url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.twblogs.net/a/5f04bc2187c12340342571e5 | 1,603,550,342,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00500.warc.gz | 934,188,356 | 11,091 | # D. Maximum Sum on Even Positions-----------------------思維(類最大子段和)
``````#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+10;
ll a[N];
int t,n;
int main()
{
cin>>t;
while(t--)
{
cin>>n;
ll ans=0;
for(int i=0;i<n;i++)
{
cin>>a[i];
if(i%2==0) ans+=a[i];
}
ll even=0,old=0,maxv=0;
for(int i=0;i<n;i+=2)
{
old=max(0ll,a[i-1]-a[i]+old);
maxv=max(maxv,old);
}
for(int i=1;i<n;i+=2)
{
even=max(0ll,a[i]-a[i-1]+even);
maxv=max(even,maxv);
}
cout<<ans+maxv<<endl;
}
}
`````` | 205 | 503 | {"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-2020-45 | latest | en | 0.113154 |
https://pastebin.com/mTLqmN8x | 1,618,695,069,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00517.warc.gz | 547,178,822 | 6,354 | # Untitled
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1. In the girl-named-Florida problem our information concerns not just the gender of the children, but also, for the girls, the name. Since our original sample space should be a list of all the possibilities, in this case it is a list of both gender and name. Denoting “girl-named-Florida” by girl-F and “girl-not-named-Florida” by girl-NF, we write the sample space this way: (boy, boy), (boy, girl-F), (boy, girl-NF), (girl-F, boy), (girl-NF, boy), (girl-NF, girl-F), (girl-F, girl-NF), (girl-NF, girl-NF), and (girl-F, girl-F).
2.
3. Now, the pruning. Since we know that one of the children is a girl named Florida, we can reduce the sample space to (boy, girl-F), (girl-F, boy), (girl-NF, girl-F), (girl-F, girl-NF), and (girl-F, girl-F). That brings us to another way in which this problem differs from the two-daughter problem. Here, because it is not equally probable that a girl’s name is or is not Florida, not all the elements of the sample space are equally probable.
4.
5. In 1935, the last year for which the Social Security Administration provided statistics on the name, about 1 in 30,000 girls were christened Florida. 5 Since the name has been dying out, for the sake of argument let’s say that today the probability of a girl’s being named Florida is 1 in 1 million. That means that if we learn that a particular girl’s name is not Florida, it’s no big deal, but if we learn that a particular girl’s name is Florida, in a sense we’ve hit the jackpot. The chances of both girls’ being named Florida (even if we ignore the fact that parents tend to shy away from giving their children identical names) are therefore so small we are justified in ignoring that possibility. That leaves us with just (boy, girl-F), (girl-F, boy), (girl-NF, girl-F), and (girl-F, girl-NF), which are, to a very good approximation, equally likely.
6.
7. Since 2 of the 4, or half, of the elements in the sample space are families with two girls, the answer is not 1 in 3—as it was in the two-daughter problem—but 1 in 2. The added information—your knowledge of the girl’s name—makes a difference.
RAW Paste Data | 587 | 2,219 | {"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-2021-17 | latest | en | 0.943035 |
https://www.datasciencecentral.com/profiles/blogs/simple-proof-of-prime-number-theorem | 1,611,842,925,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704847953.98/warc/CC-MAIN-20210128134124-20210128164124-00033.warc.gz | 684,625,798 | 27,467 | .
# Little Proof of the Prime Number Theorem
This article has been updated, and now includes two new sections at the bottom (sections 4 and 5), featuring interesting results, more accurate approximations, and nice mathematical formulas involving prime numbers and the famous Euler constant. The last formula at the bottom of this article is remarkable.
The prime number theorem is a famous result in number theory, that characterizes the asymptotic distribution of prime numbers: For instance, the fact that the n-th prime number is asymptotically equivalent to n log n. By definition, two quantities f(n) and g(n) are asymptotically equivalent, denoted as f(n) ~ g(n), if the ratio f(n) / g(n) tends to 1 as n tends to infinity.
The standard proof of the prime number theorem is extremely long and complicated, and requires knowledge of advanced mathematical theories. Here we propose a short, elementary proof that even high school students can understand. To make it rigorous, there are a number of points that require a much deeper dive. So this proof is just a sketch, but it is rather intuitive.
1. Simple proof of the prime number theorem
Let's start with the Legendre formula for n! (factorial n), easy to derive:
Here the brackets represent the floor function. Taking the logarithm, we obtain
The rightmost sum is over all primes p less than or equal to n (here the set Q(n) denotes all primes less than or equal to n.) Note that these sums are actually finite: the terms vanish respectively when p > n or k > log n / log p.
Since, based on the Stirling approximation, log n! is asymptotically equivalent to n log n, the above formula yields
Now, using the general algebraic lemma below, applied to the set Q of prime numbers, we conclude that the difference d(p) between the prime number p and the largest prime number smaller than p, is on average asymptotically equivalent to log p. It follows immediately that the n-th prime number is asymptotically equivalent to n log n. This completes the proof of the prime number theorem.
2. General Algebraic Lemma
This is a classic algebraic result that applies to many sequences of slowly increasing positive integers, not just to prime numbers. If Q is an infinite set of positive integers, with Q(n) being the subset of all integers in Q that are less than or equal to n, then under rather general conditions described below, we have
where d(p) is the difference between p and the largest element of Q that is smaller than p. I could not find a reference for this result, but it is rather intuitive, and it is probably an old theorem.The asymptotic equivalence is guaranteed if
and d(p) / p tends (on average) to zero as p tends to infinity. The above condition simply states that the difference between the left and right hand side of the asymptotic relationship, should be an order of magnitude smaller than log n as n tends to infinity. As long as this condition is met, even if there are increasingly large gaps among the elements in Q, then the result is valid. The proof is easy and based on the fact that the difference between the left and right hand side is well approximated by
especially for large values of d(p). When the gaps between the successive elements of Q are small (that is, when the d(p)'s are small) the result is even more obvious.
3. Alternate solution
Here I propose a different path to arrive at the final result, though it starts with the same Legendre formula and subsequent asymptotic development.The alternate solution is as follows. In the formula
replace the k-th prime -- let's denote it as p(k) -- by
where Ca, and b are arbitrary constants, with a > 0. Of course this assumes that p(k) must have this relatively standard approximation. Then the asymptotic formula becomes
where h(n) is the number of primes less or equal to n. Using the asymptotic relationship for p(k), by inversion, one can find an asymptotic relationship for h(n), based solely on a, b, C, and elementary functions:
Then one must prove that C = 1, a = 1, and b = 1, for the last sum to be asymptotically equivalent to log n. This last part is just a pure calculus exercise that does not involve playing with properties of integers or prime numbers. It is obvious though, that if C = 1, a = 1, and b = 1, then the asymptotic formula is true.
4. Improving the prime number theorem
So far we only used first order approximations such as log n! ~ n log n. We can obtain more accurate asymptotic formulas by taking into account second and third order terms. For instance, a better approximation for log n! is log n! ~ n log n - n. Basil Gordon showed that you could obtain an elementary proof of the prime number theorem using this approximation (reference: click here and go to page 29 for the general context; for the actual result, click here.)
There is abundant literature about more accurate formulas for the distribution of prime numbers. The most popular one is based on the logarithmic integral. Here our goal is more modest: trying to use more than the first term in the Legendre expansion in an attempt to improve accuracy. Although we do not give an answer to this question here (we show how to do it though) we stumbled upon some interesting problems and solutions.
If we include more terms both from the log n! approximation and from the Legendre formula, we get
where
The quantity e(npk) is between 0 and 1, and when averaged, it converges to a value close to 1/2; details are discussed in the next section.
This can be rewritten as
where N = N(n,p) and
with E(n,p) between 0 and 1, and indeed closer to 0.5. This can be further approximated as:
The computation of E(np) is a tricky part, though we provide some hints in the next section. In a nutshell, assuming the average value of E(n,p) is equal to a constant c (slightly below 0.5) then the last sum is asymptotically equivalent to c |Q(n)| ~ n / log n where |Q(n)| is the number of elements in Q(n). This yields the simplified formula
with 1 - c close to 0.577 (Euler's constant) based on computing the previous sum over the first million primes. See next section for details.
Can this formula be used to get a more accurate version of the prime number theorem? Maybe, but it might not be easy. Maybe it could lead to something like Dusart's inequality, stating that for n > 5, the n-th prime p(n) satisfies p(n) < n log n + n log log n.
5. Interesting problems
In this section we describe two problems related to our discussion
Average of fractional parts
Sums involving fractional parts have been studied for a long time, and interesting results have been obtained. For instance, it is well known that
where g is Euler's constant, see here for details. A similar formula for all integers (not just for the primes), involving the same 1 - g factor, also holds:
More on the subject can be found here. More on the generalized Euler constant can be found here and here. Since we are dealing here with sums such as E(n,p) that are connected to Euler's constant, it is no surprise that 1 - c in our last result, is Euler's constant, though it remains to be formally proved.
Sums involving primes
Combining the last formula in section 4 with the following well known result involving the constant B3 = 1.3325822757... (click here and here for details)
we get
Taking the limit as n tends to infinity, we eventually get the following convergent series, where the sum is over all primes:
It is quite remarkable that this last result was obtained by substracting two diverging series for which only asymptotic results are available. This is one of the few known results (with exact solution) for converging series involving primes. A more general one can be found in Don Zagier's article (Newman's Short Proof of the Prime Number Theorem, see section IV. )
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Comment | 1,809 | 7,866 | {"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} | 4.59375 | 5 | CC-MAIN-2021-04 | longest | en | 0.910785 |
http://www.studymode.com/essays/Introduction-To-Matlab-1275894.html | 1,527,451,155,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00101.warc.gz | 470,698,195 | 25,193 | Introduction to Matlab
Topics: Step response, Control theory, Systems theory Pages: 2 (251 words) Published: December 2, 2012
Lab 1
Objectives:
1. The student should be able to understand the used of Matlab simulation package for analysis. 2. The student should be able to understand the used of simulink tool to analyse a given system.
Part A: Matlab programming 1) Use MATLAB to generate the transfer function
G(s) =
5(s+15)(s+26)(s+72) s(s+55)(s+47)(s+56)(s+89)
in the following ways: The ratio of factors (i) (ii) The ratio of polynomials (b) Test the transfer function for (i) Stability Step response. Peak time? Settling time? % Overshoot? (ii) (iii) Steady state error for a unit step input
1
Part B: Simulink tool. 1) Find the transfer function T(s) = C(s)/R(s) for the system shown in Figure 1. Given that G1(s) = 1/(s+7), G2(s) = 1/(s2+2s+3), G3(s) = 1/(s+4), G4 = 1/s G5(s) = 5/(s+7), G6(s) = 1/(s2 + 5s+10), G7=3/(s+2) and G8 = 1/(s+6). {Hint : Use the connect and blkbuild commands in MATLAB's Control System Toolbox} (a) Analyse the system for stability, steady-state error and time response to a unit step input, a unit ramp input and a hyperbolic input. (b) Connect the blocks in SIMULINK and verify for the time response obtained in (a) .
G8(s)
G6(s) R(s) G1(s) + + + G3(s) + G7(s) + G2(s) + + G4(s) +
C(s)
G5(s)
Figure 1
2 | 449 | 1,352 | {"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} | 3.75 | 4 | CC-MAIN-2018-22 | latest | en | 0.73256 |
https://www.vedantu.com/question-answer/ways-to-get-a-6digit-telephone-number-whi-class-12-maths-cbse-5ee87192e8223517fbdc4908 | 1,619,176,756,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00172.warc.gz | 1,146,558,690 | 83,836 | Question
# How many ways to get a 6-digit telephone number which has at least one of their digits repeated ?
Hint: Let us find the number of ways in which no digit is repeated and then subtract that by the total number of ways of forming a 6-digit number.
As we know that we are asked to find a number of ways when at least one digit is repeated. And this can also be written as the number sum of ways when exactly one digit is repeated, when exactly two digits are repeated, when exactly three digits are repeated, when exactly four digits are repeated, when exactly five digits are repeated and when all six digits are repeated.
So, this means we had to exclude ways in which no digit is repeated from the total number of ways.
So, there are a total 10 possible digits for each of the six places and that were {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
So, total number of ways of forming 6-digit telephone number without any restriction will be 10*10*10*10*10*10 = 1000000.
And if all digits are different,
Then the number of possible digits for ${1^{st}}$ place will be 10.
Number of possible digits for ${2^{nd}}$ place will be 9.
Number of possible digits for ${3^{rd}}$ place will be 8.
Number of possible digits for ${4^{th}}$ place will be 7.
Number of possible digits for ${5^{th}}$ place will be 6.
Number of possible digits for ${6^{th}}$ place will be 5.
So, total number of ways for forming a 6-digit number in which all digits are different will be 10*9*8*7*6*5 = 151200.
So, the total number of 6-digits telephone numbers possible in which at least one digit is repeated will be 1000000 – 151200 = 848800.
Hence, the correct answer will be 848800.
Note: Whenever we come up with this type of problem where we are given that at least 1 digit is repeated then we should find the number of ways in which all the digits are different and then subtract that by the total number of possible ways without any restriction. This will be the easiest and efficient way to find the solution of the problem. | 507 | 2,005 | {"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} | 4.6875 | 5 | CC-MAIN-2021-17 | longest | en | 0.899217 |
https://www.bottleshockmovie.com/strategies-with-protection-a-mild-intro-in-order-to-crypto-stock-markets/ | 1,627,651,770,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00282.warc.gz | 683,104,566 | 8,603 | # Strategies With Protection: A Mild Intro In order to Crypto Stock markets
Allow us consider the case in point of scrambling an egg. 1st, crack the shell, pour the contents into a bowl and defeat the contents vigorously right up until you accomplished the needed result – well, a scrambled egg. This action of mixing the molecules of the egg is encryption. Considering that the molecules are mixed-up, we say the egg has accomplished a larger point out of entropy (state of randomness). To return the scrambled egg to its unique kind (which includes uncracking the shell) is decryption. Unattainable?
Even so, if we substitute the word “egg” and exchange it with “quantity”, “molecules” with “digits”, it is Attainable. This, my buddy, is the exciting planet of cryptography (crypto for short). It is a new area dominated by gifted mathematicians who uses vocabulary like “non-linear polynomial relations”, “overdefined programs of multivariate polynomial equations”, “Galois fields”, and so forth. These cryptographers utilizes language that mere mortals like us cannot pretend to understand.
In the personal computer, every little thing saved are quantities. Your MP3 file is a quantity. Your text concept is a amount. Your deal with book is a longer number. The variety sixty five represents the character “A”, ninety seven for the modest “a”, and so on.
For people, we identify quantities with the digits from to 9, the place else, the computer can only understand or one. This is the binary program which utilizes bits rather of digits. To transform bits to digits, just basically multiply the number of bits by .3 to get a great estimation. For illustration, if you have 256-bits of Indonesian Rupiah (1 of the cheapest forex denomination in the globe), Monthly bill Gates’ prosperity in comparison would be microscopic.
The hexadecimal (foundation sixteen) technique uses the 10 digits from to 9, furthermore the six added symbols from A to F. This established has sixteen diverse “digits”, hence the hexadecimal title. This notation is useful for computer employees to peek into the “real contents” stored by the pc. Alternatively, take care of these various number methods as currencies, be it Euro, Swiss Franc, British Pound and the like. Just like an object can be priced with diverse values using these currencies, a number can also be “priced” in these distinct amount methods as well.
To digress a little bit, have you at any time wondered why you had to review primary quantities in university? I am sure most arithmetic academics do not know this reply. Solution: A subbranch referred to as general public-essential cryptography which utilizes primary quantities specifically for encrypting e-mails. Over there, they are chatting of even larger figures like 2048, 4096, 8192 bits.)
When we want to encrypt something, we require to use a cipher. A cipher is just an algorithm comparable to a recipe for baking a cake. quantstamp has specific, unambiguous methods. To have out the encryption process, you need a key (some called it passphrase). A great apply in cryptography requirements the essential utilised by a cipher must be of substantial entropy to be effective.
Data Encryption Common (DES), released as a regular in the late 1970’s, was the most commonly used cipher in the 1980’s and early 1990’s. It employs a fifty six-bit essential. It was broken in the late 1990’s with specialised computers costing about US\$250,000 in 56 hours. With modern (2005) hardware, it is achievable to crack inside of a day. | 748 | 3,541 | {"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} | 2.9375 | 3 | CC-MAIN-2021-31 | latest | en | 0.913444 |
https://www.domainconverters.com/weight-and-mass/dekatonnes-conversion/ | 1,709,611,821,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00250.warc.gz | 728,748,859 | 14,709 | # Dekatonnes a Weight Unit and Conversion Chart - DomainConverters
Domainconverters > Mass and Weight Converter > Dekatonnes Conversion
## Q: What is Dekatonnes ?
Answer: It is SI multiple of weight and mass unit kilogram .
## Q: How much is 1 Dekatonnes ?
Answer: 1 Dekatonnes is 1.00E+04 times kilogram.
Symbol : not available
So As 1 dekatonnes = 1.00E+04 kilogram.
Let us look at an Example : 35 dekatonnes = 1.00E+04 X 35 kilograms.
Or we can say, 35 dekatonnes = 350000 kg.
## Dekatonnes Conversion Table & Chart
Compare one Dekatonnes with other weight units
3502626.9702277 rebahs11.023113109244 short-tons
57.101471301634 slinches685.21765561961 slugs
1574.7304441777 stones1.0E-5 teragrams
787.36522208885 tods321507.46361894 troy-ounces
26792.288778477 troy-pounds393.68261104442 trusses
1.0E+31 yoctograms1.0E-17 yottagrams
1.0E+28 zeptograms1.0E-14 zettagrams
6.0240963855422E+30 avograms220.46226218488 centals
1000000000 centigrams3.7139928394218 chaldrons
3149.6062992126 cloves111234705.22803 criths
6.0240963855422E+30 daltons100000000 decigrams
100 decitonnes1000000 dekagrams
1 dekatonnes2631578.9473684 drachmes
2705121959.071 drams1000000000 dynes
1.0977693108051E+34 electrons6.25E+22 electronvolts
1.0E-11 exagrams1.0E+22 femtograms
0.01 gigagrams1.0E-8 gigatonnes
154320987.65432 grains10000000 grams
2615062.7615063 groses100000 hectograms
1019.7162129779 hyls5.2631578947368E-24 jupiters
10000 kilos6.0221366516752E+27 kilodaltons
10000 kilograms1019.7162129779 kilograms-force
0.01 kilotonnes2.2480894387096 kips | 564 | 1,552 | {"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-2024-10 | latest | en | 0.400452 |
http://mathhelpforum.com/calculus/178523-method-least-squares.html | 1,481,185,361,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542455.45/warc/CC-MAIN-20161202170902-00026-ip-10-31-129-80.ec2.internal.warc.gz | 176,460,482 | 12,330 | # Thread: Method of Least Squares
1. ## Method of Least Squares
Question:
A capacitor C is initially charged to 10V and then connected across a resistor R. The current through R is measured at 1 milisecond intervals is:
t : 0 1 2 3 4 ms +-0.5%
i : 4.5 2.8 1.5 1.0 0.6 mA +-2.5%
(Basically this is a table where i corresponds to the t value)
Using the method of least squares, find R and C.
I need help on what equation I should use. Is it i=ie-(t/RC)?
2. Originally Posted by Salcybercat
Question:
A capacitor C is initially charged to 10V and then connected across a resistor R. The current through R is measured at 1 milisecond intervals is:
t : 0 1 2 3 4 ms +-0.5%
i : 4.5 2.8 1.5 1.0 0.6 mA +-2.5%
(Basically this is a table where i corresponds to the t value)
Using the method of least squares, find R and C.
I need help on what equation I should use. Is it i=ie-(t/RC)?
If you meant I = I0 * e^{-t/(RC)}, then yes. So take the ln of both sides to do your linear regression.
-Dan
3. Just to make sure, I0 would be the initial current when t=0, which would make the equation I = 4.5 * e^{-t/(RC)}. Am I right?
4. Originally Posted by Salcybercat
Just to make sure, I0 would be the initial current when t=0, which would make the equation I = 4.5 * e^{-t/(RC)}. Am I right?
I0 is one of the pieces of information you want to calculate. Take the ln of both sides of the equation:
$I = I_0 e^{-t/(RC)}$
$ln(I) = ln \left ( I_0 e^{-t/(RC)} \right )$
$ln(I) = ln(I_0) - \frac{1}{RC} \cdot t$
$ln(I) = -\frac{1}{RC} \cdot t + ln(I_0)$
The advantage of this form is that the equation is now in the form y = mx + b where y = ln(I) and x = t. So to do the linear regression use the t values as your x data and ln(I) values as your y data. You will come out with a slope and an intercept. The slope will be equal to -1/(RC) and the intercept will be ln(I0).
-Dan
5. Thank you for the clarification! However, I still didn't manage to get the answer. (R=2.2kohm and C = 0.89microFarad)
I've attached my workings below. I hope it's clear enough for you to understand. Do comment on any mistaken steps.
6. Originally Posted by Salcybercat
Thank you for the clarification! However, I still didn't manage to get the answer. (R=2.2kohm and C = 0.89microFarad)
I've attached my workings below. I hope it's clear enough for you to understand. Do comment on any mistaken steps.
I cheated and did the regression on my calculator. My answers are close, but different from yours. If you are allowed to use a computer to do the problem I'd recommend it.
I get m = 0.505943 x 10^3 /s and c = 1.497552. Thus I'm getting 1/(RC) = 0.505943 x 10^3 /s and I0 = 4.47073 x 10^{-3} A.
Now we have to get a bit wily. We only have one number to get RC so we need another relationship. We get that through the definition of capacitance:
C = Q/V, where Q is the initial charge on the capacitor (unknown at this point) and V is the charging potential 10 V. So...
$\frac{1}{RC} = \frac{1}{R \frac{Q}{V}} = \frac{V}{QR}$
Now recall that V = I0*R when the capacitor starts to discharge:
$\frac{1}{RC} = \frac{V}{QR} = \frac{I_0}{Q}$
and we have values of 1/(RC) and I0 from the regression. So you can find Q. From Q you can find C. From C you can find R.
Using my values I get your given answers. Your values give answers slightly off from that. I am at a loss as to what advice to give you to allow you to calculate better numbers.
-Dan
7. I must've done something wrong rounding up the numbers when calculating it manually (and no, we wouldn't be allowed to use computers during our exams )
Yes I was confused about what other equations I needed to use to compare with the RC-equation. ( I thought of using the equation of i = C (dv/dt) but I wasn't sure if the method of least squares would work with a differential equation.
But thank you for the step-to-step explanation! I got the answers
8. Originally Posted by Salcybercat
I must've done something wrong rounding up the numbers when calculating it manually (and no, we wouldn't be allowed to use computers during our exams )
Bummer. I particularly hate doing these "by hand." Glad you got the answers and good luck on the exams!
-Dan | 1,218 | 4,185 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "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-2016-50 | longest | en | 0.90118 |
http://studymoose.com/millikan-oil-drop-experiment-essay | 1,481,012,224,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541886.85/warc/CC-MAIN-20161202170901-00291-ip-10-31-129-80.ec2.internal.warc.gz | 259,102,222 | 10,372 | Haven't found the Essay You Want?
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# Millikan Oil Drop Experiment Essay
The experimental charge of electron was found to be at around x 10 ^ -17. This resulted in a very large value of error because the accepted value of the charge of the electron is at 1. 6 ^ -19. Millikan was able to draw his conclusion on the size of the charge of electron that is to be 1. 6 ^-19 when in his repeated experiments, he was able to find that the results of his experiments are explained on multiples of 1. 6 ^ -19.
It must be that there is more number of electrons that contributes to the value he was obtaining so he looked at the multiples. For this experiment, the obtained value for the multiple is at 1. 30 ^ -17. The large value of error can be explained by the instrument wear-off, inaccurate observation, incorrect incoming voltage for the plates, or worn-off plates. Answers to Questions: 1. If the apparatus would be set up in an unionizable gas, the result could be more consistent.
If we do this, there will be no changes on the gravitational and buoyant forces acting on the drop. None of the equations will be modified because in the assumption, there is neglect on the ionization in the air inside the apparatus. The ionization was not considered in the calculation of the charge. 2. We cannot perform the experiment on a vacuum because we need to apply electric field on the setup so we could measure the velocities and we need the force of gravity.
If the charge is in continuous form and infinitesimally divisible, then, the obtained values will be inconsistent with each other. We may obtain relatively large values or relatively low values that are inconsistent with the other obtained values. But, we will have a very large value for e. Learnings: I’ve realized that most of the constants we use in scientific calculations are not really exact but, actually, approximations, estimations and averages of the many trials performed.
Conclusion: Electricity has an atomic nature. It involves exchange of electrons and the atoms participate in conducting electricity. The result of this experiment is not precise with the accepted value for the charge of the electron obtained by Robert Millikan. However, the same procedure and similar setup was done as that of Millikan’s. References: Rubin, Julian. 2007. The Oil Drop Experiment. Retrieved February 22, 2009 from http://www. juliantrubin. com/bigten/millikanoildrop. html.
Essay Topics:
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We can't stand spam as much as you do No, thanks. I prefer suffering on my own | 593 | 2,716 | {"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-2016-50 | longest | en | 0.947697 |
https://functionx.com/bcb/math/Power.htm | 1,669,645,041,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00555.warc.gz | 301,092,685 | 2,602 | Arithmetic: The Power of a Number
The C++ pow() Functions
```double pow(double Source, double Raise);
long double powl(long double Source, long double Raise);```
The pow() function is used to calculate the value of one number or expression raised to the power of another number. This follows the formula:
`ReturnValue = xy`
The pow() function takes two required arguments. The first argument, x, is used as the base number to be evaluated. The second argument, y, also called the exponent, will raise x to this value. The powl() function performs the same calculation on long double numbers and returns a long double.
In the following example, a form is equipped with a Button control and an Edit control. When the user clicks the button, the constant 205.38 is raised to the power of 4.12. The result displays in the edit box:
```//---------------------------------------------------------------------------
#include <vcl.h>
#include <math.h>
#pragma hdrstop
#include "Unit1.h"
//---------------------------------------------------------------------------
#pragma package(smart_init)
#pragma resource "*.dfm"
TForm1 *Form1;
//---------------------------------------------------------------------------
__fastcall TForm1::TForm1(TComponent* Owner)
: TForm(Owner)
{
}
//---------------------------------------------------------------------------
void __fastcall TForm1::Button1Click(TObject *Sender)
{
const double Source = 205.38;
const double Exp = 4.12;
double Result = pow(Source, Exp);
Edit1->Text = Result;
}
//---------------------------------------------------------------------------```
The IntPower() function
`Extended __fastcall IntPower(Extended Base, int Exponent);`
The VCL’s IntPower() function is used to raise a number, Base, to the integral Exponent power. The first argument of this function, Base, can be an integer, a float, a double-precision number or a long double. The Exponent argument is the factor about which the Base number will be raised.
```//---------------------------------------------------------------------------
void __fastcall TForm1::Button1Click(TObject *Sender)
{
Extended Number, Base;
int Exp;
Base = StrToFloat(Edit1->Text);
Exp = StrToInt(Edit2->Text);
Number = IntPower(Base, Exp);
Edit3->Text = FloatToStr(Number);
}
//---------------------------------------------------------------------------```
The Power() Function
`Extended __fastcall Power(Extended Base, Extended Exponent);`
The Power() function takes a number (any number, including integers, floating, double or long double-precision numbers) as the Base argument and raises it to the power of the Exponent argument, which also can be any number (int, float, double, long double).
```//---------------------------------------------------------------------------
void __fastcall TForm1::Button1Click(TObject *Sender)
{
int Source = 205;
float Exp = 5.25;
double Result = Power(Source, Exp);
Edit1->Text = Result;
}
//---------------------------------------------------------------------------```
` ` | 588 | 3,033 | {"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-2022-49 | latest | en | 0.504915 |
https://mathhelpboards.com/threads/linear-speed-of-a-flywheel.6783/ | 1,632,183,019,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057119.85/warc/CC-MAIN-20210920221430-20210921011430-00653.warc.gz | 446,828,116 | 15,971 | TrigonometryLinear speed of a flywheel
karush
Well-known member
A flywheel with a $15cm$ diameter is rotating at a rate of $\displaystyle\frac{7 rad}{s}$
What is the linear speed of a point on the rim, in $\displaystyle\frac{cm}{min}$.
$s=r\theta$ so $7.5(7) = 152$cm
then $\displaystyle v=\frac{s}{t}=\frac{152cm}{s}\cdot\frac{60s}{min}=\frac{1320cm}{min}$
I am not sure just what a Radian (rad) is in this, so hope I didn't make this to simple. don't have answer so hope mine ok
MarkFL
Staff member
Your method is correct (but you have made some arithmetical errors)...I would write:
$$\displaystyle v=r\omega=\frac{15}{2}\text{ cm}\cdot7\frac{1}{\text{s}}\cdot\frac{60\text{ s}}{1\text{ min}}=?$$
karush
Well-known member
Your method is correct (but you have made some arithmetical errors)...I would write:
$$\displaystyle v=r\omega=\frac{15}{2}\text{ cm}\cdot7\frac{1}{\text{s}}\cdot\frac{60\text{ s}}{1\text{ min}}=?$$
$\displaystyle\frac{3150 cm}{min}$
MarkFL
Staff member
$\displaystyle\frac{3150 cm}{min}$
Correct. The method you used is:
$$\displaystyle v=\frac{s}{t}=\frac{r\theta}{t}=r\frac{\theta}{t}$$
Now defining the angular velocity $\omega$ to be:
$$\displaystyle \omega=\frac{\theta}{t}$$
we then have:
$$\displaystyle v=r\omega$$
That is, the linear velocity $v$ is the product of the radius of motion and the angular velocity.
Did you find the error in your previous calculations?
karush
Well-known member
let me see if this set up ok
a wheel with $30cm$ radius is rotating at a rate of $\displaystyle{3rad}{s}$ what is v in $\displaystyle\frac{m}{s}$
$\displaystyle v=r\omega$ $\displaystyle 30\text{ cm}\cdot3\frac{1}{\text{s}}\cdot \frac{m}{100\text{cm}}=$
MarkFL
Staff member
let me see if this set up ok
a wheel with $30cm$ radius is rotating at a rate of $\displaystyle{3rad}{s}$ what is v in $\displaystyle\frac{m}{s}$
$\displaystyle v=r\omega$ $\displaystyle 30\text{ cm}\cdot3\frac{1}{\text{s}}\cdot \frac{m}{100\text{cm}}=$
Yes, that is correct.
karush
Well-known member
oops just noticed the ans should be in $$\displaystyle \frac{\text {m}}{\text {min}}$$
so...
$\displaystyle 30\text{ cm}\cdot \frac{3}{\text{s}} \cdot \frac{60 \text { s}}{\text { min}} \cdot \frac{\text { m}}{100\text{ cm}}=\frac{54 \text {m}}{\text {min}}$
- - - Updated - - -
Did you find the error in your previous calculations?
yes I had 152 cm it should be 52.5 cm | 793 | 2,408 | {"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} | 4.25 | 4 | CC-MAIN-2021-39 | latest | en | 0.770581 |
https://flatearth.ws/t/newton-law-of-motion | 1,660,171,167,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571222.74/warc/CC-MAIN-20220810222056-20220811012056-00776.warc.gz | 261,745,879 | 31,685 | ## Hovering Helicopter and Earth’s Rotation
Because of inertia, we cannot travel to another location by hovering inside a helicopter and waiting for the Earth to rotate below us until we are above our destination.
Flat-Earthers claim that we cannot do such a thing as ‘proof’ of a motionless Earth. In reality, when still on the ground, the helicopter is already moving at the same velocity as Earth’s surface.
## Moving Vehicle Analogy: Inside vs Outside of the Vehicle
The moving vehicle analogy is often presented to explain how Earth’s rotation does not seem to affect an object’s motion. Similarly, a ball released inside a moving vehicle will not fall rearward because it retains its inertia.
Flat-Earthers offer the moving open-air vehicle as if it is the “correct ” analogy. In reality, with their analogy, the air moves relative to the vehicle, which exerts an additional force that does not exist in the fact being explained. Their analogy fails to represent the fact being explained.
## Airplanes and Earth’s Rotation
An airplane exists on Earth. It has the same motion as the surface of the Earth, including Earth’s rotating motion. The motion and speed of an airplane are measured relative to Earth’s surface.
Flat-Earthers claim that an airplane should not be able to fly eastward or land. In reality, an airplane is moving relative to the surface. It has Earth’s motion as a component of its motion and does not need to chase Earth’s rotation.
## Moving Vehicle Analogy and the Physical Barrier Excuse
We cannot feel the Earth rotating, and to explain the phenomenon, we usually use the moving vehicle analogy. If the vehicle is moving at a constant speed, we will not be able to sense the motion. We cannot feel the motion of Earth because of the same reason.
Flat-Earthers invented the excuse that we are standing on Earth, not inside it, to falsely reject the explanation. | 393 | 1,901 | {"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} | 2.796875 | 3 | CC-MAIN-2022-33 | longest | en | 0.938172 |
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140Bquiz3csols
# 140Bquiz3csols - 4 To get the graph of h x =-5 x-1 start...
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1 MATH 140B QUIZ#3 Solutions Date: Sept 17, 2007 1. i) True . lim t 1 - C ( t ) = 0 . 1 and lim t 1 + C ( t ) = 0 . 2 ii) True . The left and right hand limits are different numbers. iii) False . The function C is discontinuous at t = 4. Therefore, C) is the answer. 2. First note: lim xto - 3 2 x 2 + 5 x - 3 x + 3 = lim xto - 3 (2 x - 1)( x + 3) x + 3 = lim xto - 3 (2 x - 1) = - 7 , i) True . The limit is - 7 , so left-hand-limit must be - 7 . ii) True . The limit is - 7 , so right-hand-limit must be - 7 . iii) False . It is a removable discontinuity not a jump (since the C -value does not jump on either side of t = - 3 . ) Therefore, D) is the answer. 3. i) True . lim x 0 3 x tan x = 3 lim x 0 x ( sin x cos x ) = 3 lim x 0 x sin x · cos x = 3 · 1 · 1 = 3 . ii) False . lim x 0 x csc( x/ 2) = lim x 0 x sin( x/ 2) = lim x 0 x/ 2 1 2 sin( x/ 2) = 2 lim x/ 2 0 x/ 2 sin( x/ 2) = 2 · 1 = 2 . where we mutliplied top and bottom by 1/2 at the 2nd equal sign. Therefore, A) is the answer.
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Unformatted text preview: 4. To get the graph of h ( x ) =-5 x-1 , start with the graph of y = 1 /x and shift it 1 unit to the right. Then re²ect the resulting graph in the x-axis 0.5 1.0 1.5 2.0 M 60 M 40 M 20 20 40 60 i) False . lim x → 1-h ( x ) = + ∞ . ii) True . lim x → 1 + h ( x ) =-∞ . iii) False . lim x → 1 h ( x ) = DNE . Therefore, B) is the answer. 2 5. From the graph below, all statements are easy to evaluate: 1 2 3 4 M 1.5 M 1.0 M 0.5 0.5 1.0 1.5 i) True . lim x → 2-g ( x ) =-1 . ii) True . lim x → 2 + g ( x ) = 1 . iii) True . The only discontinuity is at x = 2 . Therefore, E) is the answer....
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Ask a homework question - tutors are online | 850 | 2,247 | {"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} | 3.703125 | 4 | CC-MAIN-2018-13 | latest | en | 0.704207 |
http://spotidoc.com/doc/813616/math-257-316-section-201-midterm-2-solutions | 1,545,151,735,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829542.89/warc/CC-MAIN-20181218164121-20181218190121-00270.warc.gz | 264,473,262 | 7,328 | # Math 257/316 Section 201 Midterm 2 Solutions
Math 257/316 Section 201
Midterm 2
Solutions
Problem 1.
a) [10 points] For the function f (x) = x2 on [0, 1], sketch (roughly) its odd, and even
2-periodic extensions, and find its Fourier sine series, and its Fourier cosine series,
using (if you need it) for k = 1, 2, 3, . . .,
Z
0
1
2(−1)k
x cos(kπx)dx = 2 2 ,
k π
2
1
Z
x2 sin(kπx)dx =
0
2((−1)k − 1) (−1)k
−
.
k3 π3
kπ
Using the hints, for k = 1, 2, 3, . . . , the F.S.S. and F.C.S. coefficients are (respectively)
Z
2 1 2
2((−1)k − 1) (−1)k
bk =
x sin(kπx)dx = 2
−
,
1 0
k3 π3
kπ
Z
2 1 2
4(−1)k
ak =
x cos(kπx)dx = 2 2 ,
1 0
k π
while
2
a0 =
1
so
F.S.S. of f = 2
1
Z
0
2
x2 dx = ,
3
∞ X
2((−1)k − 1)
k3 π3
k=1
F.C.S. of f =
(−1)k
−
sin(kπx),
kπ
∞
1
4 X (−1)k
+ 2
cos(kπx).
3 π
k2
k=1
b) [15 points] Solve the following problem for the heat equation with non-zero derivative
(flux) BCs:
ut = uxx
0 < x < 1, t > 0
ux (0, t) = 0, ux (1, t) = 2
t>0
u(x, 0) = 0
0≤x≤1
1
There is no steady-state, so we first seek a particular solution of the form v(x, t) =
Ax2 + Bx + Ct. To satisfy the heat equation we require
0 = vt − vxx = C − 2A
=⇒ C = 2A,
and to satisfy the BCs we require
0 = vx (0, t) = (2Ax + B)|x=0 = B,
2 = vx (1, t) = (2Ax)|x=1 = 2A,
so A = 1, B = 0, C = 2, and v(x, t) = x2 + 2t. Then we write
u(x, t) = v(x, t) + w(x, t) = x2 + 2t + w(x, t)
and notice that
wt − wxx = ut − vt − (uxx − vxx ) = ut − uxx − 2 + 2 = 0
wx (0, t) = ux (0, t) − vx (0, t) = 0 − 0 = 0,
wx (1, t) = ux (1, t) − vx (1, t) = 2 − 2 = 0
2
w(x, 0) = u(x, 0) − v(x, 0) = 0 − x ,
and we know the solution to this problem is
∞
w(x, t) =
a0 X
2 2
+
ak cos(kπx)e−k π t ,
2
k=1
where the ak are the F.C.S. coefficients of the function −x2 on [0, 1], which we he
have from part (a) (up to a minus sign). So:
u(x, t) = x2 + 2t −
∞
1
4 X (−1)k
2 2
− 2
cos(kπx)e−k π t .
3 π
k2
k=1
2
Problem 2.
a) [19 points] Solve the following equation describing diffusion with growth, subject to
non-zero BCs:
ut = α2 uxx + u
0 < x < L, t > 0
u(0, t) = 0, u(L, t) = 1
t>0
u(x, 0) = 0
0≤x≤L
but leave any Fourier coefficients in terms of integrals (i.e. do not take time to
evaluate these integrals). Hint: first find the steady-state, then use separation of
variables to find the remainder.
A steady-state v(x) must solve α2 v 00 (x) + v(x) = 0, i.e. v 00 = − α12 v, whose general
solution is v(x) = A sin(x/α) + B cos(x/α). The BCs require
0 = v(0) = B,
1
,
sin(L/α)
1 = v(L) = A sin(L/α) =⇒ A =
sin(x/α)
(provided sin(L/α) 6= 0 – let’s assume this) so v(x) = sin(L/α)
Now writing u(x, t) = v(x) + w(x, t) we find that w solves:
wt = ut = α2 uxx + u = α2 vxx + v + α2 wxx + w = α2 wxx + w
w(0, t) = u(0, t) − v(0) = 0 − 0 = 0,
w(L, t) = u(L, t) − v(L) = 1 − 1 = 0
w(x, 0) = u(x, 0) − v(x) = −v(x).
Doing separation of variables, w(x, t) = X(x)T (t) leads to
T0
1
X 00
−
=
= const. = −λ2 < 0.
α2 T
α2
X
Indeed, we know this X problem well: X 00 (x) = −λ2 X(x), X(0) = 0 = X(L) only
has non-zero solutions for negative separation constant (which is why we wrote it as
−λ2 < 0), and these are
Xk (x) = sin(kπx/L),
λk = kπ/L,
k = 1, 2, 3, . . . .
The the solutions of the corresponding T problem,
T 0 = (1−α2 λ2 )T = (1−α2 k 2 π 2 /L2 )T
are Tk (t) = e(1−α
2 k 2 π 2 /L2 )t
so the general solution is
w(x, t) =
∞
X
bk sin(kπx/L)e(1−k
k=1
3
2 α2 π 2 /L2 )t
,
(or any multiple),
and it remains to satisfy the IC:
−v(x) = w(x, 0) =
∞
X
sin(kπx/L),
k=1
a Fourier sine series. Thus
Z
Z L
2
2 L
v(x) sin(kπx/L)dx = −
sin(x/α) sin(kπx/L)dx,
bk = −
L 0
L sin(L/α) 0
and the full solution is
∞
u(x, t) =
sin(x/α) X
2 2 2
2
+
bk sin(kπx/L)e(1−k α π /L )t
sin(L/α)
k=1
with the above formula for bk .
b) [6 points] Determine the long-time (t → ∞) behaviour of the solution (considering
all possible positive values of the diffusion rate α2 ).
There are two cases:
1. α2 > L2 /π 2 : then all the exponentials are decaying to zero, and so we have
lim u(x, t) = v(x)
t→∞
(diffusion beats growth)
2. α2 < L2 /π 2 : then the first exponential is growing with t, and so u(x, t) becomes
infinite as t → ∞ (growth beats diffusion)
(recall we excluded the possibility α2 = L2 /π 2 when we assumed sin(L/α) 6= 0).
4 | 1,951 | 4,163 | {"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} | 4.40625 | 4 | CC-MAIN-2018-51 | latest | en | 0.716777 |
http://www2.macaulay2.com/Macaulay2/doc/Macaulay2-1.19/share/doc/Macaulay2/EdgeIdeals/html/_clique__Complex.html | 1,675,908,371,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00872.warc.gz | 88,777,253 | 2,443 | # cliqueComplex -- returns the clique complex of a graph
## Synopsis
• Usage:
D = cliqueComplex G
• Inputs:
• G, ,
• Outputs:
• D, , the clique complex of G
## Description
This function returns the clique complex of a graph $G$. This is the simplicial complex whose faces correspond to the cliques in the graph. That is, $F = \{x_{i_1},...,x_{i_s}\}$ is a face of the clique complex of $G$ if and only if the induced graph on $\{x_{i_1},...,x_{i_s}\}$ is a clique of $G$.
i1 : R = QQ[w,x,y,z]; i2 : e = graph {w*x,w*y,x*y,y*z} -- clique on {w,x,y} and {y,z} o2 = Graph{edges => {{w, x}, {w, y}, {x, y}, {y, z}}} ring => R vertices => {w, x, y, z} o2 : Graph i3 : cliqueComplex e -- max facets {w,x,y} and {y,z} o3 = | yz wxy | o3 : SimplicialComplex i4 : g = completeGraph R o4 = Graph{edges => {{w, x}, {w, y}, {w, z}, {x, y}, {x, z}, {y, z}}} ring => R vertices => {w, x, y, z} o4 : Graph i5 : cliqueComplex g o5 = | wxyz | o5 : SimplicialComplex | 346 | 954 | {"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} | 2.828125 | 3 | CC-MAIN-2023-06 | latest | en | 0.591275 |
https://www.techylib.com/el/view/acridbone/zeroth_law_of_thermodynamics | 1,529,309,607,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.13/warc/CC-MAIN-20180618070542-20180618090542-00436.warc.gz | 926,804,203 | 12,215 | # Zeroth Law of Thermodynamics
Μηχανική
27 Οκτ 2013 (πριν από 4 χρόνια και 8 μήνες)
84 εμφανίσεις
Zeroth Law of
Thermodynamics
and the concept of temperature and “heat flow”
Zeroth Law of Thermodynamics
There is a state function, temperature -
T
, which has the
following property:
Zeroth Law of Thermodynamics
There is a state function, temperature -
T
, which has the
following property:
!
If heat is added to a closed system,
T
will increase
.
Conversely, if heat is extracted from a closed system,
T
will decrease.
Zeroth Law of Thermodynamics
There is a state function, temperature -
T
, which has the
following property:
T
1
>
T
2
!
If two closed systems with different temperatures are
brought into thermal contact, the heat will “flow” from the
system of higher
T
to the system of lower
T
until the two
systems reach the same intermediate temperature*.
!
If heat is added to a closed system,
T
will increase
.
Conversely, if heat is extracted from a closed system,
T
will decrease.
T
2
T
1
* Together, the two
systems are an isolated
system.
Zeroth Law of Thermodynamics
There is a state function, temperature -
T
, which has the
following property:
T
final
T
1
>
T
final
>
T
2
T
final
!
If two closed systems with different temperatures are
brought into thermal contact, the heat will “flow” from the
system of higher
T
to the system of lower
T
until the two
systems reach the same intermediate temperature*.
!
If heat is added to a closed system,
T
will increase
.
Conversely, if heat is extracted from a closed system,
T
will decrease.
* Together, the two
systems are an isolated
system.
Zeroth Law of Thermodynamics
Another way of stating this is:
Two closed systems are said to be in a state of equilibrium if
when they are brought in thermal contact there is no net
change in their state functions.
T
final
T
1
>
T
final
>
T
2
T
final
* Together, the two
systems are an isolated
system.
Heat Capacity
The zeroth law does not imply that the intermediate
temperature will be an average of the two temperatures.
Actually, it doesn’t even imply a “linear” temperature scale,
only a monotonic one. Thus, a scale must be established by
other means.*
T
final
T
final
T
1
>
T
final
>
T
2
*This is established with
the 3rd Law.
Heat Capacity
For example, assume system 1 has twice the mass of
system 2 and the two system are composed of identical
materials. Then one might expect the change in temperature
for system 2 to be twice as great as for system 1.
(With the proviso the starting temperature were fairly
close.)
T
final
T
final
T
1
!
T
final
.
T
final
!T
2
Heat Capacity
For example, assume system 1 has twice the mass of
system 2 and the two system are composed of identical
materials. Then one might expect the change in temperature
for system 2 to be twice as great as for system 1.
But what if they are composed of different mass
and
different
materials?
T
final
T
final
T
1
!
T
final
.
T
final
!T
2
Heat Capacity
For example, assume system 1 has twice the mass of
system 2 and the two system are composed of identical
materials. Then one might expect the change in temperature
for system 2 to be twice as great as for system 1.
But what if they are composed of different mass
and
different
materials?
This requires the use of the concepts of heat transfer or
“heat flow” and heat capacity.
T
final
T
final
T
1
!
T
final
.
T
final
!T
2
Heat Capacity
Definition: Heat transfer or “heat flow”,
q
, is the amount of
energy transferred from one closed system to another when
brought into thermal contact.
T
final
T
final
T
1
<
T
2
Heat Capacity
Definition: Heat transfer or “heat flow”,
q
, is the amount of
energy transferred from one closed system to another when
brought into thermal contact.
Definition: Heat Capacity,
C
, is defined by the equation:
q
=
C
Ä
T
T
final
T
final
T
1
<
T
final
<
T
2
Heat Capacity
Definition: Heat transfer or “heat flow”,
q
, is the amount of
energy transferred from one closed system to another when
brought into thermal contact.
Definition: Heat Capacity,
C
, is defined by the equation:
q
=
C
Ä
T
If the system consist of a pure substance, chances are that
the heat capacity has been measured and tabulated. It is
tabulated, however, on the basis of heat capacity per gram or
heat capacity per mole of material. These are given the
symbols and names:
S=
heat capacity per gram called “specific heat”
C
= heat capacity per mole called “molar heat capacity”
Heat Capacity
A summary to this point: Definition: Heat transfer or “heat
flow”,
q
, is the amount of energy transferred from one closed
system to another when brought into thermal contact.
Definition: Heat Capacity,
C
:
q
=
C
Ä
T
S=
heat capacity per gram called “specific heat”
C
= heat capacity per mole called “molar heat capacity”
Heat Capacity
A summary to this point: Definition: Heat transfer or “heat
flow”,
q
, is the amount of energy transferred from one closed
system to another when brought into thermal contact.
Definition: Heat Capacity,
C
:
q
=
C
Ä
T
S=
heat capacity per gram called “specific heat”
C
= heat capacity per mole called “molar heat capacity”
C
, is given a subscript “p” of “v”
depending upon whether the conditions are constant
pressure or constant volume.
C
p
=
molar heat capacity at constant pressure.
C
v
=
molar heat capacity at constant volume.
In chemistry
C
p
is used most often.
Heat Capacity
Example 1: Calculate the amount of heat required to increase
the temperature of a 35.0 g piece of lead from 0.0EC to 30.0EC.
The molar heat capacity for lead:
C
p
= 26.1 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of lead:
n
= (35.0 g)/(207.2 g mol
–1
)
n
= 0.169 mol
Example 1: Calculate the amount of heat required to increase
the temperature of a 35.0 g piece of lead from 0.0EC to 30.0EC.
The molar heat capacity for lead:
C
p
= 26.1 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of lead:
n
= (35.0 g)/(207.2 g mol
–1
)
n
= 0.169 mol
2) Calculate the total heat capacity:
C
=
n
C
p
C
=
(0.169 mol)(26.1 J mol
–1
K
–1
)
C
= 4.42 J K
–1
Example 1: Calculate the amount of heat required to increase
the temperature of a 35.0 g piece of lead from 0.0EC to 30.0EC.
The molar heat capacity for lead:
C
p
= 26.1 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of lead:
n
= (35.0 g)/(207.2 g mol
–1
)
n
= 0.169 mol
Example 1: Calculate the amount of heat required to increase
the temperature of a 35.0 g piece of lead from 0.0EC to 30.0EC.
The molar heat capacity for lead:
C
p
= 26.1 J mol
–1
K
–1
3) Substitute into
q = C
Ä
T
:
q =
(4.42 J K
–1
)(30.0 - 0.0)K
q =
133 J
2) Calculate the total heat capacity:
C
=
n
C
p
C
=
(0.169 mol)(26.1 J mol
–1
K
–1
)
C
= 4.42 J K
–1
Heat Capacity
Example 1: Calculate the temperature increase when 750 J of
heat is applied to 150 g of NaCl.
The molar heat capacity for NaCl:
C
p
= 49.9 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of NaCl:
n
= (150 g)/(58.5 g mol
–1
)
n
= 2.56 mol
Example 1: Calculate the temperature increase when 750 J of
heat is applied to 150 g of NaCl.
The molar heat capacity for NaCl:
C
p
= 49.9 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of NaCl:
n
= (150 g)/(58.5 g mol
–1
)
n
= 2.56 mol
Example 1: Calculate the temperature increase when 750 J of
heat is applied to 150 g of NaCl.
The molar heat capacity for NaCl:
C
p
= 49.9 J mol
–1
K
–1
2) Calculate the total heat capacity:
C
=
n
C
p
C
=
(2.56 mol)(49.9 J mol
–1
K
–1
)
C
= 128 J K
–1
Heat Capacity
1) Calculate the number of moles of NaCl:
n
= (150 g)/(58.5 g mol
–1
)
n
= 2.56 mol
Example 1: Calculate the temperature increase when 750 J of
heat is applied to 150 g of NaCl.
The molar heat capacity for NaCl:
C
p
= 49.9 J mol
–1
K
–1
3) Substitute into
q = C
Ä
T
:
750 J
=
(128 J K
–1
T
ÄT =
5.86 K (or EC
)
2) Calculate the total heat capacity:
C
=
n
C
p
C
=
(2.56 mol)(49.9 J mol
–1
K
–1
)
C
= 128 J K
–1
Zeroth Law of
Thermodynamics
and the concept of temperature and “heat flow”
THE END | 2,470 | 7,950 | {"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-2018-26 | latest | en | 0.929476 |
https://scientific.tweetalibrarian.com/e-mathematical-constant | 1,670,070,350,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710931.81/warc/CC-MAIN-20221203111902-20221203141902-00616.warc.gz | 521,926,829 | 6,629 | # E (mathematical Constant)
If you are looking for E (mathematical Constant), simply check out our links below.
### E (mathematical Constant) - Wikipedia
The number e, also known as Euler's number, is a mathematical constant approximately equal to 2.71828 which can be characterized in many ways. [ReadMore..]
### (PDF) New Approximations To The Mathematical Constant E
PDF | Based on the Newton-Cotes and Gaussian quadrature rules, we develop several new closed form approximations to the mathematical constant e. For... | Find, read and cite all the research you need on ResearchGate PDF | Based on the Newton-Cotes and Gaussian quadrature rules, we develop several new closed form approximations to the mathematical constant e. For. [ReadMore..]
### E (mathematical Constant) - Simple English Wikipedia, The Free ...
e is a number. It is the base of natural logarithm and is about 2.71828. ... It is an important mathematical constant. The number e is occasionally called Euler's ... [ReadMore..]
### E - Euler's Number
e (Euler's Number) ; 2.7182818284590452353602874713527 (and more ...) ; e is an irrational number (it cannot be written as a simple fraction). ; e is the base of ... [ReadMore..]
### E (mathematical Constant) (e)
Dec 14, 2020 ... e (mathematical constant) (e) · This operation needs no arguments. · The data type of the values in the new column is float. ·. · Click Create ... [ReadMore..]
### Category:e (mathematical Constant) - Wikimedia Commons
Jun 20, 2021 ... Subcategories · Media in category "e (mathematical constant)". [ReadMore..]
### C++ - An Efficient Way To Compute Mathematical Constant E - Stack ...
Jun 12, 2010 ... Since it's not possible to calculate every digit of 'e', you're going to have to pick a stopping point. double precision: 16 decimal digits. [ReadMore..]
### How To Celebrate E Day, Feb. 7, 2018, With The Mathematical Constant.
The food options aren’t as good as Pi Day, but the emotional options are grand. Feb 7, 2018 ... Wednesday is E Day. That's because the date, Feb. 7, 2018, is reminiscent of the mathematical constant e, which starts out 2.718281 (and ... [ReadMore..]
### Mathematical Constants: E Pi Golden Ratio »lectronics Notes
The mathematical number e, also known as Euler's number (not to be confused with the Euler-Mascheroni constant, sometimes called simply Euler's constant) is the ... [ReadMore..]
The pages related to e (mathematical constant) are also listed below:
### Euler's Number
Euler's number is a mathematical constant with many applications in science and finance, usually denoted by the lowercase letter e. Key Takeaways · Euler's number is an important constant that is found in many contexts and is the base for natural logarithms. · An irrational number denoted by e ... [ReadMore..]
### Math.e Constant With Example In Python
Apr 19, 2019 ... Return value: float – that is the value of mathematical constant "e". Example: Input: print(math.e) Output: 2.718281828459045. Python code to ... [ReadMore..] | 747 | 3,019 | {"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-2022-49 | latest | en | 0.84162 |
https://forums.steinberg.net/t/different-normalize-algorithms-peak-and-rms-edit-bring-wavelabs-loudness-meta-normalizer-into-cubase/117634 | 1,716,879,458,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00797.warc.gz | 217,178,365 | 6,951 | # Different Normalize Algorithms - Peak and RMS.... edit: Bring Wavelabs Loudness Meta Normalizer into Cubase!!
I’m trying to adjust all my own samples to be of equal RMS so as to quickly be able swap them in and out, for instance a snare sample - I have variations of the same snare sample printed with different levels of compression for instance. Sounds with equal RMS mostly seem to sum better as well.
Having normalize with an RMS algo would speed thing up very nicely.
What you are asking is a normalize function that acts like a limiter. I guess it wouldn’t be normalizing anymore and the limiter wouldn’t know how much gain reduction to apply so you would have to adjust it manually anyways for each sample.
Normalizing only looks at the peak and raise the volume of the clip to whatever setpoint you select.
Yeah, I get what you are asking but once your peaks reach 0db you cant raise the RMS more without limiting or else it will clip.
Normalizing doesn’t have to be to 0db, it could be -6db.
Obviously it can be to whatever value you set it to be, but if you raise the RMS of your snare to -6db i guarantee you that the peak will be way above 0db and thus clipping unless you limit or soft clipping it.
I adjust peaks to -6db, then RMS from there.
If you set your peaks to -6db then the RMS is what it is. If you change the RMS by just turning volume up or down you will also change the peaks by same amounts.
Hmm, not sure if what I’m suggesting is getting across. When your normalize multiple events, it normalizes all the peaks to -6db. Some those events might be -12db, some might be, -3db, 0db - it will normalize all of these to -6db.
I want to do this based off RMS data, not peaks. If you have a snare sample for example, and you’ve taken this sample and EQ’d and compressed it a number of ways thus creating 10 new samples from the 1 original snare sample - your peaks will no longer be relevant, and if you normalize them to -6db, you will here that even though the peaks are -6db, they appear to be much much louder than the original sample or compared to some of the other processed variations because the RMS has changed. So, peak normalizing is no longer useful here.
It matters because if your RMS of those same events are like -24db, -15db and -10db with the peaks at -6, -3 and 0 then you wanna normalize the RMS to lets say -10db, then you will need to raise volume by 14db on the lowest sample and that will bring the peak from -6db to 8db and thus clipping.
Just because you have compressed your snare samples doesn’t mean the peaks doesn’t matter, it actually still matter as much as it did without compression. Specially because compression doesn’t deal with the whole transient and will let the whatever number of ms you select with attack knob through.
What I’m saying is that it makes no sense to normalize by volume using RMS value. For that you will use limiter of soft clipper.
wavelab - meta normalizer
Seems to be working fine doing it by hand… other than taking a long ass time.
One would just have to know how much headroom they have, to avoid clipping. But evening out samples based off peak, does not work - try it! As soon as one sample is shifted in frequency content from the next, their peak values are no longer relevant in terms of how loud they are as perceived by human hearing.
I guess my feature request is to bring Meta Normalizer in Cubase!
I never said they are relevant to the human hearing, I’m saying they are absolutely relevant to avoid clipping. Doesn’t matter how much you process them or how much the frequency is shifted, they still need to be below 0db to avoid clipping.
Yes of coarse, but what does clipping have to do with my suggestion? My intent was to never RMS multiple events to be 0db, and if my starting point is -6db peak, any work I do altering the RMS in this context of creating even sounding samples, is very rarely going to create a new 6.01db peak.
My action here is using a -6db peak as a starting reference point, then leveling RMS around that reference point. Clipping isn’t really a concern. | 937 | 4,099 | {"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} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.937543 |
https://europedriveguide.com/doing-homework-778 | 1,674,996,180,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00041.warc.gz | 270,444,834 | 3,380 | # Function notation solver
Here, we will show you how to work with Function notation solver. Keep reading to learn more!
## The Best Function notation solver
There are a lot of Function notation solver that are available online. Algebra is a branch of mathematics that deals with the operations between variables. These operations are known as equations. In order to solve an algebra equation, you must first understand what the symbols in the equation represent. The next step is to use these symbols to perform the operations indicated by the equation. Finally, you must simplify the equation to find the solution.
Completing the square is a process that can be used to solve certain quadratic equations. To complete the square, you need to take the equation and rearrange it so that it looks like the perfect square equation. This can be done by adding or subtracting terms from both sides of the equation until the left side is a perfect square. Once the equation is in this form, you can solve it using the quadratic equation.
There are a few things to consider when looking for algebra homework help. First, consider what type of help you need. Do you need help with specific problems, or do you need help understanding the concepts? If you need help with specific problems, consider looking for a tutor or asking a friend for help. If you need help understanding the concepts, consider looking for a study guide or watching a video tutorial. Another thing to consider is how much help you need. Do you need help
An equation solver app can be a great math tool for students of all levels. With this app, users can input an equation and the app will provide the solution. This can be helpful for homework or for solving tricky math problems. The app can also provide step-by-step instructions on how to solve the equation, which can be a great way to learn new techniques.
A math calculator with steps is a great tool for students to use when they are struggling with a math problem. By inputting the problem into the calculator, the student can see each step the calculator takes to solve the problem. This can be a great way for the student to see where they went wrong and correct their mistake. | 435 | 2,210 | {"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-2023-06 | latest | en | 0.947716 |
https://socratic.org/questions/how-do-you-solve-k-2-10-24 | 1,601,447,037,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402118004.92/warc/CC-MAIN-20200930044533-20200930074533-00453.warc.gz | 585,447,935 | 6,119 | # How do you solve k^2 = 10.24?
##### 2 Answers
Mar 9, 2018
Follow explanation. Your solution: $k = \pm 3.2$
#### Explanation:
Your equation can be arranged
${k}^{2} = 10.24$
$\sqrt{{k}^{2}} = \pm \sqrt{10.24}$
$k = \pm 3.2$
Because of the fact that $\sqrt{{k}^{2}} = \pm {\left({k}^{2}\right)}^{0.5} = \pm k$
There are two possible roots.
Mar 9, 2018
$k = \pm 3.2$
#### Explanation:
Square root both sides of the equation
sqrt(k^2= +-sqrt(10.24)
$k = \pm 3.2$
There are two possible roots.
Check solution
$+ {3.2}^{2} = 10.24$
${\left(- 3.2\right)}^{2} = 10.24$
$10.24 = 10.24$ | 239 | 596 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "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} | 4.375 | 4 | CC-MAIN-2020-40 | latest | en | 0.721626 |
https://www.aqua-calc.com/what-is/pressure/pound-force-per-hectare | 1,702,300,143,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679511159.96/warc/CC-MAIN-20231211112008-20231211142008-00786.warc.gz | 681,455,216 | 9,299 | # What is a pound-force per hectare (unit)
## The pound-force per hectare is a unit of measurement of pressure
A pound-force per hectare (lbf/ha) is a unit of pressure where a force of one pound-force (lbf) is applied to an area of one hectare.
• What is pressureInstant conversionsConversion tables
• 1 lbf/ha = 4.390053401×10-9 atm
• 1 lbf/ha = 3.336440585×10-6 Torr
• 1 lbf/ha = 0.003336440585 mTorr
• 1 lbf/ha = 3.336440585×10-6 mmHg
• 1 lbf/ha = 4.535924341×10-5 mmH2O
• 1 lbf/ha = 1.313559283×10-7 inHg
• 1 lbf/ha = 1.785796983×10-6 inH2O
• 1 lbf/ha = 4.448221607×10-6 mbar
• 1 lbf/ha = 4.448221608×10-9 b
• 1 lbf/ha = 0.0004448221608 Pa
• 1 lbf/ha = 4.448221608×10-6 hPa
• 1 lbf/ha = 4.448221608×10-7 kPa
• 1 lbf/ha = 4.4482216080729×10-10 MPa
• 1 lbf/ha = 4.4482216083761×10-28 N/pm²
• 1 lbf/ha = 4.4482216083761×10-24 N/Ų
• 1 lbf/ha = 4.4482216083761×10-22 N/nm²
• 1 lbf/ha = 4.4482216080729×10-16 N/µ²
• 1 lbf/ha = 4.4482216080729×10-16 N/µm²
• 1 lbf/ha = 4.4482216080729×10-10 N/mm²
• 1 lbf/ha = 4.448221608×10-8 N/cm²
• 1 lbf/ha = 4.448221607×10-6 N/dm²
• 1 lbf/ha = 0.0004448221607 N/m²
• 1 lbf/ha = 0.04448221607 N/dam²
• 1 lbf/ha = 444.822161 N/km²
• 1 lbf/ha = 6.4516×10-8 psi
• 1 lbf/ha = 9.290304×10-6 psf
• 1 lbf/ha = 1.0×10-28 lbf/pm²
• 1 lbf/ha = 1.0×10-24 lbf/Ų
• 1 lbf/ha = 1.0×10-22 lbf/nm²
• 1 lbf/ha = 1.0×10-16 lbf/µ²
• 1 lbf/ha = 1.0×10-16 lbf/µm²
• 1 lbf/ha = 1.0×10-10 lbf/mm²
• 1 lbf/ha = 1.0×10-8 lbf/cm²
• 1 lbf/ha = 1.0×10-6 lbf/dm²
• 1 lbf/ha = 0.0001 lbf/m²
• 1 lbf/ha = 0.01 lbf/dam²
• 1 lbf/ha = 100 lbf/km²
• 1 lbf/ha = 0.01 lbf/a
• 1 lbf/ha = 1 lbf/hm²
• 1 lbf/ha = 342.9904 lbf/nmi²
• 1 lbf/ha = 2.869814654×10-7 N/in²
• 1 lbf/ha = 4.132533103×10-5 N/ft²
• 1 lbf/ha = 4.448221608×10-23 dyn/pm²
• 1 lbf/ha = 4.448221608×10-19 dyn/Ų
• 1 lbf/ha = 4.448221608×10-17 dyn/nm²
• 1 lbf/ha = 4.448221608×10-11 dyn/µ²
• 1 lbf/ha = 4.448221608×10-11 dyn/µm²
• 1 lbf/ha = 4.448221608×10-5 dyn/mm²
• 1 lbf/ha = 0.004448221608 dyn/cm²
• 1 lbf/ha = 0.004448221608 Ba
• 1 lbf/ha = 0.4448221608 dyn/dm²
• 1 lbf/ha = 44.4822161 dyn/m²
• 1 lbf/ha = 4 448.22161 dyn/dam²
• 1 lbf/ha = 44 482 216.1 dyn/km²
• 1 lbf/ha = 4 448.22161 dyn/a
• 1 lbf/ha = 444 822.161 dyn/ha
• 1 lbf/ha = 444 822.161 dyn/hm²
• 1 lbf/ha = 152 569 731 dyn/nmi²
• 1 lbf/ha = 2.8698146526173×10-14 dyn/µin²
• 1 lbf/ha = 2.869814653×10-8 dyn/mil;²
• 1 lbf/ha = 2.869814653×10-8 dyn/thou²
• 1 lbf/ha = 0.02869814653 dyn/in²
• 1 lbf/ha = 4.1325331 dyn/ft²
• 1 lbf/ha = 180 013.142 dyn/ac
• 1 lbf/ha = 115 208 411 dyn/mi²
• 1 lbf/ha = 37.1927979 dyn/yd²
• 1 lbf/ha = 148.771192 dyn/ftm²
• 1 lbf/ha = 18 001.3142 dyn/ch²
• 1 lbf/ha = 1 800 131.42 dyn/fur²
• 1 lbf/ha = 9.9549077408569×10+23 dyn/au²
• 1 lbf/ha = 3.9812294046023×10+33 dyn/ly²
• 1 lbf/ha = 4.2353397519225×10+34 dyn/pc²
• 1 lbf/ha = 3.336440585×10-7 cmHg
#### Foods, Nutrients and Calories
ORGANIC WHOLE CHOCOLATE MILK, UPC: 881245105169 contain(s) 96 calories per 100 grams (≈3.53 ounces) [ price ]
1403 foods that contain Vitamin D3 (cholecalciferol). List of these foods starting with the highest contents of Vitamin D3 (cholecalciferol) and the lowest contents of Vitamin D3 (cholecalciferol), and Recommended Dietary Allowances (RDAs) for Vitamin D
#### Gravels, Substances and Oils
CaribSea, Freshwater, Flora Max, Original weighs 1 089.26 kg/m³ (68.00028 lb/ft³) with specific gravity of 1.08926 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 ]
Diiodomercury, beta form [HgI2] weighs 6 090 kg/m³ (380.18628 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-417A, liquid (R417A) with temperature in the range of -30°C (-22°F) to 60°C (140°F)
#### Weights and Measurements
The slug per cubic centimeter density measurement unit is used to measure volume in cubic centimeters in order to estimate weight or mass in slugs
Acceleration (a) of an object measures the object's change in velocity (v) per unit of time (t): a = v / t.
dwt/pt to mg/m³ conversion table, dwt/pt to mg/m³ unit converter or convert between all units of density measurement.
#### Calculators
Volume to Weight conversions for sands, gravels and substrates | 2,028 | 4,431 | {"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} | 3.40625 | 3 | CC-MAIN-2023-50 | latest | en | 0.341618 |
https://oeis.org/A144657 | 1,585,432,797,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493120.15/warc/CC-MAIN-20200328194743-20200328224743-00479.warc.gz | 641,735,242 | 3,762 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A144657 a(n) = Sum[Sum[(i+j)!/(i!*j!),{i,1,n}],{j,1,n}]. 2
0, 2, 14, 62, 242, 912, 3418, 12854, 48602, 184736, 705410, 2704132, 10400574, 40116572, 155117490, 601080358, 2333606186, 9075135264, 35345263762, 137846528780, 538257874398, 2104098963676, 8233430727554 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 FORMULA Recurrence: (n+1)*(12*n-19)*a(n) = 2*(30*n^2 - 24*n - 19)*a(n-1) - (48*n^2 - 9*n - 7)*a(n-2) - 2*(2*n-3)*a(n-3). - Vaclav Kotesovec, Oct 20 2012 a(n) ~ 4^(n+1)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 20 2012 MATHEMATICA Table[Sum[Sum[(i+j)!/i!/j!, {i, 1, n}], {j, 1, n}], {n, 0, 20}] (* corrected by Vaclav Kotesovec, Oct 20 2012 *) CROSSREFS Suggested by a formula in A048775. Sequence in context: A153332 A331822 A217154 * A167555 A222445 A181394 Adjacent sequences: A144654 A144655 A144656 * A144658 A144659 A144660 KEYWORD nonn AUTHOR N. J. A. Sloane, Jan 30 2009 STATUS approved
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Last modified March 28 17:51 EDT 2020. Contains 333103 sequences. (Running on oeis4.) | 576 | 1,498 | {"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} | 3.015625 | 3 | CC-MAIN-2020-16 | latest | en | 0.537661 |
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# Forum: Regression Analysis
Linear regression, linear models, nonlinear regression
1. ### What is the appropriate analysis test?
Hello everyone, I'm new to this forum. Please be kind to me. ^-^ I want to know what analysis test I can use for the following: Research...
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04-04-2013, 02:31 PM
2. ### Nested continuous Regression conditioned on dummy
Hi, Does anyone know what model to use for the following issue: I would like to estimate transaction size (continuous dependent variable) using...
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04-03-2013, 06:19 PM
3. ### Panel Data: Conducting N time-series regressions and averaging coefficients
Hi, I stumbled upon a special regression method in a published paper (A+ journal) that I hadn't heard before. In dealing with a panel data set, the...
• Replies: 3
• Views: 3,681
04-02-2013, 10:07 PM
4. ### GARCH model with insignificant arch/garch terms
Hello, I was wondering, if we run a regression where we also specify a GARCH(1,1) for the error variance, and we find that both the arch and garch...
• Replies: 2
• Views: 2,914
04-02-2013, 12:48 PM
5. ### univariate vs multivariate
Hi, I have a question about interpreting univariate vs. multivariate (linear) regression. When I'm reading journal articles often both approaches...
• Replies: 5
• Views: 40,743
04-02-2013, 05:41 AM
6. ### Variance error of estimate
Hi everyone! I need help with this question: The variance of Y is 400 and the correlation between X and Y is 0.50. a) What is the variance error...
• Replies: 0
• Views: 1,597
04-01-2013, 12:36 PM
7. ### Non constant variance in residuals?
Hi, When doing Linear Regression or Generalised Linear Modelling, and the residuals show non constant variance, how do you remove it? For example,...
• Replies: 3
• Views: 4,718
03-28-2013, 08:57 PM
8. ### Event Study where event is correlated with index
Hello, I am trying to conduct an event study using the traditional market model: StockReturn = Constant + IndexReturn + DummyVariable + Error ...
• Replies: 3
• Views: 1,519
03-28-2013, 08:54 PM
9. ### Help with understanding logit coefficient
Hi everyone, I'm very new to statistics and I would very much appreciate any help you can give me in understanding the coefficients in a logit model....
• Replies: 7
• Views: 1,959
03-28-2013, 10:19 AM
10. ### Multiple regression - assumptions
Hi, I performed experiments on two plant species, growing them together in mixtures. I repeated this experiment in 4 nutrient concentrations. I...
• Replies: 9
• Views: 2,376
03-28-2013, 06:58 AM
11. ### Regression with arctan function
Hello, I am currently working on a prepayment model for mortgage loans. For this I'd like to use a modified version of the Richard & Roll model....
• Replies: 0
• Views: 3,514
03-28-2013, 05:01 AM
12. ### Alternatives to logistic regression when the number of cases is low
Hi All I got involved in my first foray into logistic regression, but have come to realize that I probably can't use it. Let me tell you about my...
• Replies: 4
• Views: 2,204
03-27-2013, 08:43 AM
13. ### step change
Fir a step change model, is it a linear model ?
• Replies: 0
• Views: 1,103
03-27-2013, 06:31 AM
14. ### Goodness of fit versus AIC/BIC versus pseudo R squares
This is geared to Logistic regression I understand in theory you use these for different reasons. AIC for comparing non-nested models and goodness...
• Replies: 6
• Views: 4,076
03-25-2013, 10:10 PM
15. ### Error question
Okay, perhaps a silly question. When it is actually appropriate to add a normal error term to an equation that you are about to estimate? Or is...
• Replies: 0
• Views: 1,251
03-25-2013, 08:45 PM
16. ### Multiple Regression with negative constant for price
Hi there, I am carrying out a project study, in which I am studying the relationship between the price of property in three regions in the UK along...
• Replies: 2
• Views: 8,412
03-25-2013, 11:11 AM
17. ### Assessing SSE(sum of squared errors)?
Hello, I can't figure out how to solve this question: Consider these summary statistics for a random sample: Mean of x=20 Mean of y=10...
• Replies: 3
• Views: 2,031
03-24-2013, 03:27 PM
18. ### IV analysis, log-linear model, is it ok?
Hi guys, I have a problem I hope someone can help me with! The thing is, I am doing an IV analysis, 2SLS. When I do the first stage and regresses...
• Replies: 0
• Views: 1,319
03-23-2013, 09:23 AM
19. ### Observed versus Expected Slope Comparison
Hello, I am performing an analysis (for dissertation project) where I am comparing the sum of branch lengths from phylogenetic trees reconstructed...
• Replies: 2
• Views: 3,580
03-22-2013, 07:57 PM
20. ### Confidence limits for logistic regression equation
I have generated a logistic model for a disease status Y (unaffected-affected) using several independent variables on a very large (600K+) dataset. I...
• Replies: 0
• Views: 3,176
03-21-2013, 11:29 PM
21. ### How to find relationship of variables?
Hi, Here is my problem: I have 3 variables with data as below: ID Went_Out_date returned?(1=yes, 0=No) 1 3/20/2012 1...
• Replies: 0
• Views: 1,187
03-20-2013, 06:47 PM
22. ### Accuracy of prediction?
Right, I've got some data for variable X and variable Y in 2002. Using Simple Linear Regression, I've made a model which uses variable X to...
• Replies: 1
• Views: 1,567
03-20-2013, 05:51 PM
23. ### Logistic regression diagnostic questions
I have a set of 42 variables which I am trying to determine the relative importance of. Well I already did but now I am trying to determine if...
• Replies: 19
• Views: 5,634
03-20-2013, 10:40 AM
24. ### Which regression model to use with skewed & ordinal data?
I'm a bit stucked and just wondering about which regression technic to use with my dependent variable at hand. Respondents were asked to provide a...
• Replies: 3
• Views: 4,267
03-19-2013, 04:16 PM
25. ### Year dummies in Fixed Effects estimation with panel data
Quick question. I'm working with an unbalanced international dataset. I'm using fixed effects because (at least mostly because) I want to allow for...
• Replies: 0
• Views: 2,608
03-19-2013, 08:40 AM
26. ### Variable is significant in OLS regression, but insignificant in Probit marginal effec
I am investigating the determinants of wage. With the dataset I ran a pooled OLS regression, and a Probit marginal effects regression looking at...
• Replies: 2
• Views: 2,875
03-18-2013, 02:31 PM
27. ### Passing-Bablok regression
Hi all, Does anyone knows how to perform Passing-Bablok regression in SAS? I have never done it do the SAS code (with or withoutan example)...
• Replies: 0
• Views: 5,061
szm
03-18-2013, 01:56 PM
28. ### basic econometrics
explain the differences for the following paired terms with appropriate examples. (1)log-lin model and lin-log model (2)cross sectional data and...
• Replies: 2
• Views: 2,524
03-18-2013, 11:11 AM
29. ### Nested or Repeated measures analysis for ecology data?
This is a refinement of an earlier question I asked. ...
• Replies: 3
• Views: 4,232
03-18-2013, 11:03 AM
30. ### Logistic regression help
Hello, I have performed some logistic regression analysis and I was wondering if someone can tell me if I've done it and interpreted it...
• Replies: 1
• Views: 1,767
03-18-2013, 09:30 AM
31. ### Help with regression please
Hi, sorry if this post is confusing, but I'll try not to babble, for my FYP I am doing a pair of questionnaires, one has a set of questions which are...
• Replies: 3
• Views: 1,449
03-18-2013, 08:29 AM
32. ### Airline Stock Regression
Hi guys, I'm trying to run a regression with daily airline stock returns as dependent variable. I've included jet fuel price and the stock market...
• Replies: 0
• Views: 1,706
03-18-2013, 08:01 AM
33. ### Hosmer Lemeshow goodness of fit test
One of the authors I consult a lot with in logistic regression is really down with Hosmer Lemeshow goodness of fit test. He argues it has low power...
• Replies: 1
• Views: 2,641
03-15-2013, 08:51 PM
34. ### bivariate regression question
Hi everyone: I am have conducted bivariate regression analysis to examine the effect of: an (x) independent variable, age (nominal) on a (y)...
• Replies: 5
• Views: 2,689
03-15-2013, 02:52 PM
35. ### Interpreting Marginal Effect - Probit
I am using a probit model, where one of the variables is log(income). I calculated the (average) marginal effect and it resulted in 0.03. I am a...
• Replies: 1
• Views: 3,114
03-15-2013, 08:08 AM
36. ### Can I transform the principal components...?
Hi all. I am using PCA to generate PCs and then using them as independent variables in regression. I added an extra data-step after PCA which is...
• Replies: 0
• Views: 1,832
szm
03-14-2013, 01:48 PM
37. ### Large logit coefficient - what it means
I am constructing a logit model with 6 explanators, trying to predict outcomes of hockey matches using odds (the variable I am referring to below)...
• Replies: 12
• Views: 8,153
03-12-2013, 06:06 PM
38. ### Comparing Multiple Regression across multiple groups
I am looking at 4 different predictors of eating disorders in 4 different groups of participants (Caucasians, Af Americans, Lations, and Asian...
• Replies: 1
• Views: 2,642
03-12-2013, 02:58 PM
39. ### Help with sensibility analysis (event ticketing and advertising influence)
Hi everybody. I have a task to find out how usage of (advertising investment, PR etc.) different media channels influence on sales of tickets for...
• Replies: 0
• Views: 2,196
03-12-2013, 12:46 PM
40. ### Longitudinal analysis across age, when participants are all different ages from T1
Hi all! I'm looking to do a longitudinal analysis to see how a variable changes across two time points. The thing is, from T1, my participants are...
• Replies: 0
• Views: 2,476
03-12-2013, 11:21 AM
41. ### regression VS canonial correlation analysis
hello i need to know the main difference between the multivariate multiple regression and canonical correlation analysis...
• Replies: 3
• Views: 2,730
03-12-2013, 04:18 AM
42. ### Multiple regression analysis for 3 groups
Hi I am doing a multiple regresion analysis for my project. If I have 3 groups which are nominal data (nationality: PL, ENG, and PL who moved to...
• Replies: 10
• Views: 3,118
03-10-2013, 05:01 PM
43. ### bivariate analysis versus univariate logistic regresssion analysis
Hello, I have done a bivariate analysis and univariate logistic regresion analysis on the same data. Now both gave the same results (OR, 95%CI),...
• Replies: 0
• Views: 2,930
03-09-2013, 03:18 PM
44. ### Violation of linearity assumption
Hi, I'm a beginner at statistics. I'm doing a simple linear regression and I have a scatterplot of the residuals versus the predicted values and I...
• Replies: 10
• Views: 4,693
03-09-2013, 10:47 AM
45. ### Multiple Linear Regression: Outlier
Ok folks, so I am doing a multiple linear regression. I am not a statistics expert by any means, and it is part of a psychology report. The last time...
• Replies: 4
• Views: 2,408
03-07-2013, 02:16 PM
46. ### Linear Probability Model
Hi guys, Just wanted to check because the text books I am using are do not explicitly state it... It does not make sense to rely on the t-stats...
• Replies: 0
• Views: 2,641
03-07-2013, 02:03 PM
47. ### Need help interpretering results
Hello, I was wondering if anyone could help me with a few small questions! First of all i'm conducting a research and I need to compare the...
• Replies: 1
• Views: 1,822
03-05-2013, 12:27 PM
48. ### Need help interpreting regression analysis comparison results
Hi, I am using simple linear regression analysis to compare two different population groups' development over time (one dependent variable). I am...
• Replies: 2
• Views: 2,559
03-04-2013, 04:23 PM
49. ### Is Unemployment Rate important as a regressor in a cross sectional analysis?
Hello, I am doing a regression on potential sales and I want to identify the demand drivers. I was wondering if Unemployment rate could be one of...
• Replies: 2
• Views: 2,424
03-01-2013, 09:08 PM
50. ### Need advice on methods to assess nested random effects ecology data
Hi everyone, I am a PhD student working in forestry/ecology. More specifically, I am looking at the effects of harvesting on soil nutrient pools....
• Replies: 1
• Views: 2,231
03-01-2013, 12:17 PM
Use this control to limit the display of threads to those newer than the specified time frame.
Allows you to choose the data by which the thread list will be sorted.
Note: when sorting by date, 'descending order' will show the newest results first. | 3,687 | 12,874 | {"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-2017-43 | latest | en | 0.812031 |
https://proofwiki.org/wiki/Union_of_Singleton | 1,718,450,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00605.warc.gz | 430,355,496 | 11,383 | # Union of Singleton
## Theorem
Consider the set of sets $A$ such that $A$ consists of just one set $x$:
$A = \set x$
Then the union of $A$ is $x$:
$\bigcup A = x$
## Proof
Let $A = \set x$.
From the definition of set union:
$\bigcup \set x = \set {y: \exists z \in \set x: y \in z}$
from which it follows directly that:
$\bigcup \set x = \set {y: y \in x}$
as $x$ is the only set in $\set x$.
That is:
$\bigcup A = x$
$\blacksquare$ | 158 | 448 | {"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} | 3.859375 | 4 | CC-MAIN-2024-26 | latest | en | 0.816377 |
https://www.aqua-calc.com/calculate/volume-to-weight/substance/gadolinium-op-iii-cp--blank-trisulfate | 1,653,561,807,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00666.warc.gz | 710,539,601 | 7,999 | # Weight of Gadolinium(III) trisulfate
## gadolinium(iii) trisulfate: convert volume to weight
### Weight of 1 cubic centimeter of Gadolinium(III) trisulfate
carat 20.7 ounce 0.15 gram 4.14 pound 0.01 kilogram 0 tonne 4.14 × 10-6 milligram 4 140
#### How many moles in 1 cubic centimeter of Gadolinium(III) trisulfate?
There are 6.87 millimoles in 1 cubic centimeter of Gadolinium(III) trisulfate
### The entered volume of Gadolinium(III) trisulfate in various units of volume
centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
• For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.
#### Foods, Nutrients and Calories
CLASSIC RING BOLOGNA, UPC: 070919032232 contain(s) 196 calories per 100 grams (≈3.53 ounces) [ price ]
81 foods that contain Vitamin E (label entry primarily). List of these foods starting with the highest contents of Vitamin E (label entry primarily) and the lowest contents of Vitamin E (label entry primarily)
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Crystal River weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 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 ]
Iodine pentafluoride [IF5] weighs 3 190 kg/m³ (199.14519 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Sunflower oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
The tonne per cubic yard density measurement unit is used to measure volume in cubic yards in order to estimate weight or mass in tonnes
The kinematic viscosity (ν) is the dynamic viscosity (μ) divided by the density of the fluid (ρ)
troy/yd³ to g/US qt conversion table, troy/yd³ to g/US qt unit converter or convert between all units of density measurement.
#### Calculators
Calculate Ideal Body Weight and Adjusted Body Weight | 718 | 2,601 | {"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} | 3.078125 | 3 | CC-MAIN-2022-21 | longest | en | 0.697401 |
https://nz.mathematicstip.com/5520-131-a-brief-introduction-to-linear-time-invariant-sy.html | 1,638,125,281,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00296.warc.gz | 512,597,519 | 45,010 | # 13.1: A brief introduction to linear time invariant systems
Let’s start by defining our terms.
Signal. A signal is any function of time.
System. A system is some machine or procedure that takes one signal as input does something with it and produces another signal as output.
Linear system. A linear system is one that acts linearly on inputs. That is, (f_1 (t)) and (f_2 (t)) are inputs to the system with outputs (y_1 (t)) and (y_2 (t)) respectively, then the input (f_1 + f_2) produces the output (y_1 + y_2) and, for any constant (c), the input (cf_1) produces output (cy_1).
This is often phrased in one sentence as input (c_1f_1 + c_2 f_2) produces output (c_1 y_1 + c_2 y_2), i.e. linear combinations of inputs produces a linear combination of the corresponding outputs.
Time invariance. Suppose a system takes input signal (f(t)) and produces output signal (y(t)). The system is called time invariant if the input signal (g(t) = f(t - a)) produces output signal (y(t - a)).
LTI. We will call a linear time invariant system an LTI system.
Example (PageIndex{1})
Consider the constant coefficient differential equation
[3y'' + 8y' + 7y = f(t)]
This equation models a damped harmonic oscillator, say a mass on a spring with a damper, where (f(t)) is the force on the mass and (y(t)) is its displacement from equilibrium. If we consider (f) to be the input and (y) the output, then this is a linear time invariant (LTI) system.
Example (PageIndex{2})
There are many variations on this theme. For example, we might have the LTI system
[3y'' + 8y' + 7y = f'(t)]
where we call (f(t)) the input signal and (y (t)) the output signal.
## Control engineering
Durgesh Singh
is currently pursuing the Ph.D. degree in Control Engineering with IIT Madras, Chennai, India. His current research interests include system identification and performance management in cloud computing systems, distributed control, and stability analysis of cyber-physical systems.
Vijayanand is currently an M.S. Scholar in the Department of Electrical Engineering, IIT Madras.
Amol Patil is currently associated with A. C. Patil College of Engineering and pursuing the Ph.D. degree in Electronics and Telecommunication with SFIT, Mumbai, India. His current research interests include design of coordination and control algorithm in multi-agent dynamical system using algebraic graph theory .
### Course certificate
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## Signals and Systems Pdf Notes – SS Pdf Notes
Unit 1: Part A
Unit 1 – Part B
### Signals and Systems Notes Pdf – SS Notes Pdf
UNIT I
SIGNAL ANALYSIS Analogy between vectors and signals, Orthogonal signal space, Signal approximation using orthogonal functions, Mean square error, Closed or complete set of orthogonal functions, Orthogonality in complex functions, Exponential and sinusoidal signals, Concepts of Impulse function, Unit step function, Signum function.
UNIT II
FOURIER SERIES REPRESENTATION OF PERIODIC SIGNALS Representation of Fourier series, Continuous time periodic signals, properties of Fourier series, Dirichlet’s conditions, Trigonometric Fourier series and Exponential Fourier series, Complex Fourier spectrum
UNIT III
FOURIER TRANSFORMS Deriving Fourier transform from Fourier series, Fourier transform of the arbitrary signal, Fourier transform of standard signals, Fourier transform of periodic signals, properties of Fourier transforms, Fourier transforms involving impulse function and Signum function. Introduction to Hilbert Transform.
UNIT IV
SIGNAL TRANSMISSION THROUGH LINEAR SYSTEMS Linear system, impulse response, Response of a linear system, Linear time-invariant (LTI) system, Linear time variant (LTV) system, the Transfer function of an LTI system. Filter characteristics of linear systems. Distortionless transmission through a system, Signal bandwidth, system bandwidth, Ideal LPF, HPF, and BPF characteristics, Causality and Poly-Wiener criterion for physical realization, the relationship between bandwidth and rise time.
UNIT V
CONVOLUTION AND CORRELATION OF SIGNALS Concept of convolution in the time domain and frequency domain, Graphical representation of convolution, Convolution property of Fourier transforms. Cross-correlation and autocorrelation of functions, properties of the correlation function, Energy density spectrum, Parseval’s theorem, Power density spectrum, Relation between autocorrelation function and energy/power spectral density function.A relation between convolution and correlation, Detection of periodic signals in the presence of noise by correlation, Extraction of the signal from noise by filtering.
### Signals and Systems Pdf Notes – SS Pdf Notes
UNIT VI
SAMPLING Sampling theorem – Graphical and analytical proof for Band-Limited Signals, impulse sampling, Natural and Flat top Sampling, Reconstruction of the signal from its samples, effect of undersampling – Aliasing, Introduction to Band Pass sampling.
UNIT VII
LAPLACE TRANSFORMS Review of Laplace transforms, Partial fraction expansion, Inverse Laplace transform, Concept of the region of convergence (ROC) for Laplace transforms, constraints on ROC for various classes of signals, Properties of L.T’s relation between L.T’s, and F.T. of a signal. Laplace transform of certain signals using waveform synthesis.
UNIT VIII
Z–TRANSFORMS Fundamental difference between continuous and discrete time signals, discrete time signal representation using complex exponential and sinusoidal components, Periodicity of discrete time using complex exponential signal, Concept of Z- Transform of a discrete sequence. The distinction between Laplace, Fourier, and Z transforms. The region of convergence in Z-Transform, constraints on ROC for various classes of signals, Inverse Z-transform, properties of Z-transforms.
TEXTBOOKS: Signals and Systems Notes Pdf – SS Notes Pdf
1. Signals, Systems & Communications – B.P. Lathi, BS Publications, 2003.
2. Signals and systems-A.Rama Krishna Rao-2008, TMH
3. Signals and Systems – A.V. Oppenheim, A.S. Willsky and S.H. Nawab, PHI, 2nd Edn.
REFERENCES: Signals and Systems Notes – SS Notes – SS Pdf Notes
1. Signals & Systems – Simon Haykin and Van Veen, Wiley, 2nd Edition.
2. Introduction to signal and system analysis – K.Gopalan 2009, CENGAGE Learning.
3.Fundamentals of signals and systems-Michel J Robert 2008 MGH International Edition.
Convolving mask over image. It is done in this way. Place the center of the mask at each element of an image. Multiply the corresponding elements and then add them , and paste the result onto the element of the image on which you place the center of mask.
The box in red color is the mask, and the values in the orange are the values of the mask. The black color box and values belong to the image. Now for the first pixel of the image, the value will be calculated as
First pixel = (5*2) + (4*4) + (2*8) + (1*10)
Place 52 in the original image at the first index and repeat this procedure for each pixel of the image.
## Introduction to Linear Regression Analysis, 5th Edition
Introduction to Linear Regression Analysis, Fifth Edition continues to present both the conventional and less common uses of linear regression in today’s cutting-edge scientific research. The authors blend both theory and application to equip readers with an understanding of the basic principles needed to apply regression model-building techniques in various fields of study, including engineering, management, and the health sciences.
Following a general introduction to regression modeling, including typical applications, a host of technical tools are outlined such as basic inference procedures, introductory aspects of model adequacy checking, and polynomial regression models and their variations. The book then discusses how transformations and weighted least squares can be used to resolve problems of model inadequacy and also how to deal with influential observations. The Fifth Edition features numerous newly added topics, including:
• A chapter on regression analysis of time series data that presents the Durbin-Watson test and other techniques for detecting autocorrelation as well as parameter estimation in time series regression models
• Regression models with random effects in addition to a discussion on subsampling and the importance of the mixed model
• Tests on individual regression coefficients and subsets of coefficients
• Examples of current uses of simple linear regression models and the use of multiple regression models for understanding patient satisfaction data.
In addition to Minitab, SAS, and S-PLUS, the authors have incorporated JMP and the freely available R software to illustrate the discussed techniques and procedures in this new edition. Numerous exercises have been added throughout, allowing readers to test their understanding of the material.
Introduction to Linear Regression Analysis, Fifth Edition is an excellent book for statistics and engineering courses on regression at the upper-undergraduate and graduate levels. The book also serves as a valuable, robust resource for professionals in the fields of engineering, life and biological sciences, and the social sciences.
## Cyclostationarity of DMR Signals
Let’s take a brief look at the cyclostationarity of a captured DMR signal. It’s more complicated than one might think.
In this post I look at the cyclostationarity of a digital mobile radio (DMR) signal empirically. That is, I have a captured DMR signal from sigidwiki.com, and I apply blind CSP to it to determine its cycle frequencies and spectral correlation function. The signal is arranged in frames or slots, with gaps between successive slots, so there is the chance that we’ll see cyclostationarity due to the on-burst (or on-frame) signaling and cyclostationarity due to the framing itself.
## Lecture Notes
be updated during the lecture. We will read the lecture notes together and update the
text before the material is dicussed in the class.
## 13.1: A brief introduction to linear time invariant systems
We will begin with a whirlwind tour of special relativity (SR) and life in flat spacetime. The point will be both to recall what SR is all about, and to introduce tensors and related concepts that will be crucial later on, without the extra complications of curvature on top of everything else. Therefore, for this section we will always be working in flat spacetime, and furthermore we will only use orthonormal (Cartesian-like) coordinates. Needless to say it is possible to do SR in any coordinate system you like, but it turns out that introducing the necessary tools for doing so would take us halfway to curved spaces anyway, so we will put that off for a while.
It is often said that special relativity is a theory of 4-dimensional spacetime: three of space, one of time. But of course, the pre-SR world of Newtonian mechanics featured three spatial dimensions and a time parameter. Nevertheless, there was not much temptation to consider these as different aspects of a single 4-dimensional spacetime. Why not?
Consider a garden-variety 2-dimensional plane. It is typically convenient to label the points on such a plane by introducing coordinates, for example by defining orthogonal x and y axes and projecting each point onto these axes in the usual way. However, it is clear that most of the interesting geometrical facts about the plane are independent of our choice of coordinates. As a simple example, we can consider the distance between two points, given by
In a different Cartesian coordinate system, defined by x' and y' axes which are rotated with respect to the originals, the formula for the distance is unaltered:
We therefore say that the distance is invariant under such changes of coordinates.
This is why it is useful to think of the plane as 2-dimensional: although we use two distinct numbers to label each point, the numbers are not the essence of the geometry, since we can rotate axes into each other while leaving distances and so forth unchanged. In Newtonian physics this is not the case with space and time there is no useful notion of rotating space and time into each other. Rather, the notion of "all of space at a single moment in time" has a meaning independent of coordinates.
Such is not the case in SR. Let us consider coordinates ( t , x , y , z ) on spacetime, set up in the following way. The spatial coordinates ( x , y , z ) comprise a standard Cartesian system, constructed for example by welding together rigid rods which meet at right angles. The rods must be moving freely, unaccelerated. The time coordinate is defined by a set of clocks which are not moving with respect to the spatial coordinates. (Since this is a thought experiment, we imagine that the rods are infinitely long and there is one clock at every point in space.) The clocks are synchronized in the following sense: if you travel from one point in space to any other in a straight line at constant speed, the time difference between the clocks at the ends of your journey is the same as if you had made the same trip, at the same speed, in the other direction. The coordinate system thus constructed is an inertial frame .
An event is defined as a single moment in space and time, characterized uniquely by ( t , x , y , z ). Then, without any motivation for the moment, let us introduce the spacetime interval between two events:
(Notice that it can be positive, negative, or zero even for two nonidentical points.) Here, c is some fixed conversion factor between space and time that is, a fixed velocity. Of course it will turn out to be the speed of light the important thing, however, is not that photons happen to travel at that speed, but that there exists a c such that the spacetime interval is invariant under changes of coordinates. In other words, if we set up a new inertial frame ( t' , x' , y' , z' ) by repeating our earlier procedure, but allowing for an offset in initial position, angle, and velocity between the new rods and the old, the interval is unchanged:
This is why it makes sense to think of SR as a theory of 4-dimensional spacetime, known as Minkowski space . (This is a special case of a 4-dimensional manifold, which we will deal with in detail later.) As we shall see, the coordinate transformations which we have implicitly defined do, in a sense, rotate space and time into each other. There is no absolute notion of "simultaneous events" whether two things occur at the same time depends on the coordinates used. Therefore the division of Minkowski space into space and time is a choice we make for our own purposes, not something intrinsic to the situation.
Almost all of the "paradoxes" associated with SR result from a stubborn persistence of the Newtonian notions of a unique time coordinate and the existence of "space at a single moment in time." By thinking in terms of spacetime rather than space and time together, these paradoxes tend to disappear.
Let's introduce some convenient notation. Coordinates on spacetime will be denoted by letters with Greek superscript indices running from 0 to 3, with 0 generally denoting the time coordinate. Thus,
(Don't start thinking of the superscripts as exponents.) Furthermore, for the sake of simplicity we will choose units in which
we will therefore leave out factors of c in all subsequent formulae. Empirically we know that c is the speed of light, 3 × 10 8 meters per second thus, we are working in units where 1 second equals 3 × 10 8 meters. Sometimes it will be useful to refer to the space and time components of x separately, so we will use Latin superscripts to stand for the space components alone:
It is also convenient to write the spacetime interval in a more compact form. We therefore introduce a 4 × 4 matrix, the metric , which we write using two lower indices:
(Some references, especially field theory books, define the metric with the opposite sign, so be careful.) We then have the nice formula
Notice that we use the summation convention , in which indices which appear both as superscripts and subscripts are summed over. The content of (1.9) is therefore just the same as (1.3).
Now we can consider coordinate transformations in spacetime at a somewhat more abstract level than before. What kind of transformations leave the interval (1.9) invariant? One simple variety are the translations, which merely shift the coordinates:
where a is a set of four fixed numbers. (Notice that we put the prime on the index, not on the x .) Translations leave the differences x unchanged, so it is not remarkable that the interval is unchanged. The only other kind of linear transformation is to multiply x by a (spacetime-independent) matrix:
or, in more conventional matrix notation,
These transformations do not leave the differences x unchanged, but multiply them also by the matrix . What kind of matrices will leave the interval invariant? Sticking with the matrix notation, what we would like is
We want to find the matrices such that the components of the matrix are the same as those of that is what it means for the interval to be invariant under these transformations.
The matrices which satisfy (1.14) are known as the Lorentz transformations the set of them forms a group under matrix multiplication, known as the Lorentz group . There is a close analogy between this group and O(3), the rotation group in three-dimensional space. The rotation group can be thought of as 3 × 3 matrices R which satisfy
where 1 is the 3 × 3 identity matrix. The similarity with (1.14) should be clear the only difference is the minus sign in the first term of the metric , signifying the timelike direction. The Lorentz group is therefore often referred to as O(3,1). (The 3 × 3 identity matrix is simply the metric for ordinary flat space. Such a metric, in which all of the eigenvalues are positive, is called Euclidean , while those such as (1.8) which feature a single minus sign are called Lorentzian .)
Lorentz transformations fall into a number of categories. First there are the conventional rotations , such as a rotation in the x - y plane:
The rotation angle is a periodic variable with period 2. There are also boosts , which may be thought of as "rotations between space and time directions." An example is given by
The boost parameter , unlike the rotation angle, is defined from - to . There are also discrete transformations which reverse the time direction or one or more of the spatial directions. (When these are excluded we have the proper Lorentz group, SO(3,1).) A general transformation can be obtained by multiplying the individual transformations the explicit expression for this six-parameter matrix (three boosts, three rotations) is not sufficiently pretty or useful to bother writing down. In general Lorentz transformations will not commute, so the Lorentz group is non-abelian. The set of both translations and Lorentz transformations is a ten-parameter non-abelian group, the Poincaré group .
You should not be surprised to learn that the boosts correspond to changing coordinates by moving to a frame which travels at a constant velocity, but let's see it more explicitly. For the transformation given by (1.18), the transformed coordinates t' and x' will be given by
From this we see that the point defined by x' = 0 is moving it has a velocity
To translate into more pedestrian notation, we can replace = tanh -1 v to obtain
where = 1/. So indeed, our abstract approach has recovered the conventional expressions for Lorentz transformations. Applying these formulae leads to time dilation, length contraction, and so forth.
An extremely useful tool is the spacetime diagram , so let's consider Minkowski space from this point of view. We can begin by portraying the initial t and x axes at (what are conventionally thought of as) right angles, and suppressing the y and z axes. Then according to (1.19), under a boost in the x - t plane the x' axis ( t' = 0) is given by t = x tanh, while the t' axis ( x' = 0) is given by t = x /tanh. We therefore see that the space and time axes are rotated into each other, although they scissor together instead of remaining orthogonal in the traditional Euclidean sense. (As we shall see, the axes do in fact remain orthogonal in the Lorentzian sense.) This should come as no surprise, since if spacetime behaved just like a four-dimensional version of space the world would be a very different place.
It is also enlightening to consider the paths corresponding to travel at the speed c = 1. These are given in the original coordinate system by x = ± t . In the new system, a moment's thought reveals that the paths defined by x' = ± t' are precisely the same as those defined by x = ± t these trajectories are left invariant under Lorentz transformations. Of course we know that light travels at this speed we have therefore found that the speed of light is the same in any inertial frame. A set of points which are all connected to a single event by straight lines moving at the speed of light is called a light cone this entire set is invariant under Lorentz transformations. Light cones are naturally divided into future and past the set of all points inside the future and past light cones of a point p are called timelike separated from p , while those outside the light cones are spacelike separated and those on the cones are lightlike or null separated from p . Referring back to (1.3), we see that the interval between timelike separated points is negative, between spacelike separated points is positive, and between null separated points is zero. (The interval is defined to be s 2 , not the square root of this quantity.) Notice the distinction between this situation and that in the Newtonian world here, it is impossible to say (in a coordinate-independent way) whether a point that is spacelike separated from p is in the future of p , the past of p , or "at the same time".
To probe the structure of Minkowski space in more detail, it is necessary to introduce the concepts of vectors and tensors. We will start with vectors, which should be familiar. Of course, in spacetime vectors are four-dimensional, and are often referred to as four-vectors . This turns out to make quite a bit of difference for example, there is no such thing as a cross product between two four-vectors.
Beyond the simple fact of dimensionality, the most important thing to emphasize is that each vector is located at a given point in spacetime. You may be used to thinking of vectors as stretching from one point to another in space, and even of "free" vectors which you can slide carelessly from point to point. These are not useful concepts in relativity. Rather, to each point p in spacetime we associate the set of all possible vectors located at that point this set is known as the tangent space at p , or T p . The name is inspired by thinking of the set of vectors attached to a point on a simple curved two-dimensional space as comprising a plane which is tangent to the point. But inspiration aside, it is important to think of these vectors as being located at a single point, rather than stretching from one point to another. (Although this won't stop us from drawing them as arrows on spacetime diagrams.)
Later we will relate the tangent space at each point to things we can construct from the spacetime itself. For right now, just think of T p as an abstract vector space for each point in spacetime. A (real) vector space is a collection of objects ("vectors") which, roughly speaking, can be added together and multiplied by real numbers in a linear way. Thus, for any two vectors V and W and real numbers a and b , we have
Every vector space has an origin, i.e. a zero vector which functions as an identity element under vector addition. In many vector spaces there are additional operations such as taking an inner (dot) product, but this is extra structure over and above the elementary concept of a vector space.
A vector is a perfectly well-defined geometric object, as is a vector field , defined as a set of vectors with exactly one at each point in spacetime. (The set of all the tangent spaces of a manifold M is called the tangent bundle , T ( M ).) Nevertheless it is often useful for concrete purposes to decompose vectors into components with respect to some set of basis vectors. A basis is any set of vectors which both spans the vector space (any vector is a linear combination of basis vectors) and is linearly independent (no vector in the basis is a linear combination of other basis vectors). For any given vector space, there will be an infinite number of legitimate bases, but each basis will consist of the same number of vectors, known as the dimension of the space. (For a tangent space associated with a point in Minkowski space, the dimension is of course four.)
Let us imagine that at each tangent space we set up a basis of four vectors , with <0, 1, 2, 3>as usual. In fact let us say that each basis is adapted to the coordinates x that is, the basis vector is what we would normally think of pointing along the x -axis, etc. It is by no means necessary that we choose a basis which is adapted to any coordinate system at all, although it is often convenient. (We really could be more precise here, but later on we will repeat the discussion at an excruciating level of precision, so some sloppiness now is forgivable.) Then any abstract vector A can be written as a linear combination of basis vectors:
The coefficients A are the components of the vector A . More often than not we will forget the basis entirely and refer somewhat loosely to "the vector A ", but keep in mind that this is shorthand. The real vector is an abstract geometrical entity, while the components are just the coefficients of the basis vectors in some convenient basis. (Since we will usually suppress the explicit basis vectors, the indices will usually label components of vectors and tensors. This is why there are parentheses around the indices on the basis vectors, to remind us that this is a collection of vectors, not components of a single vector.)
A standard example of a vector in spacetime is the tangent vector to a curve. A parameterized curve or path through spacetime is specified by the coordinates as a function of the parameter, e.g. x (). The tangent vector V () has components
The entire vector is thus V = V . Under a Lorentz transformation the coordinates x change according to (1.11), while the parameterization is unaltered we can therefore deduce that the components of the tangent vector must change as
However, the vector itself (as opposed to its components in some coordinate system) is invariant under Lorentz transformations. We can use this fact to derive the transformation properties of the basis vectors. Let us refer to the set of basis vectors in the transformed coordinate system as . Since the vector is invariant, we have
But this relation must hold no matter what the numerical values of the components V are. Therefore we can say
To get the new basis in terms of the old one we should multiply by the inverse of the Lorentz transformation . But the inverse of a Lorentz transformation from the unprimed to the primed coordinates is also a Lorentz transformation, this time from the primed to the unprimed systems. We will therefore introduce a somewhat subtle notation, by writing using the same symbol for both matrices, just with primed and unprimed indices adjusted. That is,
where is the traditional Kronecker delta symbol in four dimensions. (Note that Schutz uses a different convention, always arranging the two indices northwest/southeast the important thing is where the primes go.) From (1.27) we then obtain the transformation rule for basis vectors:
Therefore the set of basis vectors transforms via the inverse Lorentz transformation of the coordinates or vector components.
It is worth pausing a moment to take all this in. We introduced coordinates labeled by upper indices, which transformed in a certain way under Lorentz transformations. We then considered vector components which also were written with upper indices, which made sense since they transformed in the same way as the coordinate functions. (In a fixed coordinate system, each of the four coordinates x can be thought of as a function on spacetime, as can each of the four components of a vector field.) The basis vectors associated with the coordinate system transformed via the inverse matrix, and were labeled by a lower index. This notation ensured that the invariant object constructed by summing over the components and basis vectors was left unchanged by the transformation, just as we would wish. It's probably not giving too much away to say that this will continue to be the case for more complicated objects with multiple indices (tensors).
Once we have set up a vector space, there is an associated vector space (of equal dimension) which we can immediately define, known as the dual vector space . The dual space is usually denoted by an asterisk, so that the dual space to the tangent space T p is called the cotangent space and denoted T * p . The dual space is the space of all linear maps from the original vector space to the real numbers in math lingo, if T p * is a dual vector, then it acts as a map such that:
where V , W are vectors and a , b are real numbers. The nice thing about these maps is that they form a vector space themselves thus, if and are dual vectors, we have
To make this construction somewhat more concrete, we can introduce a set of basis dual vectors by demanding
Then every dual vector can be written in terms of its components, which we label with lower indices:
In perfect analogy with vectors, we will usually simply write to stand for the entire dual vector. In fact, you will sometime see elements of T p (what we have called vectors) referred to as contravariant vectors , and elements of T p * (what we have called dual vectors) referred to as covariant vectors . Actually, if you just refer to ordinary vectors as vectors with upper indices and dual vectors as vectors with lower indices, nobody should be offended. Another name for dual vectors is one-forms , a somewhat mysterious designation which will become clearer soon.
The component notation leads to a simple way of writing the action of a dual vector on a vector:
This is why it is rarely necessary to write the basis vectors (and dual vectors) explicitly the components do all of the work. The form of (1.35) also suggests that we can think of vectors as linear maps on dual vectors, by defining
Therefore, the dual space to the dual vector space is the original vector space itself.
Of course in spacetime we will be interested not in a single vector space, but in fields of vectors and dual vectors. (The set of all cotangent spaces over M is the cotangent bundle , T * ( M ).) In that case the action of a dual vector field on a vector field is not a single number, but a scalar (or just "function") on spacetime. A scalar is a quantity without indices, which is unchanged under Lorentz transformations.
We can use the same arguments that we earlier used for vectors to derive the transformation properties of dual vectors. The answers are, for the components,
and for basis dual vectors,
This is just what we would expect from index placement the components of a dual vector transform under the inverse transformation of those of a vector. Note that this ensures that the scalar (1.35) is invariant under Lorentz transformations, just as it should be.
Let's consider some examples of dual vectors, first in other contexts and then in Minkowski space. Imagine the space of n -component column vectors, for some integer n . Then the dual space is that of n -component row vectors, and the action is ordinary matrix multiplication:
Another familiar example occurs in quantum mechanics, where vectors in the Hilbert space are represented by kets, |. In this case the dual space is the space of bras, |, and the action gives the number |. (This is a complex number in quantum mechanics, but the idea is precisely the same.)
In spacetime the simplest example of a dual vector is the gradient of a scalar function, the set of partial derivatives with respect to the spacetime coordinates, which we denote by "d":
The conventional chain rule used to transform partial derivatives amounts in this case to the transformation rule of components of dual vectors:
where we have used (1.11) and (1.28) to relate the Lorentz transformation to the coordinates. The fact that the gradient is a dual vector leads to the following shorthand notations for partial derivatives:
(Very roughly speaking, " x has an upper index, but when it is in the denominator of a derivative it implies a lower index on the resulting object.") I'm not a big fan of the comma notation, but we will use all the time. Note that the gradient does in fact act in a natural way on the example we gave above of a vector, the tangent vector to a curve. The result is ordinary derivative of the function along the curve:
As a final note on dual vectors, there is a way to represent them as pictures which is consistent with the picture of vectors as arrows. See the discussion in Schutz, or in MTW (where it is taken to dizzying extremes).
A straightforward generalization of vectors and dual vectors is the notion of a tensor . Just as a dual vector is a linear map from vectors to R , a tensor T of type (or rank) ( k , l ) is a multilinear map from a collection of dual vectors and vectors to R :
Here, "×" denotes the Cartesian product, so that for example T p × T p is the space of ordered pairs of vectors. Multilinearity means that the tensor acts linearly in each of its arguments for instance, for a tensor of type (1, 1), we have
From this point of view, a scalar is a type (0, 0) tensor, a vector is a type (1, 0) tensor, and a dual vector is a type (0, 1) tensor.
The space of all tensors of a fixed type ( k , l ) forms a vector space they can be added together and multiplied by real numbers. To construct a basis for this space, we need to define a new operation known as the tensor product , denoted by . If T is a ( k , l ) tensor and S is a ( m , n ) tensor, we define a ( k + m , l + n ) tensor T S by
(Note that the and V (i) are distinct dual vectors and vectors, not components thereof.) In other words, first act T on the appropriate set of dual vectors and vectors, and then act S on the remainder, and then multiply the answers. Note that, in general, T S S T .
It is now straightforward to construct a basis for the space of all ( k , l ) tensors, by taking tensor products of basis vectors and dual vectors this basis will consist of all tensors of the form
In a 4-dimensional spacetime there will be 4 k + l basis tensors in all. In component notation we then write our arbitrary tensor as
Alternatively, we could define the components by acting the tensor on basis vectors and dual vectors:
You can check for yourself, using (1.33) and so forth, that these equations all hang together properly.
As with vectors, we will usually take the shortcut of denoting the tensor T by its components T . . . The action of the tensors on a set of vectors and dual vectors follows the pattern established in (1.35):
The order of the indices is obviously important, since the tensor need not act in the same way on its various arguments. Finally, the transformation of tensor components under Lorentz transformations can be derived by applying what we already know about the transformation of basis vectors and dual vectors. The answer is just what you would expect from index placement,
Thus, each upper index gets transformed like a vector, and each lower index gets transformed like a dual vector.
Although we have defined tensors as linear maps from sets of vectors and tangent vectors to R , there is nothing that forces us to act on a full collection of arguments. Thus, a (1, 1) tensor also acts as a map from vectors to vectors:
You can check for yourself that T V is a vector ( i.e. obeys the vector transformation law). Similarly, we can act one tensor on (all or part of) another tensor to obtain a third tensor. For example,
is a perfectly good (1, 1) tensor.
You may be concerned that this introduction to tensors has been somewhat too brief, given the esoteric nature of the material. In fact, the notion of tensors does not require a great deal of effort to master it's just a matter of keeping the indices straight, and the rules for manipulating them are very natural. Indeed, a number of books like to define tensors as collections of numbers transforming according to (1.51). While this is operationally useful, it tends to obscure the deeper meaning of tensors as geometrical entities with a life independent of any chosen coordinate system. There is, however, one subtlety which we have glossed over. The notions of dual vectors and tensors and bases and linear maps belong to the realm of linear algebra, and are appropriate whenever we have an abstract vector space at hand. In the case of interest to us we have not just a vector space, but a vector space at each point in spacetime. More often than not we are interested in tensor fields, which can be thought of as tensor-valued functions on spacetime. Fortunately, none of the manipulations we defined above really care whether we are dealing with a single vector space or a collection of vector spaces, one for each event. We will be able to get away with simply calling things functions of x when appropriate. However, you should keep straight the logical independence of the notions we have introduced and their specific application to spacetime and relativity.
Now let's turn to some examples of tensors. First we consider the previous example of column vectors and their duals, row vectors. In this system a (1, 1) tensor is simply a matrix, M i j . Its action on a pair (, V ) is given by usual matrix multiplication:
If you like, feel free to think of tensors as "matrices with an arbitrary number of indices."
In spacetime, we have already seen some examples of tensors without calling them that. The most familiar example of a (0, 2) tensor is the metric, . The action of the metric on two vectors is so useful that it gets its own name, the inner product (or dot product):
Just as with the conventional Euclidean dot product, we will refer to two vectors whose dot product vanishes as orthogonal . Since the dot product is a scalar, it is left invariant under Lorentz transformations therefore the basis vectors of any Cartesian inertial frame, which are chosen to be orthogonal by definition, are still orthogonal after a Lorentz transformation (despite the "scissoring together" we noticed earlier). The norm of a vector is defined to be inner product of the vector with itself unlike in Euclidean space, this number is not positive definite:
(A vector can have zero norm without being the zero vector.) You will notice that the terminology is the same as that which we earlier used to classify the relationship between two points in spacetime it's no accident, of course, and we will go into more detail later.
Another tensor is the Kronecker delta , of type (1, 1), which you already know the components of. Related to this and the metric is the inverse metric , a type (2, 0) tensor defined as the inverse of the metric:
In fact, as you can check, the inverse metric has exactly the same components as the metric itself. (This is only true in flat space in Cartesian coordinates, and will fail to hold in more general situations.) There is also the Levi-Civita tensor , a (0, 4) tensor:
Here, a "permutation of 0123" is an ordering of the numbers 0, 1, 2, 3 which can be obtained by starting with 0123 and exchanging two of the digits an even permutation is obtained by an even number of such exchanges, and an odd permutation is obtained by an odd number. Thus, for example, = - 1.
It is a remarkable property of the above tensors - the metric, the inverse metric, the Kronecker delta, and the Levi-Civita tensor - that, even though they all transform according to the tensor transformation law (1.51), their components remain unchanged in any Cartesian coordinate system in flat spacetime. In some sense this makes them bad examples of tensors, since most tensors do not have this property. In fact, even these tensors do not have this property once we go to more general coordinate systems, with the single exception of the Kronecker delta. This tensor has exactly the same components in any coordinate system in any spacetime. This makes sense from the definition of a tensor as a linear map the Kronecker tensor can be thought of as the identity map from vectors to vectors (or from dual vectors to dual vectors), which clearly must have the same components regardless of coordinate system. The other tensors (the metric, its inverse, and the Levi-Civita tensor) characterize the structure of spacetime, and all depend on the metric. We shall therefore have to treat them more carefully when we drop our assumption of flat spacetime.
A more typical example of a tensor is the electromagnetic field strength tensor . We all know that the electromagnetic fields are made up of the electric field vector E i and the magnetic field vector B i . (Remember that we use Latin indices for spacelike components 1,2,3.) Actually these are only "vectors" under rotations in space, not under the full Lorentz group. In fact they are components of a (0, 2) tensor F , defined by
From this point of view it is easy to transform the electromagnetic fields in one reference frame to those in another, by application of (1.51). The unifying power of the tensor formalism is evident: rather than a collection of two vectors whose relationship and transformation properties are rather mysterious, we have a single tensor field to describe all of electromagnetism. (On the other hand, don't get carried away sometimes it's more convenient to work in a single coordinate system using the electric and magnetic field vectors.)
With some examples in hand we can now be a little more systematic about some properties of tensors. First consider the operation of contraction , which turns a ( k , l ) tensor into a ( k - 1, l - 1) tensor. Contraction proceeds by summing over one upper and one lower index:
You can check that the result is a well-defined tensor. Of course it is only permissible to contract an upper index with a lower index (as opposed to two indices of the same type). Note also that the order of the indices matters, so that you can get different tensors by contracting in different ways thus,
The metric and inverse metric can be used to raise and lower indices on tensors. That is, given a tensor T , we can use the metric to define new tensors which we choose to denote by the same letter T :
and so forth. Notice that raising and lowering does not change the position of an index relative to other indices, and also that "free" indices (which are not summed over) must be the same on both sides of an equation, while "dummy" indices (which are summed over) only appear on one side. As an example, we can turn vectors and dual vectors into each other by raising and lowering indices:
This explains why the gradient in three-dimensional flat Euclidean space is usually thought of as an ordinary vector, even though we have seen that it arises as a dual vector in Euclidean space (where the metric is diagonal with all entries +1) a dual vector is turned into a vector with precisely the same components when we raise its index. You may then wonder why we have belabored the distinction at all. One simple reason, of course, is that in a Lorentzian spacetime the components are not equal:
In a curved spacetime, where the form of the metric is generally more complicated, the difference is rather more dramatic. But there is a deeper reason, namely that tensors generally have a "natural" definition which is independent of the metric. Even though we will always have a metric available, it is helpful to be aware of the logical status of each mathematical object we introduce. The gradient, and its action on vectors, is perfectly well defined regardless of any metric, whereas the "gradient with upper indices" is not. (As an example, we will eventually want to take variations of functionals with respect to the metric, and will therefore have to know exactly how the functional depends on the metric, something that is easily obscured by the index notation.)
Continuing our compilation of tensor jargon, we refer to a tensor as symmetric in any of its indices if it is unchanged under exchange of those indices. Thus, if
we say that S is symmetric in its first two indices, while if
we say that S is symmetric in all three of its indices. Similarly, a tensor is antisymmetric (or "skew-symmetric") in any of its indices if it changes sign when those indices are exchanged thus,
means that A is antisymmetric in its first and third indices (or just "antisymmetric in and "). If a tensor is (anti-) symmetric in all of its indices, we refer to it as simply (anti-) symmetric (sometimes with the redundant modifier "completely"). As examples, the metric and the inverse metric are symmetric, while the Levi-Civita tensor and the electromagnetic field strength tensor F are antisymmetric. (Check for yourself that if you raise or lower a set of indices which are symmetric or antisymmetric, they remain that way.) Notice that it makes no sense to exchange upper and lower indices with each other, so don't succumb to the temptation to think of the Kronecker delta as symmetric. On the other hand, the fact that lowering an index on gives a symmetric tensor (in fact, the metric) means that the order of indices doesn't really matter, which is why we don't keep track index placement for this one tensor.
Given any tensor, we can symmetrize (or antisymmetrize) any number of its upper or lower indices. To symmetrize, we take the sum of all permutations of the relevant indices and divide by the number of terms:
while antisymmetrization comes from the alternating sum:
By "alternating sum" we mean that permutations which are the result of an odd number of exchanges are given a minus sign, thus:
Notice that round/square brackets denote symmetrization/antisymmetrization. Furthermore, we may sometimes want to (anti-) symmetrize indices which are not next to each other, in which case we use vertical bars to denote indices not included in the sum:
Finally, some people use a convention in which the factor of 1/ n ! is omitted. The one used here is a good one, since (for example) a symmetric tensor satisfies
and likewise for antisymmetric tensors.
We have been very careful so far to distinguish clearly between things that are always true (on a manifold with arbitrary metric) and things which are only true in Minkowski space in Cartesian coordinates. One of the most important distinctions arises with partial derivatives . If we are working in flat spacetime with Cartesian coordinates, then the partial derivative of a ( k , l ) tensor is a ( k , l + 1) tensor that is,
transforms properly under Lorentz transformations. However, this will no longer be true in more general spacetimes, and we will have to define a "covariant derivative" to take the place of the partial derivative. Nevertheless, we can still use the fact that partial derivatives give us tensor in this special case, as long as we keep our wits about us. (The one exception to this warning is the partial derivative of a scalar, , which is a perfectly good tensor [the gradient] in any spacetime.)
We have now accumulated enough tensor know-how to illustrate some of these concepts using actual physics. Specifically, we will examine Maxwell's equations of electrodynamics. In 19 th -century notation, these are
Here, E and B are the electric and magnetic field 3-vectors, J is the current, is the charge density, and × and . are the conventional curl and divergence. These equations are invariant under Lorentz transformations, of course that's how the whole business got started. But they don't look obviously invariant our tensor notation can fix that. Let's begin by writing these equations in just a slightly different notation,
In these expressions, spatial indices have been raised and lowered with abandon, without any attempt to keep straight where the metric appears. This is because is the metric on flat 3-space, with its inverse (they are equal as matrices). We can therefore raise and lower indices at will, since the components don't change. Meanwhile, the three-dimensional Levi-Civita tensor is defined just as the four-dimensional one, although with one fewer index. We have replaced the charge density by J 0 this is legitimate because the density and current together form the current 4-vector , J = (, J 1 , J 2 , J 3 ).
From these expressions, and the definition (1.58) of the field strength tensor F , it is easy to get a completely tensorial 20 th -century version of Maxwell's equations. Begin by noting that we can express the field strength with upper indices as
(To check this, note for example that F 01 = F 01 and F 12 = B 3 .) Then the first two equations in (1.74) become
Using the antisymmetry of F , we see that these may be combined into the single tensor equation
A similar line of reasoning, which is left as an exercise to you, reveals that the third and fourth equations in (1.74) can be written
The four traditional Maxwell equations are thus replaced by two, thus demonstrating the economy of tensor notation. More importantly, however, both sides of equations (1.77) and (1.78) manifestly transform as tensors therefore, if they are true in one inertial frame, they must be true in any Lorentz-transformed frame. This is why tensors are so useful in relativity - we often want to express relationships without recourse to any reference frame, and it is necessary that the quantities on each side of an equation transform in the same way under change of coordinates. As a matter of jargon, we will sometimes refer to quantities which are written in terms of tensors as covariant (which has nothing to do with "covariant" as opposed to "contravariant"). Thus, we say that (1.77) and (1.78) together serve as the covariant form of Maxwell's equations, while (1.73) or (1.74) are non-covariant.
Let us now introduce a special class of tensors, known as differential forms (or just "forms"). A differential p -form is a (0, p ) tensor which is completely antisymmetric. Thus, scalars are automatically 0-forms, and dual vectors are automatically one-forms (thus explaining this terminology from a while back). We also have the 2-form F and the 4-form . The space of all p -forms is denoted , and the space of all p -form fields over a manifold M is denoted ( M ). A semi-straightforward exercise in combinatorics reveals that the number of linearly independent p -forms on an n -dimensional vector space is n !/( p !( n - p )!). So at a point on a 4-dimensional spacetime there is one linearly independent 0-form, four 1-forms, six 2-forms, four 3-forms, and one 4-form. There are no p -forms for p > n , since all of the components will automatically be zero by antisymmetry.
Why should we care about differential forms? This is a hard question to answer without some more work, but the basic idea is that forms can be both differentiated and integrated, without the help of any additional geometric structure. We will delay integration theory until later, but see how to differentiate forms shortly.
Given a p -form A and a q -form B , we can form a ( p + q )-form known as the wedge product A B by taking the antisymmetrized tensor product:
Thus, for example, the wedge product of two 1-forms is
so you can alter the order of a wedge product if you are careful with signs.
The exterior derivative "d" allows us to differentiate p -form fields to obtain ( p + 1)-form fields. It is defined as an appropriately normalized antisymmetric partial derivative:
The simplest example is the gradient, which is the exterior derivative of a 1-form:
The reason why the exterior derivative deserves special attention is that it is a tensor , even in curved spacetimes, unlike its cousin the partial derivative. Since we haven't studied curved spaces yet, we cannot prove this, but (1.82) defines an honest tensor no matter what the metric and coordinates are.
Another interesting fact about exterior differentiation is that, for any form A ,
which is often written d 2 = 0. This identity is a consequence of the definition of d and the fact that partial derivatives commute, = (acting on anything). This leads us to the following mathematical aside, just for fun. We define a p -form A to be closed if dA = 0, and exact if A = dB for some ( p - 1)-form B . Obviously, all exact forms are closed, but the converse is not necessarily true. On a manifold M , closed p -forms comprise a vector space Z p ( M ), and exact forms comprise a vector space B p ( M ). Define a new vector space as the closed forms modulo the exact forms:
This is known as the p th de Rham cohomology vector space, and depends only on the topology of the manifold M . (Minkowski space is topologically equivalent to R 4 , which is uninteresting, so that all of the H p ( M ) vanish for p > 0 for p = 0 we have H 0 ( M ) = . Therefore in Minkowski space all closed forms are exact except for zero-forms zero-forms can't be exact since there are no -1-forms for them to be the exterior derivative of.) It is striking that information about the topology can be extracted in this way, which essentially involves the solutions to differential equations. The dimension b p of the space H p ( M ) is called the p th Betti number of M , and the Euler characteristic is given by the alternating sum
Cohomology theory is the basis for much of modern differential topology.
Moving back to reality, the final operation on differential forms we will introduce is Hodge duality . We define the "Hodge star operator" on an n -dimensional manifold as a map from p -forms to ( n - p )-forms,
mapping A to " A dual". Unlike our other operations on forms, the Hodge dual does depend on the metric of the manifold (which should be obvious, since we had to raise some indices on the Levi-Civita tensor in order to define (1.87)). Applying the Hodge star twice returns either plus or minus the original form:
where s is the number of minus signs in the eigenvalues of the metric (for Minkowski space, s = 1).
Two facts on the Hodge dual: First, "duality" in the sense of Hodge is different than the relationship between vectors and dual vectors, although both can be thought of as the space of linear maps from the original space to R . Notice that the dimensionality of the space of ( n - p )-forms is equal to that of the space of p -forms, so this has at least a chance of being true. In the case of forms, the linear map defined by an ( n - p )-form acting on a p -form is given by the dual of the wedge product of the two forms. Thus, if A (n - p) is an ( n - p )-form and B (p) is a p -form at some point in spacetime, we have
The second fact concerns differential forms in 3-dimensional Euclidean space. The Hodge dual of the wedge product of two 1-forms gives another 1-form:
(All of the prefactors cancel.) Since 1-forms in Euclidean space are just like vectors, we have a map from two vectors to a single vector. You should convince yourself that this is just the conventional cross product, and that the appearance of the Levi-Civita tensor explains why the cross product changes sign under parity (interchange of two coordinates, or equivalently basis vectors). This is why the cross product only exists in three dimensions - because only in three dimensions do we have an interesting map from two dual vectors to a third dual vector. If you wanted to you could define a map from n - 1 one-forms to a single one-form, but I'm not sure it would be of any use.
Electrodynamics provides an especially compelling example of the use of differential forms. From the definition of the exterior derivative, it is clear that equation (1.78) can be concisely expressed as closure of the two-form F :
Does this mean that F is also exact? Yes as we've noted, Minkowski space is topologically trivial, so all closed forms are exact. There must therefore be a one-form A such that
This one-form is the familiar vector potential of electromagnetism, with the 0 component given by the scalar potential, A 0 = . If one starts from the view that the A is the fundamental field of electromagnetism, then (1.91) follows as an identity (as opposed to a dynamical law, an equation of motion). Gauge invariance is expressed by the observation that the theory is invariant under A A + d for some scalar (zero-form) , and this is also immediate from the relation (1.92). The other one of Maxwell's equations, (1.77), can be expressed as an equation between three-forms:
where the current one-form J is just the current four-vector with index lowered. Filling in the details is left for you to do.
As an intriguing aside, Hodge duality is the basis for one of the hottest topics in theoretical physics today. It's hard not to notice that the equations (1.91) and (1.93) look very similar. Indeed, if we set J = 0, the equations are invariant under the "duality transformations"
We therefore say that the vacuum Maxwell's equations are duality invariant, while the invariance is spoiled in the presence of charges. We might imagine that magnetic as well as electric monopoles existed in nature then we could add a magnetic current term 4(* J M ) to the right hand side of (1.91), and the equations would be invariant under duality transformations plus the additional replacement J J M . (Of course a nonzero right hand side to (1.91) is inconsistent with F = dA , so this idea only works if A is not a fundamental variable.) Long ago Dirac considered the idea of magnetic monopoles and showed that a necessary condition for their existence is that the fundamental monopole charge be inversely proportional to the fundamental electric charge. Now, the fundamental electric charge is a small number electrodynamics is "weakly coupled", which is why perturbation theory is so remarkably successful in quantum electrodynamics (QED). But Dirac's condition on magnetic charges implies that a duality transformation takes a theory of weakly coupled electric charges to a theory of strongly coupled magnetic monopoles (and vice-versa). Unfortunately monopoles don't exist (as far as we know), so these ideas aren't directly applicable to electromagnetism but there are some theories (such as supersymmetric non-abelian gauge theories) for which it has been long conjectured that some sort of duality symmetry may exist. If it did, we would have the opportunity to analyze a theory which looked strongly coupled (and therefore hard to solve) by looking at the weakly coupled dual version. Recently work by Seiberg and Witten and others has provided very strong evidence that this is exactly what happens in certain theories. The hope is that these techniques will allow us to explore various phenomena which we know exist in strongly coupled quantum field theories, such as confinement of quarks in hadrons.
We've now gone over essentially everything there is to know about the care and feeding of tensors. In the next section we will look more carefully at the rigorous definitions of manifolds and tensors, but the basic mechanics have been pretty well covered. Before jumping to more abstract mathematics, let's review how physics works in Minkowski spacetime.
Start with the worldline of a single particle. This is specified by a map M , where M is the manifold representing spacetime we usually think of the path as a parameterized curve x (). As mentioned earlier, the tangent vector to this path is dx / d (note that it depends on the parameterization). An object of primary interest is the norm of the tangent vector, which serves to characterize the path if the tangent vector is timelike/null/spacelike at some parameter value , we say that the path is timelike/null/spacelike at that point. This explains why the same words are used to classify vectors in the tangent space and intervals between two points - because a straight line connecting, say, two timelike separated points will itself be timelike at every point along the path.
Nevertheless, it's important to be aware of the sleight of hand which is being pulled here. The metric, as a (0, 2) tensor, is a machine which acts on two vectors (or two copies of the same vector) to produce a number. It is therefore very natural to classify tangent vectors according to the sign of their norm. But the interval between two points isn't something quite so natural it depends on a specific choice of path (a "straight line") which connects the points, and this choice in turn depends on the fact that spacetime is flat (which allows a unique choice of straight line between the points). A more natural object is the line element , or infinitesimal interval:
From this definition it is tempting to take the square root and integrate along a path to obtain a finite interval. But since ds 2 need not be positive, we define different procedures for different cases. For spacelike paths we define the path length
where the integral is taken over the path. For null paths the interval is zero, so no extra formula is required. For timelike paths we define the proper time
which will be positive. Of course we may consider paths that are timelike in some places and spacelike in others, but fortunately it is seldom necessary since the paths of physical particles never change their character (massive particles move on timelike paths, massless particles move on null paths). Furthermore, the phrase "proper time" is especially appropriate, since actually measures the time elapsed on a physical clock carried along the path . This point of view makes the "twin paradox" and similar puzzles very clear two worldlines, not necessarily straight, which intersect at two different events in spacetime will have proper times measured by the integral (1.97) along the appropriate paths, and these two numbers will in general be different even if the people travelling along them were born at the same time.
Let's move from the consideration of paths in general to the paths of massive particles (which will always be timelike). Since the proper time is measured by a clock travelling on a timelike worldline, it is convenient to use as the parameter along the path. That is, we use (1.97) to compute (), which (if is a good parameter in the first place) we can invert to obtain (), after which we can think of the path as x (). The tangent vector in this parameterization is known as the four-velocity , U :
Since d = - dx dx , the four-velocity is automatically normalized:
(It will always be negative, since we are only defining it for timelike trajectories. You could define an analogous vector for spacelike paths as well null paths give some extra problems since the norm is zero.) In the rest frame of a particle, its four-velocity has components U = (1, 0, 0, 0).
A related vector is the energy-momentum four-vector , defined by
where m is the mass of the particle. The mass is a fixed quantity independent of inertial frame what you may be used to thinking of as the "rest mass." It turns out to be much more convenient to take this as the mass once and for all, rather than thinking of mass as depending on velocity. The energy of a particle is simply p 0 , the timelike component of its energy-momentum vector. Since it's only one component of a four-vector, it is not invariant under Lorentz transformations that's to be expected, however, since the energy of a particle at rest is not the same as that of the same particle in motion. In the particle's rest frame we have p 0 = m recalling that we have set c = 1, we find that we have found the equation that made Einstein a celebrity, E = mc 2 . (The field equations of general relativity are actually much more important than this one, but " R - Rg = 8 GT " doesn't elicit the visceral reaction that you get from " E = mc 2 ".) In a moving frame we can find the components of p by performing a Lorentz transformation for a particle moving with (three-) velocity v along the x axis we have
where = 1/. For small v , this gives p 0 = m + mv 2 (what we usually think of as rest energy plus kinetic energy) and p 1 = mv (what we usually think of as [Newtonian] momentum). So the energy-momentum vector lives up to its name.
The centerpiece of pre-relativity physics is Newton's 2nd Law, or = m = d / dt . An analogous equation should hold in SR, and the requirement that it be tensorial leads us directly to introduce a force four-vector f satisfying
The simplest example of a force in Newtonian physics is the force due to gravity. In relativity, however, gravity is not described by a force, but rather by the curvature of spacetime itself. Instead, let us consider electromagnetism. The three-dimensional Lorentz force is given by = q ( + × ), where q is the charge on the particle. We would like a tensorial generalization of this equation. There turns out to be a unique answer:
You can check for yourself that this reduces to the Newtonian version in the limit of small velocities. Notice how the requirement that the equation be tensorial, which is one way of guaranteeing Lorentz invariance, severely restricted the possible expressions we could get. This is an example of a very general phenomenon, in which a small number of an apparently endless variety of possible physical laws are picked out by the demands of symmetry.
Although p provides a complete description of the energy and momentum of a particle, for extended systems it is necessary to go further and define the energy-momentum tensor (sometimes called the stress-energy tensor), T . This is a symmetric (2, 0) tensor which tells us all we need to know about the energy-like aspects of a system: energy density, pressure, stress, and so forth. A general definition of T is "the flux of four-momentum p across a surface of constant x ". To make this more concrete, let's consider the very general category of matter which may be characterized as a fluid - a continuum of matter described by macroscopic quantities such as temperature, pressure, entropy, viscosity, etc. In fact this definition is so general that it is of little use. In general relativity essentially all interesting types of matter can be thought of as perfect fluids , from stars to electromagnetic fields to the entire universe. Schutz defines a perfect fluid to be one with no heat conduction and no viscosity, while Weinberg defines it as a fluid which looks isotropic in its rest frame these two viewpoints turn out to be equivalent. Operationally, you should think of a perfect fluid as one which may be completely characterized by its pressure and density.
To understand perfect fluids, let's start with the even simpler example of dust . Dust is defined as a collection of particles at rest with respect to each other, or alternatively as a perfect fluid with zero pressure. Since the particles all have an equal velocity in any fixed inertial frame, we can imagine a "four-velocity field" U ( x ) defined all over spacetime. (Indeed, its components are the same at each point.) Define the number-flux four-vector to be
where n is the number density of the particles as measured in their rest frame. Then N 0 is the number density of particles as measured in any other frame, while N i is the flux of particles in the x i direction. Let's now imagine that each of the particles have the same mass m . Then in the rest frame the energy density of the dust is given by
By definition, the energy density completely specifies the dust. But only measures the energy density in the rest frame what about other frames? We notice that both n and m are 0-components of four-vectors in their rest frame specifically, N = ( n , 0, 0, 0) and p = ( m , 0, 0, 0). Therefore is the = 0, = 0 component of the tensor p N as measured in its rest frame. We are therefore led to define the energy-momentum tensor for dust:
where is defined as the energy density in the rest frame.
Having mastered dust, more general perfect fluids are not much more complicated. Remember that "perfect" can be taken to mean "isotropic in its rest frame." This in turn means that T is diagonal - there is no net flux of any component of momentum in an orthogonal direction. Furthermore, the nonzero spacelike components must all be equal, T 11 = T 22 = T 33 . The only two independent numbers are therefore T 00 and one of the T ii we can choose to call the first of these the energy density , and the second the pressure p . (Sorry that it's the same letter as the momentum.) The energy-momentum tensor of a perfect fluid therefore takes the following form in its rest frame:
We would like, of course, a formula which is good in any frame. For dust we had T = U U , so we might begin by guessing ( + p ) U U , which gives
Fortunately, this has an obvious covariant generalization, namely p . Thus, the general form of the energy-momentum tensor for a perfect fluid is
This is an important formula for applications such as stellar structure and cosmology.
As further examples, let's consider the energy-momentum tensors of electromagnetism and scalar field theory. Without any explanation at all, these are given by
You can check for yourself that, for example, T 00 in each case is equal to what you would expect the energy density to be.
Besides being symmetric, T has the even more important property of being conserved . In this context, conservation is expressed as the vanishing of the "divergence":
This is a set of four equations, one for each value of . The = 0 equation corresponds to conservation of energy, while T k = 0 expresses conservation of the k th component of the momentum. We are not going to prove this in general the proof follows for any individual source of matter from the equations of motion obeyed by that kind of matter. In fact, one way to define T would be "a (2, 0) tensor with units of energy per volume, which is conserved." You can prove conservation of the energy-momentum tensor for electromagnetism, for example, by taking the divergence of (1.111) and using Maxwell's equations as previously discussed.
A final aside: we have already mentioned that in general relativity gravitation does not count as a "force." As a related point, the gravitational field also does not have an energy-momentum tensor. In fact it is very hard to come up with a sensible local expression for the energy of a gravitational field a number of suggestions have been made, but they all have their drawbacks. Although there is no "correct" answer, it is an important issue from the point of view of asking seemingly reasonable questions such as "What is the energy emitted per second from a binary pulsar as the result of gravitational radiation?" *****
## Electrical and Electronic Engineering
An introduction to electrical, computer and electronic systems and technology. Digital circuits and analysis techniques, computer organisation. Analog circuits and analysis techniques. Inductive power transfer, power systems and electric machines. Communication systems.
Restriction: ELECTENG 202, 204, 208, 210
### Stage II
Aims to provide a good understanding of the way electrical circuits work. It covers DC and AC circuit theorems and analysis transient analysis, including the Laplace transform transfer functions AC power calculations and time and frequency representation of signals.
Prerequisite: ELECTENG 101
Electrostatics (Coulomb’s and Gauss’s Laws, scalar potential, energy, capacitance, dielectrics), magnetostatics (Biot-Savart and Ampere’s Laws, moving conductors, magnetic forces/torques, ferromagnetic hysteresis, inductance, magnetic materials), electromagnetic induction (Faraday’s and Lenz’s Laws). Transmission lines subjected to pulse excitation, magnetic circuits and single-phase transformers. Introduction to computational electromagnetics.
Prerequisite: ELECTENG 101
Electric Circuit Analysis
Aims to provide a good understanding of the way electrical circuits work. The course covers DC and AC circuit theorems and analysis. It also introduces some semiconductor devices (diodes, transistors and operational amplifiers) and gives examples of their applications.
Prerequisite: ELECTENG 101
Restriction: ELECTENG 202 or 291
Analogue and Digital Design
Project-based course introducing the process of electrical engineering design. Students will research a diverse range of practical problems and develop solutions and prototypes, test and evaluate hardware and software solutions, and communicate the design and results.
Prerequisite: COMPSYS 201, and ELECTENG 202 or 291
Semiconductor devices and applications, diodes, bipolar junction transistors and operational amplifiers. Elementary device physics. Linear and non-linear devices, terminal characteristics, small-signal modelling and analysis. Frequency-dependent behaviour of circuits and analysis methods. Linear and non-linear circuits such as amplifiers and switching circuits. Biasing, coupling and bypass techniques. Operational amplifiers, frequency-dependence and characteristic limitations, frequency selective and non-linear switching circuits.
Prerequisite: ELECTENG 101
Fundamentals of Electrical Engineering
AC and DC circuit analysis in the context of linear electrical and electronic systems. Time and frequency domain approaches to describing and analysing electrical networks and systems.
Prerequisite: ELECTENG 101
Electronic devices and circuits for solving engineering problems. Analysis of linear and non-linear microelectronic circuits and their practical applications.
Prerequisite: ELECTENG 202 or 291
### Stage III
Introduction to linear, time-invariant, continuous-time system theory from both a time-domain and frequency domain standpoint. This leads on to the fundamental body of knowledge underlying the control and enhancement of system behaviour, with application to the analysis and control of electrical systems.
Prerequisite: ELECTENG 202
An advanced treatment of electronic circuits including a rigorous treatment of feedback, device limitations, noise effects, stability, and design considerations. Emphasis on common practical circuits taken from analog and switching applications.
Prerequisite: ELECTENG 202 or 291, and 210 or 292
Transmission lines subjected to AC excitation, the Smith chart, introduction to matching network design and introduction to antennas for radio systems. Maxwell’s equations in differential and integral form, divergence and Stokes’ theorems, skin effect and uniform plane waves (lossless/lossy media, reflection and transmission, polarisation). Case studies in computational electromagnetics.
Prerequisite: ELECTENG 204
Power Apparatus and Systems
Introduces students to three-phase electric machines and power system components. Covers theory, modelling and practical aspects for synchronous machines, induction machines, transformer connections, transmission lines and substation components.
Prerequisite: ELECTENG 204
Electrical Engineering Design 1
An appreciation of the design process as applied to various electrical and electronic engineering systems. Design skills are enhanced through a variety of engineering projects which typically introduce students to modelling, simulation and analogue and digital electronic hardware design.
Prerequisite: COMPSYS 201, and COMPSYS 209 or ELECTENG 209, and ELECTENG 202 or 291, and COMPSYS 202 or SOFTENG 281
Electrical Engineering Design 2
The formal introduction to the design process is completed by one or more open-ended projects which typically include elements of design from concept to working prototype.
Prerequisite: ELECTENG 310
Introduction to continuous-time and discrete-time signals and systems. Spectral analysis and representation of analog and digital signals, and linear, time-invariant systems. Conversion between analog and digital signals. Systems for manipulating and filtering signals in hardware and software.
Prerequisite: ELECTENG 202 or 291
Introduction to modelling in the time-domain and frequency domain. The fundamental body of knowledge underlying the control and enhancement of system behaviour, with application to the analysis and control of systems.
Prerequisite: ELECTENG 202 or 291
Research Project - Level 9
Students are required to submit a report on project work carried out on a topic assigned by the Head of Department. The work shall be supervised by a member of staff.
Prerequisite: ELECTENG 310, 311, and 30 points from ELECTENG 303, 305, 309, 331, 332
To complete this course students must enrol in ELECTENG 700 A and B
Mobile Wireless Engineering
Aspects of the design and planning of mobile radio systems. Radio propagation for mobile radio systems (multipath, narrowband and wideband channels, channel characterisation and measurements), propagation modelling (free-space, plane-earth, diffraction). Frequency reuse and interference, outage probabilities, system performance evaluation, space diversity, MIMO and millimetre-wave systems.
Prerequisite: ELECTENG 307 or 721 or 737
Advanced Power Systems - Level 9
Electricity markets: structure, pricing, optimisation, ancillary services Power system protection practices Distribution network development: Smart Grid, Demand Side participation HVDC and FACT Devices Theory and Application Renewable energy grid integration.
Prerequisite: ELECTENG 731
Advanced Control Systems - Level 9
Advanced theory of modern control systems with emphasis on optimisation techniques for both deterministic and stochastic processes. State-space modelling of dynamic systems and choice of suitable performance criteria. Adaptive, nonlinear and sliding mode control systems. Core concepts are extended by an individual research project in which a challenging control problem is analysed and solved.
Prerequisite: ELECTENG 722
Topics in Digital Signal Processing - Level 9
An advanced treatment of digital signal processing topics with an emphasis on state of the art techniques. Case studies of digital signal processing methods used to solve practical problems in science and engineering.
Prerequisite: ELECTENG 733
Studies in Electrical and Electronic Engineering 1
Advanced course on topics to be determined each year by the Head of Department.
Studies in Electrical and Electronic Engineering 2
Advanced course on topics to be determined each year by the Head of Department.
Studies in Electrical and Electronic Engineering 3
Advanced course on topics to be determined each year by the Head of Department.
Studies in Electrical and Electronic Engineering 4
Advanced course on topics to be determined each year by the Head of Department.
Studies in Electrical and Electronic Engineering 5
Advanced course on topics to be determined each year by the Head of Department.
Studies in Electrical and Electronic Engineering 6
Advanced course on topics to be determined each year by the Head of Department.
Matching networks, waveguides, transmitter/receiver design, noise, non-linear behaviour, antennas, applications in computational electromagnetics. Fundamentals of radio propagation, tropospheric effects, diffraction, link budgets, point-to-point link design, multipath propagation, introduction to area coverage (mobile radio) systems. Introduction to radar systems, the radio spectrum and exposure standards.
Prerequisite: ELECTENG 307
Restriction: ELECTENG 421, 737
State space analysis, relationship to transfer function methods, controllability and observability, multivariable plant. Computer simulation. Stability considerations. State variable feedback. Digital control system, design and realisation of digital controllers, adaptive controllers. Nonlinear systems, phase-plane and describing function techniques, Lyapunov's method of stability analysis, design of controllers for non-linear systems. Variable structure systems.
Prerequisite: ELECTENG 303 or 331 or 332
Restriction: ELECTENG 422, MECHENG 720, 724
An advanced course on topics to be determined each year by the Head of Department.
Prerequisite: Departmental approval
Advanced principles and techniques in digital transmission systems: base-band and pass-band digital systems. Geometric representation of signals: theory of orthonormal signals, correlation demodulators, optimal detector. Digital phase (PSK) and frequency (FSK) modulation. Digital communication systems with noise. Information theory, capacity theorem and applications. Signal and information coding: data compression, digital transmission, error detection and correction, block and convolutional codes. Noise, thermal noise, noise figure. Traffic theory. Digital networks and OSI model.
Prerequisite: ELECTENG 303 or 331, and 732
Restriction: ELECTENG 426, 741
Builds on the knowledge of three-phase power systems components to understand modelling, formulation and typical analysis carried out by electricity transmission, distribution and generation entities. Load flow, fault, stability and power quality. Supplemented by laboratories where students learn to use professional software to implement the theoretical aspects.
Prerequisite: ELECTENG 309
Analog AM and FM modulation. Noise in AM and FM systems. AM modulators and demodulators. Coherent and non-coherent receivers. Superheterodyne receivers. Multiplexing: FDM, TDM, CDMA. Pulse modulation. Nyquist theorem PCM modulation and multiplexing. Baseband digital transmission optimal filtering matched filter detection probability of error. Intersymbol interference, waveform coding and data compression, base-band data transmission. Introduction to digital systems and modulations.
Prerequisite: ELECTENG 303 or 331
Digital Signal Processing
Analysis and manipulation of discrete-time signals and systems. Spectral representations and analysis using the z-transform, discrete Fourier transform and fast Fourier transform. Introduction to stochastic processes. Hardware systems for processing digital signals.
Prerequisite: ELECTENG 303 or 331 or ENGSCI 311 or 313
Power Electronics - Level 9
Selected advanced concepts in power electronics are introduced through a practical and research based individual design project, utilising modern power converter topologies with supporting lectures that include: inductive power transfer and control, DC-DC converter design and control, high frequency magnetics design, semiconductor switches, practical design issues, controlled rectifiers and PWM converters with application to conventional and brushless DC motors.
Prerequisite: ELECTENG 305, 310, 311
Green Energy Technologies
Advanced green energy technologies with examples from current industry practice and cutting edge research developments. Topics include: renewable energy systems, distributed power generation, energy storage techniques, transportation electrification, power converters for renewable energy integration, soft-switched resonant converters, wireless power transfer, new semiconductor devices, motor drives, and LED lighting.
Prerequisite: ELECTENG 734
Analog and Digital Filter Synthesis
Filter concepts and network functions, a review of approximation techniques and frequency transformations, leading to a thorough treatment of passive, active and digital filter implementations.
Prerequisite: ELECTENG 303 or 331
Prerequisite: ELECTENG 307
Restriction: ELECTENG 421, 721
Selected Topics in Advanced Power Systems - Level 9
Electricity markets: structure, pricing, optimisation, ancillary services Power system protection practices Distribution Network Development: Smart Grids, Demand Side Participation, Integration of DG/renewable sources and Electric Vehicles. Core concepts are extended by an individual research project, a self-guided protection laboratory and industry engagement in advanced power system practices.
Prerequisite: ELECTENG 731
An advanced course on a topic to be determined each year by the Head of Department. Includes a substantial individual research project.
Prerequisite: Departmental approval
An advanced course on a topic to be determined each year by the Head of Department. Includes a substantial individual research project.
Prerequisite: Departmental approval
Advanced Digital Communications - Level 9
Advanced topics in modern digital communication systems and networks including: Advanced digital modulation theory and practice in single and multi-user communications systems advanced information theory including single and multiple source coding modern error control coding methods and applications traffic theory and application in communication systems and networks. Theoretical knowledge is extended by an advanced laboratory programme and research projects.
Prerequisite: ELECTENG 303 or 331, and ELECTENG 732
Restriction: ELECTENG 426, 726
Final year team exercise with students in multi-disciplinary roles, with focus on electrical and electronic engineering, integrating technical learning into realistic design outcomes. Comprehensive investigation of an open-ended, complex, real or synthetic computer, electrical and software engineering problem with simulated professional design office constraints. Includes technical, economic and environmental impact components to complete a scheme assessment report.
Prerequisite: 75 points from Part III courses listed in the BE(Hons) Schedule for the Electrical and Electronic Engineering specialisation
Students are required to submit a report on a topic assigned by the appropriate Head of Department.
## 13.1 Chromosomal Theory and Genetic Linkage
By the end of this section, you will be able to do the following:
• Discuss Sutton’s Chromosomal Theory of Inheritance
• Explain the process of homologous recombination, or crossing over
• Describe chromosome creation
• Calculate the distances between three genes on a chromosome using a three-point test cross
Long before scientists visualized chromosomes under a microscope, the father of modern genetics, Gregor Mendel, began studying heredity in 1843. With improved microscopic techniques during the late 1800s, cell biologists could stain and visualize subcellular structures with dyes and observe their actions during cell division and meiosis. With each mitotic division, chromosomes replicated, condensed from an amorphous (no constant shape) nuclear mass into distinct X-shaped bodies (pairs of identical sister chromatids), and migrated to separate cellular poles.
### Chromosomal Theory of Inheritance
The speculation that chromosomes might be the key to understanding heredity led several scientists to examine Mendel’s publications and reevaluate his model in terms of chromosome behavior during mitosis and meiosis. In 1902, Theodor Boveri observed that proper sea urchin embryonic development does not occur unless chromosomes are present. That same year, Walter Sutton observed chromosome separation into daughter cells during meiosis (Figure 13.2). Together, these observations led to the Chromosomal Theory of Inheritance , which identified chromosomes as the genetic material responsible for Mendelian inheritance.
The Chromosomal Theory of Inheritance was consistent with Mendel’s laws, which the following observations supported:
• During meiosis, homologous chromosome pairs migrate as discrete structures that are independent of other chromosome pairs.
• Chromosome sorting from each homologous pair into pre-gametes appears to be random.
• Each parent synthesizes gametes that contain only half their chromosomal complement.
• Even though male and female gametes (sperm and egg) differ in size and morphology, they have the same number of chromosomes, suggesting equal genetic contributions from each parent.
• The gametic chromosomes combine during fertilization to produce offspring with the same chromosome number as their parents.
Despite compelling correlations between chromosome behavior during meiosis and Mendel’s abstract laws, scientists proposed the Chromosomal Theory of Inheritance long before there was any direct evidence that chromosomes carried traits. Critics pointed out that individuals had far more independently segregating traits than they had chromosomes. It was only after several years of carrying out crosses with the fruit fly, Drosophila melanogaster, that Thomas Hunt Morgan provided experimental evidence to support the Chromosomal Theory of Inheritance.
Mendel’s work suggested that traits are inherited independently of each other. Morgan identified a 1:1 correspondence between a segregating trait and the X chromosome, suggesting that random chromosome segregation was the physical basis of Mendel’s model. This also demonstrated that linked genes disrupt Mendel’s predicted outcomes. That each chromosome can carry many linked genes explains how individuals can have many more traits than they have chromosomes. However, researchers in Morgan’s laboratory suggested that alleles positioned on the same chromosome were not always inherited together. During meiosis, linked genes somehow became unlinked.
#### Homologous Recombination
In 1909, Frans Janssen observed chiasmata—the point at which chromatids are in contact with each other and may exchange segments—prior to the first meiosis division. He suggested that alleles become unlinked and chromosomes physically exchange segments. As chromosomes condensed and paired with their homologs, they appeared to interact at distinct points. Janssen suggested that these points corresponded to regions in which chromosome segments exchanged. We now know that the pairing and interaction between homologous chromosomes, or synapsis, does more than simply organize the homologs for migration to separate daughter cells. When synapsed, homologous chromosomes undergo reciprocal physical exchanges at their arms in homologous recombination , or more simply, “crossing over.”
To better understand the type of experimental results that researchers were obtaining at this time, consider a heterozygous individual that inherited dominant maternal alleles for two genes on the same chromosome (such as AB) and two recessive paternal alleles for those same genes (such as ab). If the genes are linked, one would expect this individual to produce gametes that are either AB or ab with a 1:1 ratio. If the genes are unlinked, the individual should produce AB, Ab, aB, and ab gametes with equal frequencies, according to the Mendelian concept of independent assortment. Because they correspond to new allele combinations, the genotypes Ab and aB are nonparental types that result from homologous recombination during meiosis. Parental types are progeny that exhibit the same allelic combination as their parents. Morgan and his colleagues, however, found that when they test crossed such heterozygous individuals to a homozygous recessive parent (AaBb × aabb), both parental and nonparental cases occurred. For example, 950 offspring might be recovered that were either AaBb or aabb, but 50 offspring would also result that were either Aabb or aaBb. These results suggested that linkage occurred most often, but a significant minority of offspring were the products of recombination.
### Visual Connection
In a test cross for two characteristics such as the one here, can the recombinant offspring's predicted frequency be 60 percent? Why or why not?
#### Genetic Maps
Janssen did not have the technology to demonstrate crossing over so it remained an abstract idea that scientists did not widely believe. Scientists thought chiasmata were a variation on synapsis and could not understand how chromosomes could break and rejoin. Yet, the data were clear that linkage did not always occur. Ultimately, it took a young undergraduate student and an “all-nighter” to mathematically elucidate the linkage and recombination problem.
In 1913, Alfred Sturtevant, a student in Morgan’s laboratory, gathered results from researchers in the laboratory, and took them home one night to mull them over. By the next morning, he had created the first “chromosome map,” a linear representation of gene order and relative distance on a chromosome (Figure 13.4).
### Visual Connection
Which of the following statements is true?
1. Recombination of the body color and red/cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length.
2. Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/brown eye alleles and the aristae length alleles.
3. Recombination of the gray/black body color and long/short aristae alleles will not occur.
4. Recombination of the red/brown eye and long/short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color.
As Figure 13.4 shows, by using recombination frequency to predict genetic distance, we can infer the relative gene order on chromosome 2. The values represent map distances in centimorgans (cM), which correspond to recombination frequencies (in percent). Therefore, the genes for body color and wing size were 65.5 − 48.5 = 17 cM apart, indicating that the maternal and paternal alleles for these genes recombine in 17 percent of offspring, on average.
To construct a chromosome map, Sturtevant assumed that genes were ordered serially on threadlike chromosomes. He also assumed that the incidence of recombination between two homologous chromosomes could occur with equal likelihood anywhere along the chromosome's length. Operating under these assumptions, Sturtevant postulated that alleles that were far apart on a chromosome were more likely to dissociate during meiosis simply because there was a larger region over which recombination could occur. Conversely, alleles that were close to each other on the chromosome were likely to be inherited together. The average number of crossovers between two alleles—that is, their recombination frequency —correlated with their genetic distance from each other, relative to the locations of other genes on that chromosome. Considering the example cross between AaBb and aabb above, we could calculate the recombination's frequency as 50/1000 = 0.05. That is, the likelihood of a crossover between genes A/a and B/b was 0.05, or 5 percent. Such a result would indicate that the genes were definitively linked, but that they were far enough apart for crossovers to occasionally occur. Sturtevant divided his genetic map into map units, or centimorgans (cM) , in which a 0,01 recombination frequency corresponds to 1 cM.
By representing alleles in a linear map, Sturtevant suggested that genes can range from linking perfectly (recombination frequency = 0) to unlinking perfectly (recombination frequency = 0.5) when genes are on different chromosomes or genes separate very far apart on the same chromosome. Perfectly unlinked genes correspond to the frequencies Mendel predicted to assort independently in a dihybrid cross. A 0.5 recombination frequency indicates that 50 percent of offspring are recombinants and the other 50 percent are parental types. That is, every type of allele combination is represented with equal frequency. This representation allowed Sturtevant to additively calculate distances between several genes on the same chromosome. However, as the genetic distances approached 0.50, his predictions became less accurate because it was not clear whether the genes were very far apart on the same or on different chromosomes.
In 1931, Barbara McClintock and Harriet Creighton demonstrated the crossover of homologous chromosomes in corn plants. Weeks later, Curt Stern demonstrated microscopically homologous recombination in Drosophila. Stern observed several X-linked phenotypes that were associated with a structurally unusual and dissimilar X chromosome pair in which one X was missing a small terminal segment, and the other X was fused to a piece of the Y chromosome. By crossing flies, observing their offspring, and then visualizing the offspring’s chromosomes, Stern demonstrated that every time the offspring allele combination deviated from either of the parental combinations, there was a corresponding exchange of an X chromosome segment. Using mutant flies with structurally distinct X chromosomes was the key to observing the products of recombination because DNA sequencing and other molecular tools were not yet available. We now know that homologous chromosomes regularly exchange segments in meiosis by reciprocally breaking and rejoining their DNA at precise locations.
Review Sturtevant’s process to create a genetic map on the basis of recombination frequencies here.
#### Mendel’s Mapped Traits
Homologous recombination is a common genetic process, yet Mendel never observed it. Had he investigated both linked and unlinked genes, it would have been much more difficult for him to create a unified model of his data on the basis of probabilistic calculations. Researchers who have since mapped the seven traits that Mendel investigated onto a pea plant genome's seven chromosomes have confirmed that all the genes he examined are either on separate chromosomes or are sufficiently far apart as to be statistically unlinked. Some have suggested that Mendel was enormously lucky to select only unlinked genes whereas, others question whether Mendel discarded any data suggesting linkage. In any case, Mendel consistently observed independent assortment because he examined genes that were effectively unlinked.
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• Authors: Mary Ann Clark, Matthew Douglas, Jung Choi
• Publisher/website: OpenStax
• Book title: Biology 2e
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• Book URL: https://openstax.org/books/biology-2e/pages/1-introduction | 20,972 | 100,280 | {"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} | 4.15625 | 4 | CC-MAIN-2021-49 | latest | en | 0.862611 |
https://documen.tv/question/the-width-of-a-rectangle-is-4cm-less-than-the-length-if-each-dimension-were-increased-by-3cm-the-22322740-94/ | 1,721,292,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00150.warc.gz | 181,741,792 | 16,163 | ## The width of a rectangle is 4cm less than the length. If each dimension were increased by 3cm, the width of the new rectangle formed would b
Question
The width of a rectangle is 4cm less than the length. If each dimension were increased by 3cm, the width of the new rectangle formed would be 2/3 the length of the new rectangle. Find the dimensions of the original rectangle.
in progress 0
3 years 2021-08-04T10:10:32+00:00 1 Answers 18 views 0
The original length is 9 cm and the original width 5 cm
Step-by-step explanation:
The area of a rectangle A is given as the product of its’ length L and width W.
Given that the width of a rectangle is 4cm less than the length then
W = L – 4
If each dimension were increased by 3cm, then the new
Length
= L + 3
Width
= W + 3
the width of the new rectangle formed would be 2/3 the length of the new rectangle
W + 3 = 2/3(L + 3)
3W + 9 = 2L + 6
3W = 2L +6 – 9
3W = 2L – 3
Recall that W = L – 4
3(L – 4) = 2L – 3
3L – 12 = 2L – 3
3L – 2L = 12 – 3
L = 9 cm
W = 9 – 4
= 5 cm | 356 | 1,042 | {"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-2024-30 | latest | en | 0.931322 |
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# Downloading Excel for Error Log does not contain sorted data
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Senior Manager
Joined: 24 Oct 2016
Posts: 266
GMAT 1: 670 Q46 V36
### Show Tags
27 Sep 2018, 19:36
1
Hey bb
Bug
When I try to download my error log, it does not download it in the sorted format. Spreadsheet does get downloaded in the sorted format only when I had more than 2 pages of error log and I do the following steps:
1) Sort by desired column.
2) Click on only other page except 1st (i.e. Page 2)
3) Come back to Page 1.
4) Download the excel, which has sorted data.
Attachment:
Error Log (Sorted).png [ 337.5 KiB | Viewed 154 times ]
Attachment:
Downloaded Excel (Unsorted).png [ 518.49 KiB | Viewed 153 times ]
Another Suggestion:
Right now when we download the spreadsheet and want to calculate the average of timers, then we need to manually click on each cell so that they get right aligned, a format that's needed for Excel in order to use the AVERAGE function. Currently, timer column looks like the one as shown in Downloaded Excel (Unsorted).png above. If it can be in the format shown below, then would be pretty useful when one has too many rows.
Attachment:
Feature Request (Timers in correct format - Right aligned).png [ 184.58 KiB | Viewed 152 times ]
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### Show Tags
27 Sep 2018, 21:49
1
Thx. Really appreciate reporting this. This is pretty helpful. Unfortunately I’m headed to bed right now and I will reply tomorrow. Appreciate your patience with this and me
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Re: Downloading Excel for Error Log does not contain sorted data &nbs [#permalink] 27 Sep 2018, 21:49
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# Downloading Excel for Error Log does not contain sorted data
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Moderator: abhimahna
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 966 | 3,831 | {"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} | 2.84375 | 3 | CC-MAIN-2019-04 | longest | en | 0.866164 |
https://ethereum.stackexchange.com/questions/6415/eli5-how-does-a-merkle-patricia-trie-tree-work | 1,657,024,197,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104576719.83/warc/CC-MAIN-20220705113756-20220705143756-00178.warc.gz | 284,680,286 | 67,730 | # ELI5 How does a Merkle-Patricia-trie tree work?
I understand that Merkle tree are Hashes of Hashes, they have the advantage that you can verify only a subtree. But what about Patricia? What does a trie mean? And how is it used in Ethereum?
Trie (also called digital tree, prefix trie or radix trie)
An ordered tree data structure that is used to store a dynamic set or associative array where the keys are usually strings. A node's position in the tree defines the key with which it is associated. https://en.wikipedia.org/wiki/Trie
A trie for keys "A","to", "tea", "ted", "ten", "i", "in", and "inn".
Patricia - Practical Algorithm To Retrieve Information Coded In Alphanumeric (source)(orginial paper by Donald R. Morrison). A Patricia trie is a binary radix trie - binary choice at each node when traversing the trie; this is modified in Ethereum.
In ethereum, hexadecimal is used - X characters from an 16 character "alphabet". Hence nodes in the trie have 16 child nodes (the 16 character hex "alphabet") and a maximum depth of X. Note a hex character is referred to as a "nibble".
Merkle Patricia Trie
As described here, the term Merkle implies that
the root node becomes a cryptographic fingerprint of the entire data structure
Ethereum Modified Merkle Patricia Trie
The yellow paper describes a modified merkle patricia trie. This defines three different node types; extension, branch and leaf. These are descibed, using a simplified world state, in the diagram below:
• Shouldn't a key under `world-state` contain 8 nibble (32 bytes)? In your example(latest image) it contains 7 nibbles, which is 28 bytes. @atomh33ls Dec 15, 2017 at 6:57
• @Alper thanks... I shortened it to make it more concise. I think 8 nibbles is just 4 bytes... and 7 is 3.5 bytes. I'd need 64 hex chars for 32 bytes... Mar 12, 2018 at 12:56
• What do the leaf node and extension nodes of the trie do? Apr 6, 2018 at 16:46
• Thanks! How does a stale tree synchronize to an up-to-date tree? (e.g., when a client gets online after one day). Does it just download the entire up-to-date tree, or some kind of "diff" is received for the synchronization? May 26, 2018 at 22:17
• @HelinWang No problem, I think this depends on the implementation so would be better posed as a new question... May 26, 2018 at 22:24
'Trie' comes from the word retrieval, since it only uses the prefix of a word to find it in a dictionary. It is an ordered tree where the keys are usually strings ending with a terminal symbol, and each vertex represents a prefix. The root of a trie is usually an empty string, as we can see in the diagram taken from wikipedia.
For more information about the difference between a trie and a radix (Patricia) tree. | 682 | 2,719 | {"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} | 3.03125 | 3 | CC-MAIN-2022-27 | longest | en | 0.879405 |
https://www.litessayhelp.com/how-probability-distributions-affect-decisions/ | 1,726,712,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00403.warc.gz | 777,307,098 | 14,637 | ##### How Probability Distributions Affect Decisions
Imagine if you were offered a job in a different state and a major consideration for you is rent prices (assuming you planned to rent instead of buy). Your main concerns are the affordability in relation to your income and the location/condition of the property. Perhaps you would look for the cheapest rent possible within a quiet, residential community. Or, you might be willing to spend at little more than average to live in the heart of downtown. As you research the city, you learn that the mean for rents of your preferred home size are \$1,300 a month. Many people might base their decision on this number alone, but you—equipped with the knowledge of standard deviation—know there is more to that number.
If the most you could afford is \$1,100 a month in rent, then a standard deviation of \$250 might be good news because the amount you can afford is still within 1 deviation of the mean. With a standard deviation of \$75, however, you might be unwilling to make the sacrifices necessary to rent a place that you could afford. Additionally, if you were willing to spend a little more than average to live in a nice place or area, then you could easily find an amazing place with a standard deviation of \$100 but might not be able to afford the upgrade with a standard deviation of \$300.
In this Discussion, you will use the data that you gathered in the Week 1 Discussion to calculate a standard deviation and explain how this concept can affect decision making.
Locate the data that you gathered for the Week 1 Discussion.
Calculate the sample standard deviation from your cigarette price data in Week 1. Use that (and your sample average and sample size) to calculate the following (assuming a normal distribution):
Within what range would you find 90% of cigarette prices in your area?
What are the chances that someone in your area would pay 4 dollars or less per pack?
What are the chances that someone in your area would pay 10 dollars or less per pack? | 417 | 2,030 | {"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} | 3.65625 | 4 | CC-MAIN-2024-38 | latest | en | 0.95931 |
https://www.airmilescalculator.com/distance/mlx-to-dnz/ | 1,716,605,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00030.warc.gz | 545,187,599 | 30,178 | # How far is Denizli from Malatya?
The distance between Malatya (Malatya Erhaç Airport) and Denizli (Denizli Çardak Airport) is 459 miles / 739 kilometers / 399 nautical miles.
The driving distance from Malatya (MLX) to Denizli (DNZ) is 583 miles / 938 kilometers, and travel time by car is about 12 hours 11 minutes.
459
Miles
739
Kilometers
399
Nautical miles
1 h 22 min
## Distance from Malatya to Denizli
There are several ways to calculate the distance from Malatya to Denizli. Here are two standard methods:
Vincenty's formula (applied above)
• 459.222 miles
• 739.046 kilometers
• 399.053 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 458.142 miles
• 737.308 kilometers
• 398.114 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Malatya to Denizli?
The estimated flight time from Malatya Erhaç Airport to Denizli Çardak Airport is 1 hour and 22 minutes.
## Flight carbon footprint between Malatya Erhaç Airport (MLX) and Denizli Çardak Airport (DNZ)
On average, flying from Malatya to Denizli generates about 93 kg of CO2 per passenger, and 93 kilograms equals 204 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Malatya to Denizli
See the map of the shortest flight path between Malatya Erhaç Airport (MLX) and Denizli Çardak Airport (DNZ).
## Airport information
Origin Malatya Erhaç Airport
City: Malatya
Country: Turkey
IATA Code: MLX
ICAO Code: LTAT
Coordinates: 38°26′7″N, 38°5′27″E
Destination Denizli Çardak Airport
City: Denizli
Country: Turkey
IATA Code: DNZ
ICAO Code: LTAY
Coordinates: 37°47′8″N, 29°42′4″E | 538 | 1,949 | {"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} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.835758 |
http://focs.olin.edu/focswiki/2010Assignment3Faq.html | 1,506,242,105,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689900.91/warc/CC-MAIN-20170924081752-20170924101752-00295.warc.gz | 132,929,648 | 2,832 | Contents
## Sorts
Q. Should I show the heapsort as an array or as a tree?
A. Either representation is fine.
Q. In class we put people’s names into a tree using alphabetical order… was this a heap? Because wikipedia says that for a heap both children have to be less than the parent, but for what we did in class the left child is less than the parent and the right child is greater than the parent
A. We did indeed create an ordered binary tree in class. This was not a heap, although it used the tree-stored-in-an-array trick. Per the problem set:
• NB: A heap is like a partially ordered binary tree: the parent is larger than both children (this is called the "heap property"), and the children need not be ordered with respect to one another. A heap can be stored in an array using the children-at-2n+1,2n+2 trick. When a new element is inserted into the next free position, you need to re-heapify by comparing it with its parent to maintain the heap property (and so on all the way up to the root). For more details, look up heapify or heapification in Wikipedia, Cormen, or another resource (search for heapsort)...
Q. Are we supposed to sort after we have heapified the elements of the array?
A. Once the elements are heapified, you should repeat:
• remove the first element
• re-heapily
until you are done. Or, swap the first and last-unsorted elements and reheapify (which is effectively the same thing....)
## PDAs
Q. For the PDA, should we be using the ambiguous grammar, the unambiguous grammar, or just creating a PDA for the language without reference to any grammar?
A. Any PDA for the language is fine; no need to use a grammar to create one.
Q. For drawing PDAs in Assignment 3, may we used your shorthand notation (where each CFG results in a 3-state PDA), or do we have to draw out all the intermediate states?
A. You can use the (multi-push) shorthand.
## Prolog Grammar
Q. ...Prolog grammar, relationship to regular Prolog, extra arguments ...
A. You don't need to understand extra arguments to Prolog; they're there if you want to learn about them but not required in this course. You also don't need to understand about how Prolog grammar notation relates to Prolog at this point, as we haven't discussed Prolog (other than grammars) yet.
Q. ...various questions about the fidelity of the extensions to the Prolog grammar...
A. You don't need to handle everything, just a few representative cases. It is sufficient, e.g., to handle the examples shown, or choose a different adjective or prepositional phrase. For the extensions, these are for you to have fun with/explore.
Q. ...languages other than English...
A. You do not need to work on a grammar for a language other than English. However, it makes no sense to explore gender matching in English, so if you want to look at gender matching, you should switch to a language that carries gender markings.
2013-07-17 10:42 | 680 | 2,919 | {"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} | 2.6875 | 3 | CC-MAIN-2017-39 | latest | en | 0.925979 |
https://www.physicsforums.com/threads/a-rolling-cylinder-problem.896432/ | 1,508,269,877,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822480.15/warc/CC-MAIN-20171017181947-20171017201947-00087.warc.gz | 944,936,483 | 20,215 | # A rolling cylinder problem
Tags:
1. Dec 8, 2016
### doktorwho
1. The problem statement, all variables and given/known data
With the equations of motion in polar coordinates given by $r(t)=pcos{kt^2}, φ(t)=kt^2$ determine the velocity intensity of a point $M$ on the circumference of a cylinder which is rolling without friction on a horizontal plane at time $t$ is the velocity of the center of cylinder at time $t$ is $v_c$
2. Relevant equations
3. The attempt at a solution
I first drew a picture of a cylinder and noted that the angle $φ$ is the angle from the horizontal to the point $M$ and i need to use another angle which is more suited for this rolling motion and that would be the angle $\theta$ which is measured from the radius to the vertical.
From here $\theta=2φ=2kt^2$. Since i know how the angle changes, i can calculate the angular velocity and acceleration. $ω=\dot \theta = 4kt, α=\ddot \theta = 4k$. Since the function of $r$ is dependent only on $cos(kt^2)$ i gues whats in front must be the amplitude therefore the radius should be $R=p/2$? Is this correct thinking?
The velocity and acceleration are $v=wR=2pkt, a=Rα=2pk$, i can find the the $a_n=8pk^2t^2$ but how do i find the velocity at that point. Kinda stuck on that?
2. Dec 8, 2016
### haruspex
Not following that.
Where is the origin for these coordinates? Seems like it must be moving at the same velocity as the centre of the circle, and r varies between 0 and p, so I guess the origin is, at each instant, the point of contact and p is the diameter. That makes the reference frame non-inertial.
It also makes φ the angle M and the circle's centre subtend at the point of contact.
3. Dec 9, 2016
### doktorwho
The problem does not have a diagram but i tried drawing it
4. Dec 9, 2016
### haruspex
Ok, that's what I deduced about your view from what you wrote. But the equations say r=p cos(φ), so where is p in your diagram?
Seems to me φ should be the angle to the vertical and p is the diameter of the circle.
What is the angular velocity at time t? So what is vc as a function of t?
5. Dec 10, 2016
### doktorwho
Well my function of $r$ is at its peak when $cos{kt^2}=1$ so the p is a diameter and $R=p/2$, oh yeah so because of this the angle must be measure from the vertical so that the $p$ part is met? If the equation was $r(t)=psin{kt^2}$ then it would be like this right?
My angluar velocity is then, since the angle $\theta$ is now the twice of $\phi$, $w=4kt$. $v_c$ is constant at all times so it is not dependent on $t$?
6. Dec 10, 2016
### haruspex
I meant to ask what this means:
i wondered if that was a typo, and you meant without slipping (in which case vc is not constant).
It seems odd to describe it as rolling if it is slipping.
7. Dec 10, 2016
### doktorwho
Yeah xD, it just rolls without slipping, wring words sorry. It rolls so the $v_c$ is constant. But still how does this relate to the velocity of some point on the circumference? Visualising it i would think that the points on the circumference must move faster than $V_c$?
8. Dec 10, 2016
### haruspex
No, you have an increasing rotation rate, so if it is not slipping vc must be increasing.
9. Dec 10, 2016
### doktorwho
Lets imagine this. At some point in time the velocity of the center of the cyliner is $v_c$. I then conclude that the velocity of the top of the cylinder (the hightest point) must also have the velocity $+v_c$ and the lowest point ( the one where the contact is ) must have the velocity $-v_c$. Therefore the change $-2v_c$ so there must be a point of zero velocity in between. Regarding that in a angle form of our chosen system i would conclude that the velocity of the point on the circumference is $v_p=v_c*cos{\theta}$. Does that seem correct and if not, what seems wrong?
Last edited: Dec 10, 2016
10. Dec 10, 2016
### haruspex
I do not understand your reasoning. Why would it have the same velocity? What you do know is that the point of contact with the ground is instantaneously stationary.
11. Dec 11, 2016
### doktorwho
Well then is the point of contact is stationary at time $t$ and the centers speed is $v_c$ at time $t$ the only reasonable image of this would be that the highest point has a velocity of $2v_c$. Then all the other points on the circumference would have a velocity of $v_p=2v_ccosθ$ where $cosθR=h-p/2$, $h$ is the height of the point measured from the ground. This case includes the uppmost velocity of the cylinder at point $t$ which is $v_p=2v_ccos{\pi /2}$ but not sure about the velocity when $h=p/2$
Last edited: Dec 11, 2016
12. Dec 11, 2016
### haruspex
You cannot mean that. h=p cos2θ?
13. Dec 11, 2016
### doktorwho
Sorry what i meant was $R*cosθ=h-p/2$ How did you get $h=pcos^2{θ}$?
Last edited: Dec 11, 2016
14. Dec 11, 2016
### haruspex
You did not define θ so I had to guess from your vp formula.
Instantaneously, the cylinder is rotating about what point? What is its rotation rate in terms of vc and p?
15. Dec 11, 2016
### doktorwho
Instantaneously, its rotating around the point of contact. Rotation rate? $\frac{2v_c}{p}$? Im not sure what im calculating now..
16. Dec 11, 2016
### haruspex
Right, and right.
So if some point of the cylinder is at distance x from the point of contact with the ground, what is its speed?
17. Dec 11, 2016
### doktorwho
$\frac{2v_c}{p}*\frac{h}{cosθ}$?
Last edited: Dec 11, 2016
18. Dec 11, 2016
### haruspex
Right (again, guessing at your definition of θ), but you can express x in terms of θ and p, rather than θ and h. | 1,557 | 5,549 | {"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} | 3.9375 | 4 | CC-MAIN-2017-43 | longest | en | 0.908323 |
http://bestwwws.com/standard-error/calculate-standard-error-of-regression-coefficients.php | 1,537,705,769,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159359.58/warc/CC-MAIN-20180923114712-20180923135112-00448.warc.gz | 33,122,063 | 5,139 | Home > Standard Error > Calculate Standard Error Of Regression Coefficients
# Calculate Standard Error Of Regression Coefficients
## Contents
However, more data will not systematically reduce the standard error of the regression. The smaller the standard error, the more precise the estimate. For example, a materials engineer at a furniture manufacturing site wants to assess the strength of the particle board that they use. What is the formula / implementation used? have a peek at these guys
The standard error of the estimate is closely related to this quantity and is defined below: where σest is the standard error of the estimate, Y is an actual score, Y' We focus on the equation for simple linear regression, which is: ŷ = b0 + b1x where b0 is a constant, b1 is the slope (also called the regression coefficient), x Your cache administrator is webmaster. Example data. http://stats.stackexchange.com/questions/85943/how-to-derive-the-standard-error-of-linear-regression-coefficient
## Standard Error Of Regression Coefficient Formula
Browse other questions tagged standard-error inferential-statistics or ask your own question. As the sample size gets larger, the standard error of the regression merely becomes a more accurate estimate of the standard deviation of the noise. Please try the request again.
For example, type L1 and L2 if you entered your data into list L1 and list L2 in Step 1. Here is an Excel file with regression formulas in matrix form that illustrates this process. How to detect whether a user is using USB tethering? Standard Error Of Regression Coefficient Matlab Formulas for the slope and intercept of a simple regression model: Now let's regress.
You can see that in Graph A, the points are closer to the line than they are in Graph B. Standard Error Of Regression Coefficient In R The standard error for the forecast for Y for a given value of X is then computed in exactly the same way as it was for the mean model: Required fields are marked *Comment Name * Email * Website Find an article Search Feel like "cheating" at Statistics? Is there a succinct way of performing that specific line with just basic operators? –ako Dec 1 '12 at 18:57 1 @AkselO There is the well-known closed form expression for
Creating a simple Dock Cell that Fades In when Cursor Hover Over It Why was the Rosetta probe programmed to "auto shutoff" at the moment of hitting the surface? How To Calculate Standard Error Of Regression Slope The only difference is that the denominator is N-2 rather than N. You can also select a location from the following list: Americas Canada (English) United States (English) Europe Belgium (English) Denmark (English) Deutschland (Deutsch) España (Español) Finland (English) France (Français) Ireland (English) So, I take it the last formula doesn't hold in the multivariate case? –ako Dec 1 '12 at 18:18 1 No, the very last formula only works for the specific
## Standard Error Of Regression Coefficient In R
Previously, we showed how to compute the margin of error, based on the critical value and standard error. The range of the confidence interval is defined by the sample statistic + margin of error. Standard Error Of Regression Coefficient Formula The standard error of the slope coefficient is given by: ...which also looks very similar, except for the factor of STDEV.P(X) in the denominator. Standard Error Of Regression Coefficient Definition Rather, the sum of squared errors is divided by n-1 rather than n under the square root sign because this adjusts for the fact that a "degree of freedom for error″
So, if you know the standard deviation of Y, and you know the correlation between Y and X, you can figure out what the standard deviation of the errors would be More about the author How much should I adjust the CR of encounters to compensate for PCs having very little GP? The confidence interval for the slope uses the same general approach. Natural Pi #0 - Rock Creating a simple Dock Cell that Fades In when Cursor Hover Over It Is there a way to ensure that HTTPS works? Standard Error Of Regression Coefficient Excel
Find the margin of error. However... 5. Check out the grade-increasing book that's recommended reading at Oxford University! check my blog You may need to scroll down with the arrow keys to see the result.
Based on your location, we recommend that you select: . How To Calculate Standard Error In Regression Model I too know it is related to the degrees of freedom, but I do not get the math. –Mappi May 27 at 15:46 add a comment| Your Answer draft saved An unbiased estimate of the standard deviation of the true errors is given by the standard error of the regression, denoted by s.
For example, the standard error of the estimated slope is $$\sqrt{\widehat{\textrm{Var}}(\hat{b})} = \sqrt{[\hat{\sigma}^2 (\mathbf{X}^{\prime} \mathbf{X})^{-1}]_{22}} = \sqrt{\frac{n \hat{\sigma}^2}{n\sum x_i^2 - (\sum x_i)^2}}.$$ > num <- n * anova(mod)[[3]][2] > denom <- It might be "StDev", "SE", "Std Dev", or something else. Circular growth direction of hair Time waste of execv() and fork() Find Iteration of Day of Week in Month My girlfriend has mentioned disowning her 14 y/o transgender daughter more hot You can use regression software to fit this model and produce all of the standard table and chart output by merely not selecting any independent variables. | 1,190 | 5,368 | {"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} | 3.34375 | 3 | CC-MAIN-2018-39 | latest | en | 0.852471 |
https://studylib.net/doc/13664597/2.016-hydrodynamics-prof.-alex-techet-fall-2005-hw-%231 | 1,701,510,113,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100381.14/warc/CC-MAIN-20231202073445-20231202103445-00468.warc.gz | 592,825,162 | 13,079 | # 2.016 Hydrodynamics Prof. Alex Techet Fall 2005 HW #1
```2.016 Hydrodynamics
Prof. Alex Techet
Fall 2005
HW #1
Out: 13 Sept 2005
Due: 20 Sept 2005
Problem 1:
Which of the following are isotropic? Why or why not?
a)
Temperature
b) Pressure
c)
Shear stress
d) Dynamic viscosity
Problem 2:
At a particular point in the Pacific Ocean, the density of sea water increases non-linearly with depth according to
! = ! o + mz 2
where
! o is the density at the surface, z is the depth below the surface, and m is a constant. Develop an algebraic
equation for the relationship between pressure and depth.
Problem 3:
a)
Determine the horizontal and vertical forces acting on a wall sloped at angle, θ, to the horizontal seafloor as
a function of z. Assume that the water is h deep and that atmospheric pressure acts everywhere.
b) Determine the total resulting force and center of pressure.
Problem 4: Archimedes Principle on a floating vessel
Extend the results you found in problem 3 to the case of a “V” shaped vessel floating on the surface of the ocean to
show that the resulting pressure acting on the hull balances the weight of the water displaced by the vessel.
Problem 5:
A rectangular barge floats in water, ρ w. When it is empty it is immersed at depth D below the surface. Oil with
density, ρo, is poured into the barge until it is about to sink. Find a relationship for the depth of the oil at this point in
terms of the initial depth, D, the total height of the barge H and the barge width W.
Problem 6:
You are asked to design an underwater laboratory at the bottom of the Gulf of Mexico, in order to study the habits of
Migrating Whale Sharks. After attending a lecture given by Samuel Raymond, founder of Benthos, you realize that
a spherical structure would be ideal for this project. So you have decided that the laboratory will be built as a hemisphere off the sea floor and sketch the following concept design:
g
H
D/2
a)
What is the distribution of pressure over the wetted surface of the laboratory?
b)
What are the total fluid force and moment vectors exerted on the laboratory by the surrounding
fluid?
``` | 524 | 2,122 | {"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} | 3.65625 | 4 | CC-MAIN-2023-50 | latest | en | 0.90768 |
https://www.lmfdb.org/EllipticCurve/Q/5577d4/ | 1,590,540,139,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391923.3/warc/CC-MAIN-20200526222359-20200527012359-00090.warc.gz | 820,186,429 | 28,199 | # Properties
Label 5577d4 Conductor 5577 Discriminant -28216806219459 j-invariant $$\frac{9090072503}{5845851}$$ CM no Rank 1 Torsion Structure $$\Z/{2}\Z$$
# Related objects
Show commands for: Magma / SageMath / Pari/GP
## Minimal Weierstrass equation
magma: E := EllipticCurve([1, 1, 1, 7348, 83936]); // or
magma: E := EllipticCurve("5577d4");
sage: E = EllipticCurve([1, 1, 1, 7348, 83936]) # or
sage: E = EllipticCurve("5577d4")
gp: E = ellinit([1, 1, 1, 7348, 83936]) \\ or
gp: E = ellinit("5577d4")
$$y^2 + x y + y = x^{3} + x^{2} + 7348 x + 83936$$
## Mordell-Weil group structure
$$\Z\times \Z/{2}\Z$$
### Infinite order Mordell-Weil generator and height
magma: Generators(E);
sage: E.gens()
$$P$$ = $$\left(\frac{955}{9}, \frac{37424}{27}\right)$$ $$\hat{h}(P)$$ ≈ 4.73800581845
## Torsion generators
magma: TorsionSubgroup(E);
sage: E.torsion_subgroup().gens()
gp: elltors(E)
$$\left(-\frac{45}{4}, \frac{41}{8}\right)$$
## Integral points
magma: IntegralPoints(E);
sage: E.integral_points()
None
## Invariants
magma: Conductor(E); sage: E.conductor().factor() gp: ellglobalred(E)[1] Conductor: $$5577$$ = $$3 \cdot 11 \cdot 13^{2}$$ magma: Discriminant(E); sage: E.discriminant().factor() gp: E.disc Discriminant: $$-28216806219459$$ = $$-1 \cdot 3^{12} \cdot 11 \cdot 13^{6}$$ magma: jInvariant(E); sage: E.j_invariant().factor() gp: E.j j-invariant: $$\frac{9090072503}{5845851}$$ = $$3^{-12} \cdot 11^{-1} \cdot 2087^{3}$$ Endomorphism ring: $$\Z$$ (no Complex Multiplication) Sato-Tate Group: $\mathrm{SU}(2)$
## BSD invariants
magma: Rank(E); sage: E.rank() Rank: $$1$$ magma: Regulator(E); sage: E.regulator() Regulator: $$4.73800581845$$ magma: RealPeriod(E); sage: E.period_lattice().omega() gp: E.omega[1] Real period: $$0.4145491719$$ magma: TamagawaNumbers(E); sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] Tamagawa product: $$4$$ = $$2\cdot1\cdot2$$ magma: Order(TorsionSubgroup(E)); sage: E.torsion_order() gp: elltors(E)[1] Torsion order: $$2$$ magma: MordellWeilShaInformation(E); sage: E.sha().an_numerical() Analytic order of Ш: $$1$$ (exact)
## Modular invariants
#### Modular form5577.2.a.a
magma: ModularForm(E);
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
$$q - q^{2} - q^{3} - q^{4} + 2q^{5} + q^{6} - 4q^{7} + 3q^{8} + q^{9} - 2q^{10} - q^{11} + q^{12} + 4q^{14} - 2q^{15} - q^{16} - 2q^{17} - q^{18} + O(q^{20})$$
magma: ModularDegree(E); sage: E.modular_degree() Modular degree: 13824 $$\Gamma_0(N)$$-optimal: no Manin constant: 1
#### Special L-value
magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
sage: r = E.rank();
sage: E.lseries().dokchitser().derivative(1,r)/r.factorial()
gp: ar = ellanalyticrank(E);
gp: ar[2]/factorial(ar[1])
$$L'(E,1)$$ ≈ $$1.9641363885$$
## Local data
This elliptic curve is not semistable.
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
sage: E.local_data()
gp: ellglobalred(E)[5]
prime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\Delta$$) ord$$(j)_{-}$$
$$3$$ $$2$$ $$I_{12}$$ Non-split multiplicative 1 1 12 12
$$11$$ $$1$$ $$I_{1}$$ Non-split multiplicative 1 1 1 1
$$13$$ $$2$$ $$I_0^{*}$$ Additive 1 2 6 0
## Galois representations
The image of the 2-adic representation attached to this elliptic curve is the subgroup of $\GL(2,\Z_2)$ with Rouse label X13.
This subgroup is the pull-back of the subgroup of $\GL(2,\Z_2/2^2\Z_2)$ generated by $\left(\begin{array}{rr} 3 & 0 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 3 & 0 \\ 0 & 3 \end{array}\right)$ and has index 6.
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
The mod $$p$$ Galois representation has maximal image $$\GL(2,\F_p)$$ for all primes $$p$$ except those listed.
prime Image of Galois representation
$$2$$ B
## $p$-adic data
### $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(3,20) if E.conductor().valuation(p)<2]
$$p$$-adic regulators are not yet computed for curves that are not $$\Gamma_0$$-optimal.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 ordinary nonsplit ordinary ordinary nonsplit add ordinary ss ordinary ordinary ordinary ordinary ordinary ss ordinary 2 1 1 1 1 - 1 1,1 3 1 1 1 1 1,1 1 1 0 0 0 0 - 0 0,0 0 0 0 0 0 0,0 0
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2 and 4.
Its isogeny class 5577d consists of 4 curves linked by isogenies of degrees dividing 4.
## Growth of torsion in number fields
The number fields $K$ of degree up to 7 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z$ are as follows:
$[K:\Q]$ $K$ $E(K)_{\rm tors}$ Base-change curve
2 $$\Q(\sqrt{143})$$ $$\Z/4\Z$$ Not in database
$$\Q(\sqrt{-13})$$ $$\Z/4\Z$$ Not in database
$$\Q(\sqrt{-11})$$ $$\Z/2\Z \times \Z/2\Z$$ Not in database
4 $$\Q(\sqrt{-11}, \sqrt{-13})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database
We only show fields where the torsion growth is primitive. For each field $K$ we either show its label, or a defining polynomial when $K$ is not in the database. | 2,083 | 5,480 | {"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} | 3.25 | 3 | CC-MAIN-2020-24 | latest | en | 0.210857 |
http://www.gamedev.net/user/54915-kazade/?tab=posts | 1,472,025,393,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982291143.22/warc/CC-MAIN-20160823195811-00257-ip-10-153-172-175.ec2.internal.warc.gz | 461,554,279 | 18,331 | • Create Account
Member Since 01 Feb 2004
Offline Last Active Aug 20 2016 07:23 AM
### In Topic: Upward force calculation for raycast/ground collision
16 August 2016 - 03:01 PM
Randy, I'm not sure what you mean about fixing up the velocity, could you elaborate? Thanks!
### In Topic: Upward force calculation for raycast/ground collision
16 August 2016 - 01:51 PM
Thanks Randy, that's really helpful. I think I'm basically doing the right thing but something else is wrong... I realized I should probably be dividing impulse by the timestep right? (e.g. * (1/60))
### In Topic: Upward force calculation for raycast/ground collision
16 August 2016 - 07:25 AM
OK, I'm having much more trouble with this than I expected...
Say that 3 "wheels" intersect with the terrain, and I have their intersection points, penetration depths and impact normals, I assumed that I could move the car back a little bit so it wasn't intersecting, and then apply impulses at each of the 3 points (using the velocities at those points multiplied by the mass) to orient the vehicle with the terrain. In my mind that would work just as you'd expect - except that doesn't work at all - the car just flips around and flies into the air.
Can anyone give me some guidance on how to calculate the impulses to apply at each wheel? Also, what vector should I be using to move the car so that it's not intersecting? I figured I'd just move it by the greatest penetration depth of all the wheels, but then should I only apply an impulse on the wheel with the deepest penetration?
### In Topic: Upward force calculation for raycast/ground collision
14 August 2016 - 09:05 AM
Yeah, you're totally right. I should move the body first by the direction of the floor's normal * the penetration depth. Then apply an impulse of the reverse of the velocity projected onto the normal.
I was thinking I could do everything with impulses, but I'm not sure that's possible (you need to move the body from penetrating in a single step and changing the velocity isn't going to help with that).
### In Topic: Rotate to face a direction, by applying angular acceleration?
30 December 2013 - 03:02 AM
@Mike, just a few questions about that equation:
• Is "timestep" the fractional 1/60.0 or the number of steps in a second (e.g. 60)?
• Is "angle", the result of acos(desired.dot(current)) ?
• Is "angular velocity" the current angular velocity of the ship?
• What's "moment of inertia"?
Sorry if I'm being dumb
PARTNERS | 595 | 2,489 | {"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} | 3.40625 | 3 | CC-MAIN-2016-36 | longest | en | 0.954338 |
https://gitlab.cosma.dur.ac.uk/swift/swiftsim/-/commit/e2f135bf27ebe4c5766104b2240500f4bc58f671 | 1,675,820,713,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00857.warc.gz | 298,585,076 | 39,140 | Commit e2f135bf by Matthieu Schaller
### Updated the theory file with more details on the time integration algorithm as...
Updated the theory file with more details on the time integration algorithm as well as conserved quantities and the bginning of a viscosity implementation.
Former-commit-id: 66ad200345328163616a43eef5a41b1ea06618e2
parent 5154cd4e
\documentclass[a4paper,10pt]{article} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage{color} %opening \title{SPH equations} ... ... @@ -39,7 +41,8 @@ $\frac{P}{\rho\Omega}$ instead of $\Omega$ may be options worth exploring. \section{Kernel function} In what follows, we will use $r_{ij} = \vec{x_i} - \vec{x_j}$ and the kernel function given by: In what follows, we will use $\vec{r}_{ij} = \vec{x_i} - \vec{x_j}$ and $\hat{r}_{ij} = |\vec{r}_{ij}|$. The kernel function can always be decomposed as: W(\vec{x}, h) = \frac{1}{h^3}f\left(\frac{|\vec{x}|}{h}\right) ... ... @@ -63,7 +66,8 @@ where $f(q)$ is a low-order polynomial. The simplest possible choice is the cubi The constants here are NOT the constants used in GADGET as we are not following their convention of setting $h$ as the cut-off value of $W$.\\ Notice that the kernel goes to $0$ when $r_{ij} = 2h$ in this case. The constant in front of $h$ depends on the kernel Notice that the kernel goes to $0$ when $|\vec{r}_{ij}| = 2h$ in this case. The constant in front of $h$ depends on the kernel chosen and to keep it general, we should insert a constant here and say that the interaction only takes place if $r<\zeta h$ and keep $\zeta$ as a modifiable (compile time) constant. In other words, we can say that $W(x,h)$ is a function that goes to $0$ if $x > \zeta h$. \\ ... ... @@ -105,17 +109,18 @@ In the first loop of the algorithm, the secondary quantities of particle $i$ are following way: \begin{eqnarray} \rho_i &=& \sum_j m_j W(r_{ij}, h_i)\\ \rho_i &=& \sum_j m_j W(\vec{r}_{ij}, h_i)\\ h_i &=& \eta \left(\frac{m_i}{\rho_i} \right)^{1/3} \end{eqnarray} where $\eta \approx 1.2$ is a constant. These two equations can be solved iteratively using a Newton-Raphson or bisection scheme. In practice, the loop is performed over all particles $j$ which are at a distance $r_{ij}<\zeta bisection scheme. In practice, the loop is performed over all particles$j$which are at a distance$|\vec{r}_{ij}|<\zeta h$from the particle of interest. One has to iterate those two equations until their outcomes are stable.\\ Another measure of the accuracy of$h$is to use the weighted number of neighbors which (in 3D) reads N_{ngb} = \frac{4}{3}\pi \left(\zeta h\right)^3 \sum_j W(r_{ij},h_i) N_{ngb} = \frac{4}{3}\pi \left(\zeta h\right)^3 \sum_j W(\vec{r}_{ij},h_i) The (magical) value of$N_{ngb}$is a numerical parameter and its value can be expressed as a function of the more ... ... @@ -132,7 +137,7 @@ To increase the convergence rate, one can use the derivative of the density with Newton iterations: \frac{\partial \rho}{\partial h} = \sum_j m_j \frac{\partial W(r_{ij},h_i)}{\partial h} \frac{\partial \rho}{\partial h} = \sum_j m_j \frac{\partial W(\vec{r}_{ij},h_i)}{\partial h} This can also give a convergence criterion as this term must be$0$when the right value oh$h$has been found. ... ... @@ -140,7 +145,7 @@ The derivative of the kernel function has to be computed anyway to obtain a valu This term is given by \Omega_i = 1 + \frac{h_i}{3\rho_i}\sum_b m_b\frac{\partial W(r_{ij},h_i)}{\partial h} \Omega_i = 1 + \frac{h_i}{3\rho_i}\sum_b m_b\frac{\partial W(\vec{r}_{ij},h_i)}{\partial h} This concludes the first SPH loop in the standard implementation. More complicated quantities such as ... ... @@ -160,15 +165,17 @@ where$\gamma$is the polytropic index. Usually,$\gamma = \frac{5}{3}$. The second loop is used to compute the accelerations (tertiary quantities). The exact expressions are \begin{eqnarray} \vec{a} &=& - \sum_j m_j\left[\frac{P_i}{\Omega_i\rho_i^2}\vec{\nabla_r} W(r_{ij}, h_i) + \frac{P_j}{\Omega_j\rho_j^2}\vec{\nabla_r}W(r_{ij}, h_j) \right] \\ \frac{du}{dt} &=& \frac{P_i}{\rho_i^2} \sum_j m_j (\vec{v_i}-\vec{v_j})\cdot\vec{\nabla_r} W(r_{ij}, h_i) \\ \frac{dh}{dt} &=& \frac{h_i}{3}\sum_j \frac{m_j}{\rho_j} \left(\vec{v_j} - \vec{v_i} \right) \cdot\vec{\nabla_r}W(r_{ij}, \vec{a_i} &=& - \sum_j m_j\left[\frac{P_i}{\Omega_i\rho_i^2}\vec{\nabla_r} W(\vec{r}_{ij}, h_i) + \frac{P_j}{\Omega_j\rho_j^2}\vec{\nabla_r}W(\vec{r}_{ij}, h_j) \right] \label{eq:acceleration}\\ \frac{du_i}{dt} &=& \frac{P_i}{\rho_i^2} \sum_j m_j (\vec{v_i}-\vec{v_j})\cdot\vec{\nabla_r} W(\vec{r}_{ij}, h_i) \label{eq:dudt}\\ \frac{dh_i}{dt} &=& \frac{h_i}{3}\sum_j \frac{m_j}{\rho_j} \left(\vec{v_j} - \vec{v_i} \right) \cdot\vec{\nabla_r}W(\vec{r}_{ij}, h_i) \end{eqnarray} In practice the loop is here performed over all pairs of particles such that$r_{ij} < \zeta h_i$or$r_{ij} < \zeta In practice the loop is here performed over all pairs of particles such that $|\vec{r}_{ij}| < \zeta h_i$ or $|\vec{r}_{ij}| < \zeta h_j$. In general, the equations are more involved as they will contain terms to mimic the effect of viscosity or thermal conduction. These terms are pure functions of the properties of particles $i$ and $j$ and are thus very simple to insert once the code is stabilized.\\ ... ... @@ -185,7 +192,8 @@ The time step is then given by the Courant relation: \Delta t_i = C_{CFL} \frac{h_i}{c_i} where the CFL parameter usually takes a value between $0.2$ and $0.3$. The integration in time can then take place. The where the Courant parameter ($C_{DFL}$)usually takes a value between $0.2$ and $0.3$. The integration in time can then take place. The leapfrog integrator is usually used as it behaves well when coupled to gravity. \\ In the case where only one global time step is used for all particles, the minimal time step of all particles is reduced and used. \\ ... ... @@ -196,15 +204,135 @@ derivative of the smoothing length only give a rough estimate of its change. It only provides a good guess for the Newton-Raphson (or bisection) scheme. \section{Time integration} The usual scheme uses a kick-drift-kick leap-frog integrator. A full time step of size $\Delta t$ consists of the following sub-steps: \\ \textbf{first kick} Compute velocity and internal energy at half step. \begin{eqnarray*} \tilde {\vec{v}}_i &=& \vec{v}_i + \textstyle\frac{1}{2}\Delta t ~\vec{a}_i \\ \tilde u_i &=& u_i + \textstyle\frac{1}{2}\Delta t ~\frac{du_i}{dt} \end{eqnarray*} \textbf{drift} Advance time and position by a full step. \begin{eqnarray*} t &\leftarrow& t + \Delta t \\ \vec{x}_i &\leftarrow& \vec{x}_i + \Delta t \tilde {\vec{v}}_i\\ \end{eqnarray*} \textbf{prediction} Estimate velocity, internal energy and smoothing length at full step \begin{eqnarray*} \vec{v}_i &\leftarrow& \vec{v}_i + \Delta t \vec{a}_i \\ u_i &\leftarrow& u_i + \Delta t ~\frac{du_i}{dt} \\ h_i &\leftarrow& h_i + \Delta t ~\frac{dh_i}{dt} \\ \end{eqnarray*} \textbf{SPH loop 1} Compute $\rho_i$, correct $h_i$ and compute $\Omega_i$ using the first SPH loop. \\ \textbf{SPH loop 2} Compute $\vec{a_i}$ and $\frac{du_i}{dt}$ using the second SPH loop. \\ \textbf{Gravity} Compute accelerations due to gravity. \\ \textbf{Cooling} Compute the change in internal energy due to radiative cooling. \\ \textbf{second kick} Compute velocity and internal energy at end of step. \begin{eqnarray*} \vec{v}_i &=& \tilde{\vec{v}}_i + \textstyle\frac{1}{2}\Delta t ~\vec{a}_i \\ u_i &=& \tilde{u}_i + \textstyle\frac{1}{2}\Delta t ~\frac{du_i}{dt} \end{eqnarray*} \section{Conserved quantities} The following quantities are exactly conserved by the code: The energy, momentum, angular momentum and entropic function are exactly conserved by the equations: \begin{eqnarray} E &=&\sum_i m_i\left(\frac{1}{2}|\vec{v_i}|^2+u_i\right)\\ \vec{P} &=&\sum_i m_i \vec{v_i}\\ \vec{L} &=& \sum_i m_i \vec{x_i} \times \vec{v_i}\\ E &=&\sum_i m_i\left(\frac{1}{2}|\vec{v_i}|^2+u_i\right) A(s) &=& \left(\gamma -1 \right)\sum_i \frac{u_i}{\rho_i^{\gamma - 1}} \end{eqnarray} The conservation of those quantities in the code depends on the quality of the time integrator. \\ Notice that the entropic function $A(s)$ is not the physical'' entropy $s$ but is related to it through a monotonic function. It is just a convenient way to represent entropy. \section{Improved SPH equations} \textcolor{red}{WORK IN PROGRESS !!!\\} The equations \ref{eq:acceleration} and \ref{eq:dudt} correspond to a non-physical system with no viscosity and no thermal conduction. The physical model can be improved by adding some terms which vary depending on the authors. We follow here, D. Price and W. Dehnen. These terms require second derivatives of the fields, which can be expressed in terms of the second derivative of $W$. However, due to discreteness effects computing the derivatives of a field using $\partial^2_{rr}W$ is very noisy even when using high-order polynomial. For this reason a wrong second derivative is used based on the first derivative. We first introduce $F(\vec{r}_{ij},h_i)$, the scalar part of the gradient of $W$. It is defined as \nabla_r W(\vec{r}_{ij},h_i) = F(\vec{r}_{ij},h_i) \hat{r}_{ij} which in 3D implies that F_{ij}(h_i) \equiv F(\vec{r}_{ij},h_i) = \frac{1}{h_i^4}f'\left(\frac{|\vec{r}_{ij}|}{h_i}\right) where $f$ is the dimensionless part of the kernel introduced earlier. For symmetry reasons, we will use \begin{eqnarray} \bar{F}_{ij} &=& \frac{1}{2} \left(F_{ij}(h_i) + F_{ij}(h_j)\right) \\ &=& \frac{1}{2h_i^4}f'\left(\frac{|\vec{r}_{ij}|}{h_i}\right) + \frac{1}{2h_j^4}f'\left(\frac{|\vec{r}_{ij}|}{h_j}\right) \\ \end{eqnarray} in the equations. A decent place-holder for the second derivative of $W$ is then $-2F_{ij}/|\vec{r}_{ij}|$. \subsection{Artificial viscosity} Artificial viscosity can be introduced by adding a term to both equation \ref{eq:acceleration} and \ref{eq:dudt}: \begin{eqnarray*} \vec{a_i} &\stackrel{visc}{=}& 2\sum_j m_j \frac{\alpha v_{sig}\left(\vec{v}_i - \vec{v}_j\right)\cdot\hat{r}_{ij}}{\left(\rho_i + \rho_j\right)}\hat{r}_{ij}\cdot \bar{F}_{ij} \\ \frac{du_i}{dt} &\stackrel{visc}{=}& -\sum_j \frac{m_j}{(\rho_i + \rho_j)} \alpha v_{sig}\left[\left(\vec{v}_i-\vec{v}_j\right)\cdot\hat{r}_{ij}\right]^2 \bar{F}_{ij} \end{eqnarray*} where $\alpha$ is the dimensionless artificial viscosity and v_{sig} = \begin{cases} \frac{1}{2}\left[c_i + c_j - \beta\left(\vec{v}_i-\vec{v}_j\right)\cdot\hat{r}_{ij} \right] & \mbox{if} \quad \left(\vec{v}_i-\vec{v}_j\right)\cdot \hat{r}_{ij} < 0\\ 0 & \mbox{if} \quad \left(\vec{v}_i-\vec{v}_j\right)\cdot \hat{r}_{ij} > 0 \end{cases} corresponds to the maximal (average) signal speed between pairs of particles. GADGET uses $\alpha=1$ (but can be changed) and $\beta=3\alpha$ (fixed). In addition, the Balsara switch is used. Modern implementations of SPH use a variable viscosity $\alpha_i$ for each particle. The idea behind this is to switch of viscosity in the part of the flows where the fluid is dissipation-less and to switch it on in shocks. This is done by using a shock detector and then a slow decay of the viscosity with time. Following Dehen, we use \begin{eqnarray} \alpha_{loc,i} &=& \alpha_{max} \frac{h_i^2 A_i}{v_{sig,i}^2 + h_i^2 A_i} \\ \dot\alpha_i &=& \frac{2lv_{sig,i}\left(\alpha_{loc,i}-\alpha_i\right)}{h_i} \\ A_i &=& \xi_i \max\left(-\frac{d}{dt}\nabla\cdot v_i\right) \\ v_{sig,i} &=& \max\left(\frac{1}{2}(c_i+c_j) - \min(0, \left(\vec{v}_i-\vec{v}_j\right)\cdot \hat{r}_{ij})\right) \\ \end{eqnarray} The constants are usually $\alpha_{max} =2$, $l=0.05$ and $\xi_i$ = 1. \subsection{Thermal conductivity} \end{document}
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Finish editing this message first! | 3,908 | 11,706 | {"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} | 2.984375 | 3 | CC-MAIN-2023-06 | latest | en | 0.690614 |
https://www.enotes.com/homework-help/factor-equation-400139 | 1,511,286,654,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806421.84/warc/CC-MAIN-20171121170156-20171121190156-00400.warc.gz | 795,121,985 | 10,065 | # Factor the equation: 21x^2 + 22x = 8
Wilson2014 | Certified Educator
A simpler method involving the quadratic equation may be used.
`21x^2+22x=8`
`21x^2+22x-8=0`
a = 21 ; b = 22 ; c = -8
Substituting in the a, b, and c values:
`x=(-22+-sqrt(22^2-4(21)(-8)))/(2*21)`
`x=(-22+-sqrt(1156))/42`
`x=(-22+34)/42` or `x=(-22-34)/42`
Upon simplification, `x=2/7 , x= -4/3`
These are the two values that make `21x^2+22x-8=0`
In order to find the factors, we try to get all the values to one side so that the expression can equal zero.
`x=2/7 => 7x=2 => 7x-2=0`
`x=-4/3 => 3x=-4 => 3x+4=0`
Therefore, factoring the original equation results in `(7x-2)(3x+4)=0`
sciencesolve | Certified Educator
You should complete the square `21x^2 + 22x` using the following formula, such that:
`a^2 + 2ab + b^2 = (a + b)^2`
Considering `a^2 = 21x^2` and `2ab = 22x` yields:
`a = x*sqrt 21`
You need to complete the square such that:
`21x^2 + 22x + 121/21 = 8 + 121/21`
`(sqrt 21*x + 11/sqrt 21)^2 = (168 + 121)/21`
`(sqrt 21*x + 11/sqrt 21)^2 - 289/21 = 0`
Converting the difference of squares into a product yields:
`(sqrt 21*x + 11/sqrt 21 - 17/sqrt21)(sqrt 21*x + 11/sqrt 21 + 17/sqrt21) = 0`
`(sqrt 21*x - 6/sqrt21)(sqrt 21*x + 28/sqrt 21) = 0`
Hence, evaluating the factored form of the given equation yields `(sqrt 21*x - 6/sqrt21)(sqrt 21*x + 28/sqrt 21) = 0` . | 558 | 1,374 | {"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} | 4.59375 | 5 | CC-MAIN-2017-47 | latest | en | 0.601408 |
http://www.mathworks.com/help/finance/days365.html?requestedDomain=www.mathworks.com&nocookie=true | 1,519,275,580,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814002.69/warc/CC-MAIN-20180222041853-20180222061853-00483.warc.gz | 497,220,442 | 15,170 | Documentation
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days365
Days between dates based on 365-day year
Syntax
``NumDays = days365(StartDate,EndDate)``
Description
example
````NumDays = days365(StartDate,EndDate)` returns the number of days between `StartDate` and `EndDate` based on a 365-day year. All months contain their actual number of days. February always contains 28 days. If `EndDate` is earlier than `StartDate`, `NumDays` is negative. Under this convention, all months contain 30 days.Either input argument can contain multiple values, but if so, the other must contain the same number of values or a single value that applies to all. For example, if `StartDate` is an n-row character array of date character vectors, then `EndDate` must be an `N`-by-`1` vector of integers or a single integer. `NumDays` is then an `N`-by-`1` vector of date numbers. ```
Examples
collapse all
Determine the `NumDays` using date character vectors for `StartDate` and `EndDate`.
`NumDays = days365('15-jan-2000', '15-mar-2000')`
```NumDays = 59 ```
Determine the `NumDays` using a datetime array for `StartDate`.
`NumDays = days365(datetime('15-jan-2000','Locale','en_US'), '15-mar-2000')`
```NumDays = 59 ```
Determine the `NumDays` using a vector for `EndDate`.
```MoreDays = ['15-mar-2000'; '15-apr-2000'; '15-jun-2000']; NumDays = days365('15-jan-2000', MoreDays)```
```NumDays = 59 90 151 ```
Input Arguments
collapse all
Start date, specified as a scalar or an `N`-by-`1` or `1`-by-`N` vector using serial date numbers, date character vectors, or datetime arrays.
Data Types: `double` | `char` | `datetime`
End date, specified as a scalar or an `N`-by-`1` or `1`-by-`N` vector using serial date numbers, date character vectors, or datetime arrays.
Data Types: `double` | `char` | `datetime`
Output Arguments
collapse all
Number of days between two dates based on a 365-day year, returned as a scalar or an `N`-by-`1` or `1`-by-`N` vector containing the number of days.
`NumDays` returns as a double for serial date number, date character vector, or datetime inputs for `StartDate` and `EndDate`.
References
[1] Addendum to Securities Industry Association, Standard Securities Calculation Methods: Fixed Income Securities Formulas for Analytic Measures. Vol. 2, Spring 1995. | 647 | 2,404 | {"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-2018-09 | latest | en | 0.663469 |
http://www.physicsforums.com/showthread.php?t=290989 | 1,369,197,281,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701314683/warc/CC-MAIN-20130516104834-00069-ip-10-60-113-184.ec2.internal.warc.gz | 673,033,040 | 6,730 | ## A question about propagation constant in wave propagation equation
Now I'm studying acoustic waves. One question in my mind is propagation constant. It is:
$$\gamma = \alpha + j*\beta$$
while the $$\alpha$$ is attenuation constant, $$\beta$$ is phase constant. Phase constant I can understand and calculate, but how about the attenuation constant. The literature says it is: e^$$\alpha$$ = abs(A0/Ax)
where A0 is the amplitude at the source of the wave, Ax is the amplitude at some x distance of the wave.
what my question are:
Is there any other way to find out or calculate the attenuation constant. Since I'm doing a simulation of an acoustic wave, I cannot give correct parameter without knowing the attenuation constant.
PhysOrg.com physics news on PhysOrg.com >> Iron-platinum alloys could be new-generation hard drives>> Lab sets a new record for creating heralded photons>> Breakthrough calls time on bootleg booze | 201 | 932 | {"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} | 2.71875 | 3 | CC-MAIN-2013-20 | latest | en | 0.857311 |
https://phys.libretexts.org/Courses/Muhlenberg_College/Physics_122%3A_General_Physics_II_(Collett)/02%3A_Gauss's_Law/2.S%3A_Gauss's_Law_(Summary) | 1,725,846,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00376.warc.gz | 435,889,337 | 31,130 | # 2.S: Gauss's Law (Summary)
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## Key Terms
area vector vector with magnitude equal to the area of a surface and direction perpendicular to the surface cylindrical symmetry system only varies with distance from the axis, not direction electric flux dot product of the electric field and the area through which it is passing flux quantity of something passing through a given area free electrons also called conduction electrons, these are the electrons in a conductor that are not bound to any particular atom, and hence are free to move around Gaussian surface any enclosed (usually imaginary) surface planar symmetry system only varies with distance from a plane spherical symmetry system only varies with the distance from the origin, not in direction
## Key Equations
Definition of electric flux, for uniform electric field $$\displaystyle Φ=\vec{E}⋅\vec{A}→EAcosθ$$ Electric flux through an open surface $$\displaystyle Φ=∫_S\vec{E}⋅\hat{n}dA=∫_S\vec{E}⋅d\vec{A}$$ Electric flux through a closed surface $$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=∮_S\vec{E}⋅d\vec{A}$$ Gauss’s law $$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=\frac{q_{enc}}{ε_0}$$ Gauss’s Law for systems with symmetry $$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=E∮_SdA=EA=\frac{q_{enc}}{ε_0}$$ The magnitude of the electric field just outside the surface of a conductor $$\displaystyle E=\frac{σ}{ε_0}$$
## 6.2 Electric Flux
• The electric flux through a surface is proportional to the number of field lines crossing that surface. Note that this means the magnitude is proportional to the portion of the field perpendicular to the area.
• The electric flux is obtained by evaluating the surface integral
$$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=∮_S\vec{E}⋅d\vec{A}$$,
where the notation used here is for a closed surface S.
## 6.3 Explaining Gauss’s Law
• Gauss’s law relates the electric flux through a closed surface to the net charge within that surface,
$$\displaystyle Φ=∮_S\vec{E}⋅\hat{n}dA=\frac{q_{enc}}{ε_0}$$,
• where qencqenc is the total charge inside the Gaussian surface S.
• All surfaces that include the same amount of charge have the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surfaces enclose the same amount of charge.
## 6.4 Applying Gauss’s Law
• For a charge distribution with certain spatial symmetries (spherical, cylindrical, and planar), we can find a Gaussian surface over which $$\displaystyle \vec{E}⋅\hat{n}=E$$, where E is constant over the surface. The electric field is then determined with Gauss’s law.
• For spherical symmetry, the Gaussian surface is also a sphere, and Gauss’s law simplifies to $$\displaystyle 4πr^2E=\frac{q_{enc}}{ε_0}$$.
• For cylindrical symmetry, we use a cylindrical Gaussian surface, and find that Gauss’s law simplifies to $$\displaystyle 2πrLE=\frac{q_{enc}}{ε_0}$$.
• For planar symmetry, a convenient Gaussian surface is a box penetrating the plane, with two faces parallel to the plane and the remainder perpendicular, resulting in Gauss’s law being $$\displaystyle 2AE=\frac{q_{enc}}{ε_0}$$.
## 6.5 Conductors in Electrostatic Equilibrium
• The electric field inside a conductor vanishes.
• Any excess charge placed on a conductor resides entirely on the surface of the conductor.
• The electric field is perpendicular to the surface of a conductor everywhere on that surface.
• The magnitude of the electric field just above the surface of a conductor is given by $$\displaystyle E=\frac{σ}{ε_0}$$. | 2,705 | 7,851 | {"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} | 4.1875 | 4 | CC-MAIN-2024-38 | latest | en | 0.201854 |
https://edurev.in/studytube/Chapter-8-Dimensional-Analysis-Fluid-Mechanics--Me/9bf54948-cdc2-4fa7-baeb-08c53da5528b_t | 1,716,835,182,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00676.warc.gz | 187,703,521 | 66,611 | Dimensional Analysis
DIMENSIONAL ANALYSIS
• Velocity potential = [LT–1]
Stream function = [L2 T–1]
Acceleration = [LT–2]
Vorticity = [T–1]
• Total no. of variables influencing the problem is equal to the no. of independent variables plus one, one being the no. of dependent variable.
• Buckingham π theorem states that if all the n-variable are described by m fundamental dimensions, they may be grouped into (n - m) dimensions p terms.
• Selection of 3 repeating variables from the geometry of flow, fluid properties and fluid motion.
• Geometric similarity - similarity of shape
Kinematic similarity - similarity of motion
Dynamic similarity - similarity of forces
Number Equation Significance Reynolds No. Flow in closed conduit pipeFroude No. where a free surface is present, structureEulers No. In cavitation studies.Mach No. where fluid compressibility is important.Weber No. In capillary studies.
Reynolds Model Law :
(i) Velocity ratio
(ii) Time ratio
(iii) Acceleration ratio,
(iv) Force ratio
(v) Power ratio
(vi) Discharge ratio
Applications of Reynold’s Model Law :-
• Flow through small sized pipes
• Low velocity motion around automobiles and aeroplane.
• Submarines completely under water.
• Flow through low speed trubo machines.
Froude’s Model law :
(i) (Fr)prototype = (Fr)model
Vp / √ gL= Vm / √ gm Lp
It the place of model and prototype is same, then gm = gp
V = √Lr
(ii) Time scale ratio
Tr = √ Lr
(iii) Acceleration scale ratio
ar = 1
(iv) Discharge scale ratio
Qr = Lr5/2
(v) Force scale ratio
F = ρp / ρm x (L/ Lm)x (Vp / Vm)2
If the fluid used in model and prototype is same, then
Fr = Lr3
(vi) Pressure scale ratio
Pr = Lr
Applications :
• Open channels
• Notches & weirs
• Spill ways & dams
• Liquid jets from orifice
• Ship partially submerged in rough & turbulent sea
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## FAQs on Dimensional Analysis - Mechanical Engineering SSC JE (Technical)
1. What is dimensional analysis in mechanical engineering?
Ans. Dimensional analysis is a method used in mechanical engineering to analyze and understand the relationships between various physical quantities. It involves studying the dimensions and units of different variables involved in a problem to derive meaningful insights and solve engineering problems.
2. How is dimensional analysis applied in mechanical engineering?
Ans. Dimensional analysis is applied in mechanical engineering to determine the functional relationship between variables and to develop dimensionless parameters. It helps in scaling up or down physical models, predicting the behavior of systems, and validating experimental results. It also aids in simplifying complex equations and improving the efficiency of engineering designs.
3. Can dimensional analysis be used to solve complex mechanical engineering problems?
Ans. Yes, dimensional analysis can be used to solve complex mechanical engineering problems. By identifying the relevant variables and their dimensions, engineers can establish the relationships between these variables and determine the impact of each variable on the overall system. This approach can help in simplifying the problem and providing insights into the behavior of the system without the need for extensive experimentation or complex mathematical calculations.
4. What are the advantages of using dimensional analysis in mechanical engineering?
Ans. The advantages of using dimensional analysis in mechanical engineering include: 1. Simplification of complex problems: Dimensional analysis helps in simplifying complex problems by reducing the number of variables and identifying the most significant factors affecting the system. 2. Scaling and modeling: It facilitates scaling up or down physical models, allowing engineers to test and analyze the behavior of systems in a controlled and cost-effective manner. 3. Prediction and optimization: Dimensional analysis enables engineers to predict the behavior of systems based on known relationships between variables. It aids in optimizing designs and improving the efficiency of engineering solutions. 4. Experimental validation: Dimensional analysis helps in validating experimental results by identifying dimensionless parameters that should remain consistent across different scales or systems.
5. Are there any limitations or assumptions associated with dimensional analysis in mechanical engineering?
Ans. Yes, there are certain limitations and assumptions associated with dimensional analysis in mechanical engineering: 1. Linearity assumption: Dimensional analysis assumes that the relationships between variables are linear, which may not always be the case in complex systems. 2. Neglecting small effects: Dimensional analysis may neglect small effects or interactions between variables that could be significant in some cases. 3. Limited applicability: Dimensional analysis may not be suitable for all types of problems, especially those involving highly nonlinear or turbulent phenomena. 4. Dependence on accurate measurements: Dimensional analysis relies on accurate measurements of variables to establish meaningful relationships. Errors in measurements can lead to inaccurate results. Overall, while dimensional analysis is a valuable tool in mechanical engineering, it should be used in conjunction with other analytical and experimental methods to ensure accurate and reliable results.
## Mechanical Engineering SSC JE (Technical)
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; | 1,344 | 6,567 | {"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} | 3.53125 | 4 | CC-MAIN-2024-22 | latest | en | 0.545914 |
https://brainmass.com/statistics/regression-analysis/58030 | 1,480,782,123,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540975.18/warc/CC-MAIN-20161202170900-00345-ip-10-31-129-80.ec2.internal.warc.gz | 823,396,852 | 18,194 | Share
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# Underlying average of a time series
When the underlying average of a time series is very stable and there is no trend, cyclical, or seasonal influences:
a. A simple moving average forecast with n = 20 should outperform a simple moving average forecast with n = 3.
b. A simple moving average forecast with n = 3 should outperform a simple moving average forecast with n = 20.
c. An exponential moving forecast with w = 0.50 should outperform the simple moving average forecast with w = 0.01.
#### Solution Summary
The following solution involves solving concepts related to underlying average of a time series.
\$2.19 | 144 | 647 | {"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} | 3.390625 | 3 | CC-MAIN-2016-50 | longest | en | 0.909784 |
https://m.everything2.com/user/jessicaj/writeups/Arch+length?confirmop=ilikeit&like_id=2055621 | 1,726,421,442,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00843.warc.gz | 351,334,050 | 7,883 | A critical yet often overlooked foot measurement, longitudinal arch length is the distance from the heel to the ball joint of the foot. This distance is important because two people may have the same heel to longest toe measurement yet require different shoe sizes. Proper use of the Brannock device consists of taking three separate measurements. First, heel to longest toe is measured. Secondly, heel to ball of the foot is taken. If these measurements are the same that number is used to calculate width. If the numbers are different, an average is taken so width may be more accurately determined.
Suppose two women have the US measurement of seven for heel to longest toe. One woman's arch length is also seven however the second woman's arch length is a six. The first woman is a true seven, the second woman's width measurement would be calculated using size six and a half as that is the average between six and seven. What makes a shoe a certain size is the arch length distance. Most shoes flex or break at the ball joint, if your ball joint does not line up with where the shoe breaks the shoe is probably the wrong size for you.
Arch length can also used to help determine arch height. In the previous example where the second woman has a heel to toe measurement of seven but an arch length of six that length of six may indicate a high arch. Conversely, if a third woman had a heel to toe measurement of seven yet an arch length of eight she may have a flatter foot. Each foot should be measured independently of the other and considered separately before a purchase is made. Most people have one foot that is longer than the other, the same holds true for width.
Some people are reluctant to let another person measure their feet. Others are willing to let their feet be measured however become argumentative when the measurements are inconsistent with their beliefs. Feet change with the rest of your body as you age. Gravity pulls your arches down, and if they do not have anything supporting them your arches may flatten which increases foot size. Other factors such as pregnancy, weight gain, disease, and injury can also change arch length and shoe size.
When shopping for shoes if you doubt that the person measuring you has done their job correctly a good way to test them is to ask for the three measurements discussed at the beginning of this writeup: total foot length, heel to ball measurement and width. Expecting someone to remember these measurements off the top of their head may be asking a bit much, most people calculate shoe size and discard the other measurements however a quick remeasure will provide you with an opportunity to determine whether someone is using the metal slider on a Brannock to measure heel to ball of the foot. | 546 | 2,770 | {"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-2024-38 | latest | en | 0.963871 |
https://communities.sas.com/t5/Community-Matters/copy-value-from-periouvs-week-to-the-next-week/td-p/349235?nobounce | 1,532,208,392,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592778.82/warc/CC-MAIN-20180721203722-20180721223722-00212.warc.gz | 636,309,958 | 30,558 | Solved
New Contributor
Posts: 3
# copy value from periouvs week to the next week
Hi,
I have only beginning inventory for day 4.03.2017 and no ending inventory. I am able to calculate ending inventory for this daybased on the beginning invory for this day and other variables that I have. But here where the problem ibegions....I need SAS to use calculated ending inventory value from the day 4.03.2017 and have this equal to beginning inventory for dat 4.04.2017 and so on.
I am not sure where toe ven begin.
Thank you,
mag
Accepted Solutions
Solution
04-13-2017 04:16 PM
Posts: 3,059
## Re: copy value from periouvs week to the next week
``````prev_ending=lag(ending_inventory);
if missing(beginning_inventory) then beginning_inventory=prev_ending;
``````
Do yourself a huge favor and make any dates numeric and apply datetime formats to them, rather than having dates as character variables.
--
Paige Miller
All Replies
Super User
Posts: 13,583
## Re: copy value from periouvs week to the next week
It really helps to provide some example data and the desired result from that sample data.
Also, are your dates actual SAS date values or character?
New Contributor
Posts: 3
## Re: copy value from periouvs week to the next week
Hi,
The formats: plan-character, week-character, beginning_inventory, usage and orders are nuemrical.
Example of data:
Plant Week beginning_Inventory usage orders ending inventory (has to be calculated....)
h 01/14/2017 30 5 22
a 01/21/2017 .
b 01/29/2017 .
I need to calculate ending inventory: ending_inventory=beginning_inventory-usage+orders.
So for the first week it is easy to get ending_inventory whihc is 30-5+22=47 now this ending inventory from the week 01/14/2017 needs to become my beginning inventory for the week 01/29/2017 an so on...
Any idea how to approach this problem?
Thank you so much.
Solution
04-13-2017 04:16 PM
Posts: 3,059
## Re: copy value from periouvs week to the next week
``````prev_ending=lag(ending_inventory);
if missing(beginning_inventory) then beginning_inventory=prev_ending;
``````
Do yourself a huge favor and make any dates numeric and apply datetime formats to them, rather than having dates as character variables.
--
Paige Miller
New Contributor
Posts: 3
## Re: copy value from periouvs week to the next week
Thank you. It worked.
☑ This topic is solved. | 648 | 2,523 | {"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-2018-30 | latest | en | 0.716866 |
https://tunxis.commnet.edu/en/roller-coaster-design-worksheet-answers.html | 1,701,625,359,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00152.warc.gz | 672,130,994 | 7,932 | # Roller Coaster Design Worksheet Answers
Roller Coaster Design Worksheet Answers - You won't need to compute any formulas. Web one plan to secure the roller coaster is to use a chain fastened to two beams equidistant. You will design virtual roller. Web you will learn about roller coaster design, laws of motion, and about velocity and acceleration. Setting up the mathematical task • in this task, algebra 2 students will create and investigate polynomial. Web roller coaster worksheet roller coaster name: Roller coaster design worksheet e2020 answers. It is ok if the. Have each student (or each group) complete the roller coaster specifications worksheet, which asks them to identify some critical. Web you will be building a conceptual coaster using the physics concepts that are used to design real coasters.
Web students can build a rollercoaster with materials in class. Web you will learn about roller coaster design, laws of motion, and about velocity and acceleration. It is ok if the. Roller coaster design performance task. The worksheet contains space for students to plot an existing story or to. Have each student (or each group) complete the roller coaster specifications worksheet, which asks them to identify some critical. Use get form or simply click on the.
Roller Coaster Design Worksheet Answers - Web roller coaster design worksheet graphing radical functions, radical equations and extraneous roots, solving equations. You won't need to compute any formulas. Web high school students learn how engineers mathematically design roller coaster paths using the approach that a curved path can be. Roller coaster design worksheet e2020 answers. Draw a sketch of your roller coaster in the space below: Web one plan to secure the roller coaster is to use a chain fastened to two beams equidistant. Have each student (or each group) complete the roller coaster specifications worksheet, which asks them to identify some critical. Web roller coaster worksheet roller coaster name: Use get form or simply click on the. Web roller coaster design worksheet e2020 answers.
To build a roller coaster that is exciting and economical for a local. Web you will learn about roller coaster design, laws of motion, and about velocity and acceleration. This is my final project for math and i need help graphing radical functions, radical equations and extraneous roots, solving equations. The graph for model 3 should look like. You will design virtual roller.
## Web Roller Coaster Worksheet Roller Coaster Name:
Web roller coaster design worksheet graphing radical functions, radical equations and extraneous roots, solving equations. This is my final project for math and i need help graphing radical functions, radical equations and extraneous roots, solving equations. Web one plan to secure the roller coaster is to use a chain fastened to two beams equidistant. You won't need to compute any formulas.
## The Worksheet Contains Space For Students To Plot An Existing Story Or To.
To build a roller coaster that is exciting and economical for a local. You will design virtual roller. Web high school students learn how engineers mathematically design roller coaster paths using the approach that a curved path can be. Web you will be building a conceptual coaster using the physics concepts that are used to design real coasters.
## This Worksheet Makes For A Fun And Engaging Challenge For The Class.
Web terms in this set (12) step 1 define the problem. Draw a sketch of your roller coaster in the space below: Web students can build a rollercoaster with materials in class. Use get form or simply click on the.
## Setting Up The Mathematical Task • In This Task, Algebra 2 Students Will Create And Investigate Polynomial.
Web you will learn about roller coaster design, laws of motion, and about velocity and acceleration. Roller coaster design performance task. It is ok if the. Roller coaster design worksheet e2020 answers. | 773 | 3,956 | {"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-2023-50 | latest | en | 0.868884 |
https://polytope.miraheze.org/wiki/Octagrammic_duoprism?veaction=edit§ion=1 | 1,721,899,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00396.warc.gz | 413,510,506 | 12,046 | # Octagrammic duoprism
Octagrammic duoprism
Rank4
TypeUniform
Notation
Bowers style acronymStodip
Coxeter diagramx8/3o x8/3o ()
Elements
Cells16 octagrammic prisms
Faces64 squares, 16 octagrams
Edges128
Vertices64
Vertex figureTetragonal disphenoid, edge lengths 2–2 (bases) and 2 (sides)
Measures (edge length 1)
Circumradius${\displaystyle {\sqrt {2-{\sqrt {2}}}}\approx 0.76537}$
Inradius${\displaystyle {\frac {{\sqrt {2}}-1}{2}}\approx 0.20711}$
Hypervolume${\displaystyle 4(3-2{\sqrt {2}})\approx 0.68629}$
Dichoral anglesStop–4–stop: 90°
Stop–8/3–stop: 45°
Central density9
Number of external pieces32
Level of complexity12
Related polytopes
ArmyOdip, edge length ${\displaystyle {\sqrt {2}}-1}$
RegimentStodip
DualOctagrammic duotegum
ConjugateOctagonal duoprism
Abstract & topological properties
Flag count1536
Euler characteristic0
OrientableYes
Properties
SymmetryI2(8)≀S2, order 512
ConvexNo
NatureTame
The octagrammic duoprism or stodip, also known as the octagrammic-octagrammic duoprism, the 8/3 duoprism or the 8/3-8/3 duoprism, is a noble uniform duoprism that consists of 16 octagrammic prisms, with 4 meeting at each vertex.
The octagrammic duoprism can be vertex-inscribed into a sphenoverted tesseractitesseractihexadecachoron or great distetracontoctachoron.
## Vertex coordinates
The vertices of an octagrammic duoprism, centered at the origin and with unit edge length, are given by:
• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {{\sqrt {2}}-1}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {{\sqrt {2}}-1}{2}}\right)}$,
• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {{\sqrt {2}}-1}{2}},\,\pm {\frac {{\sqrt {2}}-1}{2}},\,\pm {\frac {1}{2}}\right)}$,
• ${\displaystyle \left(\pm {\frac {{\sqrt {2}}-1}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {{\sqrt {2}}-1}{2}}\right)}$,
• ${\displaystyle \left(\pm {\frac {{\sqrt {2}}-1}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {{\sqrt {2}}-1}{2}},\,\pm {\frac {1}{2}}\right)}$.
## Representations
An octagrammic duoprism has the following Coxeter diagrams: | 738 | 2,054 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "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} | 3.546875 | 4 | CC-MAIN-2024-30 | latest | en | 0.627964 |
https://metanumbers.com/25241 | 1,601,543,325,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00581.warc.gz | 487,617,325 | 7,459 | ## 25241
25,241 (twenty-five thousand two hundred forty-one) is an odd five-digits composite number following 25240 and preceding 25242. In scientific notation, it is written as 2.5241 × 104. The sum of its digits is 14. It has a total of 2 prime factors and 4 positive divisors. There are 24,612 positive integers (up to 25241) that are relatively prime to 25241.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 14
• Digital Root 5
## Name
Short name 25 thousand 241 twenty-five thousand two hundred forty-one
## Notation
Scientific notation 2.5241 × 104 25.241 × 103
## Prime Factorization of 25241
Prime Factorization 43 × 587
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 25241 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 25,241 is 43 × 587. Since it has a total of 2 prime factors, 25,241 is a composite number.
## Divisors of 25241
1, 43, 587, 25241
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 25872 Sum of all the positive divisors of n s(n) 631 Sum of the proper positive divisors of n A(n) 6468 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 158.874 Returns the nth root of the product of n divisors H(n) 3.90244 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 25,241 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 25,241) is 25,872, the average is 6,468.
## Other Arithmetic Functions (n = 25241)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 24612 Total number of positive integers not greater than n that are coprime to n λ(n) 12306 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2786 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 24,612 positive integers (less than 25,241) that are coprime with 25,241. And there are approximately 2,786 prime numbers less than or equal to 25,241.
## Divisibility of 25241
m n mod m 2 3 4 5 6 7 8 9 1 2 1 1 5 6 1 5
25,241 is not divisible by any number less than or equal to 9.
## Classification of 25241
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (25241)
Base System Value
2 Binary 110001010011001
3 Ternary 1021121212
4 Quaternary 12022121
5 Quinary 1301431
6 Senary 312505
8 Octal 61231
10 Decimal 25241
12 Duodecimal 12735
20 Vigesimal 3321
36 Base36 jh5
## Basic calculations (n = 25241)
### Multiplication
n×i
n×2 50482 75723 100964 126205
### Division
ni
n⁄2 12620.5 8413.67 6310.25 5048.2
### Exponentiation
ni
n2 637108081 16081245072521 405906706875502561 10245491188244560142201
### Nth Root
i√n
2√n 158.874 29.3338 12.6045 7.59314
## 25241 as geometric shapes
### Circle
Diameter 50482 158594 2.00153e+09
### Sphere
Volume 6.7361e+13 8.00614e+09 158594
### Square
Length = n
Perimeter 100964 6.37108e+08 35696.2
### Cube
Length = n
Surface area 3.82265e+09 1.60812e+13 43718.7
### Equilateral Triangle
Length = n
Perimeter 75723 2.75876e+08 21859.3
### Triangular Pyramid
Length = n
Surface area 1.1035e+09 1.89519e+12 20609.2
## Cryptographic Hash Functions
md5 14b7fa3d20fab4749a6ccfc6ed298684 421fdc17e6b25c275c283393bfa43939cd4e25a1 b674d43620536ff13c8f6c07477a2846fa565a4b9bd0500f48439499b6abed9d b0cd25ca39debbc144ea13e625a799f280579170238e0d5b1abb62a5ae0bceaaa50666da9c444040da2a808565a4a5c80dd9e4b4e8b8e00403cd90a0491bac66 cd421aadb3b82d47ec2f39cf771951d3f1bbdcfa | 1,461 | 4,198 | {"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-2020-40 | longest | en | 0.809493 |
https://www.fieldtriptoolbox.org/faq/how_to_interpret_the_sign_of_the_phase_slope_index/ | 1,638,227,653,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00039.warc.gz | 859,848,587 | 4,885 | Tags: faq coherence
# How to interpret the sign of the phase slope index?
The phase slope index is a bivariate measure that quantifies the consistency of the phase lag (or lead) as a function of frequency, between two signals. A value that is deviating substantially from zero, for a wider frequency range, suggests that one of the signals is consistently leading the other one, which in itself is suggestive of a ‘causal’ (at least a time-delayed) interaction between the two signals. What does the sign of the phase slope index mean, i.e. which signal is leading which?
The answer to this question is that a positive value of the phase slope index indicates that the first signal is leading the second signal. First and second signal are defined based on the order of the channels in the labelcmb field (if the data has this field). Alternatively, if the data are defined as an NxN(xNfrequency) matrix, the channel in the row is leading the channel in the column (in this case the channel names are defined in the label field of the variable.
Alternative to the answer above, if you’re unsure what’s going on, you can always simulate some data for which you know the time-lag, and see what happens with the phase slope index. If in doubt, simulate!
The following chunk of code provides data that would allow you to answer the above question yourself:
``````clear all;
% simulate some data
fsample = 1000;
nsample = fsample*30;
dat = randn(1,nsample+100);
data.trial{1}(1,:) = dat(1,1:nsample) + 0.1.*randn(1,nsample);
data.trial{1}(2,:) = dat(1,100+(1:nsample)) + 0.1.*randn(1,nsample);
data.time{1} = (1:nsample)./fsample;
data.label = {'a';'b'}; % channel 2 is leading channel 1
figure;plot(data.time{1},data.trial{1}); xlim([0 1]);
% cut into 2 second snippets
cfg = [];
cfg.length = 2;
data = ft_redefinetrial(cfg, data);
% spectral decomposition
cfg = [];
cfg.method = 'mtmfft';
cfg.output = 'fourier';
cfg.tapsmofrq = 2;
cfg.foilim = [0 100];
freq = ft_freqanalysis(cfg, data);
% connectivity estimation
cfg = [];
cfg.method = 'psi';
cfg.bandwidth = 5;
psi = ft_connectivityanalysis(cfg, freq);
% visualization
cfg = [];
cfg.parameter = 'psispctrm';
ft_connectivityplot(cfg, psi);
`````` | 582 | 2,211 | {"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-2021-49 | longest | en | 0.86349 |
https://anhngq.wordpress.com/2012/03/ | 1,653,009,128,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00224.warc.gz | 143,294,903 | 26,277 | # Ngô Quốc Anh
## March 20, 2012
### The Yamabe problem: The work by Hidehiko Yamabe
Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 20:58
Following the previous post, we are interested in solving the following equation
$\displaystyle - 4\frac{{n - 1}}{{n - 2}}{\Delta _g}\varphi + {\text{Sca}}{{\text{l}}_g}\varphi = {\text{Sca}}{{\text{l}}_{\widetilde g}}{\varphi ^{\frac{{n + 2}}{{n - 2}}}},$
where $\widetilde g=\varphi^\frac{4}{n-2}g$ (with $\varphi \in C^\infty$, $\varphi>0$) is a conformal metric conformally to $g$. In this entry, we introduce the Hidehiko Yamabe approach. His approach is variational. To keep his notation used, we rewrite the PDE as the following
$\displaystyle -\Delta \varphi + R\varphi = C_0 \varphi^\frac{n+2}{n-2}.$
Yamabe tried to minimize the following
$\displaystyle {F_q}(u) = \frac{{\displaystyle\int_M {\left( {\frac{{4(n - 1)}}{{n - 2}}|\nabla u{|^2} + R{u^2}} \right)d{v_g}} }}{{{{\left( {\displaystyle\int_M {|u{|^q}d{v_g}} } \right)}^{\frac{2}{q}}}}}$
over the Sobolev space $H^1(M)$ where $q \leqslant \frac{2n}{n-2}$. Let us say
$\displaystyle {\mu _q} = \mathop {\inf }\limits_{u \in {H^1}(M)} {F_q}(u).$
In the first stage, he showed that
Theorem B. For any $q<\frac{2n}{n-2}$, there exists a positive function $\varphi_q$ satisfying
$\displaystyle -\Delta \varphi_q + R\varphi_q = \mu_q \varphi_q^\frac{n+2}{n-2}.$
## March 10, 2012
### An integral of 1/(1+|x|^2) over the whole Euclidean space
Filed under: Giải Tích 2, Giải Tích 3, Linh Tinh — Ngô Quốc Anh @ 2:46
Sometimes, we need a precise value for following
$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}}, \quad \alpha>\frac{n}{2}.$
As such, I am going to calculate it and place the result here for future works.
In order to evaluate the above integral, we need to use the so-called co-area formula. We first write
$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \int_0^{ + \infty } {\left( {\int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} } \right)dr}.$
Note that
$\displaystyle \begin{gathered} \int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} = \frac{1}{{{{(1 + {r^2})}^\alpha }}}\int_{\partial {B_0}(r)} {dS} \hfill \\ \qquad\qquad\qquad= \frac{1}{{{{(1 + {r^2})}^\alpha }}}\text{Area}({B_0}(r)) = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}\end{gathered}.$
Therefore,
$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)}}{{\Gamma (\alpha )}}$ | 1,101 | 2,812 | {"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": 17, "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-2022-21 | latest | en | 0.634821 |
https://resources.profuturo.education/fr/ressources-educatives/tag/equations | 1,721,035,210,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00436.warc.gz | 432,876,728 | 13,127 | # Ressources
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## Distance, Velocity and Time: Equations and Relationship
Do you remember the story of the tortoise and the hare? The tortoise and the hare decided to have a race. Everybody, most of all the hare, thought the hare would win the race because he hopped much fa ...
## The Statistics of Sunspots
Sunspots are disturbed areas of the sun that look like spots because they are cooler than the surface of the sun. They have been observed and recorded since the 18th Century. This large collection of ...
Presents topics including number sense, variable expressions, statistics, decimals (including addition, subtraction, multiplication, and division), fractions, ratios, proportions, percents, geometric ...
## CK-12 Algebra II with Trigonometry Concepts (At Grade)
CK-12 Foundation’s Algebra II with Trigonometry Concepts is a collection of concepts that expands on the algebraic topics of functions, equations, and graphs for high-school students.
It is an ed ...
## CK-12 Algebra II with Trigonometr? At grade
CK-12 Foundation’s Algebra II with Trigonometry FlexBook® textbook is an expansion on the algebraic topics of functions, equations, and graphs for high-school students.
It is an educational content of ...
## CK-12 Algebra I Concepts - Honor? Advanced
CK-12 Foundation’s Algebra I Concepts - Honors is a high-level complete high school algebra course, that cover topics rigorously. This preview version is being released in stages several chapters at a ...
## CK-12 Algebra I - Second Editio? At grade
CK-12’s Algebra I Second Edition is a clear presentation of algebra for the high school student. Topics include: Equations and Functions, Real Numbers, Equations of Lines, Solving Systems of Equations ...
## CK-12 Algebra - Basi? Basic
CK-12 Foundation’s Basic Algebra FlexBook® textbook is an introduction to the algebraic topics of functions, equations, and graphs for middle-school and high-school students.
It is an educational conte ...
## Definition and general solving of quadratic equations
You'll learn the algorithms needed to solve a quadratic equations in the most general case, as well as the formula that allow to construct a quadratic equation while knowing the value of the solut ...
## Definition and how to solve linear equations
You will learn to solve simple equations of first degree and problems whose statements can be formulated as linear equations of first degree.
Definition and how to solve linear equations is a didactic ... | 531 | 2,553 | {"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-2024-30 | latest | en | 0.920796 |
https://www.prokidsedu.com/what-is-the-molar-solubility-of-pbbr2-in-0-500-m-kbr-solution | 1,670,358,391,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00147.warc.gz | 1,014,835,972 | 14,046 | # What Is The Molar Solubility Of Pbbr2 In 0.500 M Kbr Solution?
What Is The Molar Solubility Of Pbbr2 In 0.500 M Kbr Solution?. In 0.500 m kbr solution? Calculate the molar solubility of cdco3 in a buffer solution containing 0.115 m na2co3 and 0.120 m nahco3 arrow_forward calculate the molar solubility of al(oh)3 in a buffer solution with 0.100 m nh3 and 0.400 m nh4+.
What is the molar solubility of pbbr2 in pure water? What is the molar solubility of pbbr2 in purewater? We cannot speak of the concentration of a solid, which is why p bbr2 does not appear in the expression.
### What Is The Molar Solubility Of Pbbr2 In A 0.500 M Pb(No3)2 Solution?
What is the molar solubility of pbbr2 in 0.500 m kbr solution? What is the molar solubility of pbbr2 in pure water? Calculate the molar solubility of cdco3 in a buffer solution containing 0.115 m na2co3 and 0.120 m nahco3 arrow_forward calculate the molar solubility of al(oh)3 in a buffer solution with 0.100 m nh3 and 0.400 m nh4+.
### What Is The Molar Solubility Of Pbbr2 In A 0.500 M Pb(No3)2 Solution?
Experts are tested by chegg as specialists in their subject area. We cannot speak of the concentration of a solid, which is why p bbr2 does not appear in the expression. The ksp of pbbr2 is 6.60� 10�6.
### What Is The Molar Solubility Of Pbbr2 In Pure Water?
What is the molar solubility of 7,457 results, page 30 chemistry. M what is the molar solubility of pbbr, in 0.500 m kbr solution. What is the molar solubility (m) of pbbr2 in a 0.500 m pb(no3)2 solution?
### The Dissociation Response Is Stated As Follows:
Kb1 = 1.86×10^−8 kb2 = 8.33×10^−13. The molar concentrations of lead (ii) ions and bromide ions in a saturated solution of lead (ii) bromide are extremely tiny, and seem trivial when compared to the molar concentration of lead (ii) ions in a 0.500 m solution of lead (ii) nitrate, based on the value of ksp constant. We review their content and use your feedback to keep the quality high.
### In 0.500 M Kbr Solution?
What is the molar solubility of pbbr2: The interwebs give a value for the solubility product of lead bromide as ksp = 1.86 ×10−5 at 20∘c. And so ksp = 1.86 × 10−5 = [p b2+][br−]2.
See Also : What Distinguishes Active Isolated Stretching From Other Stretching Exercises | 722 | 2,284 | {"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-2022-49 | latest | en | 0.873276 |
https://www.jiskha.com/display.cgi?id=1292783003 | 1,516,661,543,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891543.65/warc/CC-MAIN-20180122213051-20180122233051-00530.warc.gz | 891,413,247 | 3,887 | # calculus - Integration
posted by .
evaluate
∫sec^7 x tanx dx
• calculus - Integration -
let u = sec x
then du = sec x tan x dx
so then we have
u^6 du
u^7/7
(1/7)sec^7 x + c
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http://7sage.com/category/logic-games/ | 1,368,914,970,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382917/warc/CC-MAIN-20130516092622-00032-ip-10-60-113-184.ec2.internal.warc.gz | 2,012,692 | 21,542 | Archive for the ‘Logic Games’ Category
I took PrepTest C (the February 2000 LSAT) under simulated testing conditions. I want to share some of my thoughts with you. This post will cover Logic Games for which I have already made the videos lessons (with links below).
Stop reading if you haven’t taken this prep test yet. It’ll ruin the test for you.
LSAT C Logic Game Summaries and Video Explanation Links
Game 1 – At a water carnival, eight lifeguards will participate in two events. One event is a boat race and the other is a rescue exercise. These lifeguards will be grouped into four two-person teams.
Game 2 – A critic ranks exactly seven restaurants. These restaurants are named Lautrec, Medici, Pastilla, Robusto, Scheherazade’s, Tantoko, and Vistula from the best – the highest rank – to the worst – the lowest rank.
Game 3 – Three parks, Jessup, Island, and Hilltop, contain attractions. The attractions are of five types: fountain, garden, museum, playground, or theater.
Game 4 – Dynamic Motors will assemble four new automobile models. Over the next three years, the models Volante, Whisper, Xavier, and Ziggurat will be assembled in its five factories F, G, H, J, and K.
Game 1 – Lifeguards at water carnival
This is a grouping game with an unusual setup. That means it’s hard initially. If you aren’t careful, it reads like an In/Out game. Once you realize that it’s not, that in fact for each of the two events, you are asked to group the eight lifeguards into four teams of two persons each, you’ll see the right game board. From there on out, this game is easy.
Game 2 – Critic ranks seven restaurants
This is a tough sequencing game. You have to be proficient in handling conditional rules in a sequencing game. The conditional rule in this game breaks up the game into, essentially, two sub-game boards. Beyond that, your proficiency with basic sequencing games will determine how quickly you can move through the questions.
Game 3 – Three parks with five attractions
This is an easy grouping game with a chart. If you setup it up with a chart, you can figure out three sub-game boards that represent all possible worlds. The questions are fast.
Game 4 – Dynamic motors assemble automobiles
This is a hard, unique game. The game board doesn’t look like any game board that we’re used to. So, you have to spend some time figuring out what game board works best to organize the information in this game. The rules are also difficult to represent visually. As with all games, spend time upfront understanding the setup, the game board, the rules, and the pieces. Otherwise, the questions will simply be a waste of time.
So, what’s the take away? Games two and four games were hard. You have to be able to move through the easy ones very quickly to save up enough time to tackle the hard ones.
______________
For more Logic Games explanations like these, hop over to our Logic Games page. There, we’ve recorded video explanations for every Logic Game going back over a decade. All in HD, with variable playback speed, and you get to ask questions. Oh, the best part: it’s completely free.
Topic: Logic Games
comment
I took PrepTest B (the February 1999 LSAT) under simulated testing conditions. I want to share some of my thoughts with you. This post will cover Logic Games for which I have already made the videos lessons (with links below).
Stop reading if you haven’t taken this prep test yet. It’ll ruin the test for you.
LSAT B Logic Game Summaries and Video Explanation Links
Game 1 – Eight boats arrive at a dock. They are named Jewel, Kashmir, Neptune, Ojibwa, Pacific, Spain, Tornado, and Valhalla.
Game 2 – A park contains at most five of seven kinds of trees. The trees are firs, laurels, maples, oaks, pines, spruces, and yews.
Game 3 - Four married couples dine at a circular table. They are named Francisco, Gabrielle, Kyoko, Lee, Olivia, Peter, Raymond, and Simone.
Game 4 – Zeno’s unfinished furniture sells five types of furniture. Footstools, hutches, sideboards, tables, and vanities. From the five, Irene will buy four. Each piece Irene buys will be made from a kind of wood: maple, oak, pine, rosewood.
Game 1 – Eight boats arrive at a dock
This is a simple, easy sequencing game. We’ve seen very similar reincarnations of this game before. You should finish this in under 5 minutes if you want to get through all the games in this set. Your proficiency with the basic sequencing chart will determine how quickly you can push through this game.
Game 2 – A park contains trees
This is a very difficult in/out game. If you do not normally have enough time to finish all the games, this is the one you should skip. The rules that make this game hard are the last two rules. One of them has an embedded conditional. Both of them demand that you represent them visually to fully understand how they control the pieces on the game board. Once you do that, you can split the game into three sub-game boards to use up these two confusing rules.
Game 3 – Married couples dine at a circular table
This is a medium difficulty spatial game. You can think of it as a circular sequencing game. It’s unusual because of the circular game board. Aside from that, this game is not very difficult. Hit the questions quick after a brief, simple game board setup. For many of the questions, you’ll have to draw sub-game boards that cater to them.
Game 4 – Zeno’s sells furniture
This is a hardish in/out game with grouping within the in group. Since there’s only one item in the out group, you should split the game board up into two sub-game boards to accomodate the two possible items that could be out. Once you do that, you can focus your attention of grouping the items within the in group. In the in group, you have to figure out what wood goes with what type of furniture. If you’re not adept with conditional logic, there is a conditional rule that could potentially be confusing.
______________
For more Logic Games explanations like these, hop over to our Logic Games page. There, we’ve recorded video explanations for every Logic Game going back over a decade. All in HD, with variable playback speed, and you get to ask questions. Oh, the best part: it’s completely free.
Topic: Logic Games, Uncategorized
comment
Did you know that there’s no such thing as a brand new Logic Game? It’s true. Every LSAT’s Logic Games are only disguised to look like they are new when in fact, they are simply reincarnations of older, existing Logic Games. Dwell on that for a second. That means you’re never going to encounter a brand new Logic Game. That means every new LSAT PrepTest you take (including the one that actually counts) will have Games that you’ve pretty much have already seen before. Not exactly the same, but very similar. Isn’t that awesome?
Think of it this way. Each Logic Game is a cookie. The LSAT’s been baking for over 20 years and they’ve baked close to 300 cookies. But, back in the kitchen, there’s actually only a few different cookie cutters that they use. Each cookie cutter cuts cookies that are very similar to the others cookies from the same cutter. So, a square cookie cutter will cut square cookies. All of these square cookies from the square cutter will resemble each other.
What does this mean for you?
You have to become acquainted with the cookie cutters (the Types of Logic Games) and not just the cookies (the Logic Games). Stop thinking that there’s 300 different Logic Games. Instead, understand that there’s only a few different types of Logic Games. Then, you have to get good at recognizing so called “new” games as old, familiar games. Old games that you’ve done already, games that you’ve already mastered through the Fool Proof Method.
You’re probably thinking “How do I know which Games are similar to which other Games?” We’re going to tell you. Right now, we are sorting all the Logic Games from LSAT PrepTest 20 (October 1996) – 68 (December 2012) into their Types (cookie cutters). We’re publishing the results as they become available.
Below, you’ll find the “In/Out” Game Type. If you’re enrolled in our online course, you’ll know that In/Out Games are the foundation of all Grouping Games (which is one of the two broad category of Logic Games, the other being Sequencing Games). In/Out Games are incredibly important to master. Here, we’ve sorted In/Out Games by similarity and difficulty.
How do I use this?
Look at the set below. Say you had trouble with the Logic Game 2 from LSAT PrepTest 33. You should do and redo (and redo and redo…) every Logic Game in its set (including itself), starting with the Games listed in its set. The ones listed in another set are less similar, though still quite similar because every Game on this page is an In/Out Game.
1. Print this list out and tape it to your wall. Games are displayed as LSAT PrepTest#.Game#.
Optional. Purchase the PDF with all the Games in the list (coming soon!)
2. Do these Game together in their set clusters using the Fool Proof Method.
3. Never miss a question on an In/Out Game again.
The Basic In/Out Games Set
PT33-Game2 | PT40-Game4 | PT45-Game3 | PT58-Game2
These are the “purest” In/Out Games. All the rules chain up very nicely. They require only an understanding of basic conditional logic.
The Basic+ In/Out Games Set
PT34-Game4 | PT41-Game3
Like the Games in the Basic group, these Games also have rules that chain up nicely. They are a little bit harder though. These Games are not immediately apparent as In/Out Games because the LSAT has disguised them.
The Easy In/Out Games Set
PT24-Game1 | PT29-Game1 | PT36-Game1 | PT48-Game1 | PT54-Game1
PT63-Game1
The Medium In/Out Games Set
PT20-Game2 | PT39-Game4 | PT47-Game2 | PT58-Game4 | PT59-Game3
The Difficult In/Out Games Set
PT31-Game2 | PT32-Game2 | PT49-Game3
These Games resemble each other less than the ones in the Basic Groups. Some of them require you to know Bi-Conditionals, De Morgan’s Law, and some are also disguised. Some of these Games have fixed their slots some did not. You also need to be aware of when conditional rules trigger and when they become irrelevant.
The In/Out with Sub-Categories Games Set
PT24-Game4 | PT26-Game4 | PT33-Game3 | PT42-Game1 | PT50-Game2
PT65-Game3
These Games contain game pieces that fall into sub-categories. At first, they are challenging, but once you learn to recognize them and draw the game board correctly, they become manageable.
The In/Out with Sequencing Games Set
PT25-Game3 | PT30-Game2 | PT32-Game1 | PT40-Game2 | PT61-Game3
These Games require you to Sequence items within the In/Out groups. You should master Sequencing Games before attempting this set. Knowledge of Conditionals, Bi-Conditionals, De Morgan’s Law are also required for some.
The Miscellaneous/Difficult In/Out Games Set
PT22-Game4 | PT23-Game2 | PT57-Game3
These Games are challenging and less similar to the other Games in the In/Out Games set.
comment
I just took PrepTest 68 (the December 2012 LSAT) under simulated testing conditions. I want to share some of my thoughts with you. This post will cover Logic Games for which I have already made the videos lessons (links below).
Stop reading if you haven’t taken this prep test yet. It’ll ruin the test for you.
LSAT 68 Logic Game Summaries and Video Explanation Links
Game 1 – A realtor is showing a prospective buyer seven houses. The first and second houses are shown in the morning. The third, fourth, and fifth are shown in the afternoon. The fifth and sixth are shown in the evening.
Game 2 – Five witnesses are scheduled to testify at a hearing. The hearing is scheduled on Monday, Tuesday, and Wednesday.
Game 3 – A maintenance company is taking service requests. Their clients are Image, Solide, and Truvest. For each client, there are two service targets – one for website requests and one for voicemail requests. The service targets are set for either 3 days, 2 days, or 1 day.
Game 4 – An editor will edit seven articles. Three articles cover finance, three cover nutrition and one covers wildlife.
Games 1 and 2
The first and second games are standard sequencing and grouping games where much of the board can be determined. I made a few sub-game boards to better visually represent the placement of items on the game board. Both games were solved in about 5 minutes each with no errors.
Game 3 – Maintenance Company
Since the first two games were quick, I expected the third and fourth games to be harder. The LSAT did not disappoint. The third one was confusing. I had to read the stimulus a couple of times to figure out how to set up the game board correctly. The stimulus used a lot of referential phrasing (“clients” and “service targets”) that made the it hard to follow. The first rule took about 4 or 5 re-readings to sink in. I thought pretty hard about what it meant for “website targets to be not longer than voicemail targets.” As is almost always the case, the time invested up front was worth it. I split up the game board into a few sub-game boards. Even with the sub-game boards drawn out, I still had to redraw them next to most of the questions to avoid careless confusion. This game took about 10 minutes with no errors.
Game 4 – Editor, Very Difficult
Going into the fourth game with about 15 minutes on the clock was comfortable. Still, I wasn’t fully prepared for how difficult it would end up being. After setting up the game board and writing down the rules, I stared at the nearly blank page for a couple of seconds. I was pretty sure there would be no point in trying to spilt the main game board up into sub-game boards. The rules were so open that I couldn’t make any inferences and so I didn’t know where to start. But with plenty of time left on the clock I just hit the questions hard and thought I would brute force my way through them.
That was risky. Most of the questions required a separate game board setup for each answer choice. That meant creating making up to 5 game boards for each question! That felt paralyzing. But you just do it. By the time I got to the last 2 questions, I was down to 2 minutes. Brute force is a highly time consuming strategy. For the second to last question, after spending about a minute on it, I eliminated the answers down to two. I had a strong feeling about one of them and just chose it so I could move onto the last question. I didn’t prove it out. I just moved onto the last question. I got lucky and got it right.
With under a minute left for the last question, I looked at answers and guessed at which one of the five probably was right given what I knew about the dynamics of the rules. Before I could even prove my only guess at the correct answer, time ran out so I had to bubble in my choice. I got lucky again.
So, what’s the take away? Two of these four games were hard. You have to be able to move through the easy ones very quickly to save up enough time to tackle the hard ones.
______________
For more Logic Games explanations like these, hop over to our Logic Games page. There, we’ve recorded video explanations for every Logic Game going back over a decade. All in HD, with variable playback speed, and you get to ask questions. Oh, the best part: it’s completely free.
Topic: Logic Games
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Here are our best LSAT Blog posts, all in one place.
General LSAT Info and Tools
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[This is a lesson excerpt from our online LSAT course, for which we invite you to enroll.]
Logic games are the hardest section of the LSAT, at first. They’re like nothing you’ve ever seen before, and they’re very confusing.
Fortunately, they’re also the easiest section to get better at. It all starts by getting a good overview of what you’re up against. That’s why we made this video introduction logic games. Enjoy!
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You may find logic games tough now. They’re difficult because they’re unfamiliar. As you practice, they get a lot easier. We promise.
Repeating logic games until you master them is the best way to get good at games. Our students have used repetition to score near perfect on games, and so can you.
Don’t worry too much about game types. You’ll find complicated classification systems in books and on the internet, but these aren’t necessary and can be distracting. We just classify games as either sequencing, grouping or a mixture of the two types.
Did you like this introduction to logic games video? It’s part of our online LSAT course. If you found the video useful, you’ll probably like the rest of our course too. You should have a look.
Topic: Lesson Excerpt, Logic Games
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October 2011 LSAT (PrepTest 64) – Section 2 (Logic Games) – Game 1
This is a very easy basic sequencing game. It’s the one about an administrator who’s supposed to assign parking spaces to employees Robertson, Souza, Togowa, Vaughn, Xu, Young. It’s from LSAT Prep Test 64, October 2011, Section 2, Questions 1-6, Logic Game 1.
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This is a great game for you to practice splitting your master game board up into a few sub-game boards. It also showcases why the LSAT rewards people who spend more time up front on setting up Logic Games and making inference before going into the questions. Watch out for the how they word before and after with a tricky use of “higher than.”
Topic: Logic Games
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Logic Games Explanations for the most recent LSAT PrepTest 66 from June 2012 is now available! Like all our other logic games explanations, these are available for free.
Game 1, Section 3, Questions 1-5 is about a chemistry class with six lab sessions over 3 days – Wednesday, Thursday, and Friday – in the morning and afternoon led by lab assistants Julio, Kevin, Lan, Nessa, Olivia, or Rebecca. Watch the video explanation below!
Topic: Logic Games
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This is one of the logic games that students hate the most. It’s the one about a music store that carries both new and used versions of jazz, opera, pop, rap, and soul CDs. It’s from LSAT PrepTest 31, June 2000, Section 1, Questions 7-13, Logic Game 2.
I love this game. The reason is because I’m a sadist. But, that’s obvious. What’s not so obvious is that this game showcases how important it is to know your conditional logic well. I don’t just mean knowing that “if” introduces a sufficient condition. That’s child’s play. What’s hard about this game is knowing which conditional rules trigger and which ones are irrelevant. Most students react to the sheer volume of rules in this game with some form paralysis. Consider the video explanation below your antidote. Watch, learn and master when conditional rules trigger and when they are irrelevant.
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For more Logic Games explanations like this one, hop over to our Logic Games page. There, we’ve recorded video explanations for every Logic Game going back over a decade. All in HD, with variable playback speed, and you get to ask questions. Oh, the best part: it’s completely free.
Topic: Lesson Excerpt, Logic Games
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This is a foundational game to master for any LSAT taker. It’s the birds in the forest game from LSAT PrepTest 33, December 2000, Section 4, Questions 6-12, Logic Game 2.
This is the first Logic Game we teach in our LSAT Course to introduce students to the Grouping Games. This one only has two groups – the “in the forest” group and the “not in the forest” group – and thus, we call it an In/Out Game. Let me repeat. You must master this game. Because this game is the acid test of whether you understand and can apply conditional logic.
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For more Logic Games explanations like this one, hop over to our Logic Games page. There, we’ve recorded video explanations for every Logic Game going back over a decade. All in HD, with variable playback speed, and you get to ask questions. Oh, the best part: it’s completely free.
Topic: Lesson Excerpt, Logic Games
comment | 4,894 | 19,935 | {"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-2013-20 | latest | en | 0.904361 |
https://www.physicsforums.com/threads/zero-average-speed-but-nonzero-average-velocity.524448/ | 1,527,281,782,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867217.1/warc/CC-MAIN-20180525200131-20180525220131-00348.warc.gz | 791,907,560 | 14,211 | # Homework Help: Zero average speed but nonzero average velocity?
1. Aug 25, 2011
### pozishunvecta
1. The problem statement, all variables and given/known data
Is it possible to have zero average speed but nonzero average velocity?
2. Relevant equations
3. The attempt at a solution
This is probably completely wrong, but I'll put it up anyway:
If one's frame of reference is also the center of an arbitrary circle, then an object that travels along the circumference of that circle will have zero average speed but nonzero average velocity.
2. Aug 25, 2011
### PeterO
It is pretty hard to travel anywhere with a zero average speed.
Having a zero average velocity but no-zero average speed is certainly possible.
Unless you are using an I-PAD or laptop, your average velocity during the time interval from when you posted this question to when you read this response will be zero - you will be back in front of your computer - so total displacement will be zero. Your average speed could be quite different, as you possibly went and got a cup of coffee while you waited. | 234 | 1,081 | {"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} | 3.25 | 3 | CC-MAIN-2018-22 | longest | en | 0.941073 |
https://www.gamedev.net/forums/topic/650095-collision-response-jerky/ | 1,563,236,213,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524290.60/warc/CC-MAIN-20190715235156-20190716021156-00258.warc.gz | 701,722,599 | 29,477 | # Collision Response Jerky
This topic is 2068 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
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I've been implementing the collision detection and response detailed in this paper:http://www.peroxide.dk/papers/collision/collision.pdf I've got it mostly working now, but the collision response is a bit jumpy. I've only implemented collision inside the triangle but for what I've testing it against, it doesn't make a difference. Here is my code:
bool physicsObject::onSameSide(glm::vec3 point, glm::vec3 p1, glm::vec3 p2, glm::vec3 p3)
{
glm::vec3 cross1 = glm::cross(p3-p2, point-p2);
glm::vec3 cross2 = glm::cross(p3-p2, p1-p2);
if (glm::dot(cross1, cross2) >= 0.0)
return true;
else return false;
}
collisionResponsePacket physicsObject::checkTriangle(unsigned int i)
{
collisionResponsePacket packetToReturn;
glm::vec3 velocity = playerESpace.vecInElipsoidSpace( glm::vec3(-camx,-camy,-camz)-glm::vec3(-lcamx,-lcamy,-lcamz));
glm::vec3 basePos = playerESpace.vecInElipsoidSpace(glm::vec3(-lcamx,-lcamy,-lcamz));
//for now, we're just going to check the one triangle.
//.obj stores vertexes in counter-clockwise order, so we're going to swap the positions.
glm::vec3 p1 =playerESpace.vecInElipsoidSpace( (glm::vec3(ModelRegistry.models[model].m.obj[i].x3,ModelRegistry.models[model].m.obj[i].y3,ModelRegistry.models[model].m.obj[i].z3)));
glm::vec3 p2 = playerESpace.vecInElipsoidSpace((glm::vec3(ModelRegistry.models[model].m.obj[i].x2,ModelRegistry.models[model].m.obj[i].y2,ModelRegistry.models[model].m.obj[i].z2)));
glm::vec3 p3 = playerESpace.vecInElipsoidSpace((glm::vec3(ModelRegistry.models[model].m.obj[i].x1,ModelRegistry.models[model].m.obj[i].y1,ModelRegistry.models[model].m.obj[i].z1)));
//get the normal
glm::vec3 normal = glm::normalize(glm::cross(p2-p1, p3-p1));
//get signed distance
float distance = glm::dot(normal, basePos) - ((normal.x*p1.x)+(normal.y*p1.y)+(normal.z*p1.z));
//normal dot velocity
float ndv = glm::dot(normal, velocity);
//times of intersection
float time1 = (1.0 - distance)/ndv;
float time2 = (-1.0 - distance)/ndv;
if (time1 > time2)
{
float temp = time2;
time2 = time1;
time1 = temp;
}
if (time1 > 1.0 || time2 < 0.0)
{
//do something to say there wasn't a collision here
}
else
{
if (time2 > 10)
time2 = 0;
if (time1 < 0.0)
time1 = 0;
//get the point where we think that the sphere is going to collide
glm::vec3 intersectionPoint = (basePos + (time1*velocity*(float)0.95)) - normal;
//now we need to check if this collided
if (onSameSide(intersectionPoint, p1, p2, p3) && onSameSide(intersectionPoint, p2, p1, p3) && onSameSide(intersectionPoint, p3, p1, p2))
{
//there was a collision inside the triangle, so wer're done here.
packetToReturn.intersectionPoint = intersectionPoint;
packetToReturn.time0 = time1;
packetToReturn.time1 = time2;
packetToReturn.collided = true;
}
else
{
//temp, I just want to be able to check more complex meshes
return packetToReturn;
//there wasn't a collision inside the triangle so now for the fin stuff, we get to check the collision for the edges and points.
//this is also called the sweep test
//first we check collisions with the vertexes
}
}
return packetToReturn;
}
void physicsObject::testCollisionMeshWithElipsoid(double time)
{
//some of the vectors we require for the response
glm::vec3 velocity = playerESpace.vecInElipsoidSpace( glm::vec3(-camx,-camy,-camz)-glm::vec3(-lcamx,-lcamy,-lcamz));
glm::vec3 basePos = playerESpace.vecInElipsoidSpace(glm::vec3(-lcamx,-lcamy,-lcamz));
glm::vec3 destination = playerESpace.vecInElipsoidSpace(glm::vec3(-camx,-camy,-camz));
std::vector<collisionResponsePacket>results;
for (int i = 0; i < ModelRegistry.models[model].m.obj.size(); i++)
//we chech each triangle
results.push_back(checkTriangle(i));
//now we find which one we are going to respond against, then do it
for (int i = 0; i < results.size(); i++)
{
if (results[i].collided == true)
{
//we found the droids we are looking for ;)
glm::vec3 newPosition = basePos + (velocity*results[i].time0*0.9f);
glm::vec3 intersectionPoint = results[i].intersectionPoint - (velocity*results[i].time0*0.9f);
glm::vec3 slingingPlaneOrigin = intersectionPoint;
glm::vec3 slidingPlaneNormal = glm::normalize(newPosition - intersectionPoint);
float distance = glm::dot(slidingPlaneNormal, destination) - ((slidingPlaneNormal.x*slingingPlaneOrigin.x)+(slidingPlaneNormal.y*slingingPlaneOrigin.y)+(slidingPlaneNormal.z*slingingPlaneOrigin.z));
destination = destination - (distance)*slidingPlaneNormal;
//convert the new point back into what the camera will take
// lcamx = results[i].intersectionPoint.x;
//lcamy = results[i].intersectionPoint.y;
//lcamz = results[i].intersectionPoint.z;
//now we can update everything
camx = -destination.x*ELIPSE_X;
camz = -destination.z*ELIPSE_Z;
camy = -destination.y*ELIPSE_Y;
break;
}
}
results.clear();
}
I know its very inefficient, but I want the functionality there first. I've looked at the sample code in the paper but I haven't been able to work out whats wrong. I'm not recursing, but when I do, the camera warps to different places. What I want to know what is wrong with the length of the velocities, etc that would cause this kind of problem. Any help or information would be appreciated.
EDIT:
I've tried dividing the distance variable, this one: "destination - (distance)*slidingPlaneNormal;" and that seems to smooth it out but it seems to cause inaccuracies in the response and causing the camera to clip through the world.
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Formatting of your code is messed up there?
Anyway, this looks bad
if (time2 > 10)
time2 = 0;
if (time1 < 0.0)
time1 = 0;
Dunno why 10 turns up there do you mean 1.0? And if it's bigger than 10 (or 1.0) don't set it to zero set it to 1.0 or 10.
##### Share on other sites
Formatting of your code is messed up there?
Anyway, this looks bad
if (time2 > 10)
time2 = 0;
if (time1 < 0.0)
time1 = 0;
Dunno why 10 turns up there do you mean 1.0? And if it's bigger than 10 (or 1.0) don't set it to zero set it to 1.0 or 10.
Thanks for pointing that out. I'm pretty sure that that shouldn't matter because, after that line nothing is being done with time2 because the collision is not fully implemented.
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• 18 | 1,820 | 6,305 | {"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} | 2.578125 | 3 | CC-MAIN-2019-30 | latest | en | 0.583011 |
https://im.kendallhunt.com/k5/teachers/grade-3/unit-4/lesson-14/lesson.html | 1,680,149,721,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949097.61/warc/CC-MAIN-20230330035241-20230330065241-00323.warc.gz | 359,189,506 | 29,039 | # Lesson 14
Ways to Represent Multiplication of Teen Numbers
## Warm-up: Notice and Wonder: Seeing Groups (10 minutes)
### Narrative
The purpose of this warm-up is to elicit the idea that while there are multiple ways to represent 2 groups of 12, some ways are more useful than others. While students may notice and wonder many things about the images, how 2 images show the groups of 12 have been organized using place value and how this type of decomposition can be helpful in finding the total are the important discussion points.
### Launch
• Groups of 2
• Display the image.
• “What do you notice? What do you wonder?”
• 1 minute: quiet think time
### Activity
• 1 minute: partner discussion
• Share and record responses.
### Student Facing
What do you notice? What do you wonder?
### Activity Synthesis
• “The image on the left is a drawing of equal groups. The other images are base-ten diagrams. What is the same and different about these representations?” (They all show 12. They all show 2 groups of the same size. In the base-ten diagrams you can see the tens easier. It’s harder to see the tens in the first drawing.)
## Activity 1: A Factor Greater than Ten (20 minutes)
### Narrative
The purpose of this activity is for students to see how, when multiplying a number larger than ten, the distributive property can be used to decompose the factor into tens and ones, creating two smaller products. Base-ten blocks are used to help students visualize what is happening when a factor is decomposed to make two more easily known products. Factors slightly larger than ten can be naturally decomposed into a ten and some ones using place value. This will be useful in subsequent lessons as students progress towards fluent multiplication and division within 100.
When students see that you can decompose a teen number into tens and ones and use this to multiply teen numbers, they look for and make use of structure (MP7).
### Required Materials
Materials to Gather
### Launch
• Groups of 2
• Give students base-ten blocks.
### Activity
• “Take a few minutes to look at Tyler’s strategy and decide if you agree or disagree with it.”
• 2–3 minutes: independent work time
• 3–4 minutes: partner discussion
• Monitor for students who connect the expressions $$7 \times 10$$ and $$7 \times 3$$ to the tens and ones portion of the place value diagram and $$7 \times 13$$ to the entire diagram.
• Have students share why they agreed or disagreed with Tyler’s strategy with a focus on using the place value diagram as a justification.
• “Where do we see the $$7 \times 13$$ in the place value diagram?”
• “Where do the 10 and the 3 come from?”
• “How could you use the place value diagram to figure out how to find the value of $$7 \times 13$$?”
• “Now use Tyler’s method on your own to find the value of $$3 \times 14$$.”
• 2–3 minutes: independent work time
• 2–3 minutes: partner discussion
### Student Facing
1. Tyler says he can use base-ten blocks to find the value of $$7 \times 13$$ because he knows $$7 \times 10$$ and $$7 \times 3$$. He says this diagram proves his thinking.
Do you agree or disagree? Explain your reasoning.
2. Use Tyler’s method to find the value of $$3 \times 14$$. Explain or show your reasoning.
### Student Response
If students say they don’t see $$7\times10$$ and $$7\times3$$ in Tyler's diagram, consider asking:
• “Where do you see $$7\times13$$ in the diagram?”
• “If we separate the tens and ones, what expression could we use to describe the tens? The ones?”
### Activity Synthesis
• Display base-ten blocks or place value diagrams that students used to solve. As students explain their work, write multiplication expressions to represent them.
• “How does this diagram (or the base-ten blocks) show how Tyler’s method could be used to multiply $$3 \times 14$$?” (We can see there are 3 tens which is 30. We can see that there are 3 groups of 4 ones which is 12. $$30 + 12$$ is 42. The whole diagram represents $$3 \times 14$$.)
• If there is time, ask students to find the value of $$4 \times 12$$ and $$5 \times 16$$ using the base-ten blocks and Tyler’s strategy.
## Activity 2: Ways to Represent (15 minutes)
### Narrative
The purpose of this activity is for students to make sense of different ways of representing multiplication of a teen number. Students analyze a gridded area diagram, base-ten blocks, and an area diagram labeled with side lengths. When they discuss how the different diagrams represent the same product, students reason abstractly and quantitatively (MP2).
MLR8 Discussion Supports. Synthesis: Show a visual display of the diagrams. As students share their observations, annotate the display to illustrate connections. For example, on each diagram, annotate the decomposition of 15 into 10 and 5 by circling the groups of 10 and the groups of 5.
Representation: Access for Perception. Begin by showing a demonstration explaining how you see the product in each of the 3 different models using a different problem to support understanding of the context.
Supports accessibility for: Conceptual Processing, Visual-Spatial Processing
### Launch
• Groups of 2
• “We’re going to look at three different ways students showed the same expression. What do you notice? What do you wonder?” (Students may notice: You can see all the squares in the first 2 diagrams, but not in the last one. The middle diagram looks like base-ten blocks. Students may wonder: Why would you choose to use one of these diagrams? What numbers were they multiplying?)
• 1 minute: quiet think time
• Share responses.
### Activity
• “Work with your partner to tell how you see the factors in each diagram and how you see the product in each diagram.”
• 5–7 minutes: partner work time
### Student Facing
Andre, Clare, and Diego represented the same expression. Their representations are shown below.
1. Where do you see the factors in each diagram?
2. Where do you see the product in each diagram?
### Activity Synthesis
• “How are these ways of representing $$3 \times 15$$ the same?” (They all represent 3 times 15. They all show the 15 being decomposed into 10 and 5. They are all shaped like a rectangle.)
• “How are these ways of representing $$3 \times 15$$ different?” (Clare used base-ten blocks, but Andre and Diego used rectangles. Diego didn’t show the squares in his rectangle, but Clare and Andre did.)
• “How could we represent the strategy shown in all the diagrams with expressions?” ($$3 \times 10$$ and $$3 \times 5$$ or $$10 \times 3$$ and $$5 \times 3$$.)
## Lesson Synthesis
### Lesson Synthesis
Display:
$$7 \times 6$$
$$(5 \times 6) + (2 \times 6)$$
$$3 \times 15$$
$$(3 \times 10) + (3 \times 5)$$
“Today we saw some different ways to represent strategies we can use to multiply teen numbers. How are the strategies we use to multiply teen numbers like the strategies we used to multiply smaller numbers in past lessons?” (We can use facts that we know to find facts that we don’t know. We can break down one of the factors into smaller parts to make it easier to multiply.) | 1,682 | 7,085 | {"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} | 4.71875 | 5 | CC-MAIN-2023-14 | longest | en | 0.887765 |
https://blog.gmatwhiz.com/category/gmat-preparation/ | 1,627,174,206,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00592.warc.gz | 151,843,932 | 37,205 | ### Is GMAT Official Guide Enough?
The most popular resource for GMAT preparation is undoubtedly the Official Guide. Designed by the makers of the GMAT test, several applicants find it customary to practice GMAT-styled questions to understand the question types that appear on the exam. But is it ideal to start your GMAT preparation with what we lovingly call the GMAT […]
### GMAT after a Break: How to go about it?
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Students taking the GMAT often ask what really matters on GMAT – is it their raw scores in Verbal and Quant sections or do their IR and AWA scores even matter? And, the most common question they ask is – how does GMAT calculate their total score? As confusing as it can get, in this […]
### When Should You Retake the GMAT?
So, you took the GMAT and you are not happy with the results? What should you do next? An obvious thing to do would be to retake the GMAT. More than 60% of all GMAT test takers retake the GMAT within two months to one year of their previous attempt. The fact that GMAC also […] | 768 | 3,473 | {"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} | 3.03125 | 3 | CC-MAIN-2021-31 | longest | en | 0.963062 |
https://www.mathworks.com/matlabcentral/cody/problems/44352-the-top-5-primes/solutions/1645391 | 1,606,630,020,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141196324.38/warc/CC-MAIN-20201129034021-20201129064021-00587.warc.gz | 763,481,110 | 19,214 | Cody
# Problem 44352. The Top 5 Primes
Solution 1645391
Submitted on 14 Oct 2018 by Athi
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
assessFunctionAbsence({'regexp','regexpi','regexprep','str2num'},'FileName','top5primes.m')
2 Pass
x = 1:10; y_correct = [7 5 3 2 NaN]; assert(isequaln(top5primes(x),y_correct))
3 Pass
x = (1:2:100).'; y_correct = [97 89 83 79 73].'; assert(isequaln(top5primes(x),y_correct))
4 Pass
x = [17 6 3 13 8 17 1 2 5 5 3 7 7 11 2 31 7 6]; y_correct = [31 11 17 17 7 7 13 3 5 7 2 3 5 NaN 2]; assert(isequaln(top5primes(x),y_correct))
5 Pass
x = interp1(magic(30).',1:5).'; y_correct = [877 733 863 719 881 829 701 751 173 769 797 139 59 157 29 89 107 43 109 13 73 NaN 11 61 NaN]; assert(isequaln(top5primes(x),y_correct))
6 Pass
rng(0); x = reshape(randperm(200,180),36,5); y_correct = [163 181 173 197 193 71 179 149 191 157 23 167 113 139 151 19 131 101 83 137 NaN 109 67 73 127]; assert(isequaln(top5primes(x),y_correct))
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http://math.stackexchange.com/questions/tagged/transformation%20eigenvalues-eigenvectors | 1,409,226,699,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500830746.39/warc/CC-MAIN-20140820021350-00084-ip-10-180-136-8.ec2.internal.warc.gz | 124,587,056 | 15,141 | # Tagged Questions
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### Constructing regular integer matrices with distinct integer eigenvalues
How can I construct matrices with positive integer values and distinct integer eigenvalues (not necessarily positive, but 0 should not be an eigenvalue). The standard-method to construct matrices ...
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### Do T and T* have the same eigenvalues with the same algebraic multiplicity?
I know that the eigenvalues of T* are the conjugates of T's eigenvalues , but how can I see each eigenvalue of T and it's conjugate , the eigenvalue of T*, have the same algebraic multiplicity?
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### find the vector $(x,y,z) \in \mathbb{R}^3$ and the constants $\lambda \in \mathbb{R}$ such that $T(x,y,z) = (\lambda x, \lambda y, \lambda z )$
Let $T : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by : $$T(x,y,z) = (x-y+4z,3x+2y-z,2x+y-z)$$ How can i find the vector $(x,y,z) \in \mathbb{R}^3$ and the constants $\lambda \in \mathbb{R}$ ...
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### Linear transformations and eigenvalues [duplicate]
Let $T: \mathbb C^n \rightarrow \mathbb C^n$ be linear. Let $\beta$ and $\gamma$ be any two ordered bases. Prove that the eigenvalues of $[T]_\beta$ and $[T]_\gamma$ are the same. Can anyone provide ... | 354 | 1,229 | {"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} | 3.21875 | 3 | CC-MAIN-2014-35 | latest | en | 0.654432 |
https://tmeastafrica.org/and-pdf/1620-dynamics-of-particles-and-of-rigid-bodies-in-plane-motion-pdf-844-843.php | 1,653,587,462,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00333.warc.gz | 640,193,066 | 8,138 | # Dynamics Of Particles And Of Rigid Bodies In Plane Motion Pdf
File Name: dynamics of particles and of rigid bodies in plane motion .zip
Size: 23947Kb
Published: 03.05.2021
The study of the motion of a rigid body on a plane RBP motion is usually one of the most challenging topics that students face in introductory physics courses.
## Engineering Systems in Motion: Dynamics of Particles and Bodies in 2D Motion
In physics , a rigid body also known as a rigid object [2] is a solid body in which deformation is zero or so small it can be neglected. The distance between any two given points on a rigid body remains constant in time regardless of external forces or moments exerted on it. A rigid body is usually considered as a continuous distribution of mass. In the study of special relativity , a perfectly rigid body does not exist; and objects can only be assumed to be rigid if they are not moving near the speed of light. In quantum mechanics , a rigid body is usually thought of as a collection of point masses. For instance, molecules consisting of the point masses: electrons and nuclei are often seen as rigid bodies see classification of molecules as rigid rotors.
## Dynamics of Rigid Bodies
Daqaq, MF. ASME Press, In Chapter 1, we learned how to find the position, velocity, and acceleration of a particle — its kinematic description — without worrying about the forces causing its motion. While this constitutes an important first step towards learning dynamics, kinematics does not lend itself naturally to obtaining the equations governing the motion of the particle. Such equations are what we are after in this book because, through their solution, we can describe and predict motion of particles. This chapter outlines the most common approach — the vectorial approach — through which the equations governing the motion of particles are derived.
Plane Motion: When all parts of the body move in a parallel planes then a rigid body said to perform plane motion. Straight Line Motion: It defines the three equations with the relationship between velocity, acceleration, time and distance travelled by the body. In straight line motion, acceleration is constant. Distance travelled in n th second:. Projectile Motion: Projectile motion defines that motion in which velocity has two components, one in horizontal direction and other one in vertical direction.
Kinematic equations relate the variables of motion to one another. In Part 3, vectors are used to solve the problem. Kinematics studies how the position of an object changes with time. This is the community to discuss both parts of Engineering Mechanics: Statics and Dynamics. Therefore, it is necessary to solve many problems independently.
A rigid body A system of particles for which the distance between the particles remain unchanged. All lines on a rigid body in its plane of motion have the same.
## Kinematics and Dynamics of Particles and Rigid Bodies in Plane Motion Study Notes
This course is an introduction to the study of bodies in motion as applied to engineering systems and structures. We will study the dynamics of particle motion and bodies in rigid planar 2D motion. This will consist of both the kinematics and kinetics of motion. Kinematics deals with the geometrical aspects of motion describing position, velocity, and acceleration, all as a function of time. Kinetics is the study of forces acting on these bodies and how it affects their motion.
Увы, у этой программы такого тщеславия нет, у нее нет инстинкта продолжения рода. Она бесхитростна и целеустремленна, и когда достигнет своей цели, то скорее всего совершит цифровое самоубийство. - Джабба театральным жестом указал на громадный экран.
### Kinematics and Dynamics of Particles and Rigid Bodies in Plane Motion Study Notes
Нет. Но если он посмотрит на монитор и увидит в окне отсчета значение семнадцать часов, то, будьте уверены, не промолчит. Стратмор задумался. - С какой стати он должен на него смотреть? - спросил. Сьюзан взглянула ему в. - Вы хотите отправить его домой.
Первой из них были предоставляемые ею данные. Бринкерхофф терпеливо ждал, пока она изучала цифры. - Хм-м, - наконец произнесла .
Dynamics & Vibrations, NAV. 3. Introduction. 5. Plane Kinematics. Particle. Rigid Body. Size. Small & Not importantBig & Important. Motion.
Девушка наконец нашла то, что искала, - газовый баллончик для самозащиты, экологически чистый аналог газа мейс, сделанный из острейшего кайенского перца и чили. Одним быстрым движением она выпрямилась, выпустила струю прямо в лицо Беккеру, после чего схватила сумку и побежала к двери. Когда она оглянулась, Дэвид Беккер лежал на полу, прижимая ладони к лицу и корчась от нестерпимого жжения в глазах.
Любой шифр можно взломать - так гласит принцип Бергофского. Она чувствовала себя атеистом, лицом к лицу столкнувшимся с Господом Богом. - Если этот шифр станет общедоступным, - прошептала она, - криптография превратится в мертвую науку.
Программист намеревался выставить ее на аукционе и отдать тому, кто больше всех заплатит. Далее в заметке сообщалось, что, хотя алгоритм вызвал громадный интерес в Японии, несколько американских производителей программного обеспечения, прослышавших о Цифровой крепости, считают эту информацию нелепой - чем-то вроде обещания превратить свинец в золото. Формула, утверждают они, - это мистификация, к которой не следует относиться серьезно. - Аукцион? - Сьюзан подняла. Стратмор кивнул: - Как раз сейчас японские компании скачивают зашифрованную версию Цифровой крепости и пытаются ее взломать.
Получить ключ было необходимо, но Стратмор отлично понимал, что посылать глухого киллера в севильский морг было бы настоящим самоубийством. И тогда он стал искать иные возможности. | 1,428 | 5,744 | {"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} | 3.5 | 4 | CC-MAIN-2022-21 | latest | en | 0.92736 |
http://www.lecpf.com/forum/175v1.php?page=enthalpy-change-for-graphite-to-diamond-2d1550 | 1,618,754,572,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00209.warc.gz | 148,421,814 | 5,683 | Given C graphite, ΔcombH=-391.25 kj; Cdiamond, ΔcombH=-393.12 kj (A) zero ( Which of following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction ? {/eq}rxn= -172.5 kJ. Which of the following set of molecules will have zero dipole moment ? The reaction is. Services, Hess's Law: Definition, Formula & Examples, Working Scholars® Bringing Tuition-Free College to the Community, {eq}\rm C\left ( s, diamond \right )= C\left ( s, graphite \right ) {/eq} 2CO2 (g) ; ΔH{eq}^o In order to solve Hess's Law problems, you manipulate the reactions with known {eq}\rm \Delta H Given C graphite, ΔcombH=-391.25 kj; Cdiamond, ΔcombH=-393.12 kj. So it means that diamond has a higher energy than graphite. This is also the form with the lowest enthalpy, so graphite has a standard enthalpy of formation equal to zero. {/eq}. Table 1 provides sample values of standard enthalpies of formation of various compounds. aus oder wählen Sie 'Einstellungen verwalten', um weitere Informationen zu erhalten und eine Auswahl zu treffen. Here, P, V, and T are pressure, volume and temperature, respectively. For which of the following reactions, $\Delta H$ is equal to $\Delta U$ ? Enthalpy change of atomization is the enegy change when one mole of gas atoms are formed from the element in its standard state. Dazu gehört der Widerspruch gegen die Verarbeitung Ihrer Daten durch Partner für deren berechtigte Interessen. To determine which form is zero, the more stable form of carbon is chosen. {/eq} values in such a way as to get the reactants of interest on the left, the products of interest on the right, and all of the other species to cancel out. This means that the change in enthalpy of the reaction depends only on the initial and final states of the reaction but doesn't depend on the pathway to get there. (a) $CO_2(g)$ is used as refrigerant for ice-cream and frozen food. If the reaction is endothermic, then it means that on a reaction pathway graph, diamond will be higher than graphite. The combustion of benzene $(l)$ gives $\ce{CO_2(g)}$ and $H_2O(l)$ . {/eq} CO2 (g) ; ΔH{eq}^o C (graphite) ---> C (diamond) By Hess's Law, you change the sign of the enthalpy for the second reaction (because you reversed it), and add it to the … All that's left is to add up the changes in enthalpy: Our experts can answer your tough homework and study questions. {/eq}rxn= -566.00 kJ, 2CO (g) {eq}\rightarrow Daten über Ihr Gerät und Ihre Internetverbindung, darunter Ihre IP-Adresse, Such- und Browsingaktivität bei Ihrer Nutzung der Websites und Apps von Verizon Media. However, equation 2 needs to be reversed in order to get the other species to cancel out. On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be: An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. We need to get the reactant and products of the three reactants given on the correct side and get the other species to cancel out. {/eq}, {eq}\rm C\left ( s, diamond \right )~+~O_{2}\left ( g \right )\rightarrow CO_{2}\left ( g \right ); \Delta H=~-395.4~kJ $\Delta_f G^{\circ}$ at 500 K for substance ‘S’ in liquid state and gaseous state are $\ce{+ 100.7 \, kcal \, mol^{-1}}$ and $\ce{+103 \, kcal \, mol^{-1}}$, respectively. What will be the enthalpy of formation of $NO_2$ from the given bond dissociation enthalpy values ? Vapour pressure of liquid ‘S’ at 500 K is approximately equal to : $\ce{(R=2 \, cal \, K^{-1} \, mol^{-1})}$, Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas ? Using Hess's Law to Calculate the Change in Enthalpy of a Reaction, Standard Enthalpy of Formation: Explanation & Calculations, Free Energy: Predicting the Spontaneity of a Reaction, Equilibrium Constant (K) and Reaction Quotient (Q), Enthalpy: Energy Transfer in Physical and Chemical Processes, Chemical Thermodynamics: Definition & Principles, Reaction Mechanisms and The Rate Determining Step, Spontaneous Process: Definition & Examples, Lattice Energy: Definition, Trends & Equation, Gibbs Free Energy: Definition & Significance, Bond Enthalpy: Definition, Calculations & Values, The Diagonal Relationship, Metallic Character, and Boiling Point, Calorimetry: Measuring Heat Transfer and Heat Capacity, LeChatelier's Principle: Disruption and Re-Establishment of Equilibrium, Molar Heat of Combustion: Definition & Calculations, UExcel Microbiology: Study Guide & Test Prep, Holt McDougal Modern Biology: Online Textbook Help, NY Regents Exam - Physics: Help and Review, CSET Science Subtest I - General Science (215): Practice & Study Guide, High School Physical Science: Help and Review, MTTC Integrated Science (Secondary)(094): Practice & Study Guide, National Eligibility Test (AIPMT): Study Guide, National Board Certification Exam - Science/Adolescence & Young Adulthood: Practice & Study Guide, UExcel Anatomy and Physiology II: Study Guide & Test Prep, Ohio Assessments for Educators - Earth & Space Science (014): Practice & Study Guide, Anatomy and Physiology: Certificate Program, Biological and Biomedical | 1,331 | 5,291 | {"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-2021-17 | latest | en | 0.758598 |
https://www.reddit.com/user/architect551 | 1,475,169,603,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661905.96/warc/CC-MAIN-20160924173741-00037-ip-10-143-35-109.ec2.internal.warc.gz | 991,679,068 | 21,540 | [–] 8 points9 points (0 children)
The Venus de Milo actually does have a crack around there. It was found in two pieces, the torso and legs, which were reattached.
[–] -1 points0 points (0 children)
It's not really the bluntness of the needle that you should care about, it's the fact that now it is not sterile.
[–] 1 point2 points (0 children)
Had one too many
[–] 0 points1 point (0 children)
Have a look at the outline of the maths bridging course ANU offers and try to base what you work on from that:
http://programsandcourses.anu.edu.au/course/ASGS1025
They use “Algebra & Trigonometry”, Sullivan 8th Edition so that might be a good place to start, although you'll need something else for calculus.
[–] -1 points0 points (0 children)
Try:
``````(Prime[#] & /@ Range[2, 10^4]) - (Prime[#-1] & /@ Range[2, 10^4])
``````
This runs quicker, still about half as slow as the first example
Edit: formatting
[–][S] 0 points1 point (0 children)
Will i be able to use a 580 for gaming at all?
[–][S] 0 points1 point (0 children)
Yea, I figured it's a stretch. Is it pretty much a bad idea to skimp any more than i have?
[–] 39 points40 points (0 children)
But what about the acceleration of the Earth towards the object where a =Gm/r2 dependant on the mass of the object.
[–] 0 points1 point (0 children)
The quantum cascade lasers they use go for a couple of thousand dollars, plus i believe they used liquid nitrogen cooled detectors which arn't exactly cheap or easy to use. Not something you can build in your garage just yet, we'll need a cheaper source of infra-red lasers first.
[–] 19 points20 points (0 children)
Ok, so 70 years is a long time. I'll assume that you drop the rock from the top of the hole which it as the surface of the Earth and that air resistance is not a problem. Since 70 years is a long time the rock will fall far enough that g will not be equal to ~9.8 as at the surface, so we must use the universal gravitational law. I'll also assume the Earth is uniform density. As the rock falls it will only feel a force due to gravity for any mass that is closer the centre of the Earth than itself. This gives us the force due to gravity of G r mR mE/rE3/2 where G is the universal gravitational constant, mR is the mass of the rock, mE is the mass of the Earth and rE is the radius of the Earth. Using newton's second law we then get: m a=G r mR mE/rE3/2 which we can solve as a differential equation to get a harmonic solution Cos[(G mE/rE3)1/2 t] . The first thing we notice is the time for this function to get to zero is 1/2 pi (R3/G mE)1/2 and plugging in the numbers thats comes out to be 21 minutes. This means that the rock will take 21 minutes to reach the centre of the Earth. Now the rock will keep going, but will decelerate and eventually turn around at the surface of the other side of the Earth and go back through the centre of the Earth all the way to the top of the hole where you dropped it. This will happen 435 000 times before 70 years have passed. So your hole must be deep enough to come completely out the other side of the Earth... oh and btw free fall is completely independent of the mass of the rock (neglecting air resistance)
Edit: because numbers
[–] 1 point2 points (0 children)
The price for hops doesn't mention the quantity?
[–] 1 point2 points (0 children)
If you try to serve a keg at room temperature you'll just get head
[–] 0 points1 point (0 children)
The video of someone being struck twice is a fake.
[–] 0 points1 point (0 children)
Care to explain how you did it?
Potentially moving to Canberra by in canberra
[–] 0 points1 point (0 children)
Choosing ANU over Melbourne will depend on the subject area and your proposed career path
[–] 0 points1 point (0 children)
A few of the imported beers you have are brewed under license in australia
[–] 3 points4 points (0 children)
The north-south pipeline planned to take water from the murray-darling basin, which also happened to be in drought at the time. You can understand the opposition when the city folk wanted to take water from a region with towns also on water restrictions.
[–] 1 point2 points (0 children)
The north was in drought too.
[–] 3 points4 points (0 children)
How about using quantum vacuum fluctuations to generate random numbers: http://qrng.anu.edu.au/
[–] 2 points3 points (0 children)
Plenty of papers around at the moment that suggest an altered theory of gravity can explain dark matter
[–] 1 point2 points (0 children)
Plonk, also in fyshwick, have them for \$40, they post orders. Not sure of the rates but give them a try. | 1,238 | 4,637 | {"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} | 3.015625 | 3 | CC-MAIN-2016-40 | latest | en | 0.928761 |
http://mathandmultimedia.com/2010/06/25/spreadsheet-and-similarity/ | 1,721,323,274,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00015.warc.gz | 14,632,007 | 14,724 | # GeoGebra Tutorial 21 – Spreadsheet and Similarity
This is the 21st tutorial of the GeoGebra Intermediate Tutorial Series. If this is your first time to use GeoGebra, you might want to read the GeoGebra Essentials Series.
In this tutorial, we investigate what happens if we multiply the coordinates of the vertices of a triangle with a constant. First, we plot three points that will be the vertices of a triangle, and then draw the triangle using the Polygon tool. Next, we will construct slider k, and see what happens if we multiply the coordinates (x1,y1), (x2,y2) and (x3,y3) of the vertices of the triangle by k. We also explore the relationship between the original triangle and the triangles with coordinates (kx1,ky1), (kx2,ky2) and (kx3,ky3).
You can view the output of this tutorial here.
Step-by-Step Instructions | 213 | 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} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.746197 |
https://economics.stackexchange.com/questions/15406/formal-definition-of-perfect-information | 1,725,847,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00394.warc.gz | 193,927,720 | 42,030 | # Formal definition of perfect information
I am looking for the formal definition of 'perfect information' in game theory.
Please direct me to a book or preferably an online paper where I can find it.
On a related note:
The Wikipedia page for the term is not very useful. It only offers an informal definition:
In game theory, an extensive-form game has perfect information if each player, when making any decision, is perfectly informed of all the events that have previously occurred
Given this definition the simultaneous move examples seem strange. The games mentioned (e.g. iterated prisoner's dilemma) could easily be altered to have sequential moves where the second mover is simply not aware of the first move. This game would have the same extensive form but would no longer fit the informal definition.
# Osborne and Rubinstein (1994)
Osborne and Rubinstein's textbook A Course in Game Theory defines (extensive) games with perfect information in three versions.
• Basic version (Def. 89.1): perfect information is the same as requiring information sets be singletons (though they don't put it this way). In the author's language, perfect information is modeled as a player function $P:H\to N$, mapping each non-terminal history $h\in H$ of previous moves to a single member in the set of players $N$. A history here is $h=(a^k)_{k=1,\dots,K}$, where $a^k$ is the action taken by the player who moves in the $k$th round, and $K$ is possibly infinite.
• Extended version 1 (perfect information with chance-moves, Sect. 6.3.1): perfect information (for players) here is basically the same as before, be the definition incorporates uncertainty in the game due to chances
• Extended version 2 (perfect information and simultaneous moves, Sect. 6.3.2): perfect information here is modeled as a player function $P$ mapping each non-terminal history to a set of players, where
[a] history in such a game is a sequence of vectors; the components of each vector $a^k$ are the actions taken by the players whose turn it is to move after the history $(a^\ell)_{\ell=1}^{k-1}$. The set of actions among which each player $i \in P(h)$ can choose after the history $h$ is $A_i(h)$; the interpretation is that the choices of the players in $P(h)$ are made simultaneously.
# Myerson (1991)
Myerson's Game Theory: Analysis of Conflict similarly defines (on page 185) an extensive form game with perfect information as information sets being singletons within each information state.
# Fudenberg and Tirole (1991)
Fudenberg and Tirole's Game Theory textbook defines perfect information informally (on page 72) as follows
We say that multi-stage game has perfect information if, for every stage $k$ and history $h^k$, exactly one player has a nontrivial choice set --- a choice set with more than one element --- and all the others have the one-element choice set "do nothing."
A simultaneous-move game is not a game of perfect information. It is a game of imperfect information.
Let me quote Gibbons (Chapter 2, p.58).
The key features of a dynamic game of complete and perfect information are that (i) the moves occur in sequence, (ii) all previous moves are observed before the next move is chosen, and (iii) the players' payoffs from each feasible combination of moves are common knowledge.
Picking up the discussion in the comments, let me again quote Gibbons, p.122, Footnote 19:
This characterization of perfect and imperfect information in terms of singleton and nonsingleton information sets is restricted to games of complete information because, as we will see in Chapter 4, the extensive-form representation of a game with perfect but incomplete information has a nonsingleton information set. In this chapter, however, we restrict attention to complete information.
• The problem with this is that it also says "dynamic game" where some moves will occur in sequence. Thus the sequential move part may not a property implied by perfect information. According to this the iterated PD example on Wikipedia is incorrect though. Commented Feb 10, 2017 at 15:31
• Seems to me that if I were to take away the dynamic part I would basically be left with 'perfect information means singleton information sets'. Is that right? Commented Feb 10, 2017 at 15:37
• Yes, perfect information means all info sets are singletons. Commented Feb 10, 2017 at 15:38
• en.wikipedia.org/wiki/Extensive-form_game (CTRL+F "singleton") - but the examples in the other wiki article you shared are indeed confusing. "Information sets are singletons" is the version I teach. I check for a "higher authority" later today. Commented Feb 10, 2017 at 15:44
• I corrected my statement, sorry for that. I thought about complete info as well. Commented Feb 10, 2017 at 16:03 | 1,079 | 4,763 | {"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} | 3.0625 | 3 | CC-MAIN-2024-38 | latest | en | 0.931383 |
https://www.chess.com/forum/view/suggestions/what-does-quotmate-in-2quot-or-quotmate-in-3quot-really-mean-on-tt | 1,529,479,559,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863489.85/warc/CC-MAIN-20180620065936-20180620085936-00451.warc.gz | 793,230,235 | 18,233 | # What does "Mate in 2" or "Mate in 3+" really mean on TT?
Consider this problem:
This is not a mate in anything (at least not less than 10 moves), yet it is tagged as such. Moreover, the "definition" of mate in 3+ given in the help topics is:
• Any puzzle/position that ends with mate in three or more moves.
This is ambiguous at best. Does it mean merely that the puzzle (including the opponent moves) happens to end in mate, or does it mean the usual definition, i.e. it is a forced mate? I think it is very confusing and deceptive to tag this a "mate in 3+" problem.
Also, no one has commented on a previous thread I started asking why TT does not include castling possibilities, which can have an effect on the correct solution to a problem:
A tag may be wrongly given (I didn't check the position), but why is the tag itself ambiguous? "mate in n+ moves" means that there is a forced mate, but it takes at least n moves.
The tag is ambiguous because it does not include the word "forced". Technically, if you take the definition as it stands, this is a mate in 3, because it is a puzzle that ends with mate, and is 3 moves long. The definition should say,
• Any puzzle/position that leads to a forced mate in 3 or more moves
You might say this is implied in the terminology, but not everyone may know this, and this might be the reason for incorrect tags.
Another reason I mention this is because there was a "puzzle of the day" the other day titled "Mate in 2" that was not, and this was pointed out, and the title never changed, and no staff even bothered to comment on the thread, "our bad, you're right, it's not mate in 2", which leads me to wonder what their definition actually means.
Yes, I'd say this is semantically implied by terminology as "mate in n" in puzzle books always means "forced mate in n".
I concede that there's a problem with people setting incorrect tags, but this is a problem throughout the web.
Agree that "mate in 3+" is so open-ended as to be meaningless (unless perhaps synonymous with "eventually winning"). Another complaint with your reference problem is that the "loser" does not play optimally after the "winner's" first move, which is a common problem in TT, and often done to illustrate a tactical point. I think a more general point (or question) is what is the practical or learning purpose of tags? If seen before doing the puzzle, they are unfair hints. If seen after doing the puzzle, they are trivial.
I think I have seen TTs that involve castling. Perhaps a tag saying whether or not castling is still permissible would be useful.
This really cleared things up for me. I thought mate in 3 was a threesome and mate in 3+ was an orgy.
Escapest_Pawn said: "Another complaint with your reference problem is that the 'loser' does not play optimally after the 'winner's' first move, which is a common problem in TT, and often done to illustrate a tactical point."
I don't have a problem with this so much, as long as the less than optimal play is not too far short of optimal.
"I think a more general point (or question) is what is the practical or learning purpose of tags? If seen before doing the puzzle, they are unfair hints. If seen after doing the puzzle, they are trivial."
I thought maybe they use the tags to collect statistics for people so they can see what types of tactics they do well on, and which they need to improve?
I thought maybe they use the tags to collect statistics for people so they can see what types of tactics they do well on, and which they need to improve?
This may well be the case and such had not occurred to me.
I just looked at the puzzle. It is in fact mate... Im not sure what the issue is. Sometimes the puzzle does not show the last move or two as there are 2 equally good moves that lead to mate in the same amount of moves, so the puzzle ends instead of giving an "alternate correct", but this is not one of them.
I am kind of confused- did you not see the pawn on b3? For the most part the tags are correct. If you feel they are incorrect, use the "analysis board" and look at the lines showed. Play the moves and see what it shows. Then give a thumbs down to tags you KNOW are incorrect. If you are not sure, do not change them.
Kphlash,
The analysis showed by the chess.com engine does not show mate:
• ({12:+5.56} 40. Rb8 Nb1 41. Rxb6+ Kxb6 42. Qxb4+ Ka7 43. Qd6 Rb7 44. Qxe6 Nc3 45. Kg4 a5 46. Kf5 Rxb3 \$18)
I also plugged the position into Houdini and it gave about +6 eval after about 10 moves, but no forced mate. Mate seems only to be forced following the blunder 40. ... Rb7 ??. Of course, maybe the analysis is wrong, and I entered in the position incorrectly on Houdini. If you have a proof of the mate, I'd be interested to see it.
I can't vote on tags, because I don't have a premium membership.
The puzzle itself is a mate in 3 due to the blunder Rb7 that allows for the Qc5#. If the person played more accurately (which is a complaint I hear alot "why didnt XX play XX instead") the puzzle wouldnt be possible.
All those other lines are generally better than the ones played (the ones played are mistakes, in order to present tactics- Rb7 in this case). If the other moves are played after Rb8 like Nb1, which is the only move shown in the analysis at that point (also the one you posted) white would be up 5.5pts roughly- houdini and the chess.com analysis are correct.
This is the full analysis sheet. Correct? You posted what black should have done after Rb8, just below that, is what actually happened. The computer did not analyze that line because Rb7 is a stupid move, but that stupid move is the point of the puzzle.
({12:+5.18} 39. ... Kb5 40. Rb8 Nc2 41. f5 exf5 42. Rxb6+ Kxb6 43. Qxf5 Nxd4+ 44. exd4 Rc3+ 45. Kf4 Rxb3 46. Qe6+ Kc7 47. Qxd5 \$18)
({12:-0.01} 39. ... Qa5 40. Qe8+ Kb7 41. Rb8+ Ka7 42. Ra8+ Kb6 43. Qb8+ Kc6 44. Qe8+ Kb6 \$10)
({12:+4.32} 39. ... Qa7 40. Ra8 Qb7 41. Qe8+ Kb6 42. Rb8 Qxb8 43. Qxb8+ Kc6 44. Qe8+ Kd6 45. f5 exf5 46. Qd8+ Kc6 47. Qf6+ \$18)
40. Rb8
({12:+5.56} 40. Rb8 Nb1 41. Rxb6+ Kxb6 42. Qxb4+ Ka7 43. Qd6 Rb7 44. Qxe6 Nc3 45. Kg4 a5 46. Kf5 Rxb3 \$18)
({12:+2.58} 40. f5 Ka5 41. Rb8 Rb7 42. Rxb7 Qxb7 43. fxe6 Nb5 44. e7 Nc7 45. e8=Q Nxe8 46. Qxe8 \$18)
({12:+0.01} 40. Rd6 Rc6 41. Rd8 Rc7 \$10)
40. ...Rb7
41. Rxb7 Qxb7
42. Qc5++
Basically, the tags show that the puzzle is a mate in 3 because, due to the mistakes made by your opponent, it is a mate in 3. not what it would have been with better play. Also, in regards to the castling, There are a few puzzles where castling is the solution.
I think the main point here, which Retgruvie is pretty famous for stating on those puzzles "do not worry about the moves you opponent makes, or what they could have done differently/better, the point of these puzzles is to recognize a mistake and capitalize on it the best possible way" sometimes, it is a piece capture, sometimes mate, and sometimes it is a simple defensive move because they missed a golden opportunity.
Kphlash,
I think you misinterpret my complaint. My complaint is not that black played less than optimally. My complaint is that the tag is incorrect.
"The puzzle itself is a mate in 3 due to the blunder Rb7 that allows for the Qc5#."
Except this completely re-defines what the traditional definition of "mate in 3" is. "White mates in 3" means, from the original position of the puzzle, white can force mate in 3 moves, but not 2. The term does not mean, "white can force mate in 3 moves, at some future unspecified position of some variation, provided black blunders or plays some particular move."
It is true that with the less than optimal play of black that presents itself in this interactive puzzle, a position is reached which is mate in 3. But if this puzzle were presented in a chess column in a newspaper with the heading "White to move and mate in 3", there would be many letters to the editor correcting them.
So I have a question. If a particular TT puzzle has less than optimal play that blunders into a mate in 1, is it correct to tag this puzzle with "mate in 1"? Is the starting position, after 1. f3, a "mate in 1" puzzle for black, because black can play 1. ... e5 and white has the opportunity to blunder into fool's mate??
This really underscores the fact that the "definitions" given in the help feature are ambiguous. Some people say the definition means what I think it means, and the tag is wrong, and someone else says the definition means something else, and the tag is correct. Obviously someone is mistaken.
It would be nice if some clarification were provided by staff, so that we all knew what we were talking about, and so that we knew if 99% of chess players in the world had to revise their definition of "mate in n" when looking at TT tags.
"If the person played more accurately (which is a complaint I hear alot 'why didnt XX play XX instead') the puzzle wouldnt be possible."
No, the puzzle would be possible, it just wouldn't be this particular puzzle. In fact, it should be possible to add another puzzle to TT that does continue with the best black response (which I believe is actually 40... Qxb8) although that puzzle would be much easier to solve, since 41. Qxb8 is easier to see. Who knows, that puzzle might already exist in TT.
Please understand me, I know why the blunder 40... Rb7 was chosen as a line... it makes a more interesting puzzle! But it does not make it a "mate in 3" puzzle. The most that can be said is that the puzzle follows a line that leads to a "mate in 3" position, but that is not what most chess players (at least not those who grew up learning tactics puzzles in actual books, not online) would understand as a "mate in 3" puzzle.
"Also, in regards to the castling, There are a few puzzles where castling is the solution."
That wasn't my concern. My concern was that there are tactics problems where whether castling is possible or not affects the outcome of the analysis. This is why all chess engines have options to check off for castling when entering a position.
Gotcha. Yes, I'm aware the puzzle is mistagged as it is not a forced mate. I thought you said the puzzle didn't end in mate (and just missed it).
The tags are often incorrect, but you cannot see the tags until after you solve the puzzle (unless there is an option I missed) so, even though the tag is not correct in a true puzzle sense, it is more of a hint to those who originally solved it incorrectly.
This "mistagging" as well as the castling issue you are concerned about seem to be solved with the same answer- It is the way the puzzle is designed.
The puzzles are designed for you to make the best move, regardless of the opponent. I forget the exact formula, but one of the Moderators explained it once- they are coded, the lower numbers (not rating) the 1st move is a mistake by your opponent, the higher numbers, the 2nd move is the mistake. All the puzzles are based on a mistake.
Just do the puzzles the best you can, and dont over analyze the tags- play them as if it were a real game.
If you have a suggestion or a comment, go here http://support.chess.com/ and contact the staff in the bottom left hand corner.
Hope that helps solve your problem better.
Kphlash,
I'll do that. The castling issue is not addressed by just saying "that's the way the puzzles are designed", though. Whether castling is allowed or not can change what the solution is. Not providing such information can make it impossible to determine the best move. | 3,005 | 11,540 | {"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} | 2.765625 | 3 | CC-MAIN-2018-26 | latest | en | 0.979075 |
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Computer Science 172 - Spring 1999 - Blum - Midterm 2
# Computer Science 172 - Spring 1999 - Blum - Midterm 2 -...
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CS172, Midterm 2, Spring 1999 CS 172, Spring 1999 Midterm Exam 2 Manual Blum This is a CLOSED BOOK examination. Calculators ARE permitted. Do all your work on the pages of this examination. For most problems, you may answer YES, NO, "I think so", "I think not", or "I don't know". Give reasons -- carefully written correct reasons -- for your YES/NO answers, and to the extent you can for your "I think such and such" answers. This examination has 7 problems, worth a total of 97 points. Problem #1 (1 pt) a) The product of binary numbers 1011 and 1101 is 1011* 1101 = _______________________________________. (10 pts: 1 pt each; double this if all are correct) b) In spaces to the right, write YES or NO (or "I think so" or "I think not" or "I don't know"). The length (in bits) of the product of an m-bit number and an n-bit number, for positive integers m and n, is: O(mn) bits_____ O(m+n) bits_____ m+n+O(1) bits _____ MAX{m,n} bits _____ min{m,n} bits _____ file:///C|/Documents%20and%20Settings/Jason%20Raft...-%20Spring%201999%20-%20Blum%20-%20Midterm%202.htm (1 of 4)1/27/2007 6:45:49 PM
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CS172, Midterm 2, Spring 1999
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https://mathoverflow.net/questions/tagged/computable-analysis | 1,708,874,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00170.warc.gz | 384,622,732 | 28,379 | # Questions tagged [computable-analysis]
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### Integration in polynomial time
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### Computability of fillability of unit cube in $\mathbb{R}^n$ by $k$ $\varepsilon$-balls
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### Are representations in computable analysis the equivalent to countably-generated condensed sets?
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### Computing the complex roots of a monic polynomial
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### Uncomputability of the identity relation on computable real numbers
Let $f_{=}$ be a function from $\mathbb{R}^{2}$ be defined as follows: (1) if $x = y$ then $f_{=}(x,y) = 1$; (2) $f_{x,y} = 0$ otherwise. I would like to have a proof for / a reference to a textbook ...
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Call a computable function a total function $\mathbb{R} \to \mathbb{R}$, for which there exists a Turing machine outputting arbitrary close approximation to $f(x)$ given arbitrary close approximation ... | 953 | 3,560 | {"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} | 2.640625 | 3 | CC-MAIN-2024-10 | latest | en | 0.819645 |
https://www.physicsforums.com/threads/r-circuit-pretty-simple-need-checked-full-working-please.417093/ | 1,508,308,257,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822747.22/warc/CC-MAIN-20171018051631-20171018071631-00375.warc.gz | 979,269,137 | 15,625 | # R circuit (pretty simple) need checked - full working - please
1. Jul 19, 2010
### pat666
R circuit (pretty simple) need checked - full working - please!!!
1. The problem statement, all variables and given/known data
see attachment
2. Relevant equations
P=IV
P_loss=I^2 R
V=IR
3. The attempt at a solution
ok
36=I^2*12
I_12Ω=√3 A
V=20.8V
That is the same for the 3 resistors in parallel (I hope)
so I18Ω=1.15A
I36Ω=0.577A
the sum of these will give me the current through the unknown resistor
Iunknown=3.46A
net voltage = 60
60=Vunknown + 20.8
Vunknown=39.2V
V=IR
R=11.3Ω
can someone please check this - wouldn't ask if I dint need it.
#### Attached Files:
• ###### resistors.png
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2. Jul 19, 2010
### rl.bhat
Re: R circuit (pretty simple) need checked - full working - please!!!
3. Jul 19, 2010
### pat666
4. Jul 19, 2010
### pat666
Re: R circuit (pretty simple) need checked - full working - please!!!
Sorry, I forgot the second half of the question,
P_loss=I^2 R
P=〖3.46〗^2*((1/12+1/18+1/36)^(-1)+11.3)
P_loss=207.6W
This seems big to me, could you please check that too.
5. Jul 20, 2010
### rl.bhat
Re: R circuit (pretty simple) need checked - full working - please!!! | 423 | 1,224 | {"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-2017-43 | longest | en | 0.825777 |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-7th-edition/chapter-1-chemical-tools-experimentation-and-measurement-section-problems-page-30/57 | 1,685,718,191,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00597.warc.gz | 862,340,004 | 12,175 | ## Chemistry (7th Edition)
(1) $7.926\times 10^3$ $mi$ (2) $7.9 \times 10^3$ $mi$
(1) Four significant figures: In 7926.381, the fourth significant figure is 6, so: 7926 is the rounded number. Now, we have to check the next number, "3"; since it is less than 5, we leave the number as 7926. Scientific notation: 1. Move the decimal point to put the number between 1 and 10 7.926 2. Now, multiply the number by 10 to the power of the number of times you moved the decimal point. If you moved to the right, the number is negative. Since we moved the decimal point by 3 times to the left: $7.926 \times 10^3$ $mi$ --------------- (2) Two significant figures: In 7926.381, the second significant figure is 9, so: 7900 is the rounded number. Now, we have to check the next number: "2". Since it is less than 5, we leave the number as 7900. Scientific notation: 1. Move the decimal point to put the number between 1 and 10 7.9 2. Now, multiply the number by 10 to the power of the number of times you moved the decimal point. If you moved to the right, the number is negative. Since we moved the decimal point by 3 times to the left: $7.9 \times 10^3$ $mi$ | 346 | 1,151 | {"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} | 3.9375 | 4 | CC-MAIN-2023-23 | latest | en | 0.888618 |
https://branemrys.blogspot.com/2012/08/euclids-common-notions.html | 1,627,377,877,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153223.30/warc/CC-MAIN-20210727072531-20210727102531-00599.warc.gz | 154,845,175 | 31,836 | ## Saturday, August 25, 2012
### Euclid's Common Notions
I was thinking about Euclid's common notions yesterday, and in particular about what we get if we focus only on them. The common notions, or axioms, if you will remember, are (in usual translation, going back to Heath):
(1) Things which are equal to the same thing are also equal to one another.
(2) If equals are added to equals, then the wholes are equal.
(3) If equals are subtracted from equals, then the remainders are equal.
(4) Things which coincide with one another equal one another.
(5) The whole is greater than the part.
What do we get if we look only at this? A very simple mereology, that is, a system of reasoning about parts and wholes. Axiom 5 tells us that a whole is distinguishable from its parts, and vice versa. The other four axioms tell us what we can conclude if parts or wholes are indistinguishable (indiscernible/equal/exactly similar) in some way. Axiom 1 tells us how things relate to each other when they are both indistinguishable from the same thing (the answer is: indistinguishably!); it's about equality with respect to equality. Axiom 2 is about equality with respect to addition; in other words, they tell us something we can conclude about the wholes when parts are joined to other parts. Axiom 3 is about equality with respect to subtraction; that is, they tell us something we can conclude about leftover parts when we remove parts from a whole.
Axiom 4 is interesting. The Greek means something like "Things that fit onto each other are alike/equal"; it tells us that if you can fit something (part or whole) onto something (part or whole) that they are indistinguishable. So this is about equality with respect to superposition, whether of parts or wholes. Axioms 2, 3, and 4, then, allow for three kinds of mereological operation: joining part to part in order to get equal wholes (addition), removing parts from wholes in order to get equal parts (subtraction), and fitting part or whole on top of part or whole in order to get equals (superposition).
It should be noted that in none of these cases do the parts and wholes actually have to be spatial parts and wholes. For instance, here is a pretty straightforward case of something that can be fit to the same mereological principles: propositions (wholes) and terms (parts). If intersubstitutable terms are predicated of intersubstitutable terms (addition), the propositions are intersubstitutable; if intersubstitutable terms are removed from intersubstitutable propositions, the leftover terms are intersubstitutable; substitution is 'coinciding' or 'fitting on top of'; propositions include more than their terms; and things intersubstitutable with the same thing are themselves intersubstitutable.
We get something geometrical in our usual sense only when we add definitions and postulates. The definitions give us kinds of objects in space, and properties of those objects; the postulates are guidelines for accurately constructing those same objects (they are things you are asked to do when drawing diagrams -- for instance, the first postulate literally asks you to bring your strokes directly from marker to marker, while the fourth asks that all your upright corners be alike). Any Euclid-style approach to geometry, then, has three elements: definitions of kinds of spatial objects and their properties; a method for constructing spatial objects; and a mereology, or system for reasoning in terms of parts and wholes. We can get variations by changing any of these. We could change the definitions; and in a sense a part of Newtonian physics is Euclid where our definitions are not of kinds of spatial objects but kinds of locomotion. We could change the methods of construction, and this is where we get all sorts of different geometries, ranging from Euclidian geometry with neusis to Riemann and others. And we could also change the mereology.
ADDED LATER
Thinking more about this, there are famously three Books of Euclid's Elements that are independent of any other book: Book I, Book V, and Book VII. These all share the same common notions (mereology) and postulates (method of construction), but they differ according to their definitions. The geometry of Book I is a comparative mereology of dimensional constructions -- dimensional here being what I meant above by 'spatial', and meaning 'having to do with length, breadth, etc.'. The definitions of Book V have to do with measured magnitudes (a magnitude is part of another when it measures the other); and those of Book VII are concerned with measured numbers (a number is part of another when it measures the other). So even within Euclidean geometry we see the way in which definitions establish the objects that are reasoned about with a simple mereology and a method of construction, in order to get a geometrical theory.
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https://forum.philosophynow.org/viewtopic.php?f=26&p=324443 | 1,519,281,105,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814036.49/warc/CC-MAIN-20180222061730-20180222081730-00666.warc.gz | 662,343,275 | 11,071 | ## help understanding Negation Introduction
What is the basis for reason? And mathematics?
Moderators: AMod, iMod
ProfAlexHartdegen
Posts: 20
Joined: Mon Jul 31, 2017 5:43 pm
### help understanding Negation Introduction
Okay, as I wrote in another thread I am reading a book called the Logic Book by Bergmann.
There are a few concepts which again I don't understand the point about which include the use of subderivations.
First, does the forum offer some way to type the symbols used in symbolic logic? Okay here is the problem.
Sorry for the crude representation but as you know the forum isn't conducive to writing symbolic logic. The subderivation is offset a bit to clarify what it is:
For :
Negation Introduction (~I)
|............| P____
|............|
|............| Q
|............|
|............| ~Q
|
|....... ~P
They give an English example:
premise (or statement?) 1: If the union votes for the contract and the management agrees to the terms of the contract, then the contract will be signed.
premise (or statement?) 2: The management agrees to the terms of the contract, but the contract will not be signed.
Conclusion?: The union does not vote for the contract.
They give a symbolized version of the argument but I'm experiencing trouble understanding how the argument works. I can't understand the use of each individual rule. I mean I know you could apply that rule but I don't see why you use the rule. The 'plan' or the 'forest from the trees' if you will.
Derive: ~U
1| (U & M) ⊃ S.................Assumption
2| M & ~S_____..............Assumption
..|
3|......|U_________..............Assumption
4|......| M.....................2 &E
5|......| U & M................3,4 &I
6|......|S.......................1,5 ⊃E
7|......|~S......................2 &E
8| ~U............................3-7 ~I
I'm sorry but I just don't see how these arguments; especially lines lines 4,5,6 and 7 lead to line 8! Can someone help me to understand? thank you so much!
Last edited by ProfAlexHartdegen on Sat Aug 12, 2017 6:58 pm, edited 1 time in total.
Harbal
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Contact:
### Re: help understanding Negative Introduction
Have you allowed for the possibility of no one having a pen?
ProfAlexHartdegen
Posts: 20
Joined: Mon Jul 31, 2017 5:43 pm
### Re: help understanding Negation Introduction
I understood the joke but I'm sure it will be more humorous after I understand the Logic .
Let's try rephrasing the argument into standard English.
The management agrees to the terms of the contract.
The union votes for the contract and the management agrees to the terms of the contract.
The contract will be signed.
The contract will not be signed.
The union does not vote for the contract.
I'm sorry I just don't understand how the argument proceeds from one line to the next or how they're linked and why.
Last edited by ProfAlexHartdegen on Sat Aug 12, 2017 6:59 pm, edited 1 time in total.
ProfAlexHartdegen
Posts: 20
Joined: Mon Jul 31, 2017 5:43 pm
### Re: help understanding Negation Introduction
Okay, I watched a video about Negation Introduction (and Negation Elimination). Here: https://www.youtube.com/watch?v=fqows_3Dwac
Is the point to take any lines derived from the Assumption that contradict to show that the negative of the assumption can be derived? That seems comprehensible to me because you do that with the Truth-Trees, which close when there is a contradiction in an atomic sentence.
Dubious
Posts: 1506
Joined: Tue May 19, 2015 7:40 am
### Re: help understanding Negation Introduction
ProfAlexHartdegen wrote:
Fri Aug 11, 2017 6:44 am
I understood the joke but I'm sure it will be more humorous after I understand the Logic .
Let's try rephrasing the argument into standard English.
The management agrees to the terms of the contract.
The union votes for the contract and the management agrees to the terms of the contract.
The contract will be signed.
The contract will not be signed.
The union does not vote for the contract.
I'm sorry I just don't understand how the argument proceeds from one line to the next or how they're linked and why.
Being miserably poor in the art of logic...as was recently pointed out, take the following for what it's worth though I hesitate to ask what that would amount to.
I think your rephrasing would make more sense if more formally given:
IF management agrees to the terms of the contract....and
IF the union votes for the contract
The contract will be signed
The contract will not be signed
I think it's misleading, confusing and redundant to again repeat in the second clause what was already given in the first.
Anyways, as IF statements we don't know whether either premise is true or not. If ONE is negative the contract will not be signed. In that sense the argument is valid since both conclusions are valid.
Harbal
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Location: Yorkshire
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### Re: help understanding Negation Introduction
I'd be interested to know why the union won't sign the contract.
Dubious
Posts: 1506
Joined: Tue May 19, 2015 7:40 am
### Re: help understanding Negation Introduction
Harbal wrote:
Sat Aug 12, 2017 10:02 pm
I'd be interested to know why the union won't sign the contract.
I'm not a member! You have to ask them. Sound logical?? Maybe its because management want's to screw them!
Harbal
Posts: 3846
Joined: Thu Jun 20, 2013 10:03 pm
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Contact:
### Re: help understanding Negation Introduction
Dubious wrote:
Sat Aug 12, 2017 10:05 pm
Maybe its because management want's to screw them!
In that case, I don't blame them for not signing, I wouldn't put up with my manager screwing me, not even if he offered me a pay increase.
Dubious
Posts: 1506
Joined: Tue May 19, 2015 7:40 am
### Re: help understanding Negation Introduction
Harbal wrote:
Sat Aug 12, 2017 10:12 pm
Dubious wrote:
Sat Aug 12, 2017 10:05 pm
Maybe its because management want's to screw them!
In that case, I don't blame them for not signing, I wouldn't put up with my manager screwing me, not even if he offered me a pay increase.
Logically what's TRUE for you can be an either/or situation for someone else. Logic confirms it!
Harbal
Posts: 3846
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Location: Yorkshire
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### Re: help understanding Negation Introduction
Dubious wrote:
Sat Aug 12, 2017 10:17 pm
Harbal wrote:
Sat Aug 12, 2017 10:12 pm
In that case, I don't blame them for not signing, I wouldn't put up with my manager screwing me, not even if he offered me a pay increase.
Logically what's TRUE for you can be an either/or situation for someone else. Logic confirms it!
The only thing that logic confirms is that you aren't employing any.
Dubious
Posts: 1506
Joined: Tue May 19, 2015 7:40 am
### Re: help understanding Negation Introduction
Harbal wrote:
Sat Aug 12, 2017 10:23 pm
Dubious wrote:
Sat Aug 12, 2017 10:17 pm
Harbal wrote:
Sat Aug 12, 2017 10:12 pm
In that case, I don't blame them for not signing, I wouldn't put up with my manager screwing me, not even if he offered me a pay increase.
Logically what's TRUE for you can be an either/or situation for someone else. Logic confirms it!
The only thing that logic confirms is that you aren't employing any.
Prove it! Show me where I went wrong...unless you're just trying to be mean and nasty.
Dubious
Posts: 1506
Joined: Tue May 19, 2015 7:40 am
### Re: help understanding Negation Introduction
Dubious wrote:
Sat Aug 12, 2017 10:41 pm
Harbal wrote:
Sat Aug 12, 2017 10:23 pm
Dubious wrote:
Sat Aug 12, 2017 10:17 pm
Logically what's TRUE for you can be an either/or situation for someone else. Logic confirms it!
The only thing that logic confirms is that you aren't employing any.
Prove it! Show me where I went wrong...unless you're just trying to be mean and nasty.
I'm still waiting!
You're the one who made the assertion. Maybe you're right. So when you make a gratuitous statement that wasn't asked for categorically denying the validity of someone's argument it's up to that person to come up with one which effectively counters it. I wouldn't in the least mind if you or anyone else could show what logic errors there may have been in what I wrote.
So again, what's your reason for saying The only thing that logic confirms is that you aren't employing any.?
wtf
Posts: 405
Joined: Tue Sep 08, 2015 11:36 pm
### Re: help understanding Negation Introduction
ProfAlexHartdegen wrote:
Thu Aug 10, 2017 10:59 pm
First, does the forum offer some way to type the symbols used in symbolic logic?
Code: Select all
``````Negation Introduction
|............| P____
|............|
|............| Q
|............|
|............| ~Q
|
|....... ~P
``````
You can line these up with the code tag. It's the </> button on the formatting palette above the Edit window.
I'm not even sure if I formatted your intent correctly. I know logic from using it in math but I never really learned to do these formal logic proofs and this notation always confuses me.
I don't remember if this site supports math markup or not, but you can always copy/paste symbols from http://math.typeit.org/.
ProfAlexHartdegen
Posts: 20
Joined: Mon Jul 31, 2017 5:43 pm
### Re: help understanding Negation Introduction
thanks for the serious reply and the tip. As for the others, I don't appreciate them hijacking my thread. I am trying to learn an academic subject and the notation and discussion is rather challenging for me too. Although I have a decent aptitude for math, advanced mathematics has been difficult for me always.
However, I've been told by several random people that I have a logical mind so I thought this book (the Logic Book) might help me to strengthen my skills in this subject of Logic as well as help to facilitate my understanding of advanced math. Logic is the underlying basis for many forms of persuasion too. As a teenager, I read the Sherlock Holmes stories and so I've been curious always about real Logic. I know there are several variants and this is one of the more modern though traditional types.
I've learned a lot more from this book than I ever would learn from your average professor who was teaching a course about it.
Still there are many points that need clarifying and restating for me to comprehend fully.
Harbal
Posts: 3846
Joined: Thu Jun 20, 2013 10:03 pm
Location: Yorkshire
Contact:
### Re: help understanding Negation Introduction
ProfAlexHartdegen wrote:
Sun Aug 13, 2017 6:32 am
As for the others, I don't appreciate them hijacking my thread.
This place is not exclusively for you, other people are allowed to take part.
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Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
Czasopismo
## Fundamenta Mathematicae
1996 | 150 | 1 | 67-96
Tytuł artykułu
### Locally constant functions
Autorzy
Treść / Zawartość
Warianty tytułu
Języki publikacji
EN
Abstrakty
EN
Let X be a compact Hausdorff space and M a metric space. $E_0(X,M)$ is the set of f ∈ C(X,M) such that there is a dense set of points x ∈ X with f constant on some neighborhood of x. We describe some general classes of X for which $E_0(X,M)$ is all of C(X,M). These include βℕ\ℕ, any nowhere separable LOTS, and any X such that forcing with the open subsets of X does not add reals. In the case where M is a Banach space, we discuss the properties of $E_0(X,M)$ as a normed linear space. We also build three first countable Eberlein compact spaces, F,G,H, with various $E_0$ properties. For all metric M, $E_0(F,M)$ contains only the constant functions, and $E_0(G,M) = C(G,M)$. If M is the Hilbert cube or any infinite-dimensional Banach space, then $E_0(H,M) ≠ C(H,M)$, but $E_0(H,M) = C(H,M)$ whenever $M ⊆ ℝ^n$ for some finite n.
Słowa kluczowe
Kategorie tematyczne
Czasopismo
Rocznik
Tom
Numer
Strony
67-96
Opis fizyczny
Daty
wydano
1996
otrzymano
1995-05-22
poprawiono
1995-11-14
Twórcy
autor
• Department of Mathematics, University of Wisconsin Madison, Wisconsin 53706 U.S.A.
autor
• Department of Mathematics, University of Wisconsin Madison, Wisconsin 53706 U.S.A.
Bibliografia
• [1] A. V. Arkhangel'skiĭ and V. V. Fedorchuk, General Topology I, Basic Concepts and Constructions, Dimension Theory, Springer, 1990.
• [2] A. Bella, A. Hager, J. Martinez, S. Woodward and H. Zhou, Specker spaces and their absolutes, I, preprint.
• [3] A. Bella, J. Martinez and S. Woodward, Algebras and spaces of dense constancies, preprint.
• [4] Y. Benyamini, M. E. Rudin and M. Wage, Continuous images of weakly compact subsets of Banach spaces, Pacific J. Math. 70 (1977), 309-324.
• [5] A. Bernard, Une fonction non Lipschitzienne peut-elle opérer sur un espace de Banach de fonctions non trivial?, J. Funct. Anal. 122 (1994), 451-477.
• [6] A. Bernard, A strong superdensity property for some subspaces of C(X), prépublication de l'Institut Fourier, Laboratoire de Mathématiques, 1994.
• [7] A. Bernard and S. J. Sidney, Banach like normed linear spaces, preprint, 1994.
• [8] M. Džamonja and K. Kunen, Properties of the class of measure separable compact spaces, Fund. Math. 147 (1995), 261-277.
• [9] P. R. Halmos, Lectures on Boolean Algebras, Van Nostrand, 1963.
• [10] T. Jech, Set Theory, Academic Press, 1978.
• [11] K. Kunen, Set Theory, North-Holland, 1980.
• [12] J. Martinez and S. Woodward, Specker spaces and their absolutes, II, Algebra Universalis, to appear.
• [13] J. van Mill, A homogeneous Eberlein compact space which is not metrizable, Pacific J. Math. 101 (1982), 141-146.
• [14] M. E. Rudin and W. Rudin, Continuous functions that are locally constant on dense sets, J. Funct. Anal. 133 (1995), 120-137.
• [15] S. J. Sidney, Some very dense subspaces of C(X), preprint, 1994.
• [6] R. Sikorski, Boolean Algebras, Springer, 1964.
Typ dokumentu
Bibliografia
Identyfikatory | 1,050 | 3,146 | {"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} | 2.515625 | 3 | CC-MAIN-2021-17 | longest | en | 0.508421 |
http://www.topperlearning.com/forums/ask-experts-19/u-had-given-an-example-of-addition-of-vectors-in-the-second-physics-motion-in-a-plane-58491/reply | 1,488,218,663,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501173405.40/warc/CC-MAIN-20170219104613-00276-ip-10-171-10-108.ec2.internal.warc.gz | 620,452,038 | 37,587 |
Question
Tue August 28, 2012 By: Muskan
u said if 2 vectors are added according to vector law of addition then R = A+ B then u gave an example about there addition i e
A = 20 units
# u had given an example of addition of vectors in the second video of this topic i have a question regarding the example
Tue August 28, 2012
in the case of vectors the addition is also dependent upon the angle between the two quantities being added.
The formula for addition of two quantities is
R2=A2+B2+2cos a
here a is the angle between the two vectors A and B
Related Questions
Mon February 20, 2017
# A ball thrown up is caught by the thrower after 4 seconds . How high did it go and with what velocity was it thrown? How far below its highest point was in 3 second after start? acceleration due to gravity is 9.8.
Mon February 20, 2017
| 217 | 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} | 3.015625 | 3 | CC-MAIN-2017-09 | longest | en | 0.980116 |
https://chrishillonline.com/bake/when-a-salt-solution-is-heated-the-temperature-continues-to-rise-even-after-it-begins-to-boil-why.html | 1,638,255,044,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00471.warc.gz | 249,341,474 | 18,878 | # When a salt solution is heated the temperature continues to rise even after it begins to boil Why?
Contents
## Why does the temperature of a boiling liquid remain constant when the heat is continuously being added?
When boiling occurs, the more energetic molecules change to a gas, spread out, and form bubbles. … In addition, gas molecules leaving the liquid remove thermal energy from the liquid. Therefore the temperature of the liquid remains constant during boiling.
## Will salt solution boil at a constant temperature?
The ions form an attraction with the solvent particles that prevents the water molecules from going into the gas phase. Therefore, the saltwater solution will not boil at 100oC. In order for the saltwater solution to boil, the temperature must be raised about 100oC.
## Why does the boiling point of water rise as salt is dissolved in it?
In order for water to boil, its vapor pressure has to equal the pressure of the atmosphere, Giddings said. … When salt is added, it makes it harder for the water molecules to escape from the pot and enter the gas phase, which happens when water boils, Giddings said. This gives salt water a higher boiling point, she said.
## Why does the temperature remain constant?
It is due to the latent heat as the heat supplied to increase the temperature of the substance is used up to transform the state of matter of the substance hence the temperature stays constant. Hence the temperature remains constant as all the heat is used up and no external heat is released or absorbed.
## Why does the temperature of water stay constant physics?
4 Answers. This is because the external pressure is constant (at one atmosphere). If you increase the pressure, e.g. by using a pressure cooker, then the temperature goes up, or likewise if you reduce the pressure the temperature goes down.
## What happens to the salt solution after boiling?
But is this actually true? Adding salt to water is going to do two things to water’s physical properties: it will raise the boiling point and it will lower the specific heat. These two changes actually work against each other. Raising the boiling point will make the water boil slower.
## What is Ebullioscopic constant?
Ebullioscopic constant (Eb) is the constant that expresses the amount by which the boiling point T b of a solvent is raised by a non-dissociating solute. Its units are K Kg mol-1. b is the molality of the solute.
## What is the effect of salt on the boiling temperature of water experiment?
It was found that adding salt to water increases the boiling time of water. The more salt you add, the higher the boiling temperature becomes therefore the solution takes a longer period of time to boil.
THIS IS USEFUL: How impurities affect boiling point?
## At what temperature does salt water boil?
When salt is present, the heat capacity of water decreases slightly. Seawater of 35 psu has a specific heat of 0.932 compared with 1.000 for pure water. Pure water freezes at 0 °C and boils at 100 °C (212 °F) under normal pressure conditions.
## What is boiling point of salt water?
For example, the boiling point of pure water at 1.0atm is 100oC while the boiling point of a 2% salt-water solution is about 102oC. Therefore, the boiling point elevation would be 2oC. | 706 | 3,298 | {"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} | 2.796875 | 3 | CC-MAIN-2021-49 | latest | en | 0.938404 |
https://fr.mathworks.com/matlabcentral/cody/problems/26-determine-if-input-is-odd/solutions/1519866 | 1,566,443,247,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316718.64/warc/CC-MAIN-20190822022401-20190822044401-00540.warc.gz | 474,167,899 | 15,431 | Cody
# Problem 26. Determine if input is odd
Solution 1519866
Submitted on 7 May 2018 by Aakash Deep
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
n = 1; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
2 Pass
n = 2; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
3 Pass
n = 28; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
4 Pass
n = 453; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
5 Pass
n = 17; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
6 Pass
n = 16; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct)) | 224 | 743 | {"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} | 2.84375 | 3 | CC-MAIN-2019-35 | latest | en | 0.448602 |
https://www.f1technical.net/forum/viewtopic.php?f=4&p=805633 | 1,545,055,635,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828507.84/warc/CC-MAIN-20181217135323-20181217161323-00125.warc.gz | 878,283,206 | 16,249 | ## Honda Power Unit Hardware & Software
All that has to do with the power train, gearbox, clutch, fuels and lubricants, etc. Generally the mechanical side of Formula One.
GhostF1
76
Joined: Tue Aug 30, 2016 3:11 am
### Re: Honda Power Unit Hardware & Software
Craigy wrote:
Thu Dec 06, 2018 5:28 pm
ian_s wrote:
Thu Dec 06, 2018 4:19 pm
Bandit1216 wrote:
Thu Dec 06, 2018 8:00 am
Will surely not even be close. I was calculating a bit with some co workers the other day. Rubber to asphalt
friction coefficient is about 1 according to the net. It might be somewhere near 1.3 with a lot of rubber on the track?? Even if you assume 1,5 coefficient and 1,5 down force to weight ratio on the NSX, which are both high assumptions IMO, they can get to 2,25 G. Not even close to F1.
I think G-force is the one thing quite different on track vs dyno though.
2G would still give better info than the maximum 1G on a dyno
We've been around this discussion before. There are dynoes that can give more than 1G on a gimbal.
What has also been discussed at length is sustained G forces. It is extremely difficult to properly simulate accurate sustained G forces on a dyno, most processes involve heavy use of computer estimations from sim programs to fill in the gaps.
This was talked about at length when Honda discovered issues with their 2017 oil tank design on the track that were not seen on the dyno which was followed by a lengthy discussion about computer simulation and replicating accurate fluid slosh dynamics and the challenges involved with that..
gruntguru
406
Joined: Sat Feb 21, 2009 6:43 am
### Re: Honda Power Unit Hardware & Software
Dynos that simulate cornering only tilt the engine so they are simulating not the g forces but the ANGLE resulting from cornering forces. 1g cornering (plus 1g gravity) results in a 45* angle and a magnitude of 1.41g. The tilting dyno can replicate the 45* angle but the magnitude is still 1g (gravity). It probably doesn't matter much - the higher force only affects things like internal stresses, moving parts in the engine and flow involving mixtures with different densities (eg bubbles). The angle matters a lot because pickups that should be submerged can suddenly be sucking air and vice versa. Under 4g cornering the angle of fluid in a reservoir will be about 76* to the horizontal - almost vertical!!!
je suis charlie
GhostF1
76
Joined: Tue Aug 30, 2016 3:11 am
### Re: Honda Power Unit Hardware & Software
Just linking another Abiteboul stab that lacks any technical fact. Basically insinuating that because TR finished behind Renault, the Honda PU is obviously worse
https://www.planetf1.com/news/renault-r ... ing-honda/
“Red Bull never tire of telling how good Honda is compared to us.
“May I state that Toro Rosso is still behind us?”
This guy is amazing.
Bandit1216
0
Joined: Fri Oct 05, 2018 3:55 pm
### Re: Honda Power Unit Hardware & Software
GhostF1 wrote:
Thu Dec 06, 2018 11:57 pm
Just linking another Abiteboul stab that lacks any technical fact. Basically insinuating that because TR finished behind Renault, the Honda PU is obviously worse
https://www.planetf1.com/news/renault-r ... ing-honda/
“Red Bull never tire of telling how good Honda is compared to us.
“May I state that Toro Rosso is still behind us?”
This guy is amazing.
He's stupid. RB is 2 secs faster than Williams. So is the renault better than the Merc then? How can you be so stupid and still be in F1.
PlatinumZealot
329
Joined: Thu Jun 12, 2008 2:45 am
### Re: Honda Power Unit Hardware & Software
Toro rosso is the fastest Honda chassis so he is well within reason!
"The true champions are also great men. They are capable of making difficult decisions, of admitting their mistakes and of pushing harder than before when they get up from a fall."
- Ferrari chairman Sergio Marchionne
GhostF1
76
Joined: Tue Aug 30, 2016 3:11 am
### Re: Honda Power Unit Hardware & Software
PlatinumZealot wrote:
Sat Dec 08, 2018 10:12 pm
Toro rosso is the fastest Honda chassis so he is well within reason!
Vague at best.. He would know that is not even remotely an accurate representation of PU performance. It's just subtle jabs.. he makes a lot of them about rivals unfortunately.
PlatinumZealot
329
Joined: Thu Jun 12, 2008 2:45 am
### Re: Honda Power Unit Hardware & Software
It was a little joke.
"The true champions are also great men. They are capable of making difficult decisions, of admitting their mistakes and of pushing harder than before when they get up from a fall."
- Ferrari chairman Sergio Marchionne
Sieper
33
Joined: Tue Mar 14, 2017 2:19 pm
### Re: Honda Power Unit Hardware & Software
I thought so already (looking at your postings overall) but indeed, on forums humor is quite easily missed. And not only humor, much of everything is usually missed, except for the insults those usually stick (or is it just me)
I can't wait for The Spain wintertests, I am really hoping for a 3 way battle this year, it might be truly possible if the Honda PU really is on the path it seems to be on.
sosic2121
16
Joined: Wed Jun 08, 2016 11:14 am
### Re: Honda Power Unit Hardware & Software
gruntguru wrote:
Thu Dec 06, 2018 11:35 pm
Under 4g cornering the angle of fluid in a reservoir will be about 76* to the horizontal - almost vertical!!!
AFAIK in Radillon corner cars are doing negativne Gs.
Snorked
43
Joined: Mon Mar 16, 2015 8:00 pm
### Re: Honda Power Unit Hardware & Software
--------------- In
2018, which was the season of the fourth year after returning, Honda I updated the power unit (PU) twice. First is "spec 2" which was introduced to the 7th round of Canada in June, and another one is " spec 3 " which appeared in the 16th round of the Russian Grand Prix in September . Regarding the characteristics of each PU and how to use it, Motoabashi explained as follows.
"First of all, I will explain the specification 1 before updating, this was a conservative (conservative) specification because last year there was a problem with PU quite frequently, a penalty due to PU exchange is not this was "been imposed
in that the battle to compete for" a few tenths, because receive the grid demotion penalty, the is a large very impact for the staff that work in the field. "
" the development side Even so, the speed of development falls by coping with troubles, and while dealing with it, the negative side comes out in another meaning that it can not take the next update so easily. "
" With Toro Rosso since 2018 season As I got together, I did not learn to run as many machines as equipped with Honda's power unit on Toro Rosso's body as much as possible Do not go there was a lot. "
"To that end, even if we closed our eyes a bit to performance, we brought conservative specifications to the pre-season test, so it was better to have higher reliability, and the opening race was almost the same with specifications."
■ " We had to use it while understanding the situation of the power unit. "In the
pre-season test, Toro Rosso Honda succeeded in aging the machine by earning the third mileage out of the 10 teams following Mercedes , Ferrari . In the opening game, although trouble of MGU - H (thermal energy regeneration system) occurred in Pierre - Guthrie 's power unit, there was no serious trouble other than that. Then, we introduced spec 2 to the 7th round Canadian Grand Prix.
"Since specification 1 ran properly, spec 2 was specifications that waved to the power, even though it is said that the current F1 has a fixed amount of fuel inflow, the specification that increased combustion efficiency "
This specification 2 was used up to the 15th round of Singapore GP in September. However, Honmohashi Deputy TD said that it was used differently depending on the race, even with the same spec 2.
"From spec 2, we used it while grasping the situation of the power unit at the site properly, because the new PU has characteristics different from the PU that we used until the end."
"Of course, its characteristics are also grasped by bench tests, but there are parts that you do not know unless you actually mount it on the car body and do not actually run.If you do not properly understand its characteristics, the original performance If you can not pull it out, you could develop into a problem in terms of trust, so we used it while grasping exactly what we do in the field. "
" That is, at first, I did not use it that way so much. I understood that I gradually attacked while understanding the characteristics, I think that I
had finished using the performance that I had at last until about the summer vacation (Hungarian GP). " Hungarian GP has entered the Q2 for the first time this year for the first time this year Toro Rosso Honda. Because Mr. Honjo Vice TD said that the communication was improved in Summary (2), Honda was able to use it considerably in specification 2 even in terms of hardware.
And specs 3 appeared in the free practice of the Russian Grand Prix at the end of September. However, with its specification 3, Honda suffered from oscillation (resonance).
"The vibration to the drive system such as the gearbox will change depending on how the engine emits torque.If there is a part where the combustion of the engine is unstable even a little unstable torque will not be stable Then it will trigger another vibration and this time it will trigger another vibration.I think that the stuffing of that side is a little sweet to use Spec 3 3 "
To respond to this problem Although it took a bit of time, it did not recur after the Mexican GP. In other words, it is not a structural problem of spec 3, but it can be said that there was a problem in how to use according to the characteristics of spec 3. As for coping with this problem, it revealed that "Improvement of oscillation has basically coped with data setting".
■ Honda and Toro Rosso staff who achieved great growth
started "more selfish" at spec 1, jumped up with spec 2, and finished the season with spec 3 in the 4th year Honda. However, what has evolved is not just the hardware called the power unit. Honda staff also grew big, along with these three specs.
PlatinumZealot
329
Joined: Thu Jun 12, 2008 2:45 am
### Re: Honda Power Unit Hardware & Software
Great story bro!
So the new combustion regimes carry their own torque characteristics and as result vibration characteristics sometimes leading to resonance and these characteristics have to be understood and alleviated with adjusted engine tuning (probably ignition, fuel injection, and VLIM? - ERS?)
"The true champions are also great men. They are capable of making difficult decisions, of admitting their mistakes and of pushing harder than before when they get up from a fall."
- Ferrari chairman Sergio Marchionne
MarcJ
3
Joined: Mon Jul 10, 2017 6:32 pm
### Re: Honda Power Unit Hardware & Software
Tommy Cookers wrote:
Fri Nov 30, 2018 11:55 am
godlameroso wrote:
Fri Nov 30, 2018 6:04 am
Prechamber or not, flame out would happen somewhere around 1.6, so for gasoline that would be a practical limit....
would conventional ignition allow F1 to run at 1.6 lambda ?
afaik we know Honda has road engines on road fuel that (with PI and preheat) run homogeneously at 2 lambda
and F1 fuel is presumably optimised for combustion qualities
don't people think that TJI-type combustion allows leaner combustion ?
or do they think it is only used for faster initiation of combustion ?
design cooling capacity surely means F1 can sustainably run sub-optimal leanness only with lower MAP than is available
yes I have suggested that briefly running less lean at max fuel rate is intended and used
given that high MAPs/exhaust pressures don't yield greater turbine recovery
Some information I know is correct.
Spark ignition only viable to lambda ~1.4 at best, above that it's about TJI or controlled auto ignition, CAI/HCCI/GDCI. TJI gives optimal TE at 1.85 in tested engines and various RCM rigs, rapid compression machines. Lean limit close to 3 with prechamber scavenging using air, ~2.6 without due some exhaust gas residuals in pre chambers.
Given the McLaren ECU predates TJI in racing and really works best using closed loop control with models, feed forward, computationally efficient hack the map not really needed with today's compute power, works well and optimised. Given they're likely using Kistler piezoelectric pressure sensing in cylinders as a exotic knock/boost sensing first knock could take them out.
Suspect surface volume ratio could have some effect on Lambda lean limit, heat loss to surfaces so the baby 266cc cylinders would be disadvantage here to the typical experience with larger cylinders.
Due to the chemistry needing to happen, SI & TJI both achieve near identical times for 0-5% fuel mass fraction burned, after this burning rate is far faster in TJI.
Hot partially burned chemistry products,radicals,etc, extinguished as they pass through orifices of prechamber is the high ignition energy advantage of TJI the included turbulence enhancement also.
Hope this information helps.
MarcJ
3
Joined: Mon Jul 10, 2017 6:32 pm
### Re: Honda Power Unit Hardware & Software
PlatinumZealot wrote:
Tue Dec 11, 2018 12:50 am
Great story bro!
So the new combustion regimes carry their own torque characteristics and as result vibration characteristics sometimes leading to resonance and these characteristics have to be understood and alleviated with adjusted engine tuning (probably ignition, fuel injection, and VLIM? - ERS?)
Pressure rise rate per crank angle with TJI, and more importantly cylinder to cylinder pressure variance, cycle to cycle CoV coefficient of variance which is solved with model controlled not map based power train control as your really controlling two chambers lambda and their interaction.
Possible mechanical solutions to crank torsional vibration, inertial variance due side thrust could be aggressive move to using
Steel pistons to improve kinematic behavior of the crank mechanism and more homogeneous surface temperature, lower crevice volume in top land and reduced blow by other advantages that would be highly advantageous operating in this combustion regime and operation.
PlatinumZealot
329
Joined: Thu Jun 12, 2008 2:45 am
### Re: Honda Power Unit Hardware & Software
MarcJ wrote:
Tue Dec 11, 2018 11:42 am
PlatinumZealot wrote:
Tue Dec 11, 2018 12:50 am
Great story bro!
So the new combustion regimes carry their own torque characteristics and as result vibration characteristics sometimes leading to resonance and these characteristics have to be understood and alleviated with adjusted engine tuning (probably ignition, fuel injection, and VLIM? - ERS?)
Pressure rise rate per crank angle with TJI, and more importantly cylinder to cylinder pressure variance, cycle to cycle CoV coefficient of variance which is solved with model controlled not map based power train control as your really controlling two chambers lambda and their interaction.
Possible mechanical solutions to crank torsional vibration, inertial variance due side thrust could be aggressive move to using
Steel pistons to improve kinematic behavior of the crank mechanism and more homogeneous surface temperature, lower crevice volume in top land and reduced blow by other advantages that would be highly advantageous operating in this combustion regime and operation.
But the article states (though not explicitly) that an edit of the engine maps solved the problem.
"The true champions are also great men. They are capable of making difficult decisions, of admitting their mistakes and of pushing harder than before when they get up from a fall."
- Ferrari chairman Sergio Marchionne
dren
226
Joined: Wed Mar 03, 2010 1:14 pm
### Re: Honda Power Unit Hardware & Software
I wonder if it was through ICE mapping or MGUK deployment.
Honda! | 3,647 | 15,850 | {"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-2018-51 | longest | en | 0.953612 |
https://programming.vip/docs/functions-student-management-system.html | 1,656,289,335,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00193.warc.gz | 536,224,108 | 5,541 | 1.1 function overview
Functions are organized snippets of code that implement a single function or associated functions. We can think of a function as a piece of code with a name, which can be called in the form of "function name ()" where necessary.
• Using function to program can make the program modular and reduce redundant code
• Make the program structure clearer
• It can improve the programming efficiency of developers
• Facilitate later maintenance and expansion.
It can realize code reuse more efficiently, and copy and paste can also realize code reuse, but the efficiency is low.
```# # Method 1: Print
# # Print a square with 2 asterisks on each side
# for i in range(2):
# for j in range(2):
# print('*', end='')
# print()
#
# # Print a square with 3 asterisks on each side
# for i in range(3):
# for j in range(3):
# print('*', end='')
# print()
#
# # Print a square with 4 asterisks on each side
# for i in range(4):
# for j in range(4):
# print('*', end='')
# print()
# Mode 2: function call
def print_t(lenth):
for i in range(lenth): # i representative bank
for j in range(lenth): # j stands for column (x-axis)
print('*', end='')
print('')
print_t(2) # function call
print_t(3)
print_t(4)
# print(print_t(2))
# print(print_t(3))
# print(print_t(4))
```
1.2 defining functions and calls
Principle: first define the function and then call the function
Definition of function
The print() function and input() used earlier are built-in functions of Python, which are defined by Python.
Developers can also define functions according to their own needs. Python uses the keyword def to define functions. Its syntax format is as follows:
Example: define a function to calculate the sum of two numbers
```# Definition of parameterless function
result = 11 + 22
print(result)
result2 = a + b
print(result2)
```
Function call
The function will not be executed immediately after the definition is completed, and will not be executed until it is called by the program.
The syntax format of function call is as follows:
Function name ([parameter list])
```def add2(a,b):
result2 = a + b
print(result2)
```
When calling the function, the computer program goes through the following four steps:
• The program pauses at the location where the function is called.
• Pass data to function parameters.
• Execute the statements in the body of the function.
• The program returns to the pause to continue.
Other functions can also be called inside a function, which is called nested call of a function.
Examples are as follows:
```# Definition of parameterless function
result = 11 + 22
print(result)
result2 = a + b
print(result2)
# Function call
```
extend
Nested definition of functions
When a function is defined, another function can be nested inside it. At this time, the nested function is called the outer function, and the nested function is called the inner function.
Examples are as follows:
```# Nested definition of functions
# Note: the outer function can not directly call the inner layer function, and can only call the inner layer function in the outer function.
result2 = a + b
print(result2)
print('test:Test of function nested definition!')
```
Note: the outer function can not directly call the inner layer function, and can only call the inner layer function in the outer function.
1.3 transfer of parameters
We usually call the parameters set when defining a function as formal parameters (formal parameters for short), and the parameters passed in when calling a function as actual parameters (actual parameters for short). Parameter transfer of function refers to the process of transferring actual parameters to formal parameters.
The transfer of function parameters can be divided into:
• Transfer of position parameters
• Keyword parameter passing
• Transfer of default parameters
• Packaging and unpacking of parameters
• Mixed transfer
Transfer of position parameters
Transfer of position parameters
Positional parameter: when calling a function, it is passed to the formal parameter in turn according to the actual parameter position defined by the function, that is, the first actual parameter is passed to the first formal parameter, and the second actual parameter is passed to the second formal parameter.
```# Transfer of position parameters
# Positional parameters: when calling a function, parameters are passed according to the parameter positions defined by the function
def user_info(name, age, gender):
print(f'The name is:{name}, Age:{age}, Gender is:{gender}')
user_info('Xiao Ming', 42, 'male')
```
Note: the order and number of parameters passed and defined must be consistent
```user_info('Xiao Ming', 'male', 42, 12) # If the number of parameters is inconsistent, an error will be reported
user_info('Xiao Ming', 'male', 42) # If the parameter order is inconsistent, no error will be reported, but the data is meaningless
```
Keyword parameter passing
Background: if the number of parameters of a function is large, it is difficult for developers to remember the role of each parameter, and it is not advisable to use location parameters.
At this time, you can pass parameters in the form of keyword parameters.
Keyword parameters are specified in the form of "key = value" or "formal parameter = argument", which makes the function clearer and eliminates the order requirements of parameters.
Note: if there are location parameters and keyword parameters at the same time, the location parameters should be put in front, and the keyword parameters still do not consider the order
```# Keyword parameter passing
# Keyword parameters are specified in the form of "key = value", which makes the function clearer and eliminates the order requirements of parameters.
# Note: if there are location parameters and keyword parameters at the same time, the location parameters should be put in front, and the keyword parameters still do not consider the order
def user_info(name, age, gender):
print(f'The name is:{name}, Age:{age}, Gender is:{gender}')
user_info(name='Xiao Ming', gender='male',age=13)
user_info('Fang Wang',gender='female', age=20)
```
Only the slash symbol "/" is restricted for position parameters, indicating that the parameters of the specified position can only pass arguments in the form of position parameters.
Examples are as follows:
```# Positional parameters only
def user_info(name, age, /, gender):
print(f'The name is:{name}, Age:{age}, Gender is:{gender}')
user_info('Fang Wang',12, gender='male')
```
Passing of default (default) parameters
Introduction to default parameters: assign default values to formal parameters when defining functions; When calling a function, you can choose whether to use the default value as needed. If not, the argument can specify a parameter to the formal parameter again
Note: the position of the default parameter must be after (at the end of) the position parameter
```# Default (default) parameter passing
# Introduction: when defining a function, assign a default value to the formal parameter; When calling a function, you can choose whether to use the default value as needed. If not, the argument can specify a parameter to the formal parameter again
# Note: the position of the default parameter must be after (at the end of) the position parameter
def user_info(name, age, gender='female'):
print(f'The name is:{name}, Age:{age}, Gender is:{gender}')
user_info('Fang Wang', 12) # Use default parameters
user_info('Fang Wang', 12, gender='male') # Modify default parameter values
```
Run the program and the output results are as follows:
Stage case - student management system
Keywords: Python
Added by scliburn on Wed, 16 Feb 2022 13:49:47 +0200 | 1,714 | 7,841 | {"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} | 2.53125 | 3 | CC-MAIN-2022-27 | longest | en | 0.811409 |
https://whatisconvert.com/28-feet-second-in-knots | 1,632,673,540,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057882.56/warc/CC-MAIN-20210926144658-20210926174658-00493.warc.gz | 641,172,791 | 7,677 | # What is 28 Feet/Second in Knots?
## Convert 28 Feet/Second to Knots
To calculate 28 Feet/Second to the corresponding value in Knots, multiply the quantity in Feet/Second by 0.59248380129641 (conversion factor). In this case we should multiply 28 Feet/Second by 0.59248380129641 to get the equivalent result in Knots:
28 Feet/Second x 0.59248380129641 = 16.589546436299 Knots
28 Feet/Second is equivalent to 16.589546436299 Knots.
## How to convert from Feet/Second to Knots
The conversion factor from Feet/Second to Knots is 0.59248380129641. To find out how many Feet/Second in Knots, multiply by the conversion factor or use the Velocity converter above. Twenty-eight Feet/Second is equivalent to sixteen point five nine Knots.
## Definition of Foot/Second
The foot per second (plural feet per second) is a unit of both speed (scalar) and velocity (vector quantity, which includes direction). It expresses the distance in feet (ft) traveled or displaced, divided by the time in seconds (s, or sec). The corresponding unit in the International System of Units (SI) is the metre per second. Abbreviations include ft/s, ft/sec and fps, and the rarely used scientific notation ft s−1.
## Definition of Knot
The knot is a unit of speed equal to one nautical mile (1.852 km) per hour, approximately 1.151 mph. The ISO Standard symbol for the knot is kn. The same symbol is preferred by the IEEE; kt is also common. The knot is a non-SI unit that is "accepted for use with the SI". Worldwide, the knot is used in meteorology, and in maritime and air navigation—for example, a vessel travelling at 1 knot along a meridian travels approximately one minute of geographic latitude in one hour. Etymologically, the term derives from counting the number of knots in the line that unspooled from the reel of a chip log in a specific time.
## Using the Feet/Second to Knots converter you can get answers to questions like the following:
• How many Knots are in 28 Feet/Second?
• 28 Feet/Second is equal to how many Knots?
• How to convert 28 Feet/Second to Knots?
• How many is 28 Feet/Second in Knots?
• What is 28 Feet/Second in Knots?
• How much is 28 Feet/Second in Knots?
• How many kt are in 28 ft/s?
• 28 ft/s is equal to how many kt?
• How to convert 28 ft/s to kt?
• How many is 28 ft/s in kt?
• What is 28 ft/s in kt?
• How much is 28 ft/s in kt? | 615 | 2,357 | {"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-2021-39 | latest | en | 0.893464 |
https://www.coursehero.com/file/8871057/returns-No-No-Is-it-ethical-not-to-take-Is-action-Weigh/ | 1,516,498,744,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00503.warc.gz | 913,288,669 | 23,141 | 3325p11hs a decision making tools
# Returns no no is it ethical not to take is action
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Unformatted text preview: returns? No No Is it ethical not to take Is action? (Weigh the harm to shareholders verses benefits to other stakeholders) stakeholders) Action outcome Yes Do it No Don’t Don’t do it do Yes Don’t Don’t do it do No Do it, Do but notify appropriate parties parties Don’t Don’t do it do A – 32 The Poker Design Process If T. J. folds, The money already in the pot EMV = (.80)(\$99,000) = \$79,200 If T. J. calls, The chance T.J. will call EMV = .20[(.45)(\$853,000) – Phillips’s bet of \$422,000] = .20[\$383,850 – \$422,000] = .20[–\$38,150] = –\$7,630 Overall EMV = \$79,200 – \$7,630 = \$71,750 A – 33 The Poker Design Process The money alreadys If T. J. folds, at i thi hthefpot st indicate in e average 70 , th of \$71,5 EMV = (.80)(\$99,000) l l EM V ny times llips’s a a The over re to be made m ough Phi analysis n th =ecision we \$79,200 is ve d large. E rk out, h o be did not w T.J. will call f would nce chance p yof If T.aJ. calls,this instaThe orrect one. in decision ure was the c d EMV p=o.20[(.45)(\$853,000) – Phillips’s bet of nd r c e a \$422,000] = .20[\$383,850 – \$422,000] = .20[–\$38,150] = –\$7,630 Overall EMV = \$79,200 – \$7,630 = \$71,750 A – 34...
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Ask a homework question - tutors are online | 486 | 1,515 | {"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-2018-05 | latest | en | 0.804174 |
https://physics.stackexchange.com/questions/692244/how-to-affirm-whether-a-frame-of-reference-is-inertial-or-non-inertial | 1,716,847,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00611.warc.gz | 393,491,729 | 50,926 | # How to affirm whether a frame of reference is Inertial or non-inertial?
As far as I know, inertial frame of reference are the ones where the all the three Newton's laws of motion hold. Having this definition we can then identify all such frames of reference which are inertial, if we have an inertial frame of reference, to begin with, to observe them by applying Newton's first law of motion i.e.,
• If S is an inertial frame of reference then we can conclude that S' is also an inertial frame of reference if velocity of S' is uniform/constant with respect to S.
Now from these, we can define a non-inertial frame of reference as a frame of reference where laws of motion are not valid in their current form and need to be modified so that they can be used (such as introduction of Fictitious force).
Now the question:
• Given a non-inertial frame of reference what is(are) the condition(s) required to affirm whether another frame of reference (being observed from the current non-inertial frame) is inertial or non-inertial?
I think a brief background to the question is required. I thought of this situation while considering the following case: suppose we are observing an observer (in space) from Earth, how may I claim that the the reference frame attached to that observer is inertial or not? Clearly earth is a non-inertial frame of reference, hence the question.
You don’t need a second frame to determine if a frame is inertial. Simply compare the coordinate acceleration in the frame to the proper acceleration measured by momentarily co-moving accelerometers. If they match then the frame is inertial. If they do not match then the frame is non-inertial and the difference between the coordinate acceleration and the proper acceleration is a fictitious force.
• I thought of this situation while considering the following case: suppose we are observing an observer (in space) from Earth, how may I claim that the the reference frame attached to that observer is inertial or not? Clearly earth is a non-inertial frame of reference, hence the question. Feb 2, 2022 at 12:29
• @shyamtripathi As I said, use accelerometers in the observer's frame. Your own frame is irrelevant.
– Dale
Feb 2, 2022 at 13:23
• I upvoted because you distinguished between proper motion and coordinate motion, and a lot of people don't see the difference. But I agree with @ShyamTripathi that you have not addressed the specific question asked.
– Arc
Feb 13, 2022 at 19:23
Given a non-inertial frame of reference what is(are) the condition(s) required to affirm whether another frame of reference (being observed from the current non-inertial frame) is inertial or non-inertial?
I think a brief background to the question is required. I thought of this situation while considering the following case: suppose we are observing an observer (in space) from Earth, how may I claim that the the reference frame attached to that observer is inertial or not? Clearly earth is a non-inertial frame of reference, hence the question.
I assume you can't go to the space observer's frame of reference, so you have to do it at a distance.
Motion is more generally composed of inertial motion (field forces) and non-inertial motion (contact forces). In your example, you are on the surface of Earth observing someone in space, so you have to separate the components.
When you are in deep-space (gravity negligible) either "standing still" or travelling with constant velocity, then you are clearly in an inertial frame. The same hapens when you are in free fall in a gravitational field (either in a parabolic trajectory over a planet's surface, or in orbit around a large mass). In both scenarios, you don't feel motion, you are weightless, because gravity is a field force and thus your accelerometers measure zero, and you are always oriented to the same direction in space, just as a gyroscope.
The inertial motion due to gravity must be measured by an external referential, say the distant stars.
Then, you have to know how much of your motion is non-inertial. Since you are being held in the surface of Earth by the normal force, then you can feel weight, because the normal force is a contact force, and thus you can measure the proper acceleration it causes using an accelerometer.
When you have determined how is your motion composed, measuring both your proper acceleration due to the normal force using accelerometers, and the coordinate acceleration due to gravity using relative position to distant stars, then you have to measure the motion of the observer in space relative to you.
Now the trick part: the observer in space motion's may also be composed by inertial and non-inertial movement, so you have to estimate what the gravity field looks like in the observer in space's position.
Once you have a) your motion decomposed into inertial and non-inertial components, b) the relative motion between you and the observer in space mapped in detail - so you know what their movement relative to the distant stars is, and c) an estimate of the gravity field around the space observer, that is, what are the geodesics in their surroundings, then you can d) subtract their inertial motion, and whatever motion is left is their non-inertial motion.
Note that this is very difficult to do in practice, not only beacuse the form of the gravitational field may be very complicated (many celestial objects, near and far, dust clouds, small but heavy meteorites, etc.), but because so many factors can cause the space observer's proper motion (small amounts of gases being expelled from it, thermal radiation emission and absorption, and even anisotropic radiation pressure - see the Pioneer anomaly). See, for example, the controversy surrounding the non-gravitational trajectory of the ʻOumuamua object.
• "The same hapens when you are in free fall in a gravitational field" No. Thats not an inertial frame. Feb 13, 2022 at 22:51
• Yes! It is! That's exactly how you tell inertial frames from non-inertial ones! There can be an inertial frame of reference that's accelerating, if the acceleration is promoted by gravity. When you free fall, an accelerometer mesaures zero, you feel no weight as you follow a geodesic of the field, there's no proper acceleration and thus its an inertial frame of reference! See Proper acceleration.
– Arc
Feb 13, 2022 at 22:56
• No, its not. If you fall freely to Earth you not in an inertial frame. Feb 13, 2022 at 22:59
• @Felicia, the main issue here is to distinguish proper acceleration from coordinate acceleration. Proper acceleration (in classical) mechanics is absolute, it does not depend on an external frame of reference, you can measure it from within, while coordinate acceleration by itself needs external sources of reference to be measured.
– Arc
Feb 13, 2022 at 23:00
• Well, then can you explain why it is not an inertial frame? Please read the proper acceleration entry, you will understand it better. The correct way to tell if a frame of reference is inertial or not is by using an accelerometer, and when you free fall your accelerometer measures zero, thus the weightlessness.
– Arc
Feb 13, 2022 at 23:02
As the OP has mentioned Newton's laws, I will not not use concepts of general relativity.
Besides the $$3$$ Newton's laws, there is also his force of gravity. So, if the planets including Earth are moving around the Sun, they are not inertial frames, they are accelerated by the gravity force. The Sun itself is not an inertial frame because it rotates around itself.
A good candidate at a first approximation (that considers the orbit of the Sun around the galaxy center as a second order effect) is a rocket that keeps the same position $$(R,\theta,\phi)$$ in a system of spherical polar coordinates with the Sun at center, and the fixed stars as angular coordinates reference. Or moving in a straight line in this frame.
The rocket should burn fuel to generate a force to balance the gravity attraction from the Sun (and other planets) at each location, so that the sum of forces are zero on it.
• This is incorrect. A planet, considering its orbit around the sun, is indeed in an inertial frame of reference. Gravity is a field force, and thus produces no internal compression, the planet simply slides through the field, You can verify this by comparing inertial rotation and non-inertial rotation: a person on a merry go round is subject to a contact force and thus is in a state of non-inertial rotation, the greater the angular velocity, and the greater the radius of rotation, the greater the centrifugal force.
– Arc
Feb 13, 2022 at 22:47
• In the case of the orbit around the sun, we don't feel centrifugal forces because gravity is a field force - else we would be completely thorn apart a long ago. Another very distinctive feature of non-inertial rotation is the direction of the rotating body: the person on the merry go round keeps spinning, say facing the center, or facing outwards the rotation, while inertial rotation always face the same orientation in space. Rotation of the planet, or the sun, around its own axis is, of course, non-inertial rotation.
– Arc
Feb 13, 2022 at 22:50
• @Arc I started my answer saying that I am talking about Newton's gravity, not GR. According to Newton, if the force of gravity is balanced by another force, the total force on the object is zero. The concept of field started in the $19^{th}$ century, probably with Faraday. Feb 13, 2022 at 23:16
• Well, agree, but the 'modern' gravity field understanding of the 19th century still predates GR, right? Newton's strict version of the laws of motion and the law of gravity actually have inconsistencies which have been elucidated along the time, but too many people still take those literally. For instance, too many people still think that any acceleration results in non-inertial movement, and that's something that I think we should try to explain better to all (the distinction between proper acceleration and coordinate acceleration).
– Arc
Feb 13, 2022 at 23:29
• @Arc notions like: orbits are inertial movements, or distinction between proper and coordinate acceleration don't have meaning out of GR, with its metrics, covariant derivatives and geodesics. Feb 14, 2022 at 1:26
This is a variation of the answer by @Felicia.
(One can also attach accelerometers to the lab, as @Dale suggests.)
From my answer to How can one tell they are accelerating? ,
have them conduct an experiment as Ivey & Hume did:
If a ball that is dropped from the top of a stand
lands at the base of the stand, then the frame is inertial.
Here are a few frames (superimposed) from Ivey and Hume's Frames of Reference video https://archive.org/details/frames_of_reference
(You can probably find it on YouTube [with slightly different timestamps]. However, this archive.org URL should be more permanent than YouTube.)
• At t=4m22s , this is a ball dropped from a cart at rest in the inertial-Lab frame. When released, there is no horizontal force on the ball, hence it has constant horizontal velocity in the Lab. It lands at the base of the stand.
At t=5m25s , this is a ball dropped from a cart in uniform motion in the inertial-Lab frame. When released, there is no horizontal force on the ball, hence it has constant horizontal velocity. It lands at the base of the stand.... just like it was at rest-and-inertial.
From your non-inertial frame, you might find it difficult to write an expression for the trajectory of the falling ball... but what you want is the result... Does the ball end up at the base of the stand?
If you don't allow me to use gravity, do a variation where the projectile is sent across the room. Did the projectile end up at the end point of a segment tangent to its initial velocity?
For accelerated cart case,
continue to How can one tell they are accelerating? In that non-inertial case, the ball does not land at the base of the stand.
The full video treats the case of a rotating frame of reference and the Foucault pendulum.
• Note the question is: how to tell if a system is not inertial at a distance. If you could instruct the space observer to do some experiment, or go there yourself, then it would be far easier to answer. The premisse of the question is that the motion of the space observer is arbitrary, can be any, including the possibility of non-inertial motion. But you must tell that by just considering your motion on Earth, and the motion of the space observer relative to Earth.
– Arc
Feb 13, 2022 at 22:36
• @Arc Observe all motions in that frame and look for the indications I suggested. Maybe there is a bouncing ball. The OP didn't set more restrictive "ground rules", so until then I think my answer and those of others are reasonable. (By the way, we make a similar assumption about the emission of light in atomic transitions in stars. We didn't setup an experiment there... we just observe some natural process and deduce what we can.) Feb 13, 2022 at 22:46
• Ok, now you explained it better, this last comment of yours connects your answer with the question (which by the way you perhaps want to add to your reasoning). You are right that the OP didn't set that kind of rule: does the system being observed has any features that might indicate non-inertial motion? If it's a star, its emitting radiation, if its a planet with children playing ball you can see the balls' trajectories, say, but if it's an asteroid with no such indicators?
– Arc
Feb 13, 2022 at 23:21
• @Arc ... for asteroid, it's harder to measure. Sure. But that's the art of experimental physics... find something to measure.... possibly after waiting for the technology and the analysis tools to be developed. In principle, there has to be at least one. (It's a different type of question if the OP asked: today, I need to measure this property of some system at a distance I have to no access to. But again, the OP made no such constraint thus far.) Feb 13, 2022 at 23:39
Tell the people in the frame you look at to go to different positions without relative motions and fixed wrt to the axes of their frame. Tell them to hold a mass. Then tell them to unleash the masses they hold. If the masses stay stationary wrt one another and the frame, the frame is inertial. If not, the frame is non-inertial.
• Again, note the question is: how to tell if a system is not inertial at a distance (think of an asteroid, for instance). If you could instruct the space observer to do some experiment, or go there yourself, then it would be far easier to answer. The premisse of the question is that the motion of the space observer is arbitrary, can be any, including the possibility of non-inertial motion. But you must tell that by just considering your motion on Earth, and the motion of the space observer relative to Earth.
– Arc
Feb 13, 2022 at 22:40
• @Arc You can see that the masses stay at the same place inside a rocket. I can look with a telescope inside your rocket. If the masses all stay at the same place wrt to the inside of the rocket then the inside of the rocket is an inertial frame. Feb 13, 2022 at 22:50
To confirm that another frame of reference $$F$$ is intertial while standing on Earth, take an accelerometer and point it in directions that shows one is accelerating, say with acceleration $$\vec{a}$$ (not explained by e.g. gravity). Measure the velocity of $$F$$ in that direction to confirm that it is accelerating with acceleration $$-\vec{a}$$. Turn the accelerometer in directions were it doesnt show acceleration. Check $$Fs$$ velocity in that direction to confirm that it is not accelerating.
If you can move into the observer's frame, your own frame doesn't matter. You simply go to the other frame, assume the frame is inertial and calculate the acceleration of some body there with the laws of motion. Then you actually measure the acceleration in the observer's frame. If they match, then you can say that the reference frame is behaving inertially in that motion you investigated.
If you can't move into the observer's frame, you would need outside information. For instance, if there's someone there saying they are static in their own frame. Do this: Measure their acceleration as seen from your own frame, next you account for the effects of the acceleration of your own frame. If the static observer is still moving with some acceleration, then the outside frame is not inertial.
• Your first paragraph not only does not address the question, it's also wrong. You have to make a distinction between proper acceleration and coordinate acceleration in order to distinguish inertial reference frames from non-inertial ones. "You simply go to the other frame, assume the frame is inertial and calculate the acceleration ..." clearly is not scientific enough to tell. Also, everyone is static on their own frames, right? And how to you "account for the effects of the acceleration of your own frame"?
– Arc
Feb 13, 2022 at 19:27
• So, unless i really slipped here i remember the title of this question being "How to tell if a system not inertial". And, per the tags, i used the Newtonian definition of an inertial system, which means a system where laws of Newton are valid. So, he knows he is in eath, an non-inertial system due to rotation, measures the relative motion on the observed space body, removes the effects of earth's own acceleration(the fictitious forces of rotation) and check if they match, if not, the space body has his own acceleration being in an non-inertial frame. Feb 13, 2022 at 22:01
• I know that this would imply that a free falling body is accelerating, which is false in relativity, but again, the tags. I can delete the answer if so desired. And notice i didn't claim that correspondence with the laws of motion is enough to determine that the reference is inertial, i said that its behaving inertially in that case, just like our everyday earth frame is approximately inertial. Feb 13, 2022 at 22:04
• I don't think you should remove the answer, perhaps improve it. You are right, no need to put relativity in the table to explain inertial frames of reference, my answer only resorts to classical mechanics. But my point is: your answer does not make a distiction between proper acceleration and coordinate acceleration, so its difficult to put it to practice. How exactly do you tell in practice if Newton's laws are valid in a given frame of reference? How do you know a priori the contribution of each fictious force?
– Arc
Feb 13, 2022 at 22:30
• You can only do that by using accelerometers and gyroscopes, and the question states that you can only do that in Earth's frame, not in the space observer's frame. Any body can be accelerated by a field force like gravity (so you are just sliding in the spacetime curvature), or by contact forces (like a person on a rocket under the normal force due to the body of the rocket), but the effects are very different. You must be able to distinguish, and that is the nice point the question raises.
– Arc
Feb 13, 2022 at 22:31 | 4,290 | 19,035 | {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-22 | latest | en | 0.916075 |
http://academic.csuohio.edu/simond/estimation/AddHinfEx1.m | 1,369,387,062,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-20/segments/1368704433753/warc/CC-MAIN-20130516114033-00091-ip-10-60-113-184.ec2.internal.warc.gz | 2,443,765 | 1,064 | function AddHinfEx1 Q = 1; R = 1; thetaMin = 0; thetaMax = 1; dtheta = 0.01; KArr = []; PArr = []; for theta = thetaMin : dtheta : thetaMax c(1) = theta^2 - theta^4 * R; c(2) = Q * theta^4 * R - Q * theta^2 + R * theta^2 - 1; c(3) = Q * (1 - 2 * theta^2 * R); c(4) = Q * R; Pall = roots(c); % Find a real positive root of the ARE that results in a stable estimator. P = inf; for i = 1 : length(Pall) if abs(theta^2 * Pall(i) - 1) < 1e-12 continue; end Pa = Pall(i) / (1 - theta^2 * Pall(i)); V = R + Pa; Fhat = 1 - Pa / V; if isreal(Pall(i)) && (Pall(i) >= 0) && (Pall(i) < P) && (abs(Fhat) < 1) P = Pall(i); K = Pa / V; end end if P == inf thetaMax = theta - dtheta; break; end PArr = [PArr P]; KArr = [KArr K]; end close all; theta = thetaMin : dtheta : thetaMax; figure; plot(theta, KArr); set(gca,'FontSize',12); set(gcf,'Color','White'); xlabel('H_\infty performance bound \theta'); ylabel('Estimator gain K'); figure; plot(theta, PArr); set(gca,'FontSize',12); set(gcf,'Color','White'); xlabel('H_\infty performance bound \theta'); ylabel('Kalman performance bound P'); | 392 | 1,075 | {"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-2013-20 | latest | en | 0.474798 |
https://www.scribd.com/document/97620774/Mathematics-Stats-Assignment-No-1 | 1,560,973,399,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999040.56/warc/CC-MAIN-20190619184037-20190619210037-00455.warc.gz | 916,066,775 | 55,021 | You are on page 1of 5
# Mathematics & Statistics for Management
Assignment No.1
Submitted by:
Gemon Kurian 11KJM113 Kristu Jayanti College Department of Management & Technology
## Q1. Definition of Statistics:
Statistics is the study of the collection, organization, analysis, and interpretation of data. It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments. A statistician is someone who is particularly well versed in the ways of thinking necessary for the successful application of statistical analysis. Such people have often gained this experience through working in any of a wide number of fields. There is also a discipline called mathematical statistics that studies statistics mathematically. The word statistics, when referring to the scientific discipline, is singular, as in "Statistics is an art." This should not be confused with the word statistic, referring to a quantity (such as mean or median) calculated from a set of data, whose plural is statistics.
## Q2. & Q3. Importance & Scopes of Statistics in different fields:
The importance of statistics in recent years cannot be over emphasized. The development in statistical studies has considerably increased its scope and importance. It is no longer regarded as the science of statecraft or a by-product of state administration. Now it embraces a host of sciences-social, physical and natural. Statistical findings have assumed unprecedented dimensions these days. Statistical thinking has become indispensable for able citizenship. Statistics is of immense use in the following cases.
1. In Planning: Planning is necessary for efficient workmanship and in formulating future policies. Statistics provides the valued interpretation of facts and figures relevant to planning. Planning depends on forecasting the future. Statistics provides the necessary tools of estimation and forecasting. So statistics is indispensable in planning. 2. In Business: Statistical knowledge is very helpful to the businessman. He formulates different plans and policies using statistics. It helps him in forecasting the future trends and tendencies. To estimate the market fluctuations, changes in the demand conditions etc. statistical techniques are often used. Hence for becoming a successful businessman, ideas in statistics are essential. 3. In Administration: In ancient times statistics was used as the science of statecraft. It was used to collect data relating to manpower, crimes, income etc. for formulating different policies. In modem times, its role has become manifold. It has become indispensable for the administrators. It is being widely used formulating and forecasting different plans and policies of the state administration.
## Mathematics & Statistics for Management
4. In Business Management: Business manager take decisions in the face of uncertainty. Statistical tools like collection classification, analysis and interpretation of data are essential in business-management. The success of modem business primarily depends on accurate forecasting of the future demand and mark trends. For this statistics is essential. Hence statistics is widely used in business management. 5. In Research Activities: One cannot think of undertaking any research activities without using statistics. Primarily, statistical techniques are used for collecting information in any research. Besides, statistical methods are used for analysis and interpretation of research findings. Thus there is hardly any branch of study where statistics is not being used. It is used in all spheres of human activities. 6. In Economics: Statistics are the straws out of which Economists have to make bricks. Statistics indispensable in economic studies. Statistical data and their analysis are used to solve a variety of economic problems such as in consumption, production, distribution of income and assets, poverty, unemployment etc. These problems are described, compared and correlated by using statistics. Use of statistics in Economics has led to the formulation of many economic laws like Engels law of consumption, Paretos Laws of distribution of income etc. Similarly statistical tools like index numbers are sued as economic barometers.
## Q4. Statistic in Managerial Decision Making:
Whether designing new products, streamlining a production process or evaluating current vs. prospective customers, todays business managers face greater complexities than ever before. Running a shop on instinct no longer suffices. Statistics provide managers with more confidence in dealing with uncertainty in spite of the flood of available data, enabling managers to more quickly make smarter decisions and provide more stable leadership to staff relying on them.
Focusing on Big Picture Statistical analysis of a representative group of consumers can provide a reasonably accurate, cost-effective snapshot of the market with faster and cheaper statistics than attempting a census of very single customer a company may ever deal with. The statistics can also afford leadership an unbiased outlook of the market, to avoid building strategy on uncorroborated presuppositions. Backing Judgments Statistics back up assertions. Leaders can find themselves backed into a corner when persuading people to move in a direction or take a risk based on unsubstantiated opinions. Statistics can provide objective goals with stand-alone figures as well as hard evidence to substantiate positions or provide a level of certainty to directions to take the company.
## Mathematics & Statistics for Management
Making Connections Statistics can point out relationships. A careful review of data can reveal links between two variables, such as specific sales offers and changes in revenue or dissatisfied customers and products purchased. Delving into the data further can provide more specific theories about the connections to test, which can lead to more control over customer satisfaction, repeat purchases and subsequent sales volume. Ensuring Quality Anyone who has looked into continuous improvement or quality assurance programs, such as Six Sigma or Lean Manufacturing, understands the necessity for statistics. Statistics provide the means to measure and control production processes to minimize variations, which lead to error or waste, and ensure consistency throughout the process. This saves money by reducing the materials used to make or remake products, as well as materials lost to overage and scrap, plus the cost of honoring warranties due to shipping defective products. Considerations Know what to measure, and manage the numbers; dont let the numbers do the managing for you, or of you. Before using statistics, know exactly what to ask of the data. Understand what each statistical tool can and cant measure; use several tools that complement one another. For example, dont rely exclusively on an "average," such as a mean rating. Customers using a fivepoint scale to rate satisfaction wont give you a 3.84; that may indicate how the audience as a group clustered, but its also important to understand the width of the spread using standard deviation or which score was used by the greatest number of people, by noting the mode. Finally, double-check the statistics by perusing the data, particularly its source, to get a sense of why the audiences surveyed answered the way they did.
## Q5. Limitation of Statistics
The important limitations of statistics are: (1) Statistics laws are true on average. Statistics are aggregates of facts. So single observation is not a statistics, it deals with groups and aggregates only. (2) Statistical methods are best applicable on quantitative data. (3) Statistical cannot be applied to heterogeneous data. (4) It sufficient care is not exercised in collecting, analyzing and interpretation the data, statistical results might be misleading. (5) Only a person who has an expert knowledge of statistics can handle statistical data efficiently. (6) Some errors are possible in statistical decisions. Particularly the inferential statistics involves certain errors. We do not know whether an error has been committed or not. | 1,483 | 8,201 | {"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} | 3.75 | 4 | CC-MAIN-2019-26 | latest | en | 0.925998 |
http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=week3.mdwn;h=39e472bf9a644c1bdac774790ed134f26ab7cf31;hp=a55659cb97b6e0f12616770115fc8680eb2ed821;hb=6a45c3860fc064728f0fa65fd7b2869f48f49878;hpb=cd20a0a226f35177c21ef48bcabfc59316e3e489 | 1,611,067,126,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00532.warc.gz | 68,982,955 | 2,235 | X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=week3.mdwn;h=39e472bf9a644c1bdac774790ed134f26ab7cf31;hp=a55659cb97b6e0f12616770115fc8680eb2ed821;hb=6a45c3860fc064728f0fa65fd7b2869f48f49878;hpb=cd20a0a226f35177c21ef48bcabfc59316e3e489 diff --git a/week3.mdwn b/week3.mdwn index a55659cb..39e472bf 100644 --- a/week3.mdwn +++ b/week3.mdwn @@ -1,3 +1,12 @@ +[[!toc]] + +##More on evaluation strategies## + +Here are notes on [[evaluation order]] that make the choice of which +lambda to reduce next the selection of a route through a network of +links. + + ##Computing the length of a list## How could we compute the length of a list? Without worrying yet about what lambda-calculus implementation we're using for the list, the basic idea would be to define this recursively: @@ -415,7 +424,7 @@ to *the tail* of the list we were evaluating its application to at the previous ##Fixed-point Combinators Are a Bit Intoxicating## - + There's a tendency for people to say "Y-combinator" to refer to fixed-point combinators generally. We'll probably fall into that usage ourselves. Speaking correctly, though, the Y-combinator is only one of many fixed-point combinators. @@ -583,11 +592,14 @@ truth and circularity](http://tinyurl.com/2db62bk) for an approach that is similar, but expressed in terms of non-well-founded sets rather than recursive functions. -HOWEVER, you should be cautious about feeling too comfortable with +##However...## + +You should be cautious about feeling too comfortable with these results. Thinking again of the truth-teller paradox, yes, -`ω` is *a* fixed point for `I`, and perhaps it has +`Ω` is *a* fixed point for `I`, and perhaps it has some a privileged status among all the fixed points for `I`, being the -one delivered by Y and all. +one delivered by Y and all (though it is not obvious why Y should have +any special status). But one could ask: look, literally every formula is a fixed point for `I`, since @@ -600,7 +612,8 @@ So the Y combinator is only guaranteed to give us one fixed point out of infinitely many---and not always the intuitively most useful one. (For instance, the squaring function has zero as a fixed point, since 0 * 0 = 0, and 1 as a fixed point, since 1 * 1 = 1, but `Y -(\x. mul x x)` doesn't give us 0 or 1.) So why in the reasoning we've +(\x. mul x x)` doesn't give us 0 or 1.) So with respect to the +truth-teller paradox, why in the reasoning we've just gone through should we be reaching for just this fixed point at just this juncture? @@ -616,6 +629,12 @@ fixed point for this referential function: if this pen cap is the referent of *this noun phrase*, then it is the referent of (3), and so for any object. +The chameleon nature of (3), by the way (a description that is equally +good at describing any object), makes it particularly well suited as a +gloss on pronouns such as *it*. In the system of +[Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/), +pronouns denote (you guessed it!) identity functions... + Ultimately, in the context of this course, these paradoxes are more useful as a way of gaining leverage on the concepts of fixed points and recursion, rather than the other way around. | 922 | 3,283 | {"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-2021-04 | latest | en | 0.897472 |
https://www.airmilescalculator.com/distance/rdm-to-mdw/ | 1,680,031,535,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00596.warc.gz | 702,400,548 | 57,337 | How far is Chicago, IL, from Redmond, OR?
The distance between Redmond (Roberts Field) and Chicago (Chicago Midway International Airport) is 1688 miles / 2717 kilometers / 1467 nautical miles.
The driving distance from Redmond (RDM) to Chicago (MDW) is 2020 miles / 3251 kilometers, and travel time by car is about 35 hours 17 minutes.
1688
Miles
2717
Kilometers
1467
Nautical miles
3 h 41 min
192 kg
Distance from Redmond to Chicago
There are several ways to calculate the distance from Redmond to Chicago. Here are two standard methods:
Vincenty's formula (applied above)
• 1688.425 miles
• 2717.256 kilometers
• 1467.201 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 1683.972 miles
• 2710.090 kilometers
• 1463.331 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
How long does it take to fly from Redmond to Chicago?
The estimated flight time from Roberts Field to Chicago Midway International Airport is 3 hours and 41 minutes.
What is the time difference between Redmond and Chicago?
The time difference between Redmond and Chicago is 2 hours. Chicago is 2 hours ahead of Redmond.
Flight carbon footprint between Roberts Field (RDM) and Chicago Midway International Airport (MDW)
On average, flying from Redmond to Chicago generates about 192 kg of CO2 per passenger, and 192 kilograms equals 423 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
Map of flight path and driving directions from Redmond to Chicago
See the map of the shortest flight path between Roberts Field (RDM) and Chicago Midway International Airport (MDW).
Airport information
Origin Roberts Field
City: Redmond, OR
Country: United States
IATA Code: RDM
ICAO Code: KRDM
Coordinates: 44°15′14″N, 121°9′0″W
Destination Chicago Midway International Airport
City: Chicago, IL
Country: United States
IATA Code: MDW
ICAO Code: KMDW
Coordinates: 41°47′9″N, 87°45′8″W | 549 | 2,174 | {"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} | 2.53125 | 3 | CC-MAIN-2023-14 | latest | en | 0.857803 |
https://zbmath.org/?q=an:06123744 | 1,627,332,683,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.92/warc/CC-MAIN-20210726183622-20210726213622-00438.warc.gz | 1,136,253,030 | 11,051 | # zbMATH — the first resource for mathematics
Belt distance between facets of space-filling zonotopes. (English. Russian original) Zbl 1275.52012
Math. Notes 92, No. 3, 345-355 (2012); translation from Mat. Zametki 92, No. 3, 381-394 (2012).
A belt of a polytope $$P$$ is the set of all facets parallel to a given $$(n-2)$$-face of $$P$$. A sequence of facets is called a belt path if every two consecutive facets in the sequence belong to the same belt. The number of different belt the facets in a belt path belong to is called the length of the path. Finally the belt distance of two facets is the length of the shortest belt path between them and accordingly, the belt diameter of a polytope $$P$$ is the maximal belt distance between any two facets of $$P$$.
The author investigates belt diameters of zonotopes that are also parallelotopes and proves an upper bound of $$\log_2(\frac45 d)$$ for $$d$$-dimensional space-filling zonotopes. To this end, he shows that it is enough to consider zonotopes whose generators lie in two conjugate sets and which behave nicely under projections in a certain direction. Afterwards he uses an inductive argument to show the statement for those special zonotopes.
The author further shows that this bound is sharp in dimensions up to $$6$$.
##### MSC:
52B11 $$n$$-dimensional polytopes 52B12 Special polytopes (linear programming, centrally symmetric, etc.) 52B05 Combinatorial properties of polytopes and polyhedra (number of faces, shortest paths, etc.)
Full Text:
##### References:
[1] G. Voronoï, ”Nouvelles applications des paramètres continus à la théorie des formes quadratiques. Deuxième mémoire. Recherches sur les paralléloèdres primitifs,” J. für Math. 136, 67–178 (1909). [2] H. Minkowski, ”Allgemeine Lehrsätze über die convexen Polyeder,”Gött. Nachr., 198–219 (1897). · JFM 28.0427.01 [3] P. McMullen, ”Convex bodies which tile space by translation,” Mathematika 27(1), 113–121 (1980). · Zbl 0432.52016 [4] B. A. Venkov, ”On a class of Euclidean polyhedra,” Vestnik Leningrad.Univ. Ser.Mat. Fiz.Him. 9(2), 11–31 (1954). [5] O. K. Zhitomirskii, ”Verschärfung eines Satzes von Voronoi,” Zh. Leningrad. Matem. Obshch. 2, 131–151 (1929). [6] R. M. Erdahl, ”Zonotopes, dicings, and Voronoi’s conjecture on parallelohedra,” European J. Combin. 20(6), 527–549 (1999). · Zbl 0938.52016 [7] A. Ordine, Proof of the Voronoi Conjecture on Parallelotopes in a New Special Case, Ph. D. Thesis (Queen’s University, Ontario, 2005). [8] B. Delaunay, ”Sur la partition régulière de l’espace à 4 dimensions.Deuxième partie,” Izv. Akad. Nauk SSSR Ser. VII. Otd. Fiz.Mat. Nauk, No. 2, 147–164 (1929). [9] M. I. Shtogrin, ”Regular Dirichlet-Voronoi partitions for the second triclinic group,” Trudy Mat. Inst. Steklov 123, 3–128 (1973) [Proc. Steklov Inst. Math. 123, 1–116 (1973)]. [10] S. S. Ryshkov and E. P. Baranovskii, ”S-types of n-dimensional lattices and five-dimensional primitive parallelohedra (with an application to covering theory),” Trudy Mat. Inst. Steklov 137, 3–131 (1976) [Proc. Steklov Inst.Math. 137, 1–140 (1976)]. · Zbl 0419.10031 [11] P. Engel, ”The contraction types of parallelohedra in $$\mathbb{E}$$ 5,” Acta Cryst. Sec. A 56(5), 491–496 (2000). · Zbl 1188.52021 [12] G. C. Shephard, ”Space-filling zonotopes,” Mathematika 21, 261–269 (1974). · Zbl 0296.52004 [13] P. McMullen, ”Space tiling zonotopes,” Mathematika 22(2), 202–211 (1975). · Zbl 0316.52005 [14] G.M. Ziegler, Lectures on Polytopes, in Grad. Texts inMath. (Springer-Verlag, NewYork, 1995), Vol. 152. [15] A. P. Poyarkov and A. I. Garber, ”On permutohedra,” Vestnik Moskov.Univ. Ser. I Mat. Mekh., No. 2, 3–8 (2006) [Moscow Univ.Math. Bull. 61 (2), 1–6 (2006)]. [16] E. S. Fedorov, Foundations of the Theory of Shapes (Emperor’s Academy of Sciences Press, St.-Petersburg, 1885) [in Russian]. [17] H. S.M. Coxeter, Regular Polytopes (Dover Publ., New York, 1973). [18] A. N. Magazinov, personal communication (2010). [19] B. A. Venkov, ”On the projections of parallelohedra,” Mat. Sb. 49(91)(2), 207–224 (1959).
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. | 1,443 | 4,413 | {"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} | 2.671875 | 3 | CC-MAIN-2021-31 | latest | en | 0.738956 |
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## Über dieses Buch
Historically, there is a close connection between geometry and optImization. This is illustrated by methods like the gradient method and the simplex method, which are associated with clear geometric pictures. In combinatorial optimization, however, many of the strongest and most frequently used algorithms are based on the discrete structure of the problems: the greedy algorithm, shortest path and alternating path methods, branch-and-bound, etc. In the last several years geometric methods, in particular polyhedral combinatorics, have played a more and more profound role in combinatorial optimization as well. Our book discusses two recent geometric algorithms that have turned out to have particularly interesting consequences in combinatorial optimization, at least from a theoretical point of view. These algorithms are able to utilize the rich body of results in polyhedral combinatorics. The first of these algorithms is the ellipsoid method, developed for nonlinear programming by N. Z. Shor, D. B. Yudin, and A. S. NemirovskiI. It was a great surprise when L. G. Khachiyan showed that this method can be adapted to solve linear programs in polynomial time, thus solving an important open theoretical problem. While the ellipsoid method has not proved to be competitive with the simplex method in practice, it does have some features which make it particularly suited for the purposes of combinatorial optimization. The second algorithm we discuss finds its roots in the classical "geometry of numbers", developed by Minkowski. This method has had traditionally deep applications in number theory, in particular in diophantine approximation.
## Inhaltsverzeichnis
### Chapter 0. Mathematical Preliminaries
Abstract
This chapter summarizes mathematical background material from linear algebra, linear programming, and graph theory used in this book. We expect the reader to be familiar with the concepts treated here. We do not recommend to go thoroughly through all the definitions and results listed in the sequel — they are mainly meant for reference.
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 1. Complexity, Oracles, and Numerical Computation
Abstract
This chapter is still of a preliminary nature. It contains some basic notions of complexity theory and outlines some well-known algorithms. In addition, less standard concepts and results are described. Among others, we treat oracle algorithms, encoding lengths, and approximation framework in which algorithms are designed and aand computation of numbers, and we analyse the running time of Gaussian elimination and related procedures. The notions introduced in this chapter constitute thenalysed in this book. We intend to stay on a more or less informal level; nevertheless, all notions introduced here can be made completely precise — see for instance Aho, Hopcroft and Ullman (1974), Garey and Johnson (1979).
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 2. Algorithmic Aspects of Convex Sets: Formulation of the Problems
Abstract
Convex sets and convex functions are typical objects of study in mathematical programming, convex analysis, and related areas. Here are some key questions one encounters frequently:
• Given a point y and a set K, is y a member of K, i. e., is y contained in K ?
• If y is not a member of K, find a hyperplane separating y from K.
• Given a linear inequality, is it valid for each vector in K ?
• Given a linear function, find a point maximizing (or minimizing) the function on K.
• Given a convex function, find its minimum.
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 3. The Ellipsoid Method
Abstract
In 1979 a note of L. G. Khachiyan indicated how an algorithm, the so-called ellipsoid method, originally devised for nonlinear nondifferentiable optimization, can be modified in order to check the feasibility of a system of linear inequalities in polynomial time. This result caused great excitement in the world of mathematical programming since it implies the polynomial time solvability of linear programming problems.
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 4. Algorithms for Convex Bodies
Abstract
We shall now exploit the ellipsoid method (the central-cut and the shallow-cut version) described in Chapter 3. In Sections 4.2, 4.3, and 4.4 we study the algorithmic relations between problems (2.1.10),..., (2.1.14), and we will prove that — under certain assumptions — these problems are equivalent with respect to polynomial time solvability. Section 4.5 serves to show that these assumptions cannot be weakened. In Section 4.6 we investigate various other basic questions of convex geometry from an algorithmic point of view and prove algorithmic analogues of some well-known theorems. Finally, in Section 4.7 we discuss to what extent algorithmic properties of convex bodies are preserved when they are subjected to operations like sum, intersection etc.
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 5. Diophantine Approximation and Basis Reduction
Abstract
As mentioned in Chapter 1, combinatorial optimization problems can usually be formulated as linear programs with integrality constraints. The geometric notion reflecting the main issues in linear programming is convexity, and we have discussed the main algorithmic problems on convex sets in the previous chapters. It turns out that it is also useful to formulate integrality constraints in a geometric way. This leads us to “lattices of points”. Such lattices have been studied (mostly from a nonalgorithmic point of view) in the “geometry of numbers”; their main application has been the theory of simultaneous diophantine approximation, i. e., the problem of approximating a set of real numbers by rational numbers with a common small denominator. We offer an algorithmic study of lattices and diophantine approximation.
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 6. Rational Polyhedra
Abstract
In most combinatorial (and real world) applications the convex sets one encounters are polyhedra. Often these polyhedra have “simple” vertices and facets. It turns out that the knowledge of such additional information on the convex sets in question extends the power of the ellipsoid method considerably. In particular, optimum solutions can be calculated exactly, boundedness and full-dimensionality assumptions can be dropped, and dual solutions can be obtained. In the case of explicitly given linear programs this was the main contribution of Khachiyan to the ellipsoid method. If the linear programs are given by some oracle — which is often the case in combinatorial optimization — then these additional goals can still be achieved, albeit with more involved techniques. In particular, we have to make use of the simultaneous diophantine approximation algorithm described in Chapter 5.
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 7. Combinatorial Optimization: Some Basic Examples
Abstract
In the remaining part of this book we apply the methods developed in the first part to combinatorial optimization. In this chapter we give some illuminating examples to explain the basic techniques of deriving polynomial time algorithms for combinatorial optimization problems. These techniques are based on combining the ellipsoid method and basis reduction with results from the field called “polyhedral combinatorics”, where combinatorial optimization problems are formulated as linear programs. Chapter 8 contains a comprehensive survey of combinatorial problems to which these methods apply. Finally, in the last two chapters we discuss some more advanced examples in greater detail.
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 8. Combinatorial Optimization: A Tour d’Horizon
Abstract
In Chapter 7 we have introduced several basic combinatorial optimization problems, and we have shown in detail how the ellipsoid method and basis reduction together with polyhedral information about these problems can be used to design polynomial time algorithms. In this chapter we give an overview about combinatorial optimization problems that are solvable in polynomial time. We also survey important theorems that provide polyhedral descriptions of the polytopes associated with these problems. We indicate how these results can be employed to derive polynomial time algorithms based on the ellipsoid method and basis reduction. The results of this chapter are presented in a condensed form, to cover as much material as possible.
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 9. Stable Sets in Graphs
Abstract
In this chapter we survey the results of the polyhedral approach to a particular NP-hard combinatorial optimization problem, the stable set problem in graphs. (Alternative names for this problem used in the literature are vertex packing, or coclique, or independent set problem.) Our basic technique will be to look for various classes of inequalities valid for the stable set polytope, and then develop polynomial time algorithms to check if a given vector satisfies all these constraints. Such an algorithm solves a relaxation of the stable set problem in polynomial time, i. e., provides an upper bound for the maximum weight of a stable set. If certain graphs have the property that every facet of the stable set polytope occurs in the given family of valid inequalities, then, for these graphs, the stable set problem can be solved in polynomial time. It turns out that there are very interesting classes of graphs which are in fact characterized by such a condition, most notably the class of perfect graphs. Using this approach, we shall develop a polynomial time algorithm for the stable set problem for perfect graphs. So far no purely combinatorial algorithm has been found to solve this problem in polynomial time.
Martin Grötschel, László Lovász, Alexander Schrijver
### Chapter 10. Submodular Functions
Abstract
The concept of a submodular function in discrete optimization appears to be in several respects analogous to that of a convex function in continuous optimization. In many combinatorial theorems and problems, submodularity is involved, in one form or another, and submodularity often plays an essential role in a proof or an algorithm. Moreover, analogous to the fast methods for convex function minimization, it turns out that submodular functions can also be minimized fast, viz. in polynomial time. However, the only method known for this is, as yet, the ellipsoid method.
Martin Grötschel, László Lovász, Alexander Schrijver
### Backmatter
Weitere Informationen | 2,236 | 10,711 | {"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} | 2.546875 | 3 | CC-MAIN-2020-16 | latest | en | 0.93305 |
http://mathoverflow.net/questions/15476/algebraic-characterization-of-transitive-spaces-of-matrices/15497 | 1,467,166,029,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397562.76/warc/CC-MAIN-20160624154957-00049-ip-10-164-35-72.ec2.internal.warc.gz | 206,534,633 | 16,725 | # Algebraic characterization of transitive spaces of matrices
Fix an integer $d \ge 2$ and let $M_d$ be the space of real $d \times d$ matrices. Let $E$ be a vector subspace of $M_d$. We say that $E$ is transitive if $E \cdot \mathbb{R}^d_* = \mathbb{R}^d$, that is, for all vectors $v \in \mathbb{R}^d_* = \mathbb{R}^d-\{0\}$ and $w \in \mathbb{R}^d$ there exists a matrix $A \in E$ such that $A \cdot v = w$.
The question is how to determine algebraically if a space of matrices is transitive or not.
More precisely, which algebraic (ie, polynomial) conditions on the entries of matrices $A_1,...,A_k$ express the fact that the space $E$ spanned by them is non-transitive?
Remarks:
1) Fix the number $k$ of generators of $E$. Let $Z$ be the subset of $\mathbb{R}^{kd^2}$ corresponding to the $k$-tuples of matrices that generate a non-transitive set. That $Z$ is the projection of an algebraic set, and therefore by Tarski-Seidenberg theorem, is a semi-algebraic set.
2) Consider the analogous problem with complex matrices and vectors in $\mathbb{C}^d$, and let $Z_C$ be the set corresponding to $Z$ above. Then $Z$ is algebraic (projectivize everything and apply the theorem that says that pprojection of algebraic is algebraic). Anyway what I'd like to see are the explicit equations for this algebraic set.
-
Transitive Lie algebras are a classically studied subject. (Google it to check.) However my space $E$ is not assumed to be a Lie subalgebra of $M_d$. – Jairo Bochi Feb 18 '10 at 13:11
A space of matrices is transitive iff its "orthogonal complement" contains no matrix of rank one.
The idea was not mine; it is I found in Sec. 4 from the paper below (See also some more modern and more readable papers that cite it):
Azoff, E.A. On finite rank operators and preannihilators. Mem. Amer. Math. Soc. 64, no. 357 (1986).
-
The condition for $E$ to be intransitive is that the determinant form is the $0$ form somewhere other than the origin. That is, every vector $v \in \mathbb R^d$ gives you an alternating $d$-form on $M_d$ and on $E$ by the determinant of the images of $v$. This form is nonzero on a vector if and only if the images of the vector by $E$ are all of $\mathbb R^d$.
Edit: You are right that the above was not a complete answer.
To be more explicit with bases for everything: Let $E$ have dimension $D$, with $d \le D \le d^2$. Let $E$ have a basis $\{E_1, ... E_D \}$ so that every element of $E$ is represented by a vector $(a_1,...,a_D)$. Represent every element of $\mathbb R^d$ by a vector $(b_1,....,b_d)$.
Then for any vector $(b_1,....,b_d)$, the determinant form is an alternating $d$-form on $E$ identified with $\mathbb R^D$. These forms have a basis of size ${D \choose d}$ given by the determinants of $d\times d$ minors, that is, project to a given $d$ coordinates, and take the determinant. To check whether the determinant form is the $0$ $d$-form, express it in terms of the basis, and see if all ${D \choose d}$ coefficients are $0$. That is, check if the ${D \choose d}$ determinants $\det [E_{f(1)}v, ..., E_{f(d)}v]$ are all $0$ for each integer-valued function $f$ with $1 \le f(1) \lt f(2) \lt ... \lt f(d) \le D$.
As we let $v$ vary but fix a basis for $E$, the coefficients of the determinant form are homogeneous polynomials of degree $d$ in the coordinates $\{b_i\}$. The variety of intersections of the zeros of those polynomials on $\mathbb R^d$ is the origin if and only if $E$ is transitive.
This gives you a test for a particular $E$ in terms of recognizing whether a variety is just a point. It still leaves the condition on $E$ in the Grassmannian as a projection of a variety.
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OK, but this does not answer the question. You suggest to take each $v\neq 0$ and then compute some associated determinant. So in theory you would have to perform infinitely many calculations: one for each $v$. __________ What I need is some computable test on the generators of $E$. – Jairo Bochi Feb 17 '10 at 9:56 | 1,146 | 3,977 | {"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} | 3.046875 | 3 | CC-MAIN-2016-26 | latest | en | 0.859283 |
http://www.chegg.com/homework-help/questions-and-answers/ac-adapter-telephone-answering-unit-uses-transformer-toreduce-line-voltage-120-v-rms-volta-q202750 | 1,472,094,598,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982292734.19/warc/CC-MAIN-20160823195812-00138-ip-10-153-172-175.ec2.internal.warc.gz | 363,161,055 | 14,156 | An AC adapter for a telephone-answering unit uses a transformer toreduce the line voltage of 120 V (rms) to a voltage of 10.0 V. The rms current delivered to the answeringsystem is 410 mA.
(a) If the primary (input) coil in thetransformer in the adapter has 240 turns, how many turns are thereon the secondary (output) coil?
turns
(b) What is the rms power delivered to the transformer? Assume anideal transformer.
(avg)input = W
### No answer yet. Submit this question to the community.
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Q:
An AC adapter for a telephone-answering unit uses a transformer toreduce the line voltage of 120 V (rms) to a voltage of 11.0 V. The rms current delivered to the answeringsystem is 360 mA.(a) If the primary (input) coil in thetransformer in the adapter has 240 turns, how many turns are thereon the secondary (output) coil? turns(b) What is the rms power delivered to the transformer? Assume anideal transformer.(avg)input = Your answer differs from the correct answerby orders of magnitude. W
Q:
An AC adapter for a telephone-answering unit uses a transformer toreduce the line voltage of 120 V (rms) to a voltage of 11.5 V. The rms current delivered to the answeringsystem is 360 mA.(a) If the primary (input) coil in thetransformer in the adapter has 240 turns, how many turns are thereon the secondary (output) coil?Your answer differs from the correct answerby 10% to 100%. turns(b) What is the rms power delivered to the transformer? Assume anideal transformer.(avg)input= W | 374 | 1,505 | {"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} | 3.125 | 3 | CC-MAIN-2016-36 | latest | en | 0.880603 |
https://www.compadre.org/nexusph/course/Interlude_4_-_Energy_The_Quantity_of_Motion_and_its_Distribution | 1,716,661,478,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00836.warc.gz | 619,965,317 | 6,881 | # Interlude 4 - Energy: The Quantity of Motion and its Distribution
You've probably heard the term "energy" for most of your lives. When you were a toddler your parents might have complained, "he/she has so much energy I just can't keep up with him/her." The term has lots of everyday meanings.
But in your science classes you certainly will have heard the term in its more technical sense — these molecules have a particular binding energy; ATP is the energy currency of the cell, and so on. But what is energy really? For a physicist, in the end everything is energy — because we have learned that mass is a form of energy. But that doesn't help very much. Here are a few key ideas to get us started.
• At beginning, the best starting point for building up the concept of energy is motion. The place where historically the idea of energy started and the place where we will start is with the sense that energy is something we will associate with moving objects — a quantification of the idea of motion.
• At the end, energy becomes the universal currency of physics — it's stuff that can, in principle, be converted into motion or that has come from motion.
The way this works is that we follow a process that has turned out to be immensely useful in organizing our thinking about physics and indeed about much of science.
1. We look at our usual "simplest possible cases" and decide how we might quantify the concept of motion.
2. We find two ways of doing this — momentum and energy. With one of them, energy, we find there is a conservation law that allows us to introduce concepts that look like transformations of the energy of motion into other forms.
3. Every time we find a situation that looks like the sum of the energies we have defined is not conserved, we try to introduce a new kind of energy in order to keep things conserved.
This process sounds circular. What good is it if every time our law fails we introduce a term to save it? Isn't this sloppy science?
Well, no. We might decide this was a useless process if we had to introduce a new kind of energy for essentially every new experiment or phenomenon we looked at. But we start with the energy of motion (kinetic energy) and quickly add the energy of interaction (potential energy). We soon discover that something being hotter can be considered a kind of energy (thermal energy) and corresponds to an increase of an internal hidden motion of an object's atoms and molecules. We then discover that there is energy stored in the structures of atomic bonding to form molecules (chemical energy) and this, in the context of quantum mechanics, can be interpreted as kinetic and potential energies distributed probabilistically. We know that light can carry energy, and finally, Einstein suggests that mass can be considered a form of energy. And that's where it now stands — sort of.
So over a period of 400 years we have invented 6 kinds of energy.* And they have served us to describe millions of experiments and situations. This is very much a part of our modeling of the world and science's ways of knowing. We create ways of thinking about phenomena that are appropriate for those phenomena (choosing a channel on cat television) and then try to stitch our understandings together (building coherence). With energy, we have developed one of the most powerful tools in the scientific arsenal — and it all begins with thinking about motion.
But finding the different kinds of forms energy can take is just a start. Because our typical environments always involve lots and lots of molecules, moving and interacting at a furious pace, energy is always being exchanged and shared. The study of how energy naturally tends to distribute itself leads to the laws of thermodynamics and a deeper understanding of what things happen spontaneously.
In the next few chapters, we will
• Develop the concepts of kinetic energy, work, and potential energy as natural consequences of looking carefully at Newton's second law;
• Build on those concepts to understand the concept of binding and the chemical bond;
• Explore the laws of internal (thermal) energy and how it is shared;
• Establish the conservation of energy theorem over multiple layers of the structure of matter — macro, molecular, and chemical;
• Develop an understanding of how energy is shared and distributes itself spontaneously (entropy and free energy).
* As of this writing, the scientific community has learned that the galaxies of our universe seem to be moving away from each other at increasing speeds -- accelerating rather than slowing down as would be expected from the fact that they attract each other gravitationally. Some source is providing huge new energies of motion to objects of galactic scale. If, as we expect, there is still a conserved total energy, this implies that there is a new kind of energy that is being transformed into kinetic energies of galaxies. We don't know what this is, but for now we are calling it "dark energy". Stay tuned!
Joe Redish 7/29/11
Article 303 | 1,024 | 5,044 | {"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-2024-22 | latest | en | 0.962229 |
https://calculator-online.net/orthocenter-calculator/ | 1,721,159,299,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00815.warc.gz | 140,028,839 | 53,642 | Pages
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# Orthocenter Calculator
Please provide coordinates of triangle vertices and the calculator will take instants to determine the coordinates of its orthocenter.
Vertex A:
Vertex B:
Vertex C:
An online Orthocenter calculator helps you to calculate the orthocenter of a triangle easily. This tool results in exact values of the coordinates of the orthocenter. Let us discuss the proper concept of the orthocenter in trigonometry.
## What is Orthocenter?
A point where the altitudes of the triangle meet is known as the point of concurrency or simply the orthocenter of the triangle.” In the above figure, AB, BC and CA are the sides of the triangle and their respective altitudes are CF, AD and BE. An altitude is simply a perpendicular line (a line drawn at 90 degree angle) that is drawn from any vertex of the triangle to its opposite side. When all right bisectors of a triangle intersect each other at a common point, that point has its own coordinates that are related to the coordinates of all the three vertices of the triangle. An orthocenter finder generates the absolute values of these coordinates within seconds.
### Properties of Orthocenter:
Let us have a focus on some of the significant properties of the orthocenter. Orthocenter of Acute Triangle: An acute triangle is the one that has all three angles (acute angles) less than 90°. In general, the orthocenter of an acute-angled triangle lies inside the triangle. You can determine the orthocenter coordinates by using free online orthocenter calculator. Orthocenter of Obtuse Triangle: The measure of one angle(obtuse angle) in an obtuse triangle is greater than 90°. For this triangle, the orthocenter always lies outside the triangle. Orthocenter of Right Triangle: For a right-angled triangle, the orthocenter lies on its vertex. The orthocenter calculator determines the orthocenter of any one of the triangles mentioned above.
### Algebraic Formulae to Calculate Orthocenter:
We will discuss all basic formulae used in finding the orthocenter of a triangle. These include: Slope of the Side: The slope of all the sides is calculated by using the following formula: m(slope) = y2 − y1/x2 − x1 Slope of Perpendicular: By means of the formula below, we can determine the slope of the perpendicular drawn on each side of the triangle. Perpendicular slope of line = −1/slope of the line =−1/m Equation of Altitudes: We need to determine the equation of the altitudes by using the expression as follows: (y - y1) = m (x - x1) After this, we need to solve the algebraic equations to find values corresponding to x and y that are the coordinates of the orthocenter. Using an orthocenter calculator creates easiness in determining the coordinates of the orthocenter for any triangle.
### How to Find the Orthocenter of a Triangle?
We will solve an example to understand the correct use of formulae in finding the orthocenter. Find the coordinates of the orthocenter of a triangle whose vertices are (2, -3) (8, -2) and (8, 6). Solution : The given points are A (2, -3) B (8, -2) and C (8, 6). Now we need to work for the slope of AC. From that we have to find the slope of the perpendicular line through B. Slope of AC = (y2 - y1) / (x2 - x1) A (2, -3) and C (8, 6) = (6 - (-3)) / (8 - 2) = 9/6 = 3/2 Slope of the altitude BE = -1/ slope of AC = -1 / (3/2) = -2/3 Equation of the altitude BE is given as : (y - y1) = m (x - x1) Here B (8, -2) and m = 2/3 y - (-2) = (-2/3) (x - 8) 3(y + 2) = -2 (x - 8) 3y + 6 = -2x + 16 2x + 3y - 16 + 6 = 0 2x + 3y - 10 = 0 Now we need to determine the slope of BC. From that we have to calculate the slope of the perpendicular line through D. Slope of BC = (y2 - y1) / (x2 - x1) B (8, -2) and C (8, 6) = (6 - (-2)) / (8 - 8) = 8/0 = undefined Slope of the altitude AD = -1/ slope of AC = -1/undefined = 0 Equation of the altitude AD is as follows: (y - y1) = m (x - x1) Here A(2, -3) and m = 0 y - (-3) = 0 (x - 2) y + 3 = 0 y = -3 By putting the value of x in the first equation: 2x + 3(-3) = 10 2x - 9 = 10 2x = 10 + 9 2x = 19 x = 19/2 x = 9.2 So, the orthocenter is (9.2,-3). You can also verify your results by putting coordinates of all the vertices in orthocenter finder.
### How orthocenter calculator works?
Absolute values for coordinates of the orthocenter can be determined by using orthocenter calculator as follows: Input:
• Insert values of all coordinates against all vertices, respectively.
• Click on ‘calculate’.
Output: The calculator calculates: Exact values of orthocenter coordinates by following each and every step.
## FAQ’s:
### Is there a different method to find orthocenter for various triangles?
No, the basic method for finding orthocenter is the same regardless the type of the triangle.
### Is there any triangle having no Orthocenter?
No, there is not any triangle whose orthocenter does not exist as it is the point of coincidence of the altitudes of the triangle.
### What does the orthocenter of a triangle indicates?
The orthocenter of a triangle show the point of concurrency of perpendicular lines.
### What is Euler’s Line?
A line passing from all the triangle centers is known as Euler’s line.
## Conclusion:
An orthocenter is an important central point of intersection for all triangles. The position of the orthocenter gives the exact idea of the type of the triangle under study. The orthocenter calculator helps you to determine the coordinates of the orthocenter within a span of seconds. Engineers make a vast use of an orthocenter calculator to make their measurements precise and accurate.
## References:
From the source of wikipedia: Cheap Flights circles and conics, Orthic triangle, Right triangle, Acute and obtuse triangles, Altitude, Trigonometric functions, equilateral triangle. From the source of Khan Academy: Common orthocenter and centroid, Proof: Triangle altitudes are concurrent (orthocenter). From the source of Lumen Learning: The Rectangular Coordinate Systems and Graphs, Cartesian Coordinate System, Graphing Equations by Plotting Points, Finding x-intercepts and y-intercepts. | 1,576 | 6,138 | {"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} | 4.0625 | 4 | CC-MAIN-2024-30 | latest | en | 0.839466 |
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# Probability & Statistics/probability question
Question
I would like to find out the probability of the following result, ideally with your showing how you derived the answer, the probability or statistical rule used, and pointing out any potential error(s) in my setting up of the scenario:
Assume that each year for 25 years, 10 different, qualified people from a given group (let's say an ethnic minority) apply for a job at a company that gets 100 applicants for every position it fills. Thus, the chance of one of these 10 persons getting hired in any given year = 10 x 1/100 = 10%.
Question: what are the odds that none ever gets hired during the 25 years?
There are two principles you can use here:
First, if an event has probability p and another event has probability q, if they are not dependent on one another, the probability of both is the product p×q.
Second, if an event has probability p, the probability that event doesn't happen is 1-p.
From this, you can say that every year, you have a 1 - 1/10 = 9/10 chance of not hiring one of these people. Then, over 25 years, the chances you never hire one would be:
9/10 × 9/10 × ... × 9/10 = (9/10)^25 ≈ 0.0718
Now, that is your answer. However, in asking this question you make two assumptions that are probably not true in real life. First, these events are probably not independent. Who you hire one year affects who you hire the next year. Second, there is no reason to believe that hiring people is random, meaning that the figure 1/10 is totally unreasonable. No part of hiring is random, really.
Questioner's Rating
Rating(1-10) Knowledgeability = 8 Clarity of Response = 8 Politeness = 8 Comment Thank you, Mr. Oliver. This was helpful. Best regards,
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#### Clyde Oliver
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Replies
Hoping someone can offer some help . . . we have a date/time field that tracks the "Last Status Change" for leads. When a Lead Status is updated, a trigger updates the date/time to "now" and then we have a formula field that calculates the number of days between now and that date/time of the last change. However, we only calculate that value for leads that have not been converted and where they don't have a Closed status. The current formula is below and it works great.
IF(AND(NOT( ISNULL( Date_Time_Last_Status_Change__c )),NOT(IsConverted),NOT(ISPICKVAL(Status,"7. Closed"))), (NOW()- Date_Time_Last_Status_Change__c ),NULL)
Now . . . here's the question. There is some talk about changing it to try and only show workdays since the change--to not calculate weekends. I have the following formula to show the workdays. BUT . . . how would I combine the conditions in the formula above and the formula below so I only get workdays since the date/time of change if the lead still meets the conditions of not being converted or closed???
CASE(MOD( DATEVALUE(Date_Time_Last_Status_Change__c) - DATE(1985,6,24),7),
0 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,2,2,3,3,4,4,5,5,5,6,5,1),
1 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,2,2,3,3,4,4,4,5,4,6,5,1),
2 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,2,2,3,3,3,4,3,5,4,6,5,1),
3 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,2,2,2,3,2,4,3,5,4,6,5,1),
4 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,1,2,1,3,2,4,3,5,4,6,5,1),
5 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c),7),1,0,2,1,3,2,4,3,5,4,6,5,0),
6 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,1,2,2,3,3,4,4,5,5,6,5,0),
999)
+
(FLOOR(( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) )/7)*5)
Looking for a strong SFDC administrator who is also able to craft solutions using tools that go beyond SFDC configurations (i.e. Apex, VisualForce Pages, HTML) while keeping focus on enabling the sales organization to be effective.
http://www.polycom.com/company/careers/career_search.html
Requisition # 285BR
Hello . . . hoping someone can provide some suggesttions to work around an issue I'm having. Basically, what I need to do is to have a field on an opportunity that can display a value from a field on the Opportunity Owner's SFDC user record. I will need that field for a few different things like reporting and also to be used for routing in a approval process. The issue is that I would expect that user record field to be readily available through a formula. But it's not. Each SFDC user has a "User Theater" field on their user record. This is the field I need to display/use on the opportunity.
If I try to create a formula field on the opportunity, I get to the "insert field" section and when I drill down I can't get to the opportunity owner's user record. The only Owner field I can get to is the Owner ID. I need to use a field on the opportunity owner's user record.
I can get to the "Created By" user record for the field I need. I can also get to the "Last Modified By" user record for the field I need . . . so why can't I get to the actual opportunity owner's user record for the field??? Created by and Last Modified by in many cases may not provide the valid info--I really need it to be based on the opportunity owner's info.
Is there any way at all to make this available on the opportunity? I'm not a developer . . . I can't write triggers. I'm hoiping there is a way to get to this data that is doable for me to implement. Any assistance would be great!
I'm struggling with another validation rule. We have a picklist where the sales reps indicate if it's a standard discount or not. Then there is a multi-select picklist where they indicate the particular type of discount they are asking for. I want to have a validation rule that says "If the Standard Discount field = Yes, then the Discount Type field can't have certain options selected. I have figured out how to do a rule to error out checking for one value in the multi-select list. But I can't seem to figure out how to check multiple options with one rule (I'd rather not have 10 separate rules if it can be done with one).
Here is my example which successfully creates an error if the Stadard Discount = Yes and then the Discount type includes "Rebate":
OR(
AND(ISPICKVAL( Standard_Discount__c , "Yes"),
INCLUDES(Discount_Type__c, "Rebate")))
So how do I take this rule and give an error if the Discount Type also includes "Trade In" or "Certification" or "Credit Return" etc.?
Could use some help . . . I created some validation rules to impose certain restrictions on a discount request functionality we have. Works great, when someone plugs in an amount of \$10k or more, the validation rule triggers and stops the save.
The problem is that we have multicurrency enabled--I was expecting the validation rule to catch the conversion rate and to prevent someone from entering what would be over the \$10k USD amount . . . that doesn't happen!!! If I switch the currency on the record to GBP and put in \$9000, the record saves. It only errors when I hit the 10000 value regardless of the currency (and 9999 GBP is way more than \$9999 USD). The formula I am using is below.
AND(ISPICKVAL( Is_this_a_10k_10percent_NST__c , "Yes"),Hardware_Incremental_DiscountUSD__c >=10000)
This is a HUGE problem . . . I need to have this rolled out by Monday and if the system is letting larger GBP/EUR discounts run through without stopping them it will be a disaster. How do you get a validation rule to take multicurrency into account??? Or can it?
Could use a little help here from the masters on this board . . . we have a discount approval process that uses an Object related to Opportunities. Our order management folks really want the Opportunity Owner's name to be reflected on the Object itself (Approval Request Form). Looking at the Opportunity itself is apparently too hard or requires extra steps. Whatever!
Anyway . . . seems like it should be an easy thing to do but all I have been able to do is create a formula field that can pull in the Opportunity "Owner ID"--which means nothing to the people who need to see it. I really need a field that can automatically show the actual Opportunity Owner Name in text on the Approval Form so order management can see "Bob Smith" instead of "00530000000diZm".
I'm sure there is probably something simple I am missing. But how in the world do I get the name to show up in a field on that Approval Form? Is there a formula function that "translates" the ID and can do that? I'm pretty stumped.
• September 15, 2009
• Like
• 0
We have a multiselect field where we can note "Special Designations" for Accounts . . . can be things like Fortune 500 or other special groups we want to associate the Account to without having tons of check boxes. We are also doing some cleanup/lockdown on some specially designated Accounts and I want to restrict users (except admins) from changing the Account Name if a particular value is selected in that Special Designations multiselect list--lets say that special designation is Fortune 500. How would that validation rule be written? I believe that multiselect lists can be included in validation rules now but I'm not sure how to actually do it.
I tried the following but I get an error trying to reference the multiselect picklist. But I am crummy with formulas so not sure what the solution is. Any suggestions would be gratefully welcomed!
AND( IF(\$User.ProfileId <>"00e70000000wI69", true, false),
IF(
Special_Designations__c
,"Fortune 500", false), IF( ISCHANGED( Name ), true, false) )
Is there any way to have a formula field that would do a simple list of the product names included on an opportunity? Example: an opportunity has Widget A, Widget B and Widget C listed in opportunity products. Would there be a way to populate a field through a formula or workflow rule just to show those product names?
The reason is that I am getting tons of questions for reports based on opportunties that do or do not have individual products or combinations of products. SFDC as far as I can tell does not really do this (like trying to get a report that will exlude ALL opportunities that contain Widget C . . . if you exclude Widget C in a filter you still get opportunities that may have Widget A, Widget B and Widget C--but it just won't show the Widjet C rows).
If there is a way to "summarize" product names in a text field then I could see a way to do this. Is there any hope? Or another way to try and do this?
Back for some formula field assistance. I am working with our IT team to set up a hierarchy structure on SFDC User records so they can trigger some dashboards off that info. So I went to create the fields that they need . . . I have lookup fields on a user record that can be used to look up other SFDC users and put them into a spot on the hierarchy. Those lookups bring back other data from that "manager user" record and shows them on the sales rep's user record (So on Sales Rep #1's user record, there is a lookup for Manager #1 (2, 3, 4, etc) and by selecting that person it returns data for that manager and puts it on the sales rep record for reference). Hope that makes sense. Anyway, it all works fine EXCEPT for one thing. I want to have a field that pulls in another Manager field--but it's a picklist on the Manager User record. And I get syntax errors when I try to do a formula field on the Sales Rep record to bring back the Manager User field value from that picklist.
So my question is . . . how do I build a formula field that can simply return the text based off a picklist driven field??? So if the Manager User record value selected from a picklist is "Country Manager", then how can I pull that Country Manager value to the Sales Rep user record? I really want to use a picklist on the Managr record, otherwise I will run into data inconsistencies.
Any suggestions? Can this be done?
We have a process that runs under a "dummy admin user" and it will periodically makes updates to a record. We want to be able to more easily track when the latest change from this user was made (and potentially generate workflow rules, etc from this field). What I am thinking of doing is just to have a field that whenever the "Last Modified" field is updated as being related to this particular user . . . . the field "Last Modified by X" field would show that last modified date. Is that possible? If so, how would that formula be written?
Here is maybe a better example of the thing I am hoping to do. "User X" makes a modification to a Lead and so the Last Modified By field shows "User X 4/8/2009". I am hoping to have a field called Last Modified by User x that would show 4/8/2009. That way we can have a view/report/workflow that can work off that field. The reason we want this is that other users could make updates to the records after User X and they become hard to find. With the formula field, we can easily see Leads this week/month that User X modifed.
Is there a way to write a validation rule that can trigger an email notification whenever ownership of a lead changes? I know I'll feel dumb after seeing it, but how would the rule be written to do this?
Leads change hands here quite a bit and people don;t always remember to check the "Send Notification" box. I'd like to see if I can have a workflow rule that will automatically trigger an alert to the new owner without that box getting checked.
I'd like to have a field that indicates when opportunities are updated from a Forecast Category . . . specifically if they are "Committed" (moved from pipeline to Commit) or "De-committed" (moved from Commit to Pipeline). Then have the top \$ decomitted deals show on a Dashboard.
My idea was to have a text field that could be populated automatically when Forecast Category was appropriately updated, maybe through a workflow rule? How would the formula need to be written to do that, if it can be done . . . any ideas??? Seems like I can't work with the field change in a report to be able to show it--so was thinking it would have to be a formula-related action.
We use opportunity workflow rules to assign tasks to people for certain approvals . . . when it's their turn to review and approve, they get a Task assigned and are sent an emial notification. However, I want to be able to set up a Task workflow rule that would see "If Task Subject contains X, Status is Not Started and the Task is Overdue by 3 days---send an email notification to remind the user that they have a task assigned to them."
I tried to do this based on criteria, but I don't see a way to define "overdue". So I am guessing that this needs to be handled with a formula. Can anyone help me with what that formula would need to look like? Do I need a custom field for Tasks that would calculate "days overdue" or is there a way to get to that without the extra field?
Thanks in advance for any assistance!!!!!!!!
Message Edited by thusker on 08-18-2008 10:13 AM
So I just noticed that when you close an Opportunity in SFDC, the Close Date does not automatically update to the day that the closed stage was selected. For some reason, I always thought (or assumed) that it did. Does anyone use validation rules or workflow rules in any way to automatically set the close date to "today" when a Stage is changed from an open to a closed value? I'd love to hear how you've done it if you have.
If the rep does not update Close Date, we can get deals falling in quarters where they do not belong and other messy data.
Is there any way to use an S-control or a formula field to look at a field value on a custom object related to an Opportunity (or to have a user click a link/button on the object) and automatically trigger an update to a field on that related opportunity??? I know in the past this was not doable . . . just wondering or hoping if something may have changed to enable this kind of thing.
Yes . . . it's me again with another formula question.
Can anyone point me in the right direction to create a formula for a validation rule that would require a field to be filled out based on Opportunity Amount and Stage? Example: Our managers want to require "Win/Loss Reason" picklist field to be filled out if an opportunity is Closed and has an Amount of \$100k or more. How would I create a formula that can look at the value of the Stage picklist and also the Amount?
I am trying to do a formula for our markeitng folks to add a field called "Month Lead Created", which will look at the Created Date for Leads and indicate the month that it was created in. Example: Lead created on 1/2/2008 = January. I thought I knew how to build the formula based on another example, but I get an error when I do the syntax check. Can anyone tell me how to fix it so it might work??
Formula is:
CASE(MONTH(CreatedDate) ,
1, "January",
2, "February",
3, "March",
4, "April",
5, "May",
6, "June",
7, "July",
8, "August",
9, "September",
10, "October",
11, "November",
12, "December", " ")
Error: Incorrect parameter for function MONTH(). Expected Date, received DateTime
Apparently, you cannot use any long text fields in Worklfow Rule conditions . . . which stinks because I need to make a field update based on a long text field not having certain things in it.
Therefore . . . is it possible to have a formula field that can return a text value (let's say "Check Errors") if a long text field is not NULL or does not contain the word "Success"?
Management wants to assign a "weighted amount" to Opportunities based on the Amount and the Probablility %. Example: if prob % = 70% weighted Amount = Amount * 25%, if prob % = 80 weighted Amount = Amount * 30% etc. Would there be a way to have a formula field in Opportnities that would make that calculation against Amount based on what the Prob % is? If yes, any suggestions on how to create it?
Again, I am not a developer or an Excel wizard unfortunately . . . but I know there are some folks out there who probably do this kind of stuff in their sleep. Any help would be welcomed!
Thanks!
Hi . . . could use some help on creating a formula (I suck at them). Management wants to open up the Probability % field to be editable . . . so reps can override the amount that is set by default through Stage selection. I think it's a bad idea, but if they insist I want to be able to restrict users to only being able to enter the % values we already have. How would I create a validation rule that would allow users to enter 0,10,50,70,80 90 or 100% but not allow them to save the record with anything else? I want to keep them from entering 39% and garbage like that.
Hoping someone can offer some help . . . we have a date/time field that tracks the "Last Status Change" for leads. When a Lead Status is updated, a trigger updates the date/time to "now" and then we have a formula field that calculates the number of days between now and that date/time of the last change. However, we only calculate that value for leads that have not been converted and where they don't have a Closed status. The current formula is below and it works great.
IF(AND(NOT( ISNULL( Date_Time_Last_Status_Change__c )),NOT(IsConverted),NOT(ISPICKVAL(Status,"7. Closed"))), (NOW()- Date_Time_Last_Status_Change__c ),NULL)
Now . . . here's the question. There is some talk about changing it to try and only show workdays since the change--to not calculate weekends. I have the following formula to show the workdays. BUT . . . how would I combine the conditions in the formula above and the formula below so I only get workdays since the date/time of change if the lead still meets the conditions of not being converted or closed???
CASE(MOD( DATEVALUE(Date_Time_Last_Status_Change__c) - DATE(1985,6,24),7),
0 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,2,2,3,3,4,4,5,5,5,6,5,1),
1 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,2,2,3,3,4,4,4,5,4,6,5,1),
2 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,2,2,3,3,3,4,3,5,4,6,5,1),
3 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,2,2,2,3,2,4,3,5,4,6,5,1),
4 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,1,2,1,3,2,4,3,5,4,6,5,1),
5 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c),7),1,0,2,1,3,2,4,3,5,4,6,5,0),
6 , CASE( MOD( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) ,7),1,1,2,2,3,3,4,4,5,5,6,5,0),
999)
+
(FLOOR(( DATEVALUE(NOW()) - DATEVALUE(Date_Time_Last_Status_Change__c) )/7)*5)
Looking for a strong SFDC administrator who is also able to craft solutions using tools that go beyond SFDC configurations (i.e. Apex, VisualForce Pages, HTML) while keeping focus on enabling the sales organization to be effective.
http://www.polycom.com/company/careers/career_search.html
Requisition # 285BR
Hi,
I am working on one workflow which sends out some notfication email to user depending on some data from opportunity.
One of the triggering rule is ...the Opportunity Amount > 15000 USD.
Since the currency in the opportunity varies, it could be USD or maybe not.
How can i conver the Opportunity Amount in other currency into USD by using formula?...
I'm struggling with another validation rule. We have a picklist where the sales reps indicate if it's a standard discount or not. Then there is a multi-select picklist where they indicate the particular type of discount they are asking for. I want to have a validation rule that says "If the Standard Discount field = Yes, then the Discount Type field can't have certain options selected. I have figured out how to do a rule to error out checking for one value in the multi-select list. But I can't seem to figure out how to check multiple options with one rule (I'd rather not have 10 separate rules if it can be done with one).
Here is my example which successfully creates an error if the Stadard Discount = Yes and then the Discount type includes "Rebate":
OR(
AND(ISPICKVAL( Standard_Discount__c , "Yes"),
INCLUDES(Discount_Type__c, "Rebate")))
So how do I take this rule and give an error if the Discount Type also includes "Trade In" or "Certification" or "Credit Return" etc.?
Could use some help . . . I created some validation rules to impose certain restrictions on a discount request functionality we have. Works great, when someone plugs in an amount of \$10k or more, the validation rule triggers and stops the save.
The problem is that we have multicurrency enabled--I was expecting the validation rule to catch the conversion rate and to prevent someone from entering what would be over the \$10k USD amount . . . that doesn't happen!!! If I switch the currency on the record to GBP and put in \$9000, the record saves. It only errors when I hit the 10000 value regardless of the currency (and 9999 GBP is way more than \$9999 USD). The formula I am using is below.
AND(ISPICKVAL( Is_this_a_10k_10percent_NST__c , "Yes"),Hardware_Incremental_DiscountUSD__c >=10000)
This is a HUGE problem . . . I need to have this rolled out by Monday and if the system is letting larger GBP/EUR discounts run through without stopping them it will be a disaster. How do you get a validation rule to take multicurrency into account??? Or can it?
Could use a little help here from the masters on this board . . . we have a discount approval process that uses an Object related to Opportunities. Our order management folks really want the Opportunity Owner's name to be reflected on the Object itself (Approval Request Form). Looking at the Opportunity itself is apparently too hard or requires extra steps. Whatever!
Anyway . . . seems like it should be an easy thing to do but all I have been able to do is create a formula field that can pull in the Opportunity "Owner ID"--which means nothing to the people who need to see it. I really need a field that can automatically show the actual Opportunity Owner Name in text on the Approval Form so order management can see "Bob Smith" instead of "00530000000diZm".
I'm sure there is probably something simple I am missing. But how in the world do I get the name to show up in a field on that Approval Form? Is there a formula function that "translates" the ID and can do that? I'm pretty stumped.
• September 15, 2009
• Like
• 0
We have a multiselect field where we can note "Special Designations" for Accounts . . . can be things like Fortune 500 or other special groups we want to associate the Account to without having tons of check boxes. We are also doing some cleanup/lockdown on some specially designated Accounts and I want to restrict users (except admins) from changing the Account Name if a particular value is selected in that Special Designations multiselect list--lets say that special designation is Fortune 500. How would that validation rule be written? I believe that multiselect lists can be included in validation rules now but I'm not sure how to actually do it.
I tried the following but I get an error trying to reference the multiselect picklist. But I am crummy with formulas so not sure what the solution is. Any suggestions would be gratefully welcomed!
AND( IF(\$User.ProfileId <>"00e70000000wI69", true, false),
IF(
Special_Designations__c
,"Fortune 500", false), IF( ISCHANGED( Name ), true, false) )
Is there any way to have a formula field that would do a simple list of the product names included on an opportunity? Example: an opportunity has Widget A, Widget B and Widget C listed in opportunity products. Would there be a way to populate a field through a formula or workflow rule just to show those product names?
The reason is that I am getting tons of questions for reports based on opportunties that do or do not have individual products or combinations of products. SFDC as far as I can tell does not really do this (like trying to get a report that will exlude ALL opportunities that contain Widget C . . . if you exclude Widget C in a filter you still get opportunities that may have Widget A, Widget B and Widget C--but it just won't show the Widjet C rows).
If there is a way to "summarize" product names in a text field then I could see a way to do this. Is there any hope? Or another way to try and do this?
Back for some formula field assistance. I am working with our IT team to set up a hierarchy structure on SFDC User records so they can trigger some dashboards off that info. So I went to create the fields that they need . . . I have lookup fields on a user record that can be used to look up other SFDC users and put them into a spot on the hierarchy. Those lookups bring back other data from that "manager user" record and shows them on the sales rep's user record (So on Sales Rep #1's user record, there is a lookup for Manager #1 (2, 3, 4, etc) and by selecting that person it returns data for that manager and puts it on the sales rep record for reference). Hope that makes sense. Anyway, it all works fine EXCEPT for one thing. I want to have a field that pulls in another Manager field--but it's a picklist on the Manager User record. And I get syntax errors when I try to do a formula field on the Sales Rep record to bring back the Manager User field value from that picklist.
So my question is . . . how do I build a formula field that can simply return the text based off a picklist driven field??? So if the Manager User record value selected from a picklist is "Country Manager", then how can I pull that Country Manager value to the Sales Rep user record? I really want to use a picklist on the Managr record, otherwise I will run into data inconsistencies.
Any suggestions? Can this be done?
I am a bit of a beginner at formulas, so hopefully one of you will think this is easy, and be able to help. :-)
Customer requirement:
Create a formula field that displays the Week Number (as Week 1, Week 2, etc), based on the Date entered on a particular record (same object).
Ex: 01/01/09 - 01/04/09 would return a value of "Week 1"
01/05/09 - 01/11/09 = Week 2
01/12/09 - 01/18/09 = Week 3
and so on.....
Is using the CASE function the way to go, and if so, what's the shortest way to create this formula?
S
• April 20, 2009
• Like
• 0
Hi,
I want to trigger a workflow rule 1 on object A, with the Evaluation Criteria as "Every time a record is created or edited".
I have another time-based workflow rule 2 which will updatea filed on Object A.
So at a certain time, the field on Object A should be updated by workflow rule 2 and then workflow rule 1 will be triggered.
However, the result turns out to be the field was updated but workflow rule 1 was not triggered at all. If I manually update that field, then I can see the expected result.
Any ideas about this? I know the update of formula filed won't be considered as a edit action but this field is just a text field.
We have a process that runs under a "dummy admin user" and it will periodically makes updates to a record. We want to be able to more easily track when the latest change from this user was made (and potentially generate workflow rules, etc from this field). What I am thinking of doing is just to have a field that whenever the "Last Modified" field is updated as being related to this particular user . . . . the field "Last Modified by X" field would show that last modified date. Is that possible? If so, how would that formula be written?
Here is maybe a better example of the thing I am hoping to do. "User X" makes a modification to a Lead and so the Last Modified By field shows "User X 4/8/2009". I am hoping to have a field called Last Modified by User x that would show 4/8/2009. That way we can have a view/report/workflow that can work off that field. The reason we want this is that other users could make updates to the records after User X and they become hard to find. With the formula field, we can easily see Leads this week/month that User X modifed.
Is there a way to write a validation rule that can trigger an email notification whenever ownership of a lead changes? I know I'll feel dumb after seeing it, but how would the rule be written to do this?
Leads change hands here quite a bit and people don;t always remember to check the "Send Notification" box. I'd like to see if I can have a workflow rule that will automatically trigger an alert to the new owner without that box getting checked.
So I just noticed that when you close an Opportunity in SFDC, the Close Date does not automatically update to the day that the closed stage was selected. For some reason, I always thought (or assumed) that it did. Does anyone use validation rules or workflow rules in any way to automatically set the close date to "today" when a Stage is changed from an open to a closed value? I'd love to hear how you've done it if you have.
If the rep does not update Close Date, we can get deals falling in quarters where they do not belong and other messy data.
Can someone please assist me with a formula to block anyone except for the Sys. Admin to close out an Oppt. to a Win. | 7,303 | 29,649 | {"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} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.72586 |
http://www.numbersaplenty.com/12103331 | 1,555,741,770,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528702.42/warc/CC-MAIN-20190420060931-20190420082931-00321.warc.gz | 274,783,351 | 3,386 | Search a number
12103331 is a prime number
BaseRepresentation
bin101110001010…
…111010100011
3211202220122112
4232022322203
511044301311
61111225535
7204606452
oct56127243
924686575
1012103331
116917459
1240782ab
13267a046
141870b99
1510e128b
hexb8aea3
12103331 has 2 divisors, whose sum is σ = 12103332. Its totient is φ = 12103330.
The previous prime is 12103279. The next prime is 12103363. The reversal of 12103331 is 13330121.
Adding to 12103331 its reverse (13330121), we get a palindrome (25433452).
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 12103331 - 218 = 11841187 is a prime.
It is a super-2 number, since 2×121033312 = 292981242591122, which contains 22 as substring.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (12103381) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a good prime.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6051665 + 6051666.
It is an arithmetic number, because the mean of its divisors is an integer number (6051666).
It is a self-describing number.
Almost surely, 212103331 is an apocalyptic number.
12103331 is a deficient number, since it is larger than the sum of its proper divisors (1).
12103331 is an equidigital number, since it uses as much as digits as its factorization.
12103331 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 54, while the sum is 14.
The square root of 12103331 is about 3478.9841908235. The cubic root of 12103331 is about 229.5981072315.
The spelling of 12103331 in words is "twelve million, one hundred three thousand, three hundred thirty-one". | 533 | 1,804 | {"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} | 3.4375 | 3 | CC-MAIN-2019-18 | latest | en | 0.880799 |
http://www.math.umd.edu/~diom/courses/673/syllabus.html | 1,516,439,119,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889542.47/warc/CC-MAIN-20180120083038-20180120103038-00153.warc.gz | 508,816,569 | 1,706 | ## Prerequisite:
Typically, MATH 411 (Advanced Calculus II), or equivalent. The course is largely self-contained; however, prior familiarity of students with analytic proofs will be very helpful. Ask for instructor's permission if you are in doubt. Handouts with reviews will be given in class.
## Focus:
Concepts and analytic techniques that permeate the rigorous theory of Partial Differential Equations (PDEs), especially PDEs that arise in applications across different disciplines. Emphasis will be placed on the existence, uniqueness and regularity of solutions, as well as special solution techniques (transforms, power series etc).
## PART I:
Introduction: Classical and weak solutions; regularity. Major linear PDEs.
Transport equation: Initial-value problem; nonhomogeneous problem.
Laplace and Poisson equations: Derivations; boundary value problems; fundamental solution; maximum principle; properties of harmonic functions; Green's function; energy methods.
Heat equation: Derivations; initial value problems; fundamental solution; properties and estimates; energy methods.
Wave equation: Derivations; initial value problems; d'Alembert formula; solution by spherical means; nonhomogeneous problem; energy methods.
Applications in classical and quantum mechanics.
## PART II:
Nonlinear first-order PDEs: Complete integrals; characteristics; calculus of variations and Hamilton-Jacobi equations; conservations laws; shock formation and entropy condition; weak solutions; the Riemann problem. Applications in gas dynamics, materials science and fluid mechanics.
## PART III:
Special representations of solutions: Fourier transform; conversion of nonlinear to linear PDEs (Hopf-Cole transform, potential functions, hodograph and Legendre transforms). Applications in fluid mechanics, statistical physics and materials science.
## PART IV:
Maximum principles for 2nd-order elliptic equations:
Weak maximum principle; strong maximum principle; Harnack's inequality. | 388 | 1,984 | {"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-2018-05 | latest | en | 0.842149 |
http://www.jiskha.com/display.cgi?id=1335324622 | 1,462,430,539,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125897.68/warc/CC-MAIN-20160428161525-00185-ip-10-239-7-51.ec2.internal.warc.gz | 616,199,006 | 3,863 | Thursday
May 5, 2016
# Homework Help: Auburn
Posted by Rose on Tuesday, April 24, 2012 at 11:30pm.
Have a sack of jelly beans-5 different colors-half are red 1/6 green 1/6 yellow 1/12 blue 6 black-how many of each color and how many in all
• Auburn - PsyDAG, Wednesday, April 25, 2012 at 10:55am
Let x = total number
x - 1/2x - 1/6x - 1/6x - 1/12x = 6
Solve for x and then the fractions.
• math - PsyDAG, Wednesday, April 25, 2012 at 10:56am
In the future, indicate your subject in the "School Subject" box, so those with expertise in the area will respond to the question. | 197 | 581 | {"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} | 2.875 | 3 | CC-MAIN-2016-18 | longest | en | 0.903776 |
https://www.allthescience.org/what-is-an-induced-current.htm | 1,702,144,288,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00769.warc.gz | 679,855,442 | 22,833 | Science
Fact Checked
What is an Induced Current?
D.R. Satori
D.R. Satori
A closed electrical circuit with energy flowing illustrates the two parts of the electromagnetic force: electricity and magnetism. Electricity is produced as electrons are pushed through a conducting wire by a source of voltage — a battery, for example. Not only does the original wire possess a flow of electrons, it also produces a magnetic field around that flow. These behaviors of electromagnetic energy come paired: the flow of electrons and the magnetic field that the flow creates. If one wire in which electrons flow is brought close to another wire, the magnetic field of the first wire will induce a flow — an induced current — along the second wire.
In 1831, Michael Faraday published his discovery that a current in one wire could induce current in another wire. In 1862, this phenomenon of induced current was described mathematically by James Clerk Maxwell; it was based on the equations of colleagues who had described other energy exchanges, such as stress flows in solids and fluid flows in liquids. Maxwell’s equation illuminates the reasons for induced current, or inductance, by showing that the flow of electricity can be measured in two ways: as a voltage drop that forces electron flow and as a magnetic flow field, originating in the flow.
Induced current can be amplified when an electrically conducting wire is forced into a tight coil in the direction of a current. A transformer works by placing coils from two circuits parallel and close to one another so that electrical energy is transferred from one circuit to the next. This inductive coupling occurs when the magnetic fields emanating from the coils intersect in phase with one another, transferring a maximum amount of energy. This exchange is similar to the push given a child on a swing: if the push is timed properly, the swing is propelled upward at the maximum speed.
When a wire with a current is coiled about an iron rod, it can produce a magnetic field that can attract or repel the magnetic field of another such electromagnet. A motor and a generator each consist of two magnets, one moving and one fixed. The moving magnet, when it contacts the stationary magnet, induces a change in direction of electron flow, which results in the magnets repelling one another. This change in direction of induced current creates an alternating push and pull, causing the moving magnet to rotate. Inductance can work in the opposite direction when mechanical energy from a rotating propeller attached to a generator magnet forces the flow of electrons into storage batteries. | 514 | 2,637 | {"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} | 3.203125 | 3 | CC-MAIN-2023-50 | longest | en | 0.958781 |
http://nathanbrixius.wordpress.com/2009/04/27/solving-traveling-salesman-problems-using-solver-foundation/ | 1,386,692,625,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164022163/warc/CC-MAIN-20131204133342-00008-ip-10-33-133-15.ec2.internal.warc.gz | 129,590,534 | 25,953 | Home > Optimization, Solver Foundation > Solving traveling salesman problems using Solver Foundation
## Solving traveling salesman problems using Solver Foundation
Update: see the comments below for some helpful hints. If you are unable to run this with your version of Solver Foundation and Gurobi, consider installing the lp_solve plugin for MSF. More details on this thread.
Here’s an example that I walked through during yesterday’s INFORMS session. Erwin has two blog postings about Solver Foundation and the traveling salesman problem, but I want to throw in my two cents because I want to emphasize a couple of points:
1. By combining C# and Solver Foundation Services it is possible to express complex models clearly and succinctly.
2. It is very easy to build powerful, reusable model libraries using C# and Solver Foundation Services.
3. Solver Foundation Services code can be used in many different application environments (ASP.Net, silverlight, DB, command line apps, WPF, …) with minimal changes.
The traveling salesman problem is a classical problem in computer science, and you should bow your head in shame if you don’t know about it (and turn in your conference badge if you happen to be in Phoenix). A salesperson needs to make a tour of a number of cities. The restrictions are that she wants to visit each city once and only once, and she wants to minimize the distance travelled. This is perhaps the definitive example of an NP-hard problem.
TSP can be solved using mixed integer programming – optimizing a linear goal with linear constraints, where some of the decision variables are integer. In this first post I will show how to formulate and solve a TSP model using Solver Foundation Services. In my second post I will show how to use the Gurobi MIP solver using SFS. There are many different ways to model the TSP – here is a nice introduction. My goal is to provide a clear, complete example – not build a “production level” TSP model, so I am going to choose a model formulation that dates back to 1960! First, I need to establish a couple of building blocks that will help me construct the data for the model. We need to know the distances between each pair of cities. Typically we are provided the coordinates of the cities and need to derive the distances. So I will introduce a Coordinate class that contains properties for the (x, y) coordinates, and properties to convert to latitude and longitude. Finally, a method that computes the distance between points.
```using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using Microsoft.SolverFoundation.Services;
namespace Microsoft.SolverFoundation.Samples {
class TravelingSalesman {
// TSP coordinate.
public class Coordinate {
public int Name { get; set; }
// X-coordinate (from TSPLIB)
public double X { get; set; }
// Y-coordinate (from TSPLIB)
public double Y { get; set; }
public Coordinate(int name, double x, double y) {
Name = name;
X = x;
Y = y;
}
public double Latitude {
get { return Math.PI * (Math.Truncate(X) + 5 * (X - Math.Truncate(X)) / 3) / 180; }
}
public double Longitude {
get { return Math.PI * (Math.Truncate(Y) + 5 * (Y - Math.Truncate(Y)) / 3) / 180; }
}
// Geographic distance between two points (as an integer).
public int Distance(Coordinate p) {
double q1 = Math.Cos(Longitude - p.Longitude);
double q2 = Math.Cos(Latitude - p.Latitude);
double q3 = Math.Cos(Latitude + p.Latitude);
// There may rounding difficulties her if the points are close together...just sayin'.
return (int)(6378.388 * Math.Acos(0.5 * ((1 + q1) * q2 - (1 - q1) * q3)) + 1);
}
}
// TSP city-city arc.
public class Arc {
public int City1 { get; set; }
public int City2 { get; set; }
public double Distance { get; set; }
}
// Burma14 from TSPLIB. Optimal tour = 3323.
private static Coordinate[] data = new Coordinate[] {
new Coordinate(0, 16.47, 96.10),
new Coordinate(1, 16.47, 94.44),
new Coordinate(2, 20.09, 92.54),
new Coordinate(3, 22.39, 93.37),
new Coordinate(4, 25.23, 97.24),
new Coordinate(5, 22.00, 96.05),
new Coordinate(6, 20.47, 97.02),
new Coordinate(7, 17.20, 96.29),
new Coordinate(8, 16.30, 97.38),
new Coordinate(9, 14.05, 98.12),
new Coordinate(10, 16.53, 97.38),
new Coordinate(11, 21.52, 95.59),
new Coordinate(12, 19.41, 97.13),
new Coordinate(13, 20.09, 94.55)
};
}```
(The data for this 14-city problem comes from the TSPLIB library). If you’ve been following my blog you know that the building blocks of a Solver Foundation model are: sets, parameters, decisions, goals, and constraints. I am going to implement a simple formulation that is centered around the following (indexed) decisions:
• Assign[i,j]: this is equal to 1 if the optimal tour contains a trip (or arc) from city i to city j.
• Rank[i]: this is equal to the number of cities visited after arriving at city i.
We have one parameter in our model:
• Distance[I,j]: the distance from city i to city j.
With that in mind, here’s the model. Explanation of the goals and constraints follow.
```
public static void Run() {
SolverContext context = SolverContext.GetContext();
context.ClearModel();
Model model = context.CreateModel();
// ------------
// Parameters
Set city = new Set(Domain.IntegerNonnegative, "city");
Parameter dist = new Parameter(Domain.Real, "dist", city, city);
var arcs = from p1 in data
from p2 in data
select new Arc { City1 = p1.Name, City2 = p2.Name, Distance = p1.Distance(p2) };
dist.SetBinding(arcs, "Distance", "City1", "City2");
// ------------
// Decisions
Decision assign = new Decision(Domain.IntegerRange(0, 1), "assign", city, city);
Decision rank = new Decision(Domain.RealNonnegative, "rank", city);
// ------------
// Goal: minimize the length of the tour.
Model.Sum(Model.ForEach(city, i => Model.ForEachWhere(city, j => dist[i, j] * assign[i, j], j => i != j))));
// ------------
// Enter and leave each city only once.
int N = data.Length;
Model.ForEach(city, i => Model.Sum(Model.ForEachWhere(city, j => assign[i, j],
j => i != j)) == 1));
Model.ForEach(city, j => Model.Sum(Model.ForEachWhere(city, i => assign[i, j], i => i != j)) == 1));
model.AddConstraint("A1", Model.ForEach(city, i => Model.Sum(Model.ForEachWhere(city, j => assign[i, j], j => i != j)) == 1));
model.AddConstraint("A2", Model.ForEach(city, j => Model.Sum(Model.ForEachWhere(city, i => assign[i, j], i => i != j)) == 1));
// Forbid subtours (Miller, Tucker, Zemlin - 1960...)
Model.ForEach(city,
i => Model.ForEachWhere(city,
j => rank[i] + 1 <= rank[j] + N * (1 - assign[i, j]),
j => Model.And(i != j, i >= 1, j >= 1)
)
)
);
Solution solution = context.Solve();
// Retrieve solution information.
Console.WriteLine("Cost = {0}", goal.ToDouble());
Console.WriteLine("Tour:");
var tour = from p in assign.GetValues() where (double)p[0] > 0.9 select p[2];
foreach (var i in tour.ToArray()) {
Console.Write(i + " -> ");
}
Console.WriteLine();
}```
In my humble opinion, the “Parameter data =” line is an awesome example of the power of LINQ data binding in Solver Foundation. We generate the 2D matrix of distances using a single LINQ expression. It would be incredibly easy to change the code to retrieve the coordinate data from a database (perhaps using a LINQ expression once again), a file, or even a user application.
The goal is straightforward: minimize the distance traveled. This is a product of the selected arcs and the distance matrix. We have two types of constraints:
• Assignment constraints: these ensure that we enter and leave each city only once.
• Subtour constraints: these ensure that we do not have any subtours. In a four city problem {A, B, C, D}, for example, we cannot have two cycles (A, B), (C, D). We need to have one tour that contains all the cities.
The assignment constraints are easy using the ForEach and ForEachWhere operations. I use ForEachWhere because I want to disallow arcs that enter and leave the same city – that doesn’t make sense. The subtour constraint is a little more complicated. It relates the “assign” and “rank” decisions. The key fact is that if there is an arc from city i to city j, rank[i] + 1 == j. Of course, if the (i, j) arc is not part of the optimal tour then all bets are off. Last note: notice that I can mix parameters, decisions, and C# variables in my expressions.
Getting the cost is very easy using goal.ToDouble(). We can get the tour using either Assign or Rank. I have chosen to use Assign because it gives me another opportunity to use LINQ. When you call GetValues() on a decision, you get arrays that contain the value along with the indexes for each decision. In this case, the last entry in the array is the one we are interested in. There are other ways to conveniently query decsision results, I’ll save that for another time.
The next post will show how we can use Solver Foundation’s plug-in model to tune the behavior of the Gurobi MIP solver.
1. 16 September 2012 at 12:49 pm | #1
Dear Nathan,
Thank you very much for the sources and explanation. I’ve got a question, the obtained solution is optimal? Solver Foundation provide exact solution and it is not a heuristic?
Best regards,
Sergey
2. 16 September 2012 at 6:07 pm | #2
Hi Sergey, thanks for your comment. Yes, the obtained solution is provably optimal. Results returned by the linear and mixed integer linear solvers are globally optimal (unless the solution status says otherwise, for example if the problem is infeasible).
Nate
• 18 September 2012 at 2:06 pm | #3
Dear Nathan,
The code works perfectly well as it gives optimal result. But the sequence seems not to be correct. E.g. optimal is: 335, but when I summarize the distances between the obtained nodes in sequence, then it is 783.
I think the optimality is found correct, I tested this code and compared results with Brute-force algorithms, for 6 points. But likely there is an error of extracting the right tour from the solver. It seems like extracted tour is not last found or so.
I would be very grateful if you could give any suggestions.
Best regards,
Sergey
3. 17 September 2012 at 9:32 pm | #4
The solution printed to Console does not appear to be correct or is at least misleading. The output:
5 -> 10 -> 15 -> …
implies you travel from city5 to city10 to city15…
But the output code appears to be written as a list of destinations ordered by origin city with no such that:
city5 comes after 0, city10 comes after 1, city15 comes after 3, etc.
-Ed
4. 18 September 2012 at 2:37 pm | #5
Dear Nathan,
I used the report command:
“Report report = solution.GetReport();
Console.Write(“{0}”, report);”
And it is possible to extract the right sequence out of there from “assign”.
Could you recommend any other elegant way?
Strange, but
” var tour = from p in assign.GetValues() where (double)p[0] > 0.9 select p[2];”
gives something wrong.
Best regards,
Sergey
• 18 September 2012 at 3:31 pm | #6
The problem is in the display of the result. This seems to work:
var tour = from p in assign.GetValues() where (double)p[0] > 0.9 select p;
foreach (var t in tour)
{
Console.WriteLine(“{0} -> {1}”, t[1], t[2]);
}
-Ed
• 18 September 2012 at 6:07 pm | #7
Ed – thanks for identifying the problem!
• 20 September 2012 at 8:15 am | #8
That works. Ed and Nathan, thanks for proposed solution!
5. 20 September 2012 at 12:30 pm | #9
I tried to run this for 25 cities, but it exceeded the Gurobi solver throttle. Is there a way to tell SFS to use a different solver? Is there a good FOSS solver out there with a plugin for SFS?
Thanks,
-Ed
6. 24 September 2012 at 5:44 pm | #15
Hi Sergey,
I have not yet been able to solve using the LPSolve plugin. If I figure it out, I’ll follow up with performance observations. Right now I’m getting the following:
“A call to PInvoke function ‘LpSolvePlugIn!LpSolveNativeInterface.lpsolve::set_timeout’ has unbalanced the stack. This is likely because the managed PInvoke signature does not match the unmanaged target signature. Check that the calling convention and parameters of the PInvoke signature match the target unmanaged signature.”
It is occuring in the plugin code at:
lpsolve.set_timeout(_lp, prms._LpSolveTimeout);
Thanks,
-Ed
7. 01 October 2012 at 2:14 pm | #18
Thanks Sergey,
I’d prefer to find/use a good open source solver and it turns out that the LPSolve is actually working marginally faster for me than Gurobi in the hacked up test case with which I’m playing. Here are two runs on 20 cities:
Version: Microsoft Solver Foundation 3.0.2.10889 Express Edition
Model Name: DefaultModel
Capabilities Applied: MILP
Solve Time (ms): 45
Total Time (ms): 466
Solve Completion Status: Optimal
Solver Selected: SolverFoundation.Plugin.LpSolve.LpSolveSolver
Version: Microsoft Solver Foundation 3.0.2.10889 Express Edition
Model Name: DefaultModel
Capabilities Applied: MILP
Solve Time (ms): 76
Total Time (ms): 534
Solve Completion Status: Optimal
Solver Selected: SolverFoundation.Plugin.Gurobi.GurobiSolver
I’m going to go back through the sample and try to identify the minimal changes to get the LPSolve performing similar to the Gurboi, but two things that may have helped are changing the distance parameter from Domain.Real to Domain.IntegerNonnegative and adding this directive:
context.Solve(new SolverFoundation.Plugin.LpSolve.LpSolveDirective());
-Ed
8. 03 November 2012 at 2:27 am | #19
Hi, Nate, how can one change the code so that it solves the classic TSP, where the salesperson begins and ends in city 0, so that we would have something like 0->3 -> 8 -> … ->0?
• 03 November 2012 at 1:57 pm | #20
Oh, now I got it. This is the piece of code I was looking for:
var tour = from p in assign.GetValues() where (double)p[0] > 0.9 select p[2];
Console.Write(“0 -> “);
int indice = 0;
int currentCity;
for (int i = 0; i < data.Length; i++)
{
currentCity = Convert.ToInt32(tour.ToArray()[indice]);
Console.Write("{0}" + ((i ” :”"), currentCity);
indice = currentCity;
}
Console.WriteLine();
The result is the correct sequence of cities:
0 -> 1 -> 13 -> 2 -> 3 -> 4 -> 5 -> 11 -> 6 -> 12 -> 7 -> 10 -> 8 -> 9 -> 0
• 11 November 2012 at 9:22 am | #21
Hi Ricardo,
But what is the difference, where to start from?
Imagine the sequences: 0-1-2-3-4-0 and 1-2-3-4-0-1
I assume that the tour is actualy the same. You can start anywhere you’d like to.
Am I right?
Best regards,
Sergey
• 11 November 2012 at 10:57 am | #22
Hi, Sergey, you are mathematically correct. But if the salesman lives in city 0, it is only natural for him to start and finish in his own city
9. 12 November 2012 at 4:55 pm | #23
Hi, Sergey, you are mathematically correct. But if the salesman lives in city 0, it is only natural for him to start and finish in his own city
After you obtain a tour, you can just make some manipulations with the order and assign any starting point you want.
10. 12 June 2013 at 3:30 pm | #24
Hi Nathan,
This is excellent code and solved something I truly over complicated. I am using this for solving production line changeover matrixes that are asymmetrical–i.e. if you are making pudding, the line can change over from vanilla to chocolate with a 15 minute wipe down but when going from chocolate to vanilla it takes 8 hours of clean up.
Anyway, my 20X20 matrix ran great and actually cut 13% off of what I thought was the optimal!
What do you think the scalability for this is. We are building up a 43X43 matrix now and want to make sure we are capable of doing 100X100.
Do you or any of the other readers think this will be a problem or have any suggestions before we attempt this?
Steve
11. 28 November 2013 at 11:37 am | #25
I tried your solution on a Graph with some not existed edges and your code did not provide any useful output. Can u please present a TSP solution for situation like I face?
1. 15 April 2012 at 1:56 am | #1
2. 26 January 2013 at 3:02 pm | #2
3. 13 November 2013 at 3:16 am | #3 | 4,164 | 15,902 | {"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-2013-48 | longest | en | 0.890304 |
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