 27 (s=9) → 36 (s=9) → 45 (s=9) → 54 (s=9) → 63 (s=9) → 72 (s=9) → 81 (s=9) → 90 (s=9) → 99 (s=18) → 117 (s=9) (c=11) (b=10)
801 (s=9) → 810 (s=9) → 819 (s=18) → 837 (s=18) → 855 (s=18) → 873 (s=18) → 891 (s=18) → 909 (s=18) → 927 (s=18) → 945 (s=18) → 963 (s=18) → 981 (s=18) → 999 (s=27) → 1026 (s=9) (c=13)
Base 2 does better:
1 → 10 (s=1) → 11 (s=2) → 101 (s=2) → 111 (s=3) → 1010 (s=2) (c=5) (b=2)
16 = 10000 (s=1) → 10001 (s=2) → 10011 (s=3) → 10110 (s=3) → 11001 (s=3) → 11100 (s=3) → 11111 (s=5) → 100100 (s=2) (c=7) (b=2)
962 = 1111000010 (s=5) → 1111000111 (s=7) → 1111001110 (s=7) → 1111010101 (s=7) → 1111011100 (s=7) → 1111100011 (s=7) → 1111101010 (s=7) → 1111110001 (s=7) → 1111111000 (s=7) → 1111111111 (s=10) → 10000001001 (s=3) (c=10) (b=2)
524047 = 1111111111100001111 (s=15) → 1111111111100011110 (s=15) → 1111111111100101101 (s=15) → 1111111111100111100 (s=15) → 1111111111101001011 (s=15) → 1111111111101011010 (s=15) → 1111111111101101001(s=15) → 1111111111101111000 (s=15) → 1111111111110000111 (s=15) → 1111111111110010110 (s=15) → 1111111111110100101 (s=15) → 1111111111110110100 (s=15) → 1111111111111000011 (s=15) → 1111111111111010010 (s=15) → 1111111111111100001 (s=15) → 1111111111111110000 (s=15) → 1111111111111111111 (s=19) → 10000000000000010010 (s=3) (c=17) (b=2)
The best sequence I found in base 3 is shorter than in base 10, but there are more sequences:
1 → 2 → 11 (s=2) → 20 (s=2) → 22 (s=4) → 110 (s=2) (c=5) (b=3)
31 = 1011 (s=3) → 1021 (s=4) → 1102 (s=4) → 1120 (s=4) → 1201 (s=4) → 1212 (s=6) → 2002 (s=4) (c=6) (b=3)
54 = 2000 (s=2) → 2002 (s=4) → 2020 (s=4) → 2101 (s=4) → 2112 (s=6) → 2202 (s=6) → 2222 (s=8) → 10021(s=4) (c=7) (b=3)
432 = 121000 (s=4) → 121011 (s=6) → 121101 (s=6) → 121121 (s=8) → 121220 (s=8) → 122012 (s=8) → 122111 (s=8) → 122210 (s=8) → 200002 (s=4) (c=8) (b=3)
648 = 220000 (s=4) → 220011 (s=6) → 220101 (s=6) → 220121 (s=8) → 220220 (s=8) → 221012 (s=8) → 221111 (s=8) → 221210 (s=8) → 222002 (s=8) → 222101 (s=8) → 222200 (s=8) → 222222 (s=12) → 1000102 (s=4) (c=12) (b=3)
And what about sequences in which digitsum(n(i)) is always greater than digitsum(n(i-1))? Base 10 is disappointing:
1 → 2 → 4 → 8 → 16 (sum=7) (count=4) (base=10)
50 (s=5) → 55 (s=10) → 65 (s=11) → 76 (s=13) → 89 (s=17) → 106 (s=7) (c=5) (b=10)
Some other bases do better:
2 = 10 (s=1) → 11 (s=2) → 101 (s=2) (c=2) (b=2)
4 = 100 (s=1) → 101 (s=2) → 111 (s=3) → 1010 (s=2) (c=3) (b=2)
240 = 11110000 (s=4) → 11110100 (s=5) → 11111001 (s=6) → 11111111 (s=8) → 100000111 (s=4) (c=4) (b=2)
1 → 2 → 11 (s=2) (c=2) (b=3)
19 = 201 (s=3) → 211 (s=4) → 222 (s=6) → 1012 (s=4) (c=3) (b=3)
58999 = 2222221011 (s=15) → 2222221201 (s=16) → 2222222022 (s=18) → 2222222222 (s=20) → 10000000201 (s=4) (c=4) (b=3)
1 → 2 → 10 (s=1) (c=2) (b=4)
4 = 10 (s=1) → 11 (s=2) → 13 (s=4) → 23 (s=5) → 100 (s=1) (c=4) (b=4)
977 = 33101 (s=8) → 33121 (s=10) → 33203 (s=11) → 33232 (s=13) → 33323 (s=14) → 100021 (s=4) (c=5) (b=4)
1 → 2 → 4 → 13 (s=4) (c=3) (b=5)
105 = 410 (s=5) → 420 (s=6) → 431 (s=8) → 444 (s=12) → 1021 (s=4) (c=4) (b=5)
1 → 2 → 4 → 12 (s=3) (c=3) (b=6)
13 = 21 (s=3) → 24 (s=6) → 34 (s=7) → 45 (s=9) → 102 (s=3) (c=4) (b=6)
396 = 1500 (s=6) → 1510 (s=7) → 1521 (s=9) → 1534 (s=13) → 1555 (s=16) → 2023 (s=7) (c=5) (b=6)
1 → 2 → 4 → 11 (s=2) (c=3) (b=7)
121 = 232 (s=7) → 242 (s=8) → 253 (s=10) → 266 (s=14) → 316 (s=10) (c=4) (b=7)
205 = 412 (s=7) → 422 (s=8) → 433 (s=10) → 446 (s=14) → 466 (s=16) → 521 (s=8) (c=5) (b=7)
1 → 2 → 4 → 10 (s=1) (c=3) (b=8)
8 = 10 (s=1) → 11 (s=2) → 13 (s=4) → 17 (s=8) → 27 (s=9) → 40 (s=4) (c=5) (b=8)
323 = 503 (s=8) → 513 (s=9) → 524 (s=11) → 537 (s=15) → 556 (s=16) → 576 (s=18) → 620 (s=8) (c=6) (b=8)
1 → 2 → 4 → 8 → 17 (s=8) (c=4) (b=9)
6481 = 8801 (s=17) → 8820 (s=18) → 8840 (s=20) → 8862 (s=24) → 8888 (s=32) → 10034 (s=8) (c=5) (b=9)
1 → 2 → 4 → 8 → 16 (s=7) (c=4) (b=10)
50 (s=5) → 55 (s=10) → 65 (s=11) → 76 (s=13) → 89 (s=17) → 106 (s=7) (c=5) (b=10)
1 → 2 → 4 → 8 → 15 (s=6) (c=4) (b=11)
1013 = 841 (s=13) → 853 (s=16) → 868 (s=22) → 888 (s=24) → 8[10][10] (s=28) → 925 (s=16) (c=5) (b=11)
1 → 2 → 4 → 8 → 14 (s=5) (c=4) (b=12)
25 = 21 (s=3) → 24 (s=6) → 2[10] (s=12) → 3[10] (s=13) → 4[11] (s=15) → 62 (s=8) (c=5) (b=12)
1191 = 833 (s=14) → 845 (s=17) → 85[10] (s=23) → 879 (s=24) → 899 (s=26) → 8[11][11] (s=30) → 925 (s=16) (c=6) (b=12)
1 → 2 → 4 → 8 → 13 (s=4) (c=4) (b=13)
781 = 481 (s=13) → 491 (s=14) → 4[10]2 (s=16) → 4[11]5 (s=20) → 4[12][12] (s=28) → 521 (s=8) (c=5) (b=13)
19621 = 8[12]14 (s=25) → 8[12]33 (s=26) → 8[12]53 (s=28) → 8[12]75 (s=32) → 8[12]9[11] (s=40) → 8[12][12][12] (s=44) → 9034 (s=16) (c=6) (b=13)
1 → 2 → 4 → 8 → 12 (s=3) (c=4) (b=14)
72 = 52 (s=7) → 59 (s=14) → 69 (s=15) → 7[10] (s=17) → 8[13] (s=21) → [10]6 (s=16) (c=5) (b=14)
1275 = 671 (s=14) → 681 (s=15) → 692 (s=17) → 6[10]5 (s=21) → 6[11][12] (s=29) → 6[13][13] (s=32) → 723 (s=12) (c=6) (b=14)
19026 = 6[13]10 (s=20) → 6[13]26 (s=27) → 6[13]45 (s=28) → 6[13]65 (s=30) → 6[13]87 (s=34) → 6[13][10][13] (s=42) → 6[13][13][13] (s=45) → 7032 (s=12) (c=7) (b=14)
1 → 2 → 4 → 8 → 11 (s=2) (c=4) (b=15)
603 = 2[10]3 (s=15) → 2[11]3 (s=16) → 2[12]4 (s=18) → 2[13]7 (s=22) → 2[14][14] (s=30) → 31[14] (s=18) (c=5) (b=15)
1023 = 483 (s=15) → 493 (s=16) → 4[10]4 (s=18) → 4[11]7 (s=22) → 4[12][14] (s=30) → 4[14][14] (s=32) → 521 (s=8) (c=6) (b=15)
1891 = 861 (s=15) → 871 (s=16) → 882 (s=18) → 895 (s=22) → 8[10][12] (s=30) → 8[12][12] (s=32) → 8[14][14] (s=36) → 925 (s=16) (c=7) (b=15)
1 → 2 → 4 → 8 → 10 (s=1) (c=4) (b=16)
16 = 10 (s=1) → 11 (s=2) → 13 (s=4) → 17 (s=8) → 1[15] (s=16) → 2[15] (s=17) → 40 (s=4) (c=6) (b=16)
1396 = 574 (s=16) → 584 (s=17) → 595 (s=19) → 5[10]8 (s=23) → 5[11][15] (s=31) → 5[13][14] (s=32) → 5[15][14] (s=34) → 620 (s=8) (c=7) (b=16)
2131 = 853 (s=16) → 863 (s=17) → 874 (s=19) → 887 (s=23) → 89[14] (s=31) → 8[11][13] (s=32) → 8[13][13] (s=34) → 8[15][15] (s=38) → 925 (s=16) (c=8) (b=16)
1 → 2 → 4 → 8 → [16] (s=16) → 1[15] (s=16) (c=5) (b=17)
1 → 2 → 4 → 8 → [16] (s=16) → 1[14] (s=15) (c=5) (b=18)
5330 = [16]82 (s=26) → [16]9[10] (s=35) → [16][11]9 (s=36) → [16][13]9 (s=38) → [16][15][11] (s=42) → [16][17][17] (s=50) → [17]2[13] (s=32) (c=6) (b=18)
1 → 2 → 4 → 8 → [16] (s=16) → 1[13] (s=14) (c=5) (b=19)
116339 = [16][18]52 (s=41) → [16][18]75 (s=46) → [16][18]9[13] (s=56) → [16][18][12][12] (s=58) → [16][18][15][13] (s=62) → [16][18][18][18] (s=70) → [17]03[12] (s=32) (c=6) (b=19)
1 → 2 → 4 → 8 → [16] (s=16) → 1[12] (s=13) (c=5) (b=20)
100 = 50 (s=5) → 55 (s=10) → 5[15] (s=20) → 6[15] (s=21) → 7[16] (s=23) → 8[19] (s=27) → [10]6 (s=16) (c=6) (b=20)
135665 = [16][19]35 (s=43) → [16][19]58 (s=48) → [16][19]7[16] (s=58) → [16][19][10][14] (s=59) → [16][19][13][13] (s=61) → [16][19][16][14] (s=65) → [16][19][19][19] (s=73) → [17]03[12] (s=32) (c=7) (b=20)
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 5,290 | 9,262 | {"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} | 3.8125 | 4 | CC-MAIN-2023-23 | latest | en | 0.531257 |
https://www.ceder.net/def/reversecutthegalaxy.php?language=czech&level=master | 1,620,798,678,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991252.15/warc/CC-MAIN-20210512035557-20210512065557-00381.warc.gz | 730,108,887 | 5,694 | Definitions of Square Dance Calls and Concepts
FAQ |
Index --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Definitions (Text Only) --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Find call:
Reverse Cut | Flip The Galaxy -- [C2]
(autor neznámý)
Z formace Galaxy.
Reverse Cut The Galaxy:
Reverse Flip The Galaxy:
Centers Phantom Run (flip away od sebe), Outsides Galaxy Circulate.
Končí v Parallel Lines.
předReverse Cut The Galaxy poCenters Trade & Spread jako Outsides Galaxy Circulate (hotovo)
předReverse Flip The Galaxy poCenters Flip Away jako Outsides Galaxy Circulate (hotovo)
Reverse Cut The Diamond [C2]: Z formace Diamond. Centers Trade & Spread, Points Diamond Circulate. Končí v Line.
Reverse Flip The Diamond [C2]: Z formace Diamond. Centers Phantom Run (Flip od sebe), Points Diamond Circulate. Končí v Line.
Reverse Cut The formation [C4]: Z dané formace. Ti, co se drží za ruce Trade and Spread, ti ostatní formation Circulate. Formation může být "O", Butterfly, (Wave-Based) Triangle, 3 By 1 Triangle, Short Six, Tall Six, "Z", atd...
Reverse Flip The formation [C4]: Z dané formace. Ti, co se drží za ruce Phantom Run, ti ostatní formation Circulate. Formation může být "O", Butterfly, (Wave-Based) Triangle, 3 By 1 Triangle, Short Six, Tall Six, "Z", atd...
Viz také Cut | Flip The Galaxy [C1].
CALLERLAB definition for Reverse Cut the Galaxy
CALLERLAB definition for Reverse Flip the Galaxy
Choreography for Reverse Cut | Flip The Galaxy
Comments? Questions? Suggestions?Page translated by David Tesař.
https://www.ceder.net/def/reversecutthegalaxy.php?language=czech&level=master
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A.
MATHEMATICS AND SURVEYING
Mathematics
1.0 Algebra
1.1 Set Theory*
1.2 Real Numbers
1.3 Algebraic Expressions and Operations
1.4 Equations and Inequalities
1.5 Roots and Powers
1.6 Linear, Quadratic and Polynomial Functions
1.7 Factoring
1.8 Roots of Algebraic Equations
1.9 System of Equations
1.10Logarithmic and Exponential Functions
1.11Arithmetic and Geometric Progressions
2.0 Trigonometry
2.1 Circular (Trigonometric) Functions
2.2 Trigonometric Identities and Equations
2.3 Solution of Triangles
2.4 Hyperbolic Functions
3.0 Analytic Geometry
3.1 Cartesian Coordinate System
3.2 Functions and Relations
3.3 Functions and their Graphs
3.4 Straight lines
3.5 Conic Sections
3.6 Polar Coordinates
3.7 Transformation of Coordinates
3.8 Parametric Equations
4.0 Calculus
4.1 Differential Equations
4.1.1 Limits and Continuity
4.1.2 Derivatives and Differentiation
4.1.3 Application of Derivatives
4.1.4 The Differential
4.1.5 Partial Derivatives
4.2 Integral Calculus
4.2.1 Theory of Integrals
4.2.2 Integration Methods
4.2.3 Definite Integrals and Applications
4.2.4 Line and Surface Integrals
4.2.5 Multiple Integrals
5.0 Differential Equations
5.1 First Order Differential Equation
5.1.1 Exact Differential Equation
5.1.2 Integrating Factors**
5.1.3 Separable Variables
5.1.4 Homogeneous Differential Equations
5.1.5 Linear Differential Equations
5.1.6 Applications
5.2 Higher Order Differential Equations
6.0 Other Topics
6.1 Infinite Series
6.1.1 Molaurin Series
6.1.2 Taylor Series
6.1.3 Fourier Series
6.2 Complex Variables**
6.3 Vector Analysis
6.4 Matrices*
6.5 Determinants*
6.6 Probability and Statistics
7.0 Engineering Economy
7.1 Present Economy Study
7.2 Time-Value Relations
7.3 Selection Among Alternatives
7.3.1 Present Worth Method
7.3.2 Annual Worth Method
7.3.3 Future Worth Method
7.3.4 Internal Rate of Return Method
7.3.5 External Rate of Return Method
Surveying
1.0 Surveying Concepts
1.1 Uses of Surveys
1.2 Operations in Surveying
1.3 Measurement and Adjustments
1.4 Field and Office Work
1.5 Surveying Instruments
2.0 Basic Surveying Measurements
2.1 Distance Measurements
2.1.1 Pacing
2.1.2 Distance Measurement with Tape
2.2 Vertical Distance Measurement; Leveling
2.3 Angle and Direction Measurement
2.3.1 Location of Points
2.3.2 Meridians
2.3.3 Bearing and Azimuth
2.3.4 Magnetic Declination
2.3.5 Instruments Used
2.3.5.1 Engineers Transit
2.3.5.2 Theodolite
2.4 Stadia and Tacheometry
2.4.1 Principles of Stadia
2.4.2 Plane Table and Alidade
3.0 Survey Operations
3.1 Traverse
3.1.1 Deflection Angle Traverse
3.1.2 Interior Angle Traverse
3.1.3 Traverse by Angle to the Right
3.1.4 Azimuth Traverse
3.1.5 Compass Traverse
3.1.6 Stadia Traverse
3.1.7 Plane Table Traverse
3.2 Calculation of Areas of Land
3.2.1 Area by Triangle
3.2.2 Area by Coordinates
3.2.3 Area by Double Meridian Distance
(DMD) and Latitude
3.2.4 Irregular Boundaries (Simpsons
and Trapezoidal Rules)
3.3 Triangulation and Trilateralization
3.3.1 Horizontal Control System
3.3.2 Triangulation Figures and
Procedures
3.3.3 Error Propagation
3.3.4 Trilateralization
3.4 Astronomical Observation
3.4.1 Celestial Sphere
3.4.2 Equator System
3.4.3 The PZS Triangle
3.4.4 Azimuth and Hour Angle at
Elongation
3.4.5 Time
3.4.6 Solar Observation
3.4.7 Stellar Observation
4.0 Engineering Surveys
4.1 Topographic Survey
4.1.1 Horizontal Control
4.1.2 Vertical Control (contours)
4.1.3 Location of Details
4.2 Route Surveying
4.2.1 Horizontal Curves
4.2.1.1 Simple Curves
4.2.1.2 Compound Curves
4.2.1.3 Superelations
4.2.1.4 Spiral Curves
4.2.2 Vertical Curves
4.2.3 Earthwork Operations
4.2.3.1 Methods of Determining
Earthwork Volumes
4.2.3.2 Borrow Pits
4.3 Hydrographic Surveys
4.3.1 Datum
4.3.2 Soundings
B. HYDRAULICS
1.0 Fluid Mechanics
1.1 Properties of Fluids
1.2 Fluid Statics
1.3 Fluid Flow Concepts and Basic Equations
1.4 Dimensionally Analysis and Dynamic
Similitude
1.5 Viscous Flow and Fluid Resistance
1.6 Ideal Fluid Flow
1.7 Steady Flow in Closed Conduits
1.8 Steady Flow in Open Channels
2.0 Hydrology
2.1 Hydrologic Cycle
2.1.1 Precipitation
2.1.2 Stream-flow
2.1.3 Evaporation
2.1.4 Transpiration
2.2 Hydrograph Analysis
2.2.1 Runoff
2.2.2 Storage Routing
2.3 Groundwater
3.0 Hydraulics, System and Structure
3.1 Reservoirs
3.2 Dams
3.3 Spillways, Gates, and Outlet Works
3.4 Open Channels
3.5 Pressure Conduits
3.6 Hydraulics Machinery
4.0 Irrigation, Flood Control and Drainage
4.1 Irrigation
4.1.1 Water Requirement
4.1.2 Soil-Water Relation
4.1.3 Water Quality
4.1.4 Methods
4.1.5 Structures
4.2 Flood Control
4.2.1 Design Flood
4.2.2 Flood Control Structures
4.3 Drainage
4.3.1 Estimate of Flow
4.3.2 Storm Drainage
4.3.3 Land and Highway Drainage
4.3.4 Culvets and Bridges
4.3.5 Drainage Structures
5.0 Water Supply and Sewerage
5.1 Fundamental Concept
5.1.1 Mathematics of Growth
(Population Forecasting)
5.1.2 Environmental Chemistry
5.1.3 Mass and Energy Transfer
5.2 Water Supply and Treatment
5.2.1 Components of Water Supply
System
5.2.1.1 Water Reservoir and
Storage
5.2.1.2 Water Distribution System
5.2.1.3 Water Containment
Structures
5.2.2 Water Consumptions Periods of
Design
5.2.3 Pre-treatment Methods
5.2.4 Principles of Sedimentation
5.2.5 Sedimentation Tank Design
5.2.6 Coagulation-Sedimentation
5.2.7 Slow Sand Filtration
5.2.8 Rapid Sand Filtration
5.2.9 The Rapid Sand Filter
5.2.10 Underdrain System
5.2.11 Wash Troughs
5.2.12 The Washing Process
5.2.13 Clear Well and Plant Capacity
5.2.14 Water Disinfection
5.3 Waste Water Treatment
5.3.1 Quantity
5.3.2 Methods
5.3.3 Theory of Activated Sludge
5.3.4 Aration Tank
5.3.5 Biokinetic Parameters*
5.3.6 Clarifiers
C. DESIGN AND CONSTRUCTION
1.0 Statics of Rigid Bodies
1.1 Force System
1.1.1 Concurrent and Non-current
Force System
1.1.2 Parallel and Non-parallel Force
System
1.1.3 Planar and Three Dimensional
Force System
1.1.4 Distributed Forces
1.1.5 Frictional Forces
1.2 Equilibrium of Forces
1.2.1 Reactions
1.2.2 Free Body Diagram
1.2.3 Two Force Bodies
1.2.4 Three Force Bodies
1.3 Truss Analysis
1.3.1 Method of Joints
1.3.2 Method of Sections
1.3.3 Graphical Methods
1.4 Beams and Frames
1.4.1 Reactions
1.4.2 Shear Diagrams
1.4.3 Bending Moment Diagrams
1.5 Related Topics
1.5.1 Moment of Lines and Areas
1.5.2 Centroids
1.5.3 Moments of Inertia
1.5.4 Center of Mass
1.5.5 Center of Forces
2.0 Dynamics of Rigid Bodies
2.1 Kinematics of Particles
2.1.1 Rectilinear Motion
2.1.2 Curvilinear Motion
2.2 Kinetics of Particles
2.2.1 Newtons Second Law
2.2.2 Dynamic Equilibrium
2.2.3 Work and Energy Principle
2.2.4 Kinetic and Potential Energy
2.2.5 Impulse and Momentum Principle
2.3 Kinematics of Rigid Bodies
2.3.1 Translation
2.3.2 Rotation
2.3.3 General Plane Motion
2.4 Kinetics of Rigid Bodies
2.4.1 D' Lambert's Principle
2.4.2 Work and Energy Principle
2.4.3 Impulse and Momentum Principle
3.0 Mechanics
3.1 Stresses and Strains
3.2 Material Properties
3.3 Axially Loaded Members
3.4 Thin Walled Pressure Vessels
3.5 Torsional Stresses
3.6 Internal Forces and Stresses in Beams
3.6.1 Flexural Stress
3.6.2 Shear Stress
3.6.3 Combined Stresses
3.6.4 Principal Stresses
3.6.5 Unsymmetrical Banding
3.7 Deflections
3.7.1 Double Integration Methods
3.7.2 Area Moment Method
3.7.3 Conjugate Beam Method
3.8 Statistically Indeterminate Beams
3.9 Shear Center
3.10 Curved Beams
3.11 Non-homogenous Beams
3.12 Impact Loading
3.13 Stress Concentration
3.14 Repeated Loading
3.15 Elastic Instability (Buckling)
3.16 Analysis of Connections
3.16.1 Riveted and Bolted Connections
3.16.2 Welded
4.0 Structural Analysis
4.1 Loadings
4.1.1 Vertical Loads (dead and live
loads)
4.1.2 Lateral Loads (Wind and
Earthquake Loads)
4.1.3 Impact Loads
4.2 Energy Methods for Deformation
Analysis
4.2.1 Castiglianos Theorem
4.2.2 Virtual Work Method (Unit Load)
4.3 Influence Lines
4.4 Frame Analysis
4.4.1 Approximate Methods
4.4.2 Exact Methods*
4.4.3 Moment Distribution
4.5 Stiffness and Flexibility Methods of
Analysis**
4.5.1 Trusses
4.5.2 Beams
4.5.3 Frames
5.0 Design of Timber Structures
5.1 Properties of Wood
5.2 Design of Tension Member
5.3 Design of Bending Members
5.3.1 Laterally Supported Beams
5.3.2 Laterally Unsupported Beams
5.4 Design of Compression Members
5.4.1 Short Columns
5.4.2 Slender Columns
5.4.3 Spaced Columns
5.5 Timber Connections
6.0 Design of Steel Structures
6.1 General
6.1.1 Properties of Structural Steel
6.1.2 Design Philosophy
6.1.2.1 Allowable Stress Design
6.1.2.2 Load and Resistance
Factor Design
6.2 Tension Members
6.3 Connections
6.3.1 Bolted
6.3.2 Welded
6.4 Compression Members
6.5 Beams
6.5.1 Compact Sections
6.5.2 Non-compact Sections
6.6 Beam Columns
6.7 Plastic Analysis and Limit Design*
6.8 Composite Steel and Concrete
7.0 Reinforced Concrete Structures
7.1 General
7.1.1 Properties of Concrete Materials
7.1.2 Design Philosophies and
Procedures
7.2 Flexural Analysis and Design
7.3 Shear and Diagonal Tension
7.4 Bond, Anchorage Development Lengths
7.5 Serviceability Requirements
7.5.1 Crack Control
7.5.2 Deflections
7.6 Columns
7.6.1 Short Columns
7.6.2 Slender Columns
7.7 Slabs
7.8 Footings
7.9 Retaining Wall
7.10 Prestressed Concrete
8.0 Soil Mechanics and Foundation
8.1 Soil Properties
8.2 Soil Classification
8.3 Flow of Water in Soils
8.3.1 Permeability
8.3.2 Seepage
8.3.3 Effective and Porewater Pressure
8.4 Soil Strength
8.4.1 Shear Strength
8.4.2 Bearing Capacity
8.5 Compressibility of Soils
8.5.1 Elastic Settlement
8.5.2 Consolidation Settlement
8.6 Soil Improvement
8.6.1 Compaction
8.6.2 Soil Stabilization
8.7 Earth Pressures and Retaining Wall
8.8 Slope Stability Analysis
9.0 Design of Civil Engineering Structures and
Systems
9.1 Transportation Engineering**
9.1.1 Highway and Urban Transportation
Planning and Economics**
9.1.2 Driver, Vehicle, Traffic and Road
Characteristics**
9.1.3 Highway Design
9.1.4 Traffic Engineering and Highway
Operations**
9.1.5 Road and Pavement Design
9.2 Airport Engineering**
9.3 Ports and Harbors
9.4 Containment Structures (Tanks, soils, storage
tanks)
9.5 Bridges
10.0 Construction and Management
10.1 Engineering Relations and Ethics**
10.2 Contracts & Specifications
10.3 Construction Project Organization
10.4 Planning and Scheduling (PERT/CPM)
10.5 Construction Estimates
10.6 Construction Methods & Operations
10.7 Construction Equipment Operations and
Maintenance | 3,415 | 10,329 | {"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} | 3.1875 | 3 | CC-MAIN-2020-05 | latest | en | 0.607041 |
http://www.jiskha.com/display.cgi?id=1299191685 | 1,496,142,941,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463615093.77/warc/CC-MAIN-20170530105157-20170530125157-00131.warc.gz | 685,528,522 | 4,151 | # trigonometry (please double check this)
posted by on .
Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only.
a.)
( I got confused doing this 1 can you help me with it.)
3 sin theta - 4 cos theta = 2
b.)
(can you correct this 1 for me.)
tan (x+15)= 3 tan x
tan(x+15) = (tanx + tan15)/(1-tanx tan15)
(tanx + tan15)/(1-tanx tan15) = 3 tanx
tanx + tan 15 = 3 tanx - 3tan2x tan15
3 tan2x tan15 - 2tanx + tan15 = 0
tan15 = 0.27, denote tanx as t
0.81 t2 - 2t + 0.27 = 0
D = 0.1252
t1 = 1.67
t2 = 0.80
tan (x1) = 1.67
x1 = arctan (1.67) + ¥ðn = 59.087 + 180n = 59¨¬ 5'+ 180n, n=0,1,2,..
tan(x2) = 0.80
x2 = arctan(0.80) + ¥ðn = 38.66 + 180n = 38¨¬ 40' + 180n, n=0,1,2,..
Answer: 59¨¬ 5', 239¨¬ 5', 38¨¬ 40', 218¨¬ 40'
• trigonometry (please double check this) - ,
Answer for b. 29deg 5',239deg 5',38deg 40',218deg 40'.
can you plz check if i messed anything.
• trigonometry (please double check this) - ,
kjskjdVKNSDKBvkSJJNb ksd | 453 | 1,076 | {"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} | 3.84375 | 4 | CC-MAIN-2017-22 | latest | en | 0.659894 |
https://homes.cs.washington.edu/~jrl/teaching/cse599Qau22/homeworks/hw5.html | 1,675,361,251,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00613.warc.gz | 303,809,227 | 3,774 | # CSE 599Q: Homework #5
## Quantum probability and information
Due: Sun, Dec 11 @ 11:59pm
## 1. Positive semidefinite matrices [12 points]
A Hermitian matrix $M \in \C^{d \times d}$ is said to be positive semidefinite, denoted $M \succeq 0$ if it holds that $\bra{u}M\ket{u} \geq 0$ for all $\ket{u} \in \C^d$.
1. [3 pts] Prove that $M \succeq 0$ if and only if $\bra{u}M\ket{u} \geq 0$ holds for all unit vectors $\ket{u} \in \C^d$.
2. [3 pts] Let $M \in \C^{d \times d}$ be a diagonal matrix. Verify that $M$ is Hermitian if and only if all its diagonal entries are real and $M \succeq 0$ if and only if all diagonal entries is nonnegative.
3. [3 pts] Consider any matrix $A \in \C^{d \times d'}$. Show that $A^* A$ is Hermitian and $A^*A \succeq 0$.
4. [3 pts] Recall the Frobenius inner product on matrices $A,B \in \C^{d\times d}$:
$\langle A,B\rangle \seteq \tr(A^*B).$
Prove that if $A \succeq 0$ and $B \succeq 0$, then $\langle A,B\rangle \geq 0$.
You may use the fact that every Hermitian matrix $M \in \C^{d \times d}$ can be written as $M = \sum_{i=1}^d \lambda_i \ket{u_i}\bra{u_i}$ where $\lambda_1,\ldots,\lambda_d$ are the real eigenvalues of $M$ and $\ket{u_1},\ldots,\ket{u_d}$ is an orthornormal basis of eigenvectors for $M$.
## 2. Purification of quantum states [10 points]
Suppose that $\rho^A$ is a $d \times d$ density matrix. Write $\rho^A$ in its eigenbasis:
$\rho^A = \sum_{j=1}^n \lambda_j \ket{v_j}\bra{v_j}\,.$
Let $\mathbb{C}^B$ denote a $d$-dimensional Hilbert space with basis $\ket{1},\ket{2}\,\ldots,\ket{d}$ and define
\begin{align*} \ket{u^{AB}} & \seteq \sum_{j=1}^n \sqrt{\lambda_j} \ket{v_j} \ket{j} \\ \rho^{AB} &\seteq \ket{u^{AB}} \bra{u^{AB}} \end{align*}
Show that $\rho^A = \tr_B(\rho^{AB})$. In other words, the mixed state $\rho^A$ can be seen as arising from taking a joint system in the pure state $\ket{u^{AB}}$ and then discarding the $B$-part of the space.
## 3. Quantum uncertainty splits evenly [10 points]
Recall that if $X$ is a classical random variable such that $p_i \seteq \Pr[X=i]$ for $i=1,2,\ldots,d$, then the Shannon entropy of $X$ is defined by
$H(X) \seteq \sum_{i=1}^d p_i \log \frac{1}{p_i}.$
This is a measure of the uncertainty of $X$ measured in bits (or measured in “nats” if we use the natural logarithm).
Define the von Neumann entropy of a $d \times d$ density matrix $\rho$ by
$\mathcal{S}(\rho) \seteq \sum_{j=1}^d \lambda_j \log \frac{1}{\lambda_j}\,.$
Suppose that $\rho = \ket{u^{AB}} \bra{u^{AB}}$ is a pure state and $\rho^A = \tr_B(\rho), \rho^B = \tr_A(\rho)$. Show that
$\mathcal{S}(\rho^A) = \mathcal{S}(\rho^B)\,.$
Note that the two states don’t necessarily have the same dimension, so they could each have a different number of eigenvalues.
[ Hint: Show first that if $U$ is a $d \times d$ matrix, then $UU^*$ and $U^*U$ have the same non-zero eigenvalues. ]
## 4. Negative conditional entropy [10 points]
1. [5 pts] In classical probability theory, if $A$ and $B$ are two random variables, one defines the entropy of $A$ conditioned on $B$ by the formula
$H(A \mid B) \seteq \sum_{x} \Pr[B=x] \cdot H(A \mid \{B=x\})\,,$
where $A \mid \{B=x\}$ is the random variable $A$ condition on $X$. This quantity is nonnegative because the entropy $H(A \mid \{B=x\})$ is always nonnegative.
Prove that
$H(A \mid B) = H(A,B) - H(B)\,.$
In particular, right-hand side is always nonnegative, and therefore
$H(B) \leq H(A,B)\,.$
This asserts the relatively obvious fact that the pair of random variables $\{A,B\}$ has more uncertainty than the single random variable $B$. In other words, it is easier to predict $B$ than to simultaneously predict both $A$ and $B$.
2. [5 pts] You will show that this fails dramatically in the quantum setting where the conditional entropy can be negative! Using Problem 2, Show that there is a state $\rho^{AB}$ with
$\mathcal{S}(\rho^B) > \mathcal{S}(\rho^{AB})\,,$
where $\rho^B \seteq \tr_A(\rho^{AB})$. In other words, the entropy of the subsystem is actually bigger than the entropy of the full system. | 1,408 | 4,056 | {"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": 3, "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} | 3.484375 | 3 | CC-MAIN-2023-06 | longest | en | 0.696406 |
https://www.physicsforums.com/threads/moment-of-inertia-for-a-thick-spherical-shell.861915/ | 1,532,189,206,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592636.68/warc/CC-MAIN-20180721145209-20180721165209-00107.warc.gz | 940,473,447 | 15,112 | # Homework Help: Moment of Inertia for a Thick Spherical Shell
1. Mar 13, 2016
### b100c
1. The problem statement, all variables and given/known data
A)
Consider a hollow sphere of uniform density with an outer radius $R$ and inner radius $\alpha R$, where $0\leq\alpha\leq1$. Calculate its moment of inertia.
B) Take the limit as $\lim_{\alpha\to1}$ to determine the moment of inertia of a thin spherical shell.
2. Relevant equations
Moment of Inertia: $I = \int r^2 dm$
3. The attempt at a solution
$dm = \rho dV$. Where rho is density. The volume element for a sphere is $$dV=r^2sin\theta d\theta d\varphi dr$$
So I think I would integrate over a sphere but instead from inner radius to the outer radius? $$I = \rho \int_{\alpha R}^{R} r^4 dr \int_{0}^{\pi} sin\theta d\theta \int_{0}^{2\pi} d\varphi$$
Which yields $$\frac{4\pi}{5} \rho (R^5 - \alpha R^5)$$
If $$\rho = \frac{m}{V} = \frac{m}{\frac{4\pi (R^3 - \alpha R^3)}{3}}$$
Then the equation for moment of inertia becomes
$$I = \frac{3}{5}mR^2 \frac{1-\alpha^5}{1-\alpha^3}$$
The problems is now when I take the limit as alpha approaches 1, and apply l'Hopital's rule, I get that moment of inertia is $mR^2$, when there should be a factor of $\frac{2}{3}$?
2. Mar 13, 2016
### BvU
In your I calculation, the r in your relevant equation is not the distance from the origin, but the distance from the axis of rotation ...
3. Mar 13, 2016
### b100c
Thanks BvU, that was a stupid mistake on my part. I replaced r with rsin(theta) in spherical coordinates and I got the correct answer.
4. Mar 13, 2016
well done! | 512 | 1,579 | {"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} | 3.921875 | 4 | CC-MAIN-2018-30 | latest | en | 0.81929 |
https://www.mathworks.com/matlabcentral/cody/problems/32-most-nonzero-elements-in-row/solutions/217786 | 1,508,688,019,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825308.77/warc/CC-MAIN-20171022150946-20171022170946-00466.warc.gz | 977,279,966 | 11,517 | Cody
Problem 32. Most nonzero elements in row
Solution 217786
Submitted on 15 Mar 2013 by Tomas Lindquist Olsen
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
%% a = [ ... 1 2 0 0 0 0 0 5 0 0 2 7 0 0 0 0 6 9 3 3]; r_correct = 4; assert(isequal(fullest_row(a),r_correct))
2 Pass
%% a = [ ... 1 2 0 0 0 0 5 0 0 6 9 -3 2 7 0 0 0 0 0 0]; r_correct = 3; assert(isequal(fullest_row(a),r_correct))
3 Pass
%% a = [ ... 1 0 0 0 0 0 0 0 0 0 0 0 0 2 3]; r_correct = 5; assert(isequal(fullest_row(a),r_correct))
4 Pass
%% a = [ ... 0 0 0 -3 0 0]; r_correct = 4; assert(isequal(fullest_row(a),r_correct)) | 304 | 720 | {"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} | 2.8125 | 3 | CC-MAIN-2017-43 | latest | en | 0.578959 |
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``` plz explain right hand thumb rule in detail giving illustrative examples..
```
6 years ago
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``` Dear student,
Right Hand Thumb Rule:
If a current carrying conductor is imagined to be held in your right hand such that the thumb points along the direction of current, then the direction of the wrapped fingers will give the direction of magnetic field lines.
Accordingly if current in wire is vertically upward, the magnetic field lines are anticlockwise, while if current in wire is vertically downward, the magnetic field lines are clockwise.
Some of the Right Hand Thumb Rule:
Right Hand Palm Rule No 1:
If we stretch our right hand palm in such a way that the thumb points along the direction of current (I), the stretched fingers towards the observation point P, then the direction of magnetic field will be along the outward drawn perpendicular to palm.
All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Sagar Singh
B.Tech, IIT Delhi
```
6 years ago
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More Questions On Magnetism | 1,447 | 6,003 | {"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} | 2.5625 | 3 | CC-MAIN-2017-04 | longest | en | 0.848199 |
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# In a study of 250 adults, the mean heart rate was 70 beats per minute. Assume the population of heart rates is known to be approximately normal with a standard deviation of 12 beats per minute. What is the 99% confidence interval for the mean beats per minute? 68.9 − 76.3 70 − 72 61.2 − 72.8 68 − 72
Accepted Solution
A:
The z score for a confidence level of 99% is 2.575.. You need to know this to complete this type of problem.. the other two commonly used confidence levels are 90% and 95% having z scores of 1.645 and 1.96 respectively. You need to memorize this!
To find the confidence interval, it will be defined as the sample mean plus/minus the margin of error.. Now to find the margin of error, divide the standard deviation of 12 by the square root of the number of elements in your sample(√250). Then take that result and multiply it by the Z score I mentioned above for a 99% confidence level.
In this case: the sample mean is 70 and the margin of error is approximately 1.95
So to calculate the confidence interval, do the following:
70 - 1.95 = 68.05 rounded to the nearest whole number is 68
70 + 1.95 = 71.95 rounded to the nearest whole number is 72
Looks like that would be the last choice. hope this helps.. it's been awhile since my work with statistics :-) | 334 | 1,291 | {"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} | 4.4375 | 4 | CC-MAIN-2024-26 | latest | en | 0.912973 |
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CHAPTER 9 INVENTORY COSTING AND CAPACITY ANALYSIS 9-16 (30 min.) Variable and absorption costing, explaining operating-income differences. 1. Key inputs for income statement computations are April May Beginning inventory Production Goods available for sale Units sold Ending inventory 0 500 500 350 150 150 400 550 520 30 The budgeted fixed cost per unit and budgeted total manufacturing cost per unit under absorption costing are April May (a) Budgeted fixed manufacturing costs (b) Budgeted production (c)=(a)÷(b) Budgeted fixed manufacturing cost per unit (d) Budgeted variable manufacturing cost per unit (e)=(c)+(d) Budgeted total manufacturing cost per unit \$2,000,000 500 \$4,000 \$10,000 \$14,000 \$2,000,000 500 \$4,000 \$10,000 \$14,000 (a) Variable costing April 2011 May 2011 Revenues a \$8,400,000 \$12,480,000 Variable costs Beginning inventory \$ 0 \$1,500,000 Variable manufacturing costs b 5,000,000 4,000,000 Cost of goods available for sale 5,000,000 5,500,000 Deduct ending inventory c (1,500,000 ) (300,000 ) Variable cost of goods sold 3,500,000 5,200,000 Variable operating costs d 1,050,000 1,560,000 Total variable costs 4,550,000 6,760,000 Contribution margin 3,850,000 5,720,000 Fixed costs Fixed manufacturing costs 2,000,000 2,000,000 Fixed operating costs 600,000 600,000 Total fixed costs 2,600,000 2,600,000 Operating income \$1,250,000 \$3,120,000 a \$24,000 × 350; \$24,000 × 520 c \$10,000 × 150; \$10,000 × 30 b \$10,000 × 500; \$10,000 × 400 d \$3,000 × 350; \$3,000 × 520 9-1
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(b) Absorption costing April 2011 May 2011 Revenues a \$8,400,000 \$12,480,000 Cost of goods sold Beginning inventory \$ 0 \$2,100,000 Variable manufacturing costs b 5,000,000 4,000,000 Allocated fixed manufacturing costs c 2,000,000 1,600,000 Cost of goods available for sale 7,000,000 7,700,000 Deduct ending inventory d (2,100,000) (420,000) Adjustment for prod.-vol. variance e 0 400,000 U Cost of goods sold 4,900,000 7,680,000 Gross margin 3,500,000 4,800,000 Operating costs Variable operating costs f 1,050,000 1,560,000 Fixed operating costs 600,000 600,000 Total operating costs 1,650,000 2,160,000 Operating income \$1,850,000 \$ 2,640,000 a \$24,000 × 350; \$24,000 × 520 d \$14,000 × 150; \$14,000 × 30 b \$10,000 × 500; \$10,000 × 400 e \$2,000,000 – \$2,000,000; \$2,000,000 – \$1,600,000 c \$4,000 × 500; \$4,000 × 400 f
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Jill Tulane University ‘16, Course Hero Intern | 1,053 | 3,621 | {"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} | 3.046875 | 3 | CC-MAIN-2018-26 | latest | en | 0.792009 |
https://jeopardylabs.com/print/math-models-20 | 1,725,975,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00118.warc.gz | 303,513,807 | 4,178 | Scale Factor
Pythagorean Theorem
Slope
Solving Linear Equations
No calculator
100
True or False? If the scale factor is less than 1, the scale will cause the image or figure to grow.
False
100
What is the Pythagorean theorem formula?
a2+b2=c2
100
What is the slope-intercept equation?
y=mx+b
100
46=12x+490
Solve for x.
x=-37
100
12x+49=70y-11
Solve for y
y=(12x+60)/(70)
200
What is a mortgage payment?
200
True or false? A triangle does not need to be a right triangle to use Pythagorean theorem.
False
200
What is the slope of this equation?
y=17x-21
17
200
3(x+6)+(6x+22)=21-12x
Solve for x.
Round to the nearest tenth.
-.9
200
A triangle has one angle that is 36 degrees and another that is 6 degrees. What is the angle measure of the remaining angle?
138 degrees
300
What is the annual average income for workers in Texas?
52-57 thousand
300
A ten-foot ladder is leaning up against a building and the base of the ladder is 6 feet away from the buildings wall. How high up is the ladder on the building in inches?
96 inches
300
What is the slope of the following equation? (Round to the nearest tenth)
472x+17y=1067-6y+.5
-20.5
300
What is the first step to completing this problem?
7x+21-72x=67-x+228
solve for x
Combine like terms
300
10(2+42) -7(62+3)
Simplify the expression
-93
400
If triangle ABC is
400
Three squares come together to form a right triangle. The largest square has side lengths of 18 inches. One of the other squares has side lengths of 12 inches. What is the perimeter of the remaining square in inches?
Round to the nearest hundredth.
53.67 inches
400
The slope of a line is 3 and it goes through the point (2,3). Another point on the line goes through the x-coordinate 1. Find the missing y-coordinate.
0
400
4x+8x=-9+17x-1
Solve for x.
x=2
400
A right triangle's hypotenuse has a length of 20. One of the legs on the triangle has a length of 8. What is the length of the remaining leg?
Square root of 336
500
You have a monthly income of \$3,200. After paying rent (\$1,200), utilities (\$150), and groceries (\$300), you want to save 20% of your remaining income. How much money do you have left after these expenses, and how much should you save?
Money left \$1,550
Savings goal \$310
500
A ship takes off from a port heading West and travels 1245 miles. It then travels south 129 miles. How far is the ship from the port it left from in miles?
Truncate to the nearest Hundredth.
1251.66
500
Suppose that the water level of a river is 34 feet and that it is receding at a rate of 1/2 feet per day. Write an equation for the water level, y, after x days. In how many days will the water level be 26 feet?
16 days
500
2(w+3)−10=6(32−3w)
solve for w.
Round to the nearest tenth.
w=9.8
500
A right triangle has an area of 472. The height of the triangle is 14. What is the length of base of the triangle?
Round to the nearest tenth
67.4 | 840 | 2,951 | {"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} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.892427 |
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# Questions tagged [encoding-scheme]
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32 questions with no upvoted or accepted answers
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41 views
### Messy Representation Encoding example
I am currently working through Metaheuristics by El-Ghazali Talbi where he discusses encodings of algorithms. "Messy representations: In linear representations of fixed length, the semantics of ...
• 21
69 views
### Intuitive explanation on why stochastic encoding performs better in channel coding
I am a little confused about stochastic encoding in channel coding. For example, in the identification problem (R. Ahlswede and G. Dueck, “Identification via channels”), the authors claim that we can ...
85 views
### What is "orthogonality" in the context of Instruction Encoding?
What does it mean by "orthogonality" in the context of Instruction Encoding? Why CISC Architecture is orthogonal while RISC is not?
52 views
### How can I optimise a 8b10b encoding for maximum alignment?
Imagine you encode an 8 bit symbol as a 10 bit symbol that is sent sequentially over a wire. The goal at the receiver is to detect the byte boundary. Since there are 4 times more encoded symbols than ...
• 133
47 views
### Card Shuffling, Bounding Mixing time using Paths and Flows
I've been struggling with a problem that is very similar to a 2014 question posted here. The question in particular is 3(1) and 3(2). To paraphrase, we are supposed to use paths and an encoding ...
• 473
66 views
### Error detection code supporting arithmetic
I am looking for error detection codes, which support addition in the encoded domain and are separate (a tuple of ($N$, $R(N)$), where $N$ denotes the functional value and $R(N)$ its redundancy). So ...
• 21
992 views
### Encoding Turing machines in binary in a simple way that gurarentee unique readability
I may come across something like (let $X$ be a binary representation of a Turing machine $M$), I know that we can easily encode the description of the machine as sequence of numbers (by assigning a ...
• 403
58 views
### Typed representation of a memory model
Assume a simple procedural language, where statements write and read from local memory via references and procedures accept arguments of n different scalar types (say floats, ints and strings) and ...
• 610
1 vote
96 views
### Succint data structure for storing n numbers using a mask-based partition
I've a set $N$ of $n$ distinct numbers of $d$-bits each; call them words ($1\leq n\leq 2^d$, with $d\geq 1$). I want to store $N$ using as least space as possible, by partitioning $N$ and factoring ...
• 529
1 vote
97 views
### Initialization of embedding space value for VQ-VAE
I am trying to study a research paper about VQ-VAE: Neural Discrete Representation Learning, Aaron van den Oord, Oriol Vinyals, Koray Kavukcuoglu, NeurIPS 2017. I have difficulty to understand the ...
• 123
1 vote
55 views
### How do I compress data vectors in broadcast messages?
Let us model a wireless broadcast network as an undirected graph $G(V,E)$ where there is an edge between every pair of nodes ${i,j}\in V$ if they are in transmission range of each other. $w_{i\to j}$ ...
• 21
1 vote
133 views
### Base-N encoding with smallest output
I have a set of bytes (utf-8), and need to encode them into the smallest dataset possible using a Base N encoding scheme. Is it simply the higher the Base N encoding is (ie something like Base85 ...
• 111
1 vote
96 views
### Difference between a regular and a stationary source?
As far as I understand a stationary source is a regular source but it's not necessarily true the other way around. And a stationary source is a source for which its distribution is unaffected by a "...
• 333
1 vote
129 views
### Byte Stuffing, escaping a flag at maximum frame
I want to encode a message using byte stuffing where every frame delimited by a flag byte 'F' at both ends and escape code equal to the character E. I am trying to encode the message 'DEEP-FRIED-...
• 11
1 vote
212 views
### How to encode each possible b-tree of a sequence of n numbers?
Lehmer codes can be used to encode each possible permutation of a sequence of n numbers. Often the main goal is just to map a range of numbers from 1 to x to the possible permutations of a sequence of ...
• 133
1 vote
101 views
### Is the halting problem a matter of an encoding scheme ?
I was reading about the halting problem recently, there is a video on youtube where it tries to explain the halting problem easily (since it is complicated to explain). So, (A,C & H) have ...
• 143
18 views
### A specification of all possible languages for representing graphs?
There are a few commonly used markup languages for specifying graph structures. I am interested in discovering alternative graph notations that may be counterintuitive yet more compact or useful in ...
52 views
### Kolmogorov complexity of tuples
Exercise 14.1 of Elements of Information Theory asks us to prove that there is a constant c such that $$K(x,y)\leq K(x) +K(y)+c$$ for all binary strings $x,y$. Intuitively this seems true. Just write ...
11 views
### Determining the benefit from increasing bits available in base 32 checksum
I'm looking at using something like Crockford Base-32 encoding in a situation where people have to manually write IDs in a very compact space. Crockford Base-32 describes a simple checksum algorithm ...
• 181
71 views
### Calculation of compression ratio using arithmetic encoding?
Arithmetic encoding is one of the most famous entropy encoding techniques, and I am using it to encode an image. For this, I am using the built-in function of Matlab that also gives other values such ...
345 views
### File conversion to binary format to save storage space
Our test log files are stored in .asc format. Each file contains around 5000 rows of these Logging Strings as shown in the image. I would like to reduce the size of these files.They are around 4-5 MBs ...
158 views
### Epsilon balanced Code
A linear code is termed as an $\epsilon -$balanced code if all the codewords are having fractional hamming weight $\in (1/2-\epsilon,1/2+\epsilon)$. I want to show that for every $\epsilon\in (0,1/2)$,...
• 246
199 views
I have $m$ equations of the following form: $$x_1+x_2+\cdots+x_n=s,$$ where each variable is either 1 or 0, and the total number of variables is $m\approx3{,}000$. So I’m thinking of modeling each ...
• 960
29 views
### LDPC:How to detect total number of errors in a codeword using minimum distance?
Currently, i working with LDPC for FPGA implementation. I have formed H matrix of size 102X204 (N=204,K=102,M=102,R=0.5). Doubt in Error detection and correction for bigger matrix. I know that code'...
• 101
44 views
• 191 | 1,644 | 6,815 | {"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} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.902265 |
https://blogs.thebitx.com/index.php/2021/11/03/python-__ipow__-magic-method-finxter/ | 1,643,095,306,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304798.1/warc/CC-MAIN-20220125070039-20220125100039-00251.warc.gz | 195,102,999 | 43,170 | # Python __ipow__() Magic Method | Finxter
0
37
## Syntax
`object.__ipow__(self, other)`
The Python `__ipow__()` magic method implements the in-place exponentiation `x **= y` that calculates the result of the power function `x ** y`, and assignsit to the first operands’ variable `x`. This type of in-place operation is also called augmented arithmetic assignment. The method simply returns the new value to be assigned to the first operand.
• When you call `x **= y`, Python first attempts to call `x.__ipow__(y)`.
• If this is not implemented, it tries the normal exponentiation operation `x.__pow__(y)`.
• If this is not implemented either, it tries reverse exponentiation operation `y.__rpow__(x)` with swapped operands.
The result is then assigned to the first operand `x`. If none of those operations is implemented, Python raises a `TypeError`.
We call this a “Dunder Method” for Double Underscore Method” (also called “magic method”). To get a list of all dunder methods with explanation, check out our dunder cheat sheet article on this blog.
## Basic Example Overriding __ipow__
In the following code example, you create a class `Data` and define the magic method `__ipow__(self, other)`.
• The “self” argument is the default argument of each method and it refers to the object on which it is called—in our case, the first operand of the in-place operation.
• The “other” argument of the in-place method refers to the second operand, i.e., `y` in the in-place operation `x **= y`.
The return value of the operation returns a dummy string `'finxter 42'` to be assigned to the first operand. In practice, this would be the result of the in-place exponentiation operation.
```class Data:
def __ipow__(self, other):
return 'finxter 42'
x = Data()
y = Data()
x **= y
print(x)
# finxter 42
```
## In-Place Exponentiation **= without __ipow__()
To support the in-place power function on a custom class, you don’t have to overwrite the `__ipow__()` method. Because if the method is not defined, Python will fall back to the normal `__pow__()` method and assign its result to the first operand.
Here’s an example:
```class Data:
def __pow__(self, other):
return 'finxter 42'
x = Data()
y = Data()
x **= y
print(x)
# finxter 42
```
Even though the `__ipow__()` method is not defined, the in-place exponentiation operation `x **= y` still works due to the `__pow__()` “fallback” magic method!
## In-Place Exponentiation **= without __ipow__() and __pow__()
To support in-place exponentiation `x **= y` on a custom class, you don’t even have to overwrite any of the `x.__ipow__(y)` or `x.__pow__(y)` methods. If both are not defined, Python falls back to the reverse `y.__rpow__(x)` method and assigns its result to the first operand.
Here’s an example where you create a custom class for the first operand that doesn’t support the exponentiation operation. Then you define a custom class for the second operand that defines the `__rpow__()` method. For the in-place operation, Python falls back to the `__rpow__()` method defined on the second operand and assigns it to the first operand `x`:
```class Data_1:
pass
class Data_2:
def __rpow__(self, other):
return 'finxter 42'
x = Data_1()
y = Data_2()
x **= y
print(x)
# finxter 42
```
## TypeError: unsupported operand type(s) for **=
If you try to perform in-place exponentiation `x **= y` but neither `x.__ipow__(y)`, nor `x.__pow__(y)`, nor `y.__rpow(x)` is defined, Python raises a “`TypeError: unsupported operand type(s) for **="`. To fix this error, simply define any of those methods before performing the in-place operation.
```class Data:
pass # ... you should define __ipow__ here to prevent error! ... #
x = Data()
y = Data()
x **= y```
Output:
```Traceback (most recent call last):
File "C:UsersxcentDesktopcode.py", line 8, in <module>
x **= y
TypeError: unsupported operand type(s) for ** or pow(): 'Data' and 'Data'```
References:
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Join the free webinar now! | 1,213 | 4,843 | {"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} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.730176 |
https://studylib.net/doc/18631271/620-ans.-ans.-aa-%3D-%7B-3.00-i---1.75-j%7D-ft%3Es2-aa-%3D--1i---1.. | 1,560,684,052,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00064.warc.gz | 624,208,681 | 41,650 | # 620 Ans. Ans. aA = {-3.00 i + 1.75 j} ft>s2 aA = -1i - 1.25 j
```91962_06_s16_p0513-0640
6/8/09
3:00 PM
Page 620
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
z
16–139. The man stands on the platform at O and runs out
toward the edge such that when he is at A, y = 5 ft, his mass
center has a velocity of 2 ft>s and an acceleration of 3 ft>s2,
both measured relative to the platform and directed along
the positive y axis. If the platform has the angular motions
shown, determine the velocity and acceleration of his mass
center at this instant.
v ⫽ 0.5 rad/s
a ⫽ 0.2 rad/s2
O
y ⫽ 5 ft
vA = vO + Æ * rA>O + (vA>O)xyz
x
vA = 0 + (0.5k) * (5j) + 2j
vA = {-2.50i + 2.00j} ft>s
#
aA = aO + Æ * rA>O + Æ * (Æ * rA>O) + 2Æ * (vA>O)xyz + (aA>O)xyz
Ans.
aA = 0 + (0.2k) * (5j) + (0.5k) * (0.5k * 5j) + 2(0.5k) * (2j) + 3j
aA = -1i - 1.25j - 2i + 3j
aA = {-3.00i + 1.75j} ft>s2
Ans.
620
A
y
```
##### Related flashcards
Number theory
27 Cards
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14 Cards | 484 | 1,200 | {"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} | 3.3125 | 3 | CC-MAIN-2019-26 | longest | en | 0.853633 |
http://www.jiskha.com/display.cgi?id=1233891122 | 1,495,934,182,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00154.warc.gz | 678,640,162 | 3,783 | # math 115
posted by on .
A state sales tax rate is 6.5%. If the tax on a purchase was \$16.51, what was the price of the purchase.
6.5%/\$16.51=\$254.00
Check this out. Help me !
• math 115 - ,
You're right.
• math 115 - ,
you are right
6.5 % = \$16.51
100 % = x
16.51 x 1oo= 1651 / 6.5 = 254
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# 78 now consider a conservative force field in one
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Now consider a conservative force field in one dimension described by a potential V ( x ) so that F = - V ( x ). Then ˙ x = y, ˙ y = - 1 m V ( x ) . Equilibrium points occur where V ( x 0 ) = 0 and y 0 = 0 (so they are all on the x -axis). At such a point, J = 0 1 - V ( x 0 ) /m 0 , so T = 0 and Δ = V ( x 0 ) /m . Hence, using the summary table in the previous section, all equilibrium points must be either saddles (if V ( x 0 ) < 0) or centres (if V ( x 0 ) > 0). This analysis agrees with the stability analysis of § 4.2, with the following phase plane diagrams locally (i.e., close to the equilibrium point): We also have the energy equation 1 2 my 2 + V ( x ) = E. This equation, for different values of the constant E , defines the trajectory curves in the ( x, y )-plane. Trajectories for this system are therefore symmetric in y (which enables us to draw the diagrams above, knowing that the directions given by the eigenvectors must also be reflectionally symmetric in the x -axis). On a given trajectory, at any value of x there are just two values of y , one positive and one negative. If a trajectory is bounded then it must have y = 0 at each end, at which points it joins up. Trajectories for a conservative system must therefore be either closed (i.e., they come back to where they started) or unbounded . Example: the Duffing oscillator for a particle of unit mass, ¨ x = x - x 3 . This corresponds to V ( x ) = - 1 2 x 2 + 1 4 x 4 . We will have a saddle at x = 0 and centres at x = ± 1. The phase portrait is as follows: 79
Example: a simple pendulum of length l (as in § 2.5) has ¨ θ = - g l sin θ. Let y = ˙ θ , so that ˙ θ = y, ˙ y = - g l sin θ. The stable equilibria are at θ = 2 and the unstable ones at θ = (2 n + 1) π ( n Z ), by considering J = 0 1 - ( g/l ) cos θ 0 . The phase diagram has this form: Around stable equilibria, the pendulum oscillates back and forth (e.g., curve A ). This motion is known as libration . If the pendulum has sufficiently large energy then it can instead undergo rotation (e.g., curve C ) where it always has the same sign of ˙ θ . The curves which join the saddle points (e.g., B ) are known as separatrices because they separate the phase plane into regions containing these two different kinds of motion. They correspond physically to the (highly unlikely) motion where the pendulum starts vertically upwards then executes precisely one revolution, ending vertically upwards again. 80
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11.4 Damped Systems Consider a simple pendulum with damping: ¨ θ = - g l sin θ - k ˙ θ (11.3) where k is a small positive constant. Letting y = ˙ θ we have ˙ θ = y, ˙ y = - g l sin θ - ky. The equilibrium points are still where sin θ = 0; we have J = 0 1 - ( g/l ) cos θ - k . At θ = 2 , T = - k and Δ = g/l , corresponding to a stable focus (from the table in § 11.2, assuming that k is small enough that k 2 < 4 g/l ). At θ = (2 n + 1) π , T = - k and Δ = - g/l so we have a saddle. Because this system is not conservative (the force depends on ˙ θ as well as θ , prohibiting the existence of a potential V ( θ )), the solution curves are not symmetric in y and are not closed. The phase portrait is as follows: Note that if we define energy in the same way as for an undamped pendulum, E = 1 2 ml 2 ˙ θ 2 - mgl cos θ, then d E d t =
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Jill Tulane University ‘16, Course Hero Intern | 1,250 | 4,668 | {"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} | 3.921875 | 4 | CC-MAIN-2018-47 | latest | en | 0.883578 |
http://slidegur.com/doc/33321/7--direction---activity | 1,521,354,518,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645538.8/warc/CC-MAIN-20180318052202-20180318072202-00296.warc.gz | 264,380,741 | 8,228 | ### 7- DIRECTION - Activity
```DIRECTION ON A MAP
1. On the compass rose below label the 4 cardinal points, the 12 ordinal points, and
the appropriate degrees for each point.
2. Answer the following question using the compass rose.
a) 90 from N is:
e) between N and NW is:
b) 45 from E is:
f) between S and NE is:
c) 180 from NE is:
g) between NE and SW is:
d) 45 from SW is:
h) between SW and NE is:
3. State the compass direction midway between the following points.
a) N and S:
d) SSW and WSW:
b) S and N:
e) N and SW:
c) SW and NW:
f) WSW and NNW:
4. Name the opposite direction for each of the following compass points
a) Southwest:
d) West:
b) North northwest:
e) North northeast:
c) East southeast:
f) South:
5. In what direction are you traveling when flying from:
a) Toronto to Montreal:
b) Vancouver to Victoria:
f) Regina to Winnipeg:
c) Victoria to Iqaluit:
g) St. John’s to Halifax:
d) Yellowknife to Victoria:
h) Sudbury to Thunder Bay:
6. Describe the location of each city within its province following the example given (ie.
VANCOUVER: Vancouver is located in the southwest corner of British Columbia).
a) Winnipeg:
b) Thunder Bay:
c) Edmonton:
d) St. John’s:
e) Regina:
7. Describe the location of each city using nearby features as a reference (ie. TORONTO:
Toronto is located on the northwest shore of Lake Ontario).
a) Hamilton:
b) Calgary:
c) Halifax:
d) Quebec:
e) Yellowknife:
f) Georgetown:
DIRECTION ON A MAP - answers
1. On the compass rose below label the 4 cardinal points, the 12 ordinal points, and
the appropriate degrees for each point.
2. Answer the following question using the compass rose.
a) 90 from N is: E
e) between N and NW is: NNE, NE,
ENE, E, ESE, SE, SSE, S, SSW, SW,
WSW, W, WNE
b) 45 from E is: SE
f) between S and NE is: SSW, SW,
WSW, W, WNW, NW, NNW, N, NNE
c) 180 from NE is: SW
g) between NE and SW is: ENE, E, ESE,
SE, SSE, S, SSW
d) 45 from SW is: W
h) between SW and NE is: WSW, W,
WNW, NW, NNW, N, NNE
3. State the compass direction midway between the following points.
a) N and S: E
d) SSW and WSW: SW
b) S and N: W
e) N and SW: ESE
c) SW and NW: W
f) WSW and NNW: WNW
4. Name the opposite direction for each of the following compass points
a) Southwest: NE
d) West: E
b) North northwest: SSE
e) North northeast: SSW
c) East southeast: WNW
f) South: N
5. In what direction are you traveling when flying from:
a) Toronto to Montreal: NE
b) Vancouver to Victoria: SSW
f) Regina to Winnipeg: E
c) Victoria to Iqaluit: NE
g) St. John’s to Halifax: SW
d) Yellowknife to Victoria: SW
h) Sudbury to Thunder Bay: WNW
6. Describe the location of each city within its province following the example given (ie.
VANCOUVER: Vancouver is located in the southwest corner of British Columbia).
a) Winnipeg: SSE corner of Manitoba, NNE of Brandon
b) Thunder Bay: N of Lake Superior, SW Ontario, NW of Toronto
c) Edmonton: E Alberta (Central)
d) St. John’s: W coast of Newfoundland | 888 | 2,920 | {"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} | 3.0625 | 3 | CC-MAIN-2018-13 | latest | en | 0.880475 |
http://commoncoregeometry.blogspot.com/2018/02/high-school-geometry-performance-task.html | 1,547,691,386,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00036.warc.gz | 51,142,249 | 17,010 | ## Tuesday, February 27, 2018
### High School -- Geometry Performance Task: Properties of Quadrilaterals (Day 117)
Today I subbed in another math class -- at a different high school from yesterday. This time I subbed in a Geometry class. I definitely want to do the "Day in the Life" for today, since this is, after all, a Geometry blog.
And so here is the "Day in the Life" for today -- Day 109 in this district:
7:00 -- Rise and shine -- this is yet another "first period" (zero period) class.
The Geometry students are still working on similarity -- Lesson 7-3 of the Glencoe text. In fact, this is what I wrote three years ago about today's lesson (from a tutor's perspective):
Last night I tutored my geometry student again. Section 7-3 of the Glencoe text is on similar triangles, including all three similarity statements: AA, SAS, and SSS.
Well, my student is still slightly confused with solving and setting up proportions. But he was more concerned with a question from Section 7-2 of the Glencoe text, on similar polygons. There is a group of similarity problems in the Glencoe text that are flawed.
Three years ago I ended up writing more about Lesson 7-2 than 7-3. But unfortunately, one of the questions on today's Lesson 7-3 worksheet also contains a (different) flaw, as I'll soon find out.
Most of the first few problems have the students calculate missing lengths in similar triangles. Notice that since one of the side lengths is missing, SSS Similarity goes right out the door. Thus all of the problems on the first side of the worksheet use either AA~ or SAS~.
7:55 -- First period leaves and second period begins. They begin the Lesson 7-3 worksheet.
8:50 -- Second period leaves and third period begins. They begin the Lesson 7-3 worksheet.
9:45 -- This is the same school I subbed at a week ago today -- so now it's tutorial time. Several students arrive for tutorial, but many of them are working on subjects other than math. The two Geometry students I see have a different teacher and are working in a different lesson. I do help these two briefly.
10:25 -- Tutorial ends -- and believe it or not, so does my school day! The teacher I'm subbing for isn't merely a coach -- he's the athletic director. And so in order to allow him to organize and plan all the school sports, he's given only a three-class load that ends at nutrition time.
So this is the shortest "Day in the Life" I've ever done. But I still have plenty to say, about both the curriculum and my New Year's Resolutions.
Yesterday I covered the first resolution, but my only chance to follow it was during sixth period. As this is such an important resolution, let's continue to focus on Resolution #1 for today's half-day:
1. Implement a classroom management system based on how students actually think.
It's often tricky to implement classroom management when students are working on different assignments or finishing a multi-day assignment. Students can always claim "I'm done, so let me do whatever I want."
Today, on the other hand, everyone is working on the Lesson 7-3 worksheet. This means that I can manage the classroom strongly -- just make sure that everyone is working on it.
When first period begins, I do problem #1 on the board. Later on, I do #2 as well. I'm hoping that the students can finish at least the first side of the paper. But many students fail to finish -- and those who do the work skip questions or jump to the other side. I end up having to leave the names of two students who do nothing other than copy #1 and #2 from the board.
In second period, I leave Questions #1 and #2 on the board and start on Question #3. Once I try the problem, it's no longer any wonder why students keep skipping it -- the question is flawed.
Here is the Question #3 from the Glencoe Lesson 7-3 worksheet:
3. Identify the similar triangles. Then find the measure of QR.
In Triangle QSV, point R is between Q and S, while T is between V and S. The following lengths are given on the diagram:
QS = VS = 36
QV = 36sqrt(2)
TS = 24
QR = x
I tell my students another story from my tutoring days, which I've also chronicled on the blog:
I do admit that students might be tricked by this sort of question. I remember once when I was teaching or tutoring a student who was in the similarity chapter of the text. Upon seeing the two triangles in a question much like this one, he immediately started writing a proportion. I told him that a proportion can be set up only if the triangles are similar, and so I asked him, how did he know that the triangles are similar? His response was, of course the triangles were similar because we were in Chapter 7, the similarity chapter of the text!
Yes, my student cleverly figured out that in the similarity chapter, nearly every problem would have a pair of similar triangles. But this problem illustrates what's wrong with his thinking
And here's the problem on the worksheet with today's students -- the given info isn't sufficient to conclude that the triangles are similar!
If you don't believe me, then try to prove that Triangle QSV ~ RST. We've already eliminated SSS~ as a possibility for any question on this worksheet, but what about SAS~? The two triangles have Angle S in common, but one of the adjacent sides, RS, has length 36 - x. Since the unknown value that we;re solving for is part of one of the sides, this rules out SAS~ as a possibility as well.
This leaves only AA~. But no angle information is given, so we cannot conclude that Angles Q and TRS are congruent, nor Angles V and RTS. Notice that in the next three questions, we're given that two of the sides are parallel (which would give us congruent corresponding angles), but no parallel lines are stated in this question.
Notice that if we could conclude that Triangle QSV ~ RST, then x = 12. But what's to stop us from choosing another point on QS, calling it R', and then claiming that QR' = 12 instead of QR?
Ironically, here are the instructions for the first four questions on the other side of the worksheet:
"Determine whether each pair of triangles is similar. If so, write a similarity statement. If not, what would be sufficient to prove the triangles similar? Explain your reasoning."
And doubly ironically, all four pairs of triangles have enough info to conclude they're similar. (There are two SAS~ questions and one each of SSS~ and AA~.) The only pair of triangles not to contain sufficient info is on the wrong side of the worksheet as #3.
Well, let's answer this question for the wrong #3. Here is what we need to conclude that Triangles QSV and RST are indeed similar. Any one of the following is sufficient:
• Angle Q = TRS
• Angle V = RTS
• QV | | RT
• RT = 24sqrt(2)
This last one needs some explanation. The side lengths of QSV are 36, 36, 36sqrt(2). This should be recognizable as the sides of a 45-45-90 triangle, with the right angle at S. Then knowing RT will allow us to conclude that the triangles are similar by HL. For some strange reason, HL Similarity never appears in any texts, yet it's obvious that HL~ should work. The same proof for the other similarity theorems works for HL~ as well.
There's one more sufficient statement that will allow us to conclude similarity -- RS = 24, then the triangles are similar by SAS~. But if we had RS, then we could find x = QR by the Betweenness Theorem (Segment Addition) as QR = QS - RS, without using similarity at all.
In second period, I end up spending so much time on Question #3 that I rush through the final problem on the front of the worksheet -- Question #7, a word problem on shadow lengths. Students, of course, try to avoid word problems, and so many don't finish Question #7. (Some students also try to avoid Question #3 because of the square root, which isn't the real problem with it.)
But returning to classroom management, I do catch one student who doesn't even have his worksheet on his desk. I can't catch his name, so I must return to identifying him by backpack. Since it's still wintertime, he's wearing a coat, which I also use for identification. Students wear different clothes everyday, but they often wear the same jacket -- and of course, they don't change their backpack.
In third period, I do Questions #1, #2, and then #4-7 on the board, saving a discussion about Question #3 for last. Students still have some trouble with the word problem in #7, but this time I'm able to get them through it. I don't need to write any names.
During tutorial, another management issue turns up. This is one that has come up with several other teachers I've spoken to during breaks. A student texts someone on a cell phone, then claims that the recipient is her mother about to undergo surgery. Indeed, this is what the other sub I saw in the lounge had to kick out a student for yesterday. That student yelled at the sub that some things, like life and death, are more important than school, and that exceptions to the no phone rule should always, always, be granted during emergencies, no questions asked.
From my perspective, this is a blatant attempt to neuter the no phone rule. Students who claim that they are texting a parent are almost always texting another teenager, and the topic isn't an emergency, but entertainment. Their idea is that since teachers can never be certain that a text isn't an emergency, the only 100% fair thing to do is to let students text whenever they want to whomever they want, even if it interferes with their education. To them, letting kids do what they want is the only fair thing, and everything else is unfair. To them, if you want to be considered a "good" or "fair" teacher, it's very, very important to let them have fun, to the exclusion of everything else.
And returning to the girl in my classroom, the second time she takes out her phone, she and a group of girls started laughing -- odd behavior for someone whose mom is undergoing surgery. This clinches it, of course -- she's texting another teen to discuss entertainment. The problem, of course, is that this is tutorial, so it's impossible to get any identifying information for a student who might not even be in any of his classes. Of course, she claims that she's finished all her written work. In other words, to her, the only fair thing to do is let her have a free 40-minute texting and entertainment period everyday (since she can claim she has no work everyday) -- nothing less is fair.
Last year at my middle school, only three students claimed that emergencies should override the no phone rule. Two of the claims turned out to be genuine -- one was the special scholar (January 6th post) whose mother was making arrangements to pick up her elementary school sister, and the other had a death in her family. (She and her younger brother were eventually picked up from school.) To this day, I still believe the third student was lying about the emergency.
If I were a parent, I would never text my children during the school day, not even if a death in the family happens. If a relative were to die one minute after school starts, I would wait seven full hours, until one minute after the final dismissal bell, to text my children. I'd be too shocked by the news even to consider texting anyone. And besides, I'd assume that my kids would have their phones confiscated if they were to text -- the last thing they need is to lose a relative and a phone the same day. If it's absolutely necessary to contact them, I'd make every effort to do so without texting them -- including contacting or driving up to the school. And if I can't find any way to contact them without a text, I'd tell them not to respond until after the final dismissal bell. And if I were a student, I might not even want to text my parents anyway! (Note: with shootings in the news, here I assume that if an active shooter arrives at my kids' school, all classes are cancelled and students can safely text without phone confiscation.)
OK, that's enough about classroom management for today. I do wish to write a little more about teaching the Glencoe text. If I were a regular teacher with the Glencoe text, how would I pace it.
Since the Glencoe text has a Chapter 0, it's tempting to follow the same digit pattern that I've established for the U of Chicago text. But this is a little too fast -- it forces Chapter 7 on similarity into the first semester. I like the idea of starting the second semester with similarity -- and apparently this is exactly what's done in this district.
Of course, to be still in Chapter 7 five weeks into the semester is a bit slow, especially if the goal is to reach Chapter 13 by the end of the year. A better pace would be to start the Big March with Chapter 8, on trigonometry. (That's right -- trig at the Big March again!) Apparently, the Geometry students I see in tutorial are already in Chapter 8. They are studying special right triangles -- and they are able to recognize the 36-36-36sqrt(2) still written on the board as a special 45-45-90 triangle.
In the Glencoe text, Chapter 4, on congruence, appears before Chapter 9, on transformations. I find an old copy of the first semester final in the classroom, and apparently, Chapter 9 is pushed up into the first semester. This fits the Common Core definition of congruence. And it also appears that Lesson 9-6 is combined with Chapter 7. This makes sense as Lesson 9-6 is on dilations.
I also find a copy of a Performance Task, similar to what students may find on the SBAC. There's a hole in my U of Chicago pacing plan since Chapter 11 has only six sections. And I've never posted a Performance Task before, despite this being a Common Core blog.
And so I post this task as an activity for today and tomorrow. Actually, I create my own version of the problem rather than the district copy. This is to block out the name of the district since I don't post identifying information (and to avoid any issues with SBAC, which might mistake this practice question for a real test question). Also, I changed it from one day to two days. On the actual SBAC, students are typically given a two-hour block to complete the Performance Task, so they should have two days to complete it.
The question is about the coordinates of a square. It fits perfectly with Chapter 11 of the U of Chicago text -- and indeed it's similar to the Lesson 11-1 activity from two weeks ago.
In the Glencoe text, the Performance Task fits with Chapter 6, on quadrilaterals. Glencoe Chapter 6 is similar to Chapter 5 in the U of Chicago text (and Chapter 6 in the Third Edition), except that coordinates appear early. Indeed, I lamented three years ago that the Distance Formula appears as early as Chapter 1 in the Glencoe text! | 3,375 | 14,733 | {"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} | 3.453125 | 3 | CC-MAIN-2019-04 | longest | en | 0.950044 |
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# Should zero be considered a number?
Question #51985. Asked by Buck540. (Oct 27 04 10:55 AM)
Stew54
I think it depends what you mean by "number". If you are just talking about the basic numbers used for counting things then possibly it isn't, because you never start counting a group of things from zero. For any other defintion of "number" for mathematical purposes though, zero has as many functions as any other and so it is.
[Oct 27 04 2:18 PM] Stew54 writes:
It is, for example, the number which when added to x leaves x unchanged, as 1 is the number which when multiplied by x leaves x unchanged.
Oct 27 04, 11:24 AM
Legolaschic
Mathematically speaking zero is not a member of the set of counting/natural numbers but it is a member of the sets: whole numbers, integers, rational numbers, real numbers, and complex numbers.
Oct 27 04, 1:47 PM
kaylofgorons
If zero isn't a number, what goes between 1 and -1? Another mathematic purpose for it is anything to the power of zero equals 1. I'm with Legolaschic, it's an integer and belongs with the rest of those sets. By the way, I've heard that the idea of zero is a sign of intellectual achievement of ancient cultures.
Oct 27 04, 4:36 PM
peasypod
In my opinion zero is the symbol representing the absence of any magnitude. It is also the cardinality of an empty set. I suppose it could be called a number whose sum with any other number is that other number....
[Oct 27 04 10:15 PM] peasypod writes:
Another interesting tidbit....
In terms of arithemtic, zero is the identity of addition (any number a has the property a + 0 = a) in the same way that 1 is the identity of multiplication. Hence zero is a very necessary number.
Oct 27 04, 5:19 PM
gmackematix
As for zero, it is a number that has some quirky properties but quite a few numbers do. It has as much right to its place on the number line as -17 or pi.
If you are going to say zero doesn't belong because adding it to a number doesn't change that number, then we could argue one isn't a number because it doesn't change things it multiplies. Zero and one are known as the identities of the functions of addition and multipliation respectively.
Basically, a number is a property we ascribe to sets of objects. With a set of sheep, say, we count them using what are known as the natural numbers. By saying there are 3 sheep we are using a symbol, 3 (called a numeral), to indicate that this set can be paired off one-to-one with any other set with the property we call 3 (such as corners of a triangle). Any set that can be matched off with another set in this way is said to have the same cardinality. Clearly by removing one object from from each of two sets with the same cardinality they will still have the same cardinality (in our sheep example it will drop from 3 to 2). If we keep removing an object from sets with the same cardnality we will end up with empty sets. These have the cardinality zero (represented by 0). This is why zero is often given a pride of place at the start of the counting or natural numbers.
If the set is something like pounds in debt or credit then we need to include negative numbers and we have the set of whole numbers or integers with zero in the middle. To describe yet more complcated sets such as points on a line we need another sort of number called a real number, but yes, zero is included in that set of numbers too.
Oct 27 04, 7:18 PM
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Anonymous
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does 0.9999999.......... equal 1
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Voting Style: Open Point System: 7 Point Started: 7/14/2018 Category: Philosophy Updated: 3 years ago Status: Debating Period Viewed: 772 times Debate No: 116576
Debate Rounds (4)
20 comments have been posted on this debate. Showing 1 through 10 records.
Posted by A1DS 3 years ago
No it's not because 0.999... isn't infinity, it's a real number (see previous comment). The key distinction is the infinite repeating digits are after the decimal place instead of before it, and therefore 0.999... doesn't fit the definition of infinity as I can list a number (e.g. 2) which is larger than 0.999... so it's an entirely different scenario. In the case of infinity, it is impossible to list a number of greater value, so clearly 0.999... represents a different scenario.
Posted by DeletedUser 3 years ago
But the thing is your equation was doing the exact same thing
Posted by A1DS 3 years ago
Which, by definition makes it infinity, so you cannot manipulate it with the algebra you have been trying to use. You'll never see a proper, valid mathematical proof using ...X.Y because it's not essential, any such "number" (and I use that term loosely because infinity is a more of a concept than a number) can be represented as infinity. Your algebra isn't actually mathematically valid.
Posted by DeletedUser 3 years ago
I didn't mean "add an infinite string of 9's to it". I meant right out the rest of the number which happens in this case to be an infinite string of 9's.
Posted by A1DS 3 years ago
It may help you to look into what constitutes, real, rational and irrational numbers. Rational numbers is the set of numbers that can be represented by an integer fraction. 0.999... falls into this category as it can be represented by 9/9 (which is equal to 1 funnily enough). Irrational numbers cannot be represented this way, and include examples such as pi and the square root of 2. Both of these can include examples of numbers with an infinite number of decimals places. Such as 0.333... and 0.999... from the rational numbers and both of the aforementioned examples for the irrationals. Together, the rational and irrational numbers form a set that we refer to as the real numbers. You can google the mathematical axioms on which the reals are defined if you'd like to learn more, but they get quite complicated quite quickly.
Posted by A1DS 3 years ago
Reflectively, I think where your problem is actually coming from Masterful is you don't know what a real number is. Something with an infinite number of decimals is still a real number. Pi is a real number, the square root of 2 is a real number, just because decimals go on infinitely doesn't mean these numbers behave like infinity. So while you've correctly identified that infinity is a concept rather than a number, you're incorrectly applying that same assumption to real numbers.
Hope this clears things up for you.
Posted by A1DS 3 years ago
Limit theory* sorry I ran out of characters.
Posted by A1DS 3 years ago
Okay so first I'm going to address what jackgilbert had to say:
You can't just "add an infinite string of nines to the left" it's not mathematically valid. Any such number with an infinite string of non-zero digits "to the left" is just represented as infinity. Your maths doesn't hold up because you fail to realise that infinity is a concept, not a number, so the algebraic techniques you're attempting to apply to it don't work. I'll wager the person who made that video actually doesn't have a particularly good understanding of mathematics because what he's doing is wrong.
Both examples are exactly the same as both are just equal to infinity. In neither case is your algebra consistent because in neither case are you correctly applying algebraic principles to actual 'numbers', you can consider it like trying to multiply "limit" by 10, this is obviously nonsensical because you cannot multiply a concept by 10.
0.999... isn't infinity? It does not behave like infinity and is a real number. I can say 0.999... is less than 2 and more than 0.9. I can use it in equations, I can apply a variety of completely valid operations to it precisely because it is a real number. I'm not entirely sure what your argument is and it seems largely nonsensical to me. First of all "this truth" does not create horrors for those who believe 0.999...=1, quite conversely it is essential for proving it, and I'd suggest entirely contradictory to your statement, the vast majority of people who recognise 0.999...=1 will do so on the basis of a firm understanding of mathematics, including this concept of infinity. I think you misunderstand what a real number is? Real numbers include basically any number that isn't the square root of a negative (the imaginary numbers). So the irrational numbers (like pi), integers etc. are all considered real numbers.
Hope this has cleared things up, but I think the best thing for you guys is to at limi
Posted by DeletedUser 3 years ago
x=........99999999.9999999..........
Multiply both sides by ten we get
10x=........99999999.9999999......
We get exactly the same number as before because infinity goes on forever in both ways. This proves that if you multiply infinity by 10 it does nothing
10x=x
9x=0
x=0
The funny thing is it is actually consistent with the other equations because the algebra is telling me .......99999999 is -1 and 0.999999..... is 1 and .......999999999.9999999....... is 0 so 1+-1=0.
Posted by DeletedUser 3 years ago
the original x is infinitely many 9's going off to the left. Multiplying it by ten takes the last 9 away and replaces it with 0. Infinity times 10 is still infinity. Let me right out an equation for you
x=..........9999999999
Multiply both sides by 10
10x=........99999999990
But look at ....9999990. It is exactly the same as the original x except minus the final 9.
10x=x-9
9x=-9
x=-1
It is exactly the same thing you did with your equation.
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1. Population
What is the population of the city with 3% annual growth, if in 10 years the city will have 60,000 residents?
2. Geometric sequence 5
About members of geometric sequence we know: ? ? Calculate a1 (first member) and q (common ratio or q-coefficient)
3. Family parcels
In father will he divided the land so that the older son had three bigger part than younger son. Later elder son gave 2.5 ha field to younger and they had both the same. Determine the area of family parcel.
4. MO Z6-6-1
Write integers greater than 1 to the blanks in the following figure, so that each darker box was product of the numbers in the neighboring lighter boxes. What number is in the middle box?
5. Pyramid Z8–I–6
Each brick of pyramid contains one number. Whenever possible, the number in each brick is lowest common multiple of two numbers of bricks lying directly above it. That number may be in the lowest brick? Determine all possibilities.
6. Diagonals of diamond
Find the area and circumference of the diamond ABCD with 15m and 11m diagonals.
7. Cost reduction
Two MP3 players whose price was equal to originally have been discounted the first by 20%, the second by 35%. After the price reduction was the difference in their prices 750, - CZK. What was the original price of each of the two players?
8. How old
The student who asked how many years he answered: "After 10 years I will be twice as old than as I was four years ago. How old is student?
9. Two numbers
The sum of the two numbers is 1. Find both numbers if you know that half of the first is equal to one seventh of the second number.
10. Cuboid edges in ratio
Cuboid edges lengths are in ratio 2:4:6. Calculate their lengths if you know that the cuboid volume is 24576 cm3.
5 of the same bread has the same weight as three bread and 4 kg of fruit. What weight has one bread?
12. Rabbits 3
Viju has 40 chickens and rabbits. If in all there are 90 legs. How many rabbits are there with Viju?
13. Ruler
How far from Peter stands 2m hight John? Petr is looking to John over ruler that keeps at arm's distant 60 cm from the eye and on the ruler John measured the height of 15 mm.
14. Two numbers
We have two numbers. Their sum is 140. One-fifth of the first number is equal to half the second number. Determine those unknown numbers.
15. Land - isosceles trapezoid
Calculate the content and perimeter of the building plot in the form of an isosceles trapezoid with bases 120m, 95m and height 50m.
16. Unknown number 5
Daniel think an integer. When he change this number at a ratio of 2:5 he got number 2.8. Determine what number think Daniel.
17. Tree
Between points A and B is 50m. From A we see a tree at an angle 18°. From point B we see the tree in three times bigger angle. How tall is a tree?
18. Glass
Trader ordered from the manufacturer 200 cut glass. The manufacturer confirmed the order that the glass in boxes sent a kit containing either four or six glasses. Total sent 41 boxes. a) How many boxes will contain only 4 glasses? b) How many boxes will co
19. VCP equation
Solve the following equation with variations, combinations and permutations: 4 V(2,x)-3 C(2,x+ 1) - x P(2) = 0
20. Factory
In the factory workers work in three shifts. In the first inning operates half of all employees in the second inning and a third in the third inning 200 employees. How many employees work at the factory?
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posted by .
If a negative charge is initially at rest in an electric field, will it move towards a region of higher potential or lower potential? What about a positive charge? How does the potential energy of the charge change in each case?
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• Dec 6th 2010, 11:02 PM
tintin2006
Lagrange Multipliers
Find the dimensions of the closed right circular cylindrical can of smallest surface area whose volume is $\displaystyle 16pi\;cm^3$.
I managed to find r = 2cm, h = 4cm which gives the surface area of $\displaystyle 24pi \,cm^2$ using the method $\displaystyle \nabla f=\lambda \nabla g \; and \; g(x,y,z)=0$.
My question is that, how come the final result of r and h is min and not max. What if the question ask for the largest surface area in this case?
• Dec 6th 2010, 11:52 PM
FernandoRevilla
You needn't Lagrange multipliers:
$\displaystyle \pi r^2h=16\Leftrightarrow h=16/\pi r^2$
now, substitute $\displaystyle h$ in $\displaystyle S$.
Regards.
Fernando Revilla
• Dec 7th 2010, 07:23 PM
tintin2006
Thanks but the chapter is Lagrange multipliers so... and what's that S you're talking about?
• Dec 7th 2010, 08:35 PM
Soroban
Hello, tintin2006!
Quote:
$\displaystyle \text{Find the dimensions of the closed right circular cylindrical can}$
$\displaystyle \text{of smallest surface area whose volume is }16\pi\text{ cm}^3$.
$\displaystyle \text{My question: how come the final result of }r\text{ and }h\text{ is min and not max?}$
$\displaystyle \text{What if the question asked for the }largest\text{ surface area?}$
Consider a lump of stuff about the size of a Campbell's soup can.
. . And let's say it has a volume of $\displaystyle 16\pi$ cm$\displaystyle ^2.$
We can measure the surface area of that cylinder.
Now take a rolling pin and flatten the stuff like a pizza.
We will still have a cylinder.
. . It will have a large circular top and bottom . . . and a very small height.
You can "see" that it has a much larger surface area.
In theory, we can continue to flatten it and make a larger and larger pizza.
It may be only one molecule thick, but its top and bottom may cover an acre.
Flatten it even more until its thickness is that of an atom.
By then the pizza dough may cover, say, Canada.
My point is that there is no maximum surface area for the cylinder. | 593 | 2,065 | {"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} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.858544 |
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Problem solving is a very useful skill for every learner like you. With thorough understanding and constant practice, this would challenge you to think logically and eventually find it an enjoyable activity.
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What is 6 wholes and 7 over 32 as a decimal?
Updated: 4/28/2022
Wiki User
12y ago
The number 6. 7/32 can be written in decimal as 6.21875.
Wiki User
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Research Article | Open Access
Volume 2018 |Article ID 7289092 | https://doi.org/10.1155/2018/7289092
François Dubeau, Calvin Gnang, "Fixed Point and Newton’s Methods in the Complex Plane", Journal of Complex Analysis, vol. 2018, Article ID 7289092, 11 pages, 2018. https://doi.org/10.1155/2018/7289092
# Fixed Point and Newton’s Methods in the Complex Plane
Accepted12 Dec 2017
Published29 Jan 2018
#### Abstract
We revisit the necessary and sufficient conditions for linear and high order of convergence of fixed point and Newton’s methods in the complex plane. Schröder’s processes of the first and second kind are revisited and extended. Examples and numerical experiments are included.
#### 1. Introduction
In this paper, we revisit fixed point and Newton’s methods to find a simple solution of a nonlinear equation in the complex plane. This paper is an adapted version of [1] for complex valued functions. We present only proofs of theorems we have to modify compared to the real case. We present sufficient and necessary conditions for the convergence of fixed point and Newton’s methods. Based on these conditions we show how to obtain direct processes to recursively increase the order of convergence. For the fixed point method, we present a generalization of Schröder’s method of the first kind. Two methods are also presented to increase the order of convergence of the Newton’s method. One of them coincide with the Schröder’s process of the second kind which has several forms in the literature. The link between the two Schröder’s processes can be found in [2]. As for the real case, we can combine methods to obtain, for example, the super-Halley process of order and other possible higher order generalizations of this process. We refer to [1] for details about this subject.
The plan of the paper is as follows. In Section 2, we recall Taylor’s expansions for analytic functions and the error term for truncated expansions. In Section 3 we consider the fixed point method and its necessary and sufficient conditions for convergence. These results lead to a generalization of the Schröder’s process of the first kind. Section 4 is devoted to Newton’s method. Based on the necessary and sufficient conditions, we propose two ways to increase the order of convergence of the Newton’s method. Examples and numerical experiments are included in Section 5.
#### 2. Analytic Function
Since we are working with complex numbers, we will be dealing with analytic functions. Supposing is an analytic function and is in its domain, we can writefor any . Then, for we havewhere is the analytic function:Moreover, the series for and have the same radius of convergence for any , andfor .
#### 3. Fixed Point Method
A fixed point method use an iteration function (IF) which is an analytic function mapping its domain of definition into itself. Using an IF and an initial value , we are interested by the convergence of the sequence . It is well known that if the sequence converges, it converges to a fixed point of .
Let be an IF, be a positive integer, and be such that the following limit exists: Let us observe that for we have We say that the convergence of the sequence to is of (integer) order if and only if , and is called the asymptotic constant. We also say that is of order . If the limit exists but is zero, we can say that is of order at least .
From a numerical point of view, since is not known, it is useful to define the ratio:
Following [3], it can be shown that
We say that is a root of of multiplicity if and only if for , and . Moreover, is a root of of multiplicity if and only if there exists an analytic function such that and .
We will use the big notation and the small notation , around , respectively, when and , when
For a root of multiplicity of , it is equivalent to write or . Observe also that if is a simple root of , then is a root of multiplicity of . Hence is equivalent to .
The first result concerns the necessary and sufficient conditions for achieving linear convergence.
Theorem 1. Let be an IF, and let stand for its first derivative. Observe that although the first derivative is usually denoted by , one will write to maintain uniformity throughout the text.(i)If , then there exists a neighborhood of such that for any in that neighborhood the sequence converges to .(ii)If there exists a neighborhood of such that for any in that neighborhood the sequence converges to , and for all , then .(iii)For any sequence which converges to , the limit exists and .
Proof. (i) By continuity, there is a disk such that . Then if , we haveand . Moreoverand the sequence converges to because .
(ii) If , there exists a disk , with , such that . Let us suppose that the sequence is such that for all . If and , then we have Let , and suppose are in . Because eventually and . Then the infinite sequence cannot converge to .
(iii) For any sequence which converges to we have
For higher order convergence we have the following result about necessary and sufficient conditions.
Theorem 2. Let be an integer and let be an analytic function such that . The IF is of order if and only if for , and . Moreover, the asymptotic constant is given by
Proof. (i) The (local) convergence is given by part (i) of Theorem 1. Moreover we haveand hence(ii) If the IF is of order , assume that for with . We havewhereButand henceSo .
It follows that, for an analytic IF and , the limit exists if and only if for .
As a consequence, for an analytic IF we can say that (a) is of order if and only if , or, equivalently, if and , and (b) if is a simple root of , then is of order if and only if , or, equivalently, if and .
Schröder’s process of the first kind is a systematic and recursive way to construct an IF of arbitrary order to find a simple zero of . The IF has to fulfill at least the sufficient condition of Theorem 2. Let us present a generalization of this process.
Theorem 3 (see [1]). Let be a simple root of , and let be an analytic function such that . Let be the IF defined by the finite series:where are such that for Then is of order , and its asymptotic constant is
For in (22), we recover the Schröder’s process of the first kind of order [47], which is also associated with Chebyshev and Euler [810]. The first term could be seen as a preconditioning to decrease the asymptotic constant of the method, but its choice is not obvious.
#### 4. Newton’s Iteration Function
Considering and in (22), we obtain which is Newton’s IF of order to solve . The sufficiency and the necessity of the condition for high-order convergence of the Newton’s method are presented in the next result.
Theorem 4. Let and let be an analytic function such that and . The Newton iteration is of order if and only if for , and . Moreover, the asymptotic constant is
Proof. (i) If for , and we haveButIt follows thatso(ii) Conversely, if is of order we have for , and . Hence is a root of multiplicity of and we can write We also haveButso we obtainwhereIt follows that is a root of multiplicity of . Hence for , and .
We can look for a recursive method to construct a function which will satisfy the conditions of Theorem 4. A consequence will be that will be of order , and . A first method has been presented in [11, 12]. The technique can also be based on Taylor’s expansion as indicated in [13].
Theorem 5 (see [11]). Let be analytic such that and . If is defined by then , , for . It follows that is of order at least .
Let us observe that in this theorem it seems that the method depends on a choice of a branch for the th root function. In fact the Newton iterative function does not depend on this choice because we haveIn fact the next theorem shows that a branch for the th root function is not necessary.
Theorem 6 (see [12]). Let be given by (36); one can also write where
Unfortunately, there exist no general formulae for and its asymptotic constant exists. However, the asymptotic constant can be numerically estimated with (7).
A second method to construct a function which will satisfy the conditions of Theorem 4 is given in the next theorem.
Theorem 7 (see [1]). Let be a simple root of . Let be defined bywhere and are two analytic functions such that for . Then is of order , with
Let us observe that if we set with given by (22), then verifies the assumptions of Theorem 7.
Remark 8. For a given pair of and in Theorem 7, the linearity of expression (42) with respect to and for computing ’s allows us to decompose the computation for in two computations, one for the pair and and the other for the pair and , and then add the two ’s hence obtained.
#### 5. Examples
Let us consider the problem of finding the roots of unity: for which we have . Hence we would like to solveforAs examples of the preceding results, we present methods of orders and obtained from Theorems 3, 5, and 7. For each method, we consider also presenting the basins of attraction of the roots.
The drawing process for the basins of attraction follows Varona [14]. Typically for the upcoming figures, in squares , we assign a color to each attraction basin of each root. That is, we color a point depending on whether within a fixed number of iteration (here ) we lie with a certain precision (here ) of a given root. If after 25 iterations we do not lie within of any given root we assign to the point a very dark shade of purple. The more there are dark shades of purple, the more the points have failed to achieve the required precision within the predetermined number of iteration.
##### 5.1. Examples for Theorem 3
We start with iterative methods of order . From Theorem 3, we first want . We observe that the simplest such function is . Such a choice has the advantage that derivative of higher order than 2 of this function will be 0, thus simplifying further computation. This is in fact the choice of function which leads to Newton’s method and Chebyshev family of iterative methods. We observe however that it is generally possible to consider different choices of functions, although most might be numerically convenient as we will illustrate here. We need , in such we can also look at where . In the examples that follow we will look at such functions .
In Table 1, we have considered functions of this kind. We have developed explicit expressions for . Figure 1 presents different graphs for the basins of attraction for these methods. We observe that some of them have a lot of purple points.
Now let us consider method of order with with . In this case we obtainand its asymptotic constant is Examples of basins of attraction are given in Figure 2 for . The smallest asymptotic constant is for .
##### 5.2. Examples for Theorem 5
Gerlach’s process described in Theorems 5 and 6 leads to Newton’s method for and Halley’s method for . For our problem we have These methods are well known standard methods. For comparison, their basins of attraction are given in Figure 3.
##### 5.3. Examples for Theorem 7
To illustrate Theorem 7, we set and for , and let us consider methods of orders and to solve . Table 2 presents the quantities , , , and for for this example.
2 3
We observe that the asymptotic constant of the method of order for is zero; it means that this method is of an order of convergence higher than , and in fact it corresponds to Halley’s method which is of order . We observe that methods of order for the values of and both correspond to Halley’s method for our specific problem. Examples of basins of attraction are given in Figure 4 for methods of order and in Figure 5 for methods of order using values of .
#### 6. Concluding Remarks
In this paper we have presented fixed point and Newton’s methods to compute a simple root of a nonlinear analytic function in the complex plane. We have pointed out that the usual sufficient conditions for convergence are also necessary. Based on these conditions for high-order convergence, we have revisited and extended both Schröder’s methods of the first and second kind. Numerical examples are given to illustrate the basins of attraction when we compute the third roots of unity. It might be interesting to study the relationship, if there is any between the asymptotic constant and the basin of attraction for such methods.
#### Conflicts of Interest
The authors declare that they have no conflicts of interest.
#### Acknowledgments
This work has been financially supported by an individual discovery grant from NSERC (Natural Sciences and Engineering Research Council of Canada) and a grant from ISM (Institut des Sciences Mathématiques).
#### References
1. F. Dubeau and C. Gnang, “Fixed point and Newton's methods for solving a nonlinear equation: from linear to high-order convergence,” SIAM Review, vol. 56, no. 4, pp. 691–708, 2014. View at: Publisher Site | Google Scholar | MathSciNet
2. F. Dubeau, “Polynomial and rational approximations and the link between Schröder's processes of the first and second kind,” Abstract and Applied Analysis, vol. 2014, Article ID 719846, 5 pages, 2014. View at: Publisher Site | Google Scholar
3. F. Dubeau, “On comparisons of chebyshev-halley iteration functions based on their asymptotic constants,” International Journal of Pure and Applied Mathematics, vol. 85, no. 5, pp. 965–981, 2013. View at: Publisher Site | Google Scholar
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5. E. Schröder, “On Infinitely Many Algorithms for Solving Equations,” in Institute for advanced Computer Studies, G. W. Stewart, Ed., pp. 92–121, University of Maryland, 1992. View at: Google Scholar
6. J. F. Traub, Iterative Methods for the Solution of Equations, Prentice-Hall, NJ, USA, Englewood Cliffs, 1964.
7. A. S. Householder, The Numerical Treatment of a Single Nonlinear Equation, McGraw-Hill Book Co., NY, USA, 1970. View at: MathSciNet
8. E. Bodewig, “On types of convergence and on the behavior of approximations in the neighborhood of a multiple root of an equation,” Quarterly of Applied Mathematics, vol. 7, pp. 325–333, 1949. View at: Publisher Site | Google Scholar | MathSciNet
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14. J. L. Varona, “Graphic and numerical comparison between iterative methods,” The Mathematical Intelligencer, vol. 24, no. 1, pp. 37–46, 2002. View at: Publisher Site | Google Scholar | MathSciNet
Copyright © 2018 François Dubeau and Calvin Gnang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | 3,709 | 15,667 | {"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} | 2.625 | 3 | CC-MAIN-2020-50 | longest | en | 0.885822 |
https://stats.stackexchange.com/questions/495546/is-it-possible-for-two-random-variables-to-be-negatively-correlated-but-both-be?noredirect=1 | 1,726,840,615,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00538.warc.gz | 503,767,040 | 41,473 | # Is it possible for two random variables to be negatively correlated, but both be positively correlated with a third r.v.?
Would it be possible for two variables to be negatively correlated with one another, yet be positively correlated with a third variable? Are there any concrete examples?
• Commented Nov 8, 2020 at 15:31
• Random variables form a vector space, in which you're asking whether it's possible for two vectors to have negative dot product, while both having positive dot product with a third vector. The answer to that is yes. In fact, given any two variables/vectors, as long as they are not directly opposite each other, their angular bisector will have positive dot product with each of them. Commented Nov 9, 2020 at 0:29
• Example - having umbrella and being wet from rain are negatively correlated but both are positively correlated with rain. Commented Nov 9, 2020 at 13:31
• Another natural example: the indicator random variables for the events "Smith wins the competition", "Jones wins the competition", and "Either Smith or Jones wins the competition". Commented Nov 30, 2020 at 1:02
Certainly. Consider multivariate normally distributed data with a covariance matrix of the form
$$\begin{pmatrix} 1 & - & + \\ - & 1 & + \\ + & + & 1 \end{pmatrix}.$$
As an example, we can generate 1000 such observations with covariance matrix
$$\begin{pmatrix} 1 & -0.5 & 0.5 \\ -0.5 & 1 & 0.5 \\ 0.5 & 0.5 & 1 \end{pmatrix}$$
in R as follows:
library(mixtools)
set.seed(1)
xx <- rmvnorm(1e3,mu=rep(0,3),
sigma=rbind(c(1,-.5,.5),c(-.5,1,.5),c(.5,.5,1)))
cor(xx[,c(1,2)])
cor(xx[,c(1,3)])
cor(xx[,c(2,3)])
The first two columns are negatively correlated ($$\rho=-0.5$$), the first and the third and the second and the third are positively correlated ($$\rho=0.5$$).
• My first thought was to do a simulation like this, which certainly proves that such random variables exist. However, I still struggle to think of a practical place where such a relationship would exist.
– Dave
Commented Nov 8, 2020 at 14:17
• @Dave this particular example is the case when two variables are negatively correlated and a third variable is a sum of the two. Commented Nov 8, 2020 at 14:21
• @Dave The motivation for my asking the question is a good example. This question was actually motivated by portfolio construction - I was wondering if it’ll be possible to find e.g. two stocks whose returns are negatively correlated, despite them perhaps both being correlated to the broader asset class of stocks in general. So the simplest case of where this might happen would be in an equally weighted two-stock portfolio, where the returns of the two stocks are negatively correlated, but each would be positively correlated to portfolio returns Commented Nov 9, 2020 at 6:35 | 713 | 2,777 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-38 | latest | en | 0.91594 |
https://dividedby.org/994000-divided-by-11 | 1,713,662,829,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00374.warc.gz | 189,936,150 | 23,953 | Home » Divided by 11 » 994000 Divided by 11
# 994000 Divided by 11
Welcome to 994000 divided by 11, our post which explains the division of nine hundred ninety-four thousand by eleven to you. 🙂
The number 994000 is called the numerator or dividend, and the number 11 is called the denominator or divisor.
The quotient (integer division) of of 994000 and 11, the ratio of 994000 and 11, as well as the fraction of 994000 and 11 all mean (almost) the same:
994000 divided by 11, often written as 994000/11.
Read on to find the result in various notations, along with its properties.
## Calculator
Show Steps
90363.6363636363
=
90363 Remainder 7
The Long Division Steps are explained here. Read them now!
## What is 994000 Divided by 11?
We provide you with the result of the division 994000 by 11 straightaway:
994000 divided by 11 = 90363.63
The result of 994000/11 is a non-terminating, repeating decimal.
The repeating pattern above, 63, is called repetend, and denoted overlined with a vinculum.
This notation in parentheses is also common: 994000/11 = 90363.(63): However, in daily use it’s likely you come across the reptend indicated as ellipsis: 994000 / 11 = 90363.63… .
• 994000 divided by 11 in decimal = 90363.63
• 994000 divided by 11 in fraction = 994000/11
• 994000 divided by 11 in percentage = 90363.63%
Note that you may use our state-of-the-art calculator above to obtain the quotient of any two integers or whole numbers, including 994000 and 11, of course.
Repetends, if any, are denoted in ().
The conversion is done automatically once the nominator, e.g. 994000, and the denominator, e.g. 11, have been inserted.
To start over overwrite the values of our calculator.
## What is the Quotient and Remainder of 994000 Divided by 11?
The quotient and remainder of 994000 divided by 11 = 90363 R 7
The quotient (integer division) of 994000/11 equals 90363; the remainder (“left over”) is 7.
994000 is the dividend, and 11 is the divisor.
In the next section of this post you can find the additional information in the context of nine hundred ninety-four thousand over eleven, followed by the summary of our information.
Observe that you may also locate many calculations such as 994000 ÷ 11 using the search form in the sidebar.
The result page lists all entries which are relevant to your query.
Give the search box a go now, inserting, for instance, nine hundred ninety-four thousand divided by eleven, or what’s 994000 over 11 in decimal, just to name a few potential search terms.
Further information, such as how to solve the division of nine hundred ninety-four thousand by eleven, can be found in our article Divided by, along with links to further readings.
## Conclusion
To sum up, 994000/11 = 90363.(63). The indefinitely repeating sequence of this decimal is 63.
As division with remainder the result of 994000 ÷ 11 = 90363 R 7.
You may want to check out What is a Long Division?
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For a list of our similar sites check out the sidebar of our home page. – Article written by Mark, last updated on December 10th, 2023 | 917 | 3,616 | {"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} | 4.46875 | 4 | CC-MAIN-2024-18 | latest | en | 0.927657 |
https://3jack.blogspot.com/2009/10/3jacks-translation-of-tgm-part-9h.html?showComment=1256618492963 | 1,722,841,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640434051.17/warc/CC-MAIN-20240805052118-20240805082118-00287.warc.gz | 61,082,699 | 15,130 | ## Monday, October 26, 2009
### 3Jack's Translation of TGM: Part 9H
10-11 (Pressure Point Combinations)
The corresponding Chapter 7 translation can be found HERE.
There are 4 pressure points.
#1 Pressure Point
The lifeline in the right hand where it touches the left thumb or the clubshaft. It is 'actively' used when 'hitting' and passively used when 'swinging.' Even with my right hand chop motion on the downswing, the #1 PP is a favorite of mine to use and the main goal for myself is for pressure on my left thumb to be at its maximum at impact.
#2 Pressure Point
The last 3 fingers in the left hand. Used by the 'swinger' who uses the 'Rope Handle technique.' This is the golfer imagining at the top of the swing they are pulling the club straight down, much like a person pulling the rope to ring a large bell from the medievel times.
#3 Pressure Point
The BASE JOINT of the right index finger. Here's Mr. Hogan explaining the #3 PP without even knowing that Homer Kelley called it such.
There was some talk on this blog of the pressure point moving, from the base joint to the middle knuckle, but it DOES NOT MOVE. In fact, it's very dangerous if the golfer tries to use the middle knuckle on the right index finger as the #3 PP because when you flip with the right hand, it's the right index finger that does the work, in particular the middle knuckle.
In fact, Mr. Hogan called the right index finger and right thumb 'swing wreckers' mostly because flippers usually have the pressure in the wrong part of the right thumb and right index finger which causes them to flip.
The rest of the #3 Pressure Point in this section has Homer talking about what Lynn Blake is discussing in this video.
So the #3 PP can be used by both hitters and swingers, they just have a different location as to where the #3 PP is in relation to the clubshaft.
#4 Pressure Point
The part of the upper left arm where it touches the left side of the body. Famous golf coach Jimmy Ballard teaches a 'left side connection' which is basically using the #4 PP to hit the ball.
This can be active for both the swinger (for normal swinger procedures) and for the 4-barrel hitter.
Usually the pressure point coincides with the power accumulator. Meaning, usually if you use the #1 Power Accumulator, then you use the #1 PP. However, there is some interchangeability that can be done here. Such as the #1 PP could drive the #2 PA if wanted. In that scenario, the #2 PA still has the same motion and application, it's just driven by the #1 PP instead of the #2 PP.
3JACK
Unknown said...
Rich,
I think I was the one you were refering to when you mentioned that someone stated that the #3 pressure point moved in the downswing.
My TGM instructor told me that a swinger feels it in the crease of the middle knuckle or first pad until the release. Then if you are real good you should feel it move slightly out towards the end of the finger near and through impact as the right hand rolls/swivals through impact.
Does that a conflict with what you feel as a swinger? (I know you are a hitter)
Rich H. said...
I just disagree with it, although feels are subjective. It states pretty clearly in the book that it's the base joint of the right index finger and nowhere does it state that it's the middle knuckle. Watch Lynn's video and it shows that the #3 PP does not change for the swinger. In fact, here's what HK says about it 'Remember, with Swinging, PP #3 must have a Feel of being rotated a quarter turn at the Top with Standard Wrist Action, just and only because ofthe Loading Action direction -- no actual movement of anything. So from the Top to Release, the Loading puts the top side of the Clubshaft against the first knuckle of the forefinger.'
The first knuckle is not the middle knuckle.
And personally, if you want the exact translation of the book, Lynn is the guy to go to because nobody, outside of maybe Alex Sloan, knows exactly what Homer was teaching and knows TGM better.
Again, if that's what it feels to the golfer to get the correct alignments, great. But I think it's dangerous to feel that because it can cause a flip with the right hand.
James Bermingham said...
Ive recently started using & focusing on #PP1 on my left thumb "at Maximum", and I have to say its spot on, and is really helping me keeping both arms straight after impact with Exstensor action.
I,ve never read a blog more thoroughly than yours.
Top work. A real treasure find | 1,023 | 4,469 | {"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} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.92351 |
http://www.stanford.edu/class/ee263/quizzes/eig.html | 1,397,866,591,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00567-ip-10-147-4-33.ec2.internal.warc.gz | 668,642,009 | 1,732 | $\newcommand{\ones}{\mathbf 1}$
Suppose $A\in {\mathbf R}^{n \times n}$ has real eigenvalue $\lambda$.
$A$ has a right eigenvector $v$ and a left eigenvector $w$ associated with $\lambda$.
$\chi(\lambda) = 0$, where $\chi(s) = \det (sI-A)$.
$w^Tv = 0$.
$w^Tv = 1$.
Suppose $A \in {\mathbf R}^{n \times n}$ is diagonalizable.
The eigenvalues of $A$ are distinct.
$A^T$ is diagonalizable.
$A^2$ is diagonalizable.
Suppose $A\in {\mathbf R}^{n \times n}$ is defective.
$A^2$ is defective.
$A$ does not have an independent set of $n$ eigenvectors.
$A$ has repeated eigenvalues.
$A$ is singular.
If the continuous-time system $\dot x(t) = Ax(t)$ is stable then the discrete-time system $y(t+1) = e^Ay(t)$ is stable. (In both cases, stable means all trajectories converge to zero as $t\to\infty$.)
If the discrete-time system $y(t+1) = e^Ay(t)$ is stable, then the continuous-time system $\dot x(t) = Ax(t)$ is stable. | 310 | 928 | {"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} | 2.96875 | 3 | CC-MAIN-2014-15 | latest | en | 0.842453 |
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# Algebra I STAAR/EOC Objective 4 Review
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If you are preparing students for the Algebra I STAAR Test, then this resource is for you. This product covers all of the Objective 4 TEKS. You will find 3 pages of test-like questions with grid answers, fill in the blank and multiple choice. The student will practice calculator skills for finding the line of best fit and the correlation coefficient. The final page is a 3 question quiz. Topics covered in objective 4:
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Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TpT’s content guidelines. | 530 | 2,217 | {"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} | 3.296875 | 3 | CC-MAIN-2021-04 | latest | en | 0.772888 |
https://www.entropy.energy/scholar/node/projectors | 1,638,650,578,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363006.60/warc/CC-MAIN-20211204185021-20211204215021-00344.warc.gz | 826,484,441 | 3,878 | # Projectors
## Alexei Gilchrist
Projectors are linear operators with the property that the square of the operator is the operator.
### 1 Subspaces
Given a vector space, a subset of vectors that form a vector space is called a subspace (so it must be closed, have inverses etc). Since a vector space has to have a null vector $$|\textbf{0}\rangle$$, all subspaces must contain this null vector.
Definition: Direct-sum
For subspaces $$\mathbb{V}_1$$ and $$\mathbb{V}_2$$ of $$\mathbb{V}$$, then the direct-sum $$\mathbb{V}_1\oplus\mathbb{V}_2$$ is all linear combinations of vectors from $$\mathbb{V}_1$$ and $$\mathbb{V}_2$$.
Definition: Orthogonality for subspaces
Two subspaces are orthogonal if every vector from one subspace is orthogonal to every vector in the other subspace.
Definition: Orthogonal complement
The orthogonal complement, $$\mathbb{V}_1^\perp$$, to a vector subspace $$\mathbb{V_1}$$ in $$\mathbb{V}$$, is every vector $$|v\rangle\in\mathbb{V}$$ for which $$\langle v| u\rangle=0$$ holds for all $$|u\rangle\in\mathbb{V}_1$$.
For any subspace $$\mathbb{V}_1$$ in $$\mathbb{V}$$, then $$\mathbb{V}_1\oplus\mathbb{V}_1^\perp = \mathbb{V}$$.
One useful decomposition to bear in mind is that every vector $$|x\rangle\in\mathbb{V}$$ can be uniquely written as $$|x\rangle = |v\rangle+|v^\perp\rangle$$ where $$|v\rangle\in\mathbb{V}_1$$ for a given subspace $$\mathbb{V}_1$$, and $$|v^\perp\rangle\in\mathbb{V}_1^\perp$$ (to be proved).
### 2 Projectors
Definition: Projector
Say $$\mathbb{H}_a$$ is a subspace of $$\mathbb{H}$$ and $$\mathbb{H}_b = \mathbb{H}_a^\perp$$ is the orthogonal complement. For any state $$|\psi\rangle=|\psi_a\rangle+|\psi_b\rangle$$ where $$|\psi_a\rangle\in\mathbb{H}_a$$ and $$|\psi_b\rangle\in\mathbb{H}_b$$ (and of course $$\langle \psi_a| \psi_b\rangle=0$$), the projector $$\mathbb{P}_a$$ onto subspace $$\mathbb{H}_a$$ is defined by the action on such states
$\mathbb{P}_a|\psi\rangle = |\psi_a\rangle$
Projectors are Hermitian. For some other vector $$|\phi\rangle=|\phi_a\rangle+|\phi_b\rangle$$ where $$|\phi_a\rangle\in\mathbb{H}_a$$ and $$|\phi_b\rangle\in\mathbb{H}_b$$, in addition to $$|\psi\rangle$$ above. Then
\begin{align*}(|\phi\rangle, \mathbb{P}_a |\psi\rangle) &= (|\phi_a\rangle+|\phi_b\rangle, |\psi_a\rangle) = (|\phi_a\rangle,|\psi_a\rangle) \\ (\mathbb{P}_a|\phi\rangle, |\psi\rangle) &= (|\phi_a\rangle,|\psi_a\rangle+|\psi_b\rangle) = (|\phi_a\rangle,|\psi_a\rangle)\end{align*}
The signature feature of projectors is $$\mathbb{P}^2=\mathbb{P}$$:
$\mathbb{P}_a^2|\psi\rangle = \mathbb{P}_a\mathbb{P}_a|\psi\rangle = \mathbb{P}_a|\psi_a\rangle = |\psi_a\rangle$
Obviously this extends to any positive interger power.
Since projectors are Hermitian, their eigenvalues must be real, and since $$\mathbb{P}^2=\mathbb{P}$$, the eigenvalues must either be 1 or 0: say $$|\lambda\rangle$$ is an eigenvector of projector $$\mathbb{P}$$ with eigenvalue $$\lambda$$, then
\begin{align*}\mathbb{P}\mathbb{P}|\lambda\rangle &= \mathbb{P}|\lambda\rangle = \lambda |\lambda\rangle \\ \mathbb{P}\mathbb{P}|\lambda\rangle &= \lambda\mathbb{P}|\lambda\rangle = \lambda^2 |\lambda\rangle.\end{align*}
So we must have $$\lambda^2 = \lambda$$ and this is only satisfied if $$\lambda=0$$ or $$\lambda=1$$. | 1,129 | 3,280 | {"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} | 4.375 | 4 | CC-MAIN-2021-49 | longest | en | 0.660779 |
https://study.com/academy/answer/let-g-be-a-graph-whose-vertices-are-the-integers-1-through-8-and-let-the-adjacent-vertices-of-each-vertex-be-given-by-the-table-below-vertex-adjacent-vertices-1-2-3-4-2-1-3-4-3-1-2-4-4.html | 1,580,197,457,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251776516.99/warc/CC-MAIN-20200128060946-20200128090946-00163.warc.gz | 678,607,498 | 24,324 | # Let G be a graph whose vertices are the integers 1 through 8, and let the adjacent vertices of...
## Question:
Let G be a graph whose vertices are the integers 1 through 8, and let the adjacent vertices of each vertex be given by the table below.
1 (2,3,4)
2 (1,3,4)
3 (1,2,4)
4 (1,2,3,6)
5 (6,7,8)
6 (4,5,7)
7 (5,6,8)
8 (5,7)
Assume that, in a traversal of G, the adjacent vertices of a given vertex are returned in the same order as they are listed in the above table.
a. Draw G.
b. Order the vertices as they are visited in a DFS traversal starting at vertex 1.
c. Order the vertices as they are visited in a BFS traversal starting at vertex 1.
## Searching Graphs:
The two ways to search the nodes of graphs are Depth First Search (DFS) and Breadth First Search (BFS). DFS dives as deep as it can to children of children before backtracking at each dead end. BFS looks across each level, then the next level of children.
a. Draw G.
b. Give the sequence of vertices of G visited using a DFS traversal starting at vertex 1.
Depth First Search visits all the...
Become a Study.com member to unlock this answer! Create your account | 307 | 1,146 | {"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} | 2.703125 | 3 | CC-MAIN-2020-05 | longest | en | 0.911938 |
http://nasawavelength.org/resource-search?facetSort=1&topicsSubjects=Mathematics&materialsCost=Over+%2420&educationalLevel=High+school | 1,527,473,526,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00224.warc.gz | 192,073,641 | 13,905 | Narrow Search
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Measuring the Heat Capacity of Greenhouse Gases
This quantitative experiment involves lab teams in comparing a sample of room air with one of the greenhouse gases - carbon dioxide, nitrous oxide, or methane - and measuring their heat capacity. The activity requires an infrared heat source, such... (View More)
Audience: High school
Materials Cost: Over \$20 per group of students
Human Sling Shot
This activity demonstrates Newton’s Second Law (F=ma), and helps show the relationship between potential and kinetic energy. Students sit on a skateboard in a sling shot configuration, and are accelerated down the hall. Potential energy from the... (View More)
Audience: High school
Materials Cost: Over \$20 per group of students
Calculator Controlled Robots: Hands-on Mathematics and Science Discovery Educator Guide
Using a graphing calculator and a Norland Research calculator robot, students create programs in TI-BASIC to direct their robot through a variety of tasks. Ten robot missions and three exploration extensions are included in this lesson booklet.... (View More)
Freshwater Macroinvertebrates Protocol
This activity guides students through sampling, identification and counting of macroinvertebrates sampled in a GLOBE hydrology study site, and understand how the taxa composition found in the sample can be an indicator of water quality and ecosystem... (View More)
Design Challenge: How to Keep Items Cool in Boiling Water?
This is a design challenge about heat transfer and insulation. Learners will apply the scientific method to design and build a container that will keep items cool when placed in boiling water. They will practice collaboration in team-building and in... (View More)
Audience: High school
Materials Cost: Over \$20 per group of students
Polarimeter: Do NOT Use this Device to Measure a Polar Bear!
This is an activity about polarized light. Learners will use a polarizing filter to build and calibrate a simple polarimeter, use the constructed polarimeter to find sources of polarized light, and measure the angle of polarization of polarized... (View More)
Colorgraphs: See the world in a new light
In this activity, learners will explore the properties of color filters and filter bandpass by observing light sources using diffraction grating and color filters and create a graph of percent transmission versus wavelength to characterize the... (View More)
Sunspot Tracker: There's a Little Dark Spot on the Sun Today
This is an activity about the movement of sunspots. Learners will project an image of the Sun using a telescope, binoculars, or a pinhole projector, observe and record sunspots over the course of several days, and calculate the speed of the observed... (View More)
1 | 638 | 3,096 | {"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} | 2.609375 | 3 | CC-MAIN-2018-22 | latest | en | 0.864604 |
http://www.thinkib.net/mathstudies/page/10674/probability-trees | 1,490,695,500,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189686.56/warc/CC-MAIN-20170322212949-00270-ip-10-233-31-227.ec2.internal.warc.gz | 727,562,945 | 16,122 | Probability Trees
'Use what you already to know to see just how and why probability tree diagrams work'
Probability is a topic that moves very quickly from that which is easily understood and intuitive to something seemingly much more complex. Sample space diagrams, probability trees, fractions decimals, 'and' rules, 'or' rules and so on..... It is easy to lose track of what it all means and to understand how all of these things are inter related. In this activity the aim is for you to demonstrate to yourselves the link between all of the things that you know. In doing so you should see how a sample space diagram and a tree diagram effectively do the same thing and understand why sometimes you multiply and sometimes you add probabilities. The activity demands some revision of fraction manipulation and hopefully demonstrates why a tree diagram, once understood, is the more versatile tool for lots of problems. The activity might take between 2 - 3 hours to complete and involves puzzles, challenges, and practice!
Aims
This activity aims to bridge the gap between sample space and tree diagrams for combined events.
Resources
There are quite a few resources required for this activity and they are all saved as Google docs linked to below. (Documents are set to print in A4 - The 'For computers' links at the end link to Google docs that can be saved locally for editing). Below there are some previews of some of the worksheets. There are also some Probability Trees teacher notes
Part 1 - Sample Space - Practise working out the possible outcomes for combined events. For computers
Part 2 - Fractions and Multiplying - Here are some tasks with a set of fractions that relevant to the problem. This will need printing, copying and cutting out. (unless you work on computers) For computers
Part 3 - Fractions and adding - More related problems with the fractions. For Computers
Part 4 - The tree Diagram - Link it all together, use the sample space diagram to help you build a tree diagram and see the links between the previous activities. For computers
Part 5 - Practice! - Put what you have learned in to practice with these fill in the gap probability tree diagrams. (you are given a little bit less information each time. For computers
Activity in action
Here are some photos of the activity in action.
Probability Tress - Part 2
Multiplying Fractions - The 'and' rule.
Probability Tress - Part 5
This worksheet represents the 'End Goal' of the activity.
This links directly to sections new - 3.6 and 3.7, previous - 3.3, 3.8 and 3.9 of the syllabus.
Description
Here follows an outline if what the task is. If students are not reading this page then the teacher will need to show and give this overview.
The activity is outlined clearly on the different sheets linked to above.
• Part 1 - Sample Space
• Part 2 - Fractions and Multiplying
• Part 3 - Fractions and adding
• Part 4 - Tree diagrams
• Part 5 - Probability trees (practice)
I did it my way!
As a practising maths teacher I know that most of us like to give activities our own little twist and do them 'our way'. It would be great to add a little collect of 'twists' from users. You can either add your twist to the comments section below or e-mail them directly to me at jamesn@inthinlking.co.uk In time, some of these twists may appear here....
All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. Any unauthorised copying or posting of materials on other websites is an infringement of our copyright and could result in your account being blocked and legal action being taken against you. | 787 | 3,686 | {"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} | 4.4375 | 4 | CC-MAIN-2017-13 | latest | en | 0.945496 |
https://sweetcode.io/simple-predictions-with-random-forest-algorithm/ | 1,701,789,279,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100551.2/warc/CC-MAIN-20231205140836-20231205170836-00689.warc.gz | 629,759,951 | 24,774 | # Make Simple Predictions with the Random Forest Algorithm
5535 VIEWS
Many machine learning algorithms require predicting the class of given data. You can make simple predictions with the Random Forest Algorithm. It is one of the most popular machine learning algorithms used for such a task.
The Random Forest Algorithm is able to predict the class of a given observation through the implementation and use of a ‘Random Forest’. ‘Random Forest’ is a term to describe several ‘decision trees’ that are working together to produce a collective output.
## Decision Tree
A decision tree is a tree data structure that determines the distinct classes of data within some given data sample.
Consider the following example that further explains the concept of a decision tree:
In Fig. 1 there are seven characters. Some of these characters are red in colour while others are blue. Also note that some of the characters are underlined while others are not.
The colour of the characters in the data identify the two distinct children nodes. So, we can observe in Fig. 2 that the red characters are placed in the left child node and the blue characters are placed in the right child node.
Although the data is split into two categories, certain features in the left child node differ. Some of the characters are underlined and others are not underlined.
Therefore, the data in the left child node of the root node are further split into two child nodes. The underlined characters are placed in the resulting left child node and the characters that are not underlined are placed in the resulting right child node. This can be seen below in Fig 3:
Fig. 3
In the formulation of decision trees, we always ask the following questions at any given node in the tree:
Which feature in the data of this node is going to allow me to
1) Split the data into groups which are as distinct as possible.
2) Split the data into groups such that the data that are within the resulting groups are as similar to each other as possible.
## Random Forest Classifier
To make simple predictions, the Random Forest Classifier performs the task of classification. It is based on the idea that multiple decision trees can operate as an ‘ensemble’. The concept of an ‘ensemble’ means that each tree in the Random Forest spits out a class prediction. The class with the most votes becomes the Random Forest Classifier’s prediction.
Below in Fig.4 is an image showcasing how the Random Forest Classifier works:
Prerequisites for the Random Forest Classifier to perform well
To enable a Random Forest Classifier to perform at high levels of accuracy, two main prerequisites should be satisfied. They are:
1) The features of the dataset that are selected to be used to discriminate between the data values should be good enough.
2) The predictions (and therefore the errors) made by the individual trees need to have low correlations with each other.
You can achieve these two prerequisites the the application of two processes: “bagging,” and “feature randomness.”
## Bagging
The Random Forest Algorithm uses “bagging” to make simple predictions. This is the process of training each decision tree in the random forest. You base the training on a random selection of data samples from the given training dataset with replacement.
In the process of bagging, we are not drawing subsets from the training dataset and training each decision tree on a different subset. Rather, if we have a training dataset of size N, we train each decision tree on a dataset of size N. That dataset consists of data samples drawn at random from the training dataset with replacement.
Consider the following example: If our training dataset was [1,2,3,4,5,6] then we might train one of the decision trees in the random forest with the following list [1,2,2,3,6,6]. Notice the size of the dataset in this example. The size of the dataset used to train the decision tree is the same size as the original training dataset. What is the difference? Two ‘2’s and two ‘6’s make up the dataset, which trains the decision tree. That’s because the data samples chosen to form a dataset that is used to train a decision tree are chosen at random and with replacement.
## Feature Randomness
You may want to split a parent code into children nodes in a normal decision tree. Evaluate every possible feature in the training set. Then, apply the feature that produces the most separation between the left and right child nodes.
However, in the random forest classifier, each decision tree can only pick from a random subset of the main features of the training dataset. This brings about more variation among the decision trees in the model. It ultimately results in a lower correlation across the decision trees in the random forest.
The image in Fig. 5, below, demonstrates this idea of feature randomness for making predictions in a way that can be easily understood.
## Putting It All Together
1) Obtain all of the training data that is going to be used to build decision trees in the random forest.
2) Determine the number of decision trees that you would like to have in the random forest.
3) For each decision tree in the random forest, apply the concept of bagging to obtain the dataset that would be used to train it.
4) Train the decision trees in the random forest with their respective datasets.
5) Once the decision trees are done training, obtain the test data sample and supply it to all of the decision trees in the random forest.
6) Count the number of occurrences of each distinct prediction.
7) The prediction with the most number of occurrences is provided as the final prediction of the random forest.
Code for the Random Forest Classifier
The code for building a Random Forest classifier to make simple predictions can be written by following the code snippets in this Sweetcode post
Conclusion:
I hope that through this tutorial, you can now build your own random forest classifier to make simple predictions.
David Sasu is a senior studying Computer Science in Ashesi University. He is passionate about understanding technology and using it to solve important problems. He is currently working on the creation of information systems for under-funded orphanages in his country, Ghana. He hopes to specialize in the fields of Artificial Intelligence and cybersecurity to enable him to create systems to help safeguard and improve the African continent. | 1,281 | 6,453 | {"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} | 3.625 | 4 | CC-MAIN-2023-50 | latest | en | 0.933089 |
https://mathswithdavid.com/ks4-number-long-multiplication-and-division/ | 1,679,393,920,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943695.23/warc/CC-MAIN-20230321095704-20230321125704-00028.warc.gz | 466,847,775 | 24,181 | # KS4. Number. Long Division and Word Problems
We need a good technique for long division that will work every time no matter how difficult the numbers. Around the world there are various different techniques – ask your teacher to show you an example technique, and if you have an alternative technique, perhaps you could share it with the class too? Cambridge are happy for you to use any technique as long as you master a formal technique that consistently works.
Note that with the non-integer part of a number, a long division technique can give you the answer with a fractional part a with a decimal part. Both of these can be useful (in the exercise below we will write the answers with a decimal part, including no more than 2 decimal places)
Example
• Calculate
• 981 ÷ 3
• 637 ÷ 7
• 600 ÷ 7
• 241 ÷ 5
Exercise
Calculate the following without using a calculator:
(1.) 672 ÷ 21 (2.) 425 ÷ 17 (3.) 576 ÷ 32 (4.) 247 ÷19 (5.) 875 ÷ 25 (6.) 574 ÷26 (7.) 806 ÷ 34 (8.) 748 ÷ 41 (9.) 666 ÷ 24 (10.) 707 ÷ 52 (11.) 951 ÷ 27 (12.) 806 ÷ 34 (13.) 2917 ÷ 42 (14.) 2735 ÷ 18 (15.) 56274 ÷ 19
Solutions
(1.) 31, (2.) 25, (3.) 18, (4.) 13, (5.) 35, (6.) 22.08, (7.) 23.71, (8.) 18.24, (9.) 27.75, (10.) 13.6, (11.) 35.22, (12.) 23.71, (13.) 69.45, (14.) 151.94, (15.) 2961.79
Word Problems
In practice mathematics is useful because it can be applied to solve problems. Throughout our mathematics course we must develop our ability to quickly read questions, understand the way in which we can apply mathematics to answer them and then use the relevant mathematics.
Worked Examples
Let’s identify which types of arithmetic we need to use and then find the answer for each of the following questions:
Exercise
Let’s try this for the exercise below:
Solutions
(1.) \$47.04, (2.) 46, (3.) 7592, (4.) 12 with 17c change, (5.) 8, (6.) \$80.64, (7.) \$14m, (8.) \$85, (9.) \$21,600 | 609 | 1,886 | {"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} | 4.6875 | 5 | CC-MAIN-2023-14 | latest | en | 0.757577 |
https://www.mail-archive.com/sympy@googlegroups.com/msg31989.html | 1,524,417,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945624.76/warc/CC-MAIN-20180422154522-20180422174522-00616.warc.gz | 826,248,775 | 3,698 | [sympy] Re: symbolic Four vectors
```Thank you so much I am goig to foloow your suggestion
El viernes, 6 de abril de 2018, 1:02:31 (UTC-5), Moises Zeleny escribió:
>
> Hi, everyone
>
> I am a new sympy user and usually I work with Jupyter notebook. I would
> like construct a new symbolic class named FV (Four vector), I have made this
>
> from sympy import *
> init_printing()
>
> class FV:
>
> def __init__(self,p0,p1,p2,p3):
> self.p0 = p0
> self.p1 = p1
> self.p2 = p2
> self.p3 = p3
>
> def __str__(self):
> return '({a},{b},{c},{d})'.format(a=self.p0, b=self.p1, c=self.p2,
> d=self.p3)
>
> def __repr__(self):
> return self.__str__()
>
> p0,p1,p2,p3 = self.p0,self.p1,self.p2,self.p3
> k0,k1,k2,k3 = other.p0,other.p1,other.p2,other.p3
> return FV(p0+k0, p1+k1, p2+k2, p3+k3)
>
> def __mul__(self,other):
> p0,p1,p2,p3 = self.p0,self.p1,self.p2,self.p3
> k0,k1,k2,k3 = other.p0,other.p1,other.p2,other.p3
> return FV(p0*k0,-p1*k1,-p2*k2,-p3*k3)
>
> def __abs__(self):
> from sympy import sqrt
> p0,p1,p2,p3 = self.p0,self.p1,self.p2,self.p3
> return sqrt(p0**2 - p1**2 - p2**2 - p3**2)
>
> def __eq__(self,other):
> return self.p0 == other.p0 and self.p1 == other.p1 and self.p2 ==
> other.p2 and self.p3 == other.p3
>
> def __neq__(self,other):
> return not self.__eq__(other)
>
> but this works while I use float or int python objects for the components
> in FV, when I use symbols I obtained
>
> p = {i:symbols('{{p^{a}}}'.format(a=i)) for i in range(4)}
> pmu = FV(p[0],p[1],p[2],p[3])
> pmu
>
> ({p^0},{p^1},{p^2},{p^3})
>
>
> this output do not used a init_printing(). Can someone help me please?
>```
```
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To unsubscribe from this group and stop receiving emails from it, send an email | 697 | 1,957 | {"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} | 2.90625 | 3 | CC-MAIN-2018-17 | latest | en | 0.583529 |
https://docs.google.com/spreadsheets/d/1se_fyVYei4c25IdlBXxXF6dNfrzj1oqKrf07SO2iLE4/edit?usp=sharing | 1,568,750,034,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573105.1/warc/CC-MAIN-20190917181046-20190917203046-00024.warc.gz | 442,502,660 | 49,685 | Suggested Schedule for ECON 202 MACRO online Fall2019
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ABCDEFGHIJKLMNOPQRSTU
1
Suggested Schedule for ECON 202 Macro online, Summer 2019 YOU SHOULD: Suggested Calendar Days to study each UnitYOU SHOULD: Suggested Target Date to Complete this UnitYOU MUST: Hard DEADLINE dates in red are fixed deadlinesPoints Available in this UnitGraded
Assignments
Cumulative
Total Points Available
since start of course through this point
Approx. % of Coursework Completed up to and including this Unit
2
Semester Starts8/22/2019
3
Unit 17Aug-29202 Forums, Quiz205%
4
5
Unit 28Sep-620Worksheet, Quiz4010%
6
Unit 38Sep-1420Worksheet, Quiz6015%
7
Unit 48Sep-2220Worksheet, Quiz8020%
8
Unit 58Sep-3025
Worksheet, Forum, Quiz
10526%
9
Unit 66Oct-610Quiz11529%
10
Midterm Test (units 1-6)
1Oct-720online test13534%
11
12
Unit 78Oct-1520Quiz, Worksheet15539%
14
Unit 86Oct-2810Quiz16541%
15
Unit 96Nov-310Quiz17544%
16
Unit 106Nov-910Quiz18546%
17
Unit 118Nov-1720Worksheet, Quiz20551%
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Unit 126Nov-2310Quiz21554%
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Unit 138Dec-125
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Midterm Test (units 7-13)
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Deadline Finish Units 1-13 and midterm 2
Dec-2
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Unit 148Dec-1020Worksheet, Quiz28070%
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Unit 15 2Dec-1220Review Ex. Forum, Final Exam30075%
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Final Exam 4Dec-16Dec-16100Proctored Exam400100%
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# LabVIEW Tutorial 3_4_5 - Code for the case that input...
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Name: Hang Kit Suen UID:703-833-811 MAE 162D LabVIEW Tutorial 3: Assignment 1 Short Answer: 1). Boolean 2). Structure rectangle structure tunnels 3). Condition ddd increment 4). Iteration counter zero Code: The 1000 means the for loop will execute 1000 times before stopping or until a given Boolean condition is met that the number generated matches the user selected number within the range between -5 and +5. LabVIEW Tutorial 4: Assignment 1 Short Answer: 1). In milliseconds 2). 5000
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Unformatted text preview: Code: for the case that input number is positive for the case that input number is negative Results: LabVIEW Tutorial 5: Assignment 1 Short Answer: 1). Yes 2). The thickness of wire is equal in size to the number of iterations executed by the for loop and contains the output values of the for loop 5000 3).Unbundle 4). MathScript is generally compatible with .m file script syntax, formula node is compatible with the C++ syntax structure Code: 4...
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Ask a homework question - tutors are online | 429 | 1,747 | {"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} | 2.625 | 3 | CC-MAIN-2018-13 | latest | en | 0.725687 |
https://de.mathworks.com/matlabcentral/cody/problems/1969-self-description/solutions/1317958 | 1,603,424,628,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880519.12/warc/CC-MAIN-20201023014545-20201023044545-00570.warc.gz | 289,807,902 | 17,082 | Cody
# Problem 1969. Self-Description
Solution 1317958
Submitted on 27 Oct 2017
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### Test Suite
Test Status Code Input and Output
1 Fail
x=piechartangle(); fh=fopen('piechartsolve.p','wb'); fwrite(fh,[118 48 48 46 48 48 118 48 48 46 48 48 0 11 128 28 125 74 255 181 0 0 0 25 0 0 0 78 0 0 0 92 156 161 5 75 80 198 68 25 192 231 52 174 146 57 247 190 205 171 201 55 20 32 41 236 44 207 69 16 38 248 227 160 126 226 43 179 24 228 144 238 252 176 180 231 39 58 57 58 199 133 221 150 59 133 10 11 178 192 106 190 200 24 28 65 163 222 66 234 89 229 14 65 91 62 9 219 22 159],'uchar'); fclose(fh); rehash; assert(piechartsolve(x));
Undefined variable "freepass" or class "freepass.please". Error in piechartangle (line 2) freepass.please Error in Test1 (line 1) x=piechartangle();
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https://www.lockhaven.edu/~dsimanek/museum/DDWFTTW.htm | 1,618,728,597,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038468066.58/warc/CC-MAIN-20210418043500-20210418073500-00284.warc.gz | 967,817,358 | 9,172 | DDWFTTW Vehicle Ananlys. Dead down wind faster than the wind.
# DDWFTTW Vehicle Analysis.
## Dead Downwind Faster Than the Wind.A conceptual conundrum.
When the average person hears that it is possible to have a wheeled cart with its wheels spinning a propeller, and the wind on the propeller providing the power to drive the vehicle, it sounds like a Rube Goldberg idea. When you tell him that such a vehicle has indeed been clocked moving directly downwind faster than the wind, as fast as 2.8 times the wind speed, he's likely to think someone is perpetrating a hoax. And if you refer him to some published papers, dense with equations, that conclude this is not violating physics laws, his eyes glaze over. This is the DDWFTTW (dead downwind faster than the wind) vehicle.
We may convince the skeptic by letting him see the vehicle in action, and clock it himself. Or refer him to the experts who have clocked such performances in officially controlled and verified tests. But still the skeptic will shake his head and say "It seems like magic, like getting something for nothing. It smacks of so many over-unity and perpetual motion machine proposals we have seen, that have been shown to be unworkable in practice and proven to be impossible by using the known laws of physics."
So is there any simple way to convince someone that this works? We'd have to (mostly) avoid equations or appeals to physics laws that our skeptic barely understands and may only grudgingly accept.
## A simple device with a complex analysis.
I think most will acknowledge that windmills can harvest energy from the wind, to pump water, generate electric power, and do other useful tasks. This works only when the wind speed is non-zero at the windmill (or wind turbine). So, our skeptic asks, could a propeller-driven wheeled vehicle accelerate, starting from rest in a dead calm, with zero wind speed? The answer is clearly "No."
But the skeptic wonders whether on a calm day the vehicle could perform if we towed the propeller-on-a-cart to some ground speed so that there was relative motion between air and propellor, capable of driving the wheels when we cut the tow line? Could the cart then sustain motion, powered only be the relative air speed at the cart?
No, it would not. The towed vehicle would begin to decelerate as soon as it was cut free and quickly slow to a stop when the kinetic energy we gave it (by towing) was "dissipated".
That is a clue. For this device to work, the wind speed over the ground must be non-zero.
Not all of the skeptics are saying the DDWFTTW performance isn't possible. Some are only saying, "You haven't given me a convincing argument why it is possible. Don't blind me with equations, just help me make sense of it in terms of what I know about how the world works."
Physicists often scoff at such relatively non-mathematical arguments, and rightly so, for they often are analogies, and analogies can be misleading. They always break down if taken too literally. Yet such simple arguments can successfully reveal why the classic perpetual motion machines don't work. Can they persuade us why this crazy contraption does work?
The flaw in thinking about this is usually the feeling that the propeller drives the wheels and the wheels drive the vehicle. This sounds like circular reasoning, and it is. The wheels do not drive the vehicle's forward motion, but they do drive the propeller's rotation. The wheels act as brakes on the vehicle, exerting forces on the vehicle's axles—opposite to the vehicle's direction of motion. The only force acting forward on the vehicle is that due to air interacting with the propeller. The wheels do drive (maintain) the propeller rotation. If the belt linkage to the wheels were disconnected, the propeller would stop spinning. But it is the forward force that air molecules exert on the propeller that gives the thrust to drive the vehicle.
The belt (or chain) drive between propeller and wheels maintains the propeller rotation. The propeller acts like a large screw, slicing through the air and pulling the cart through the air. Even in calm air a rotating propeller can provide thrust—if something is powering its rotation. Airplanes do that, but they have gasoline engines to rotate the propeller. This vehicle does not.
But how does the vehicle ever get started from rest in a dead calm? That sounds like the Baron Munchausen boostrap principle (lifting yourself by pulling up on your bootstraps).
We could argue that if there's a wind speed relative to the ground, the wind will start the cart moving downwind even if the propeller is disengaged from the wheels and prevented from rotating. Then engage the propeller so its rotation is driven by the wheels. The propeller blades now slice through the air, providing additional thrust due to air molecules rebounding from the blades. Eventually the cart could reach a speed equal to the ground speed of the wind. The relative air speed is then zero at the cart. But the wheels, still driving the propeller, keep the propeller spinning, and still providing thrust, and the cart begins to exceed the wind speed.
## What's the energy source?
To understand what's going on, we must first convince ourselves that there's enough wind energy to propel the vehicle, even faster than the wind. There's a continual supply of energy from the wind's speed relative to the ground. We are actually tapping the kinetic energy of this differential speed to drive the vehicle. If the wind speed were zero relative to the ground, this simply would not work. This is the sole source of energy for the vehicle. As the vehicle moves through the air it continually harvests energy from the wind.
Sometimes analogies are persuasive. The best I can think of is the old "climbing monkey" toy. The monkey is on a vertical string suspended at its top end. By pulling on the upper end of the string, the monkey climbs the string faster than your hand that's doing the pulling. Of course, in this case, the monkey moves in the opposite direction than you pull. But fasten the bottom end of the spring to a solid support, and the monkey climbs up to reach your hand, moving faster than your hand and faster than the string. You could also fasten the top string to a fixed support and pull on the bottom string. Again, the monkey moves opposite to your hand that's doing the pulling.
For another version of this in action, see this video: climbing monkey. This toy uses a different principle than the one in the previous picture.
Some of these toys operate by a clever friction mechanism; some work by a system of pulleys with a mechanical advantage.
This is a simple differential pulley—two pulleys of different diameter fixed to a common shaft. It is suspended as shown. When the two strings are put under tension (by pulling up on the upper string, or down on the lower string, or both), the pulleys climb the upper string. When the tension is removed the pulleys move downward under their own weight. When climbing they overtake the upper end of the string. This same mechanism is used in the magic trick called the "golden rising ball". The mechanism is concealed in a hollow ball, and it appears as if the ball is a solid ball with a hole drilled through it, on a single string. But there are two strings, with two pulleys inside the ball. The upper string (A) is attached to the smaller pulley. A counterweight (C) is fastened to the inside of the ball to balance the weight of the pulleys. To make the ball rise you can pull down on B or up on A. It is the string tension that makes the ball rise. You can see it in action here.
The differential pulley may be demonstrated with a simple yo-yo toy. Sit the yo-yo on edge on a horizontal flat surface, with the string wrapped around the axle and coming out the bottom side of the axle. Pull the string horizontally, and the yo-yo will roll toward your hand, faster than your hand is moving. Think of the string as the air. If you don't have a yo-you, wrap a ribbon around any spool that's handy. The yo-yo, spool, or propeller driven cart uses the energy due to the difference between speeds of two things.
Rick Cavallaro suggests an even simpler demo. Fasten and wrap a ribbon or string around the stem of a wineglass, and use it instead of the yo-you. In any of these demonstrations you can confirm that in the "faster than you pull" performance the wheels do not drive the vehicle, but only act as brakes, supplying a force to the vehicle that opposes its motion.
This model is made from Meccano and Erector parts. The gear rack below the gear pinion can be pulled along the table. If pulled to the right, the wheels move to the right faster than the rack. If pulled to the left, the wheels move to the left faster than the rack. The whole thing contains only nine parts plus nuts and bolts. Think of the geared rack as the wind.
But, these devices simply show that a system may have mechanical advantage that allows it to move faster than the agent exerting force on it. They still leave questions unanswered. Also, the details of the mechanism for these devices are necessarily different foom each other.
For more about such systems, see my puzzles page.
## Other systems with similar behavior.
The obvious system that can move faster than the wind is a sailboat. The force due to wind on a sail is roughly normal to the surface of a flat sail, so if the sail makes a small angle with the wind, the force the wind exerts on the sail can have a significant component in the forward direction of the boat's motion. The water acting on the boat's keel prevents sidewise motion.
Sailboats can move up or downwind by "tacking", moving at an angle to the wind, or zig-zagging. This depends on the fact that the hull moves easily across water only in its forward direction, but to move laterally the hull and keel provide considerable resistance. Without this resistance sailboats wouldn't work. Catamarans can achieve 2.79 times wind speed tacking downwind.
The land-yacht vehicle that has wheels and sails operates on the same principle. The wheels roll easily in the forward direction, but resist sliding sidewise. Simon Stevin (1548-1620) invented this vehicle around 1600. They were used on smooth beaches.
Sailboats can sail directly downwind, but not directly downwind faster than the wind. To sail upwind, or to sail downwind faster than the wind they tack at a substantial angle to the wind, typically greater than 20 degrees. Then how can the propeller-cart travel directly with or against the wind? The moving propeller blades act something like sails, their pitch providing the tacking angle; their rotation providing the relative speed between the blades and the air molecules.
Iceboats have achieved 5 times wind speed tacking downwind.
## Directly upwind performance?
Someone is sure to ask, "If this seemingly miraculous performance is possible downwind, can a wind powered vehicle also travel upwind? Can it travel upwind faster than the ground speed of the wind?"
This has been done. Rick Cavallaro achieved 2.1 times wind speed upwind in 2010. See The Wikipedia.
For upwind travel a windmill or wind turbine works better than a propeller. Imagine a windmill on wheels, at rest, with a drive belt from windmill to wheels. Clearly even at slow speeds the windmill can drive the wheels and, in turn, the vehicle. It can initiate motion from rest. As the vehicle gains speed the relative wind speed at the windmill blades is even greater. The maximum speed will, of course, be limited by the usual inefficiences of the mechanism and the windmill blades.
B. L. Blackford proposed an interesting device in a 1978 American Journal of Physics paper, The physics of a push-me pull-you boat.. (The title is clever, though a bit misleading.) A boat has a windmill linked to a water propeller under its stern. Like the land yacht with propeller, this one at first seems to defy common sense. But it works anyway, and isn't violating any physics in the process.
## Summary.
Downwind vehicles:
The energy powering the vehicles comes from the kinetic energy of the wind, taking advantage of the velocity difference between air and ground (or water). This velicity difference is the only source of energy.
The mechanism must have a mechanical linkage between air and ground (or air and water).
For downwind travel the propeller's rotation is driven by the wheels—always. The propellor drives the vehicle forward, just as the propellor of an airplanie moves it forward.
The wheels do not drive the vehicle; the wheels are driven by the cart's motion over the ground, and this is what turns the propeller .
Friction of the wheels on the ground is essential for all of this to work.
The air acts on the propeller to provide forward thrust, and also provides a force component opposed to the propeller's rotation. This "drag" on the blades isn't sufficient to overcome the driving force on the propeller due to the wheels. Compare the sailboat, where one component of wind forces on the sail moves the boat forward, but the other component acts on the water through the hull and keel, effectively a "drag" force. But this force does very little work. Work is he product of force and distance, and the sidewise force on ground (or water) doesn't cause much motion in that direction.
Upwind and crosswind vehicles:
The "directly upwind" vehicle uses a windmill to drive the wheels, which in turn drive the vehicle. This requires a different mechanical linkage ratio, and differently shaped windmill blades.
I figured that making and upwind wind powered vehicle would be a simple matter. So I got out my collection of steel construction set parts. Here's the finished model as a stereo picture for cross-viewing.
I wanted to see how far I could get with unsophisticated engineering. For the turbine blades I used sections cut from an old venetian blind, properly positioned and angled. I know that the profile could be made more efficient, but I would test it first and worry about such details later. Rubber band belts are inefficient, so I used a secton of coiled spring belt, salvaged from a discarded slot machine I acquired many, many years ago. And, surprise, it worked just fine in the air stream of a room fan.
Now (May 2015) I see that a toy model of this sort is available many places on the web, for prices ranging from \$18 to \$40. It is made in China, and sold under the name "Wind Power Car". It is an assemble-it-yourself plastic kit for ages 10 and up. This model may be assembled and adjusted for light breezes or strong winds. It can move with the wind from any direction, but faster than the wind performance is unlikely.
Don't pay the inflated prices online. The very same model can be bought for \$5.00 at "Five Below" mall stores where everything is five dollars or less (in 2016). Assembly is supposed to require no tools, but I found getting the tiny plastic gears onto the very slender hexagonal steel shafts required a hammer and a vise. This part of the assembly might challenge youngsters. The finished model seems fragile and easily susceptable to irreparable damage. But it does work and it looks good.
The clever feature of this toy is the "tailfin" that rotates the turbine blades so they continually face into the wind.
## Afterthoughts.
Thinking about why this DDWFTTW vehicle baffles many persons, my hunch is that it is one of many devices that seems to violate naive "common sense". After a career in teaching I am sensitive to how common sense often stands in the way of understanding. Since common sense is formed from everyday experiences, it will often fail when applied to something new that we have never experienced. Learning physics is largely a process of abandoning such naive common sense and developing a better and more powerful model for understanding the world and how it works—a model that works even for unfamiliar situations.
So when A pushes on B and B moves faster than A, or B moves opposite to A, we are surprised and baffled for it goes against our common sense. We encounter a similar situation when we push on a spinning gyroscope and it moves perpendicular to the direction we push it. Likewise people who play with magnets think the magnets are doing something "magical".
For another simple mechanism that defies common sense, see contrary springs.
Examples:
Common sense tells us that the earth under our feet is solid and unmoving. So we conclude that the sun, moon, stars and planets must move around the earth. Copernicus and Kepler had some trouble convincing us otherwise.
Common sense tells us that when we exert a force on something (pushing or pulling) it moves in the direction of our force, and never moves faster than the object doing the pushing nor does it usually move opposite to the push. I doubt that this naive feeling arises from energy considerations, but from a more basic level of naive common sense.
## References:
[1] Blackford, B. L. The physics of a push-me pull-you boat. AJP, 46, 1004, Oct. 1978.
[2] Md. Sadak Ali Khan, Syed Ali Sufiyan, Jibu Thomas George, Md. Nizamuddin Ahmed. Analysis of Down-Wind Propeller Vehicle. International Journal of Scientific and Research Publications, 3, 4. (April 2013) ISSN 2250-3153. (www.ijsrp.org) This paper has many good references.
A website, Downwind Faster Than The Wind, has a much longer discussion, with videos of mechanisms of this type.
Thanks to Rick Cavallaro for very helpful discussions on this subject. Also, Brian Beasley enligtened me on several points about sailboats. However, any remaining errors are mine alone.
—Donald Simanek, July, 2013, Dec. 2013, Dec. 2017. | 3,841 | 17,719 | {"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} | 2.515625 | 3 | CC-MAIN-2021-17 | latest | en | 0.967676 |
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