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http://www.next.gr/circuits/QRD1114IR-Sensor-l28914.html | 1,638,392,198,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360951.9/warc/CC-MAIN-20211201203843-20211201233843-00558.warc.gz | 125,476,291 | 18,589 | # QRD1114IR Sensor
Posted on Feb 5, 2014
Initially I chose another infrared sensor for test, PIR Motion Sensor, which was to detect any motion in front of the sensor but it turened out not so sensitive. So that I looked at this QRD1114 IR Sensor for substitute. In this test, I used both as digital and analog input to see what difference it has and how sensitive it is. Object or surfaces must be within 0. 5cm. This circuit will not distinguish
between white and black objects but it will let you know when you are at the edge of a surface. If you want to detect the difference between white or black surfaces, in the circuit the Input pin on the Micro should be an analog to digital converter or some other device that can utilize variable voltage levels. A black surface will give a voltage some where between 0V and 5V and white surfaces will give a voltage of 5V. Determining the voltage level for your black and white surfaces will require experimentation. I have tested object/surface detection and stiil try to figure out Black/White difference detection. Here is the schematic of how I hooked up QRD1114 IR Sensor to PIC18F452. Portd. 1 is used for digital input from QRD1114 IR Sensor`s signal while portb. 7 is for output to light a LED. Here I use PWM to let LED light gradually. The lighness depends on the variable "adcVar", which is in direct-ratio (code 1) or inverse-ratio (code 2) to the distance between reflecting surface and the sensor. In code one, when the reflecting surface is out of the range (0. 5cm), the sensor will read adcVar in the maximum (It should be 1023 in theory but it`s 1016 in fact. Also, the minimum is 30 instead of 0. DEFINE OSC 20 DEFINE ADC_BITS 10 ` Set number of bits in result DEFINE ADC_CLOCK 3 ` Set clock source (3=rc) DEFINE ADC_SAMPLEUS 50 ` Set sampling time in uS TRISA = %11111111 ` Set PORTA to all input ADCON1 = %10000010 ` Set...
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https://stats.stackexchange.com/questions/13078/what-is-the-correct-posterior-when-data-are-sufficient-statistics | 1,716,051,150,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00892.warc.gz | 486,363,301 | 45,321 | # What is the correct posterior when data are sufficient statistics?
Say you have N observations that are iid.
$$\forall i, \quad p(X_i=x_i|\mu,\sigma,I) = \frac{1}{\sqrt{2\pi}\sigma} \exp\left(-\frac{1}{2\sigma^2}(x_i-\mu)^2\right)$$ then $$p(x_1,\dots,x_N|\mu,\sigma,I) = \frac{1}{(\sqrt{2\pi}\sigma)^N} \exp\left(-\frac{N}{2\sigma^2}[(\bar{x}-\mu)^2 + \bar{\sigma}^2]\right)$$ with $\bar{x}$ the sample mean and $\bar{\sigma}^2$ the sample variance. Say instead of the observations, you are given their sufficient statistics: e.g. sample mean, sample variance and number of points. You want to estimate the true mean, variance and number of points. What is the correct posterior (or likelihood) for this case, assuming that the data follow a gaussian distribution? So rigorously speaking, $$p(\bar{x},\bar{\sigma}^2,N|I)=p(x_1,\dots,x_N|\mu,\sigma,I) J(\bar{x},\bar{\sigma}^2,N)$$ where $J$ is the jacobian of the transformation from the $x_i$ to the sufficent statistics. But I can drop the Jacobian when I write my likelihood function of the sufficient statistics, because the Jacobian does not depend on the parameters to be estimated. Right?
Now assume you are only given sample mean and standard deviation. Is there any way to estimate the number of points involved? I feel like not but I have difficulties expressing why.
• $N$ is not know? Could you please confirm? Jul 15, 2011 at 12:05
• not in the last question. Jul 15, 2011 at 15:56
When N is known, you can use the fact that the sample mean and variance are independent (conditionally on $\mu, \sigma$) and have known distributions. You can justify this approach because you already known that the sample mean and variance are sufficient statistics and the likelihood has to factor this way (your comment about the Jacobian is correct as well). Or like @jpillow says, you could also derive the result by starting with the full likelihood and charging through the algebra. You'll end up at the same place
But the sample mean and variance aren't sufficient for $\mu, \sigma$ if $N$ is unknown. It isn't be hard to show mathematically. Start with the likelihood you derived earlier, replacing $N$ with $N(\boldsymbol{x})$ to make the dependence explicit :
$$p(x_1,\dots,x_{N(\boldsymbol{x})}|N=N(\boldsymbol{x}),\mu,\sigma,I) = \frac{1}{(\sqrt{2\pi}\sigma)^{N(\boldsymbol{x})}} \exp\left(-\frac{N(\boldsymbol{x})}{2\sigma^2}[(\bar{x}-\mu)^2 + \bar{\sigma}^2]\right)$$
Now if $\bar x, \bar\sigma$ were sufficient, the above would have to factor as $h(\boldsymbol{x})g_{\mu, \sigma}(\bar \mu, \bar\sigma)$ where $h$ doesn't involve $\mu, \sigma$ and $g$ only depends on the data through the sufficient statistics. Because of the leading term $\frac{1}{(\sqrt{2\pi}\sigma)^{N(\boldsymbol{x})}}$ you can't get there.
I suppose the intuition is that if I give you a sample mean and variance but not the sample size you can't be sure if it came from 5 observations or 5 million...
• great! almost there! now how do you write the likelihood if you're given the sample mean and variance? Cause you have to compute the Jacobian this time, no? Or you simply assume it's a gaussian? Jul 15, 2011 at 16:08
• @yannick I'm not sure what you mean; are you looking for the likelihood in the case of unknown $N$? When you put down the likelihood it's conditional on any unknown parameters, including $N$, so the above expression should work. But you have more fundamental problems in that the data (that is, the sample mean and variance) carry no information about the sample size. You need more information about $N$ coming in through either your prior on $N$, or from some other data source.
– JMS
Jul 15, 2011 at 16:32
• precisely, if the data don't give you N, you have no hope of getting any reliable estimate if your prior is uninformative. But you still want an estimate of the true mean and standard deviation. So how should you write the likelihood to estimate these? Just a one-dimensional gaussian, like I've seen a lot of times? But then you are making an assumption on the value of N(x)! Jul 15, 2011 at 17:37
• @yannick I'm still not sure I understand your point. You can specify the likelihood $p(\bar\mu, \bar\sigma^2 | \mu, \sigma, N)$ based on the sampling distribution of $\bar\mu, \bar\sigma^2$ since you've assumed that $x_i\sim N(\mu, \sigma^2)$. What you want is the marginal posterior for $\mu, \sigma^2$. Taking $D$ as the data and slightly abusing notation, this would involve computing $\sum_{i=1}^{n=\infty} p(\mu, \sigma^2, N=i|D) = \sum_{i=1}^{n=\infty} p(\mu, \sigma^2| N=i, D)p(N=i|D)$.
– JMS
Jul 15, 2011 at 19:21
• But the data have no information about $N$ so $p(N=i|D) = p(N=i)$ and you're just integrating over your prior. So yes, you need an informative prior on $N$. But any posterior inferences are going to be so heavily driven by your prior on $N$ that the results are basically useless unless you can (very strongly) justify your choice of prior.
– JMS
Jul 15, 2011 at 19:24
If you don't know the value of $N$, then we simply integrate it out, the same as any other nuisance parameter. We know that $N$ is discrete, and as long as your standard deviation is not zero, then you know that you have at least $2$ observations, so $N\geq 2$. So we write out the posterior as:
$$p(\mu,\sigma|\overline{x},s^2,I)=\sum_{N=2}^{\infty}p(\mu,\sigma,N|\overline{x},s^2,I)$$ $$=\sum_{N=2}^{\infty}p(N|\overline{x},s^2,I)p(\mu,\sigma|N,\overline{x},s^2,I)=\frac{\sum_{N=2}^{\infty}p(N|\overline{x},s^2,I)p(\mu,\sigma|N,\overline{x},s^2,I)}{\sum_{N=2}^{\infty}p(N|\overline{x},s^2,I)}$$
The last form is written to show that we basically have a weighted average of $p(\mu,\sigma|N,\overline{x},s^2,I)$ for different $N$. This term is the standard posterior distribution you get in most cases, so I won't discuss it here. I will focus on the first term, which is the weights. Now we can re-arrange it using Bayes theorem to get:
$$p(N|\overline{x},s^2,I)\propto p(N|I)p(\overline{x},s^2|N,I)$$
From @JMS comment, he seems to think that $p(\overline{x},s^2|N,I)$ does not depend on $N$, so that we are basically weighting by the prior $p(N|I)$. This is incorrect, unless we have very little prior information about $\mu$ and $\sigma$. I will show this with a specific example, followed by my attempt at explaining why this is the case. Now we can further decompose this as:
$$p(\overline{x},s^2|N,I)=\iint p(\overline{x},s^2,\mu,\sigma|N,I)\,d\mu\, d\sigma$$ $$=\iint p(\overline{x},s^2|\mu,\sigma,N,I)p(\mu,\sigma|N,I)\,d\mu\,d\sigma$$ $$=\iint p(\overline{x}|\mu,\sigma,N,I)p(s^2|\mu,\sigma,N,I)p(\mu,\sigma|N,I)\,d\mu\,d\sigma$$ The last line follows from the know fact that for normal with known mean and variance, sample mean is independent of sample standard deviation. We also have:
$$p(\overline{x}|\mu,\sigma,N,I)=\sqrt{\frac{N}{2\pi\sigma^2}}\exp\left(-\frac{N}{2\sigma^{2}}[\overline{x}-\mu]^2\right)$$ $$\frac{Ns^2}{\sigma^2}\sim \chi^2_{N-1}\implies p(s^2|\mu,\sigma,N,I)=\frac{1}{\Gamma[\frac{N-1}{2}]}\left(\frac{Ns^2}{2}\right)^{(N-1)/2}\sigma^{-(N-1)}\exp\left(-\frac{Ns^2}{2\sigma^2}\right)$$
Now if I was to specialise to the prior $p(\mu,\sigma|N,I)\propto \sigma^{-\alpha-1}\exp\left(-\frac{\beta}{\sigma^2}\right)$ we get:
$$p(\overline{x},s^2|N,I)\propto\frac{\Gamma[\frac{N-1}{2}+\alpha]}{\Gamma[\frac{N-1}{2}]}\left(\frac{Ns^2}{2}\right)^{(N-1)/2}\left(\frac{Ns^2}{2}+\beta\right)^{-(N-1+\alpha)/2}$$
Where factors independent of $N$ have been absorbed into the "$\propto$" symbol. Note that this still depends on $N$ unless we use the Jeffrey's prior $\alpha=\beta=0$. Hence, if you know something about the parameters of the normal distribution, you can use this information to estimate $N$ from $s^2$ and $\overline{x}$. Now where is the intuition behind this? I don't really know, but I'll have a go, as I was initially quite surprised myself at this result, but on a bit of reflection it does make sense.
We know from the chi-square pivotal quantity, that we can roughly expect $Ns^2\approx (N-1)\sigma^2$. Now if we use an informative prior for $\sigma^2$ like I did then we have $\sigma^2\approx \frac{\alpha}{2\beta}$ (this is the prior mean for $\sigma^2$). But we can then use this to solve for $N$, and we get $$N\approx\left(1-\frac{2s^2\beta}{\alpha}\right)^{-1}=\left(1-\frac{s^2}{\hat{\sigma}^{2}_{PRIOR}}\right)^{-1}$$
Now why doesn't this work for the Jeffreys prior? Well, when we use the Jeffreys prior, the prior estimate is undefined so the above equation is arbitrary for Jeffreys prior. Additionally, we essentially estimate $\sigma^2\approx s^2$ with the Jeffreys prior and so we already have $Ns^2\approx (N-1)\sigma^2$ independently of the value of $N$ - hence we cannot use this to help us estimate $N$. You can see this also in the form of $p(\overline{x},s^2|N,I)$ as it essentially has the heuristic form:
$$\frac{\text{normalisation with }\mu,\sigma\text{ estimated from N obs and prior information}}{\text{normalisation with }\mu,\sigma\text{ estimated from N obs}}$$
So if you are in some sort of sequential situation, you can get estimates for $N$ after the receiving the first mean and variance.
• that's a neat contribution! so one can estimate N if one uses a normal-inverse gamma prior for the mean and variance parameters. However, I feel that your estimate is strongly affected by the choice of the prior, and in that sense is not very robust. Basically, your data sets $(N-1)/N\sigma$ and your prior sets $\sigma$. So you can make inferences on $N$, but they don't come from the data, they're only due to the prior. Aug 20, 2011 at 16:32
• @yannick - yes the prior information matters - but this hardly calls for a "non-robust" label - this gives the impression that it must be wrong. If the prior information does matter, then it should only be called "non-robust" if that prior information is incorrectly specified. If you have prior information, why would you want your inference to be independent of that information? That would be just as crazy as ignoring the data. And remember, my explanation was an attempt at intuitively understanding why the posterior was not equal to the prior. Aug 21, 2011 at 21:03
• ...(cont'd)... there may be other features in the posterior which make more accurate predictions. Aug 21, 2011 at 21:04
You can get there more easily if you just collect terms:
$P(\{X_i\}|\mu,\sigma^2) = \prod_{i=1}^N \frac{1}{\sqrt{2\pi \sigma^2}} \exp (-\frac{(x_i-\mu)^2}{2\sigma^2}) = \frac{1}{\sqrt{|2\pi \sigma^2 I|}}\exp (-\frac{\sum (x_i-\mu)^2}{2\sigma^2})$
If you massage the terms in that exponent (start by multiplying out the quadratics), you should be able to get to the desired expression. (This is probably easier than computing the Jacobian explicitly). Actually, it's probably easier if you plug in explicit expressions for sufficient statistics $\bar \mu$ and $\bar \sigma$ and work backward to the expression here.
• i think OP's problem is more complicated than usual because $N$ is also unknown. I am not sure if the question makes sense. $N$ must be known, right? Jul 15, 2011 at 12:05
• hmm not sure because the prefactor does not have the right exponent. What's "I" by the way? And still, if you change from the $X_i$ to the sufficient statistics, you have to compute a Jacobian. Jul 15, 2011 at 16:10
• Sorry, that was just laziness--$I$ is the $N\times N$ identity matrix; this is another way of writing multivariate normal. It's the same as writing $2 \pi \sigma^2$ to the power $N/2$. Jul 15, 2011 at 17:01
• And no--no need to compute the Jacobian here (unless you want to for some reason). The likelihood here is just the product of the likelihoods from each observation. With a little effort you can collect the terms in the product of likelihoods and rewrite them in terms of the sufficient statistics. (Note, as JMS says, you DO need to know $N$). Jul 15, 2011 at 17:13 | 3,509 | 11,885 | {"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} | 3.671875 | 4 | CC-MAIN-2024-22 | latest | en | 0.882687 |
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# RD Sharma Class 12 Solutions Chapter 23 - Algebra of Vectors (Ex 23.5) Exercise 23.5 - Free PDF
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## Class 12 Solutions Free PDF for Maths Written by Experts of Vedantu
Free PDF download of RD Sharma Class 12 Solutions Chapter 23 - Algebra of Vectors Exercise 23.5 solved by Expert Mathematics Teachers on Vedantu.com. All Chapter 23 - Algebra of Vectors Ex 23.5 Questions with Solutions for RD Sharma Class 12 Maths to help you to revise the complete Syllabus and Score More marks.
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The sections of Chapter 23 and its sub-exercises do not only help students to search for the particular section they are looking for but will also help them to concentrate on only a single exercise while studying. This provides a much-improved concentration level than when you see a large number of questions being mentioned. Hence with the sectioning and improved concentration students tend to finish off the topics mentioned in the RD Sharma Class 12 Solutions Chapter 23 - Algebra of Vectors with much more ease.
5. Does the topic mentioned in the RD Sharma Class 12 Solutions Chapter 23 - Algebra of Vectors Ex 23.5 carry any importance for the Class 12 board exam?
Yes, in fact, all of the topics and the questions that are mentioned in the RD Sharma Class 12 Solutions Chapter 23 - Algebra of Vectors Ex 23.5 carry high importance in the main Class 12 board exams and hence making sure you study them well is important. You will also be needed to solve some Vedantu sample papers for CBSE Class 12 Maths which allows you to get an idea of the questions that are involved from Chapter 23 of the Algebra subject. | 1,228 | 5,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-33 | latest | en | 0.912077 |
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# Matt Offredi
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Bit Reversal
Given an unsigned integer _x_, convert it to binary with _n_ bits, reverse the order of the bits, and convert it back to an inte...
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Given an unsigned integer x, find the largest y by rearranging the bits in x
Given an unsigned integer x, find the largest y by rearranging the bits in x. Example: Input x = 10 Output y is 12 ...
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Find the longest sequence of 1's in a binary sequence.
Given a string such as s = '011110010000000100010111' find the length of the longest string of consecutive 1's. In this examp...
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Converting binary to decimals
Convert binary to decimals. Example: 010111 = 23. 110000 = 48.
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Convert given decimal number to binary number.
Convert given decimal number to binary number. Example x=10, then answer must be 1010.
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Check if sorted
Check if sorted. Example: Input x = [1 2 0] Output y is 0
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Calculate area of sector
A=function(r,seta) r is radius of sector, seta is angle of sector, and A is its area. Area of sector A is defined as 0.5*(r^2...
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Pizza!
Given a circular pizza with radius _z_ and thickness _a_, return the pizza's volume. [ _z_ is first input argument.] Non-scor...
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Make roundn function
Make roundn function using round. x=0.55555 y=function(x,1) y=1 y=function(x,2) y=0.6 y=function(x,3) ...
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Rounding off numbers to n decimals
Inspired by a mistake in one of the problems I created, I created this problem where you have to round off a floating point numb...
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Matlab Basics - Rounding III
Write a script to round a large number to the nearest 10,000 e.g. x = 12,358,466,243 --> y = 12,358,470,000
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Matlab Basics - Rounding II
Write a script to round a variable x to 3 decimal places: e.g. x = 2.3456 --> y = 2.346
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Check that number is whole number
Check that number is whole number Say x=15, then answer is 1. x=15.2 , then answer is 0. <http://en.wikipedia.org/wiki/Whole...
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MATLAB Basic: rounding IV
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MATLAB Basic: rounding III
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MATLAB Basic: rounding II
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MATLAB Basic: rounding
Do rounding near to zero Example: -8.8, answer -8 +8.1 answer 8
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Sum the Digits of a Number
Given an integer, sum the digits repeatedly until you end up with a single value less than 10. For example, if you add the di...
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Given a 4x4 matrix, swap the two middle columns
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Get the elements of diagonal and antidiagonal for any m-by-n matrix
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Specific Element Count
Given a vector _v_ and a element _e_, return the number of occurrences of _e_ in _v_. Note: NaNs are equal and there may be n...
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Getting logical indexes
This is a basic MATLAB operation. It is for instructional purposes. --- Logical indexing works like this. thresh = 4...
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Replicate elements in vectors
Replicate each element of a row vector (with NaN) a constant number of times. Examples n=2, A=[1 2 3] -> [1 1 2 2 3 3] ...
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Set the array elements whose value is 13 to 0
Input A either an array or a vector (which can be empty) Output B will be the same size as A . All elements of A equal to 13...
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Find the largest value in the 3D matrix
Given a 3D matrix A, find the largest value. Example >> A = 1:9; >> A = reshape(A,[3 1 3]); >> islargest(A) a...
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Back to basics 21 - Matrix replicating
Covering some basic topics I haven't seen elsewhere on Cody. Given an input matrix, generate an output matrix that consists o...
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Reverse the vector
Reverse the vector elements. Example: Input x = [1,2,3,4,5,6,7,8,9] Output y = [9,8,7,6,5,4,3,2,1]
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Sum all integers from 1 to 2^n
Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10.
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Magic is simple (for beginners)
Determine for a magic square of order n, the magic sum m. For example m=15 for a magic square of order 3.
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Make a random, non-repeating vector.
This is a basic MATLAB operation. It is for instructional purposes. --- If you want to get a random permutation of integer...
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2021-12-02, 17:28 #100 bur Aug 2020 79*6581e-4;3*2539e-3 1E516 Posts How to proceed (sieve, LLR, P-1) depends on which method removes candidates faster. For sieving that's a bit hard to determine since it removes an n=10M almost as fast as n=1M while the LLR of the former takes 100 times more. For P-1 vs LLR it's easy though. Just time how long LLR takes to remove 10 candidates and how long P-1 would take in the same range. I strongly suspect P-1 doesn't make sense. It's only useful for Mersennes because they are so large and have those specially formed factors including the exponent. Here's how I do it: do a few LLR tests and write down the average time. Then sieve for 1-2 days and see if that removes candidates faster or not. If yes, continue sieving, if not, continue LLR until an LLR test again takes longer to remove a candídate than sieving did.
2021-12-02, 18:25 #101
ValerieVonck
Mar 2004
Belgium
5·132 Posts
Quote:
Originally Posted by bur How to proceed (sieve, LLR, P-1) depends on which method removes candidates faster. For sieving that's a bit hard to determine since it removes an n=10M almost as fast as n=1M while the LLR of the former takes 100 times more. For P-1 vs LLR it's easy though. Just time how long LLR takes to remove 10 candidates and how long P-1 would take in the same range. I strongly suspect P-1 doesn't make sense. It's only useful for Mersennes because they are so large and have those specially formed factors including the exponent. Here's how I do it: do a few LLR tests and write down the average time. Then sieve for 1-2 days and see if that removes candidates faster or not. If yes, continue sieving, if not, continue LLR until an LLR test again takes longer to remove a candídate than sieving did.
1. I have retested the primes found by gd_barnes : they are confirmed
2. On my i3 I am sieving the input from gd_barnes to 260T: eta 03/22
3. On my i5 imac I am llr’ing the range 1.3m to 1.4m each candidate takes 733 seconds
4. Sieving on that latest machine I am eliminating a candidate each 500 - 550 seconds
4.1. On a sidenote gd_barnes Saïd to me you are missing factors due to a wrong setting in NewPgen, my file ranges from 1.3m to 50m and I already sieved to 260T … taking this up from the beginning will take also take time
5. I will take a stab at p-1 when my current range is done or will try it on my i7 portable (laptop)
Ftm I am recovering from a mild Covid infection
Kind regards,
Valerie
2021-12-03, 01:51 #102 VBCurtis "Curtis" Feb 2005 Riverside, CA 10100001001112 Posts If you're going to LLR test the entire file, then one should sieve until the factor-removal rate is half as fast (twice the seconds per factor) as an LLR test takes for the smallest candidate in the file. This is due to the way the sieve scales- when you take candidates out of the sieve for LLR, the sieve speed does not change linearly. Since there is not much speed to be gained by taking candidates out, we leave the small candidates in longer than common-sense might suggest. A rough scaling example: If you take out 10% of the candidates from the sieve file, the sieve only runs ~5% faster ("faster" as measured by p/sec).
2021-12-08, 05:20 #103 ValerieVonck Mar 2004 Belgium 5·132 Posts I gathered a few statistics: - Sieving yields a factor each 552 seconds - An llr test takes 735 seconds for the range 1.3M -> 1.4M (which I am currently busy testing) - A P-1 test takes 30 minutes, with bounds B1 = 1M - An ECM test takes 30 minutes for each curve, with bounds B1 = 1M For the moment, I will complete the range 1.3M -> 1.4M Kind regards, Valerie
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Sun Jan 23 00:33:08 UTC 2022 up 183 days, 19:02, 0 users, load averages: 1.09, 1.23, 1.32 | 1,104 | 3,836 | {"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-2022-05 | latest | en | 0.912722 |
https://gishomework.com/intermediate-algebra-4-questions/ | 1,701,897,363,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100603.33/warc/CC-MAIN-20231206194439-20231206224439-00169.warc.gz | 317,792,910 | 8,481 | # Intermediate Algebra 4 Questions
1.)Write the following as an inequality.
Y is less than 4 and greater than or equal to −3
Use y only once in your inequality.
2.)Kira deposits \$2000 into an account that pays simple interest at a rate of 4%
per year. How much interest will she be paid in the first 5 years?
3.)Translate this sentence into an equation.
The product of Raj’s height and 2 is 24.
Use the variable r to represent Raj’s height.
4.)Graph the inequality below on the number line.
<b6
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• Money-back guarantee | 153 | 630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-50 | latest | en | 0.899766 |
http://mathoverflow.net/questions/148731/for-which-n-is-there-only-one-group-of-order-n/148734 | 1,469,385,564,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824133.26/warc/CC-MAIN-20160723071024-00094-ip-10-185-27-174.ec2.internal.warc.gz | 170,453,901 | 18,032 | # For which $n$ is there only one group of order $n$?
Let $f(n)$ denote the number of (isomorphism classes of) groups of order $n$. A couple easy facts:
1. If $n$ is not squarefree, then there are multiple abelian groups of order $n$.
2. If $n \geq 4$ is even, then the dihedral group of order $n$ is non-cyclic.
Thus, if $f(n) = 1$, then $n$ is a squarefree odd number (assuming $n \geq 3$). But the converse is false, since $f(21) = 2$.
Is there a good characterization of $n$ such that $f(n) = 1$? Also, what's the asymptotic density of $\{n: f(n) = 1\}$?
-
(In case people want some data and known results, oeis.org/A000001) – Andrés E. Caicedo Nov 13 '13 at 4:32
Presumably the density of $n$ with $f(n)=1$ is zero, because there are lots of semidirect products. – Lucia Nov 13 '13 at 4:40
Since we always have the cyclic group of order $n$, then $f(n)=1$ if and only if $n$ is a cyclic number. The cyclic numbers are well known: they are the square-free integers $n=p_1\cdots p_r$, where $p_1\lt p_2\lt\cdots\lt p_r$ in which $p_i$ does not divide any of $p_j-1$ for all $j\neq i$. See e.g. Pete Clark's answer here and references cited there. – Arturo Magidin Nov 13 '13 at 6:06
I'd rather have thought this question would be a good candidate for migration to Math.SE ... . – Stefan Kohl Nov 13 '13 at 14:04
@StefanKohl: The first part (values of $n$ for which $f(n)=1$), definitely yes; the second part (asymptotic density of cyclic numbers) seems a better fit for MO, though. – Arturo Magidin Nov 13 '13 at 17:47
$f(n)=1$ if and only if $\gcd(n,\phi(n))=1$, where $\phi$ is the Euler phi-function. These $n$ are tabulated at http://oeis.org/A003277
The result is found in Tibor Szele, Über die endichen Ordnungszahlen, zu denen nur eine Gruppe gehört, Comment. Math. Helv. 20 (1947) 265–267, MR0021934 (9,131b).
-
Let $G(x)$ denote the number of $n \leq x$ such that there is exactly $1$ isomorphism class of groups of order $n$. Then: $$G(x) \sim e^{-\gamma}\frac{x}{\log\log\log(x)}$$ where $\gamma$ is Euler's constant. This is a result of Erdos, Murty and Murty. Their paper also contains other interesting results on the distribution of values of the group order function.
-
Thanks — that's an interesting paper. To clarify, does $\gamma$ denote the Euler–Mascheroni constant? – Daniel Hast Nov 13 '13 at 5:08
@Daniel, Yes. It arises (essentially) through the use of Merten's formula. – Mark Lewko Nov 13 '13 at 5:18
The original question has already been answered, so I thought I would provide a slightly more general version.
The short paper http://www.math.ku.dk/~olsson/manus/three-group-numbers.pdf describes those orders for which there are precisely 1, 2 or 3 groups of the given order.
-
If $p$ is the smallest prime dividing $n$, then if $n=pm$ and $p|\phi(m)$ then there exists a semidirect product of the cyclic group of order $p$ and the cyclic group of order $m$. So $f(n)$ is not $1$ for such $n$. Now given a prime $p$, most values of $m$ that are odd and coprime to $p$ will have $\phi(m)$ being a multiple of $p$ (all we need is some prime factor of $m$ to be $1\pmod p$, and usually $m$ will have some such factors). Since most numbers won't be coprime to all small primes, this will give a proof that the density of numbers with $f(n)=1$ is zero.
Note: Mark Lewko posted the interesting reference to Erdos, Murty & Murty while I was writing the answer above. Comparing our answers, one can see that the numbers with $f(n)=1$ are closely related to the numbers $n$ having no prime factor below $\log \log n$.
- | 1,091 | 3,558 | {"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} | 3.484375 | 3 | CC-MAIN-2016-30 | latest | en | 0.824597 |
https://www.examfocus.com/gre-test-structure/ | 1,719,235,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00858.warc.gz | 667,017,738 | 13,484 | # GRE Test Format
## Everything You Need to Know About GRE Test Structure
The GRE examination score is of paramount importance for all candidates seeking admission in graduate schools of the USA. Thus, as you start preparing for the examination, you will find that knowledge of the GRE test structure is essential. Prior information regarding the types of questions asked, number of sections in the test, format of the examination etc. will help you prepare for the test efficiently. Understanding the GRE StructureThe GRE is designed in such a manner that it evaluates your merit, as a prospective graduate school student, from all aspects. In other words, the GRE score that you receive not only represents your knowledge, but also your critical and analytical skills. You will understand the nature of this examination if you go through the various features of the test, some of which are as follows:The test format of the GRE is such that it incorporates all the subjects that you have studied in high school. The examination consists of 3 parts, viz, Verbal Reasoning, Quantitative Reasoning and Analytical Writing. An outline of the types of questions that you will come across in each section is as follows:
• The first section comprises questions like, reading comprehension, word meanings, drawing inferences from a given text etc.
• The second part, as the name suggests, consists of mathematics questions. Here you will have to solve mathematical problems using your knowledge of arithmetic, algebra, statistics, geometry and probability.
• In the final section of the GRE, you will have to write two essays on the given topics. One of the essays will be analytical in nature and the other has to be argumentative.
If you need more information regarding the types of questions asked in the GRE then click on the following link:http://www.ets.org/gre/revised_general/about/content/. Once you have understood the question types, you must understand the design of the GRE question paper.The Question Paper FormatAnother essential feature that you need to know is regarding the format of the question paper. Each section has a separate structure, since; they consist of a different number of questions and different timings, as well. The composition of each section of the GRE is as follows:
• The Verbal Reasoning section consists of two sub-sections. Each sub-section consists of 20 questions. You will get 60 minutes to solve the entire section.
• The Quantitative Reasoning section also consists of 2 sub-sections, each comprising of 20 questions. You will get a total of 70 minutes to solve all 40 questions.
• The Analytical Writing section consists of 2 writing tasks, which have to be completed in 60 minutes.
If you require more information on the question paper format then click on the following link:http://www.ets.org/gre/revised_general/about/content/cbt/ Knowledge of the GRE registration process completes your understanding of the GRE test structure. This has been discussed in the following section of the article.Thorough Understanding of the Test Registration ProcessWhile gathering knowledge regarding the test structure for GRE, ensure that you give sufficient importance to the test registration process. GRE registration can become confusing, since it requires a number of personal information. Moreover, early registration will make certain that you do not have to pay a hefty late fees fine. So go online and check the registration process and register as early as possible.Thus, once you have thorough knowledge of the GRE test structure, you will be able to develop a preparatory scheme. You will be able to understand the sections that require more attention. Once you have understood that, you can prepare a schedule, based on which you can study for the examination, because proper preparation alone will ensure good results. | 746 | 3,869 | {"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-2024-26 | latest | en | 0.93092 |
https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-16-integrals-and-vector-fields-section-16-4-green-s-theorem-in-the-plane-exercises-16-4-page-979/32 | 1,679,892,481,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00496.warc.gz | 886,856,934 | 13,109 | ## Thomas' Calculus 13th Edition
The integral is $0$ irrespective of the curve $C$.
The tangential form for Green's Theorem can be computed as: The counterclockwise Circulation is : $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy$ $\implies \oint_C x^3 \ dx -y^3 \ dy= \iint_{R} (\dfrac{\partial (-y^3) }{\partial x}-\dfrac{\partial (x^3) }{\partial y}) \ dx \ dy$ $\implies \oint_C x^3 \ dx -y^3 \ dy= \iint_{R} [0-0] \ dx \ dy$ $\implies \oint_C x^3 \ dx -y^3 \ dy=\oint_C x^3 \ dx -y^3 \ dy=0$ This means that the integral is $0$ irrespective of the curve $C$. | 231 | 619 | {"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} | 3.375 | 3 | CC-MAIN-2023-14 | latest | en | 0.630779 |
https://pastecode.io/s/rg805pmr | 1,713,015,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816734.69/warc/CC-MAIN-20240413114018-20240413144018-00036.warc.gz | 441,846,950 | 35,710 | # k-th smallest number
user_0443324754
python
2 years ago
507 B
58
Indexable
Never
```from heapq import *
n, k = tuple(map(int,input().split()))
A = list(map(int,input().split()))
result = ""
h = []
for i in range(k + 1): #generate the A[1,k];
heappush(h, -A[i])
result= result+str(-h[0])+" "
for j in range(k,n): #compare the A[k+1] with h[0](smallest) and ,until k+1 = n;
if A[j] < -h[0]: #if A[k+1] smaller, change the order.(replace A[k+1] with h[0])
-heapreplace(h,-A[j])
result= result+str(-h[0])+" "
print(result.strip())``` | 193 | 563 | {"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-2024-18 | latest | en | 0.515551 |
http://mathforum.org/library/drmath/view/51956.html | 1,498,177,216,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319933.33/warc/CC-MAIN-20170622234435-20170623014435-00385.warc.gz | 250,013,399 | 3,775 | Associated Topics || Dr. Math Home || Search Dr. Math
Calculating the Distance Between a Point and a Plane
```
Date: 10/12/95 at 19:57:12
From: Anonymous
Subject: Distance between a point and a plane
I am trying to calculate the minimum distance between
a point, located in 3-dimensional space, and a plane.
I define the plane by determining the equation
Ax + By + Cz + D = 0, from three points, each with
x,y,z coordinates.( I think I can do this!)
Next I want to determine the minimum distance from this
plane to a point which has x,y,z coordinates. I believe
the minimum distance from the plane to the point is the
perpendicular distance from the plane to the point.
I am unsure of the equations to perform this.
Thanks
Chris
```
```
Date: 10/15/95 at 10:50:45
From: Doctor Ethan
Subject: Re: Distance between a point and a plane
The equation is: Ax + By + Cz + D
_________________
_________________
\/ A^2 + B^2 + C^2
In this equation the x,y,z are the point in question and the A,B,C,D
are the values in the standard expression for the plane.
It is a lot like the equation for the distance from a line.
Now we can figure out how to derive it.
Let us sart by finding a vector normal to the plane.
Certainly the most convienient is {A,B,C}.
Did you know that it will always be normal to the plane?
I will leave it to you to check that truth.
Now the derivation is pretty simple.pick any point on the plane, examine the
vector between it and the desired point.
Let's call the point in question P and the point we have choen on the plane
D. Then we can make a side view drawing that looks like this.
P
**
* *
* *
* *
* *
D******
Now what we are look ing for is that veritcal distance. One way to find it,
would be to find that bottom point and then use the Dot Product to find the
distance between them.
However we can use that Normal vector to do something slicker. I am going to
In this drawing I an going to but the tail of the vector {A,B,C} at point D.
And I will call its head N.
So we have...
P
**
* *
* *
N * *
* * *
* * *
* * *
* * *
** *
D*********O (Call it O just so we can talk about it)
Now again since we know the points D and P, and we know the vector {A,B,C},
we can do a neat trick.
We know from Geometry that
Length of PO
------------ = Cos (Angle P)
Length of DP
Which is
Length of PO = (Length of DP) Cos (Angle P)
[I will refer to this later as equation 1]
But the Cos (Angle P) equals Cos (Angle NDP)
And you may recall that the Dot product of two vectors is also their lengths
times the Cos of the angle in between them.
So {A,B,C} . (P-D) = Length of {A,B,C} * Length of DP * Cos (Angle NDP)
That is pretty close to what we had in equation 1.
All we have to do is divide by length of {A,B,C}
So we have this
{A,B,C} . (P-D)
Length of PO = ---------------
Length of {A,B,C}
Now if we choose P to be {x,y,z} and let D be {0,0,-D/C}
You should see how we got the above answer.
Hope that helps......
-Doctor Ethan, The Geometry Forum
```
Associated Topics:
College Linear Algebra
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 915 | 3,471 | {"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} | 4.21875 | 4 | CC-MAIN-2017-26 | longest | en | 0.944043 |
http://teacher.scholastic.com/activities/wwatch/tguide/teaching_35.htm | 1,369,462,410,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705575935/warc/CC-MAIN-20130516115935-00089-ip-10-60-113-184.ec2.internal.warc.gz | 260,175,708 | 5,767 | Lesson 2: Operation Weather Station
Lesson Introduction: Students work in cooperative groups to create a working weather station that they use to collect data from over a one-week period. During that time, students also collect actual weather information from newspapers or online sources. Students then compare the two sets of data and draw conclusions about weather patterns.
Duration
Seven class sessions
Student Objectives
Students will
1. Understand when exact measurements are preferred over estimates.
2. Learn to use different instruments to report wind speed, wind direction, air pressure, snow/rainfall, and temperature
3. Work cooperatively with classmates to create an operational weather station
Materials
1. Copies of the Weather Data Sheet — two copies per student sheets
2. Daily weather reports (e.g., from daily newspaper or online at sites such as www.weather.com, www.noaa.com, or Web site of local news)
3. Pencils
4. See the materials needed to build each weather tool
Directions
Part 1:
1. Assess students' prior knowledge through discussion and by asking questions such as:
What is weather?
Why is the sky blue?
How can meteorologists predict tomorrow's weather or that a snowstorm is possible in three days?
2. Discuss the difference between guessing or predicting and taking an exact measurement. Ask students to think about instances when you want an exact measurement instead of a prediction. For example: If you were sick, would you want the doctor to guess what your temperature was? Or if your parents are driving a car, should they guess how fast they are going or use a tool to measure their exact miles per hour?
3. Ask students if they know of any tools that can be used for measuring weather. These include an anemometer, barometer, thermometer, rain gauge, and wind vane. Write the names of each of these on the board or in a word bank as you discuss it so students may refer to them later.
4. Have students go to the Gather Data section of Weather Watch to investigate the six different weather forecasting tools.
5. Explain to students they will be creating a weather station with the forecasting tool, which they will use to learn about the weather for the next several days. (Depending on current weather conditions, the weather stations can include either a rain gauge or a snow gauge.)
Part 2
1. Set up a supply station where students can easily pick up all the materials they need to build their weather tools.
2. Divide students into either five or ten groups. Each group will be responsible for one of the tools needed for the weather station (so you will be able to set up one or two complete stations).
3. Give each group a copy of the directions to build an anemometer, wind vane, barometer, thermometer, or rain or snow gauge.
4. Allow approximately 30 minutes for students to build their tool. After all of the tools are completed, have each group explain what their tool does and how it works.
5. Have students set up their weather stations in an area that is far from walls, shrubs, and trees.
Part 3
1. Distribute two copies of the Weather Data Sheet to each student. On the top of one have them write, "Weather Station Data" and atop the other, "Actual Weather Data."
2. Next, have students visit the weather station and follow the directions for experimenting with an anemometer, wind vane, barometer, thermometer, or rain or snow gauge. Have them record the data they collect on the "Weather Station Data" sheet.
3. Once students have come back inside, go over their findings together. Ask students if they believe their results are accurate. Discuss ways you might get a weather forecast that is accurate (e.g., on television, on the radio, by calling a weather station, going online).
4. Using a reliable print or online weather source, find a current weather report. Have students record this data on the "Actual Weather Data" sheet. Compare and contrast any differences between the real and actual data.
5. Repeat steps 2-4 for the next four days.
6. After five days of data collecting, have students look over their results. Discuss how the weather pattern changed over the week. What were the highest and lowest temperatures? Did temperature remain consistent? How did temperature and wind speeds compare? Does one affect the other? Who observed the most rain/snow? How can location determine if an area gets rain or snow? What did you observe about air pressure? When the pressure went up, what happened to your other observations?
Assessment and Evaluation
1. Were students able to follow the directions in order to build their tool properly?
2. Could students explain what each tool did?
3. Did students perform the experiments accurately in order to collect data?
4. Did students correctly record data on both charts?
5. Were students able to see differences between real and estimated weather readings?
6. Could students make connections between certain weather conditions and others such as temperature change and wind direction?
7. Was there anything you could have done differently to make this lesson more successful?
Lesson Extensions
Have students record their daily readings in a line graph for one or more areas of the forecast. Advise students to color code their graph for easy reading. The student readings could be in red while the professional readings could be in blue. | 1,094 | 5,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2013-20 | longest | en | 0.9317 |
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Posted 12 Apr 2016
# Combinations in C++
, 12 Apr 2016 CPOL
Rate this:
An article on finding combinations
## Introduction
Combination is the way of picking a different unique smaller set from a bigger set, without regard to the ordering (positions) of the elements (in the smaller set). This article teaches you how to find combinations. First, I will show you the technique to find combinations. Next, I will go on to explain how to use my source code. The source includes a recursive template version and a non-recursive template version. At the end of the article, I will show you how to find permutations of a smaller set from a bigger set, using both `next_combination()` and `next_permutation()`.
Before all these, let me first introduce to you the technique of finding combinations.
## The Technique
• n: n is the larger sequence from which the r sequence is picked.
• r: r is the smaller sequence picked from the n sequence.
• c: c is the formula for the total number of possible combinations of r, picked from n distinct objects: n! / (r! (n-r)!).
• The ! postfix means factorial.
### Explanation
Let me explain using a very simple example: finding all combinations of 2 from a set of 6 letters {A, B, C, D, E, F}. The first combination is AB and the last is EF.
The total number of possible combinations is: n!/(r!(n-r)!)=6!/(2!(6-2)!)=15 combinations.
Let me show you all the combinations first:
```AB
AC
AE
AF
BC
BD
BE
BF
CD
CE
CF
DE
DF
EF```
If you can't spot the pattern, here it is:
```AB | AB
A | AC
A | AE
A | AF
---|----
BC | BC
B | BD
B | BE
B | BF
---|----
CD | CD
C | CE
C | CF
---|----
DE | DE
D | DF
---|----
EF | EF```
The same thing goes to combinations of any number of letters. Let me give you a few more examples and then you can figure them out yourself.
Combinations of 3 letters from {A, B, C, D, E} (a set of 5 letters).
The total number of possible combinations is: 10
```A B C
A B D
A B E
A C D
A C E
A D E
B C D
B C E
B D E
C D E```
Combinations of 4 letters from {A, B, C, D, E, F} (a set of 6 letters).
The total number of possible combinations is: 15.
```A B C D
A B C E
A B C F
A B D E
A B D F
A B E F
A C D E
A C D F
A C E F
A D E F
B C D E
B C D F
B C E F
B D E F
C D E F```
I'm thinking if you would have noticed by now, the number of times a letter appears. The formula for the number of times a letter appears in all possible combinations is n!/(r!(n-r)!) * r / n == c * r / n. Using the above example, it would be 15 * 4 / 6 = 10 times. All the letters {A, B, C, D, E, F} appear 10 times as shown. You can count them yourself to prove it.
## Source Code Section
Please note that all the combination functions are now enclosed in the `stdcomb` namespace.
### The Recursive Way
I have made a recursive function, `char_combination()` which, as its name implies, takes in character arrays and processes them. The source code and examples of using `char_combination()` are in char_comb_ex.cpp. I'll stop to mention that function. For now, our focus is on `recursive_combination()`, a template function, which I wrote using `char_combination()` as a guideline.
The function is defined in combination.h as below:
```// Recursive template function
template <class RanIt, class Func>
void recursive_combination(RanIt nbegin, RanIt nend, int n_column,
RanIt rbegin, RanIt rend, int r_column,int loop, Func func)
{
int r_size=rend-rbegin;
int localloop=loop;
int local_n_column=n_column;
//A different combination is out
if(r_column>(r_size-1))
{
func(rbegin,rend);
return;
}
//===========================
for(int i=0;i<=loop;++i)
{
RanIt it1=rbegin;
for(int cnt=0;cnt<r_column;++cnt)
{
++it1;
}
RanIt it2=nbegin;
for(int cnt2=0;cnt2<n_column+i;++cnt2)
{
++it2;
}
*it1=*it2;
++local_n_column;
recursive_combination(nbegin,nend,local_n_column,
rbegin,rend,r_column+1,localloop,func);
--localloop;
}
}```
The parameters prefixed with '`n`' are associated with the n sequence, while the r-prefixed one are r sequence related. As an end user, you need not bother about those parameters. What you need to know is `func`. `func` is a function defined by you. If the combination function finds combinations recursively, there must exist a way the user can process each combination. The solution is a function pointer which takes in two parameters of type `RanIt` (stands for Random Iterator). You are the one who defines this function. In this way, encapsulation is achieved. You need not know how `recursive_combination()` internally works, you just need to know that it calls `func` whenever there is a different combination, and you just need to define the `func()` function to process the combination. It must be noted that `func()` should not write to the two iterators passed to it.
The typical way of filling out the parameters is `n_column` and `r_column` is always 0, `loop` is the number of elements in the r sequence minus that of the n sequence, and `func` is the function pointer to your function (`nbegin` and `nend`, and `rbegin` and `rend` are self-explanatory; they are the first iterators and the one past the last iterators of the respective sequences).
Just for your information, the maximum depth of the recursion done is r+1. In the last recursion (r+1 recursion), each new combination is formed.
An example of using `recursive_combination()` with raw character arrays is shown below:
```#include <iostream>
#include <vector>
#include <string>
#include "combination.h"
using namespace std;
using namespace stdcomb;
void display(char* begin,char* end)
{
cout<<begin<<endl;
}
int main()
{
char ca[]="123456";
char cb[]="1234";
recursive_combination(ca,ca+6,0,
cb,cb+4,0,6-4,display);
cout<<"Complete!"<<endl;
return 0;
}```
An example of using `recursive_combination()` with a vector of integers is shown below:
```#include <iostream>
#include <vector>
#include <string>
#include "combination.h"
typedef vector<int>::iterator vii;
void display(vii begin,vii end)
{
for (vii it=begin;it!=end;++it)
cout<<*it;
cout<<endl;
}
int main()
{
vector<int> ca;
ca.push_back (1);
ca.push_back (2);
ca.push_back (3);
ca.push_back (4);
ca.push_back (5);
ca.push_back (6);
vector<int> cb;
cb.push_back (1);
cb.push_back (2);
cb.push_back (3);
cb.push_back (4);
recursive_combination(ca.begin (),ca.end(),0,
cb.begin(),cb.end(),0,6-4,display);
cout<<"Complete!"<<endl;
return 0;
}```
### The Non-Recursive Way
If you have misgivings about using the recursive method, there is a non-recursive template function for you to choose (actually there are two).
The parameters are even simpler than the recursive version. Here's the function definition in combination.h:
```template <class BidIt>
bool next_combination(BidIt n_begin, BidIt n_end,
BidIt r_begin, BidIt r_end);
template <class BidIt>
bool next_combination(BidIt n_begin, BidIt n_end,
BidIt r_begin, BidIt r_end, Prediate Equal );```
And its reverse counterpart version:
```template <class BidIt>
bool prev_combination(BidIt n_begin, BidIt n_end,
BidIt r_begin, BidIt r_end);
template <class BidIt>
bool prev_combination(BidIt n_begin, BidIt n_end,
BidIt r_begin, BidIt r_end, , Prediate Equal );```
The parameters `n_begin` and `n_end` are the first and the last iterators for the n sequence. And, `r_begin` and `r_end` are iterators for the r sequence. `Equal` is the predicate for comparing equality.
You can peruse the source code for these two functions in combination.h and its examples in next_comb_ex.cpp and prev_comb_ex.cpp, if you want.
A typical way of using `next_combination` with raw character arrays is as below:
```#include <iostream>
#include <vector>
#include <string>
#include "combination.h"
using namespace std;
using namespace stdcomb;
int main()
{
char ca[]="123456";
char cb[]="1234";
do
{
cout<<cb<<endl;
}
while(next_combination(ca,ca+6,cb,cb+4));
cout<<"Complete!"<<endl;
return 0;
}```
A typical way of using `next_combination` with a vector of integers is as below:
```#include <iostream>
#include <vector>
#include <string>
#include "combination.h"
template<class BidIt>
void display(BidIt begin,BidIt end)
{
for (BidIt it=begin;it!=end;++it)
cout<<*it<<" ";
cout<<endl;
}
int main()
{
vector<int> ca;
ca.push_back (1);
ca.push_back (2);
ca.push_back (3);
ca.push_back (4);
ca.push_back (5);
ca.push_back (6);
vector<int> cb;
cb.push_back (1);
cb.push_back (2);
cb.push_back (3);
cb.push_back (4);
do
{
display(cb.begin(),cb.end());
}
while(next_combination(ca.begin (),ca.end (),cb.begin (),cb.end()) );
cout<<"Complete!"<<endl;
return 0;
}```
### Certain Conditions Must Be Satisfied In Order For next_combination() to Work
1. All the objects in the n sequence must be distinct.
2. For `next_combination()`, the r sequence must be initialized to the first r-th elements of the n sequence in the first call. For example, to find combinations of r=4 out of n=6 {1,2,3,4,5,6}, the r sequence must be initialised to {1,2,3,4} before the first call.
3. As for `prev_combination()`, the r sequence must be initialised to the last r-th elements of the n sequence in the first call. For example, to find combinations of r=4 out of n=6 {1,2,3,4,5,6}, the r sequence must be initialsed to {3,4,5,6} before the first call.
4. The n sequence must not change throughout the process of finding all the combinations, else results are wrong (makes sense, right?).
5. `next_combination()` and `prev_combination()` operate on data types with the `==` operator defined. That is to mean if you want to use `next_combination()` on sequences of objects instead of sequences of POD (Plain Old Data), the class from which these objects are instantiated must have an overloaded `==` operator defined, or you can use the predicate versions.
When the above conditions are not satisfied, results are undetermined even if `next_combination()` and `prev_combination()` may return `true`.
### Return Value
When `next_combination()` returns `false`, no more next combinations can be found, and the r sequence remains unaltered. Same for `prev_combination()`.
### Some Information About next_combination() and prev_combination()
1. The n and r sequences need not be sorted to use `next_combination()` or `prev_combination()`.
2. `next_combination()` and `prev_combination()` do not use any `static` variables, so it is alright to find combinations of another sequence of a different data type, even when the current finding of combinations of the current sequence have not reached the last combination. In other words, no reset is needed for `next_combination()` and `prev_combination()`.
Examples of how to use these two functions are in next_comb_ex.cpp and prev_comb_ex.cpp.
## So What Can We Do With next_combination()?
With `next_combination()` and `next_permutation()` from STL algorithms, we can find permutations!!
The formula for the total number of permutations of the r sequence, picked from the n sequence, is: n!/(n-r)!
We can call `next_combination()` first, then `next_permutation()` iteratively; that way, we will find all the permutations. A typical way of using them is as follows:
```sort(n.begin(),n.end());
do
{
sort(r.begin(),r.end());
//do your processing on the new combination here
do
{
//do your processing on the new permutation here
}
while(next_permutation(r2.begin(),r2.end()))
}
while(next_combination(n.begin(),n.end(),r.begin(),r.end() ));```
However, I must mention that there exists a limitation for the above code. The n and r sequences must be sorted in ascending order in order for it to work. This is because `next_permutation()` will return `false` when it encounters a sequence in descending order. The solution to this problem for unsorted sequences is as follows:
```do
{
//do your processing on the new combination here
for(cnt i=0;cnt<24;++cnt)
{
next_permutation(r2.begin(),r2.end());
//do your processing on the new permutation here
}
}
while(next_combination(n.begin(),n.end(),r.begin(),r.end() ));```
However, this method requires you to calculate the number of permutations beforehand.
## So How Do I Prove They Are Distinct Permutations?
There is a `set` container class in STL we can use. All the objects in the `set` container are always in sorted order, and there are no duplicate objects. For our purpose, we will use this `insert()` member function:
`pair <iterator, bool> insert(const value_type& _Val);`
The `insert()` member function returns a pair, whose `bool` component returns `true` if an insertion is made, and `false` if the `set` already contains an element whose key had an equivalent value in the ordering, and whose iterator component returns the address where a new element is inserted or where the element is already located.
proof.cpp is written for this purpose, using the STL `set` container to prove that the permutations generated are unique. You can play around with this, but you should first calculate the number of permutations which would be generated. Too many permutations may take ages to complete (partly due to the working of the `set` container), or worse, you may run out of memory!
If you are interested, you can proceed to read the second part of the article: Combinations in C++, Part 2.
## History
• 14th September 2009 - Added the example code
• 21st February 2008 - Added the finding combinations of vectors in the source code
• 26th November 2006 - Source code changes and bug fixes
• All functions are enclosed in the `stdcomb` namespace
• Solved a bug in `prev_combination` that `!=` operator must be defined for the custom class, unless the data type is a POD
• `next_combination` and `prev_combination` now run properly in Visual C++ 8.0, without disabling the checked iterator
• `next_combination` and `prev_combination` have a predicates version
• 30th July 2003 - First release on CodeProject
## Share
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Article Copyright 2016 by Shao Voon Wong | 4,074 | 15,568 | {"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-2017-13 | latest | en | 0.818356 |
https://jumbotutors.com/psy-870-module-7-problem-set-optimism-and-longevity-47698/ | 1,643,100,679,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304798.1/warc/CC-MAIN-20220125070039-20220125100039-00355.warc.gz | 375,351,766 | 11,746 | # Psy 870 module 7 problem set – optimism and longevity 47698
Question
Module 7 Problem Set
This problem set introduces you to the use of SPSS for analyzing data with multiple predictor variables and one continuous scale DV to investigate comparison of means. You will perform a multiple regression analysis on the data and report your output.
General Requirements:
Use the following information to ensure successful completion of the assignment:
· Download the SPSS/PASW data set file “Module 7 SPSS Data File,” and use it for this assignment.
· Download the “Module 7 Problem Set” file and use it for this assignment.
Directions:
Perform the following tasks to complete this assignment:
1. Conduct necessary analyses using SPSS so you can answer the questions listed in the exercise.
2. Submit your responses to the exercise questions as a Word document.
3. Submit the SPSS Output files showing the analyses you performed in SPSS to compute the answers for related questions. NOTE: You will need to copy the SPSS file to a Word doc for submission.
Module 7 Problem Set
Optimism and Longevity
A cancer specialist from the Los Angeles County General Hospital (LACGH) rated patient optimism in 20- to 40-year-old male patients with incurable cancer in 1970. In 1990, the researcher examined hospital records to gather the following data:
· Socioeconomic status (1–7 rating of occupation; higher ratings indicate higher levels of SES)
· Age in 1970
· Optimism in 1970 (1–100 rating, higher scores indicate higher levels of optimism)
· Longevity (years lived after the 1970 diagnosis)
Using the SPSS data file for Module 7 (located in Topic Materials), calculate a simultaneous multiple regression with SES, age, and optimism as the independent variables and longevity as the dependent variable.
1. Do the independent variables correlate statistically significantly and practically with the dependent variable?
2. Is collinearity between the independent variables a concern?
3. What is the Rand adjusted R-square for all independent variables entered simultaneously?
4. What variable(s) provide a significant unique contribution(s)?
5. Compose a results section for this statistical analysis. | 455 | 2,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-05 | latest | en | 0.851582 |
https://www.physicsforums.com/threads/proof-by-induction-problem.540561/ | 1,550,929,798,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249501174.94/warc/CC-MAIN-20190223122420-20190223144420-00344.warc.gz | 907,762,936 | 12,005 | Proof by Induction Problem
1. Oct 15, 2011
Ryuky
1. The problem statement, all variables and given/known data
So we have to prove that $\frac{(n+1)(n+2)(n+3)...(2n)}{1*3*5...*(2n-1)}$ = 2n
2. The attempt at a solution
I. For n=1, obviously the proposition is true. (2*1/(2-1) = 2^1 = 2)
II. Let n=k and assume $\frac{(k+1)(k+2)(k+3)...(2k)}{1*3*5...*(2k-1)}$ = 2k.
Now, for n=k+1 we have: 2k * $\frac{(2k +2)}{2k+1)}$ = 2k+1 → 2(k+1)*$\frac{(k+1)}{2k+1)}$ = 2k+1. Which is not true.
So, I cannot figure out what am I doing wrong. | 243 | 538 | {"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 | 4 | CC-MAIN-2019-09 | latest | en | 0.835394 |
https://brainmodels.readthedocs.io/en/latest/_modules/brainmodels/synapses/dual_exp.html | 1,643,225,798,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304961.89/warc/CC-MAIN-20220126192506-20220126222506-00683.warc.gz | 198,498,557 | 6,261 | # Source code for brainmodels.synapses.dual_exp
# -*- coding: utf-8 -*-
import brainpy as bp
import brainpy.math as bm
from .base import Synapse
__all__ = [
'DualExpCUBA', 'DualExpCOBA',
]
[docs]class DualExpCUBA(Synapse):
r"""Current-based dual exponential synapse model.
**Model Descriptions**
The dual exponential synapse model [1]_, also named as *difference of two exponentials* model,
is given by:
.. math::
g_{\mathrm{syn}}(t)=g_{\mathrm{max}} \frac{\tau_{1} \tau_{2}}{
\tau_{1}-\tau_{2}}\left(\exp \left(-\frac{t-t_{0}}{\tau_{1}}\right)
-\exp \left(-\frac{t-t_{0}}{\tau_{2}}\right)\right)
where :math:\tau_1 is the time constant of the decay phase, :math:\tau_2
is the time constant of the rise phase, :math:t_0 is the time of the pre-synaptic
spike, :math:g_{\mathrm{max}} is the maximal conductance.
However, in practice, this formula is hard to implement. The equivalent solution is
two coupled linear differential equations [2]_:
.. math::
\begin{aligned}
&g_{\mathrm{syn}}(t)=g_{\mathrm{max}} g \\
&\frac{d g}{d t}=-\frac{g}{\tau_{\mathrm{decay}}}+h \\
&\frac{d h}{d t}=-\frac{h}{\tau_{\text {rise }}}+ \delta\left(t_{0}-t\right),
\end{aligned}
The current onto the post-synaptic neuron is given by
.. math::
I_{syn}(t) = g_{\mathrm{syn}}(t).
**Model Examples**
.. plot::
:include-source: True
>>> import brainpy as bp
>>> import brainmodels
>>> import matplotlib.pyplot as plt
>>>
>>> neu1 = brainmodels.neurons.LIF(1)
>>> neu2 = brainmodels.neurons.LIF(1)
>>> syn1 = brainmodels.synapses.DualExpCUBA(neu1, neu2, bp.connect.All2All())
>>> net = bp.Network(pre=neu1, syn=syn1, post=neu2)
>>>
>>> runner = bp.StructRunner(net, inputs=[('pre.input', 25.)], monitors=['pre.V', 'post.V', 'syn.g', 'syn.h'])
>>> runner.run(150.)
>>>
>>> fig, gs = bp.visualize.get_figure(2, 1, 3, 8)
>>> plt.plot(runner.mon.ts, runner.mon['pre.V'], label='pre-V')
>>> plt.plot(runner.mon.ts, runner.mon['post.V'], label='post-V')
>>> plt.legend()
>>>
>>> plt.plot(runner.mon.ts, runner.mon['syn.g'], label='g')
>>> plt.plot(runner.mon.ts, runner.mon['syn.h'], label='h')
>>> plt.legend()
>>> plt.show()
**Model Parameters**
============= ============== ======== ===================================================================================
**Parameter** **Init Value** **Unit** **Explanation**
------------- -------------- -------- -----------------------------------------------------------------------------------
delay 0 ms The decay length of the pre-synaptic spikes.
tau_decay 10 ms The time constant of the synaptic decay phase.
tau_rise 1 ms The time constant of the synaptic rise phase.
g_max 1 µmho(µS) The maximum conductance.
============= ============== ======== ===================================================================================
**Model Variables**
================ ================== =========================================================
**Member name** **Initial values** **Explanation**
---------------- ------------------ ---------------------------------------------------------
g 0 Synapse conductance on the post-synaptic neuron.
s 0 Gating variable.
pre_spike False The history spiking states of the pre-synaptic neurons.
================ ================== =========================================================
**References**
.. [1] Sterratt, David, Bruce Graham, Andrew Gillies, and David Willshaw.
"The Synapse." Principles of Computational Modelling in Neuroscience.
Cambridge: Cambridge UP, 2011. 172-95. Print.
.. [2] Roth, A., & Van Rossum, M. C. W. (2009). Modeling Synapses. Computational
Modeling Methods for Neuroscientists.
"""
[docs] def __init__(self, pre, post, conn, delay=0., g_max=1., tau_decay=10.0, tau_rise=1.,
method='exp_auto', name=None):
super(DualExpCUBA, self).__init__(pre=pre, post=post, conn=conn, method=method, name=name)
self.check_pre_attrs('spike')
self.check_post_attrs('input')
# parameters
self.tau_rise = tau_rise
self.tau_decay = tau_decay
self.delay = delay
self.g_max = g_max
# connections
self.pre_ids, self.post_ids = self.conn.require('pre_ids', 'post_ids')
# variables
self.g = bm.Variable(bm.zeros( len(self.pre_ids)))
self.h = bm.Variable(bm.zeros( len(self.pre_ids)))
self.pre_spike = bp.ConstantDelay(self.pre.num, delay, dtype=pre.spike.dtype)
@property
def derivative(self):
dg = lambda g, t, h: -g / self.tau_decay + h
dh = lambda h, t: -h / self.tau_rise
return bp.JointEq([dg, dh])
[docs] def update(self, _t, _dt):
self.pre_spike.push(self.pre.spike)
delayed_pre_spikes = self.pre_spike.pull()
self.g.value, self.h.value = self.integral(self.g, self.h, _t, dt=_dt)
self.h.value += bm.pre2syn(delayed_pre_spikes, self.pre_ids)
self.post.input += self.g_max * bm.syn2post(self.g, self.post_ids, self.post.num)
[docs]class DualExpCOBA(DualExpCUBA):
"""Conductance-based dual exponential synapse model.
**Model Descriptions**
The conductance-based dual exponential synapse model is similar with the
current-based dual exponential synapse model <./brainmodels.synapses.DualExpCUBA.rst>_,
except the expression which output onto the post-synaptic neurons:
.. math::
I_{syn}(t) = g_{\mathrm{syn}}(t) (V(t)-E)
where :math:V(t) is the membrane potential of the post-synaptic neuron,
:math:E is the reversal potential.
**Model Examples**
.. plot::
:include-source: True
>>> import brainpy as bp
>>> import brainmodels
>>> import matplotlib.pyplot as plt
>>>
>>> neu1 = brainmodels.neurons.HH(1)
>>> neu2 = brainmodels.neurons.HH(1)
>>> syn1 = brainmodels.synapses.DualExpCOBA(neu1, neu2, bp.connect.All2All(), E=0.)
>>> net = bp.Network(pre=neu1, syn=syn1, post=neu2)
>>>
>>> runner = bp.StructRunner(net, inputs=[('pre.input', 5.)], monitors=['pre.V', 'post.V', 'syn.g', 'syn.h'])
>>> runner.run(150.)
>>>
>>> fig, gs = bp.visualize.get_figure(2, 1, 3, 8)
>>> plt.plot(runner.mon.ts, runner.mon['pre.V'], label='pre-V')
>>> plt.plot(runner.mon.ts, runner.mon['post.V'], label='post-V')
>>> plt.legend()
>>>
>>> plt.plot(runner.mon.ts, runner.mon['syn.g'], label='g')
>>> plt.plot(runner.mon.ts, runner.mon['syn.h'], label='h')
>>> plt.legend()
>>> plt.show()
**Model Parameters**
============= ============== ======== ===================================================================================
**Parameter** **Init Value** **Unit** **Explanation**
------------- -------------- -------- -----------------------------------------------------------------------------------
delay 0 ms The decay length of the pre-synaptic spikes.
tau_decay 10 ms The time constant of the synaptic decay phase.
tau_rise 1 ms The time constant of the synaptic rise phase.
g_max 1 µmho(µS) The maximum conductance.
E 0 mV The reversal potential for the synaptic current.
============= ============== ======== ===================================================================================
**Model Variables**
================ ================== =========================================================
**Member name** **Initial values** **Explanation**
---------------- ------------------ ---------------------------------------------------------
g 0 Synapse conductance on the post-synaptic neuron.
s 0 Gating variable.
pre_spike False The history spiking states of the pre-synaptic neurons.
================ ================== =========================================================
**References**
.. [1] Sterratt, David, Bruce Graham, Andrew Gillies, and David Willshaw.
"The Synapse." Principles of Computational Modelling in Neuroscience.
Cambridge: Cambridge UP, 2011. 172-95. Print.
"""
[docs] def __init__(self, pre, post, conn, delay=0., g_max=1., tau_decay=10.0, tau_rise=1.,
E=0., method='exp_auto', name=None):
super(DualExpCOBA, self).__init__(pre, post, conn, delay=delay, g_max=g_max,
tau_decay=tau_decay, tau_rise=tau_rise,
method=method, name=name)
self.check_post_attrs('V')
# parameters
self.E = E
[docs] def update(self, _t, _dt):
self.pre_spike.push(self.pre.spike)
delayed_pre_spikes = self.pre_spike.pull()
self.g.value, self.h.value = self.integral(self.g, self.h, _t, dt=_dt)
self.h.value += bm.pre2syn(delayed_pre_spikes, self.pre_ids)
post_g = bm.syn2post(self.g, self.post_ids, self.post.num)
self.post.input += self.g_max * post_g * (self.E - self.post.V) | 2,202 | 8,614 | {"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} | 2.96875 | 3 | CC-MAIN-2022-05 | latest | en | 0.521703 |
https://www.letsstudytogether.co/reasoning-ability-high-level-puzzles-set-27/ | 1,679,744,259,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945323.37/warc/CC-MAIN-20230325095252-20230325125252-00286.warc.gz | 980,036,809 | 36,118 | Welcome to Online Reasoning Section in letsstudytogether.co Here we are providing a set of Reasoning Ability Quiz for RBI Grade B, NICL AO and upcoming IBPS exams on High-level Puzzles.
## High-level Puzzles
[maxbutton id=”23″]
Directions (Q. 1-5): Read the following information carefully and answer the questions.
[maxbutton id=”20″]
[maxbutton id=”21″]
1.Which of the following game is played by D?
A. Hockey
B. Cricket
E. None of these
2.The player who got 1st rank in Basketball game, lives on which of the following floor?
A. 7th floor
B. 2nd floor
C. 6th floor
D. 4th floor
E. None of floor
3.If A is related to Kabaddi in the same way as Y is related to volleyball, then how is W related to, following the same pattern?
A. Cricket
D. Hockey
E. None of these
4.Who got 3rd rank in volleyball game?
A. The player, who plays Basketball
B. C
C. The player, who plays Kabaddi.
D. Y
E. None of these
5.Four of the five are following a same pattern, and then which of the following option is not
following the same pattern?
A. B
B. The player, who plays Hockey.
C. A
D. The player, who plays Kho-Kho.
E. W
2. D. 4th floor
3. D. Hockey
4. C. The player, who plays Kabaddi.
5. C. A
Floor Players Games 9 B Cricket 8 Y Badminton 7 W Basketball 6 A Volleyball 5 X Hockey 4 D Kabaddi 3 Z Football 2 E Shooting 1 C Kho-Kho
Games/ Players A W D Y Volleyball II IV III I Kabaddi I III IV II Basketball III II I IV Badminton IV I II III
## Descriptive Book Vol – 1 : NICL AO
### NICL AO Mains 2017 study material (Based on Latest pattern )
[maxbutton id=”4″]
[maxbutton id=”1″]
[maxbutton id=”6″] [maxbutton id=”7″]
[maxbutton id=”9″] [maxbutton id=”10″] | 515 | 1,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-14 | latest | en | 0.898127 |
https://www.qualitygurus.com/measurements-of-dispersion-variation/ | 1,726,036,479,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00847.warc.gz | 884,697,808 | 65,483 | # Measurements of Dispersion (Variation)
• /
• Blog
• /
• Measurements of Dispersion (Variation)
In descriptive statistics, several common variation measurements are used to describe the spread or dispersion of a group of observations. Some of the most common measurements of variation include:
1. Range: The range is the difference between the largest and smallest values in a group of observations. It is a simple and easy-to-understand measure of variation, but it is sensitive to outliers and can be affected by extreme values.
2. Interquartile range (IQR): The interquartile range is the difference between the third and the first quartile in a group of observations. It is calculated by subtracting the value of the first quartile (Q1) from the value of the third quartile (Q3). The IQR is a more robust measure of variation than the range because it is not affected by extreme values.
3. Standard deviation: The standard deviation measures the observations' average distance from the data's mean. It is calculated by taking the square root of the average squared differences between each observation and the mean. The standard deviation is a commonly used measure of variation because it is not affected by extreme values and considers the variability of all the observations in the group. The formula for the standard deviation of a sample is:
$$s = \sqrt{\frac{\sum(x_i - \overline{x})^2}{n-1}}$$. The formula for the standard deviation of the population is calculated using this formula: $$\sigma = \sqrt{\frac{\sum(x_i - \mu)^2}{N}}$$
4. Variance: The variance measures the average squared difference of the observations from the mean of the data. It is calculated by taking the average squared differences between each observation and the mean. The variance is the square of the standard deviation.The formula for the variance of a sample is: $$s^2 = \frac{\sum(x_i - \overline{x})^2}{n-1}$$ The formula for the standard deviation of the population is calculated using this formula: $$\sigma^2 = \frac{\sum(x_i - \mu)^2}{N}$$
## Conclusion
Dispersion, also known as variation, refers to the spread or distribution of values in a data set. It is an essential concept in statistics, as it allows us to understand how much the values in a data set vary. Several measurements of dispersion can be used to quantify the amount of variation in a data set. Some of the most common measurements include the range, interquartile range, variance, and standard deviation. Understanding and using these measurements can help understand the characteristics of a data set and make informed decisions based on the data.
## Relationship between Quality and Reliability
49 Courses on SALE! | 582 | 2,691 | {"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} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.912394 |
http://perplexus.info/show.php?pid=9583&cid=55221 | 1,550,287,055,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479838.37/warc/CC-MAIN-20190216024809-20190216050809-00097.warc.gz | 202,155,445 | 4,577 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
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Harmonic mean and constant determination puzzle (Posted on 2015-06-13)
In a triangle ABC , ∠BAC = 90o. Point D lies on the side BC, and satisfies
∠BDA = k*∠BAD, where k is a real constant.
Find the value of k, given that length of AD is the harmonic mean between the respective lengths of BD and CD.
No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
A different approach Comment 2 of 2 |
Using the intersecting chord theorem, it follows that the circle
with diameter BC and centre O, passes through A (because
of the right angle) and a point E where ADE is a straight line
with |DE| = |BC|/2, which is the radius of the circle.
/BDA = /EDO (vert opp)
= /EOD (angles of isosceles triangle EDO)
= 2 /BAD (angles at centre and circumf)
So k = 2
Posted by Harry on 2015-06-16 10:16:51
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Forums (1) | 291 | 1,021 | {"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} | 3.328125 | 3 | CC-MAIN-2019-09 | longest | en | 0.836795 |
http://www.ck12.org/geometry/Triangle-Identification-as-Similar-Congruent-or-Neither/lesson/Triangle-Identification-as-Similar-Congruent-or-Neither-MSM6/ | 1,493,587,250,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125849.25/warc/CC-MAIN-20170423031205-00544-ip-10-145-167-34.ec2.internal.warc.gz | 479,887,380 | 32,590 | # Triangle Identification as Similar, Congruent, or Neither
## Understand the terms similar and congruent and apply them to triangles.
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Triangle Identification as Similar, Congruent, or Neither
### Source: https://commons.wikimedia.org/wiki/File%3AReglas.svg License: CC BY-NC 3.0 [Figure1]
Gabriel is in geometry. He uses different tools to measure different items. One day he sits a couple of his tools together and he notices that two of his tools are triangles. His class has recently compared triangles and identified them as similar, congruent or neither. The class is supposed to be using their measuring tools to measure items around the classroom and determine if they are similar, congruent or neither. Gabriel takes a good look at the two shapes then decides to identify them so that he can add them to his list. Are Gabriel's triangles similar, congruent or neither?
In this concept, you will learn to distinguish between triangles.
### Identifying Triangles as Similar, Congruent, or Neither
Congruent means being exactly the same. When two line segments have the same length, they are congruent. When two figures have the same shape and size, they are congruent.
These two triangles are congruent. They are exactly the same in every way. They are the same size and the same shape. Their side lengths are the same and that their angle measures are the same.
Sometimes, two figures will be similar. Similar means that the figures have the same shape, but not the same size. Similar figures are not congruent.
These two triangles are similar. They are the same shape, but they are not the same size.
### Examples
#### Example 1
Earlier, you were given a problem about Gabriel and his measuring tools.
He notices that two of his tools are triangles. Are his triangles similar, congruent or neither?
First, check to see if the triangles are the same size.
No
Next, check to see if the triangles are the same shape.
No
Then, identify the triangles.
The answer is that the triangles are neither similar, nor congruent. Gabriel will list the triangles as neither.
#### Example 2
Identify the following triangles as similar, congruent or neither.
First, check to see if the triangles have the same size.
Yes
Next, check to see if the triangles have the same shape.
Yes
Then, identify the triangles.
Congruent
The answer is that the triangles are congruent.
#### Example 3
Identify the following triangles as similar, congruent, or neither.
First, check to see if the triangles are the same size.
No
Next, check to see if the triangles are the same shape.
No
Then, identify the triangles.
Neither congruent, nor similar
The answer is that the triangles are neither congruent, nor similar.
#### Example 4
Identify the following triangles as similar, congruent, or neither.
First, check to see if the triangles are the same size.
No
Next, check to see if the triangles are the same shape.
Yes
Then, identify the triangles.
Similar
The answer is that the triangles are similar.
#### Example 5
Identify the following triangles as similar, congruent, or neither.
First, check to see if the triangles are the same size.
Yes
Next, check to see if the triangles are the same shape.
Yes
Then, identify the triangles.
Congruent
The answer is that the triangles are congruent.
### Review
Identify the given triangles as visually similar, congruent or neither.
Answer each of the following questions.
1. Triangles and are congruent. Does this mean that their angle measures are the same? Why?
2. True or false. If triangles and are similar, then the side lengths are different but the angle measures are the same.
3. True or false. Similar figures have exactly the same size and shape.
4. True or false. Congruent figures are exactly the same in every way.
5. Triangles and are similar. If this is true, then the side lengths are the same, true or false.
6. True or false. To figure out if two figures are similar, then their side lengths form a proportion.
7. Define similar figures
8. Define congruent figures.
9. Use a ruler to draw a congruent pair of triangles.
10. Use a ruler to draw a pair of triangles that is similar.
To see the Review answers, open this PDF file and look for section 9.15.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Congruent Congruent figures are identical in size, shape and measure.
Similar Two figures are similar if they have the same shape, but not necessarily the same size. | 1,048 | 4,770 | {"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": 1, "texerror": 0} | 4.75 | 5 | CC-MAIN-2017-17 | latest | en | 0.953027 |
https://www.esaral.com/q/statement-1-the-temperature-dependence-of-resistance-is-usually-given-as-42837 | 1,674,887,726,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00056.warc.gz | 746,478,026 | 38,050 | Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
# Statement-1 : The temperature dependence of resistance is usually given as $Question: Statement-1 : The temperature dependence of resistance is usually given as$\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{t})$. The resistance of a wire changes from$100 \Omega$to$150 \Omega$when its temperature is increased from$27^{\circ} \mathrm{C}$to$227^{\circ} \mathrm{C}$. This implies that$\alpha=2.5 \times 10^{-3} /{ }^{\circ} \mathrm{C}$. Statement$-2: R=R_{0}(1+\alpha \Delta t)$is valid only when the change in the temperature$\Delta \mathrm{T}$is small and$\Delta \mathrm{R}=\left(\mathrm{R}-\mathrm{R}_{0}\right) \ll \mathrm{R}_{0}$. 1. Statement$-1$is true, Statement$-2$is true; Statement$-2$is not the correct explanation of Statement$-1$2. Statement-1 is false, Statement$-2$is true 3. Statement-1 is true, Statement-2 is false 4. Statement$-1$is true, Statement$-2$is true; Statement$-2$is the correct explanation of Statement$-1\$
Correct Option: 1
Solution:
Solution is not required | 326 | 1,074 | {"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} | 3.296875 | 3 | CC-MAIN-2023-06 | longest | en | 0.510679 |
http://omega.albany.edu:8008/mat214dir/hmwk4/p798-8.html | 1,544,473,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823442.17/warc/CC-MAIN-20181210191406-20181210212906-00301.warc.gz | 213,612,826 | 2,918 | Calc III Question ?
Find the Domain D and the Range R for the function: f(x,y)=sqrt(x-y). D = R = positive real numbers All the real numbers D = {(x,y) Î R2 : x ³ y }, R = R+ D does not exist but the range is all the reals. D = {(x,y) Î R2 : x ³ y } = R
Carlos Rodriguez <carlos@math.albany.edu>
This problem was contributed by a student. It is offered as it is with no warranty of any kind | 118 | 393 | {"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-2018-51 | latest | en | 0.920403 |
http://kishoreathrasseri.wordpress.com/tag/nitc/ | 1,368,945,346,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696384181/warc/CC-MAIN-20130516092624-00025-ip-10-60-113-184.ec2.internal.warc.gz | 146,738,165 | 13,827 | ## NITC Past and Present
I was going over the area of NITC in Wikimapia, when I suddenly realized that the satellite images of the place were somewhat old. In fact, it shows the campus in its state even before I joined the college, but not much older, since the only difference is the absence of the Central Computer Centre and the new Mini Canteen.
Over the last four years I’ve seen acres of forest give way to concrete structures and I thought I’d just highlight the changes I’ve seen on the map. These estimates are very rough and have been made from my countless walks around the campus. They are very conservative estimates and I suspect a lot more area has been cleared, as I have not even been to each and every corner of the campus.
## Nothing Short of a Miracle!
That’s what happened today. The occasion was a cultural night arranged by the Staff Club of NITC (no one knew one even existed, till today!). Some of the faculty got together and wrote a drama to be staged tonight. I learnt about it from Deepak sir’s blog a few days ago, and that in itself seemed like a miracle. But the actual performance was nothing short of unbelievable, to say the least! It was fantastic. About the drama itself, I keep that for another day. I’m just too dazed by that performance to analyze it critically. Besides, Deepak sir has promised to make the script and video available.
I always knew that the faculty members were fantastic people outside the classroom, but hats off to them for this wonderful performance!
## DSP Lab – Week 1
### Constructing the Complex Plane
Suppose we have a sampled signal defined by the sequence $h(n)$, $n=0,1,2,...,N-1$
Its Z- transform is given by $H(z) = \sum_{n=0}^{N-1} h(n)z^{-n}$ .
It maps the original sequence into a new domain, which is the complex plane $z=e^{sT}$ where $s=\sigma+j\omega$ is the parameter in the Laplace domain and $T$ is the sampling period.
The $j\omega$ axis in the $s$-plane maps onto the unit circle with centre at the origin in the $z$-plane. So the value of $H(z)$ at different points on the unit circle actually gives the contribution of the frequency component given by $\angle z$, in the original signal.
This, in effect, gives the Discrete Fourier Transform of the sequence. Consider the following example:
%original sequence
h = [1,2,3,4];
%number of chosen points on the unit circle
N = 64;
%define the chosen points
z = complex(cos(2*pi/N*(0:N-1)),sin(2*pi/N*(0:N-1)));
%evaluate H(z) at each point
for i = 1:N
H(i) = 1+2*z(i)^-1+3*z(i)^-2+4*z(i)^-3;
end
%plot the unit circle
plot(z)
%plot the value of H(z) along the unit circle
figure
plot(abs(H))
%plot the N-point DFT of h(n)
figure
plot(abs(fft(h,64)))
This example computes the value of $H(z)$ at 64 uniformly spaced points on the unit circle and compares it with the 64 point DFT. We can see that both (fig. b & c) are identical.
unit circle
value of H(z)
abs(fft(h))
## Digital Signal Processing Lab
We are very lucky to have Abhilash sir in charge of DSP lab this semester. I really used to enjoy his classes during the Signals and Systems course in the third semester. In fact it’s the only course so far, in which I’ve felt like paying attention. I think I will document whatever we do in the lab, on my blog, when I get time.
## Freedom Walk at NITC
It was almost midnight when the Freedom Walkers arrived here on Wednesday. They were thouroughly worn out from the long long walk from Thamarassery. They had dinner from our mini canteen, and then I led them to the rooms in PG-2 hostel which Sandeep sir from the Electrical department had booked.
Next morning, a few S3 guys and I went to meet them. We had a small gathering in their room and discussed what all we could do to spread Free Software here at NITC. Since all of us except one were from Electronics, Jemshid suggested that we could get started on some Embedded GNU/Linux work. We can think of conducting workshops to get people interested in it.
Prasad talked about the Freedom Toaster they had made, and suggested that we could try to make one with a vending machine, as a project. They said we would have their support if someone is ready to take it up.
It’s too bad we couldn’t organize a more elaborate meeting with the Freedom Walkers, because of the exams. But we’ve got some pointers to think about, when we sit down to make a concrete plan regarding the FOSS Cell activities.
## RMS lecture at NIT Calicut
As a part of this year’s Tathva, the technical fest of NIT Calicut, there will be a video conference with Richard Stallman.
Topic: The Free Software Movement and the GNU/Linux Operating System
Date: 25th October
Time: 9.00 am to 11 am
See http://tathva.org/lectures.gl.php?PHPSESSID=1426bc5eea73716979ab0c62978e053f for more details.
## GNU/Linux Install Fest at NITC
As the first activity of the upcoming FOSS Cell NITC, we organized a GNU/Linux install fest on the occasion of Software Freedom Day. Around twenty people turned up during the day. The only undesirable part was that a couple of laptops, after installing Ubuntu, couldn’t boot Windows. Got to sort out their issues soon. We’ve set up a technical support mailing list for people to post their problems.Considering that it was the first ever event by our FOSS Cell, it didn’t go too badly.
## Software Freedom Day at NITC
We are planning to celebrate the Software Freedom Day through an install fest and demos. There was a meeting today to get some volunteers for the event, and around twenty S3 students turned up. Only some of them have used GNU/Linux before, and they have been given the task of familiarising the others with it before the event. We are also hopeful of launching our FOSS Cell officially on that day. More about the event as it materializes…
## Exams… (again)
It’s something I’ve written about a lot in the past, and am sure to write a lot in the future, as well! Well, this time the occasion is the end of the first Interim tests of the fifth semester. This semester is better, in that we have only five core courses, the sixth being Environmental Studies.
Last semester I got into the habit of doing something I’d never before done in my life- cramming junk into my head just before the exams. The level of my motivation was so low, and those of my ignorance and indifference to the subjects were so high that whatever grades I got that sem were due to this last minute mugging up. (I should mention that I had got through the previous three semesters without disaster, mainly because of the fundamentals acquired while at school, rather than any effort on my part.) But it was sheer mental torture.
I’ve decided that I’d do nothing of the sort this time round. That I’d let myself explore the topics at my pace and discretion. Of course, it doesn’t guarantee good grades-the converse, quite possibly. But I think that’s the only way ahead for me. Learning everything taken in class exhaustively is out of the question. This is not school. That would probably drive me mad. Learning something for an examination is mental torture. And it’s against the purpose of exams.
At least now, I love all subjects and feel a real thirst for knowledge. That’s a big improvement from last sem. As for grades, I couldn’t care less so long as I pass.
## First Meeting
Fifteen of us, including Deepak sir and Murali sir met today for what was the first meeting of the FOSS cell. It was notable that there were six S3 guys, from Electronics, Computer Science, Electrical and Civil, all of them motivated from last year’s FOSSMeet. We got to know each other, and decided on the things to be done immediately for setting up a group. It’s a new beginning and I’m looking forward to it. The S3 guys seem to be an enthusiastic bunch and there is enough reason to be optimistic about this new venture. | 1,859 | 7,825 | {"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": 12, "codecogs_latex": 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} | 3.28125 | 3 | CC-MAIN-2013-20 | latest | en | 0.9608 |
https://www.physicsforums.com/threads/kinematics-interesting-problem-velocity-as-a-function-of-distance.303509/ | 1,529,764,626,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865081.23/warc/CC-MAIN-20180623132619-20180623152619-00472.warc.gz | 886,138,473 | 15,235 | # Homework Help: Kinematics: Interesting problem (velocity as a function of distance)
1. Mar 29, 2009
### leden
I have spent a long time stuggling with an interesting problem which was set up by my brother some time ago:
Consider an object that starts moving along the line according to the equation below:
$$v(s)=1+{s}^{2}$$
where s is the total distance travelled by the object and v is the velocity of the object as a function of the distance travelled. respectively.
Notice that the object starts moving at velocity 1 m/s and the velocity continuously increases as it travels.
The question is, what is the velocity of the object at some time t?
$$v(t)=?$$
What is the distance travelled at some time t?
$$s(t)=?$$
I would greatly appreciate if someone posted a detailed solution to the above problem.
2. Mar 29, 2009
### tiny-tim
Welcome to PF!
Hi leden! Welcome to PF!
erm … that's not the way this forum works! :rofl:
Show us what you've tried, and where you're stuck, and then we'll know how to help!
3. Mar 29, 2009
### Staff: Mentor
Hint: v = ds/dt. Set up the integral and solve for s(t) first.
4. Mar 29, 2009
### leden
Re: Welcome to PF!
I am aware of that..
But the problem is that am not so much familiar with calculus and when it comes to physics I don't know what to do when integrating/differentiating some functions which is dependent on many variables...
The first things that bothers me is whether it is correct that
$$v(s)=ds/dt$$.
Ok, have tried the following...
$$v(s)=\frac{ds}{dt} = 1 + {s}^{2}$$
$$ds=dt+{s}^{2}dt$$
$$\int ds=\int dt + \int{s}^{2}dt$$
$$s=t+\int{s}^{2}dt$$
I don't know what to do know, beacause I'm going into cycles...
Last edited: Mar 29, 2009
5. Mar 29, 2009
### Staff: Mentor
Re: Welcome to PF!
OK.
Nah... do this:
$$\frac{ds}{1 + {s}^{2}} = dt$$
(Now you can integrate.)
6. Mar 29, 2009
### leden
Thanks,
let me try to solve now...
$$\frac{ds}{1+{s}^2}=dt$$
$$\int\frac{ds}{1+{s}^2}=\int dt$$
$$\arctan s=t$$
$$s(t)=\tan t$$
Now
$$v(t)=\frac{ds}{dt}=\frac{\tan t}{dt}=\frac{1}{{\cos}^{2}t}$$
This is strange, because it seems that velocity is practically infinity at point t = pi/2.
Last edited: Mar 29, 2009
7. Mar 30, 2009
### tiny-tim
So is s … what's strange about that?
8. Mar 30, 2009
### Domestikus
OK but how is it possible that
$$v(s)=\frac{ds}{dt}$$
and
$$v(t)=\frac{ds}{dt}$$
Velocity, by the definition, is the first derivative of position, but as the function of time.
So you can't say that velocity as the function of position is also the first derivative of position.
Correct me if I'm wrong, but that doesn't make any sense to me :)
9. Mar 30, 2009
### Staff: Mentor
I'm not seeing the problem. v ≡ ds/dt, which can be expressed as a function of time or of distance. So? | 856 | 2,777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 3.953125 | 4 | CC-MAIN-2018-26 | latest | en | 0.887051 |
http://download.plt-scheme.org/doc/4.2.5/html/reference/sets.html | 1,369,399,594,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704658856/warc/CC-MAIN-20130516114418-00093-ip-10-60-113-184.ec2.internal.warc.gz | 82,401,485 | 4,801 | #### 3.16Sets
A set represents a set of distinct elements. For a given set, elements are equivalent via equal?, eqv?, or eq?. Two sets are equal? when they use the same element-comparison procedure (equal?, eqv?, or eq?) and have equivalent elements. A set can be used as a sequence (see Sequences).
Operations on sets that contain elements that are mutated are unpredictable in much the same way that hash table operations are unpredictable when keys are mutated.
(require scheme/set)
The bindings documented in this section are provided by the scheme/set library, not scheme/base or scheme.
(set? v) → boolean? v : any/c
Returns #t if v is a set, #f otherwise.
(set-eqv? set) → boolean? set : set?
Returns #t if set compares elements with eqv?, #f if it compares with equal? or eq?.
(set-eq? set) → boolean? set : set?
Returns #t if set compares elements with eq?, #f if it compares with equal? or eqv?.
(set v ...) → set? v : any/c
(seteqv v ...) → set? v : any/c
(seteq v ...) → set? v : any/c
Creates a set that uses equal?, eq?, or eqv?, respectively, to compare elements. The given vs are added to the set. The elements are added in the order that they appear as vs, so in the first two cases, an earlier element that is equal? or eqv? but not eq? to a later element takes precedence over the later element.
(set-empty? set) → boolean? set : set?
Returns #t if set has no members, #f otherwise.
(set-member? set v) → boolean? set : set? v : any/c
Returns #t if v is in set, #f otherwise.
(set-add set v) → set? set : set? v : any/c
Like operations on immutable hash tables, “constant time” set operations actually require O(log N) time for a set of size N.
Produces a set that includes v plus all elements of set. This operation runs in constant time.
(set-remove set v) → set? set : set? v : any/c
Produces a set that includes all elements of set except v. This operation runs in constant time.
(set-union set ...+) → set? set : set?
Produces a set that includes all elements of all given sets, which must all use the same equivalence predicate (equal?, eq?, or eqv?). This operation runs in time proportional to the total size of all given sets except for the largest.
(set-intersect set ...+) → set? set : set?
Produces a set that includes only the elements in all of the given sets, which must all use the same equivalence predicate (equal?, eq?, or eqv?). This operation runs in time proportional to the total size of all given sets except for the largest.
(set-subtract set ...+) → set? set : set?
Produces a set that includes all elements the first sets that are not present in any of the other given setss. All of the given sets must use the same equivalence predicate (equal?, eq?, or eqv?). This operation runs in time proportional to the total size of all given sets except the first one.
(set-map set proc) → (listof any/c) set : set? proc : (any/c . -> . any/c)
Applies the procedure proc to each element in set in an unspecified order, accumulating the results into a list.
(set-for-each set proc) → void? set : set? proc : (any/c . -> . any)
Applies proc to each element in set (for the side-effects of proc) in an unspecified order.
(in-set set) → sequence? set : set?
Explicitly converts a set to a sequence for use with for and other forms. | 922 | 3,301 | {"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} | 3.46875 | 3 | CC-MAIN-2013-20 | latest | en | 0.624963 |
https://jinjaconnection.com/qa/question-what-should-be-the-distance-between-two-cars.html | 1,606,757,421,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216897.58/warc/CC-MAIN-20201130161537-20201130191537-00497.warc.gz | 365,312,176 | 10,541 | # Question: What Should Be The Distance Between Two Cars?
## What is the 3 second rule in food?
After being dropped on the floor dried fruit was found to have klebsiella.
Biscuits proved to be a food relatively safe to eat after being dropped on the floor for three seconds, five seconds or ten seconds, due to their low water content.
## How many feet do you have to stay behind a car?
500 feetStay at least 500 feet behind any moving emergency vehicle (fire truck, ambulance, patrol car) displaying flashing warning lights and sounding a siren.
## What are 3 things you can do to avoid a collision?
There is almost always something you can do to avoid an impending collision or reduce its severity. Depending on the situation, you can do one of these 3 things to prevent a collision: stop, steer away or speed up.
## How many feet is 45 mph?
Convert 45 Miles per Hour to Feet per Secondmphfps45.006645.0166.01545.0266.02945.0366.04496 more rows
## How many car lengths is 3 seconds?
Three seconds distance is equivalent to 50 metres. As most cars are between 4 and 5 metres long, perhaps the easiest way to gauge this is 10 car lengths. The age-old method of judging is to begin counting as the car in front passes a landmark (tree, post …).
## What is the 4 second rule?
You should apply the four-second rule when it’s wet, frosty or when you are towing a trailer. The four-second rule means that you leave four seconds between you and the vehicle in front. It gives you more time to react and more time to stop.
## What is the 1 second rule?
The 2 Second rule. Page 1. The 2 Second rule. The two-second rule is a rule by which a driver may maintain a safe following distance at any speed. The rule is that a driver should ideally stay at least two seconds behind any vehicle that.
## How many feet is 70 mph?
Now, to convert units without converting the value, we multiply by 1 (which is called the multiplicative identity for that reason). Now, to remember when driving, 70 mph is 102 2/3 ft/sec.
## Which is faster 60 miles an hour or 60 feet a second?
Example: What is 60 mph in fps? A mile has 5,280 feet, and an hour has 3,600 seconds, so 60 miles per hour is: 60 x 5,280 / 3,600 = 88 fps.
## What is the standard length of a car?
around 4500mmWhat is the Average Car Length? The average car length is around 4500mm or 14,7 feet. For example, this is the length of a car in the mid-size class like Audi A4, which should give you a perspective of the length of any car models. Of course, some vehicles are both much more extended and much shorter.
## What is the length of car in rain?
If the road is wet or icy, this will significantly increase braking distances. Double the gap between your car and the car in front when it’s wet. Leave an even bigger gap if it’s icy – some advice says 10 times bigger.
## What should be the distance between two vehicles?
The rule is that a driver should ideally stay at least two seconds behind any vehicle that is directly in front of his or her vehicle. It is intended for automobiles, although its general principle applies to other types of vehicles.
## How many car lengths is 2 seconds?
Remaining at least 2 seconds from the vehicle in front will provide a distance of one car length per 5 mph, at which ever speed you drive. The 2 second rule is used regardless of speed because the distance between your vehicle and the one in front will extend the faster you travel.
## What is the 3 to 6 second rule?
Double and Triple the 3-Second Rule If you are driving in heavy traffic, driving at night or in weather conditions that are not ideal, such as rain or fog, consider doubling the 3-second rule to six seconds as a safety precaution.
## Should you brake while turning?
Brake before entering the turn. … You shouldn’t brake while turning as this can cause skidding. Basically, asking your tires to slow down and turn at the same time may exceed their traction. The same is true for accelerating while turning.
## How far does a car travel in 1 second?
Here’s some food for thought. At 55 mph, your vehicle is traveling at about 80 feet per second. Feet-per-second is determined by multiplying speed in miles-per-hour by 1.47 (55 mph x 1.47 = 80 feet per second.) With this in mind, let’s add the perception and reaction distance to the formula.
## When driving what is the 3 second rule?
Simply leave 3 seconds worth of room between you and the vehicle you are following. Just watch the vehicle in front of you pass a road sign or other inanimate object on the side of the road and count out “One Massachusetts, Two Massachusetts, Three Massachusetts” before your vehicle passes that same object.
## What is an easy way to know if you are maintaining the safe driving distance?
Safe Following DistanceLocate a fixed point ahead. … When the vehicle ahead of you passes that fixed point, count to yourself, “one thousand one, one thousand two, one thousand three.” If your vehicle passes the same fixed point when you say “one thousand three,” then you have a three-second safe following distance.More items…•
## Is it illegal to drive close to another car?
The tailgating (or preventing) vehicle will drive as close as possible to another leading vehicle to prevent the side vehicle from cutting in. Like all forms, this practice of tailgating is illegal and attempts to force the side vehicle to slow down and get into the line of traffic behind the tailgating vehicle.
## What is a safe driving distance?
What’s good about the “3 second rule” is that it helps you keep a safe following-distance at any speed. Using the “3 second rule” gives you a bigger following-distance the faster you drive. Generally speaking, you should allow more than a 3 second following-distance in rain, fog and on icy roads. | 1,325 | 5,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-50 | latest | en | 0.962538 |
https://cs.nyu.edu/pipermail/fom/2021-October/022927.html | 1,709,078,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00218.warc.gz | 182,128,891 | 3,895 | # 905: Foundations of Large Cardinals/3
Harvey Friedman hmflogic at gmail.com
Wed Oct 13 15:17:10 EDT 2021
```1. Conceptual Remarks concerning 903 - #2
2. Second Thoughts about Elementary Extensions versus Elementary
Equivalence - #2
3. Two Universe Elementary Extension - #2
4. Two Extended Universe Restricted Elementary Extension - here
5. Double Completion Elementary Extension - foundations of ZF - #4
6. Multiple Universes Elementary Extension - next posting - #5
7. Two Universe Elementary Extension/complex - #6
4. Two Extended Universe Restricted Elementary Extension
Here as in section 3, V_0,V_1 are universes in the sense of ranks on
strongly inaccessible cardinals kappa < lambda. We got a weak form of
strongly Mahlo cardinals in section 3, from elementary extension, and
failed to achieve a strongly Mahlo cardinal. Here we stay with two
universes, rather than go to three universes, and take the
intermediate step of using classes of V_0 and classes of V_1. I.e.,
using V(kappa + 1) and V(lambda + 1).
Now obviously V(lambda + 1) cannot be an elementary extension of
(V(kappa + 1), because of the parameter kappa. But we can insist that
the parameters be from V(kappa) - just like they are when we used
V(kappa),(lambda) in section 3.
We will get second order indescribable cardinals from this, much
bigger than weakly compact which is much bigger than strongly Mahlo -
but only in the constructible universe L. Thus we might as well just
assume lambda < kappa are ordinals and derive that they are strongly
inaccessible cardinals in L.
THEOREM 4.1. Let V(lambda + 1) be a proper elementary extension of
V(kappa + 1) with parameters from V(kappa), where kappa < lambda are
ordinals.
i. kappa and lambda are strong limit cardinals, and strongly
inaccessible cardinals in L.
ii. kappa and lambda are definably second order indescribable
cardinals in V(kappa), V(lambda) respectively.
iii. kappa is a second order indescribable cardinal in L(lambda).
iv. In L(kappa), L(lambda), there are arbitrarily large second order
indescribable cardinals.
In ii, definably refers to allowing *elements* of V(kappa) and
L(lambda) as parameters in the definition of the subset of V(kappa)
and V(lambda) being reflected downwards, and where the definitions are
over V(kappa +1) and V(lambda + 1), respectively..
Proof: kappa <= omega is immediate from even elementary equivalence
between V(kappa) and V(lambda). kappa, lambda are limit ordinals by
proper elementary extension. Suppose f:V(alpha) onto kappa, where
alpha < kaap. We get a well ordering of a subset of V(alpha) of type
kappa. The well ordering can be used as a parameter in elementary
extension, obtaining also that this well ordering is of order type
lambda, which is impossible. So kappa is a strong limit cardinal.
Hence lambda is a strong limit cardinal. However proving regularity is
an issue. But not in L. Suppose kappa is not regular in L. Let f:alpha
into kappa be strictly increasing and unbounded, in L. Assume f is
constructibly least. Then f is definable with parameter alpha over
V(kappa + 1), and so it retains this property over V(lambda + 1),
which is impossible.
For ii, let A contained in V(kappa) be definable over V(kappa) with
parameters from V(kappa) where (V(kappa),A) satisfies phi, defined
over V(kappa + 1). Use restricted elementary extension to extend to A'
contained in V(lambda) with the same definition, also (V(lambda +
1),A') satisfying phi. Then reflecting down to V(kappa), we see that
according to V(lambda + 1), A' reflects down to A for phi. This is a
statement about V(lambda + 1) and hence by restricted elementary
extension, A also reflects down somewhere for phi. This establishes
that kappa is definably second order indescribable in V(kappa). By
restricted elementary extension, lambda is also definably second order
indescribable in V(lambda).
For iii, let n be given. Suppose kappa is not Pi-1-n indescribable in
the sense of L(lambda). Let A contained in V(kappa) be constructibly
least such that A does not reflect downwards. Then A is definable over
L(kappa + 1) with no parameters. This contradicts i.
For iv, let alpha be such that in L(kappa), there are no second order
indescribable cardinals > alpha. By elementary extension, in
L(lambda), there are no second order indescribable cardinals > alpha.
COROLLARY 4.2. ZFC + there exists arbitrarily large second order
indescribable cardinals is interpretable in KP + there exists ordinals
alpha < beta such that V(beta + 1) is an elementary extension of
V(alpha + 1) with parameters from V(alpha).
We cannot go much farther because third order indescribability dominates.
THEOREM 4.3. (Known) Let mu be the least third order indescribable
cardinal. There exists strongly inaccessible cardinals kappa < lambda
< mu such that V(lambda + 1) is an elementary extension of V(kappa +
1) with parameters from V(kappa). We can take kappa to be arbitrarily
large below mu.
Proof: Let A contained in V(mu) be the set of all (phi,x1,...,xn) such
that V(mu) satisfies phi(x1,...,xn), where phi is a second order
formula with n free variables, and x1,...,xn in V(mu). Let psi be the
third order statement about A that A is as described with mu replaced
by the union of A. Then psi holds of A intersect V(alpha) for
arbitrarily large alpha < mu, and these alpha must be strongly
inaccessible. QED
It is clear that we can go further using V(alpha + n), V(beta + n),
for interpreting (many) (n+1)-th order indescribable cardinals, n >=
1.
To make the next big leap with the V(alpha)'s, we need to consider
three or more V(alpha)'s at once with the proper way of handling
elementary extensions for pairs of V(alpha)'s. This is taken up in
section 6.
Meanwhile in section 5 we work with two relations instead of two set
theoretic universes. We regard this as a kind of foundations for ZF.
##########################################
My website is at https://u.osu.edu/friedman.8/ and my youtube site is at
This is the 905th in a series of self contained numbered
postings to FOM covering a wide range of topics in f.o.m. The list of
previous numbered postings #1-899 can be found at | 1,557 | 6,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-10 | latest | en | 0.812029 |
https://www.scribd.com/document/273589657/BS-Example-Steel1 | 1,563,961,542,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195532251.99/warc/CC-MAIN-20190724082321-20190724104321-00094.warc.gz | 814,299,288 | 63,063 | You are on page 1of 7
# Design Equations
Page 1 of 7
International Codes
British Codes - Design Per British Cold Formed Steel Code
2E.4 Design Equations
Tensile Strength
The allowable tensile strength, as calculated in STAAD as per BS5950-5, section 7 is
described below.
The tensile strength, P t of the member should be determined from clause 7.2.1
Where
Ae
py
## is the net area An determined in accordance with cl.3.5.4
is the design strength
## Combined bending and tension
As per clause 7.3 of BS 5950-5:1998 members subjected to both axial tension and
bending should be proportioned such that the following relationships are satisfied at the
ultimate limit state
And
and
1
Where
Ft
Pt
## is the applies tensile strength
is the tensile capacity determined in accordance with clause 7.2.1 of the subject
code
M z ,M y ,M cz ,M cy are as defined in clause 6.4.2 of the subject code
Compressive Strength
The allowable Compressive strength, as calculated in STAAD as per BS5950-5, section 6
Design Equations
Page 2 of 7
is described below.
For sections symmetrical about both principal axes orclosed cross-sections which are not
subjected to torsionalflexural buckling, the buckling resistance under axial load, Pc, may
be obtained from the following equation as per clause 6.2.3 of the subject code
For Sections symmetrical about a single axis and which are not subject to torsional
flexural buckling, the buckling resistance under axial load, Pc, may be obtained from the
following equation as per clause 6.2.4 of the subject code
Where the meanings of the symbols used are indicated in the subject clauses.
Torsional flexural buckling
Design of the memberswhich have at least one axis of symmetry, and which are subject to
torsional flexural buckling should be done according to the stipulations of the clause
6.3.2 using factored slenderness ratio aL E /r in place of actual slenderness ratio while
reading Table 10 for the value of Compressive strength(p c ).
Where
= 1 , otherwise
Where the meanings of the symbols used are indicated in the subject clause.
Combined bending and compression
Members subjected to both axial compression and bending should be checked for local
capacity and overall buckling
Local capacity check as per clause 6.4.2 of the subject code
## Overall buckling check as per clause 6.4.3 of the subject code
For Beams not subjected to lateral buckling, the following relationship should be satisfied
Design Equations
Page 3 of 7
For Beams subjected to lateral buckling, the following relationship should be satisfied
Fc
P cs
Mz
My
M cz
## is the applied axial load
is the short strut capacity as per clause 6.2.3
is the applied bending moment about z axis
is the applied bending moment about y axis
is the moment capacity in bending about the local Z axis in the absence of F c and
M y , as per clause 5.2.2 and 5.6
M cy
is the moment capacity in bending about the local Y axis, in the absence of F c
and M z ,as per clause 5.2.2 and 5.6
Mb
P Ez
P Ey
C bz ,C by
## is the lateral buckling resistance moment as per clause 5.6.2
is the flexural buckling load in compression for bending about the local Z axis
is the flexural buckling load in compression for bending about the local Y axis
are taken as unity unless their values are specified by the user
The Mcz, Mcy and Mb are calculated from clause numbers 5.2.2 and 5.6 in the manner
described hereinbelow.
Calculation of moment capacities
For restrained beams, the applied moment based on factored loads should not be greater
then the bending moment resistance of the section, M c
Mcz = SzzX po
Mcy = Syy X po
Where
is the Moment resistance of the section in z axis
M cz
M cz
is the Moment resistance of the section in z axis
po
is the limiting stress for bending elements under stress gradient and should not
greater then design strength p y
For unrestrained beams the applied moment based on factored loads should not be greater
Design Equations
Page 4 of 7
than the smaller of the bending moment resistance of the section , M c , and the buckling
resistance moment of the beam, M b
Then buckling resistance moment, M b, may be calculated as follows
MY
is the yield moment of the section , product of design strength p y and elastic
modules of the gross section with respect to the compression flange Zc
ME
is the elastic lateral buckling resistance as per clause 5.6.2.2
## is the Perry coefficient
Please refer clause numbers 5.2.2 and 5.6 of the subject code for a detailed discussion
regarding the parameters used in the abovementioned equations.
Shear Strength
The maximum shear stress should not be greater then 0.7 p y as per
clause 5.4.2
The average shear stress should not exceed the lesser of the shear yield strength, p v or the
shear buckling strength, q cr as stipulated in clause 5.4.3 of the subject code.
The parameters are calculated as follows :pv = 0.6 X p y
Pv= A*Min(pv,qcr)
Where
Pv
py
t
D
## is the shear capacity in N/mm^2
is the design strength in N/mm^2
is the web thickness in mm
is the web depth in mm
## Combined bending and Shear
For beam webs subjected to both bending and shear stresses the member should be
designed to satisfy the following relationship as per the stipulations of clause 5.5.2 of the
subject code
Design Equations
Page 5 of 7
Where
Fv
M
Mc
## is the shear force
is the bending moment acting at the same section as F v
is the moment capacity determined in accordance with 5.2.2
The next table contains the input parameters for specifying values of design variables and selection of
design options.
Parameter Name
BEAM
Default Value
Description
1.0
## When this parameter is set
to 1.0 (default), the
is determined by checking
a total of 13 equally
spaced locations along the
length of the member. If
the BEAM value is 0.0, the
13 location check is not
checking is done only at
the locations specified by
the SECTION command
details. For TRUSS
members only start and
end locations are
designed.
CMZ
1.0
Coefficient of equivalent
uniform bending Cb. See
BS:5950-5:1998,5.6. Used
and bending design.
CMY
1.0
Coefficient of equivalent
uniform bending Cb. See
BS:5950-5:1998,5.6. Used
Design Equations
Page 6 of 7
CWY
1.0
## Specifies whether the cold
work of forming
strengthening effect
should be included in
resistance computation.
See BS:5950-5:1998,3.4
Values: 0 effect should
not be included
1 effect should
be included
FLX
Specifies whether
torsional-flexural buckling
restraint is provided or is
not necessary for the
member. See BS:59505:1998, 5.6
Values:
0 Section not subject to
torsional flexural
buckling
1 Section subject to
torsional flexural buckling
FU
FYLD
KX
430 MPa
## Ultimate tensile strength of
steel in current units.
MPa
1.0
## Effective length factor for torsional buckling. It is a fraction
and is unit-less. Values can range from 0.01 (for a column
completely prevented from buckling) to any user specified
large value. It is used to compute the KL/R ratio for twisting
for determining the capacity in axial compression.
KY
1.0
Effective length factor for overall buckling about the local Yaxis. It is a fraction and is unit-less. Values can range from
0.01 (for a column completely prevented from buckling) to any
user specified large value. It is used to compute the KL/R
ratio for determining the capacity in axial compression.
KZ
1.0
## Effective length factor for overall buckling in the local Z-axis. It
is a fraction and is unit-less. Values can range from 0.01 (for
a member completely prevented from buckling) to any user
specified large value. It is used to compute the KL/R ratio for
determining the capacity in axial compression.
LX
## Member Unbraced length for twisting. It is input in the current units of
Design Equations
Page 7 of 7
length length. Values can range from 0.01 (for a member completely
prevented from torsional buckling) to any user specified large
value. It is used to compute the KL/R ratio for twisting for
determining the capacity in axial compression.
LY
## Member Effective length for overall buckling in the local Y-axis. It is
length input in the current units of length. Values can range from
0.01 (for a member completely prevented from buckling) to
any user specified large value. It is used to compute the KL/R
ratio for determining the capacity in axial compression.
LZ
## Member Effective length for overall buckling in the local Z-axis. It is
length input in the current units of length. Values can range from
0.01 (for a member completely prevented from buckling) to
any user specified large value. It is used to compute the KL/R
ratio for determining the capacity in axial compression.
MAIN
## 0 Check slenderness ratio
0 Do not check slenderness ratio
NSF
DMAX
RATIO
TRACK
1.0
## Net section factor for tension members
2540.0
Maximum allowable depth. It is input in the current units of
cm.
length.
1.0
0
## Permissible ratio of actual to allowable stresses
This parameter is used to control the level of detail in which
the design output is reported in the output file. The allowable
values are:
0 - Prints only the member number, section name, ratio, and
PASS/FAIL status.
1 - Prints the design summary in addition to that printed by
TRACK 1
2 - Prints member and material properties in addition to that
printed by TRACK 2. | 2,394 | 9,336 | {"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-2019-30 | latest | en | 0.916844 |
https://brightkite.com/essay-on/fewgewf | 1,575,872,557,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517557.43/warc/CC-MAIN-20191209041847-20191209065847-00200.warc.gz | 302,868,842 | 16,054 | # Fewgewf Essay
1119 words - 4 pages
New Era UniversityCollege of Engineering and TechnologyEE211- EE Technology 2Thursday (1:00-5:00)Laboratory Job No.1:"Voltage and Current Checking of an AC Generator"Salac, Aljon F. Rating:2nd Year BSEEReynaldo Dela CruzINSTRUCTOREE211EE Technology 2Thursday (1:00-5:00)Laboratory Job No.2"Voltage and Current Checking of an AC Generator"ObjectiveTo familiarized with AC generator.To check the voltage and current output of an AC generator.Equipment1-DWM1-VOM4-Connecting wires alligator clips (big wire)4-Connecting wires banana plug (big wire)ProcedureCheck all connecting wires for open and check the calibration of all measuring instrument.Connect the circuit shown in Fig 1.1.Turn ON the prime mover (3 induction motor).Close the main circuit breaker and turn ON load switch no.1.Using DWM, measure and record the total voltage, current and power.Turn ON load switch no.1 and no.2 and repeat procedure no.5Using procedure no.5 for load switches 1,2,3; load switches 124 and 1-5 and complete table 1.1.From the name plate of AC generator, find and record all rated values like KW, KVA, volt, amp, RPM, etc.From current date, solve for the branch current of each load I1, I2, I3, I4 and I5.Compute and record the average of the voltage.Compute and record the average of current.Compute and record the average of VA.Circuit DiagramData and ResultTABLE 1.1Measured Values
V (volt)
I (amp)
P (watt)
1
268V
0.34A
89W
1,2
270V
1.32A
345W
1,2,3
271V
2.45A
644W
1,2,3,4
266V
4.60A
1208W
1,2,3,4,5
269V
6.13A
1648W
Computed Values
Remark
a)
b)
c)
Question/ProblemBy Internet research: research aboutAC generatorTypes of AC generatorManufacturer and SpecificationAn alternating current generator, or AC generator, produces an alternating current, which means the voltage produced alternately reverses from positive to negative polarity, producing a corresponding change in the direction of current flow.Much like a DC generator, an AC generator requires a coil to cut across the force lines of a magnetic field. This coil is attached to two slip rings, which deliver the current to and from the load destination, thus completing the circuit. Alternating current generators are often called alternators.During the first half turn, the coil cuts across the field near the magnet's north pole. Electrons travel up the wire, and the lower slip ring becomes positively charged. When the coil cuts near the South Pole of the wire during the second half turn, the lower slip ring becomes negatively charged, and electrons move down the wire. The faster the coil turns, the faster the electrons move, increasing the frequency (in Hertz) of the current produced by the generator.INDUCTION GENERATORSAn induction generator or asynchronous generator is a type of AC electrical generatorthat uses the principles of induction motors to produce power. Induction generators operate by mechanically turning their rotor faster than the synchronous...
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# According to public health officials, in 1998 Massachusetts
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According to public health officials, in 1998 Massachusetts became the first state in which more babies were born to women over the age of thirty than under it.
A. than
B. than born
C. than they were
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19 Jul 2009, 07:08
A is the best choice for me.
B: 'than born' sounds awkward.
C: who were born? the women or the babies? not very clear.
D & E: just wrong.
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19 Jul 2009, 14:32
OA is A.
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19 Jul 2009, 14:44
i would go for B
D&E -- tense change --error
out of A,B,C -
option c - 'they' is ambiguous ....can refer to women ..changing meaning ....".......women over the age of thirty than they were under it.
out of A&B -
option A - lacks born req for ||el structure
hence option B
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07 Jan 2011, 10:38
could anybody further elaborate on what's wrong with B?
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Re: Massachusetts Babies [#permalink] 07 Jan 2011, 10:38
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Amplitude of sound Rating: 5,1/10 135 reviews
How Are Amplitude and Loudness Related?
This could represent two sound waves, for example bass and treble. It can be installed on your computer for easy reference, and includes all the sounds, text searching, bookmarking, and many printing options. Unfortunately, this has the effect of modulating the loudness of the sound as well. How do waves differ based upon their shapes? This chain continues until the particles run out of energy. Thanks for the info, bye. Journal of the American Association for Laboratory Animal Science. So, now can we find out the speed of a wave? See how it divides the rectangle bounding it into equal halves? All types of amplitudes are equally valid for describing sound waves mathematically, but pressure amplitudes are the one we humans have the closest connection to.
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Amplitude Calculator
The amplitude of this wave is 0. Sound cannot be produced without the vibration of objects. Plus, you are confusing loudness a psychoacoustic perception with level a logarithmic ratio of intensities. One time when I went to a concert there were so many people shouting and screaming that even when I covered my ears the sound waves still came through. We can represent their relationship with a simple equation: This equation represents the relationship between frequency and period. Period and Frequency The period is the time it takes a wave to complete one cycle. A magnitude 5 earthquake represents 100 times the ground motion and 900 times the energy released of a magnitude 3 earthquake.
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What is the amplitude of sound waves?
I would think it would be! Pulse amplitude is measured with respect to a specified reference and therefore should be modified by qualifiers, such as average, instantaneous, peak, or root-mean-square. Thank you for this phenomenal wonder. Start with the equation that relates intensity to displacement amplitude. For example if you change the frequency of a sound wave from 500Hz to 15000Hz, this won't necessarily be perceived as a change in loudness. A magnitude based on energy radiated by an earthquake, M e, can now be defined.
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What is amplitude in sound waves
I think that it's crazy that some people were doing that and they didn't even realize it! Thank you again for this wonderful wonder!!! Instead, amplitude measurements are almost always used as the raw data in some computation. I didn't know that this chain reaction continued until the air particles run out of energy and we call this a sound wave. This value is the sound intensity level right at the speaker. We've seen the picture above before. Really I think she thought I was talking gibberish. So there is no upper limit for amplitude.
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Physics for Kids: Sound Wave Characteristics
I really enjoyed this wonder's video and it was cool how they used household objects to make music. After settling into bed, you may hear your blood pulsing through your ears. And a quadrupling of the amplitude of a wave is indicative of a 16-fold increase in the amount of energy transported by the wave. I learned the last question in this wonder tonight. Sound intensity levels are quoted in decibels dB much more often than sound intensities in watts per meter squared. Ripples in water, sound traveling in air, and coordinated vibrations of objects are examples of waves you have probably encountered in your life.
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Sound Properties: Amplitude, period, frequency, wavelength (video)
I haven't ever played a guitar, but I understand the concept of sound energy and the transfer of it because last night our science homework was to read in our textbook about sound energy and the transfer with instruments. On the other hand, if you the same string with a lot of force, the note will be much louder. Wave speed is found by multiplying the wavelength and the frequency. The larger the amplitude, the more energy a wave has. It is reasonable to assume that sound is transmitted into a stethoscope 100 times as effectively compared with transmission though the air.
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Wave Parameters: Wavelength, Amplitude, Period, Frequency & Speed
Just visualize the cosine squared curve traced out over one cycle. When your partner hit the spoon with another one the results rocked! There are lots of ways your ears can get damaged, Alex! We are using a lower case for pressure to distinguish it from power, denoted by above. Sound is a variation in pressure. The unit of frequency measurement is Hertz Hz for short. Now I have just a few questions.
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What is the Unit of Amplitude Of sound?
We can't say for sure when the video was made, but that's when it first got posted online. Dogs could hear this note, though. It's easiest to find the wavelength by measuring the spatial distance between two wave crests. A vibration can cause a disturbance to travel through a medium, transporting energy without transporting matter. In school, we're also learning about sound energy. Anyways, thanks for the information. How do we know how much energy it carries? So, you see - period and frequency are reciprocals of each other.
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What is the amplitude of sound waves?
I visualized drawing the ocean waves and it helped me understand crest and troughs and that helped me understand amplitude. Children can hear up to around 20000 Hz, but the loudness drops rapidly above about 12000 Hz. Intensity is defined to be the power per unit area carried by a wave. Thus, sound intensity levels in decibels fit your experience better than intensities in watts per meter squared. We use the letter capital T to represent the period. ¥Question: A 20 Hz sound would have to be at what level to seem as loud as a 1,000 Hz sound at 20dB? Determine the amplitude, period, and wavelength of such a wave. It is only the amount radiated from the earthquake as seismic waves, which ought to be a small fraction of the total energy transferred during the earthquake process.
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What Makes Sounds Louder?
Â¥ Critical Distance: the distance from the source along the line of principal radiation, where the level of the direct sound and reverberant sound is equal. These hairs are responsible for receiving sound waves and sending sound information to the brain. They are described below: Period of sound A period can be said to be the time taken to do something. I've learned a lot from this wonder! Typical sounds have frequencies in the 100s or even 1000s of hertz. One of the more striking things about the intensities in is that the intensity in watts per meter squared is quite small for most sounds. In science, we use a lowercase v to symbolize speed because it's also called velocity.
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Contributor
## Pick ith element from concat or similar
Hi, I wish to pick the i'th element of a concatenated selection
=pick(1, concat( [Field], ', ' ))
I would have thought that I should be possible however, currently the second argument is interpreted as a single argument and not as many.
Is my approach possible and am I just missing something?
1 Solution
Accepted Solutions
MVP
## Re: Pick ith element from concat or similar
Hi Christoph,
Try this. You can use distinct inside the concat function, if you have duplicate values in the input field.
=Pick(1, \$(=Concat( Chr(39) & [Field] & Chr(39), ', ')))
22 Replies
MVP
## Re: Pick ith element from concat or similar
You can use something like below..
=SubField(Concat(DISTINCT [Field],', '),',',1)
//The above one is for picking 1st item
=SubField(Concat(DISTINCT [Field],', '),',',2)
// This is for 2nd item and so on
MVP
## Re: Pick ith element from concat or similar
Hi Christoph,
Try this. You can use distinct inside the concat function, if you have duplicate values in the input field.
=Pick(1, \$(=Concat( Chr(39) & [Field] & Chr(39), ', ')))
Contributor
## Re: Pick ith element from concat or similar
Yes. This was the combination I was looking for.
Is there a distinction between the two?
MVP
## Re: Pick ith element from concat or similar
Sorry. Which two you are asking?
Contributor
## Re: Pick ith element from concat or similar
=Pick(1, \$(=Concat( Chr(39) & [Field] & Chr(39), ', ')))
vs
=SubField(Concat(DISTINCT [Field],', '),',',1)
Contributor
## Re: Pick ith element from concat or similar
Iterating on that, I would like to use that further so that I can use the single value in some elaborate expression:
Something like:
=only({\$<[Field]={\$(=SubField(GetFieldSelections( [Field],', ',3),',',1))}>} [Field])
But I do not understand the parsing of expressions good enough yet.
MVP
## Re: Pick ith element from concat or similar
Both will show the same result. If you want to check how it is working, you can create a straight table and leave the dimension tab blank and add your expression like below. This way you can easily identify how the expression is parsing the value and debug your expression if required.
BTW, you have to add single quotes.
=only({\$<[Field]={'\$(=SubField(GetFieldSelections( [Field],', ',3),',',1))'}>} [Field])
MVP | 636 | 2,450 | {"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-2018-51 | latest | en | 0.8005 |
https://dlmf.nist.gov/search/search?q=uniform%20asymptotic%20expansions%20for%20large%20order | 1,716,794,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00594.warc.gz | 178,515,299 | 8,062 | # uniform asymptotic expansions for large order
(0.007 seconds)
## 1—10 of 38 matching pages
##### 3: 10.20 Uniform Asymptotic Expansions for Large Order
###### §10.20 UniformAsymptoticExpansions for LargeOrder
10.20.5 $Y_{\nu}\left(\nu z\right)\sim-\left(\frac{4\zeta}{1-z^{2}}\right)^{\frac{1}{4}% }\left(\frac{\operatorname{Bi}\left(\nu^{\frac{2}{3}}\zeta\right)}{\nu^{\frac{% 1}{3}}}\sum_{k=0}^{\infty}\frac{A_{k}(\zeta)}{\nu^{2k}}+\frac{\operatorname{Bi% }'\left(\nu^{\frac{2}{3}}\zeta\right)}{\nu^{\frac{5}{3}}}\sum_{k=0}^{\infty}% \frac{B_{k}(\zeta)}{\nu^{2k}}\right),$
10.20.9 $\rselection{{H^{(1)}_{\nu}}'\left(\nu z\right)\\ {H^{(2)}_{\nu}}'\left(\nu z\right)}\sim\frac{4e^{\mp 2\pi i/3}}{z}\left(\frac{% 1-z^{2}}{4\zeta}\right)^{\frac{1}{4}}\*\left(\frac{e^{\mp 2\pi i/3}% \operatorname{Ai}\left(e^{\pm 2\pi i/3}\nu^{\frac{2}{3}}\zeta\right)}{\nu^{% \frac{4}{3}}}\sum_{k=0}^{\infty}\frac{C_{k}(\zeta)}{\nu^{2k}}+\frac{% \operatorname{Ai}'\left(e^{\pm 2\pi i/3}\nu^{\frac{2}{3}}\zeta\right)}{\nu^{% \frac{2}{3}}}\sum_{k=0}^{\infty}\frac{D_{k}(\zeta)}{\nu^{2k}}\right),$
For asymptotic properties of the expansions (10.20.4)–(10.20.6) with respect to large values of $z$ see §10.41(v).
##### 4: 10.1 Special Notation
For older notations see British Association for the Advancement of Science (1937, pp. xix–xx) and Watson (1944, Chapters 1–3).
##### 5: 10.41 Asymptotic Expansions for Large Order
###### §10.41(ii) UniformExpansions for Real Variable
10.41.4 $K_{\nu}\left(\nu z\right)\sim\left(\frac{\pi}{2\nu}\right)^{\frac{1}{2}}\frac{% e^{-\nu\eta}}{(1+z^{2})^{\frac{1}{4}}}\sum_{k=0}^{\infty}(-1)^{k}\frac{U_{k}(p% )}{\nu^{k}},$
##### 6: 10.24 Functions of Imaginary Order
For mathematical properties and applications of $\widetilde{J}_{\nu}\left(x\right)$ and $\widetilde{Y}_{\nu}\left(x\right)$, including zeros and uniform asymptotic expansions for large $\nu$, see Dunster (1990a). …
##### 7: 10.45 Functions of Imaginary Order
For properties of $\widetilde{I}_{\nu}\left(x\right)$ and $\widetilde{K}_{\nu}\left(x\right)$, including uniform asymptotic expansions for large $\nu$ and zeros, see Dunster (1990a). …
##### 10: Bibliography O
• F. W. J. Olver (1959) Uniform asymptotic expansions for Weber parabolic cylinder functions of large orders. J. Res. Nat. Bur. Standards Sect. B 63B, pp. 131–169. | 927 | 2,322 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 41, "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-2024-22 | latest | en | 0.479795 |
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10.7: Surface Area and Volume of Cylinders
Difficulty Level: At Grade Created by: CK-12
Introduction
The Bean Containers
Jillian’s grandmother loves to cook. One day in between sewing projects, she takes Jillian to the grocery store and comes home with a big bag of assorted beans. Grandma’s baked beans are Jillian’s favorite, and she is thrilled that Grandma is going to cook them for dinner.
Jillian takes two different jars from the cupboard. One is long and thin and one is wide.
“Which jar should I use?” Jillian asks her grandmother.
“Use whichever one will hold the most,” her grandmother says.
Jillian looks at the two jars. Here is what they look like.
Jillian has measured each jar to try to figure out which one will hold the most. She just isn’t sure what to do now.
This is your task. Jillian will need to figure out the volume of each cylinder. This lesson will teach you all about calculating volume. Calculate the volume of each cylinder and then you will know which one will hold the most beans.
What You Will Learn
By the end of this lesson you will be able to demonstrate the following skills:
• Identify surface area of cylinders as the sum of the areas of faces using nets.
• Find surface area of cylinders using formulas.
• Identify volume of cylinders as the sum of volumes of layers of unit cubes.
• Find volumes of cylinders using formulas.
Teaching Time
I. Identify Surface Area of Cylinders as the Sum of the Areas of Faces Using Nets
In the last lesson, you learned about how to calculate the surface area and volume of different prisms. In this section, you will learn about calculating the surface area and volume of cylinders. Let’s start with calculating the surface area of a cylinder.
Here is a cylinder. Notice that it has two parallel congruent circular bases. The face of the cylinder is one large rectangle. In fact, if you were to “unwrap” a cylinder here is what you would see.
This is what the net of a cylinder looks like.
Just like when we were working with prisms, we can use the net of a cylinder to calculate the surface area of the cylinder.
How can we calculate the surface area of a cylinder using a net?
To calculate the surface area of a cylinder using a net, we need to figure out the area of the two circles and the area of the rectangle too.
Let’s think back to how to find the area of a circle. To find the area of a circle, we use the following formula.
There are two circular bases in the cylinder, so we can multiply the area of the circle by two and have the sum of the two areas.
The radius of the circles in the net above is 3 inches. We can substitute this given value into the formula and figure out the area of the two circles.
Next, we need to figure out the area of the curved surface. If you look at the net, the curved surface of the cylinder is rectangular in shape.
The length of the rectangle is the same as the circumference of the circle. Huh? Let’s look at the net. Since the length of the rectangle wraps around the circle rim, it is the same length as the circumference of the circle. To find the area of the curved surface, we need the circumference times the height.
Now we can add up the areas.
The surface area of the cylinder is .
II. Find Surface Areas of Cylinders Using Formula
Looking at the work that we just did, can we combine these steps to write a formula?
You can see that we have gotten a “thumbs up” on this idea! Let's write a formula and use that formula to find the surface area of a cylinder.
The formula for finding the surface area of a cylinder combines the formula for the area of the top and bottom circles with the formula for finding the area of the rectangular 'wrap' around the side. Remember that the wrap has a length equal to the circumference of the circular end, and a width equal to the height of the cylinder. Here it is.
Let’s apply this formula with an example.
Example
We work this problem through by substituting the given values into the formula. 4 centimeters is the radius of the circular bases. 8 centimeters is the height of the cylinder.
The surface area of the cylinder is . Notice that this works well whether you have a net or a picture of a cylinder. As long as you use the formula and the given values, you can figure out the surface area of the cylinder.
Practice a few of these on your own. Find the surface area of each cylinder using a formula.
Take a few minutes to check your work with a friend. Did you notice something in number 3?
III. Identify Volume of Cylinders as the Sum of Volumes of Layers of Unit Cubes
Now that you have learned how to figure out the surface area of a cylinder, let’s look at volume.
Volume is the amount of space contained within a solid figure. Since cylinders often contain liquid, you can imagine that the volume of cylinders often has to do with some kind of liquid. In the case of cylinders, you can think of volume as capacity.
Here is a cylinder that is probably used in a science lab. Here volume would be compared with capacity of liquid.
Here is a picture of a boy swimming in a pool. When you think about volume in this case, it is the capacity of the pool. The volume would be the amount of water in the pool.
This paint can is a cylinder. If we wanted to figure out the volume of this cylinder, we would need to figure out the amount of space inside the paint can. This would be the volume of the cylinder.
How can we think about the volume of a cylinder?
We can think about the volume of a cylinder as we would think about the volume of a prism. We can use unit cubes to fill a cylinder. Here is an example.
You can see that we have started to fill this cylinder with cubes to calculate the volume. The problem is that the cubes don’t fit perfectly inside the cylinder. To calculate the volume of a cylinder accurately, we need to use a formula.
One thing is sure though, no matter which formula we use, the volume of a cylinder is calculated in cubic units.
IV. Find Volumes of Cylinders Using Formulas
Which formula can we use to calculate the volume of a cylinder?
To calculate the volume of a cylinder, we need to calculate the area of the circular base. That will give us a measure for the number of unit cubes that can fit across the bottom of the cylinder. The height of the cylinder will show us how high cubes can be stacked inside the cylinder.
Here is the formula for finding the volume of a cylinder.
Now let’s apply this formula when working with an example.
Example
The radius of the circular base is 2 inches. The height of the cylinder is 7 inches. If we take both of these given measures and substitute them into the formula, we can solve for the volume of the cylinder.
The volume of the cylinder is .
Use the formula to find the volume of the following cylinders.
Take a few minutes to check your work with a partner.
Real Life Example Completed
The Bean Containers
Here is the original problem once again. Use what you have learned about the volume of cylinders to help Jillian problem solve this dilemma.
Jillian’s grandmother loves to cook. One day in between sewing projects, she takes Jillian to the grocery store and comes home with a big bag of assorted beans. Grandma’s baked beans are Jillian’s favorite, and she is thrilled that Grandma is going to cook them for dinner.
Jillian takes two different jars from the cupboard. One is long and thin and one is wide.
“Which jar should I use?” Jillian asks her grandmother.
“Use whichever one will hold the most,” her grandmother says.
Jillian looks at the two jars. Here is what they look like.
Jillian has measured each jar to try to figure out which one will hold the most. She just isn’t sure what to do now.
First, let’s go back and reread the problem. Underline any important information in red.
Jillian needs to figure out the volume of each cylinder. She can use the formula below to do this. Jillian suspects that the wide jar will hold more. What do you think?
Let’s start with the long thin jar. The diameter of the jar is 8 inches. We need the radius of the jar, so we can divide the diameter in half. The radius of this jar is 4 inches.
Wow! That jar sure does hold a lot. Let’s work on the wide jar now. The diameter of this jar is 12 inches, so the radius is 6 inches.
Jillian is amazed. The long, thin jar holds more volume than the wide jar does. Jillian takes the beans and puts them into the jar.
Sometimes volume can be tricky! What looks like it holds sometimes doesn’t!
Vocabulary
Volume
the amount of space inside a three-dimensional figure.
Surface Area
the entire outer covering or surface of a three-dimensional figure. It is calculated by the sum of the areas of each of the faces and bases of a solid.
Cylinder
a three-dimensional figure with two congruent parallel circular bases and a curved flat surface connecting the bases.
the measure of the distance halfway across a circle.
Technology Integration
Other Videos:
1. http://www.mathplayground.com/mv_volume_cylinders.html – This is a video by Brightstorm about how to find the volume of a cylinder.
1. http://www.mathplayground.com/mv_surface_area_cylinders.html – This is a Brightstorm video on how to calculate the surface area of a cylinder.
Time to Practice
Directions: Calculate the surface area of each of the following cylinders using nets.
1.
2.
3.
4.
5.
Directions: Find the surface area and volume of each of the following cylinders.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
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Subjects: | 2,151 | 9,769 | {"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} | 4.5625 | 5 | CC-MAIN-2017-13 | latest | en | 0.925089 |
https://www.edificemagazine.ch/40130/31UAW12/dimensional_figure_of_jaw_crusher.html | 1,631,862,346,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055601.25/warc/CC-MAIN-20210917055515-20210917085515-00369.warc.gz | 783,899,182 | 7,252 | #### detail dimensional figures of 2012 jaw crusher
detail dimensional figures of 2012 jaw crusher Quantitative three-dimensional microtextural analyses of tooth wear For lower pharyngeal jaws, three-dimensional microtextural data were Details of statistical methods, and results of additional analyses, are included in the Figure 1.
#### Detail Dimensional Figures Of Crusher Muzex
Gyratory Crusher Hydraulic Three Dimensional Map. Gyratory Crusher Hydraulic Three Dimensional Map 48 fc cone crusher capacity a gyratory crusher is similar in basic concept to a jaw crusher consisting of a concave cone crushers 36 fc find complete details about cone crushers from jw hewitrobins 3048 grizzly king jaw crusher 48 x 16 vibrating
#### Jaw Crusher ScienceDirect
2016-1-1 Figure 4.4. Jaw Crusher Operating Geometry. The Motion of the Jaw is Approximately Parallel . The distance, h, between A and B is equal to the distance the particle would fall during half a cycle of the crusher eccentric, provided the cycle frequency allows sufficient time for the particle to do so. If ν is the number of cycles per minute
#### Jaw crusher kinematics simulation and analysis
2015-3-31 2.3.1 device simulation model jaw crusher at work Establishing dual-chamber jaw crusher by using three-dimensional models UG shown in Figure 1, and by paeasolid format output. Open the ADAMS software, just import the output of a three-dimensional map.
#### Jaw Crusher an overview ScienceDirect Topics
Rose and English [9] determined the capacity of a jaw crusher by considering the time taken and the distance travelled by the particles between the two plates after being subjected to repeat crushing forces between the jaws. Therefore, dry particles wedged between level A and level B (Figure 4.4) would leave the crusher at the next reverse movement of the jaw.
#### “Computer Aided Design of Jaw crusher”
2010-5-14 Figure 1.2 2) Dodge Type Jaw Crusher The movable jaw is pivoted at the bottom and connected to an eccentric shaft. The universal crushers are pivoted in the middle so that the jaw can swing at the top and the bottom as well. Maximum amplitude of motion is obtained at the top of the crushing plates. Dodge type crushers
#### Jaw crusher kinematics simulation and analysis
Jaw crusher kinematics simulation and analysis crusher ijres 97 Page Figure 2 double cavity jaw crusher drive and restraint 2.3.2 impose constraints and drive the simulation model Before the establishment of a good geometry, motion analysis and
#### Jaw crusher SlideShare
2014-12-9 Figure 1 8. • Working Principle Of A Jaw Crusher • A jaw crusher uses compressive force for breaking of particle. • A Jaw Crusher reduces large size rocks or ore by placing the rock into compression. A fixed jaw, mounted in a "V" alignment is the stationary breaking surface, while the movable jaw
#### Jaw crusher kinematics simulation and analysis
2015-3-31 2.3.1 device simulation model jaw crusher at work Establishing dual-chamber jaw crusher by using three-dimensional models UG shown in Figure 1, and by paeasolid format output. Open the ADAMS software, just import the output of a three-dimensional map.
#### (PDF) A fundamental model of an industrial-scale jaw
In this study, an analytical perspective is used to develop a fundamental model of a jaw crusher. Previously, jaw crushers were modelled in regard to certain aspects, for example, energy
#### Dimensional Stone Crushing Bahamas
Dimensional Stone Crushing Bahamas. Product capacity : 5-2200t/h. Max Feeding Size : 125-1500mm. Output Size : 10-400mm . This series of jaw crusher belongs to stone crushing equipment which is widely used in the works of metallurgy, mining, cement, chemistry, refractory and ceramics as well as highway construction and water conservancy.
#### Jaw crusher kinematics simulation and analysis
Jaw crusher kinematics simulation and analysis 1. International Journal of Research in Engineering and Science (IJRES) ISSN (Online): 2320-9364, ISSN (Print): 2320-9356 ijres Volume 3 Issue 3 ǁ March. 2015 ǁ PP.95-99 ijres 95 Page Jaw crusher kinematics simulation and analysis Xiaodong Guo, Tao Yang, Caixia Lv, Longmei Dong School of Mechanical Engineering Inner Mongolia
#### Double toggle jaw crusher mechanism Manufacturer
Crusher mechanism jaw laurastownshiptourscoza. Dynamic Analysis of Double Toggle Jaw Crusher Using Pro quality of jaw crusher, this paper takes full advantage of the Function module of the Pro Engineer platform to make model simulation and dynamic analysis on the actual jaw crusher mechanism, and provided the updated path for the design and manufacture of Jaw Crusher
#### Jaw plate kinematical analysis for single toggle jaw
Oduori et al. [19] analysed the kinematics of the single toggle jaw crusher, as modelled in Figure 1, and found the following expression: Cao et al. [20] used the dimensional data for a PE 400
#### Analysis of the Single Toggle Jaw Crusher Force
The single toggle jaw crusher mechanism can be modelled as a planar mechanism, as shown in Figure 1. However, in this mechanism, it is the coupler that is the output link and the transmission angle, as defined in the above cited literature, fails to be a suitable indicator of the efficacy of force transmission.
#### Steady Transmission Limestone Potassium Barite
Raymond Mill Joyalcrusher, The purpose of the raymond mill is to grind materials in the fields of building materials mining metallurgy and chemical industry the materials must be nonflammable and nonexplosive materials such as limestone calcite barite dolomite potassium feldspar ma Steady Transmission Limestone Potassium Barite Raymond Mill
#### granite processing in quarry ME Mining Machinery
Granite Dimensional Stone Quarrying and Processing: 3.3.1 granite quarry operations 4 3.3.2 granite processing operations 5 4 lci results 5 references 24 list of figures figure 1. process flow diagram for granite quarrying operations 2 figure 2. process flow diagram for granite processing operations 3 list of tables 6 6 7 13 22 table 1. gross energy to produce one ton of granite products table
#### antimony ore jig for titanium mine in romania 10
Mobile crusher can work in the places with complex environments, so it is particularly used to process construction waste. Similarly, it also includes mobile jaw crusher, mobile cone crusher and mobile impact crusher, etc. kinds of equipment, which provides many choices for customers.
#### Dimensional Stone Crushing Bahamas
Dimensional Stone Crushing Bahamas. Product capacity : 5-2200t/h. Max Feeding Size : 125-1500mm. Output Size : 10-400mm . This series of jaw crusher belongs to stone crushing equipment which is widely used in the works of metallurgy, mining, cement, chemistry, refractory and ceramics as well as highway construction and water conservancy.
#### Computer Aided Design and Analysis of Swing Jaw Plate
2015-10-1 computerized. The design and modeling jaw plates of crusher is accomplished by using CATIA, by using this package three dimensional model of jaw plates jaw crusher has been developed. Finite Element Analysis of jaw plates are carried out by using ANSYS12 programming. This work is
#### Double toggle jaw crusher mechanism Manufacturer
Crusher mechanism jaw laurastownshiptourscoza. Dynamic Analysis of Double Toggle Jaw Crusher Using Pro quality of jaw crusher, this paper takes full advantage of the Function module of the Pro Engineer platform to make model simulation and dynamic analysis on the actual jaw crusher mechanism, and provided the updated path for the design and manufacture of Jaw Crusher
#### Jaw crusher fixed jaw casting process Qiming
Jaw crusher is mainly used in mine production. Jaw crusher fixed jaw is a key jaw crusher spare parts in this series and belongs to the newly developed product series of our company. This product is a large-scale box-type steel casting, which is mainly produced by integral casting.
#### Jaw crusher fixed jaw casting process
Jaw crusher fixed jaw casting process . Jaw crusher is mainly used in mine production. Jaw crusher fixed jaw is a key jaw crusher spare parts in this series and belongs to the newly developed product series of our company. This product is a large-scale box-type steel casting, which is
#### Steady Transmission Limestone Potassium Barite
Raymond Mill Joyalcrusher, The purpose of the raymond mill is to grind materials in the fields of building materials mining metallurgy and chemical industry the materials must be nonflammable and nonexplosive materials such as limestone calcite barite dolomite potassium feldspar ma Steady Transmission Limestone Potassium Barite Raymond Mill
#### Powder metallurgy basics & applications
2016-5-4 They are used to ensure close dimensional tolerances, good surface finish, increase density, corrosion resistance etc. Different types of grinding equipments/methods are shown in the figure Jaw crusher Gyratory crusher Roll crusher Ball Mill Vibratory
#### granite processing in quarry ME Mining Machinery
Granite Dimensional Stone Quarrying and Processing: 3.3.1 granite quarry operations 4 3.3.2 granite processing operations 5 4 lci results 5 references 24 list of figures figure 1. process flow diagram for granite quarrying operations 2 figure 2. process flow diagram for granite processing operations 3 list of tables 6 6 7 13 22 table 1. gross energy to produce one ton of granite products table
#### antimony ore jig for titanium mine in romania 10
Mobile crusher can work in the places with complex environments, so it is particularly used to process construction waste. Similarly, it also includes mobile jaw crusher, mobile cone crusher and mobile impact crusher, etc. kinds of equipment, which provides many choices for customers.
#### Quartering an overview ScienceDirect Topics
Sangeeta Mukhopadhyay, Subodh K. Maiti, in Bio-Geotechnologies for Mine Site Rehabilitation, 2018. 23.4.3 Collection and Analysis of Mine Soil Samples. At each site, six quadrates (10 m × 10 m) were laid down, and from each quadrate five mine soil samples were collected (0–10 cm depth) and mixed thoroughly and reduced the weight approximately to 0.5 kg by coning– quartering method to | 2,214 | 10,240 | {"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-2021-39 | latest | en | 0.873012 |
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# Source Gmatclub posted by goalsnr Doctor: Research shows
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Source Gmatclub posted by goalsnr Doctor: Research shows [#permalink]
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09 Dec 2010, 11:33
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Source Gmatclub posted by goalsnr
Doctor: Research shows that adolescents who play video games on a regular basis are three times as likely to develop carpal tunnel syndrome as are adolescents who do not play video games. Federal legislation that prohibits the sale of video games to minors would help curb this painful wrist condition among adolescents.
The doctor’s conclusion depends on which of the following assumptions?
(A) The majority of federal legislators would vote for a bill that prohibits the sale of video games to minors.
(B) Not all adolescents who play video games on a regular basis suffer from carpal tunnel syndrome.
(C) Playing video games is the only way an adolescent can develop carpal tunnel syndrome.
(D) Most parents would refuse to purchase video games for their adolescent children.
(E) The regular playing of video games by adolescents does not produce such beneficial effects as better hand-eye coordination and improved reaction time.
[Reveal] Spoiler:
[Reveal] Spoiler: OA
D
The guy who posted thisw question on the forum said the OA is D.
I agree on D but I feel C serves equally well.
If the carpal syndrome can be caused by things other than video games.then why will be the syndrome be curbed.
Curbed means stopped.
To undermine D , what if the adolscents are able to borrow the games from some of the older persons.
[Reveal] Spoiler: OA
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Re: Carpal tunnel [#permalink]
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09 Dec 2010, 12:03
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Great question - thanks for posting this one (and for the invitation to chime in!).
This is a terrific example of two crucial strategies for Assumption questions:
1) You have to identify the exact conclusion of any CR strengthen/weaken/assumption question
2) The assumption is something that is necessary to the argument; not merely something that would strengthen it
Here, the conclusion is that "prohibiting the sale of video games to minors would help curb carpal tunnel syndrome among adolescents. That's entirely different than something that would eliminate it.
The problem with choice C as it pertains to this conclusion is that the conclusion doesn't require video games to be the only or even a primary reason that kids develop this problem. Even if only 5% of carpal tunnel cases were related to video games, eliminating video games would still "help curb" the problem, however slightly. We don't need for video games to be the only reason that kids get this problem...we just need it to be one reason.
Now, with D, we need that to be the case - prohibiting "the sale of games to minors" doesn't necessarily mean that minors won't still play the games. In order to curb the problem we need to stop kids from playing the games, not simply from buying the games. If D were untrue - if it were the opposite - then we could say that parents will just buy these games for their kids anyway, so the kids won't stop playing the games. Therefore, we need D to be true - it's required by the argument otherwise the conclusion (that this law will help reduce the problem) isn't supported at all.
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Re: Source Gmatclub posted by goalsnr Doctor: Research shows [#permalink]
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07 Feb 2015, 19:18
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I have seen similar GMAT questions like this one many times; it looks like the GMAT wants us to distinguish between what the law says and what people actually do. as in this example, the law is prohibiting certain action, but the conclusion went as far as to conclude that the regulations will prevent the happening of the action. well, if the laws by themselves were capable of stopping people from doing bad things, we wouldn't need courts or jails...
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Re: Carpal tunnel [#permalink]
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09 Dec 2010, 12:00
mundasingh123 wrote:
Source Gmatclub posted by goalsnr
Doctor: Research shows that adolescents who play video games on a regular basis are three times as likely to develop carpal tunnel syndrome as are adolescents who do not play video games. Federal legislation that prohibits the sale of video games to minors would help curb this painful wrist condition among adolescents.
The doctor’s conclusion depends on which of the following assumptions?
(A) The majority of federal legislators would vote for a bill that prohibits the sale of video games to minors.
(B) Not all adolescents who play video games on a regular basis suffer from carpal tunnel syndrome.
(C) Playing video games is the only way an adolescent can develop carpal tunnel syndrome.
(D) Most parents would refuse to purchase video games for their adolescent children.
(E) The regular playing of video games by adolescents does not produce such beneficial effects as better hand-eye coordination and improved reaction time.
[Reveal] Spoiler:
[Reveal] Spoiler: OA
D
The guy who posted thisw question on the forum said the OA is D.
I agree on D but I feel C serves equally well.
If the carpal syndrome can be caused by things other than video games.then why will be the syndrome be curbed.
Curbed means stopped.
To undermine D , what if the adolscents are able to borrow the games from some of the older persons.
I am agree with you mundasingh123 , the answer should be "C" instead of "D''
First of all, it is asked to find the supposition between evidence and conclusion and if D were the supposition then why would the legislature ban the video games. Parents themselves be able to reduce the carpel tunnel syndrome.
Secondly, since it used in the stimulus "video games to minors would help curb this painful..." in the last line of the stimulus, we can take it as legislation can completely tackle this problem with taking such step. Therefore i believe that the given evidence completely supports the supposition given in option "C", which leads to the given conclusion.
Hence, My answer is "C"
Last edited by vyassaptarashi on 09 Dec 2010, 12:04, edited 1 time in total.
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Re: Source Gmatclub posted by goalsnr Doctor: Research shows [#permalink]
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04 Jan 2015, 16:56
Hello from the GMAT Club VerbalBot!
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Re: Source Gmatclub posted by goalsnr Doctor: Research shows [#permalink]
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04 Jan 2015, 17:00
The key to answer this question is in this sentence... "Federal legislation that prohibits the sale of video games to minors would help curb this painful wrist condition among adolescents"
Thus if the parents are still willing to buy the games for their kids then the legislation is a failure. So, the answer is D.
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Re: Source Gmatclub posted by goalsnr Doctor: Research shows [#permalink]
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12 Jul 2016, 05:31
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: Source Gmatclub posted by goalsnr Doctor: Research shows [#permalink] 12 Jul 2016, 05:31
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1 Doctor: Research shows that adolescents who play video games 20 01 Nov 2009, 02:38
Doctor: Research shows that adolescents who play video games 14 10 Jul 2009, 09:09
61 Doctor: Research shows that adolescents who play video games 54 04 Aug 2008, 16:29
Doctor: Research shows that adolescents who play video games 1 16 Jul 2007, 18:01
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# Source Gmatclub posted by goalsnr Doctor: Research shows
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Powered by phpBB © phpBB Group and phpBB SEO 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®. | 2,621 | 10,932 | {"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-2017-04 | latest | en | 0.89602 |
https://www.physicsforums.com/threads/the-real-line-is-misleading.895311/ | 1,638,024,127,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358189.36/warc/CC-MAIN-20211127133237-20211127163237-00510.warc.gz | 1,056,876,679 | 17,691 | # The real line is misleading?
Real analysis shows that the real line is complete. All the gaps between the rationals are filled. It is perfectly continuous. Of course, all of this is based off of certain axioms that make it work mathematically.
Are those axioms now sound? According to quantum, the plank length is the smallest unit of space allowable in existence. So if the idea of continuity is flawed in reality, then the current state of much of mathematics is also on shaky grounds.
DennisN
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According to quantum, the plank length is the smallest unit of space allowable in existence.
With kindness I challenge you to find a credible source that says/demonstrates that the Planck length is the smallest unit of space allowable in existence.
There is no experimental evidence of any smallest unit of space.
jtbell, vanhees71 and PeroK
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Real analysis shows that the real line is complete. All the gaps between the rationals are filled. It is perfectly continuous. Of course, all of this is based off of certain axioms that make it work mathematically.
Are those axioms now sound? According to quantum, the plank length is the smallest unit of space allowable in existence. So if the idea of continuity is flawed in reality, then the current state of much of mathematics is also on shaky grounds.
This is about the 4th post in the last week where it's claimed that the Planck length is the "smallest unit of length". This is a misinterpretation. See.
https://www.physicsforums.com/insights/hand-wavy-discussion-planck-length/
Moreover, mathematics is in no way dependent on the physical realities of time and space! How you can apply mathematics may be dependent on that, but not mathematics itself.
vanhees71
This is about the 4th post in the last week where it's claimed that the Planck length is the "smallest unit of length". This is a misinterpretation. See.
https://www.physicsforums.com/insights/hand-wavy-discussion-planck-length/
Moreover, mathematics is in no way dependent on the physical realities of time and space! How you can apply mathematics may be dependent on that, but not mathematics itself.
ah touche.
I know that math is independent of physical reality. But at the same time, we are using that math to prove things about the real world. I was watching a lecture where the quantization of momentum was derived using the periodic property of the momentum wave function, which is a form of euler's formula. These are continous trig functions. Then I thought that the plank length might also derived in a similar manner: using continuous sines and cosines. It would be problematic if continuous functions based off of continuous space is used to prove space is quantized; that would be a contradiction.
But after reading the link, I see that space isn't really quantized.
With kindness I challenge you to find a credible source that says/demonstrates that the Planck length is the smallest unit of space allowable in existence.
There is no experimental evidence of any smallest unit of space.
Understood. I'm new to quantum. I believe my misconception came of popular science books and my knowledge of the descrete energies and momentums and thought plank length may be similar.
DennisN
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2020 Award
Understood. I'm new to quantum. I believe my misconception came of popular science books and my knowledge of the descrete energies and momentums and thought plank length may be similar.
Excellent . You are definitely not alone in thinking what you thought - it is quite easy to make that thought leap from Planck length to thinking of it as a smallest unit of space. Yet, I should add that there could be a smallest unit of space - but to test this is very, very hard, with our technological capabilities of today.
PeroK
Homework Helper
Gold Member
2020 Award
ah touche.
I know that math is independent of physical reality. But at the same time, we are using that math to prove things about the real world. I was watching a lecture where the quantization of momentum was derived using the periodic property of the momentum wave function, which is a form of euler's formula. These are continous trig functions. Then I thought that the plank length might also derived in a similar manner: using continuous sines and cosines. It would be problematic if continuous functions based off of continuous space is used to prove space is quantized; that would be a contradiction.
But after reading the link, I see that space isn't really quantized.
There is a good, elementary example of the relationship between mathematics and physics in terms of a bouncing ball (there's a homework post about this at the moment). You can model a bouncing ball as an infinite geometric series of increasingly smaller bounces and, if you sum the infinite series you get a finite time at which the ball stops. In reality, of course, you do not have an infinite series of bounces and there comes a point where the mathematical bounces are so small that they are immeasurable, indistinguishable from the internal kinetic behaviour of the ball and, of course, smaller than the Planck length.
The mathematics stands up even though, in reality, you can never take the sum of an infinite series and there can only be in reality a finite number of bounces.
FallenApple | 1,133 | 5,358 | {"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 | 0.936877 |
https://www.doodlemaths.com/dmsc-article-yr2-meas-7-2-tell-and-write-the-time-to-quarter-to-the-hour/ | 1,601,119,900,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400241093.64/warc/CC-MAIN-20200926102645-20200926132645-00742.warc.gz | 807,265,246 | 13,514 | # Tell and write the time to quarter to the hour
Introduced in the Year 2 curriculum as: "Tell and write the time to five minutes, including quarter past/to the hour and draw the hands on a clock face to show these times"
When the time is 'quarter to', it is 15 minutes before the next hour. The
long hand (minute hand) points to the 9. The short hand (hour hand)
points just before the hour.
Example 1:
The time is quarter to 11
(Or 10:45)
Example 2:
The time is quarter to 6
(Or 5:45)
Example 3:
The time is quarter to 8
(Or 7:45)
Example 4:
The time is quarter to 2
(Or 1:45)
Example 5:
Which of these clocks shows 'quarter to' the hour?
Only the first clock shows 'quarter to' the hour. The long hand points to the 9. It's time is quarter to 4 (or 3.45). | 225 | 777 | {"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-2020-40 | latest | en | 0.86639 |
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Topic: spectral derivation of aperiodic function
Replies: 2 Last Post: Feb 11, 2013 5:37 PM
Messages: [ Previous | Next ]
Nasser Abbasi Posts: 6,677 Registered: 2/7/05
Re: spectral derivation of aperiodic function
Posted: Feb 11, 2013 5:37 PM
On 2/11/2013 3:51 PM, TideMan wrote:
>>
>>
>>> I wonder if there exist other spectral methods
>>
>>> to calculate the derivative on an aperiodic domain or if my only hope is
>>
>>> to use the finite difference method?
>>
>>
>>
>> no, it's likely your 'other hope' is to spend some time thinking more carefully about your problem(s).
>
> Yes, I was bemused by this post also.
> It implies that Kolmogorov's -5/3 law of turbulence and all other such work is rubbish.
> Perhaps OP is confusing non-periodic with non-stationary?
> FFT is not good for non-stationary data.
>
Actually, what I was confused about is not that saying that
FFT can't be used for aperiodic signals, but the need to use
FFT to find derivatives.
At school the teacher did not talk about this part of FFT.
He just told us FFT is used to find frequencies on signals.
I mean, if one wants to find derivative of a discrete
sequence in Matlab, why not use the command diff() ?
http://www.mathworks.com/help/matlab/ref/diff.html
or 3 points centered difference, or such schemes.
So I think there is still more to this issue than meets
the eyes.
--Nasser
Date Subject Author
2/11/13 Derek Goring
2/11/13 Nasser Abbasi | 431 | 1,648 | {"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-2018-13 | latest | en | 0.933671 |
http://security.stackexchange.com/questions/20941/how-can-i-decode-a-message-that-was-encrypted-with-a-one-time-pad?answertab=oldest | 1,462,519,411,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861735203.5/warc/CC-MAIN-20160428164215-00126-ip-10-239-7-51.ec2.internal.warc.gz | 256,322,011 | 22,213 | # How can I decode a message that was encrypted with a one-time pad?
How can I decrypt a message that was encrypted with a one-time pad? Would I need to get that pad, and then reverse the process? I'm confused here.
-
One time pads can be "decrypted" into anything, making them unbreakable. They are used by sovereign head of states (including USA) to communicate. Their one "vulnerability" is distribution of the key. – November Oct 2 '12 at 18:50
One-Time Pad is unbreakable, assuming the pad is perfectly random, kept secret, used only once, and no plaintext is known. This is due to the properties of the exclusive-or (xor) operation.
Here's its truth table:
``````A xor B = X
A | B | X
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 0
Number of 0s in column A = 2
Number of 1s in column A = 2
Number of 0s in column B = 2
Number of 1s in column B = 2
Number of 0s in column X = 2
Number of 1s in column X = 2
``````
Note that it introduces no bit-skew - the number of 0s and 1s in the inputs are equal to the number of 0s and 1s in the output, i.e. two of each. Furthermore, if you know only one element from a row, you cannot predict the values of the other two, since they are equally probable.
For example, let's say we know that X is 0. There's an equal probability that A = 0 and B = 0, or A = 1 and B = 1. Now let's say we know that X is 1. There's an equal probability that A = 0 and B = 1, or A = 1 and B = 0. It's impossible to predict. So, if you only know one element, you cannot possibly determine any information about A or B.
The next interesting property is that it is reversible, i.e.
``````A xor A = 0
B xor B = 0
A xor 0 = A
B xor 0 = B
A xor B xor B = A xor 0 = A
A xor B xor A = B xor 0 = B
``````
So, if we take any value and xor it with itself, the result is cancelled out and it always results in 0. This means that, if we xor a value A with a value B, then later xor that result with either A or B, we get B or A respectively. The operation is reversible.
This lends well to cryptography, because:
• xor introduces no bitskew
• xor has equally probable inputs for any given output
• given any two of A, B, X we can compute the third
As such, the following is perfectly secure:
``````ciphertext = message xor key
``````
but only if message is the same length as key, key is perfectly random, key is only used once, and only one element is known to an attacker. If they know the ciphertext, but not the key or message, it's useless to them. They cannot possibly break it. In order to decrypt the message, you must know the entire key and the ciphertext.
Keep in mind that the key must be completely random, i.e. every bit must have an equal probability of being 1 or 0, and be completely independent of all other bits in the key.
This actually turns out to be rather impractical, for a few reasons:
• Generating perfectly random keys is hard. Software generators (and many hardware generators) often have minuscule biases and odd repeating properties. It's almost impossible to gain truly random data in anything but tiny amounts.
• If the attacker knows the ciphertext and can correctly guess parts of the message (e.g. he knows it's a Windows executable, and therefore must start with `MZ`) he can get the corresponding key bits for the known range. These bits are useless for decrypting other parts of the message, but can reveal patterns in the key if it's poorly generated.
• You must be able to distribute the key, and your key must be equally as long as your message. If you can keep your key 100% secret between those of you who are authorised to read the message, why not just keep your message 100% secret instead?
The weak link here is your random number generator. The security of the one time pad is entirely limited by the security of your generator. Since a perfect generator is almost impossible, a perfect one-time pad is almost impossible too.
The final problem is that the key can only be used once. If you use it for two different messages, and the attacker knows both ciphertexts, he can xor them together to get the xor of the two plaintexts. This leaks all sorts of information (e.g. which bits are equal) and completely breaks the cipher.
So, in conclusion, in a perfect one-time pad you need to know the ciphertext and key in order to decrypt it, but perfect one-time pads are almost impossible.
-
One-time pads are extremely hard to break, in fact they are still used in some situations as if they are done correctly then they are essentially unbreakable. In a one-time pad system every character is changed by a stream of random data which is shared by both sides, without a copy of the pad you will not be able to break the code.
One of the few weaknesses in the system is the random data source. In WWII British one-time pads were being broken and they traced it to a worker whose job it was to pull random numbered balls out of a drum. The way it was supposed to work was that the worker would spin the drum, pull out a ball at random without looking at it, spin again, pick again, etc, etc. The worker started taking shortcuts by pulling out more than one ball after each spin and looking at the numbers, picking out favorites. It introduced patterns which enabled the opposition to break the pads, and lives were lost as a result.
The same is true today with pseudo-random number sources. Encryption protocols that should take millennia to break will really only last for years or decades without a true random data source.
-
Do you have a reference for that interesting WWII story? – m_t_ Mar 17 at 16:31
Assuming the encryption was competently done, the only way to decrypt is to get the pad.
-
If you don't have the key ...
Brute force the ciphertext (permute it) and analyze the output for potential cleartext in the expected language.
Statistical analysis can assist in the detection of potential human-generated language in the output (maybe chi-squared?)
-
You can't brute-force a one-time pad. If you generate every possible pad, you'd find that it also decrypts the message into every possible output text, so there's no way to tell which one is correct. – Brendan Long Oct 2 '12 at 21:01
+1 Yes, that is rather my point. – adric Oct 2 '12 at 21:15
But in your answer you say that you could permute the key and analyse the output but you can't. For example, say I sent the message `kill the king` and you intercept the 13 character (104 bit) ciphertext. You then somehow generate all 2^104 possible one-time pads (good luck doing that on a normal computer). Unfortunately, you find that along with `kill the king` as a possible output, there's also `save the king`, `save the duck`, `feed me ducks`, (consider padding/compression), etc., and there's absolutely no way to tell which one is correct without knowing something about the key. – Brendan Long Oct 2 '12 at 22:16
+1 again. You are doing a much better job of explaining why this is unlikely to succeed than I did, cheers. – adric Oct 3 '12 at 15:51
While one-time-pad encryption is provably impossible to break, note that it is also extraordinarily rare.
Part of the definition of OTP is that the pad must contain truly random data, and truly random data can be hard to come by for computers. Instead, often the pad is composed of pseudo-random data, which is generally what you get when you ask a computer for a random number.
In this case, you no longer have unbreakable encryption. Instead, you have a message XORed with the output of a deterministic and often reversible algorithm, which can be attacked by reproducing the same pseudo-random string. This attack is particularly devastating if you know what algorithm is used and even more so if you know how the seed is generated. In such a case, the code can be cracked in a matter of seconds. But even without knowing the PRNG seed, often the pattern can be derived from the message itself.
Also, OTP encryption is particularly unwieldy because the key is as long as the encrypted message, and must somehow be transmitted to the recipient without being intercepted. Any software that claims to use OTP encryption that does not require you to exchange a key block as large as your message (or larger) is not using truly random data, and is therefore not using OTP encryption.
Also, one of the key features of OTP is that you cannot reuse old pads. If you reuse an old one even once, that can be enough to allow the code to be broken.
-
one time pad could de easily cracked if you know the output format. Such as audio headers, gifs images, etc. By reverse engineering the stream back. It could be good for some sort of characters but not for some other fields where the output format could be easily identify by other means. Nothing is 100% secure nowadays.
-
This is simply incorrect. – Xander Dec 5 '13 at 16:40
The question was, how can you decrypt a message that was created from one time pad, pretty sure that hasn't been answered yet.
Originally the one time pad was used with just characters, and is very basic. Let's say you have a message 'killtheking' (taken from one response), and you want to encrypt it so you need a key. For the key you need at least the same number of characters as the message, so you roll a dice which has 26 sides (one for each letter) 11 times, because there are 11 characters in the message.
Assume the result is aqheivlekrw
now we subtract the position in the alphabet of the letter in they key, from the position in the alphabet of the letter in the message, and the resulting number is the position in the alphabet of the cypher text. Wrap any letters that create a minus result as below (like if the result is 30, then use 26 - 30 = 4 or 'd'.
eg:
``````k - a = (11 - 1) = 10 = j
``````
'j' is the 10th letter in the alphabet so we use that.
for the rest of the message we get:
``````i - g = (9 - 7) = 2 = b
l - h = (12 - 8) = 4 = d
l - e = (12 - 5) = 7 = g
t - i = (16 - 9) = 7 = g
h - v = (8 - 22) = -14 (26 - 14 = 12) = l
e - l = (5 - 12) = - 7 (26 - 7 = 19) = s
k - e = (11 - 5) = 6 = f
i - k = (9 - 11) = -2 (26 - 2 = 24) = x
n - r = (14 - 18) = -4 (26 - 4 = 22) = v
g - w = (7 - 23) = -16 (26 - 16 = 10) = j
``````
so in the end
``````your message = killtheking
you key = agheivlekrw
``````
pass the cypher text to whomever you want to recieve it and they do the opposite of what you have done to encrypt it, they have to add the key to the cypher text:
``````j + a = (10 + 1 = 11) k
b + g = (2 + 7 = 9) i
d + h = (4 + 8 = 12) l
g + e = (7 + 5 = 12) l
...
``````
and so on.
Computers use XOR because it does the same thing and is a lot quicker and easier to implement, but to understand how it works, I found that it is good to see the 'human' usable example. So on a computer you can encrypt the message by doing:
Cypher = Message XOR Key
and decrypt it again by using:
Message = Cypher XOR Key
It is as simple as that.
-
This does, of course, assume the die is perfectly random. For the example, we can probably assume this safely. But in reality getting a perfectly unpredictable sequence is not that easy. – S.L. Barth Mar 17 at 7:22
yes, using a loaded dice would kind of defeat the purpose, however, the encryption/decryption method is the same regardless of how effectively your random source is. – Neil H Mar 17 at 23:42 | 2,916 | 11,338 | {"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} | 3.609375 | 4 | CC-MAIN-2016-18 | latest | en | 0.931221 |
https://www.physicsforums.com/threads/help-with-contour-integrals.258209/ | 1,516,271,254,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887224.19/warc/CC-MAIN-20180118091548-20180118111548-00772.warc.gz | 997,867,072 | 15,732 | # Help with contour integrals!
1. Sep 21, 2008
### quasar_4
1. The problem statement, all variables and given/known data
Show that $$\int$$ $$\frac{ln(x) dx}{x^{3/4} (1+x)}$$ = - $$\sqrt{2}$$ Pi^2
2. Relevant equations
Residue theorem - integral = 2*Pi*i * sum of residues
3. The attempt at a solution
I am so lost. I don't even know where to start. I don't understand how to construct my contour. I'm not sure what I'm supposed to avoid. I don't understand how they're finding the residue. I don't get anything about this. Except that I think (!) I can show that |z*f(z)| goes to 0 as z--> 0 and --> 0 as z --> infinity. They do this in the book, but I don't really understand why (something to do with contour radii disappearing in extreme limits). I really need a better text (we have a review book that has a PAGE on this, so it's not enough, but I don't know what books are good).
All I can think of is do some kind of substitution where z = exp(i*theta), but I don't know beyond that... the examples in my book are very confusing. I really need help. My professor is out of the country for a few weeks and gave us this to work on. I don't know what to do without being able to get help
Any help, even just pointing to a good lucid text, would be so appreciated. This is a physics class, not a math class, so it would have to be reading that was clear enough to someone without complex analysis background. Thanks! | 378 | 1,429 | {"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} | 2.875 | 3 | CC-MAIN-2018-05 | longest | en | 0.976502 |
https://www.askiitians.com/forums/Analytical-Geometry/the-number-of-values-of-b-for-which-there-is-an-is_119726.htm | 1,696,461,521,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00440.warc.gz | 681,552,502 | 42,941 | # The number of values of b for which there is an isosceles triangle with sides of length b + 5, 3b – 2 and 6 – b is 1. 0 2. 2 3. 1 4. 3
Grade:11
## 3 Answers
Ananya Verma
20 Points
8 years ago
putting different values,
when b=2
property of isosceles triangle is proved.
kothapallisagarvivek
12 Points
6 years ago
Case (I)
b + 5 = 3b - 2
=> b = 7/2
So sides are 5/2 , 17/2 , 17/2
Case (II)
b + 5 = 6 - b = b = 1/2
Sides 11/2 , -1/2 , 11/2 is Not possible
Case (III)
3b - 2 = 6 - b
4b = 8
b = 2
7, 4, 4
two cases are possible
therefore answer is 2
kothapallisagarvivek
12 Points
6 years ago
Case (I)
b + 5 = 3b - 2
=> b = 7/2
So sides are 5/2 , 17/2 , 17/2
Case (II)
b + 5 = 6 - b = b = 1/2
Sides 11/2 , -1/2 , 11/2 is Not possible
Case (III)
3b - 2 = 6 - b
4b = 8
b = 2
7, 4, 4
two cases are possible
therefore answer is 2.
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# Answer to Question #63289 in Macroeconomics for Aisha
Question #63289
Given a hypothetical consumption function of the form: Y = C + I0 + G0 C = α + β Where: = Y – T Y = Income T = Taxes and that: Compute the equilibrium level of income and consumption
1
2016-11-17T11:47:09-0500
Y = α + β + (Y – T) + G0
Y = C
α + β + (Y – T) + G0 = α + β
Y = C = T – G0
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for any assignment or question with DETAILED EXPLANATIONS! | 185 | 571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-10 | latest | en | 0.840224 |
http://www.math-problem-solving.com/attractive-rich-famous.html | 1,493,429,747,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123172.42/warc/CC-MAIN-20170423031203-00496-ip-10-145-167-34.ec2.internal.warc.gz | 593,535,649 | 9,852 | # Attractive Rich Famous
## Which one would you like to be?
#### +1 Math Problem Solving
Page
Site
Attractive, Rich, Famous...Which one would you like to be?
This was an interesting post made on Facebook and proved to be very popular. Everybody liked the idea of being at least one of these things.
What's this got to do with maths? Hold your horses...Were getting there.
A lot of people thought that having one attribute would automatically bring about another. Or hope that it would.
Opinions differed but at the end of the day most people were happy with having just one of these attributes.
The most popular and obvious of them...Rich of course!
But, what brings about this question on our facebook wall? What prompted me to ask this question. Do you all think I post things nilly willy for the fun of it?
A-la..no! There is a reason behind every question. For this one, I wanted to know what was behind everyone's motive or goal in life. What one dream, if possible, would people want in life.
If there was one thing that could change peoples lives, would it be to be as famous as Madonna or Obama?
Maybe, if we had all the money in the world, then it would make us as rich, famous and attractive as them.
However way you want to look at it, the bottom line is that.. that extra bit of cash, the stuff that buys all our luxuries in life. The stuff that makes us look good, feel good and uber fabulosity goodness, actually makes our heart and feelings harder.
Yes it does. The more money we make, the more it makes us numb to hurtful comments. The more it makes us numb to pain. Various test have been done to prove this.
People have been experimented and made to keep their hand in freezing cold water. The ones that were given a lump sum of money managed to keep their hands in the cold water the longest. However, the poor people, they only managed a miserly few seconds.
Give me a wad of cash and I will gladly go swimming in freezing cold water any time ;)
But, how does all this relate to mathematics I hear you screaming.
Well it does and it doesn't.
For a start, do you know whats the one thing an employer looks at when he/she get's a job application? Do you think they are interested in your extra curricular activities?
Do you think they want to know how well versed you are at Gym or Business studies? Maybe if you were applying for a sports instructor or a business marketing dude, then yes.
However, the first thing an employer wants to know, is how well versed you are at solving problems. Can you confidently, independently solve sticky situations. Whether it be customer service or the payroll accounts. Can you? Can you?
Yes, the first thing they look at is your math grade. Is it up to par standard? If its not good, then your general solving problems is also not up to par.
Its brutal.
Its harsh.
But its the God's honest truth and sometimes the truth hurts. So, if you haven't got that Math grade, maybe you should take a moment and think about brushing up your skills.
If you think you can do well (or have already done well) and can make lots of cash without those math grades, then good luck to you :) just remember to drop us a line.
I know that one thing for sure, there's hundreds of people out there who have managed to secure savings without having to do any dreaded maths. Yes, they are self employed and have successful online businesses. Its not a secret. Anyone can do it. It takes a bit of hard work, but tell me one thing that's worth obtaining that is not hard. Certainly your math qualification for a start is hard work.
We may not be able to control some things in our life such as attractiveness. But we can control some things.
And we all like to see pretty people in this world. As we all very well know, beauty is skin deep and what counts on the inside is a million times more precious.
More precious of course than any gorgeous, superficial being who is horrible in this world.
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For Updates on Free Online Math Tutoring | 1,308 | 5,689 | {"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-2017-17 | longest | en | 0.985259 |
https://www.david-cook.org/can-i-use-an-op-amp-with-a-single-supply/ | 1,679,943,992,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00321.warc.gz | 783,916,933 | 14,351 | # Can I use an op amp with a single supply?
## Can I use an op amp with a single supply?
One of the most common applications questions on opera- tional amplifiers concerns operation from a single supply voltage. “Can the model OPAxyz be operated from a single supply?” The answer is almost always yes. Operation of op amps from single supply voltages is useful when negative supply voltages are not available.
Why does op amp need dual power supply?
Operational amplifiers have two power supply rails because they usually need to swing bipolar – output voltages that go either positive or negative in response to the normal range of input signals.
### What is single power supply?
With a single supply op amp, the V+ terminal of the op amp receives a positive voltage and the V- terminal connects to ground. A signal into the op amp can only swing as far as the power supply allows. Therefore, any input signal fed into the op amp can only swing from the positive voltage supply to ground.
Does op amp need negative voltage?
Here is a fact: Op-amps that are expected to handle ground referenced AC signals in a linear manner must see a negative supply voltage in respect to their input pins. This configuration sets both the input and the output points to average ground.
## What is dual supply opamp?
Dual Supply opamp has two supply rails with reference to GND to an opamp i.e +VCC and -VCC rails. Your applied voltage can swing between these two voltage levels. Hence, the output signal can swing only between these voltage(+VCC and -VCC) limits and they cannot exceed above these levels.
Can you split power supply?
Analog circuits often need a split-voltage power supply to achieve a virtual ground at the output of an amplifier. These split-voltage power supplies are generally low power supplies supporting tens of milliamps of differential current loads.
### Can opamp work without power supply?
lm741 without simetric power supply With a single positive supply then an input does not work if it is grounded, it must be at least +2v for it to work properly. If you bias the inputs at half the supply voltage then any opamp will work with a single supply voltage.
Does an op amp need a power supply?
An op amp needs a power supply because internally it is composed of a number of transistors. , you see the enormous amount of transistors which makes up an op amp.
## What is a DC single power supply?
What are single output power supplies? In AC-DC Single Output Power Supplies, a specific DC output is obtained from a 120VAC or 240VAC input using a combination of transformers, diodes and transistors. AC-DC Single Output Power Supplies can be of two types: regulated power supplies, and unregulated power supplies.
What is split supply?
Often a circuit requires a power supply that provides negative voltage as well as positive voltage. By reversing the direction of the diode and the capacitor (if it is polarized), the half-wave rectification circuit with low-pass filter provides a negative voltage.
### What type of power supply is required for op amp?
Op-amps use a DC supply voltage, typically anywhere from a few volts on up to 30 V or more. If the power supply is a perfect DC voltage source (that is, it gives the same voltage no matter what happens), the op-amp’s output would be solely governed by its inputs.
Which type of power supply is used in op amp?
## Can you split 12V power supply?
Feed the input 12V directly to the output 12V loads. Feed the 12V input to the input of a switching regulator module to convert the 12V to 9V output. Make sure the module is capable of supplying all the 9V load current required. Feed the 12V input to the input of the second switching regulator.
Are single supply op amps any good?
New op amps, such as the TLC247X, TLC07X, and TLC08X have excellent single-supply parameters. When used in the correct applications, these op amps yield almost the same performance as their split-supply counterparts. The single-supply op amp design normally requires some form of biasing. Introduction
### How to design a single-supply op amp?
The key to single-supply design is in remembering that voltage potentials are meaningful only when taken relative to other potentials. For an op amp circuit, the bottom line is this: the signal should be somewhere around mid supply.
What is the gain of a single supply op-amp?
Because you have a single supply op-amp configuration, you would also need to bias the inverting input of your op-amp to +4.5V (Where did you connect R3?) The gain will be (R3+R4)/R3 which is +11 for the resistor values shown here.
## What are the applications of single supply op-amp circuits?
Although it is advantageous to implement op-amp circuits with balanced dual supplies, there are many practical applications where, for energy conservation or other reasons, single-supply operation is necessary or desirable. For example, battery power, in automotive and marine equipment, provides only a single polarity. | 1,075 | 5,012 | {"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-2023-14 | longest | en | 0.914806 |
http://www.chegg.com/homework-help/questions-and-answers/consider-2-parallel-metallic-plates-seperated-distance-5mm-plate-areas-20-square-meters-to-q958337 | 1,386,267,014,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163047052/warc/CC-MAIN-20131204131727-00084-ip-10-33-133-15.ec2.internal.warc.gz | 275,931,353 | 7,812 | # gauss law
0 pts ended
consider 2 parallel metallic plates seperated by a distance of 5mm.. each plate has an areas of 20 square meters, and one has a total positive charge of 5 uC and the other a charge of -5uC
draw a diagram illustrating
a) use gauss law to calculate the electric field between the two plates
b) use the value of E you just obtained to calcuate the potential difference between the two plates
c) calculate the speed of a 10gm particle of a charge +2uC if it is released from the positive plate as it arrives at the negative plate
show all equations used and work | 138 | 584 | {"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-2013-48 | latest | en | 0.934424 |
https://www.mathnasium.com/ca/blog/20221228-card-playing-day | 1,721,654,007,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00264.warc.gz | 753,443,113 | 12,285 | Get Started by Finding a Local Centre
# Card Playing Day!
Dec 28, 2022
It’s Card Playing Day! A day for you and your child to take advantage of the entertainment and learning that a simple deck of cards offers, after a busy holiday season.
Not only are playing cards fun, but they are helpful instructional tools. They offer tactile learning for children and are primarily used to help introduce math facts, reinforce them, or extend student thinking.
Traditional card decks provide engaging and fun practice to help students master concepts such as basic computation. At Mathnasium, we have two special sets of playing cards to teach arithmetic skills, geometric shape properties, patterns, probability, critical thinking, and problem-solving. Our numbered set contains 52 cards with the face cards removed and replaced with the numbers 0-12, whereas our shape set includes 53 cards with geometrical figures.
Playing cards are useful math tools! Use numbered sets to compare card values and their attributes (i.e., greater than, less than, equal to; ordering consecutive numbers; match similar cards) and to practice basic computational facts (i.e., find pairs of numbers to make ten; build addition facts; compare products of pairs of numbers; solve for missing parts of addition/multiplication statements). Use shape sets to help your child recognize and compare/contrast geometric shapes.
So while your family and friends are still gathered together for the holiday season, invite your loved ones of all ages to join in on the fun! And look at the different card games we shared this year in our annual holiday gift guide: SKYJO, SET, and Blink!
For more information on math-teaching tools, contact your nearest Mathnasium Learning Centre. Happy card playing!
## OUR METHOD WORKS
Mathnasium meets your child where they are and helps them with the customized program they need, for any level of mathematics. | 399 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.904386 |
http://www.oxfordscholarship.com/oso/viewoxchap/10.1093$002facprof:oso$002f9780199229611.001.0001$002facprof-9780199229611-chapter-5 | 1,368,998,484,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698080772/warc/CC-MAIN-20130516095440-00013-ip-10-60-113-184.ec2.internal.warc.gz | 636,274,608 | 34,290 | ## Kiichiro Itsumi
Print publication date: 2009
Print ISBN-13: 9780199229611
Published to Oxford Scholarship Online: January 2009
DOI: 10.1093/acprof:oso/9780199229611.001.0001
# Aeolic Phrases
Chapter:
(p. 24 ) 5. Aeolic Phrases
Source:
Pindaric Metre: The ‘Other Half’
Publisher:
Oxford University Press
DOI:10.1093/acprof:oso/9780199229611.003.0005
# Abstract and Keywords
This chapter describes the typology of various aeolic phrases and their characteristics. The main topics are the aeolic base, the second position of reversed dodrans, catalexis, prolongation, and resolution. Statistics are provided so as to make clear the tendency of Pindar. Difference from the general tendency found in a particular verse provides the criteria by which stanza-forms are classified later in Chapter 8.
# A. The Basic Structure
Aeolic cola are characterized by the presence of one or other of two ‘asymmetrical’ phrases:
—⌣⌣—⌣— dodrans (dod) —⌣—⌣⌣— reversed dodrans (rdod)
Both phrases have six positions, and both have a double short and a single short flanked by longs. The difference lies in the order. They are mirror images of each other.
From the structural point of view, these two basic phrases, the dodrans and the reversed dodrans, can be preceded by ‘aeolic base’ (hereafter called ‘full base’, ∘∘), or by single anceps (‘half‐base’, ×). 40
no base —⌣⌣—⌣— —⌣—⌣⌣— dodrans (dod) reversed dodrans (rdod) half‐base ×—⌣⌣—⌣— ×—⌣—⌣⌣— telesillean (tel) heptasyllable (hepta) full base ∘∘—⌣⌣—⌣— ∘∘—⌣—⌣⌣— glyconic (gl) wilamowitzianum (wil)
## 1.The second position of reversed dodrans
The second position of the reversed dodrans is shown above as short. In the mainstream of Greek metre it is anceps (—×—⌣⌣—). But in Pindaric usage the position is hardly ever realized as long. Of 63 examples which contain the reversed dodrans, counting together the reversed dodrans proper and those prolonged by half or full base, none has the second position realized as long (i.e. ———⌣⌣—) consistently throughout all the repetitions. In most examples (56 phrases), a long syllable never appears and a short syllable always fills the anceps position. Even in the other seven, short anceps is dominant, and long anceps is often (but not always) caused by proper nouns. Apparently in (p. 25 ) Pindar, the long anceps is an occasional licence. But it must be noted that not all the long syllables can be eliminated as exceptional. There are a few examples which undoubtedly have anceps at the position, like aeolic cola in tragedy. These defy easy emendation, and attempts to emend them metri causa should be rejected. As I shall argue later, these examples are judged to belong to stanza‐forms in rigidly aeolic style (Class I). For more details, see E below.
## 2.Full base
‘Aeolic base’ consists of two ancipitia in Lesbian poetry. Thus the notation ×× is appropriate there, and a base in this form is a distinctive characteristic of aeolic metre. But in Pindaric usage, as in Attic poetry, the base is actually occupied by the following combinations of syllables: ——, —⌣, ⌣—, and ⌣⌣⌣. 41 For a base in this form, I use the symbol ∘∘.
There are considerable differences in the frequency of each form of the base. Pindar predominantly employs —⌣. Responsion between different forms is not entirely free, being admitted only between —— and —⌣ (therefore the notation —× is applicable for the scheme for particular passages), 42 but the other two forms (⌣— and ⌣⌣⌣) correspond only with themselves, except in one or two cases. So, while the notation ×— can be used for tragedy and elsewhere, it is not appropriate for Pindar. 43 This peculiarity of Pindar is shared without exception by all the phrases starting with the base. For the statistical detail, see C below.
## (p. 26 ) 3.Half‐base
The aeolic base can be reduced to a single anceps: ‘half‐base’. It can be —, ⌣, or ⌣⌣. The last form, ⌣⌣, stands in the same relationship to — or ⌣ as ⌣⌣⌣ at full base does to —— etc. Responsion is possible between — and ⌣, but not between ⌣⌣ and — or ⌣. I use the symbol × to cover all these forms (including ⌣⌣) for convenience’s sake. For the statistics, see D below.
## 4.Catalexis
The final ⌣— of the dodrans can be changed into —; this process is called catalexis. This is an a priori definition of catalexis, and it implies that the final position of, for example, pherecratean is not anceps but triseme (equal to three morae, ⨼). 44
The following are catalectic cola:
Catalexis is naturally not applicable to the reversed dodrans. 45
Some scholars posit phrases truncated yet further; for example:
∘∘—⌣⌣—
Incidentally, this form subsumes as one of its realizations the reversed dodrans (—⌣—⌣⌣—). The doubly truncated phrase can be dispensed with, so far as Pindaric examples are concerned: —⌣—⌣⌣— and ⌣⌣⌣—⌣⌣— are the reversed dodrans, while ⌣——⌣⌣— is acephalous e (∧e)+d. 46 It is not necessary to assume (p. 27 ) the existence of still shorter phrases, since ×—⌣⌣— and —⌣⌣— are freer D/e phrases (× d and d).
## 5.Prolongation
Just as phrases may be abbreviated (catalexis), so they may be prolonged. Prolongation is possible in both dodrans and reversed dodrans; for example,
∘∘—⌣⌣—⌣—— hipponactean (hipp)
is a glyconic prolonged by —. Presumably the value of this — is not triseme (nor true long) but anceps (or perhaps the long similar to the final long in the dactylic hexameter), comparable with that of link anceps in D/e, in Pindaric metre. Admittedly, the hipponactean could be seen as a catalectic form of ∘∘—⌣⌣—⌣—⌣—, but in practice this phrase is extremely rare, and is not found with a hipponactean as clausula. The same applies to the hagesichorean. Some phrases prolonged by — have no name. This type of phrase is designated by the notation +1, e.g. wil+1 (see the chart below). I analyse the aristophanean
—⌣⌣—⌣——
as dodrans prolonged by — (dod+1). The aristophanean is certainly the catalectic form of —⌣⌣—⌣—⌣— in Attic poetry, but it is not found with that function in Pindar. 47 Rather, one of the two examples functions as prolonged dodrans (dod+1):
O1e7 ⌣— —⌣— —⌣⌣—⌣— —⌣⌣—⌣——ǁ ∧e e dod ar
(p. 28 ) Beside prolongation by — (+1) there are longer ones like
×—⌣—⌣⌣—⌣— hepta+2 48 ∘∘—⌣⌣—⌣—⌣—— gl+3 (phalaecian)
Outside Pindaric metre, the ending ⌣—— can be taken as bacchiac. The name ‘bacchiac’ is not neutral since it inevitably implies ‘catalectic iambic’. ‘Phalaecian’ may therefore be less tendentious than gl+ba. But whether the last postion is created by catalexis or not is uncertain. It may be anceps like — (+1). I use the notation +3 without any implication. For the same reason, the ending ⌣— (+2) must be differentiated from the spondee following an aeolic colon.
In fact, we cannot always distinguish between catalexis and prolongation, although such a distinction may have been made in performance, i.e. audibly by music and visibly by dance. For example, some examples of —⌣—⌣⌣—— could in theory be rdod+1 (prolongation), but in practice we have no way to distinguish it from pherecratean with base in the form —⌣ (catalexis). The most delicate case is
N3e2 —⌣—⌣⌣— —⌣—⌣⌣——ǁ
where it is very tempting to suppose that reversed dodrans is repeated, with the second dodrans followed by anceps. Also, the pherecratean in the apparent priapean dicolon at O1s1 may not be pherecratean, but in fact rdod+1 (for the analysis of O1s1, see further 7. 5). On the other hand, so delicate a classification risks producing too many borderline cases. So hereafter I shall treat every example as pherecratean. The same is the case with rdod+2 (=gl) and rdod+3 (=hipp). 49
Below I set out all the theoretically possible forms. The forms in square brackets are the ones which can or should be analysed in different ways, and * is attached to forms which do not occur in the eighteen majors. (p. 29 )
dodrans reversed dodrans prolonged by — —⌣⌣—⌣—— [—⌣—⌣⌣—— ] aristophanean (ar) [rdod+1 = ph] ×—⌣⌣—⌣—— ×—⌣—⌣⌣—— hagesichorean (hag) *hepta+1 ∘∘—⌣⌣—⌣—— ∘∘—⌣—⌣⌣—— hipponactean (hipp) wil+1 ⌣— [—⌣⌣—⌣—⌣— ] [—⌣—⌣⌣—⌣— ] [dod+2 = d ⌣ e] 50 [rdod+2 = gl] [×—⌣⌣—⌣—⌣— ] ×—⌣—⌣⌣—⌣— [tel+2 = × d ⌣ e] hepta+2 [ —⌣—⌣⌣—⌣—⌣— ] ∘∘—⌣—⌣⌣—⌣— *[gl+2 = rdod ⌣ e] 51 wil+2 ⌣—— [—⌣⌣—⌣—⌣—— ] [—⌣—⌣⌣—⌣—— ] *[dod+3 = d ⌣ e —] [rdod+3 = hipp] ×—⌣⌣—⌣—⌣—— ×—⌣—⌣⌣—⌣—— tel+3 hepta+3 ∘∘—⌣⌣—⌣—⌣—— ∘∘—⌣—⌣⌣—⌣—— gl+3 (phalaecian) wil+3
Although examples are very scarce—indeed only one, P8e7, and that a forced analysis—the following phrase (a type of dodecasyllable) is taken as a single aeolic phrase:
hepta+2+3 ×—⌣—⌣⌣—⌣—⌣——
The enneasyllabic phrase, hepta+2, is prolonged in the same manner as the glyconic in gl+3 by ⌣——. Its theoretical correlative
wil+2+3 ∘∘—⌣—⌣⌣—⌣—⌣——
does not in fact occur in the eighteen majors.
# (p. 30 ) B. Classification by Ending and Frequency
All the aeolic phrases are arranged below in a different synoptic chart according to the number of positions preceded by the choriamb. In the following chart the number of occurrences is given in the rightmost column:
zero ending reversed dodrans —⌣—⌣⌣— 23 heptasyllable ×—⌣—⌣⌣— 12 wilamowitzianum ∘∘—⌣—⌣⌣— 15 +1 ending adonean —⌣⌣—— 2 reizianum ×—⌣⌣—— 9 pherecratean ∘∘—⌣⌣—— 11 *hepta+1 ×—⌣—⌣⌣—— 0 52 wil+1 ∘∘—⌣—⌣⌣—— 1 +2 ending dodrans —⌣⌣—⌣— 10 telesillean ×—⌣⌣—⌣— 29 glyconic ∘∘—⌣⌣—⌣— 49 hepta+2 ×—⌣—⌣⌣—⌣— 3 wil+2 ∘∘—⌣—⌣⌣—⌣— 4 +3 ending aristophanean —⌣⌣—⌣—— 2 hagesichorean ×—⌣⌣—⌣—— 1 hipponactean ∘∘—⌣⌣—⌣—— 2 hepta+3 ×—⌣—⌣⌣—⌣—— 2 wil+3 ∘∘—⌣—⌣⌣—⌣—— 2 [+4 ending] 53 +5 ending tel+3 ×—⌣⌣—⌣—⌣—— 4 gl+3 ∘∘—⌣⌣—⌣—⌣—— 3 hepta+2+3 ×—⌣—⌣⌣—⌣—⌣—— 1 *wil+2+3 ∘∘—⌣—⌣⌣—⌣—⌣—— 0
(p. 31 ) There are 145 phrases with zero or +2 ending against 40 with +1, +3, or +5. At the moment we should not jump to the conclusion that blunt aeolic endings are preferred to pendent, because phrases with zero ending and phrases with +2 can stand at the middle of the verse as well as at verse‐end and, consequently, have more chance of being employed. The frequency of blunt ending will be discussed at greater length later (8. A. 4).
# C. The Full‐Base Group
The following table consists of the same material as the table above, but in a different arrangement.
pherecratean ∘∘—⌣⌣—— 11 glyconic ∘∘—⌣⌣—⌣— 49 hipponactean ∘∘—⌣⌣—⌣—— 2 gl+3 ∘∘—⌣⌣—⌣—⌣—— 3 wilamowitzianum ∘∘—⌣—⌣⌣— 15 wil+1 ∘∘—⌣—⌣⌣—— 1 wil+2 ∘∘—⌣—⌣⌣—⌣— 4 wil+3 ∘∘—⌣—⌣⌣—⌣—— 2
The full‐base group can be further divided according to the nature of the base.
—⌣ ⌣⌣⌣ ⌣— —— —× ×— ⩊̲× (⩊̲—) 54 total gl 12 12 8 3 11 1 1 (1) 49 ph 6 1 1 3 11 hipp 2 2 gl+3 1 1 1 3 wil 9 1 2 3 15 wil+1 1 1 wil+2 1 2 1 4 wil+3 2 2 32 15 12 6 19 1 1 (1) 87
The four left‐hand columns after the stub indicate the number of the phrases in which exact responsion is rigidly kept throughout all the (p. 32 ) repetitions. There are in total 65 instances (32+15+12+6). 55 The proportion of these is remarkably high (65/87 = 74.7%). In other words, the anceps does not work as its name suggests. Another characteristic is the rarity of examples of —— (6 in total), compared with 31 of —⌣ and 15 of ⌣⌣⌣. It should be remembered that —— is the most usual form in tragedy. 56 In contrast, Pindar seems usually to have avoided the ‘heavy’ full base (——).
## 1.The base —×
Pindar’s preference for repeating the same form of full base is still more evident when one examines each repetition of the 19 examples of —× (for the terms ‘example’ and ‘repetition’, see Key). 57 Two peculiar groups emerge. First, in six cases —— occurs only in a limited number of repetitions, in most cases one repetition only, while the others are of the shape —⌣, and, moreover, all the —— without exception involve proper nouns. For example, at six out of seven repetitions of glyconic in I8s5c and I8s6, aeolic base is consistently occupied by —⌣, the exceptions being in s5c at v. 55c (= v. 56 Sn.) and in s6 v. 16 (= v. 16a Sn.). At both v. 55c and v. 16, Αἴγινα, a proper noun and a key word of I8, occupies the base ——, and consequently the scheme is given as —×—⌣⌣—⌣—. The same is true in the other four examples (P2e5, N4s5, N4s6, I7e5). Secondly, there are also some examples where a strong preference is observable, but where, nevertheless, there is a single deviation from the norm which (p. 33 ) does not involve a proper noun. Thus, at O9s2, P10s1, and P10e1, —⌣ is found in every repetition of the verse except one, but no proper noun is involved. This tendency is a manifestation of the ‘All‐but‐One’ rule (see Part III, C). At O9s5 and P8e3/4, —⌣ being the norm, —— is used more than once, in all instances but one with a proper noun. Conversely, at O9s8, —— is the norm, but —⌣ occurs once, without proper noun.
So in 65 examples the aeolic base is of identical form throughout and in another 12 examples it deviates from the chosen norm only very rarely. Yet Pindar does not always show the same attachment to a single form. In seven verses for which the notation —× is appropriate, no such strong preference is observable: O9s3 (3 longs/8 repetitions), P2s2 (4/8), P8s5 (4/10), P8e2 (2/5), N2s3 (3/5), N2s4 (2/5), N4s3 (8/12). In these cases the ‘anceps’ is really employed as anceps.
The situation is intriguing. A simplistic metrical rule is not appropriate. An explanation is to be sought not in a general metrical theory but in the examination of each metrical context; in other words, of the style of each stanza‐form as a whole. Interestingly, the occurrence of —× is concentrated in a limited number of stanza‐forms or odes: O9s (4 examples), P2e (1), P2s (1), P8s (1), P8e (2), P10s (1), P10e (1), N2s (2), N4s (3), I7e (1), I8s (2). Adding to them the examples of —— (6 in total), it is reasonable to suppose that Pindar employed ——, whether in responsion with —⌣ or not, only in some limited stanza‐forms. As will be argued later, the metrical context of these stanza‐forms is aeolic, not freer D/e. Thus the presence of —— is to be accepted as one of the criteria of Class I (aeolic) stanza‐forms.
Here emerges an important admonitory remark. It is dangerous to emend the text by introducing a word which has a long syllable at the second position of a glyconic (or equivalents), without consideration of the nature of the stanza‐form as a whole. Thus, for example, I am dubious about the transposition of πλαγχθǷντϵς at N7s8 (v. 37) proposed by Boeckh and accepted by most editors. 58
## (p. 34 ) 2.The base ⌣⌣⌣
Pindar uses ⌣⌣⌣ frequently as full base. He is perhaps the first Greek poet to introduce resolution in the aeolic base; the Lesbian poets and Anacreon did not use it, nor did Aeschylus except once. 59 It becomes common in later tragedy, especially in Euripides. Theoretically, ⌣⌣⌣ can be taken as a resolved form of —⌣ or ⌣—; but as a rule ⌣⌣⌣ does not correspond with the other forms, even in tragedy. The sole exception in Pindar is found at the beginning of P5e9:
P5e9 ⩊̲⌣—⌣⌣—⌣— —⌣⌣— ⌣ ⩊̲⌣— —⌣—ǁ gl de e
There, correspondence occurs between ⌣⌣⌣ ὕδατι (31) and —⌣ ὅϕ´ρα (62) and ϵὔχομαι (124) and —— σκυρωτάν (93). Perhaps the peculiarity of the style of P5 as a whole may be related to this irregularity (see Part II, ad loc.).
## 3.The base ⌣—
The base ⌣— is common neither in Pindar nor in tragedy. Of 12 examples in total, four are found in P2e, and two in I8s. This scarcity is partly explained by the fact that Pindar does not use glyconic or other phrases with full base κατὰ στίχον (P2e1a–4 are a rare exception). Anacreon and the dramatic poets are fond of the regularity produced by repetition, but with slight variation of the base. That is not Pindar’s manner. Thus O1s1 (but see below, 7. 6) and P8e6 are exceptional:
O1s1 ⌣——⌣⌣—⌣— —⌣—⌣⌣——ǁ gl ph P8e6 ⌣——⌣⌣—⌣— —⌣—⌣⌣—⌣—ǁ gl gl
In other words, since the phrase ⌣——⌣⌣—⌣— and the like are not invariably employed next to an unambiguous aeolic phrase starting with full base, there are cases in which it is not easy to tell aeolic base ⌣— from acephalous e (⌣—). For example, the first phrase of
P5e2 ⌣——⌣⌣—⌣— ⌣ —⌣— —⌣—⌣⌣—ǀ
may be not a glyconic, but ∧e+dodrans followed by short anceps+e+rdod, since a number of verses in P5 start with ∧e (Part II, ad loc.). Similarly, (p. 35 )
N7s1 ⌣——⌣⌣—⌣— —⌣—⌣—ǁ gl e 2
may be ∧e+dod+e 2, taking into consideration that the resemblance between the phrases —⌣⌣—⌣— and —⌣—⌣— is fully exploited in this stanza‐form (see Part II, ad loc.). As for the wilamowitzianum, I analyse
I8s5a ⌣——⌣—⌣⌣— —⌣—⌣⌣— —⌣—⌣⌣—ǀ
as wil+rdod+rdod. But it may be ⌣— (∧e)+threefold —⌣—⌣⌣—. But at the same time it must not be forgotten that there are some examples of ⌣——⌣⌣—⌣— which are, judging from the context, most certainly glyconic: for example, N2s1, and three consecutive verses of P2e (e1b–3). It is impracticable to take all the examples of ⌣— as ∧e. For the ambiguity, see further 7. 4.
# D. The Half‐Base Group
reizianum ×—⌣⌣—— 9 telesillean ×—⌣⌣—⌣— 29 hagesichorean ×—⌣⌣—⌣—— 1 tel+3 ×—⌣⌣—⌣—⌣—— 4 heptasyllable ×—⌣—⌣⌣— 12 hepta+2 ×—⌣—⌣⌣—⌣— 4 hepta+3 ×—⌣—⌣⌣—⌣—— 3 hepta+2+3 ×—⌣—⌣⌣—⌣—⌣—— 1
— ⌣ ⌣⌣ × total tel 14 3 6 6 29 reiz 4 3 2 9 hag 1 1 tel+3 2 2 4 hepta 3 1 8 12 hepta+2 1 1 1 3 hepta+3 2 2 hepta+2+3 1 1 24 9 8 20 61
(p. 36 ) Generally speaking, long is preferred to short in half‐base: 24 examples against 9. 60 Especially when an aeolic phrase makes up a verse on its own, long is dominant. P10s2 starts with short anceps, and is the sole exception, while long anceps occurs at every repetition in nine phrases (= verses). Besides, at explicit anceps (×, 20 examples) where both long and short are freely employed, long is dominant. Every repetition being counted, long occupies the anceps position 74 times, short 60 times. In some verses the anceps strongly tends to be either long or short: e.g. N4s3 (long is used at 11 out of 12 repetitions; ‘All‐but‐One’ is a manifestation of extreme cases of this tendency) or N4s1 (2 out of 10). These verses, and others of N4, indicate that it is neither a metrical rule nor the metrical context that regulates the realization of the anceps. N4 is a very simple aeolic stanza (monostrophic), and its first three verses start with hepta or hepta+2. But the ratio of long to short in each verse varies considerably: 2/10 (s1), 9/3 (s2), 11/1 (s3). 61 This seems to be a matter of aesthetic preference in the individual context.
## 1.Position of half‐base in verse
Half‐base is used either at the beginning of a verse (in cases where the verse starts with the phrase in question) or in the middle (in cases where the phrase is preceded by others). The dominance of long is clearer at the beginning of the verse. (p. 37 )
— ⌣ ⌣⌣ × total beginning 18 3 6 10 37 middle 6 6 2 10 24
Long anceps is exclusively used in half the verses (18 out of 37). However, there are 3 verses which invariably start with short anceps in all the repetitions: O9s10 (tel), P8e1 (hepta+2), P10s2a (hepta). At the 10 explicit ancipitia at the beginning of the verse, no strong preference is evident: long occurs in a total of 46 repetitions, short in 31. Just one verse has a high occurrence of short syllables: N4s1 (hepta+2, 10 short out of 12).
There are 6 examples of half‐base in mid‐verse where anceps is invariably short. All these phrases are preceded by…—⌣—⌣—, …—⌣⌣—⌣— or …—⌣⌣—, as the final part of the preceding phrase, so that the whole verse makes a sequence in which single short or double short alternates with single long without intervening long or anceps:
…—⌣—⌣—⌣—⌣⌣—…
…—⌣⌣—⌣—⌣—⌣⌣—…
…—⌣⌣—⌣—⌣⌣—…
These verses make a group of particularly Pindaric colour. The examples will be collected and discussed in I. 8.
## 2.Half‐base ⌣⌣
In 8 cola, ⌣⌣ occupies the base. All the cola are telesillean or its prolonged form (tel+3). I do not classify
O9e3, O13s1 ⌣⌣—⌣⌣——ǁ
as reizianum, nor the first phrase of
O1e5 ⌣⌣—⌣—⌣⌣— —⌣—⌣—ǁ
as ‘heptasyllable’ starting with ⌣⌣: O9e3, O13s1 is acephalous D and anceps, and O1e5 is ∧dd. 62
(p. 38 ) ⌣⌣ is never in responsion with —, let alone ⌣. Even outside Pindar, I do not know any certain example of ⩊̲—⌣⌣—⌣—. Whether the two double shorts in the phrase
⌣⌣—⌣⌣—⌣—
are equal, and if so, whether they are distinguishable from those of ∧D
⌣⌣—⌣⌣—
is an interesting question.
# E. The First Two Positions of Reversed Dodrans
As already demonstrated above (A. 1), there is no example which has ———⌣⌣— at all the repetitions. Thus reversed dodrans itself and phrases which include it are classified into three groups: —⌣—⌣⌣—, —×—⌣⌣—, ⌣⌣⌣—⌣⌣—.
—⌣ —× ⌣⌣⌣ total dod 13 1 9 23 hepta 10 1 11 wil 13 2 15 wil+1 1 1 hepta+2 2 2 4 wil+2 4 4 hepta+3 2 2 wil+3 1 1 2 hepta+2+3 1 1 46 7 10 63
## 1.Reversed dodrans starting with —×
The form —×—⌣⌣—, of course, permits initial ——. Of the seven examples of —×, four (O9s6/7, N4s1, N4s2, N4s6) occur in those stanza‐forms where aeolic characteristics are most evident, which I shall call Class I. This tendency is obviously related to the fact that —— tends to be used at the full base in the Class I stanza‐forms (see C. 1 above). But even here, the repetitions with long anceps (p. 39 ) remain very much in the minority. 63 Besides these four examples, there remain a further three. In
P10s2a ⌣—×—⌣⌣—ǀ hepta
long anceps is found only at v. 8 (ἀμϕκτιόνων). If this word can be treated as a proper noun, then the irregularity would be mitigated. The sixth example,
P11s4 —×⩊̲⌣⌣— — —⌣⩊ —⌣⌣—ǁ rdod × e d
is most irregular in that not only is long anceps used in the second position, but there is resolution to the ‘left’ of aeolic nucleus in half the repetitions, and long medial anceps occurs after rdod. 64 However, even here, there are some limitations: in repetitions where resolution occurs, the preceding anceps is always short. Thus there is no example of ——⩊⌣⌣—, its avoidance agrees with the general rule that long anceps does not precede resolved long; cf. 6. C (iii). The last (P8e7) is a curious verse as a whole, and, whatever analysis is chosen, its colometry will inevitably be a forced one. I offer the following provisionally:
P8e7 —— ×—×—⌣⌣—⌣—⌣——ǁ sp hepta+2+3
The anceps position of the (assumed) reversed dodrans (hepta = × rdod) is filled with long at 3 repetitions out of 5. See Part II, ad loc.
## 2.Reversed dodrans starting with ⌣⌣⌣
Ten verses have resolution of the initial position of the reversed dodrans (⌣⌣⌣—⌣⌣—). 65 There is no phrase in which resolved position is in responsion with unresolved. Of the nine examples of resolved reversed dodrans, five stand at the beginning of the verse. One is tempted to ask whether reversed dodrans in this form sounded identical in performance with the first part of a glyconic with base in the form ⌣⌣⌣. (p. 40 )
⌣⌣⌣—⌣⌣—
⌣⌣⌣—⌣⌣—⌣—
We cannot answer this question since we do not know the time‐value of anceps. It is, however, a reasonable hypothesis, and one which I propose to adopt, that the two phrases did sound identical.
Compare the following verses, which make a spectrum:
P2e4 ⌣⌣⌣—⌣⌣—⌣——⌣—ǁ gl e P6s3 ⌣⌣⌣—⌣⌣—⌣⩊̅—⌣—ǁ gl e N7e4 ⌣⌣⌣—⌣⌣—⌣⩊—⌣—ǁ gl e P2s7 ⌣⌣⌣—⌣⌣—⩊⌣——⌣—ǁ rdod e e O1e6b ⌣⌣⌣—⌣⌣——⌣—ǁ rdod e
The first two (P2e4, P6s3) are unambiguously gl+e. Following these two verses, I analyse N7e4 as gl+e, not as rdod+e 2. However, the next, P2s7, is not glyconic followed by long anceps+e, because there is no certain case at all of gle. 66 On the other hand, P2s7 is similar to O1e6b, where the first three shorts unambiguously belong to rdod.
The metrical context must be taken into consideration: reversed dodrans of the form in ⌣⌣⌣—⌣⌣— occurs in a limited number of stanza‐forms: O1e (1 example), P2s (2 examples), P2e (2 examples), P5e (2 examples), N6s (1 example), N7s (1 example), N7e (1 example). These are classified as Class II (freer D/e) or Class III (amalgamated), not as Class I (aeolic), except for P2e (an ambiguous case). On the other hand, the aeolic base in the forms —— or —× is a mark of Class I. It may not have been harmonized with resolved dodrans.
# F. Resolution
Apart from ⌣⌣⌣ at full base and ⌣⌣ at half‐base, a long position in aeolic phrase is sometimes filled by two short syllables. This can reasonably be regarded as resolution. 67 Every long position of dodrans and reversed dodrans can be resolved. For example, the following forms of glyconic are found: (p. 41 )
Left of the choriambic nucleus N6e2 —⌣⩊⌣⌣—⌣—ǀ resolved Right of the choriambic nucleus P8s2 —⌣—⌣⌣⩊⌣—ǁ resolved resolved Final position resolved I8s5c —×—⌣⌣—⌣⩊̲ —⌣——ǁ P6s3 ⌣⌣⌣—⌣⌣—⌣⩊̅ —⌣—ǁ
Resolution must be attested by responsion with unresolved position. Examples of resolved long position corresponding with unresolved are found in four verses:
P11s4 —⌣̄⩊̲⌣⌣— — —⌣⩊ —⌣⌣—ǁ rdod × e d N7s7 —⩊̲⌣⌣—⌣— ⩊⌣—⌣—ǁ tel e 2 P6s3 ⌣⌣⌣—⌣⌣—⌣⩊̅ —⌣—ǁ gl e I8s5c —⌣̄—⌣⌣—⌣⩊̲ —⌣— —ǁ gl e—
In the first two (P11s4, N7s7), resolution is found at the left‐hand long position of the choriambic nucleus (⩊̲⌣⌣—). Of these, the case of N7s7 is not surprising, since resolution involves a proper noun (v. 70 Εὐξένιδα), and may be taken as a special licence. But P11s4 is extraordinary. Of its eight repetitions, four have resolution, and none of the resolutions involves a proper noun (v. 9 θέμιν, v. 41 τὸ δὲ, v. 52 ἀνὰ, v. 57 καλλίονα). 68 In the other two (P6s3, I8s5c), resolution occurs at the final position of the glyconic, that is, of the dodrans (—⌣⌣—⌣⩊̲). At P6s3, the position is unresolved in only one repetition (v. 48 ἥβαν). And out of seven repetitions at I8s5c, three are resolved (v. 25c = v. 26 Sn. πινυτοί τϵ, v. 35c = v. 36 Sn. λϵχέων, v. 45c = v. 46 Sn. ἐπέων). Here the situation is similar to P11s4 in that resolved positions and unresolved are freely used in responsion.
There are 16 examples in which resolution is present at every repetition (i.e. responsion between resolved and unresolved is (p. 42 ) absent). Of these, 10 are the initial of reversed dodrans, mentioned in section E above. The others are: the final of the dodrans, N7s2, N7e3, N7e4; the left of the nucleus, N6e2 (gl); the right of the nucleus, P8s2 (gl), P11s2b (gl). In some verses it is difficult to decide which position is resolved. The sequence…⌣⌣⌣—⌣— is typically confusing. For example, see the phrases at the end of
N7e3 —⌣—⌣⌣—⌣⌣⌣—⌣—ǁ N7e4 ⌣⌣⌣—⌣⌣—⌣⌣⌣—⌣—ǁ
These are resolved forms of either (i)…⌣——⌣— or (ii)…—⌣—⌣—. If (i) is chosen, the verses are glyconic+cretic (e). On the other hand, if (ii) is chosen, the verses are reversed dodrans+e 2. I choose (i) on the analogy of P6s3 above, but there is no certainty here; see further Part II, ad loc.
# G. Acephaly
Besides the phrases above, there is one more phrase that is probably aeolic, judging from its context:
O13s5 ⌣⌣—⌣— —⌣—⌣⌣——ǁ ∧dod ph
The first half of this verse is explicable as acephalous dodrans. Acephaly is rather common, as will be demonstrated, in freer D/e, but to what extent it occurs in aeolic cola in general is not certain. 69 At least there are no other examples in the eighteen majors. Acephalous dodrans (⌣⌣—⌣—) is familiar as a constituent of the third verse of the Attic skolion, but it is doubtful whether such an extra‐generic comparison is meaningful. Rather, the proper comparandum is the Simonidean poem 542 P. Here acephalous dodrans is most certainly attested. The poem is monostrophic, with a seven‐verse stanza: 70 (p. 43 )
1 —⌣⌣—⌣⌣—⌣—⌣——ǀ 2/3 ⌣⌣— — —⌣—⌣⌣—⌣— —⌣—⌣⌣—⌣— × —⌣—ǀ 4/5 ⌣⌣—⌣— —⌣—⌣⌣—⌣— —⌣—⌣⌣—⌣—ǁ 6 ⌣⌣—⌣— —⌣—⌣⌣—ǀ 7 — —⌣— —⌣—⌣⌣—ǁ 8 ⌣— —⌣— —ǀ 9+10 —⌣—⌣⌣— × —⌣⌣—⌣——ǀ
Although there remain some uncertainties in the reconstruction of the strophe, it is certain that the phrase is situated at the beginning of the two verses 4/5 and 6. The basic structure of the former (4/5) is the repetition of —⌣—⌣⌣—⌣—. It is natural to extrapolate from it that ⌣⌣—⌣— is equal to —⌣—⌣⌣—⌣— minus the initial —⌣—. The latter (6) is almost identical with O13s5. These two Simonidean verses and O13s5 incorporate the palindromic movement (…⌣⌣—⌣— —⌣—⌣⌣…) found in O1s1 and elsewhere (see 8. A. 6 below).
Readers accustomed to Snell’s analysis may wonder why I exclude from aeolic consideration of ‘dactylic expansion’ (e.g. ∘∘—⌣⌣—⌣⌣—⌣—). There are, however, problems of ambiguity here, which I prefer to discuss following my definition of freer D/e; see below, 7. 6.
## Notes:
(40) The historical process is not taken into account here. Rather, the glyconic is the most common of all the aeolic cola in Pindaric metre as elsewhere.
(41) Some editors accept the responsion between ⌣⌣— and —— at N6e8. If the text there is correct, we must admit not only ⌣⌣— as a variation of aeolic base quite possibly unique in Greek poetry, but also the responsion between it and ——. This is outrageous. There is a notorious glyconic starting with ⌣⌣—: Aristoph. Ra. 1322 (πϵρίβαλλ᾽, ὠ̑ τέκνον, ὠλένας). Bacchylides 18 uses ⌣⌣——⌣⌣—⌣— in the environment of glyconics. I leave provisionally the transmitted text and the responsion ⩊̲—, but this does not mean that I accept it. See further, Part II, ad loc.
(42) But even this responsion is, in fact, restricted to Class I stanza‐forms. See below, 5. C.
(43) For the responsion in tragedy (with statistics), see Itsumi, ‘Glyconic’, 67–8.
(44) Cf. Parker, ‘Catalexis’, 15. Ancient metricians use the term ‘catalexis’ for mechanical amputation of the last syllable of a colon. That was a product of their conception of ‘final anceps’, which is now seen to be invalid. Cf. E. Rossi, RFIC 91 (1963), 52–71.
(45) Thus the latter phrase of the eupolidean dicolon
∘∘—×—⌣⌣— ∘∘—×—⌣—
is not the catalectic form of the former; contr. West, GM 95.
(46) Whether an actual phrase —⌣—⌣⌣— or ⌣⌣⌣—⌣⌣— should be taken (i) as the base (∘∘) + —⌣⌣—, or (ii) as reversed dodrans itself (of which the first position is resolved in the latter case) is, in a sense, a purely academic problem. A more important difference lies in the implication: if (i) is accepted, it means that ⌣——⌣⌣— is a correlative form of —⌣—⌣⌣— or ⌣⌣⌣—⌣⌣— and even that responsion between them is theoretically possible. There are 5 cases of ⌣——⌣⌣—, which do not correspond with other forms, in the eighteen majors (O1e2, P2e6, P5e5, P10s2b, P10s4). Judging from the metrical context and consistency with other related ones, it is better to take the phrase as acephalous e (∧e) + d (i.e. a freer D/e phrase); For acephale in general, see further 6, B and for the ambiguity between aeolic full base ⌣— and ∧e, see 7, (4). Thus I prefer (ii) to (i).
(47) Even in tragedy, aristophaneans are found repeated in synaphea; see Eur. Ba. 105 ff. cited in n. 28 above. Stinton supposes that link anceps occasionally occurs in aeolo‐choriambics (—⌣⌣—⌣—× in our case); at least, he admits the difficulties about this type of colon. See Stinton, ‘Pause and Period’, esp. ‘Postscript’, 64–6 (= Collected Papers, 358–61).
(48) This aeolic enneasyllable has been regarded as anceps plus glyconic by some scholars. This interpretation entails the unlikely supposition that there is a colon in Greek metre which starts with three ancipitia. And the so‐called ‘Barrett’s scheme’ (Hippolytos, Appendix I, p. 422) is refuted by Dale, LM 153 ff.
(49) Although the argument is circular, I am inclined to suppose that pherecratean in Class I stanza‐forms (aeolic) is really pherecratean (i.e. catalectic glyconic) but only seems to be so in Class III (amalgamated), being in fact rdod+1. So are other related phrases. I shall come back to this problem in the final chapter.
(50) The metrical contexts of actual examples refute the aeolic analysis of de and ⌣ de. They are collected at 8. C. 7.
(51) The analysis, rdode, is concordant with that of
—⌣—⌣⌣—⌣—⌣—⌣—
which is easily accepted as glyconic + link anceps (⌣) + e. Note that the anceps is always short in these cases. Examples are collected and discussed at 8. B. 5.
(52) There are two examples outside the eighteen majors: Pae6e10, e11.
(53) There is no phrase with +4 ending because I do not analyse phrases as aeolic that could theoretically be included in such a category. Thus —⌣⌣—⌣—⌣— in the list above I analyse as de, although it is equal to dod+2.
(54) The last form in parentheses, ⩊̲—, is based on a certainly corrupt text (N6e8); cf. n.41 above, but is included in the calculation.
(55) —⌣ (gl) P5s3, P8s2, P8e6, P11s2b, N3s4, N4s4, N6s1b, N6s2, N6e2, N7e3, I7e1, I8s4; (ph) O1s1, O1s4, O13s2, O13s5, P5e4 N3e2; (hipp) P2e8, N7s8 (wil) P5s2, P5e1, P6s1/2, P6s4, P6s5, N6s2, I8s1/2, I8s3, I8s4; (wil+2) N3s7; (wil+3) P10s5, N7s4.
⌣⌣⌣ (gl) P2e1a, P2e4, P6s1/2, P6s3, P6s6, P8s1, P11e4, N2s4, N3s3, N4s7, N7e4, N7e5; (ph) P11e2; (gl+3) N7e5; (wil) O9s6/7.
⌣— (gl) O1s1, P2e1b, P2e2, P2e3, P5e2, P8e6, N2s1, N7s1; (wil) I8s1/2, I8s5a; (wil+2) P2e8, P5s8
—— (gl) O9s4, N2s4, I7s5a; (ph) P8e3/4; (gl+3) I7s3/4; (wil+1) P6s7/8
(56) Itsumi, ‘Glyconic’, 67–8.
(57) —× (gl) O9s3, O9s5, O9s8, P2s2, P2e5, P8e3/4, P10e1, N2s3, I7e5, I8s5c, I8s6; (ph) P8e2, P10s1, N2s4; (gl+3) O9s2; (wil) N4s3, N4s5, N4s6; (wil+2) P8s5. The examples of the other two forms (×—, ⩊̲×) are too rare to be taken into account. For the detail of ×—, see Part II, O10s6. ⩊̲× (P5e9) will be illustrated later.
(58) The paradosis of v. 37 does not make good metre and must be emended; but Boeckh’s transposition is not as easy a solution as has been supposed. See further Part II ad loc.
(59) Itsumi, ‘Glyconic’, 68.
(60) * means that half‐base is used in the middle of a verse.
— (tel) O1e6a, *O9e4, O10e6, P11e3, *P2s2, P2s8, P5s7b, N3s1, N7s6, N7s7, N7s8, *I7s5a, I7e2, I7e3; (reiz) *O9s4, *O9s5, *O9s6/7, O9s9; (tel+3) N2s2, N4s8; (hepta) P2e7, P8s3, P11s2a; (hepta+2) P11s5.
⌣ (tel) O9s10, *I8s3, *I8s7; (reiz) *O1e4, *N4s4, *N4s6; (hag) *P2s8; (hepta) P10s2a; (hepta+2) P8e1.
⌣⌣ (tel) O9s1, *P2s4, P10s6, N3s8, N3e4, *N3e4; (tel+3) I7s1, I7e4.
× (tel) O9s2 (long at 7 repetitions out of 8), *P10e6 (1/4), *N4s5 (2/12), I7s2 (2/6), *I7s3/4 (1/6), *I7e5 (2/3); (reiz) *O9s3 (7/8), *N2s3 (2/5); (hepta) *O9e8 (2/4), P8s6 (4/10), P8e2 (3/5), P10e3 (2/4), P10e4 (3/4), N4s2 (9/12), N4s3 (11/12), *I7s5a (5/6); (hepta+2) N4s1 (2/12); (hepta+3) *O9e8 (3/4), P10e5 (3/4); (hepta+2+3) *P8e7 (3/5).
(61) Or 10/2 (s2) and 12 /0 (s3), if the initial syllable of ὁ χρυσός (v. 82) is scanned long, and if Οὐλυμπίᾳ(v. 75) is read instead of Ὀλυμπίᾳ.
(62) Some of the telesilleans, ⌣⌣—⌣⌣—⌣—, may be interpreted in a different manner. See 7. 2.
(63) O9s6/7 (3 long out of 8), N4s1 (2/12), N4s2 (3/12), N4s6 (1/12). Moreover, four of these long ancipitia are in proper nouns.
(64) Resolution at aeolic nucleus is discussed below, 5. F. For the mid‐long anceps see 6. D.
(65) (rdod) O1e6b, P2s2, P2s7, P2e2, P2e3, P5e3, P5e6, N6s3, N7e2; (wil+3) N7s4.
(66) An aeolic phrase is, in general, not followed by a long anceps. There are only two exceptions (P10s6, P11s4). See below, 6. 3.
(67) Resolution of longs in aeolic verses is rare in Attic drama; cf. Itsumi, ‘Glyconic’. It is almost restricted to later Euripides. The unique example of Aeschylus is Cho. 317: τύχοιμ᾽ ἂν ἕκαθϵν οὐρίσας (⌣—⩊⌣⌣—⌣—). It worries many (e.g. Dale, BICSSuppl. 21.2 (1981), 15: ‘resolution in 317 very ugly and hardly possible in Aeschylus’), but it may be accepted by referring to N6e2 and, especially, P11s4. It is noteworthy that P11 treats the myth of Clytaemestra and Orestes. Perhaps Aeschylus is influenced by Pindar. Further, Cho. 315 = 332 (the first verse of the same strophe) may betray another influence of Pindar. —⌣⌣—⌣⌣—⌣— is, most certainly, not glyconic in Aeschylus (see Itsumi, ‘Glyconic). Rather it may be related to P11s1 (for this verse, see 7. 6 below).
(68) For the avoidance of long anceps preceding the resolution, see 6. C above.
(69) Note that O13s is the only stanza‐form where the metre shifts from non‐D/e to D/e in the middle. The alternative interpretation is: ⌣⌣ + e (⌣⌣ being substituted for anceps). See Part II, ad loc. Outside the eighteen majors, ∧dod occurs at Parth1s1; see Part II, Appendix B.
(70) I follow Page in the reconstruction of the text, but the colometry is my own. West, GM 66 gives an analysis different from both Page’s and mine. | 13,832 | 35,881 | {"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-2013-20 | longest | en | 0.867478 |
http://bootmath.com/product-of-spheres-embeds-in-euclidean-space-of-1-dimension-higher.html | 1,532,090,063,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591596.64/warc/CC-MAIN-20180720115631-20180720135631-00618.warc.gz | 58,365,601 | 6,056 | # Product of spheres embeds in Euclidean space of 1 dimension higher
This problem was given to me by a friend:
Prove that $\Pi_{i=1}^m \mathbb{S}^{n_i}$ can be smoothly embedded in a Euclidean space of dimension $1+\sum_{i=1}^m n_i$.
The solution is apparently fairly simple, but I am having trouble getting a start on this problem. Any help?
#### Solutions Collecting From Web of "Product of spheres embeds in Euclidean space of 1 dimension higher"
EDIT: I cannot delete this post as it’s been already accepted 🙁
The proof is carried out by induction on $m.$
$m=1$ is trivial by choosing coordinates $(x^{(1)}_0,x^{(1)}_1,…,x^{(1)}_{n_1})$ where $\sum_{j=0}^{n_1} (x_j^{(1)})^2=1$, so let $m=2,$ then similarly $S^{n_1} \times S^{n_2}$ is embedded in $\mathbb{R}^{n_1+n_2+2}$ which also lies in the hypersurface $H$ with equation $\sum_{j=0}^{n_1} (x_j^{(1)})^2+\sum_{j=0}^{n_2} (x_j^{(2)})^2=2.$ In fact, $H$ is diffeomorphic to $S^{n_1+n_2+1}.$ Now, the embedding is missing at least a point, for example $p=(0,…,0,1,1) \in H,$ so by stereographic projection $S^{n_1+n_2+1} \setminus \{p\}$ is diffeomorphic to $\mathbb{R}^{n_1+n_2+1}.$
Suppose that the assertion holds for $<m,$ then $(S^{n_1}\times…\times S^{n_{m-1}}) \times S^{n_m}$ is embedded diffeomorphically in $\mathbb{R}^{n_1+…+n_{m-1}+1} \times \mathbb{R}^{n_m+1}\cong \mathbb{R}^{n_1+…+n_{m}+2}$ hence following the same argument you can reduce the dimension by $1,$ so the result.
• Note first that $\mathbb{R}\times\mathbb{S}^n$ smoothly embeds in $\mathbb{R}^{n+1}$ for each $n$, via $(t,\textbf{p})\mapsto e^t\textbf{p}$.
• Taking the Cartesian product with $\mathbb{R}^{m-1}$, we find that $\mathbb{R}^m\times\mathbb{S}^n$ smoothly embeds in $\mathbb{R}^{m}\times\mathbb{R}^n$ for each $m$ and $n$.
• By induction, it follows that $\mathbb{R}\times\prod_{i=1}^m \mathbb{S}^{n_i}$ smoothly embeds in a Euclidean space of dimension $1+\sum_{i=1}^m n_i$.
The desired statement follows. | 733 | 1,963 | {"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} | 3.640625 | 4 | CC-MAIN-2018-30 | latest | en | 0.758763 |
https://www.acmicpc.net/problem/17971 | 1,600,946,780,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400217623.41/warc/CC-MAIN-20200924100829-20200924130829-00174.warc.gz | 706,897,255 | 10,410 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 512 MB 75 41 35 64.815%
문제
Consider a situation that we need to distribute n gifts to n people fairly and randomly. For this purpose one ancient technique is popular in Asia, and is usually used to represent a random permutation. Chinese call it Ghost Leg (畫鬼腳), Japanese Budda Lots (あみだくじ) and Korean Ladder Game (사다리타기). We first start with some terms. This ladder consists of several vertical poles and horizontal bars connecting two adjacent vertical poles. From the top of each vertical pole, a path is traced through the ladder using the following three steps:
1. When tracing a vertical pole, continue downwards until an end of the first bar is reached, then continue along the bar.
2. When tracing a bar, continue along it until the end of the bar is reached, then continue down the vertical pole.
3. Repeat Step 1 and Step 2 until the bottom of a vertical pole is reached.
Figure D.1(a) shows a ladder L with three vertical poles and three bars. The vertical poles are numbered with 1, … , n from left to right. The paths of tracing these three vertical poles are shown in (b), (c), and (d). The input (1, 2, 3) is permuted finally to πL = (3, 2, 1) by the ladder L. Note that we do not allow the case that two immediately adjacent horizontal bars meet at some point (like w in Figure D.1(e)) because there is no unique way in tracing at the point.
You are given a ladder L which achieves a permutation πL. If the permutation πL is unchanged after removing a set of bars containing a particular bar, then the bar is said to be redundant for πL. Your task is to find all redundant bars for πL. This is equivalent to constructing a minimal ladder by removing all redundant bars from L. In the minimal ladder of L, the removal of any non-empty set of bars from the ladder gives a different permutation from πL.
입력
Your program is to read from standard input. The input starts with a line containing one integers n (3 ≤ n ≤ 50), where n is the number of vertical poles. The vertical poles are ordered from left to right. Let di,j denote the depth of the j-th bar between the i-th and (i + 1)-th poles from the top, which is an integer between 1 and 1,000. In the following n − 1 lines, the i-th line contains the sequence of depths di,j, where 1 ≤ i < n, j ≥ 1 and di,j < di,j+1. Notice that this depth sequence is ended with zero (0), which is not a depth but just a marker to indicate the end of the sequence. See Figure D.2 for illustration.
(a) Depth di,j of the bars (b) Sample input 1 (c) Sample input 2
Figure D.2: Depths of the bars between two adjacent poles in the ladder.
출력
Your program is to write to standard output. Print a set of the bars to be remained in a minimal ladder for πL. The first line should contain k, the number of the bars remained in the minimal ladder. Then, each line of the following k lines should contain two indices i and j of di,j of the bar in the minimal ladder. Note that the minimal ladder is not unique.
예제 입력 1
6
2 5 8 0
6 0
1 10 0
4 6 11 0
5 7 9 0
예제 출력 1
8
3 1
4 1
5 1
2 1
4 2
1 3
3 2
4 3
예제 입력 2
5
2 6 9 0
3 7 8 11 0
2 4 9 0
6 10 0
예제 출력 2
6
1 1
3 1
2 1
4 1
3 3
2 4
출처
• 스페셜 저지를 만든 사람: alex9801
• 잘못된 조건을 찾은 사람: jh05013 | 945 | 3,233 | {"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} | 3.015625 | 3 | CC-MAIN-2020-40 | longest | en | 0.89333 |
https://www.globalsino.com/EM/page4933.html | 1,721,321,939,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00477.warc.gz | 662,925,116 | 3,862 | Dependence of Collection Angle on Objective Aperture in TEM Imaging Mode
- Practical Electron Microscopy and Database -
- An Online Book -
https://www.globalsino.com/EM/
This book (Practical Electron Microscopy and Database) is a reference for TEM and SEM students, operators, engineers, technicians, managers, and researchers. ================================================================================= In TEM imaging mode, if no objective aperture is used then the collection angle is very large (> ~ 100 mrads) and need not be calculated accurately for EELS analysis because small differences in a large β do not affect the EELS spectrum or subsequent quantification. To calculate β in image mode with no aperture inserted, one needs to know the magnification of the diffraction pattern (DP) in the back-focal plane of the projector lens (which is the front-focal plane of the spectrometer), given by β = r0/L ------------------------------------------ [4933a] here, r0 is the maximum radius of the DP in the focal plane of the spectrometer (e.g. typical value of ~ 5 µm), and L is camera length defined by Magnification versus Camera Length. The typical value of β is ~ 100 mrad. When an objective aperture is inserted, the collection angle can be easily calculated by considering the geometrical size and focal length of the objective lens (as shown in Figure 4933) and using Equation [4929], given by, β = R0/f ------------------------------------------ [4933b] here, R0 is the radius of objective aperture and f the focal length of the objective lens. The reason why we do not need to consider the entrance aperture of EELS/GIF setup is that all the electrons through the objective aperture to the spectrum setup fall into the entrance aperture. However, some considerable error due to chromatic aberration can occur. Figure 4933. Schematic diagram showing β in TEM image mode is determined by the dimensions of the objective aperture.
=================================================================================
The book author (Yougui Liao) welcomes your comments, suggestions, and corrections, please click here for submission. If you let book author know once you have cited this book, the brief information of your publication will appear on the “Times Cited” page. | 461 | 2,316 | {"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 | 0.843691 |
https://calculla.com/amount_of_substance | 1,713,770,363,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818081.81/warc/CC-MAIN-20240422051258-20240422081258-00600.warc.gz | 134,476,144 | 218,878 | Amount of substance units converter
Converts amount of substance from one unit to another e.g. from millimoles (mmol) to number of particles (atoms or molecules depending on substance) or vice versa.
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Symbolic algebra
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Inputs data - value and unit, which we're going to convert#
Value Unit yottamole [Ymol]zettamole [Zmol]examole [Emol]petamole [Pmol]teramole [Tmol]gigamole [Gmol]megamole [Mmol]kilomole [kmol]hektomole [hmol]mole [mol]decimole [dmol]centimole [cmol]milimole [mmol]micromole [µmol]nanomole [nmol]pikomole [pmol]femtomole [fmol]attomole [amol]zeptomole [zmol]yoctomole [ymol]number of particles [particles] Decimals 0123456789
SI#
Unit Symbol Symbol(plain text) Value as symbolic Value as numeric Notes Unit conversion formula yottamole Show source$Ymol$ Ymol Show source$\text{...}$ - Derived amount of substance unit in SI system. One yottamole is equal to septylion of moles: $1\ Ymol= 10^{24}\ mol$ Show source$...$ zettamole Show source$Zmol$ Zmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One zettamole is equal to sextillion of moles: $1\ Zmol= 10^{21}\ mol$ Show source$...$ examole Show source$Emol$ Emol Show source$\text{...}$ - Derived amount of substance unit in SI system. One examole is equal to quintillion of moles: $1\ Emol= 10^{18}\ mol$ Show source$...$ petamole Show source$Pmol$ Pmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One petamole is equal to quadrillion of moles: $1\ Pmol= 10^{15}\ mol$ Show source$...$ teramole Show source$Tmol$ Tmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One teramole is equal to trillion of moles: $1\ Tmol= 10^{12}\ mol$ Show source$...$ gigamole Show source$Gmol$ Gmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One gigamole is equal to billion of moles: $1\ Gmol= 10^{9}\ mol$ Show source$...$ megamole Show source$Mmol$ Mmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One megamole is equal to million of moles: $1\ Mmol=1000000\ mol= 10^{6}\ mol$ Show source$...$ kilomole Show source$kmol$ kmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One kilomole is equal to thausand of moles: $1\ kmol=1000\ mol= 10^{3}\ mol$ Show source$...$ hektomole Show source$hmol$ hmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One hektomole is equal to hundred of moles: $1\ hmol=100\ mol= 10^{2}\ mol$ Show source$...$ mole Show source$mol$ mol Show source$\text{...}$ - The basic amount of substance unit in the SI system. One mole corresponds to the amount of substance that contains $6.023 \times 10^{23}$ particles (→ see Avogadro's number). Depending on the type of substance, it can be the number of atoms, ions or chemical molecules.$1\ mol = N_A\ \text{particles} = 6.023 \times 10^{23}\ \text{particles}$ Because definition and properties of the single particle depends on type of substance there is no constant relation between number of moles and mass. It means that, for example, one mole of water has different mass than one mole of atomic helium. Show source$...$ decimole Show source$dmol$ dmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One decimole is equal to one tenth of mole: $1\ dmol=0.1\ mol= 10^{-1}\ mol$ Show source$...$ centimole Show source$cmol$ cmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One centimole is equal to one hundredth of mole: $1\ cmol=0.01\ mol= 10^{-2}\ mol$ Show source$...$ milimole Show source$mmol$ mmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One milimole is equal to one thousandth of mole: $1\ mmol=0.001\ mol= 10^{-3}\ mol$ Show source$...$ micromole Show source$\mu mol$ µmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One micromole is equal to one millionth of mole: $1\ \mu mol=0.000001\ mol= 10^{-6}\ mol$ Show source$...$ nanomole Show source$nmol$ nmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One nanomole is equal to one billionth of mole: $1\ nmol= 10^{-9}\ mol$ Show source$...$ pikomole Show source$pmol$ pmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One pikomole is equal to one trillionth of mole: $1\ pmol= 10^{-12}\ mol$ Show source$...$ femtomole Show source$fmol$ fmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One femtomole is equal to one quadrillionth of mole: $1\ fmol= 10^{-15}\ mol$ Show source$...$ attomole Show source$amol$ amol Show source$\text{...}$ - Derived amount of substance unit in SI system. One attomole is equal to one quintillionth of mole: $1\ amol= 10^{-18}\ mol$ Show source$...$ zeptomole Show source$zmol$ zmol Show source$\text{...}$ - Derived amount of substance unit in SI system. One zeptomole is equal to one sextillionth of mole: $1\ zmol= 10^{-21}\ mol$ Show source$...$ yoctomole Show source$ymol$ ymol Show source$\text{...}$ - Derived amount of substance unit in SI system. One yoctomole is equal to one septillionth of mole: $1\ ymol= 10^{-24}\ mol$ Show source$...$
other#
Unit Symbol Symbol(plain text) Value as symbolic Value as numeric Notes Unit conversion formula number of particles Show source$particles$ particles Show source$\text{...}$ - Number of particles in the probe. Depending on the type of substance, it can be the number of atoms, ions or chemical molecules. Number of particles equals to Avogadro's number is one mole of substance.$N_A\ \text{particles} = 6.023 \times 10^{23}\ \text{particles} = 1\ mol$ Show source$...$
Some facts#
• The basic SI unit of amount of substance is one mole.
• One mole of substance contains the same number of molecules (or atoms in the case of free elements that do create molecules) as 12 grams of carbon isotope 12C.
• In one mole there is 6,022140857 (74) × 10 23 particles (atoms, molecules, ions, etc.). This number is often called Avogadro number:
$N_A = 6,022140857(74) \times 10^{23}$
• One mole of substance may correspond to different mass. For example, one mole of water weighs 18,01528 g, but one mole of carbon dioxide 44,01 g. The mass of one mole of substance is called molar mass and is substance specific. You can read more about molar mass (including molar masses of selected substances) visiting our another calculator: Molar mass.
• One mole of perfect gas under normal conditions (temperature 273K, pressure 1023 hPa) occupies a volume of 22.42 dm3. You can find more about the volume occupied by various gases in our another calculators: Molar volume of gases and Clapeyron's equation.
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Tags to Polish version: | 2,232 | 7,931 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 87, "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-2024-18 | latest | en | 0.740199 |
https://datascienceparichay.com/article/python-convert-integer-to-hexadecimal/ | 1,713,184,721,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00070.warc.gz | 162,112,728 | 36,417 | # Python – Convert Integer to Hexadecimal
In this tutorial, we will look at how to convert an integer to hexadecimal in Python with the help of some examples.
## How to convert integer to hexadecimal in Python?
You can use the Python built-in `hex()` function to convert an integer to its hexadecimal form in Python. Pass the integer as an argument to the function. The following is the syntax –
```# convert int i to hexadecimal
hex(i)```
It returns the integer’s representation in the hexadecimal number system (base 16) as a lowercase string prefixed with `'0x'`.
## Examples
Let’s look at some examples of using the above function to convert int to hex.
Pass the integer as an argument to hex function.
```# int variable
num = 15
# int to hex
num_hex = hex(num)
# display hex and type
print(num_hex)
print(type(num_hex))```
Output:
```0xf
<class 'str'>```
You can see that we get the hexadecimal representing the integer as a lowercase string with `'0x'` prefix.
If you do not want the prefix, you can use the string slice operation to remove the prefix from the returned hex string.
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Let’s now apply the same function to a negative integer.
```# int variable
num = -15
# int to hex
num_hex = hex(num)
# display hex and type
print(num_hex)
print(type(num_hex))```
Output:
```-0xf
<class 'str'>```
### Integer to uppercase hexadecimal string
Alternatively, you can use the Python built-in `format()` function to convert an integer to its hexadecimal form. You can customize the format of the returned string.
For example, to convert an integer to an uppercase hexadecimal string, use the format string `'#X'` or `'X'`.
```# int variable
num = 15
# int to hex
num_hex = format(num, '#X')
# display hex and type
print(num_hex)
print(type(num_hex))```
Output:
```0XF
<class 'str'>```
We get the hexadecimal for the integer as an uppercase string.
With the `format()` function you can customize the returned hexadecimal string – with or without the prefix, uppercase or lowercase, etc.
```# int variable
num = 15
# int to hex
print(f"Lowercase with prefix: {format(num, '#x')}")
print(f"Lowercase without prefix: {format(num, 'x')}")
print(f"Uppercase with prefix: {format(num, '#X')}")
print(f"Uppercase without prefix: {format(num, 'X')}")```
Output:
```Lowercase with prefix: 0xf
Lowercase without prefix: f
Uppercase with prefix: 0XF
Uppercase without prefix: F```
You might also be interested in – | 692 | 2,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-18 | longest | en | 0.73547 |
https://www.java-forums.org/new-java/45815-using-variables-loop.html | 1,490,545,393,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189242.54/warc/CC-MAIN-20170322212949-00312-ip-10-233-31-227.ec2.internal.warc.gz | 941,382,086 | 15,794 | # Thread: Using variables in a loop
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## Using variables in a loop
Java Code:
```import javax.swing.*;
public class MortgageApp2 {
public static void main(String args[])
{
String loanamt,
interest_peryr,
yearstot,
pay_amt;
double loan,
interest_yr,
years,
interest_mo,
loan_pmts,
payments,
new_balance,
old_balance,
interest,
principle,
payamt,
balance;
loanamt= JOptionPane.showInputDialog("Enter Loan Amount");
interest_peryr=JOptionPane.showInputDialog("Enter the Interest Percentage: Example 5.6");
yearstot=JOptionPane.showInputDialog("Enter Loan Period in Years");
loan=Double.parseDouble(loanamt);
interest_yr=Double.parseDouble(interest_peryr);
years=Double.parseDouble(yearstot);
interest_mo = interest_yr/12/100; // Turns percent whole numbers into decimals
loan_pmts= years * 12;
payments = (loan*interest_mo/(1- Math.pow((1+interest_mo),-loan_pmts)));
JOptionPane.showMessageDialog (null, "Your payment is: \$ " + payments);
principle= payments - (loan *interest_mo);
JOptionPane.showMessageDialog(null, "your prinicple paid is: \$ " + principle);
interest= payments-principle ;
JOptionPane.showMessageDialog(null, "your interest paid is: \$ " + interest);
for(int month=yearstot ;month<=(yearstot*12);month++){
}
System.exit(0);
}
}```
So in the last line of my code i am trying to call the years of the loan entered by the user to do a loop in months, so it would be year * 12 so i can get the total amount of months to increment the loop, and then from there i want to add the calculation to do principle and so on, but how do i get the loops to run bu using user defined years?
Thanks!!
2. how do i get the loops to run bu using user defined years
Are you talking about the yearstot variable?
Your code shows that its value is being read in from the user.
Can you explain what is wrong with your current code? It appears that the loop is controlled by a value entered by the user.
What values do you want for the month variable inside of the loop?
3. Please make an effort to indent your code consistently. This will make it easier to read. Not only for us but also for you. As a learner sometimes it is difficult to see things when the code is all over the place like a mad womans knitting.
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Ok so i made a mistake in the code, it is now as follows :
for(int month=yearstot ;month<=(yearstot*12);month++)
So i need the loop to start at 1(the first month payment) and end at yearstot*12 (the last month payment)
So that each month it will increment until the loops hits the last month of the mortgage, from there will put im all the calculations.
Right now i am getting a operator cannot be applied error with the above code :/ not to good with loops yet
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opps the code is for(int month=1 ;month<=(yearstot*12);month++) // this is the code that is throwing operator exception
6. yearstot is a String. Please tell me what the result of "hello" * 2 will be.
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oooooooo!! so if i change it to years it worked!! ahahah! i been going at this all day, starting to miss things! Thank you!
8. code that is throwing operator exception
When you get an error, please copy and paste the full text of the error here.
What data type is yearstot? Can you multiply it by 12?
Look at the Integer class's methods if you need to convert to a integer.
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now to figure out how to get the calculations to work, I will probably be back in a little bit. :)
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Java Code:
```import javax.swing.*;
public class MortgageApp2 {
public static void main(String args[])
{
String loanamt,
interest_peryr,
yearstot,
pay_amt;
double loan,
interest_yr,
years,
interest_mo,
loan_pmts,
payments,
new_balance,
old_balance,
interest,
principle,
payamt,
balance;
loanamt= JOptionPane.showInputDialog("Enter Loan Amount");
interest_peryr=JOptionPane.showInputDialog("Enter the Interest Percentage: Example 5.6");
yearstot=JOptionPane.showInputDialog("Enter Loan Period in Years");
loan=Double.parseDouble(loanamt);
interest_yr=Double.parseDouble(interest_peryr);
years=Double.parseDouble(yearstot);
interest_mo = interest_yr/12/100; // Turns percent whole numbers into decimals
loan_pmts= years * 12;
payments = (loan*interest_mo/(1- Math.pow((1+interest_mo),-loan_pmts)));
JOptionPane.showMessageDialog (null, "Your payment is: \$ " + payments);
principle= payments - (loan *interest_mo);
JOptionPane.showMessageDialog(null, "your prinicple paid is: \$ " + principle);
interest= payments-principle ;
JOptionPane.showMessageDialog(null, "your interest paid is: \$ " + interest);
for(int month=1 ;month<=(years*12);month++){
balance= balance -(loan-principle);
//new_balance=bala
System.out.println("balance is " + balance);
}
System.exit(0);
}
}```
ok so now that i got the loop to work.... Now, but all it is doing is repeating the same thing over and over again. I am new to mortgages and loops, still in college, and 2nd week of my first java class. How do i get the prgram to take a payment off each time the loop goes threw a month?
So the first one takes the principle off the loan, but then how to i get the program to take take the new balance into account and subtract another payment?
11. how to i get the program to take take the new balance into account and subtract another payment?
Write down the steps needed to do the processing. How would you do it with paper and pencil?
You must understand the algorithm before you try to code it.
12. Your problem now is a Math one and nothing to do with Java. You need to step away from the computer, grab a pen and some paper and work out how to calculate your sums. Once you have got it clear in your head (and on paper) you can then easily write the code to do what you want.
SNAP!
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• | 1,515 | 6,107 | {"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} | 2.78125 | 3 | CC-MAIN-2017-13 | longest | en | 0.723682 |
https://www.studypool.com/discuss/518488/how-many-bicycles-were-produced-from-day-15-to-21?free | 1,495,757,094,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.80/warc/CC-MAIN-20170525233846-20170526013846-00605.warc.gz | 948,313,809 | 13,945 | ##### How many bicycles were produced from day 15 to 21?
Calculus Tutor: None Selected Time limit: 1 Day
May 6th, 2015
t = 1
because from 15 - 21 it is one week.
80 + 0.1 . 1^2 - 0.7 . 1
80 + 0.1 - 0.7
79.4
Hence 79 is the correct answer
May 6th, 2015
May 6th, 2015
...
May 6th, 2015
...
May 6th, 2015
May 26th, 2017
check_circle | 148 | 341 | {"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} | 3.359375 | 3 | CC-MAIN-2017-22 | longest | en | 0.973022 |
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Airplane-related Math Problem.
Steinwayartist From United States of America, joined Aug 2001, 66 posts, RR: 0Posted Tue Aug 28 2001 19:30:04 UTC (14 years 9 months 3 days 14 hours ago) and read 2911 times:
My math teacher (I'm dropping the course, btw, but not because of this ) gave us a problem that started out like this: "An airplane is flying at 350 mi/h and one mile high..." Well, we all know that's a big no-no, seeing as how there's a (287.5mph) 250Kts<-->10,000ft.MSL limit I scribbled "illegal" and all sortsa stuff on the page, explaining why it's inconceivable, etc., after which we had to turn the paper in. I wonder what she'll have to say... Just a little humor -Nick
David L From United Kingdom, joined May 1999, 9645 posts, RR: 42 Reply 1, posted Tue Aug 28 2001 19:35:52 UTC (14 years 9 months 3 days 14 hours ago) and read 2878 times:
Did the question state which country the aircraft was flying over? Signed, A. Killjoy
Toady From United Kingdom, joined Feb 2001, 724 posts, RR: 0 Reply 2, posted Tue Aug 28 2001 19:52:12 UTC (14 years 9 months 3 days 14 hours ago) and read 2870 times:
Why is it "inconceivable & illegal"? The question is "An airplane..." not "A civil airliner...". You'll end up with egg on your face if your teacher tells you that the problem refers to a military plane flying in time of war.
Iainhol From , joined Dec 1969, posts, RR: Reply 3, posted Tue Aug 28 2001 19:55:09 UTC (14 years 9 months 3 days 14 hours ago) and read 2867 times:
Probably not the best thing to do, I used to do things like that is school, I corrected my English Teacher about flying over Disneyland at night, Bio teacher he cliaim gas was not a fluid (LOL!), and so on. All is does is make them grader tougher when it comes to your work! Iain
Steinwayartist From United States of America, joined Aug 2001, 66 posts, RR: 0 Reply 4, posted Tue Aug 28 2001 20:29:06 UTC (14 years 9 months 3 days 13 hours ago) and read 2844 times:
You don't have to bash me like that! Waahaaa lol, but yeah, you're right. I have a history of doodling. Long story short, I'm bored in the class. Anyway, I should've known that there's an exception for everything, and that posting stuff about planes here was bound to be challenged. :-p Oh well, let's go back to watching the US Open... -Nick
N863DA From United States of America, joined Sep 2004, 48 posts, RR: 5 Reply 5, posted Tue Aug 28 2001 22:50:00 UTC (14 years 9 months 3 days 11 hours ago) and read 2808 times:
Slightly off topic, but I had a final, last semester for Aviation Principles of Management at ERAU, and as a bonus question at the end of the exam, the question read something along the lines of: If a wide-body 767 jet, scheduled to depart Los Angeles at 14:45 and arrive Honolulu at 18:30, actually departs LAX at 17:00 with a headwind of 50kts over a distance of 2554 miles, a) what year was the aircraft built and b) what color(s) is/are the tail of the aircraft? The answer, of course, being the Delta geek-orama that I am, was a) 2000 and b) (semi-trick question) Red, Royal Blue & Light Blue. And I got the credit for it!!!! It made the difference between a B and an A for the test! For some things, it pays to be alert and answer the trick question literally - EG 'it's not legal' - as sometimes that's the answer to some of the questions I've had in my recent finals. But for regular non-aviation based math, just go with the problem as it was meant... FLY DELTA JETS and sail UNITED STATES LINES N 8 6 3 D A
David L From United Kingdom, joined May 1999, 9645 posts, RR: 42 Reply 6, posted Tue Aug 28 2001 22:52:01 UTC (14 years 9 months 3 days 11 hours ago) and read 2803 times:
Sorry. But you're right, you should have known
AA61hvy From United States of America, joined Nov 1999, 13977 posts, RR: 55 Reply 7, posted Wed Aug 29 2001 02:47:57 UTC (14 years 9 months 3 days 7 hours ago) and read 2782 times:
i hate when teachers do that... i had a problem on my algebra exam last year, and the final answer said it took 6 hours to fly 300 miles... stupid math teachers!!
Go big or go home
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Further 380 Delay Can Create Acute Problem For EK posted Sat May 3 2008 05:28:00 by Aviationbuff | 1,595 | 5,554 | {"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-2016-22 | latest | en | 0.905554 |
https://stats.stackexchange.com/questions/24023/eta2-and-paired-t-tests?noredirect=1 | 1,638,657,288,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00235.warc.gz | 607,402,669 | 34,166 | # $\eta^2$ and paired t-tests
When doing a paired t-test, is it appropriate to use the paired t value to calculate $\eta^2$?
I ask because it is not appropriate to calculate Cohen's d based on the paired t value.
Thanks.
$$d_{paired}=\frac{\bar{D}}{s_{\bar{D}}}$$
What you can't do is directly compare the paired version of Cohen's d with the Cohen's d for independent data. As is pointed out here, using Cohen's d is probably better than $\eta^2$. (That's a really good answer, by the way, it may be worth your time to read if you're interested in these issues.)
As soon as you entered the paired test arena the question arises, "what should I do with that pesky within observational unit (subjects etc) variance". Your effect size is bound to reflect some opinion in this regard. $d_{paired}$ as in gung's answer contains an assumption that the effect size should be calculated disregarding the variance in observational units. The equivalent assumption is made in a partial $\eta^2$ calculation. In short, there is no reason not to calculate $\eta^2$ or $d_{paired}$. Remember your paired samples-t is the same as a one-sample t-test (this is why you can use the one-sample t equation for Cohen's d as in gung's answer) or a 1 factor 2 level within subjects ANOVA. So, effect sizes which are appropriate for those statistics can also be appropriate for your paired samples t. | 331 | 1,381 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-49 | latest | en | 0.92977 |
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
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Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
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Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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• linear equation formulas for inequalities | 914 | 3,809 | {"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} | 2.71875 | 3 | CC-MAIN-2018-13 | latest | en | 0.91472 |
https://socratic.org/questions/how-do-you-solve-the-equation-32y-2-24y-0 | 1,576,148,390,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540543252.46/warc/CC-MAIN-20191212102302-20191212130302-00501.warc.gz | 537,064,114 | 6,060 | # How do you solve the equation: -32y^2-24y=0?
##### 1 Answer
Jun 28, 2015
The solutions are:
color(blue)(y=0, y=-3/4
#### Explanation:
$- 32 {y}^{2} - 24 y = 0$
$- 8 y$ is common to both terms:
$= - 8 y \left(4 y + 3\right) = 0$
Equating the terms with zero to obtain the solution:
• $- 8 y = 0$
color(blue)(y=0
• $4 y + 3 = 0$
color(blue)(y = -3/4 | 155 | 359 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 4.3125 | 4 | CC-MAIN-2019-51 | longest | en | 0.711554 |
http://mathhelpforum.com/trigonometry/101687-help-formula.html | 1,529,495,984,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00544.warc.gz | 193,382,923 | 9,702 | 1. ## Help with Formula
I have a formula that I need to solve for "h" and am having some difficulty.
The formula is:
3V=h([a+{2h/tan}]^2 + a^2 + a*[a+{2h/tan}])
Can anyone help?
Thanks!
2. Originally Posted by HARRENSTEIN
I have a formula that I need to solve for "h" and am having some difficulty.
The formula is:
3V=h([a+{2h/tan}]^2 + a^2 + a*[a+{2h/tan}])
Can anyone help?
Thanks!
We need a variable assigned to $\displaystyle \tan$ before we can proceed. $\displaystyle \tan{(?)}$
3. tan(theta) | 167 | 507 | {"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} | 3.28125 | 3 | CC-MAIN-2018-26 | latest | en | 0.918106 |
http://easy-ciphers.com/parries | 1,656,686,429,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00440.warc.gz | 21,310,142 | 19,690 | Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: parries
cipher variations: qbssjft rcttkgu sduulhv tevvmiw ufwwnjx vgxxoky whyyplz xizzqma yjaarnb zkbbsoc alcctpd bmdduqe cneevrf doffwsg epggxth fqhhyui griizvj hsjjawk itkkbxl jullcym kvmmdzn lwnneao mxoofbp nyppgcq ozqqhdr
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: parries Cipher: kziirvh
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
```a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
```
Plain: parries Cipher: ABBBA AAAAA BAAAA BAAAA ABAAA AABAA BAAAB
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: parries
cipher variations:
qbssjft
ubaaznd
ybiipvn
cbqqfdx
gbyyvlh
kbggltr
sbwwrjl
wbeehrv
abmmxzf
ebuunhp
ibccdpz
mbkktxj
rcttkgu
vcbbaoe
zcjjqwo
dcrrgey
hczzwmi
lchhmus
tcxxskm
xcffisw
bcnnyag
fcvvoiq
jcddeqa
nclluyk
sduulhv
wdccbpf
edsshfz
idaaxnj
mdiinvt
udyytln
ydggjtx
cdoozbh
gdwwpjr
kdeefrb
odmmvzl
tevvmiw
xeddcqg
bellsyq
fettiga
jebbyok
nejjowu
vezzumo
zehhkuy
deppaci
hexxqks
leffgsc
pennwam
ufwwnjx
yfeedrh
cfmmtzr
gfuujhb
kfcczpl
ofkkpxv
wfaavnp
afiilvz
efqqbdj
ifyyrlt
mfgghtd
qfooxbn
vgxxoky
zgffesi
dgnnuas
hgvvkic
lgddaqm
pgllqyw
xgbbwoq
bgjjmwa
fgrrcek
jgzzsmu
nghhiue
rgppyco
whyyplz
ahggftj
ehoovbt
ihwwljd
mheebrn
qhmmrzx
yhccxpr
chkknxb
ghssdfl
khaatnv
ohiijvf
shqqzdp
xizzqma
bihhguk
fippwcu
jixxmke
niffcso
rinnsay
ziddyqs
dilloyc
hittegm
libbuow
pijjkwg
tirraeq
yjaarnb
cjiihvl
gjqqxdv
kjyynlf
ojggdtp
sjootbz
ajeezrt
ejmmpzd
ijuufhn
mjccvpx
qjkklxh
ujssbfr
zkbbsoc
dkjjiwm
hkrryew
lkzzomg
pkhheuq
tkppuca
bkffasu
fknnqae
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rkllmyi
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elkkjxn
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mlaapnh
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klwwhjp
oleexrz
slmmnzj
wluudht
bmdduqe
fmllkyo
jmttagy
nmbbqoi
rmjjgws
vmrrwec
dmhhcuw
hmppscg
lmxxikq
pmffysa
tmnnoak
xmvveiu
cneevrf
gnmmlzp
knuubhz
onccrpj
snkkhxt
wnssxfd
eniidvx
inqqtdh
mnyyjlr
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unoopbl
ynwwfjv
doffwsg
honnmaq
lovvcia
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tolliyu
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rohhauc
voppqcm
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epggxth
ipoonbr
mpwwdjb
qpeetrl
upmmjzv
ypuuzhf
gpkkfxz
kpssvfj
opaalnt
spiibvd
wpqqrdn
apyyhlx
fqhhyui
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rqffusm
vqnnkaw
zqvvaig
hqllgya
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pqbbmou
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xqrrseo
bqzzimy
griizvj
krqqpdt
oryyfld
srggvtn
wroolbx
arwwbjh
irmmhzb
mruuxhl
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urkkdxf
craajnz
hsjjawk
lsrrqeu
pszzgme
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jsnniac
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vslleyg
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itkkbxl
mtssrfv
qtaahnf
utiixvp
ytqqndz
ctyydlj
ktoojbd
otwwzjn
steeprx
wtmmfzh
atuuvhr
etcclpb
jullcym
nuttsgw
rubbiog
vujjywq
zurroea
duzzemk
luppkce
puxxako
tuffqsy
xunngai
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fuddmqc
kvmmdzn
ovuuthx
svccjph
wvkkzxr
avsspfb
evaafnl
mvqqldf
qvyyblp
uvggrtz
yvoohbj
cvwwxjt
gveenrd
lwnneao
pwvvuiy
twddkqi
xwllays
bwttqgc
fwbbgom
nwrrmeg
rwzzcmq
vwhhsua
zwppick
dwxxyku
hwffose
mxoofbp
qxwwvjz
uxeelrj
yxmmbzt
cxuurhd
gxcchpn
oxssnfh
wxiitvb
axqqjdl
exyyzlv
ixggptf
nyppgcq
ryxxwka
vyffmsk
zynncau
dyvvsie
hyddiqo
pyttogi
tybbeos
xyjjuwc
byrrkem
fyzzamw
jyhhqug
ozqqhdr
szyyxlb
wzggntl
azoodbv
ezwwtjf
izeejrp
qzuuphj
uzccfpt
yzkkvxd
czsslfn
gzaabnx
kziirvh
parries
tazzymc
xahhoum
bappecw
faxxukg
jaffksq
ravvqik
zallwye
dattmgo
habbcoy
lajjswi
The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: parries Cipher: cneevrf
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: parries Cipher: 53112424425134
Extended Methods:
Method #1
Plaintext: parries
method variations: ufwwokx zlbbtpc eqggyuh kvmmdzn
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```p a r r i e s
5 1 2 2 4 5 3
3 1 4 4 2 1 4 ```
They are then read out in rows:
51224533144214
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: parries Cipher: egynqiq
Method #3
Plaintext: parries
method variations: cfitwly fitwlyc itwlycf twlycfi wlycfit lycfitw ycfitwl
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: parries
all 5040 cipher variations:
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Read more ...[1] , [2] , [3] | 23,705 | 49,501 | {"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-2022-27 | latest | en | 0.828345 |
https://nrich.maths.org/public/topic.php?code=32&cl=3&cldcmpid=5611 | 1,586,363,018,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00098.warc.gz | 589,674,625 | 8,994 | # Resources tagged with: Multiplication & division
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### There are 46 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### A Chance to Win?
##### Age 11 to 14 Challenge Level:
Imagine you were given the chance to win some money... and imagine you had nothing to lose...
### 3388
##### Age 11 to 14 Challenge Level:
Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
### What Is the Question?
##### Age 11 to 16 Challenge Level:
These pictures and answers leave the viewer with the problem "What is the Question". Can you give the question and how the answer follows?
### A First Product Sudoku
##### Age 11 to 14 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Galley Division
##### Age 14 to 16 Challenge Level:
Can you explain how Galley Division works?
### Oh! Hidden Inside?
##### Age 11 to 14 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Repeaters
##### Age 11 to 14 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Integrated Product Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.
##### Age 11 to 14 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### The Remainders Game
##### Age 7 to 14 Challenge Level:
Play this game and see if you can figure out the computer's chosen number.
### Multiples Sudoku
##### Age 11 to 14 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
##### Age 11 to 14 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
### Countdown
##### Age 7 to 14 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Powerful Factorial
##### Age 11 to 14 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Eminit
##### Age 11 to 14 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### One O Five
##### Age 11 to 14 Challenge Level:
You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .
##### Age 11 to 14 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Factoring Factorials
##### Age 11 to 14 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Vedic Sutra - All from 9 and Last from 10
##### Age 14 to 16 Challenge Level:
Vedic Sutra is one of many ancient Indian sutras which involves a cross subtraction method. Can you give a good explanation of WHY it works?
### Two Many
##### Age 11 to 14 Challenge Level:
What is the least square number which commences with six two's?
### Ones Only
##### Age 11 to 14 Challenge Level:
Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.
### Times Right
##### Age 11 to 16 Challenge Level:
Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?
### Like Powers
##### Age 11 to 14 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### Long Multiplication
##### Age 11 to 14 Challenge Level:
A 3 digit number is multiplied by a 2 digit number and the calculation is written out as shown with a digit in place of each of the *'s. Complete the whole multiplication sum.
### Missing Multipliers
##### Age 7 to 14 Challenge Level:
What is the smallest number of answers you need to reveal in order to work out the missing headers?
### Factorial
##### Age 14 to 16 Challenge Level:
How many zeros are there at the end of the number which is the product of first hundred positive integers?
### Skeleton
##### Age 11 to 14 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### Expenses
##### Age 14 to 16 Challenge Level:
What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?
### Remainders
##### Age 7 to 14 Challenge Level:
I'm thinking of a number. My number is both a multiple of 5 and a multiple of 6. What could my number be?
### Going Round in Circles
##### Age 11 to 14 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### As Easy as 1,2,3
##### Age 11 to 14 Challenge Level:
When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .
### A One in Seven Chance
##### Age 11 to 14 Challenge Level:
What is the remainder when 2^{164}is divided by 7?
### X Marks the Spot
##### Age 11 to 14 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Largest Number
##### Age 11 to 14 Challenge Level:
What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.
##### Age 7 to 14 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### More Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Thirty Six Exactly
##### Age 11 to 14 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Diggits
##### Age 11 to 14 Challenge Level:
Can you find what the last two digits of the number $4^{1999}$ are?
### Charitable Pennies
##### Age 7 to 14 Challenge Level:
Investigate the different ways that fifteen schools could have given money in a charity fundraiser.
### Number Rules - OK
##### Age 14 to 16 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### More Mods
##### Age 14 to 16 Challenge Level:
What is the units digit for the number 123^(456) ?
### Round and Round and Round
##### Age 11 to 14 Challenge Level:
Where will the point stop after it has turned through 30 000 degrees? I took out my calculator and typed 30 000 ÷ 360. How did this help?
### Slippy Numbers
##### Age 11 to 14 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### So It's Times!
##### Age 7 to 14 Challenge Level:
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
### 2010: A Year of Investigations
##### Age 5 to 14
This article for teachers suggests ideas for activities built around 10 and 2010. | 2,163 | 8,676 | {"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} | 4.21875 | 4 | CC-MAIN-2020-16 | latest | en | 0.84109 |
https://support.nag.com/numeric/nl/nagdoc_27cpp/flhtml/e04/e04rmf.html | 1,702,138,045,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00407.warc.gz | 596,911,462 | 7,122 | # NAG FL Interfacee04rmf (handle_set_nlnls)
## 1Purpose
e04rmf is a part of the NAG optimization modelling suite and defines the number of residuals in a sum of squares objective function (nonlinear least squares problems) and, optionally, the sparsity structure of their first derivatives.
## 2Specification
Fortran Interface
Subroutine e04rmf ( nres,
Integer, Intent (In) :: nres, isparse, nnzrd, irowrd(nnzrd), icolrd(nnzrd) Integer, Intent (Inout) :: ifail Type (c_ptr), Intent (In) :: handle
#include <nag.h>
void e04rmf_ (void **handle, const Integer *nres, const Integer *isparse, const Integer *nnzrd, const Integer irowrd[], const Integer icolrd[], Integer *ifail)
The routine may be called by the names e04rmf or nagf_opt_handle_set_nlnls.
## 3Description
After the initialization routine e04raf has been called and unless the objective function has already been defined, e04rmf may be used to declare the objective function of the optimization problem as a sum of squares. It will typically be used in data fitting or calibration problems of the form
$minimize x∈ℝn fx= ∑ j=1 mr rj x 2 subject to lx≤x≤ux ,$
where $x$ is an $n$-dimensional variable vector and ${r}_{i}\left(x\right)$ are nonlinear residuals (see Section 2.2.3 in the E04 Chapter Introduction). The values of the residuals, and possibly their derivatives, will be communicated to the solver by a user-supplied function. e04rmf also allows the structured first derivative matrix
$∂rjx ∂xi i=1,…,n , j=1,…,mr$
to be declared as being dense or sparse. If declared as sparse, its sparsity structure must be specified here.
See Section 3.1 in the E04 Chapter Introduction for more details about the NAG optimization modelling suite.
None.
## 5Arguments
1: $\mathbf{handle}$Type (c_ptr) Input
On entry: the handle to the problem. It needs to be initialized by e04raf and must not be changed before the call to e04rmf.
2: $\mathbf{nres}$Integer Input
On entry: ${m}_{r}$, the number of residuals in the objective function.
If ${\mathbf{nres}}=0$, no objective function will be defined and irowrd and icolrd will not be referenced.
Constraint: ${\mathbf{nres}}\ge 0$.
3: $\mathbf{isparse}$Integer Input
On entry: is a flag indicating if the nonzero structure of the first derivative matrix is dense or sparse.
${\mathbf{isparse}}=0$
The first derivative matrix is considered dense and irowrd and icolrd will not be referenced. The ordering is assumed to be column-wise, namely the routine will behave as if ${\mathbf{nnzrd}}=n×{m}_{r}$ and the vectors irowrd and icolrd filled as:
• ${\mathbf{irowrd}}=\left(1,2,\dots ,n,1,2,\dots ,n,\dots ,1,2,\dots ,n\right)$;
• ${\mathbf{icolrd}}=\left(1,1,\dots ,1,2,2,\dots ,2,\dots ,{m}_{r},{m}_{r},\dots ,{m}_{r}\right)$.
${\mathbf{isparse}}=1$
The sparsity structure of the first derivative matrix will be supplied by nnzrd, irowrd and icolrd.
Constraint: ${\mathbf{isparse}}=0$ or $1$.
4: $\mathbf{nnzrd}$Integer Input
On entry: the number of nonzeros in the first derivative matrix.
Constraint: if ${\mathbf{isparse}}=1$ and ${\mathbf{nres}}>0$, ${\mathbf{nnzrd}}>0$.
5: $\mathbf{irowrd}\left({\mathbf{nnzrd}}\right)$Integer array Input
6: $\mathbf{icolrd}\left({\mathbf{nnzrd}}\right)$Integer array Input
On entry: arrays irowrd and icolrd store the sparsity structure (pattern) of the first derivative matrix as nnzrd nonzeros in coordinate storage (CS) format (see Section 2.1.1 in the F11 Chapter Introduction). The matrix has dimensions $n×{m}_{r}$. irowrd specifies one-based row indices and icolrd specifies one-based column indices. No particular order of elements is expected, but elements should not repeat and the same order should be used when the first derivative matrix is evaluated for the solver.
Constraints:
• $1\le {\mathbf{irowrd}}\left(\mathit{l}\right)\le n$, for $\mathit{l}=1,2,\dots ,{\mathbf{nnzrd}}$;
• $1\le {\mathbf{icolrd}}\left(\mathit{l}\right)\le {\mathbf{nres}}$, for $\mathit{l}=1,2,\dots ,{\mathbf{nnzrd}}$.
7: $\mathbf{ifail}$Integer Input/Output
On entry: ifail must be set to $0$, $-1$ or $1$ to set behaviour on detection of an error; these values have no effect when no error is detected.
A value of $0$ causes the printing of an error message and program execution will be halted; otherwise program execution continues. A value of $-1$ means that an error message is printed while a value of $1$ means that it is not.
If halting is not appropriate, the value $-1$ or $1$ is recommended. If message printing is undesirable, then the value $1$ is recommended. Otherwise, the value $-1$ is recommended since useful values can be provided in some output arguments even when ${\mathbf{ifail}}\ne {\mathbf{0}}$ on exit. When the value $-\mathbf{1}$ or $\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
The supplied handle does not define a valid handle to the data structure for the NAG optimization modelling suite. It has not been initialized by e04raf or it has been corrupted.
${\mathbf{ifail}}=2$
The Hessians of nonlinear functions have already been defined, a nonlinear objective cannot be added.
The problem cannot be modified in this phase any more, the solver has already been called.
${\mathbf{ifail}}=3$
The objective function has already been defined.
${\mathbf{ifail}}=6$
On entry, ${\mathbf{isparse}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{isparse}}=0$ or $1$.
On entry, ${\mathbf{nnzrd}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{nnzrd}}>0$.
On entry, ${\mathbf{nres}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{nres}}\ge 0$.
${\mathbf{ifail}}=8$
On entry, $i=〈\mathit{\text{value}}〉$, ${\mathbf{icolrd}}\left(\mathit{i}\right)=〈\mathit{\text{value}}〉$ and ${\mathbf{nres}}=〈\mathit{\text{value}}〉$.
Constraint: $1\le {\mathbf{icolrd}}\left(\mathit{i}\right)\le {\mathbf{nres}}$.
On entry, $i=〈\mathit{\text{value}}〉$, ${\mathbf{irowrd}}\left(\mathit{i}\right)=〈\mathit{\text{value}}〉$ and $n=〈\mathit{\text{value}}〉$.
Constraint: $1\le {\mathbf{irowrd}}\left(\mathit{i}\right)\le n$.
On entry, more than one element of first derivative matrix has row index $〈\mathit{\text{value}}〉$ and column index $〈\mathit{\text{value}}〉$.
Constraint: each element of first derivative matrix must have a unique row and column index.
${\mathbf{ifail}}=-99$
See Section 7 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 9 in the Introduction to the NAG Library FL Interface for further information.
Not applicable.
## 8Parallelism and Performance
e04rmf is not threaded in any implementation.
None.
## 10Example
In this example, we demonstrate how to declare a least squares problem through e04rmf and solve it with e04fff on a very simple example. Here $n=2$, ${m}_{r}=3$ and the residuals are computed by:
$r1x = x1+ x2- 0.9 r2x = 2x1+ x2- 1.9 r3x = 3x1+ x2- 3.0$
The expected result is:
$x=0.95,0.10$
with an objective value of $0.015$.
### 10.1Program Text
Program Text (e04rmfe.f90)
None.
### 10.3Program Results
Program Results (e04rmfe.r) | 2,327 | 7,623 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 76, "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-2023-50 | latest | en | 0.757604 |
https://www.scribd.com/document/170567003/2006CM | 1,571,313,318,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986673538.21/warc/CC-MAIN-20191017095726-20191017123226-00398.warc.gz | 1,062,227,511 | 63,584 | You are on page 1of 3
# QUALIFYING EXAMINATION Classical Mechanics August 30, 2006 9 -11 am Three small beads, two of mass m and one
of mass 2m, are confined to move on a ring of radius R as shown. Each is connected to its nearest neighbors by stretched springs of spring constant k. When in equilibrium, they are symmetrically located about the ring. A. Find the frequencies of small oscillations. B. Describe qualitatively the motion that corresponds to = 0. (25 points)
1.
2.
If the polar ice caps were to melt, the level of the seas would rise, the increase in height being h 60m. Estimate the change that would ensue in the length of the day. Describe your simplifying assumptions.
(25 points)
## Qualifying Examination ______________________________________________________Classical Mechanics Aug. 30, 2006
3.
Consider an equilateral triangle ABC made of three metallic rods, each of mass m and length a. The triangle is suspended in a horizontal plane by three inextensible strings of negligible mass, each of length l. The strings are attached at each vertex of the triangle (points A, B, and C) and are connected to fixed points (A, B, and C) lying in a plane above the triangle. The plane A BC is always parallel to the plane of the triangle. At equilibrium, each string is vertical. Consider small rotational oscillations about the z-axis (shown on the diagram), which goes through the center of the triangle (marked O). The rotation angle and the direction of gravity (vector g) are shown in the figure. (a) Find the moment of inertia of the equilateral triangle about the vertical z-axis passing through the center of the triangle. You may use without proof the result that the moment of inertia of a rod about the vertical axis passing through the center of mass of the rod and perpendicular to the rod is ma2 /12. (b) Calculate the vertical displacement z of the triangle in the case when the rotation angle is changed from zero to . (c) For small angle rotations, show that the Lagrangian for this system can be written in & 2 , and find the constants, C , C . the form L = C1 + C 2 2 + C3 2 3 (d) Using Lagranges equation (or otherwise) find the frequency of small oscillations tr of the suspended triangle in terms of the constants C2 and C3. (25 points)
## Qualifying Examination ______________________________________________________Classical Mechanics Aug. 30, 2006
4.
A uniform rod of mass M and length L is held flat on a horizontal table and at right angles to an edge of the table. The center of mass C of the rod projects a distance d beyond the table edge at A. The Rod is released at rest. It starts to rotate about A and eventually slides off the table. The cocfficient of static friction is . The moment of inertia of a rod of length L and mass M about an axis perpendicular to the rod and passing through the center of mass is ML2 /12 (a) Calculate the angular velocity of the rod as a function of the rotation angle before sliding occurs. (b) The force exerted by the table edge on the rod has a component N in a direction perpendicular to the rod and the table edge. Calculate N as a function of before sliding occurs. (c) Calculate the angle 0 when sliding begins.
(25 points) | 731 | 3,223 | {"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-2019-43 | latest | en | 0.916679 |
https://mathoverflow.net/questions/250855/identity-involving-a-sum-over-all-partitions-of-n/250862#250862 | 1,708,514,305,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00795.warc.gz | 410,225,378 | 27,126 | # Identity involving a sum over all partitions of $n$
In some work I've been doing on the cohomology of the moduli space of curves, the following identity has come up:
$$\prod_{i=1}^n \frac{x^{i-1}}{x^i-1} = \sum_{(a_1^{r_1},\ldots,a_{\ell}^{r_{\ell}}) \vdash n} \left(\prod_{j=1}^{\ell} \frac{1}{a_i^{r_i} (x^{a_i}-1)^{r_i} (r_i)!}\right).$$
Here $x$ is a formal variable and the sum on the RHS is over all partitions of $n$. By $(a_1^{r_1},\ldots,a_{\ell}^{r_{\ell}}) \vdash n$, I mean a partition of the form
$$r_1 a_1 + r_2 a_2 + \cdots + r_{\ell} a_{\ell} = n$$
with $r_1,\ldots,r_{\ell} \geq 1$ and $a_1>a_2>\cdots>a_{\ell} \geq 1$.
I have verified this identity with Mathematica for $1 \leq n \leq 20$. However, I cannot figure out how to prove that it is always true. Can anyone help me?
This reminds me a little bit of the identity in this question, and I've tried without success to use the tools discussed in the answers to that question to solve it.
EDIT: In case anyone is interested, a version of this identity now appears as Lemma 5.2 in my paper "The high dimensional cohomology of the moduli space of curves with level structures" (joint w/ Neil Fullarton), which can be downloaded from my webpage here. Thanks to Lucia for telling me how to prove it!
• Nice! Food for thought: $n! / \prod_{j=1}^\ell\left(a_i^{r_i} r_i!\right)$ is the number of permutations in $S_n$ having cycle type $\left(a_1^{r_1},\ldots,a_\ell^{r_\ell}\right)$. Thus, the right hand side is probably better regarded as an average over $S_n$. Sep 27, 2016 at 18:10
• This follows if you use the interpretation in terms of cycle decompositions (as in that question) together with a simple combinatorial identity for partitions (see en.wikipedia.org/wiki/Q-Pochhammer_symbol and the section on combinatorial interpretation, which is pretty much your identity). Sep 27, 2016 at 18:22
• The RHS is the coefficient of $t^n$ in $$\exp (\sum_{n \ge 1} \frac{t^n}{n} (x^n-1)^{-1})$$ Sep 27, 2016 at 18:33
• @Lucia: I'm having trouble figuring out your argument. Can you give a few more details? I'm sorry for being slow -- I'm just a simple topologist, and this kind of combinatorics is far outside my comfort zone. Sep 27, 2016 at 18:37
• @AndyPutman: Hope the quick sketch below helps. Will look later if there are still any issues. Sep 27, 2016 at 18:47
Here's a quick sketch (since I'm pressed for time). Multiply both sides of the identity by $t^n$ and sum over $n$ from $0$ to infinity. From the cycle decomposition identity (Polya's formula) the right side becomes $$\exp \Big( \sum_{i=1}^{\infty} \frac{t^i}{i (x^i-1)} \Big)= \exp\Big( -\sum_{i=1}^{\infty} \frac{t^i}{i} \sum_{j=0}^{\infty} x^{ji} \Big) = \exp\Big( \sum_{j=0}^{\infty} \log (1-t x^j) \Big) = \prod_{j=0}^{\infty} (1-tx^j).$$ The RHS is also known as a Pochhammer symbol (see the Wikipedia article linked in my comment): it is $(t;x)_{\infty}$. The wikipedia article already describes the combinatorial identity (simple partition relation) $$(t;x)_{\infty} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n(n-1)/2} }{(x;x)_n} t^n,$$ where $$(x;x)_n = (1-x) (1-x^2) \cdots (1-x^n).$$ This matches what you get from multiplying your LHS by $t^n$ and summing.
• Also, for clarification: usually Polya's formula (cycle decomposition) is written as a sum over permutation $S_n$. From this, you get to the RHS in the above problem by converting to "cycle types" so that sum run though permutation $\lambda\vdash n$. Sep 28, 2016 at 0:16 | 1,148 | 3,485 | {"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} | 3.375 | 3 | CC-MAIN-2024-10 | latest | en | 0.892821 |
https://socratic.org/questions/5a54b0577c014956253c9a3b | 1,596,719,638,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736962.52/warc/CC-MAIN-20200806121241-20200806151241-00499.warc.gz | 514,129,297 | 6,120 | # Question #c9a3b
See Below.
#### Explanation:
$2 {m}^{2} + 2 m - 12 = 0$
$\Rightarrow 2 {m}^{2} + \left(6 - 4\right) m - 12 = 0$ [First factorise the left hand expression]
$\Rightarrow 2 {m}^{2} + 6 m - 4 m - 12 = 0$ [I choose 6 and 4 because $6 \cdot 4 = 2 \cdot 12$, and $6 - 4 = 2$]
$\Rightarrow 2 m \left(m + 3\right) - 4 \left(m + 3\right) = 0$ [Group the like terms]
$\Rightarrow \left(m + 3\right) \left(2 m - 4\right) = 0$ [Take the common part]
$\Rightarrow 2 \left(m + 3\right) \left(m - 2\right) = 0$
$\Rightarrow \left(m + 3\right) \left(m - 2\right) = 0$
If two numbers are multiplied and the answer is zero, then either of them must be zero or both of them.
So, Either, $m + 3 = 0$
$\Rightarrow m = - 3$
Or, $m - 2 = 0$
$\Rightarrow m = 2$
The solution of the quadratic equation is $m = - 3 , 2$ | 329 | 825 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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} | 4.5625 | 5 | CC-MAIN-2020-34 | latest | en | 0.543493 |
https://math.stackexchange.com/questions/1123099/estimating-a-convolution-type-maximal-function | 1,563,666,826,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526799.4/warc/CC-MAIN-20190720235054-20190721021054-00405.warc.gz | 467,584,070 | 36,172 | # estimating a convolution type maximal function
Let $\phi : \mathbb{R}^n \rightarrow \mathbb{R}_{+}$ be a $C^1$ function with $supp(\phi) \subset B(0,1)$ and $\int \phi = 1$. Define $$\phi_t(x) := t^{-n} \phi({x/t})$$ and set $$M_{\phi} f(x) := \sup_{t > 0} |f \ast \phi_t (x)|$$ for locally integrable $f$. Now take a function $a$ which is an $\infty$-atom, i.e. with support contained in some cube $Q$ of radius (half the length of a side) $r$ and center $x_0$ and with $||a||_{\infty} \leq \frac{1}{|Q|}$ where $|Q|$ denotes the Lebesgue measure of the cube. I need to show that there exists a constant $c$ depending only on $n$ and $\phi$ such that for all $x \not\in 2Q$ (the cube twice as big as $Q$ and with the same center) we have $$M_{\phi}a(x) \leq \frac{cr}{||x-x_0||^{n+1}}$$.
My attempt: previous point of the exercise shows that $M_{\phi}(f)$ can be bounded from above by the classical centred maximal function of Hardy Littlewood, hence I tried using that, but it didn't work out. If I use the maximal function, then using the fact the the $L^1$ norm of $a$ can be at most $1$ and taking $x$ far away from $x_0$ we can take a ball of radius $||x-x_0|| +r$, then the average over this ball will be something like $\frac{1}{||x-x_0||^{n}}$ which is much larger than the estimate given in the problem, at least for $x$ far away from $x_0$.
It sounds like you're not exploiting the zero mean of atoms and the compact support of $\phi$. First, observe that we can write the cube $Q$ as $Q=r[-1,1]^{n}+x_{0}$. I claim that it suffices to consider the case $r=1$ and $x_{0}=0$. Indeed, the maximal operator $M_{\phi}$ is translation invariant and dilation invariant (check this). So if we have proven the assertion for $\infty$-atoms supported in the cube $[-1,1]^{n}$, we can apply our result to $(\tau^{-x_{0}}a)_{1/r}$ to obtain \begin{align*} M_{\phi}(a)(x)=M_{\phi}(a)\left(r\dfrac{x-x_{0}}{r}+x_{0}\right)&=r^{-n}M_{\phi}(\tau^{-x_{0}}a)_{1/r}\left(\dfrac{x-x_{0}}{r}\right)\\ &\lesssim_{n,\phi}r^{-n}\left|\dfrac{x-x_{0}}{r}\right|^{-n-1}\\ &=\dfrac{r}{\left|x-x_{0}\right|^{n+1}} \end{align*} for all $\left|x\right|\notin 2Q$.
Using the hypothesis that $a$ has mean zero, we have the estimate $$\left|\int_{\mathbb{R}^{n}}a(y)\phi_{t}(x-y)dy\right|=\left|\int_{\mathbb{R}^{n}}a(y)\left[\phi_{t}(x-y)-\phi_{t}(x)\right]dy\right|\leq\int_{\mathbb{R}^{n}}\left|a(y)\right|\left|\phi_{t}(x-y)-\phi_{t}(x)\right|dy$$ Since $\phi$ is $C^{1}$ and compactly supported, we may use the mean value inequality to obtain the estimate for the RHS above $$\leq t^{-n}\int_{\mathbb{R}^{n}}\left|a(y)\right|\left\|\nabla\phi\right\|_{L^{\infty}}t^{-1}\left|y\right|dy$$ Since $\phi$ has compact support, we see that $\left|x-y\right|/t\leq C$, for some $C>0$. Since $a$ is supported in $Q$, the integrand vanishes for $y\notin Q$, and simple geometry show that $\left|x-y\right|\geq\left|x\right|/2$. Whence, for $x$ fixed, the integrand vanishes unless $$t\geq\dfrac{\left|x\right|}{2C}$$ Using this inequality, we conclude the estimate $$\left|(a\ast\phi_{t})(x)\right|\leq\dfrac{\left\|\nabla\phi\right\|_{L^{\infty}}(2C)^{n+1}}{\left|x\right|^{n+1}}\int_{\mathbb{R}^{n}}\left|a(y)\right|dy\leq\dfrac{\left\|\nabla\phi\right\|_{L^{\infty}}(2C)^{n+1}}{\left|x\right|^{n+1}},\quad\forall t>0$$ Taking the supremum of the LHS over all $t>0$, we conclude that $$M_{\phi}a(x)\leq\dfrac{\left\|\nabla\phi\right\|_{L^{\infty}}(2C)^{n+1}}{\left|x\right|^{n+1}}$$ | 1,284 | 3,462 | {"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} | 3.34375 | 3 | CC-MAIN-2019-30 | latest | en | 0.782271 |
https://discourse.pymc.io/t/effective-selection-of-prior-distributions/4314 | 1,656,619,690,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00714.warc.gz | 257,884,633 | 4,918 | # Effective selection of prior distributions
Hoping someone might be able to provide a “cheat sheet” for effective selection of prior distributions, e.g. a list of available distributions with accompanying sentence explaining when each should be used. Or, alternatively, recommend a good resource. I have some general questions but also a specific model I am trying to specify. General questions:
1. When should HalfNormal be chosen instead of HalfCauchy? Why would you choose one over the other?
2. How might one choose between Normal, Gamma, Exponential and Poission prior distributions?
3. I understand it is desirable for all random variables in the statistical model to have similar magnitude. If this is not possible for a particular parameter, modelling the logarithm of the parameter can be effective. Are there other ways of achieving this if using the logarithm is not suitable?
The model I am trying to specify is the following:
\begin{align} H(t) = \begin{cases} \frac{H_{ult}}{2} \big( 1- \exp(-B \: t^C) \big) & t < t_2 \\ \frac{H_{ult}}{2} \big( 1- \exp(-B \: t^C) \big) + \frac{H_{ult}}{2} \frac{t-t_2}{t-t_2+D} & t_2 \leq t \leq t_{end} \end{cases} \end{align}
I have data for H at four time values and wish to estimate the five parameters B, C, D, t_2, H_{ult}. I am using a continuous switch point for the change of curve as seen in Model with conditional parameters causing mass matrix to contain zeros on diagonal.
1. All parameters are positive values. Is it therefore effective to use HalfNormal or HalfCauchy prior distributions?
2. As t_2 lies somewhere between 0 and t_{end}, is it effective to use the Beta prior distribution with \alpha=1 and \beta=1 and scale it up by t_{end}?
3. Should C have a special prior distribution as it is an exponent?
I usually start with https://github.com/stan-dev/stan/wiki/Prior-Choice-Recommendations. And then you should follow https://arxiv.org/abs/1709.01449 to use prior predictive sampling to check your prior (and likelihood)
4 Likes
In addition to Junpeng’s recommendations, if you’ve got McElreath’s Rethinking, there is a whole section about the different kinds of priors (chapter on entropy and GLMs).
3 Likes | 557 | 2,191 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-27 | latest | en | 0.811019 |
https://maker.pro/forums/threads/favourite-circuits-for-cc-generator.24137/page-2 | 1,695,539,676,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00620.warc.gz | 417,825,444 | 62,060 | # Favourite circuits for CC generator ?
W
#### Winfield Hill
Jan 1, 1970
0
ChrisGibboGibson wrote...
Sorry, folks, I meant to say the TLV431, which controls its HI pin:
| +V o current sink out
| | |
| 100k ||-' 2N7000
| | ||<- 2N7002 (smt)
| +---||-+
| \__|__ | 1.24
| / \_____| I = ---- = 1.0 mA
| TLV431 /___\ | R R
| | 1.2k
| | |
| GND ---+------'
Yeah I'm still working on that and making no progress.
Easy. :>)
Use the LM385 here (it controls its LO pin), with a p-type mosfet.
| +V ---+------,
| | | R
| \__|__ 1.2k 1.24
| / \_____| I = ---- = 1.0 mA
| | |
| +---||-+ VP0106N
| | ||-> bss110
| 100k ||-, bss84 (smt)
| | |
| | o current source out
| ===
| GND
F
#### Fred Bartoli
Jan 1, 1970
0
Winfield Hill said:
ChrisGibboGibson wrote...
Sorry, folks, I meant to say the TLV431, which controls its HI pin:
| +V o current sink out
| | |
| 100k ||-' 2N7000
| | ||<- 2N7002 (smt)
| +---||-+
| \__|__ | 1.24
| / \_____| I = ---- = 1.0 mA
| TLV431 /___\ | R R
| | 1.2k
| | |
| GND ---+------'
Easy. :>)
Use the LM385 here (it controls its LO pin), with a p-type mosfet.
| +V ---+------,
| | | R
| \__|__ 1.2k 1.24
| / \_____| I = ---- = 1.0 mA
| | |
| +---||-+ VP0106N
| | ||-> bss110
| 100k ||-, bss84 (smt)
| | |
| | o current source out
| ===
| GND
Bummer! Guess what : I've just discovered I've always assumed the 385-adj
controled its top pin like the 431, and never bothered to check the DS, just
Still I win on transistor count and yours is vastly over engineered
C
#### ChrisGibboGibson
Jan 1, 1970
0
Winfield said:
ChrisGibboGibson wrote...
Sorry, folks, I meant to say the TLV431, which controls its HI pin:
| +V o current sink out
| | |
| 100k ||-' 2N7000
| | ||<- 2N7002 (smt)
| +---||-+
| \__|__ | 1.24
| / \_____| I = ---- = 1.0 mA
| TLV431 /___\ | R R
| | 1.2k
| | |
| GND ---+------'
Easy. :>)
Use the LM385 here (it controls its LO pin), with a p-type mosfet.
| +V ---+------,
| | | R
| \__|__ 1.2k 1.24
| / \_____| I = ---- = 1.0 mA
| | |
| +---||-+ VP0106N
| | ||-> bss110
| 100k ||-, bss84 (smt)
| | |
| | o current source out
| ===
| GND
Now that I do like.
Cheers
Gibbo
J
#### John Woodgate
Jan 1, 1970
0
I read in sci.electronics.design that Winfield Hill
Sorry, folks, I meant to say the TLV431, which controls its HI pin:
Do these arrangements produce a predictable source resistance or is it
just 'very, very high'?
F
#### Fred Bartoli
Jan 1, 1970
0
John Woodgate said:
I read in sci.electronics.design that Winfield Hill
Do these arrangements produce a predictable source resistance or is it
just 'very, very high'?
These source resistances are very, very, very, very very, very high.
Even the simple one with a single transistor is over 1G.
You still can render them predictable with and additional resistor
R
#### Rich Grise
Jan 1, 1970
0
Bummer! Guess what : I've just discovered I've always assumed the 385-adj
controled its top pin like the 431, and never bothered to check the DS,
Still I win on transistor count and yours is vastly over engineered
But I thought it was a transient response contest!
Cheers!
Rich
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J | 1,198 | 3,232 | {"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-2023-40 | longest | en | 0.745381 |
https://de.maplesoft.com/support/help/addons/view.aspx?path=DifferentialGeometry%2FLessonsAndTutorials%2FDifferentialGeometry%2FUtilities | 1,708,767,095,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474526.76/warc/CC-MAIN-20240224080616-20240224110616-00394.warc.gz | 211,055,493 | 41,221 | Lesson 3: Utilities - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
# Online Help
###### All Products Maple MapleSim
DifferentialGeometry Lessons
Lesson 3: Some DifferentialGeometry Utilities
Overview
DifferentialGeometry includes a powerful set of utilities for performing linear algebra computations on spaces of vectors, differential forms, and tensors (VFT).
In this lesson, you will learn to do the following:
– Extract the set of coefficients of a DifferentialGeometry VFT.
– Extract specific coefficients of a DifferentialGeometry VFT.
– Determine if a given VFT is a linear combination of a set of VFT.
– Create a spanning list of independent VFT (a basis) from a list of VFT.
– Generate a basis for the space of p-forms.
– Generate a collection of tensors.
– Extend a given set of independent VFT to a basis for a subspace.
– Find the dual basis for the cotangent space from a basis for the tangent space.
– Find the annihilator subspace for a given set of vectors or forms.
DGinfo/CoefficientSet, DGinfo/CoefficientList
The command DGinfo can be used to extract all or some of the coefficients of a vector, differential form or tensor. Exercises 11 illustrates the use of these commands in programming differential geometry applications.
> with(DifferentialGeometry): with(Tools):
> DGsetup([x, y ,z], "M");
${\mathrm{frame name: M}}$ (2.1)
Define a vector X = aD_x + bD_y + bD_z
M > X := evalDG(a*D_x + b*D_y + b*D_z);
${X}{:=}{a}{}{\mathrm{D_x}}{+}{b}{}{\mathrm{D_y}}{+}{b}{}{\mathrm{D_z}}$ (2.2)
Find the set of coefficients of X.
M > DGinfo(X, "CoefficientSet");
$\left\{{a}{,}{b}\right\}$ (2.3)
Find the list of all coefficients of X.
M > DGinfo(X, "CoefficientList", "all");
$\left[{a}{,}{b}{,}{b}\right]$ (2.4)
Find the coefficients of D_x and D_y in X (Method 1).
M > DGinfo(X, "CoefficientList", [[1], [2]]);
$\left[{a}{,}{b}\right]$ (2.5)
Find the coefficients of D_x and D_y in X (Method 2).
M > DGinfo(X, "CoefficientList", [D_x, D_y]);
$\left[{a}{,}{b}\right]$ (2.6)
Define a type (1,1) tensor T.
M > T := evalDG(a*dx &t D_x + b*dy &t D_z + c*dz &t D_x);
${T}{:=}{a}{}{\mathrm{dx}}{}{\mathrm{D_x}}{+}{b}{}{\mathrm{dy}}{}{\mathrm{D_z}}{+}{c}{}{\mathrm{dz}}{}{\mathrm{D_x}}$ (2.7)
Find the coefficient of dx D_x in T.
M > DGinfo(T, "CoefficientList", [[1, 1]]);
$\left[{a}\right]$ (2.8)
Find the coefficients of dy D_z and dz D_z in T.
M > DGinfo(alpha, "CoefficientList", [dy &tensor D_z, dz &tensor D_z]);
$\left[{\mathrm{α}}\right]$ (2.9)
GetComponents
The command GetComponents provides a very useful set of procedures for determining if a given VFT or list of VFT can be expressed as a linear combination of a set of VFT.
M > with(DifferentialGeometry): with(Tools):
M > DGsetup([x, y ,z], "M"):
Example 1. Express the vector X as a linear combination of the vectors in the list B. Check the result with DGzip.
M > X := evalDG(2*D_x + D_y - D_z);
${X}{:=}{2}{}{\mathrm{D_x}}{+}{\mathrm{D_y}}{-}{\mathrm{D_z}}$ (3.1)
M > B := evalDG([D_x - D_y, D_y - D_z, D_z + D_x]);
${B}{:=}\left[{\mathrm{D_x}}{-}{\mathrm{D_y}}{,}{\mathrm{D_y}}{-}{\mathrm{D_z}}{,}{\mathrm{D_x}}{+}{\mathrm{D_z}}\right]$ (3.2)
M > C := GetComponents(X, B);
${C}{:=}\left[{1}{,}{2}{,}{1}\right]$ (3.3)
M > DGzip(C, B, "plus");
${2}{}{\mathrm{D_x}}{+}{\mathrm{D_y}}{-}{\mathrm{D_z}}$ (3.4)
Example 2. GetComponents returns an empty list if the VFT is not a linear combination of the given list of VFT. For example, the 2-form alpha is not a linear combination of the 2-forms in the list S.
M > alpha := dy &wedge dz;
${\mathrm{α}}{:=}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}$ (3.5)
M > S := [dx &wedge dy, dx &wedge dz];
${S}{:=}\left[{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}\right]$ (3.6)
M > GetComponents(alpha, S);
$\left[{}\right]$ (3.7)
Example 3. The first argument to GetComponents can also be a list of VFTs, in which case the list of lists of components is returned. In this example, we find all the components C of all the vectors in the basis A as linear combinations of the vectors in the basis B. We find the change of basis Matrix P relating the two bases A and B.
M > A := evalDG([D_x + D_y + 2*D_z, D_y + D_z, 2*D_z]);
${A}{:=}\left[{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{2}{}{\mathrm{D_z}}{,}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{2}{}{\mathrm{D_z}}\right]$ (3.8)
M > B := evalDG([D_x - D_y, D_y - D_z, D_z + D_x]);
${B}{:=}\left[{\mathrm{D_x}}{-}{\mathrm{D_y}}{,}{\mathrm{D_y}}{-}{\mathrm{D_z}}{,}{\mathrm{D_x}}{+}{\mathrm{D_z}}\right]$ (3.9)
M > C := GetComponents(A, B);
${C}{:=}\left[\left[{-}{1}{,}{0}{,}{2}\right]{,}\left[{-}{1}{,}{0}{,}{1}\right]{,}\left[{-}{1}{,}{-}{1}{,}{1}\right]\right]$ (3.10)
M > P := LinearAlgebra[Transpose](Matrix(C));
Example 4. With the optional argument method = "real", the GetComponents command will determine if the linear combination can be found with real numbers as coefficients -- (no functions allowed). Compare the results of the following commands.
M > GetComponents(x &mult D_x, [D_x]);
$\left[{x}\right]$ (3.11)
M > GetComponents(x &mult D_x, [D_x], method = "real");
$\left[{}\right]$ (3.12)
M > GetComponents(x &mult D_x, [D_x, x &mult D_x, (x^2) &mult D_x], method = "real");
$\left[{0}{,}{1}{,}{0}\right]$ (3.13)
GenerateForms, GenerateTensors, GenerateSymmetricTensors
The utilities GenerateForms, GenerateTensors, GenerateSymmetricTensors are used to generate bases for different spaces of differential forms and tensors.
M > with(DifferentialGeometry): with(Tools): with(Tensor):
M > DGsetup([u, v, w, x, y], E5);
${\mathrm{frame name: E5}}$ (4.1)
Example 1. Define a list Omega of four 1-forms.
E5 > Omega := [du, dv, dw, dx];
${\mathrm{Ω}}{:=}\left[{\mathrm{du}}{,}{\mathrm{dv}}{,}{\mathrm{dw}}{,}{\mathrm{dx}}\right]$ (4.2)
Find a basis for the space of all 2-forms generated by Omega.
E5 > GenerateForms(Omega, 2);
$\left[{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dv}}{,}{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dw}}{,}{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dx}}{,}{\mathrm{dv}}{}{\bigwedge }{}{\mathrm{dw}}{,}{\mathrm{dv}}{}{\bigwedge }{}{\mathrm{dx}}{,}{\mathrm{dw}}{}{\bigwedge }{}{\mathrm{dx}}\right]$ (4.3)
Find a basis for the space of all 4-forms generated by Omega.
E5 > GenerateForms(Omega, 4);
$\left[{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dv}}{}{\bigwedge }{}{\mathrm{dw}}{}{\bigwedge }{}{\mathrm{dx}}\right]$ (4.4)
Example 2. Find a basis for the space of all rank 3 tensors whose first components are taken from the list S1, whose second components come from S2 and whose third components come from S3.
E5 > S1 := [D_x, D_y];
${\mathrm{S1}}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}\right]$ (4.5)
E5 > S2 := [du, dv];
${\mathrm{S2}}{:=}\left[{\mathrm{du}}{,}{\mathrm{dv}}\right]$ (4.6)
E5 > S3 := [dw, dx];
${\mathrm{S3}}{:=}\left[{\mathrm{dw}}{,}{\mathrm{dx}}\right]$ (4.7)
E5 > GenerateTensors([S1, S2, S3]);
$\left[{\mathrm{D_x}}{}{\mathrm{du}}{}{\mathrm{dw}}{,}{\mathrm{D_x}}{}{\mathrm{du}}{}{\mathrm{dx}}{,}{\mathrm{D_x}}{}{\mathrm{dv}}{}{\mathrm{dw}}{,}{\mathrm{D_x}}{}{\mathrm{dv}}{}{\mathrm{dx}}{,}{\mathrm{D_y}}{}{\mathrm{du}}{}{\mathrm{dw}}{,}{\mathrm{D_y}}{}{\mathrm{du}}{}{\mathrm{dx}}{,}{\mathrm{D_y}}{}{\mathrm{dv}}{}{\mathrm{dw}}{,}{\mathrm{D_y}}{}{\mathrm{dv}}{}{\mathrm{dx}}\right]$ (4.8)
Example 3. Find a basis for the space of all rank 2 symmetric tensors generated by Omega.
E5 > GenerateSymmetricTensors(Omega, 2);
$\left[{\mathrm{du}}{}{\mathrm{du}}{,}\frac{{1}}{{2}}{}{\mathrm{du}}{}{\mathrm{dv}}{+}\frac{{1}}{{2}}{}{\mathrm{dv}}{}{\mathrm{du}}{,}\frac{{1}}{{2}}{}{\mathrm{du}}{}{\mathrm{dw}}{+}\frac{{1}}{{2}}{}{\mathrm{dw}}{}{\mathrm{du}}{,}\frac{{1}}{{2}}{}{\mathrm{du}}{}{\mathrm{dx}}{+}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{du}}{,}{\mathrm{dv}}{}{\mathrm{dv}}{,}\frac{{1}}{{2}}{}{\mathrm{dv}}{}{\mathrm{dw}}{+}\frac{{1}}{{2}}{}{\mathrm{dw}}{}{\mathrm{dv}}{,}\frac{{1}}{{2}}{}{\mathrm{dv}}{}{\mathrm{dx}}{+}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dv}}{,}{\mathrm{dw}}{}{\mathrm{dw}}{,}\frac{{1}}{{2}}{}{\mathrm{dw}}{}{\mathrm{dx}}{+}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dw}}{,}{\mathrm{dx}}{}{\mathrm{dx}}\right]$ (4.9)
DGbasis
The utility DGbasis will select a maximal collection of independent VFT from a list of VFT. With method = "real", the program finds a maximal collection of independent VFT over the real numbers, otherwise independence is taken with respect to the ring of functions on the manifold.
E5 > with(DifferentialGeometry):
E5 > DGsetup([x, y, z], E3):
Example 1. The 2nd and 3rd vectors in S1 define a basis for the subspace spanned by the vectors in the list S1.
E3 > S1 := evalDG([D_x, 2*D_x, D_x - D_y, D_y]);
${\mathrm{S1}}{:=}\left[{\mathrm{D_x}}{,}{2}{}{\mathrm{D_x}}{,}{\mathrm{D_x}}{-}{\mathrm{D_y}}{,}{\mathrm{D_y}}\right]$ (5.1)
E3 > DGbasis(S1);
$\left[{\mathrm{D_x}}{,}{\mathrm{D_x}}{-}{\mathrm{D_y}}\right]$ (5.2)
Example 2. Over the ring of functions, the 2nd and 3rd vectors in the list S2 are linearly dependent on the 1st vector. Over the real numbers, the vectors in S2 are linearly independent.
E3 > S2 := evalDG([D_x, x*D_x, x^2*D_x]);
${\mathrm{S2}}{:=}\left[{\mathrm{D_x}}{,}{x}{}{\mathrm{D_x}}{,}{{x}}^{{2}}{}{\mathrm{D_x}}\right]$ (5.3)
E3 > DGbasis(S2);
$\left[{\mathrm{D_x}}\right]$ (5.4)
E3 > DGbasis(S2, method = "real");
$\left[{\mathrm{D_x}}{,}{x}{}{\mathrm{D_x}}{,}{{x}}^{{2}}{}{\mathrm{D_x}}\right]$ (5.5)
ComplementaryBasis
Let S1 be a subspace of S2 (a vector space of vectors, differential forms, or tensors) with basis B1 and B2 respectively. The command ComplementaryBasis calculates a complementary subspace C to S1 in S2, that is, S1 + C = S2 (direct sum). The vectors defining the basis for C are chosen from B2.
E3 > with(DifferentialGeometry):
E3 > DGsetup([x, y, z, w], E4):
Example 1. To extend the set of basis vectors B1 to a basis for the span of B2, we need to include the vector D_w.
E4 > B1 := evalDG([D_x, D_y]);
${\mathrm{B1}}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}\right]$ (6.1)
E4 > B2 := [D_x, D_y, D_w];
${\mathrm{B2}}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{\mathrm{D_w}}\right]$ (6.2)
E4 > ComplementaryBasis(B1, B2);
$\left[{\mathrm{D_w}}\right]$ (6.3)
Example 2. We calculate a complement to the subspace of 2-forms spanned by B1 in the space of all 2-forms on E4.
E4 > B1 := [dx &wedge dy, dx &wedge dz, dx &wedge dw];
${\mathrm{B1}}{:=}\left[{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dw}}\right]$ (6.4)
E4 > Omega := Tools:-DGinfo("FrameBaseForms");
${\mathrm{Ω}}{:=}\left[{\mathrm{dx}}{,}{\mathrm{dy}}{,}{\mathrm{dz}}{,}{\mathrm{dw}}\right]$ (6.5)
E4 > B2 := Tools:-GenerateForms(Omega, 2);
${\mathrm{B2}}{:=}\left[{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dw}}{,}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}{,}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dw}}{,}{\mathrm{dz}}{}{\bigwedge }{}{\mathrm{dw}}\right]$ (6.6)
E4 > ComplementaryBasis(B1, B2);
$\left[{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}{,}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dw}}{,}{\mathrm{dz}}{}{\bigwedge }{}{\mathrm{dw}}\right]$ (6.7)
DualBasis
If X[i], i = 1 ... n is a basis for a vector space V, then the dual basis for the dual space V^* consists of the 1-forms theta[j], j = 1 ... n, satisfying theta[j](X[i]) = delta[i, j], where delta[i, j] is the Kronecker delta.
E4 > with(DifferentialGeometry):
E4 > DGsetup([x, y, z], E3):
Example 1. The vectors in the list B define a basis for the tangent space to E3 at each point.
E3 > B := evalDG([D_x + D_y + D_z, D_y + D_z, D_z]);
${B}{:=}\left[{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{\mathrm{D_z}}\right]$ (7.1)
Calculate the basis Theta for the cotangent space which is dual to B.
E3 > Theta := DualBasis(B);
${\mathrm{Θ}}{:=}\left[{\mathrm{dx}}{,}{-}{\mathrm{dx}}{+}{\mathrm{dy}}{,}{-}{\mathrm{dy}}{+}{\mathrm{dz}}\right]$ (7.2)
Let us check this result using the procedure Hook.
E3 > Matrix(3, 3, (i, j) -> Hook(B[i], Theta[j]));
Example 2. The command DualBasis can also be applied to a list of 1-forms. In this example, the dual basis to the 1-forms Theta are the vectors B we started with in Example 1.
E3 > DualBasis(Theta);
$\left[{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{\mathrm{D_z}}\right]$ (7.3)
Annihilator
Let A be a set of vectors in a vector space V. Then the annihilating space B is the subspace of covectors alpha in the dual space V^* such that alpha(X) = 0 for all X in A. The dimension of B is dim(B) = dim(V) - dim(span(A)).
E3 > with(DifferentialGeometry):
E3 > DGsetup([x, y, z, u, v], "M"):
Example 1. Find the annihilating subspace B to the set of vectors A and check the result.
M > A := [D_x, D_y];
${A}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}\right]$ (8.1)
M > B := Annihilator(A);
${B}{:=}\left[{\mathrm{dv}}{,}{\mathrm{du}}{,}{\mathrm{dz}}\right]$ (8.2)
M > Matrix(2, 3, (i, j) -> Hook(A[i], B[j]));
Example 2. Find the annihilating subspace B to the set of 1-forms A.
M > A := evalDG([dx + dy, dy - du]);
${A}{:=}\left[{\mathrm{dx}}{+}{\mathrm{dy}}{,}{\mathrm{dy}}{-}{\mathrm{du}}\right]$ (8.3)
M > B := Annihilator(A);
${B}{:=}\left[{\mathrm{D_v}}{,}{-}{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{\mathrm{D_u}}{,}{\mathrm{D_z}}\right]$ (8.4)
M > Matrix(3, 2, (i, j) -> Hook(B[i], A[j]));
Exercises
Exercise 1
For what values of the parameter a are the differential 1-forms in the set S linearly independent?
Hint: A list of 1-forms are independent if and only if the their wedge product is non-zero.
M > with(DifferentialGeometry):
M > DGsetup([x, y, z, w], "M"):
M > S := evalDG([dx + a*dy, dx - a*dy + dz, dx + dy - a*dz]);
${S}{:=}\left[{\mathrm{dx}}{+}{a}{}{\mathrm{dy}}{,}{\mathrm{dx}}{-}{a}{}{\mathrm{dy}}{+}{\mathrm{dz}}{,}{\mathrm{dx}}{+}{\mathrm{dy}}{-}{a}{}{\mathrm{dz}}\right]$ (9.1.1)
Solution
Compute the wedge product of the forms in S (Method 1).
M > Omega := S[1] &wedge S[2] &wedge S[3];
${\mathrm{Ω}}{:=}\left({a}{-}{1}{+}{2}{}{{a}}^{{2}}\right){}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}$ (9.1.1.1)
Compute the wedge product of the forms in S (Method 2).
M > Omega := DGzip(S, "wedge");
${\mathrm{Ω}}{:=}\left({a}{-}{1}{+}{2}{}{{a}}^{{2}}\right){}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}$ (9.1.1.2)
Get the coefficient of Omega, set it to zero, and solve for a.
M > Eq := Tools:-DGinfo(Omega, "CoefficientSet");
${\mathrm{Eq}}{:=}\left\{{a}{-}{1}{+}{2}{}{{a}}^{{2}}\right\}$ (9.1.1.3)
M > solve(Eq, {a});
$\left\{{a}{=}\frac{{1}}{{2}}\right\}{,}\left\{{a}{=}{-}{1}\right\}$ (9.1.1.4)
The 3 1-forms in the set S are linearly independent except when a = 1/2 and a = -1.
Exercise 2
Write a program ChangeCoefficient that will change the value of a prescribed coefficient in a vector, differential form, or tensor. Take as input for the program a VFT, a "monomial" VFT representing the term in the VFT that is to be changed, and the new value of the coefficient. Use your program to change the coefficient of dx^dz in Omega from 2 to K.
M > with(DifferentialGeometry):
M > DGsetup([x, y, z, w], "M"):
M > Omega := evalDG(dx &w dy + 2*dx &w dz + 3*dy &w dz);
${\mathrm{Ω}}{:=}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{+}{2}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}{+}{3}{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}$ (9.2.1)
Solution
M > ChangeCoefficient := proc(X, T, a)
M > local b;
M > b := DifferentialGeometry:-Tools:-DGinfo(X, "CoefficientList", [T])[1];
M > X &plus ((a - b) &mult T);
M > end:
Example 1. Change the coefficient of dx^dz in Omega from 2 to K.
M > Omega := evalDG(dx &w dy + 2*dx &w dz + 3*dy &w dz);
M > ChangeCoefficient(Omega, dx &wedge dz, K);
${\mathrm{Ω}}{:=}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{+}{2}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}{+}{3}{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}$
${\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{+}{K}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}{+}{3}{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}$ (9.2.1.1)
Example 2. Change the coefficient of dx dx D_y in T from 1 to 4.
M > T := evalDG(dx &t dy &t D_z + 2*dx &t dz &t D_x + 3*dy &t dz &t D_y);
${T}{:=}{\mathrm{dx}}{}{\mathrm{dy}}{}{\mathrm{D_z}}{+}{2}{}{\mathrm{dx}}{}{\mathrm{dz}}{}{\mathrm{D_x}}{+}{3}{}{\mathrm{dy}}{}{\mathrm{dz}}{}{\mathrm{D_y}}$ (9.2.1.2)
M > ChangeCoefficient(T, dx &t dx &t D_y, 4);
${4}{}{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{D_y}}{+}{\mathrm{dx}}{}{\mathrm{dy}}{}{\mathrm{D_z}}{+}{2}{}{\mathrm{dx}}{}{\mathrm{dz}}{}{\mathrm{D_x}}{+}{3}{}{\mathrm{dy}}{}{\mathrm{dz}}{}{\mathrm{D_y}}$ (9.2.1.3)
Exercise 3
M > with(DifferentialGeometry):
M > DGsetup([x, y, z, w], "M"):
[i] Express the vector field X as a linear combination of the vector fields B1. Check your result using DGzip.
M > X := evalDG(D_x + 3*D_y - D_z);
${X}{:=}{\mathrm{D_x}}{+}{3}{}{\mathrm{D_y}}{-}{\mathrm{D_z}}$ (9.3.1)
M > B1 := evalDG([D_x - 2*D_y, D_y + D_z, D_x - D_y + 2*D_z]);
${\mathrm{B1}}{:=}\left[{\mathrm{D_x}}{-}{2}{}{\mathrm{D_y}}{,}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{\mathrm{D_x}}{-}{\mathrm{D_y}}{+}{2}{}{\mathrm{D_z}}\right]$ (9.3.2)
[ii] Express the differential 2-form alpha as a linear combination of the 2-forms B2. Check your result using DGzip.
M > alpha := evalDG(2*dx &w dy);
${\mathrm{α}}{:=}{2}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}$ (9.3.3)
M > B2 := evalDG([dx &w dy + dz &w dw, dx &w dz + dy &w dw, dx &w dw + dy &w dz, dx &w dy + dx &w dz, dx &w dz + dy &w dz, dy &w dz + dy &w dw]);
${\mathrm{B2}}{:=}\left[{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{+}{\mathrm{dz}}{}{\bigwedge }{}{\mathrm{dw}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}{+}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dw}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dw}}{+}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{+}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}{+}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}{,}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}{+}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dw}}\right]$ (9.3.4)
[iii] Express the rank 3 tensor T as a linear combination of the tensors B3. Check your result using DGzip.
M > T := evalDG(dx &t D_y &t D_z - dx &t D_z &t D_x + 2*dy &t D_z &t D_z - 2*dz &t D_z &t D_y);
${T}{:=}{\mathrm{dx}}{}{\mathrm{D_y}}{}{\mathrm{D_z}}{-}{\mathrm{dx}}{}{\mathrm{D_z}}{}{\mathrm{D_x}}{+}{2}{}{\mathrm{dy}}{}{\mathrm{D_z}}{}{\mathrm{D_z}}{-}{2}{}{\mathrm{dz}}{}{\mathrm{D_z}}{}{\mathrm{D_y}}$ (9.3.5)
M > B3 := evalDG([dx &t D_y &t D_z - dz &t D_z &t D_y, dx &t D_z &t D_x + dz &t D_z &t D_y, dy &t D_z &t D_z]);
${\mathrm{B3}}{:=}\left[{\mathrm{dx}}{}{\mathrm{D_y}}{}{\mathrm{D_z}}{-}{\mathrm{dz}}{}{\mathrm{D_z}}{}{\mathrm{D_y}}{,}{\mathrm{dx}}{}{\mathrm{D_z}}{}{\mathrm{D_x}}{+}{\mathrm{dz}}{}{\mathrm{D_z}}{}{\mathrm{D_y}}{,}{\mathrm{dy}}{}{\mathrm{D_z}}{}{\mathrm{D_z}}\right]$ (9.3.6)
Solution
[i] Express the vector field X as a linear combination of the vector fields B1. Check your result using DGzip.
M > C1 := GetComponents(X, B1);
${\mathrm{C1}}{:=}\left[{7}{,}{11}{,}{-}{6}\right]$ (9.3.1.1)
M > X &minus DGzip(C1, B1, "plus");
${0}{}{\mathrm{D_x}}$ (9.3.1.2)
[ii] Express the differential 2-form alpha as a linear combination of the 2-forms B2. Check your result using DGzip.
M > C2 := GetComponents(alpha, B2);
${\mathrm{C2}}{:=}\left[{0}{,}{-}{1}{,}{0}{,}{2}{,}{-}{1}{,}{1}\right]$ (9.3.1.3)
M > alpha &minus DGzip(C2, B2, "plus");
${0}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}$ (9.3.1.4)
[iii] Express the rank 3 tensor T as a linear combination of the tensors B3. Check your result using DGzip.
M > C3 := GetComponents(T, B3);
${\mathrm{C3}}{:=}\left[{1}{,}{-}{1}{,}{2}\right]$ (9.3.1.5)
M > T &minus DGzip(C3, B3, "plus");
${0}{}{\mathrm{dx}}{}{\mathrm{D_x}}{}{\mathrm{D_x}}$ (9.3.1.6)
Exercise 4
Show that the vector X is a linear combination of the vectors with variable coefficients but not with constant coefficients.
M > with(DifferentialGeometry):
M > DGsetup([x, y, z], "M"):
M > X := D_z;
${X}{:=}{\mathrm{D_z}}$ (9.4.1)
M > B := evalDG([x*D_x + y*D_y + z*D_z, x*D_y - y*D_x, x*D_z - z*D_x, z*D_y - y*D_z]);
${B}{:=}\left[{x}{}{\mathrm{D_x}}{+}{y}{}{\mathrm{D_y}}{+}{z}{}{\mathrm{D_z}}{,}{-}{y}{}{\mathrm{D_x}}{+}{x}{}{\mathrm{D_y}}{,}{-}{z}{}{\mathrm{D_x}}{+}{x}{}{\mathrm{D_z}}{,}{z}{}{\mathrm{D_y}}{-}{y}{}{\mathrm{D_z}}\right]$ (9.4.2)
Solution
M > GetComponents(X, B);
$\left[\frac{{z}}{{{x}}^{{2}}{+}{{y}}^{{2}}{+}{{z}}^{{2}}}{,}{-}\frac{\left({{\mathrm{_t15}}}_{{1}{,}{1}}{}{{x}}^{{2}}{+}{{y}}^{{2}}{}{{\mathrm{_t15}}}_{{1}{,}{1}}{+}{y}{+}{{\mathrm{_t15}}}_{{1}{,}{1}}{}{{z}}^{{2}}\right){}{z}}{{x}{}\left({{x}}^{{2}}{+}{{y}}^{{2}}{+}{{z}}^{{2}}\right)}{,}\frac{{{x}}^{{2}}{+}{{y}}^{{2}}{+}{y}{}{{\mathrm{_t15}}}_{{1}{,}{1}}{}{{x}}^{{2}}{+}{{y}}^{{3}}{}{{\mathrm{_t15}}}_{{1}{,}{1}}{+}{y}{}{{\mathrm{_t15}}}_{{1}{,}{1}}{}{{z}}^{{2}}}{{x}{}\left({{x}}^{{2}}{+}{{y}}^{{2}}{+}{{z}}^{{2}}\right)}{,}{{\mathrm{_t15}}}_{{1}{,}{1}}\right]$ (9.4.1.1)
The presence of the parameter _t[1, 1] indicates that the vectors in B are not linearly independent and that X can be expressed in terms of the vectors in B in infinitely many ways.
To see if X is a constant coefficient linear combination of the vectors in B, use GetComponents with the option method = "real".
M > GetComponents(X, B, method = "real");
$\left[{}\right]$ (9.4.1.2)
Exercise 5
Extend the set of 1 forms S to a basis for the cotangent space of M.
M > with(DifferentialGeometry):
M > DGsetup([x, y, z, w], "M"):
M > S := evalDG([dx - dy + dz, dy + dw]);
${S}{:=}\left[{\mathrm{dx}}{-}{\mathrm{dy}}{+}{\mathrm{dz}}{,}{\mathrm{dy}}{+}{\mathrm{dw}}\right]$ (9.5.1)
Solution
Use the ComplementaryBasis command with the standard basis for the cotangent space as the second argument.
M > C := ComplementaryBasis(S, [dx, dy, dz, dw]);
${C}{:=}\left[{\mathrm{dx}}{,}{\mathrm{dy}}\right]$ (9.5.1.1)
The 1-forms S and C together form a basis for the cotangent space of M.
Exercise 6
Find a maximally independent set of VFT from each of the following lists of VFT.
M > with(DifferentialGeometry):
M > DGsetup([x, y, z, w], "M"):
[i]
M > B1 := evalDG([ D_x, D_y, D_z, D_z, D_y, D_w]);
${\mathrm{B1}}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{\mathrm{D_z}}{,}{\mathrm{D_z}}{,}{\mathrm{D_y}}{,}{\mathrm{D_w}}\right]$ (9.6.1)
[ii]
M > B2 := evalDG([D_x &t dx, D_y &t dy, D_x &t dx + D_y &t dy, D_y &t dx]);
${\mathrm{B2}}{:=}\left[{\mathrm{D_x}}{}{\mathrm{dx}}{,}{\mathrm{D_y}}{}{\mathrm{dy}}{,}{\mathrm{D_x}}{}{\mathrm{dx}}{+}{\mathrm{D_y}}{}{\mathrm{dy}}{,}{\mathrm{D_y}}{}{\mathrm{dx}}\right]$ (9.6.2)
[iii] Most of the linear algebra utilities in DifferentialGeometry also work with Vectors and Matrices.
M > B3 := [Matrix([[1, 0], [0, 1]]), Matrix([[1, 2], [0, 1]]), Matrix([[1, 1], [0, 1]]), Matrix([[1, -1], [0, -1]])];
Solution
Part [i]
M > B1, DGbasis(B1);
$\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{\mathrm{D_z}}{,}{\mathrm{D_z}}{,}{\mathrm{D_y}}{,}{\mathrm{D_w}}\right]{,}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{\mathrm{D_z}}{,}{\mathrm{D_w}}\right]$ (9.6.1.1)
Part[ii]
M > B2, DGbasis(B2);
$\left[{\mathrm{D_x}}{}{\mathrm{dx}}{,}{\mathrm{D_y}}{}{\mathrm{dy}}{,}{\mathrm{D_x}}{}{\mathrm{dx}}{+}{\mathrm{D_y}}{}{\mathrm{dy}}{,}{\mathrm{D_y}}{}{\mathrm{dx}}\right]{,}\left[{\mathrm{D_x}}{}{\mathrm{dx}}{,}{\mathrm{D_y}}{}{\mathrm{dy}}{,}{\mathrm{D_y}}{}{\mathrm{dx}}\right]$ (9.6.1.2)
Part[iii]
M > B3, DGbasis(B3);
Exercise 7
Show that the vectors in the list B are linearly independent over the real numbers but not over the ring of real-valued functions on M.
M > with(DifferentialGeometry):
M > DGsetup([x, y], "M"):
M > B := evalDG([D_x, D_y, x*D_x, y*D_x, x*D_y, y*D_y]);
${B}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{x}{}{\mathrm{D_x}}{,}{y}{}{\mathrm{D_x}}{,}{x}{}{\mathrm{D_y}}{,}{y}{}{\mathrm{D_y}}\right]$ (9.7.1)
Solution
Note that there are 6 vectors in the list B.
M > nops(B);
${6}$ (9.7.1.1)
We apply the DGbasis procedure with the option method = "real".
M > C1 := DGbasis(B, method = "real");
${\mathrm{C1}}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{x}{}{\mathrm{D_x}}{,}{y}{}{\mathrm{D_x}}{,}{x}{}{\mathrm{D_y}}{,}{y}{}{\mathrm{D_y}}\right]$ (9.7.1.2)
M > nops(C1);
${6}$ (9.7.1.3)
Since the number of vectors in B equals the number of vectors in C1, the vectors in B must be independent over the real numbers.
However, over the ring of functions, there are only two independent vectors.
M > C2 := DGbasis(B);
${\mathrm{C2}}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}\right]$ (9.7.1.4)
Exercise 8
a) Generate a basis of all 3-forms for the manifold N with coordinates z1, z2, z3, z4, z5.
b) Generate all symmetric rank 4 contravariant tensors for the manifold P with coordinates u, v.
c) Generate all type (2, 1) tensors (covariant rank 2, contravariant rank 1) which are skew-symmetric in the first two arguments for the manifold Q with coordinates x, y, z.
Solution
a)
M > with(DifferentialGeometry): with(Tools): DGsetup([z1, z2, z3, z4, z5], "N"):
N > Omega := DGinfo("FrameBaseForms");
${\mathrm{Ω}}{:=}\left[{\mathrm{dz1}}{,}{\mathrm{dz2}}{,}{\mathrm{dz3}}{,}{\mathrm{dz4}}{,}{\mathrm{dz5}}\right]$ (9.8.1.1)
N > GenerateForms(Omega, 3);
$\left[{\mathrm{dz1}}{}{\bigwedge }{}{\mathrm{dz2}}{}{\bigwedge }{}{\mathrm{dz3}}{,}{\mathrm{dz1}}{}{\bigwedge }{}{\mathrm{dz2}}{}{\bigwedge }{}{\mathrm{dz4}}{,}{\mathrm{dz1}}{}{\bigwedge }{}{\mathrm{dz2}}{}{\bigwedge }{}{\mathrm{dz5}}{,}{\mathrm{dz1}}{}{\bigwedge }{}{\mathrm{dz3}}{}{\bigwedge }{}{\mathrm{dz4}}{,}{\mathrm{dz1}}{}{\bigwedge }{}{\mathrm{dz3}}{}{\bigwedge }{}{\mathrm{dz5}}{,}{\mathrm{dz1}}{}{\bigwedge }{}{\mathrm{dz4}}{}{\bigwedge }{}{\mathrm{dz5}}{,}{\mathrm{dz2}}{}{\bigwedge }{}{\mathrm{dz3}}{}{\bigwedge }{}{\mathrm{dz4}}{,}{\mathrm{dz2}}{}{\bigwedge }{}{\mathrm{dz3}}{}{\bigwedge }{}{\mathrm{dz5}}{,}{\mathrm{dz2}}{}{\bigwedge }{}{\mathrm{dz4}}{}{\bigwedge }{}{\mathrm{dz5}}{,}{\mathrm{dz3}}{}{\bigwedge }{}{\mathrm{dz4}}{}{\bigwedge }{}{\mathrm{dz5}}\right]$ (9.8.1.2)
b)
N > with(DifferentialGeometry): with(Tensor): DGsetup([u, v], "P"):
P > GenerateSymmetricTensors([D_u, D_v], 4);
$\left[{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{,}\frac{{1}}{{4}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{+}\frac{{1}}{{4}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{+}\frac{{1}}{{4}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{+}\frac{{1}}{{4}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{,}\frac{{1}}{{6}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{+}\frac{{1}}{{6}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{+}\frac{{1}}{{6}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{+}\frac{{1}}{{6}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{+}\frac{{1}}{{6}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{+}\frac{{1}}{{6}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{,}\frac{{1}}{{4}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{+}\frac{{1}}{{4}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{+}\frac{{1}}{{4}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{}{\mathrm{D_v}}{+}\frac{{1}}{{4}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{}{\mathrm{D_u}}{,}{\mathrm{D_v}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}{}{\mathrm{D_v}}\right]$ (9.8.1.3)
c)
P > with(DifferentialGeometry): with(Tensor): DGsetup([x, y, z], "Q"):
P > T1 := GenerateSymmetricTensors([dx, dy, dz], 2);
${\mathrm{T1}}{:=}\left[{\mathrm{dx}}{}{\mathrm{dx}}{,}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dy}}{+}\frac{{1}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dx}}{,}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dz}}{+}\frac{{1}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dx}}{,}{\mathrm{dy}}{}{\mathrm{dy}}{,}\frac{{1}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dz}}{+}\frac{{1}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dy}}{,}{\mathrm{dz}}{}{\mathrm{dz}}\right]$ (9.8.1.4)
Q > T2 := [D_x, D_y, D_z];
${\mathrm{T2}}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{\mathrm{D_z}}\right]$ (9.8.1.5)
Q > GenerateTensors([T1, T2]);
$\left[{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{D_x}}{,}{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{D_y}}{,}{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{D_z}}{,}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dy}}{}{\mathrm{D_x}}{+}\frac{{1}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dx}}{}{\mathrm{D_x}}{,}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dy}}{}{\mathrm{D_y}}{+}\frac{{1}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dx}}{}{\mathrm{D_y}}{,}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dy}}{}{\mathrm{D_z}}{+}\frac{{1}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dx}}{}{\mathrm{D_z}}{,}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dz}}{}{\mathrm{D_x}}{+}\frac{{1}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dx}}{}{\mathrm{D_x}}{,}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dz}}{}{\mathrm{D_y}}{+}\frac{{1}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dx}}{}{\mathrm{D_y}}{,}\frac{{1}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dz}}{}{\mathrm{D_z}}{+}\frac{{1}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dx}}{}{\mathrm{D_z}}{,}{\mathrm{dy}}{}{\mathrm{dy}}{}{\mathrm{D_x}}{,}{\mathrm{dy}}{}{\mathrm{dy}}{}{\mathrm{D_y}}{,}{\mathrm{dy}}{}{\mathrm{dy}}{}{\mathrm{D_z}}{,}\frac{{1}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dz}}{}{\mathrm{D_x}}{+}\frac{{1}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dy}}{}{\mathrm{D_x}}{,}\frac{{1}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dz}}{}{\mathrm{D_y}}{+}\frac{{1}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dy}}{}{\mathrm{D_y}}{,}\frac{{1}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dz}}{}{\mathrm{D_z}}{+}\frac{{1}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dy}}{}{\mathrm{D_z}}{,}{\mathrm{dz}}{}{\mathrm{dz}}{}{\mathrm{D_x}}{,}{\mathrm{dz}}{}{\mathrm{dz}}{}{\mathrm{D_y}}{,}{\mathrm{dz}}{}{\mathrm{dz}}{}{\mathrm{D_z}}\right]$ (9.8.1.6)
Exercise 9
Find a complementary basis in the space of rank 3 contravariant tensors to the space of symmetric rank 3 contravariant tensors for the manifold P with coordinates u, v.
Q > with(DifferentialGeometry):
Q > with(Tensor):
Q > DGsetup([u, v], Q):
Solution
Q > Omega := [du, dv]:
Use the GenerateSymmetricTensors command to find a basis for the rank 3 symmetric covariant tensors.
Q > T1 := GenerateSymmetricTensors(Omega, 3);
${\mathrm{T1}}{:=}\left[{\mathrm{du}}{}{\mathrm{du}}{}{\mathrm{du}}{,}\frac{{1}}{{3}}{}{\mathrm{du}}{}{\mathrm{du}}{}{\mathrm{dv}}{+}\frac{{1}}{{3}}{}{\mathrm{du}}{}{\mathrm{dv}}{}{\mathrm{du}}{+}\frac{{1}}{{3}}{}{\mathrm{dv}}{}{\mathrm{du}}{}{\mathrm{du}}{,}\frac{{1}}{{3}}{}{\mathrm{du}}{}{\mathrm{dv}}{}{\mathrm{dv}}{+}\frac{{1}}{{3}}{}{\mathrm{dv}}{}{\mathrm{du}}{}{\mathrm{dv}}{+}\frac{{1}}{{3}}{}{\mathrm{dv}}{}{\mathrm{dv}}{}{\mathrm{du}}{,}{\mathrm{dv}}{}{\mathrm{dv}}{}{\mathrm{dv}}\right]$ (9.9.1.1)
Use the GenerateTensors command to find a basis for the rank 3 covariant tensors.
Q > T2 := GenerateTensors([Omega, Omega, Omega]);
${\mathrm{T2}}{:=}\left[{\mathrm{du}}{}{\mathrm{du}}{}{\mathrm{du}}{,}{\mathrm{du}}{}{\mathrm{du}}{}{\mathrm{dv}}{,}{\mathrm{du}}{}{\mathrm{dv}}{}{\mathrm{du}}{,}{\mathrm{du}}{}{\mathrm{dv}}{}{\mathrm{dv}}{,}{\mathrm{dv}}{}{\mathrm{du}}{}{\mathrm{du}}{,}{\mathrm{dv}}{}{\mathrm{du}}{}{\mathrm{dv}}{,}{\mathrm{dv}}{}{\mathrm{dv}}{}{\mathrm{du}}{,}{\mathrm{dv}}{}{\mathrm{dv}}{}{\mathrm{dv}}\right]$ (9.9.1.2)
Q > Answer := ComplementaryBasis(T1, T2);
${\mathrm{Answer}}{:=}\left[{\mathrm{du}}{}{\mathrm{du}}{}{\mathrm{dv}}{,}{\mathrm{du}}{}{\mathrm{dv}}{}{\mathrm{du}}{,}{\mathrm{du}}{}{\mathrm{dv}}{}{\mathrm{dv}}{,}{\mathrm{dv}}{}{\mathrm{du}}{}{\mathrm{dv}}\right]$ (9.9.1.3)
Exercise 10
Find the dual basis for each of the following. Check your result.
Q > with(DifferentialGeometry):
Q > DGsetup([x, y, u, v], M):
[i]
M > A := evalDG([D_x + D_y + D_u + D_v, D_y + D_u + D_v, D_u + D_v, D_v]);
${A}{:=}\left[{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{\mathrm{D_u}}{+}{\mathrm{D_v}}{,}{\mathrm{D_y}}{+}{\mathrm{D_u}}{+}{\mathrm{D_v}}{,}{\mathrm{D_u}}{+}{\mathrm{D_v}}{,}{\mathrm{D_v}}\right]$ (9.10.1)
[ii]
M > B := evalDG([dx + dy, dy + du, du + dv, dv - dx]);
${B}{:=}\left[{\mathrm{dx}}{+}{\mathrm{dy}}{,}{\mathrm{dy}}{+}{\mathrm{du}}{,}{\mathrm{du}}{+}{\mathrm{dv}}{,}{-}{\mathrm{dx}}{+}{\mathrm{dv}}\right]$ (9.10.2)
Solution
Part [i]
M > Answer1 := DualBasis(A);
${\mathrm{Answer1}}{:=}\left[{\mathrm{dx}}{,}{-}{\mathrm{dx}}{+}{\mathrm{dy}}{,}{-}{\mathrm{dy}}{+}{\mathrm{du}}{,}{-}{\mathrm{du}}{+}{\mathrm{dv}}\right]$ (9.10.1.1)
M > Matrix(4, 4, (i, j) -> Hook(A[i], Answer1[j]));
Part [ii]
M > Answer2 := DualBasis(B);
${\mathrm{Answer2}}{:=}\left[\frac{{1}}{{2}}{}{\mathrm{D_x}}{+}\frac{{1}}{{2}}{}{\mathrm{D_y}}{-}\frac{{1}}{{2}}{}{\mathrm{D_u}}{+}\frac{{1}}{{2}}{}{\mathrm{D_v}}{,}{-}\frac{{1}}{{2}}{}{\mathrm{D_x}}{+}\frac{{1}}{{2}}{}{\mathrm{D_y}}{+}\frac{{1}}{{2}}{}{\mathrm{D_u}}{-}\frac{{1}}{{2}}{}{\mathrm{D_v}}{,}\frac{{1}}{{2}}{}{\mathrm{D_x}}{-}\frac{{1}}{{2}}{}{\mathrm{D_y}}{+}\frac{{1}}{{2}}{}{\mathrm{D_u}}{+}\frac{{1}}{{2}}{}{\mathrm{D_v}}{,}{-}\frac{{1}}{{2}}{}{\mathrm{D_x}}{+}\frac{{1}}{{2}}{}{\mathrm{D_y}}{-}\frac{{1}}{{2}}{}{\mathrm{D_u}}{+}\frac{{1}}{{2}}{}{\mathrm{D_v}}\right]$ (9.10.1.2)
M > Matrix(4, 4, (i, j) -> Hook(Answer2[i], B[j]));
Exercise 11
Write a program TestIndependence (without using DGbasis) that will return true if a given list of VFT are linearly independent and false otherwise.
This exercise introduces some important coding techniques which are used repeatedly in the DifferentialGeometry procedures and which are often needed to create new applications.
Hint: Use DGzip to form a VFT X from the given list with unknown coefficients. Extract the coefficients of X to obtain a set of equations Eq. Use LinearAlgebra:-GenerateMatrix to obtain the coefficient matrix A for the system of linear equations defined by Eq. Use LinearAlgebra:-Rank to calculate the rank of the matrix A.
Test you program on the list of vectors B1 and the list of 2-forms B2.
M > with(DifferentialGeometry):
M > DGsetup([x, y, z], M);
${\mathrm{frame name: M}}$ (9.11.1)
M > B1 := evalDG([D_x - D_y, D_y + D_z, D_x - 2*D_y + 3*D_z]);
${\mathrm{B1}}{:=}\left[{\mathrm{D_x}}{-}{\mathrm{D_y}}{,}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{\mathrm{D_x}}{-}{2}{}{\mathrm{D_y}}{+}{3}{}{\mathrm{D_z}}\right]$ (9.11.2)
M > B2 := evalDG([dx &w dy, dx &w dz, dx &w dy + dz &w dx]);
${\mathrm{B2}}{:=}\left[{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}{,}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{-}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dz}}\right]$ (9.11.3)
Solution
M > TestIndependence := proc(VFTlist)
M > local A, X, n, vars, Eq, r, answer;
M > n := nops(VFTlist);
M > vars := [seq(_a||i, i = 1 .. n)];
M > X := DGzip(vars, VFTlist, "plus");
M > Eq := Tools:-DGinfo(X, "CoefficientSet");
M > A := LinearAlgebra:-GenerateMatrix(Eq, vars)[1];
M > r := LinearAlgebra:-Rank(A);
M > if r < n then answer := false else answer := true fi;
M > answer;
M > end:
M >
> TestIndependence(B1);
${\mathrm{true}}$ (9.11.1.1)
M > TestIndependence(B2);
${\mathrm{false}}$ (9.11.1.2)
To watch the program at work, use the Maple debug command.
M > debug(TestIndependence);
${\mathrm{TestIndependence}}$ (9.11.1.3)
M > TestIndependence(B1);
{--> enter TestIndependence, args = [_DG([[vector, M, []], [[[1], 1], [[2], -1]]]), _DG([[vector, M, []], [[[2], 1], [[3], 1]]]), _DG([[vector, M, []], [[[1], 1], [[2], -2], [[3], 3]]])]
${n}{:=}{3}$
${\mathrm{vars}}{:=}\left[{\mathrm{_a1}}{,}{\mathrm{_a2}}{,}{\mathrm{_a3}}\right]$
${X}{:=}\left({\mathrm{_a3}}{+}{\mathrm{_a1}}\right){}{\mathrm{D_x}}{-}\left({2}{}{\mathrm{_a3}}{-}{\mathrm{_a2}}{+}{\mathrm{_a1}}\right){}{\mathrm{D_y}}{+}\left({3}{}{\mathrm{_a3}}{+}{\mathrm{_a2}}\right){}{\mathrm{D_z}}$
${\mathrm{Eq}}{:=}\left\{{\mathrm{_a3}}{+}{\mathrm{_a1}}{,}{3}{}{\mathrm{_a3}}{+}{\mathrm{_a2}}{,}{-}{2}{}{\mathrm{_a3}}{+}{\mathrm{_a2}}{-}{\mathrm{_a1}}\right\}$
${r}{:=}{3}$
${\mathrm{answer}}{:=}{\mathrm{true}}$
${\mathrm{true}}$
<-- exit TestIndependence (now at top level) = true}
${\mathrm{true}}$ (9.11.1.4)
There is one change in the program that should be made -- there is no assurance that the variables _a1, _a2, _a3 are not defined by the user, in which case, the program will not work. We can avoid this problem by using a little utility program from the Maple tools package. This program is guaranteed to generate a sequence of unassigned names.
M > tools/_Qn(p,[],4);
${\mathrm{p1}}{,}{\mathrm{p2}}{,}{\mathrm{p3}}{,}{\mathrm{p4}}$ (9.11.1.5)
M > p1 := 3: p5 := 0:
M > tools/_Qn(p,[],4);
${\mathrm{p2}}{,}{\mathrm{p3}}{,}{\mathrm{p4}}{,}{\mathrm{p6}}$ (9.11.1.6)
Therefore the line vars := [seq(_a||i, i = 1 .. n)]; should be replaced by vars := [tools/_Qn(_a, [], n)].
Exercise 12
A p-form omega is called decomposable if it is the wedge product of p 1-forms. If we define A = {X : Hook(X, omega) = 0}, then it is not difficult to prove that omega is decomposable if and only if dim(A) = n - p, where n is the dimension of the ambient manifold. The space A is called the retracting space for omega. Moreover, if omega is decomposable, then it is a multiple of the wedge product of the 1-forms in Annihilator(A). (See Hook, Annihilator)
[i] Write a program RetractingSpace that will compute the retracting space for a p-form omega.
Hint: Follow Example 11 and use LinearAlgebra:-NullSpace.
[ii] Write a program DecomposeForm that will determine if a p-form is decomposable and, if so, return a list of p 1-forms whose wedge product is the given p-form. If the form is not decomposable return an empty list.
Test your programs on the forms alpha, beta, theta.
M > with(DifferentialGeometry):
M > DGsetup([x, y, z, u, v], M):
M > alpha := dx &wedge dy &wedge dz;
${\mathrm{α}}{:=}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}$ (9.12.1)
M > beta := evalDG(dx &w dy + du &w dv);
${\mathrm{β}}{:=}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{+}{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dv}}$ (9.12.2)
M > theta := evalDG(9*dx &w dy &w du - 13*dx &w dy &w dv + 5*dx &w du &w dv + 2*dy &w du &w dv);
${\mathrm{θ}}{:=}{9}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{du}}{-}{13}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dv}}{+}{5}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dv}}{+}{2}{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dv}}$ (9.12.3)
Solution
M > RetractingSpace := proc(omega0)
M > local omega, A, X, n, vars, alpha, Eq, B, N, p, C, c, i;
M > omega := evalDG(omega0);
M > A := Tools:-DGinfo("FrameBaseVectors");
M > n := Tools:-DGinfo("FrameBaseDimension");
M > vars := [tools/_Qn(_a,[],n)];
M > X := DGzip(vars, A,"plus");
M > alpha := Hook(X, omega);
M > Eq := Tools:-DGinfo(alpha, "CoefficientSet");
M > B := LinearAlgebra:-GenerateMatrix(Eq, vars)[1];
M > N := LinearAlgebra:-NullSpace(B);
M > p := nops(N);
M > C := Array(1 .. p);
M > for i to p do
M > c := convert(N[i], list);
M > C[i] := DGzip(c, A, "plus");
M > od;
M > convert(C, list);
M > end:
M >
M > DecomposeForm := proc(omega0)
M > local A, omega, n, p, q, B, C, alpha, Eq, soln, newC1, k;
M > omega := evalDG(omega0);
M > n := Tools:-DGinfo("FrameBaseDimension");
M > p := Tools:-DGinfo(omega, "FormDegree");
M > A := RetractingSpace(omega);
M > q := nops(A);
M > if q <> n - p then return [] fi;
M > B := Tools:-DGinfo("FrameBaseForms");
M > C := Annihilator(A, B);
M > alpha := k &mult DGzip(C, "wedge");
M > Eq := Tools:-DGinfo(omega0 &minus alpha, "CoefficientSet");
M > soln:= solve(Eq, {k});
M > newC1 := eval(k &mult C[1], soln);
M > [newC1, op(C[2 .. -1])];
> end:
Example 1.
M > alpha;
${\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}$ (9.12.1.1)
M > RetractingSpace(alpha);
$\left[{\mathrm{D_v}}{,}{\mathrm{D_u}}\right]$ (9.12.1.2)
M > DecomposeForm(alpha);
$\left[{-}{\mathrm{dy}}{,}{\mathrm{dx}}{,}{\mathrm{dz}}\right]$ (9.12.1.3)
Example 2.
M > beta;
${\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{+}{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dv}}$ (9.12.1.4)
M > RetractingSpace(beta);
$\left[{\mathrm{D_z}}\right]$ (9.12.1.5)
M > DecomposeForm(beta);
$\left[{}\right]$ (9.12.1.6)
Example 3.
M > theta;
${9}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{du}}{-}{13}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dv}}{+}{5}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dv}}{+}{2}{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{du}}{}{\bigwedge }{}{\mathrm{dv}}$ (9.12.1.7)
M > RetractingSpace(theta);
$\left[{-}\frac{{2}}{{9}}{}{\mathrm{D_x}}{+}\frac{{5}}{{9}}{}{\mathrm{D_y}}{+}\frac{{13}}{{9}}{}{\mathrm{D_u}}{+}{\mathrm{D_v}}{,}{\mathrm{D_z}}\right]$ (9.12.1.8)
M > F := DecomposeForm(theta);
${F}{:=}\left[{-}{9}{}{\mathrm{dx}}{-}{2}{}{\mathrm{dv}}{,}\frac{{13}}{{2}}{}{\mathrm{dx}}{+}{\mathrm{du}}{,}\frac{{5}}{{2}}{}{\mathrm{dx}}{+}{\mathrm{dy}}\right]$ (9.12.1.9)
M > theta &minus DGzip(F, "wedge");
${0}{}{\mathrm{dx}}{}{\bigwedge }{}{\mathrm{dy}}{}{\bigwedge }{}{\mathrm{dz}}$ (9.12.1.10)
M >
> | 16,532 | 41,478 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 129, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-10 | latest | en | 0.684194 |
http://www.sawaal.com/statement-and-conclusions-questions-and-answers/all-snakes-are-trees-some-trees-are-roads-all-roads-are-mountains--conclusions-nbsp--1-some-moun_2817 | 1,529,583,398,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00529.warc.gz | 491,046,802 | 17,147 | 16
Q:
# All snakes are trees. Some trees are roads. All roads are mountains.Conclusions: 1. Some mountains are snakes2. Some roads are snakes3. Some mountains are trees
A) Only 1 follows B) Only 2 follows C) Only 3 follows D) Both 1 and 2 follow
Explanation:
All snakes are trees. Some trees are roads.
Since the middle term is not distributed even once in the premises,so no conclusion follows
Since one primise is particular,the conclusion must be particularand should'nt contain the middle term. So,it follows that 'Some trees are mountains'. 3 is the converse of this statement and so it holds.
All snakes are trees. Some trees are mountains.
Since the middle term is not distributed even once in the preises, so no definite conclusion follows
Q:
In each question given below four conclusions are followed by statements. You have to take the three given statements to be true (even if they seem to be at variance from the commonly known facts). Read the conclusions and decide which of the following set of statement indicate that conclusion is logically follows.
Statements :
1. All chairs are tents.
2. No chairs is jugs.
3. No jugs is glasses.
4. No glasses is pots.
Conclusions :
I. All pots are tents.
II. All glasses are chair.
III. Some jugs are tents.
A) Only conclusion I follows B) Both conclusion II & III follows C) None follows D) Only conclusion II follows
Explanation:
3 275
Q:
Find which of the two conclusions I and II given below is/are definitely true from the the given statements.
Statements: I > W > T > N, I = F = G = C
Conclusions: I. W > I
II. C > N
A) Only Conclusion I is true B) Only Conclusion II is true C) Both conclusions I & II are true D) Neither conclusion I nor II is true
Explanation:
From the given Statements :
1. I > W > T > N
2. F = G = C
Conclusions are :
I. W > I (False)
II. C > N (True)
7 272
Q:
Study the question and all the three statements given and decide whether any information provided in the statements are redundand and can be dispensed with while answering the questions.
Question :
What is the rate of interest per year ?
Statements :
1. The amount invested is Rs. 6000.
2. The amount becomes Rs. 6741.60 in 2 years at compound interest.
3. The difference between compound interest and simple interest in 2 years is Rs. 21.60.
A) Only 3 B) Both 2 & 3 C) Any two of (1, 2 & 3) D) None
Answer & Explanation Answer: C) Any two of (1, 2 & 3)
Explanation:
Using Both 1 & 2 statements we get the rate of interest as in (1) we have given principle amount and in (2) compound interest for 2 years. By this data we get the rate%.
Using Both 1 & 3 statements we can get the rate% as we have principle amount & difference between compound interest and simple interest in 2 years.
Using Both 2 & 3 statements we get compound interest & simple interest by which we get principle amount. So that we can calculate %rate.
Hence by using any two of the three statements(1,2&3) we get rate of interest.
5 341
Q:
Statements :-
All kids are God.
No God is a Human.
Some Human are Son.
All Son are Men.
Conclusions :-
(a) Some Son are not God.
(b) Some Men are Human.
A) Only A follows B) Only B follows C) Both (A) and (B) follows D) Neither (A) nor (B) follows
Explanation:
7 501
Q:
In the following question, two equations numbered 1 and 2 are given. You have to solve both the equations and determine the relation between a and b.
1. $5{a}^{2}-18a+9=0$
2. $3{b}^{2}+5b-2=0$
A) a >= b B) a <= b C) a < b D) a > b
Explanation:
From solving 1 and 2 we get,
1.
5a(a-3)-3(a-3) = 0
(5a-3)(a-3) = 0
a = 3 or 3/5
2.
3b(b+2)-1(b+2) = 0
b = -2 or b = 1/3
Here when a = 3, a > b for b = -2 and b = 1/3
when a = 3/5. a > b for b = -2 and b = 1/3.
Hence, it is clear that a > b.
4 406
Q:
Assuming the given statements to be true, find which of the two conclusions A and B given below is/are definitely true.
Statements: C < T < L= P > Q, N > C > Y
Conclusions: A. P > Y B. Y < P
A) only conclusion B is true. B) only conclusion A is true. C) neither conclusion I nor II is true. D) either conclusion I or II is true.
Explanation:
From given statements, we can conclude that
N > C < T < L= P > Q .....(1)
Here given that C > Y but in eq(1) we got that C < T < T <= P => Y is definitely less than P.
So only conclusion B is True.
3 366
Q:
Q : Which train did Harish catch to go to office ?
Statements :
A. Harish missed his usual train of 4.15 p.m. A train comes in every 15 minutes.
B. Harish did not catch the 4.45 p.m. train or any train after that time.
A) If statement B alone is sufficient but statement A alone is not sufficient. B) If statement A alone is sufficient but statement B alone is not sufficient. C) If both statement together are sufficient, but neither statement alone is sufficient. D) If statement A and B together are not sufficient.
Answer & Explanation Answer: D) If statement A and B together are not sufficient.
Explanation:
From both statements we cannot conclude the train catched by Harish
Since he missed at 4.15 and train coes at 4.30, 4.45, 5.00,...
But in B given that he didn't catch the train at 4.45 and after that.
So both statements A & B together are not sufficient to answer the question.
3 421
Q:
What is the value of KL ?
Statement A: ${K}^{2}$= 4.
Statement B: L = 0.
A) Only A is sufficient B) Only B is sufficient C) Both (A) and (B) are sufficient D) None of the above | 1,529 | 5,470 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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} | 4.21875 | 4 | CC-MAIN-2018-26 | latest | en | 0.89774 |
https://myhomeworkwriters.com/effective-annual-rate-assignment-homework-for-you-8/ | 1,679,406,023,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00403.warc.gz | 483,151,375 | 18,921 | # Effective Annual Rate Assignment | Homework For You
Parker & Stone, Inc., is looking at setting up a new manufacturing plant in South Park to produce garden tools. The company bought some land six years ago for \$4.5 million in anticipation of using it as a warehouse and distribution site, but the company has since decided to rent these facilities from a competitor instead. If the land were sold today, the company would net \$4.8 million. The company wants to build its new manufacturing plant on this land; the plant will cost \$12 million to build, and the site requires \$720,000 worth of grading before it is suitable for construction. What is the proper cash flow amount to use as the initial investment in fixed assets when evaluating this project? (Enter your answer in dollars, not millions of dollars, e.g. 1,234,567.)
Cash flow amount Lang Industrial Systems Company (LISC) is trying to decide between two different conveyor belt systems. System A costs \$224,000, has a four-year life, and requires \$71,000 in pretax annual operating costs. System B costs \$318,000, has a six-year life, and requires \$65,000 in prelax annual operating costs. Suppose LISC always needs a conveyor belt system; when one wears out, it must be replaced. Assume the tax rate is 30 percent and the discount rate is 9 percent. Calculate the EAC for both conveyor belt systems. (Negative amounts should be indicated by a minus sign. Do not round intermediate calculations and round your answers to 2 decimal places, 0.g. 32.16.)
EAC System A System B Which conveyor belt system should the firm choose? System B System A. Get Finance homework help today
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Claim my 15% OFF Order in Chat | 1,175 | 5,045 | {"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-2023-14 | latest | en | 0.943737 |
http://nrich.maths.org/public/leg.php?code=3&cl=3&cldcmpid=664 | 1,444,789,340,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443738095178.99/warc/CC-MAIN-20151001222135-00247-ip-10-137-6-227.ec2.internal.warc.gz | 225,565,813 | 9,269 | # Search by Topic
#### Resources tagged with Integers similar to In Particular:
Filter by: Content type:
Stage:
Challenge level:
### There are 39 results
Broad Topics > Numbers and the Number System > Integers
### In Particular
##### Stage: 4 Challenge Level:
Write 100 as the sum of two positive integers, one divisible by 7 and the other divisible by 11. Then find formulas giving all the solutions to 7x + 11y = 100 where x and y are integers.
### Upsetting Pitagoras
##### Stage: 4 and 5 Challenge Level:
Find the smallest integer solution to the equation 1/x^2 + 1/y^2 = 1/z^2
### Our Ages
##### Stage: 4 Challenge Level:
I am exactly n times my daughter's age. In m years I shall be exactly (n-1) times her age. In m2 years I shall be exactly (n-2) times her age. After that I shall never again be an exact multiple of. . . .
### Rudolff's Problem
##### Stage: 4 Challenge Level:
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?
### Coffee
##### Stage: 4 Challenge Level:
To make 11 kilograms of this blend of coffee costs £15 per kilogram. The blend uses more Brazilian, Kenyan and Mocha coffee... How many kilograms of each type of coffee are used?
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Whole Number Dynamics IV
##### Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
### Whole Number Dynamics I
##### Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Latin Numbers
##### Stage: 4 Challenge Level:
Let N be a six digit number with distinct digits. Find the number N given that the numbers N, 2N, 3N, 4N, 5N, 6N, when written underneath each other, form a latin square (that is each row and each. . . .
### Dalmatians
##### Stage: 4 and 5 Challenge Level:
Investigate the sequences obtained by starting with any positive 2 digit number (10a+b) and repeatedly using the rule 10a+b maps to 10b-a to get the next number in the sequence.
### Fracmax
##### Stage: 4 Challenge Level:
Find the maximum value of 1/p + 1/q + 1/r where this sum is less than 1 and p, q, and r are positive integers.
### Not a Polite Question
##### Stage: 3 Challenge Level:
When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...
### Base Puzzle
##### Stage: 3 Challenge Level:
This investigation is about happy numbers in the World of the Octopus where all numbers are written in base 8 .Octi the octopus counts.
### Never Prime
##### Stage: 4 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Times Right
##### Stage: 3 and 4 Challenge Level:
Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?
### Whole Numbers Only
##### Stage: 3 Challenge Level:
Can you work out how many of each kind of pencil this student bought?
### Mini-max
##### Stage: 3 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
### Diophantine N-tuples
##### Stage: 4 Challenge Level:
Take any whole number q. Calculate q^2 - 1. Factorize q^2-1 to give two factors a and b (not necessarily q+1 and q-1). Put c = a + b + 2q . Then you will find that ab+1 , bc+1 and ca+1 are all. . . .
### Euler's Squares
##### Stage: 4 Challenge Level:
Euler found four whole numbers such that the sum of any two of the numbers is a perfect square. Three of the numbers that he found are a = 18530, b=65570, c=45986. Find the fourth number, x. You. . . .
### Lesser Digits
##### Stage: 3 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
### Phew I'm Factored
##### Stage: 4 Challenge Level:
Explore the factors of the numbers which are written as 10101 in different number bases. Prove that the numbers 10201, 11011 and 10101 are composite in any base.
### Arrange the Digits
##### Stage: 3 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### Two Much
##### Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
### Even Up
##### Stage: 3 Challenge Level:
Consider all of the five digit numbers which we can form using only the digits 2, 4, 6 and 8. If these numbers are arranged in ascending order, what is the 512th number?
### AB Search
##### Stage: 3 Challenge Level:
The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?
### 1 Step 2 Step
##### Stage: 3 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
### Score
##### Stage: 3 Challenge Level:
There are exactly 3 ways to add 4 odd numbers to get 10. Find all the ways of adding 8 odd numbers to get 20. To be sure of getting all the solutions you will need to be systematic. What about. . . .
### Seven Up
##### Stage: 3 Challenge Level:
The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?
### Six Times Five
##### Stage: 3 Challenge Level:
How many six digit numbers are there which DO NOT contain a 5?
### Pairs
##### Stage: 3 Challenge Level:
Ann thought of 5 numbers and told Bob all the sums that could be made by adding the numbers in pairs. The list of sums is 6, 7, 8, 8, 9, 9, 10,10, 11, 12. Help Bob to find out which numbers Ann was. . . .
### What Are Numbers?
##### Stage: 2, 3, 4 and 5
Ranging from kindergarten mathematics to the fringe of research this informal article paints the big picture of number in a non technical way suitable for primary teachers and older students.
### Aba
##### Stage: 3 Challenge Level:
In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.
### Slippy Numbers
##### Stage: 3 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### The Patent Solution
##### Stage: 3 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
### Sissa's Reward
##### Stage: 3 Challenge Level:
Sissa cleverly asked the King for a reward that sounded quite modest but turned out to be rather large...
### As Easy as 1,2,3
##### Stage: 3 Challenge Level:
When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .
### Eight Dominoes
##### Stage: 2, 3 and 4 Challenge Level:
Using the 8 dominoes make a square where each of the columns and rows adds up to 8 | 2,121 | 8,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2015-40 | longest | en | 0.850466 |
https://www.interviewarea.com/q-and-a/how-much-do-you-have-to-make-a-year-to-afford-a-400000-house | 1,702,098,301,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00386.warc.gz | 896,408,913 | 13,437 | # How much do you have to make a year to afford a \$400000 house?
The annual salary needed to afford a \$400,000 home is about \$165,000. Over the past two years, home prices have skyrocketed amid the combined impacts of a global pandemic and housing inventory shortages. Between 2020 and 2022, home prices soared 30%, according to Freddie Mac.
## How much salary do you need for a 400k mortgage?
What income is required for a 400k mortgage? To afford a \$400,000 house, borrowers need \$55,600 in cash to put 10 percent down. With a 30-year mortgage, your monthly income should be at least \$8200 and your monthly payments on existing debt should not exceed \$981.
## How much should you make a year to afford a \$300000 house?
How much do I need to make to buy a \$300K house? To purchase a \$300K house, you may need to make between \$50,000 and \$74,500 a year. This is a rule of thumb, and the specific salary will vary depending on your credit score, debt-to-income ratio, the type of home loan, loan term, and mortgage rate.
## How much do I need to make to afford a 350k house?
You need to make \$129,511 a year to afford a 350k mortgage. We base the income you need on a 350k mortgage on a payment that is 24% of your monthly income. In your case, your monthly income should be about \$10,793. The monthly payment on a 350k mortgage is \$2,590.
## How much house can I afford if I make 120k a year?
Safe debt guidelines
If you make \$50,000 a year, your total yearly housing costs should ideally be no more than \$14,000, or \$1,167 a month. If you make \$120,000 a year, you can go up to \$33,600 a year, or \$2,800 a month—as long as your other debts don't push you beyond the 36 percent mark.
## How much can I spend on a house if I make 70K a year?
If you're an aspiring homeowner, you may be asking yourself, “I make \$70,000 a year: how much house can I afford?” If you make \$70K a year, you can likely afford a home between \$290,000 and \$360,000*. That's a monthly house payment between \$2,000 and \$2,500 a month, depending on your personal finances.
## Can I afford a 300k house on a 60k salary?
To afford a house that costs \$300,000 with a down payment of \$60,000, you'd need to earn \$44,764 per year before tax. The monthly mortgage payment would be \$1,044. Salary needed for 300,000 dollar mortgage.
## How much house can I afford with a salary of \$100000?
A 100K salary means you can afford a \$350,000 to \$500,000 house, assuming you stick with the 28% rule that most experts recommend. This would mean you would spend around \$2,300 per month on your house and have a down payment of 5% to 20%.
## What mortgage can I get with 100K salary?
If you're earning \$100,000 per year, your average monthly (gross) income is \$8,333. So, your mortgage payment should be \$2,333 or less.
## How much is a 500k house monthly payment?
The average mortgage rate for a \$500,000, 30-year fixed-rate loan is around 5.4% for those with good credit. So, your monthly payment would be around \$2250 without taxes and fees.
## How much should I make to afford a 500k house?
You need to make \$185,016 a year to afford a 500k mortgage. We base the income you need on a 500k mortgage on a payment that is 24% of your monthly income. In your case, your monthly income should be about \$15,418.
## How much house can I afford on \$65,000 a year?
I make \$65,000 a year. How much house can I afford? You can afford a \$195,000 house.
## How much house can I afford on a 200K salary?
How much house can I afford if I make \$200K per year? A mortgage on 200k salary, using the 2.5 rule, means you could afford \$500,000 (\$200,00 x 2.5). With a 4.5 percent interest rate and a 30-year term, your monthly payment would be \$2533 and you'd pay \$912,034 over the life of the mortgage due to interest.
## How much mortgage can I afford if I make 80000 a year?
For the couple making \$80,000 per year, the Rule of 28 limits their monthly mortgage payments to \$1,866. Ideally, you have a down payment of at least 10%, and up to 20%, of your future home's purchase price.
## How much house can I afford if I make 75000 a year?
If you're making \$75,000 each year, your monthly earnings come out to \$6,250. To meet the 28 piece of the 28/36 rule, that means your monthly mortgage payment should not exceed \$1,750. And for the 36 part, your total monthly debts should not come to more than \$2,250.
## How much do I need to make to afford a 450k house?
To finance a 450k mortgage, you'll need to earn roughly \$135,000 – \$140,000 each year. We calculated the amount of money you'll need for a 450k mortgage based on a payment of 24% of your monthly income. Your monthly income should be around \$11,500 in your instance. A 450k mortgage has a monthly payment of \$2,769.
## Is \$100 K enough to buy a house?
A \$100K salary puts you in a good position to buy a home
But you'll need more than a good income to buy a house. You will also need a strong credit score, low debts, and a decent down payment. With all these factors and \$100K of income per year, most doors in the mortgage world will be open to you.
## Can I afford a million dollar home?
Experts suggest you might need an annual income between \$100,000 to \$225,000, depending on your financial profile, in order to afford a \$1 million home. Your debt-to-income ratio (DTI), credit score, down payment and interest rate all factor into what you can afford.
## How much is 50k a year hourly?
An average person works about 40 hours per week, which means if they make \$50,000 a year, they earn \$24.04 per hour.
## Can I afford a 300k house on a 70k salary?
On a \$70,000 income, you'll likely be able to afford a home that costs \$280,000–380,000. The exact amount will depend on how much debt you have and where you live — as well as the type of home loan you get.
## How much is 70k a year hourly?
A salary of \$70,000 equates to a monthly pay of \$5,833, weekly pay of \$1,346, and an hourly wage of \$33.65.
## What is a good salary in America?
According to the US Bureau of Labor Statistics (BLS), the median annual wage across all occupations in 2021 was \$58,260 [1]. For a person living in Phoenix, Arizona, where the median wage is \$56,610, earning above the national average may be considered very good.
## How much house can I afford with \$36,000 a year?
If you make \$3,000 a month (\$36,000 a year), your DTI with an FHA loan should be no more than \$1,290 (\$3,000 x 0.43) — which means you can afford a house with a monthly payment that is no more than \$900 (\$3,000 x 0.31). FHA loans typically allow for a lower down payment and credit score if certain requirements are met. | 1,769 | 6,708 | {"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-2023-50 | latest | en | 0.951219 |
https://egvideos.com/video/missouri/grade-1/math/1.gm.a.2/composite-shapes | 1,652,716,718,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510138.6/warc/CC-MAIN-20220516140911-20220516170911-00074.warc.gz | 284,106,548 | 9,493 | # Missouri - Grade 1 - Math - Geometry and Measurement - Composite Shapes - 1.GM.A.2
### Description
Compose and decompose two-and three-dimensional shapes to build an understanding of part-whole relationships and the properties of the original and composite shapes.
• State - Missouri
• Standard ID - 1.GM.A.2
• Subjects - Math Common Core
### Keywords
• Math
• Geometry and Measurement
## More Missouri Topics
Use addition and subtraction within 20 to solve problems.
Use properties as strategies to add and subtract.
1.RA.B.6 Demonstrate that subtraction can be solved as an unknown-addend problem. 1.RA.C.7 Add and subtract within 20. 1.RA.C.8 Demonstrate fluency with addition and subtraction within 10.
1.RA.A.3 Develop the meaning of the equal sign and determine if equations involving addition and subtraction are true or false.
1.RA.A.4 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers.
1.NBT.A.1 Understand that 10 can be thought of as a bundle of 10 ones – called a “ten”.
1.NBT.A.2 Understand two-digit numbers are composed of ten(s) and one(s). | 264 | 1,124 | {"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} | 3.890625 | 4 | CC-MAIN-2022-21 | longest | en | 0.873941 |
http://oeis.org/A105059 | 1,586,259,128,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371700247.99/warc/CC-MAIN-20200407085717-20200407120217-00535.warc.gz | 125,606,785 | 3,666 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A105059 Numbers n such that 100000n - 1 is prime. 0
2, 6, 8, 12, 14, 18, 26, 27, 30, 35, 48, 50, 56, 68, 78, 81, 99, 102, 111, 116, 119, 125, 134, 135, 137, 144, 159, 170, 182, 189, 194, 200, 210, 212, 221, 225, 228, 240, 242, 249, 251, 261, 266, 267, 278, 281, 294, 300, 302, 303, 312, 314, 321, 327, 338, 341 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS EXAMPLE If n=2, then 100000 * n - 1 is prime. If n=78, then 100000 * n - 1 is prime. MATHEMATICA Select[ Range[ 343], PrimeQ[100000# - 1] &] (* Robert G. Wilson v, Apr 05 2005 *) PROG (PARI) is(n)=isprime(100000*n-1) \\ Charles R Greathouse IV, Jun 13 2017 CROSSREFS Sequence in context: A191965 A173064 A111367 * A108187 A102166 A024894 Adjacent sequences: A105056 A105057 A105058 * A105060 A105061 A105062 KEYWORD nonn,easy AUTHOR Parthasarathy Nambi, Apr 04 2005 EXTENSIONS More terms from Robert G. Wilson v, Apr 05 2005 STATUS approved
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Last modified April 7 04:20 EDT 2020. Contains 333292 sequences. (Running on oeis4.) | 532 | 1,446 | {"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-2020-16 | latest | en | 0.625022 |
https://apsprostore.com/determine-your-own-personal-possibilities-associated-with-earning-typically-the-lottery-brilliant-a-huge-number-powerball/ | 1,624,400,435,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00568.warc.gz | 115,818,572 | 11,156 | # Determine Your own personal Possibilities associated with Earning typically the Lottery Brilliant A huge number Powerball
Do you know just how to calculate the chances of winning the lottery, which includes the Florida Lottery? You are able to calculate each set associated with prospects for each various lottery activity you participate in. With the help connected with a small hand organised online car loan calculator or with the free finance calculator on your own personal computer, you just increase in numbers this numbers together in addition to add one particular division approach when “the order” associated with your chosen statistics is definitely not required for a new particular lottery game.
Just what you “need to be able to know” is the number regarding complete balls that typically the winning quantities are pulled from….. is it 59, 56, 49, 49, or perhaps 39? If there is a secondary drawing regarding the single extra soccer ball, such as the “red ball” together with Powerball or the Mega Millions’ “gold ball” you need to be able to know how many balls are usually in this team mainly because well. Are there forty-nine or maybe 39?
It won’t matter if it is definitely the Florida, Ohio, Arizona, PA or maybe NJ Lotto. This strategy or perhaps formula gives you the real probabilities. Florida Lottery is usually 6/53. New York Lotto is 6/59. The Kentkucky Lottery, Massachusetts Lottery, Wisconsin Lottery, as well as State regarding Washington Lotto carry a 6/49 lottery numbers ratio. Illinois Lottery carries some sort of 6/52.
When you have this information correctly in front of you along with your calculator around hand, you can start off functioning the formulas. You need to decide on 5 regular balls and one extra ball correctly equalled to the winning drawn statistics to win the multi-million dollar jackpot that best individuals dream about receiving someday.
From the first instance there are 56 balls in the first class and fouthy-six balls within the secondary group. In get to gain the Lottery jackpot you need to match all these balls (5 and up. 1) exactly, nonetheless not necessarily necessarily in order. The Ca Lottery’s Smart Fetta And also is 47/27. The big drum is usually spinning with the original part of the getting. You have a 1/56 chance to match your number to this 1st ball.
Along with one soccer ball removed as soon as the first amount has been drawn, anyone now have a 1/55 potential for matching one other a person of your figures to the second ball driven. With each drawn amount a ball is removed lowering the number of remaining balls by a new total of one.
Chances of you correctly coordinating the number on the particular third baseball to end up being drawn is now 1/54 from the total quantity of tennis balls remaining around the drum. Together with the third ball removed from often the drum and sitting while using other two winning numbers, your odds of correctly corresponding the fourth soccer ball is diminished to 1/53.
As you can observe if a ball is published through the drum the possibilities are lowered by one. You began with the 1/56 chance, after that using each new being successful number it is decreased for you to 1/55, 1/54, 1/53, together with with the fifth soccer ball you might have the odds regarding 1/52 appropriately matching that fifth earning number. This is the first section of the formula of how to help calculate your current odds associated with winning the lotto, as well as the Lakewood ranch Lottery.
Now take these five odds representing the five being successful numbers (1/56, 1/55, 1/54, 1/53, and 1/52). Typically the “1” on top associated with the portion represents your own personal one and only possible opportunity to properly match the used number.
Now you get your finance calculator and increase in numbers all top numbers (1x1x1x1x1) equal a person (1). So next you multiply all of the underside numbers (56x55x54x53x52). Accurately came into and multiplied an individual learn the total is 458, 377, 920. The different small fraction becomes 1/458, 377, 920. This is some sort of 458 , 000, 000 to one chance to win. In the event that you where required to pick this numbers to be able just similar to they are drawn, next these would be the odds versus you for you to win this Go with 5/56 ball lottery match.
Luckily or even unfortunately, you will be not really required to pick often the numbers in the precise order they are really drawn. Often the second step in the formula will decrease the odds, which in turn permits you to match up these five winning amounts in any order. Inside this action you is going to multiply the number connected with paintballs drawn — five (1x2x3x4x5). With calculator inside hand the thing is that often the total equals a hundred and twenty.
To be able to give you the right to choose your own personal 5 matching numbers in a purchase, you create these kinds of odds by dividing 120/417, 451, 320. You definitely need a calculator for that a single. 120/458, 377, 920 drops your odds of being successful this lottery to 1/3, 819, 816. These are usually over 3. a few , 000, 000 to one odds towards you of winning that Pick 5/56 ball lotto game.
If this have been often the Mega Millions Lottery, you have to add the “gold ball” to these 5 winning drawn balls in order to win this Multi-Million Dollar Jackpot. The only gold ball is measured as a 1/46 likelihood of related this properly, and since you will be pulling just one number it must be an exact match. Once more, you only have that “1” possiblity to do it best. Now you need for you to increase 3, 819, 816 by 46.
Grab the calculator and do the particular propagation. Your own personal final odds against you winning typically the Mega Thousands Jackpot are usually calculated being 175, 711, 536 or even clearly reported 175 mil, 711 million, 5 hundred thirty-six thirty-six to one (175, 711, 536 to 1). Now you know how to compute the odds involving being successful the Mega A huge number Lotto.
The Powerball Lottery calculations are based on the 1/59 for the primary five white colored balls and even 1/39 to the “red” electricity ball. In your first set of multipliers is 59x58x57x56x55. This collection totals six-hundred, 766, 320. Now separate 600, 766, 360 by 120 (1x2x3x4x5). Your new total is definitely 5, 006, 386. Presently there is a 1/39 likelihood to get the “red” ball. 39 x your five, 006, 386 gives you the real odds involving succeeding the Powerball Jackpot, specifically 195, 249, 054 to at least one.
Another 5 +1 Lotto that looks to be all around the United Expresses is the “Hot Lotto” which has a 39/19 count. Its played in 15 various States. POWER Lottery, Delaware Lottery, Florida Lottery, New jersey Lottery, Kansas Lottery, Maine Lottery, Minnesota Lottery, Montana Lottery, Brand new Hampshire Lotto, New Mexico Lottery, Upper Dakota Lottery, Oklahoma Lotto, South Dakota Lottery, Vermont Lottery, and even the West Va Lotto. The final odds of winning the minimum \$1 Million Lottery jackpot is 15, 939, 383 to 1.
A Pick 6/52 baseball Lottery game formula appears to be like this: (1/52, 1/51, 1/50, 1/49, 1/48, 1/47) for a total of 14, 658, 134, 400 split by 720 (1x2x3x4x5x6) for any odds of 1/20, 358, 520. Your likelihood to win typically the 6/52 Lottery is over 14. 5 million to one for you to win, such like the Illinois Estrazione.
The particular Hoosier Lottery that will uses Indiana State’s play name, offers a 6/48. Michigan Lottery is 6/47, Arizona ( az ) Lotto and Missouri Lotto are usually 6/44, Maryland Lotto is definitely 6/43, and Carm�n Lottery is 6/42. keputusan 4d hari ini Compare this particular to the California Lottery.
A new Pick 5/39 golf ball Lottery game formulation appears like this specific: (1/39, 1/38, 1/37, 1/36, 1/35) regarding a total regarding 69, 090, 840 separated by simply 120 (1x2x3x4x5) for the probabilities of 1/575, 757 involving winning the Jackpot such as Illinois Little Lotto. Various other States with the similar 5/39 lottery numbers consist of the NC Lottery, Georgia plus Florida Lottery Wonderland 5, and Tennessee Lottery’s Get 5. Virginia Lottery’s Funds 5 carries the 5/34 range.
Now, isn’t this better to pick a Lottery game with lower odds against you? | 1,995 | 8,299 | {"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} | 2.921875 | 3 | CC-MAIN-2021-25 | latest | en | 0.94957 |
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## Statistics
describe Probability using Z-score
Find the probabilities that a random mutable Z having a standard normal distribution will take on a value-
1) Between 0.87 and 1.28
2) Between -0.34 and 0.62
3) Greater than 0.85
3) Greater than -0.65
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M920224
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Describe what you learned about the impact of economic, social, and demographic trends affecting the US labor environmen | 999 | 4,400 | {"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} | 3.328125 | 3 | CC-MAIN-2017-04 | longest | en | 0.932288 |
https://groupprops.subwiki.org/wiki/Lcm_of_degrees_of_irreducible_representations | 1,660,141,184,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571190.0/warc/CC-MAIN-20220810131127-20220810161127-00725.warc.gz | 287,905,166 | 8,789 | Lcm of degrees of irreducible representations
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This article defines an arithmetic function on groups
View other such arithmetic functions
This term is related to: linear representation theory
View other terms related to linear representation theory | View facts related to linear representation theory
Definition
For a group over a field
Suppose $G$ is a group and $K$ is a field. The lcm of degrees of irreducible representations of $G$ is defined as the least common multiple of all the degrees of irreducible representations of $G$ over $K$.
Typical context: finite group and splitting field
The typical context is where $G$ is a finite group and $K$ is a splitting field for $G$. In particular, the characteristic of $K$ is either zero or is a prime not dividing the order of $G$, and every irreducible representation of $G$ over any extension field of $K$ can be realized over $K$.
Note that the lcm of degrees of irreducible representations depends (if at all) only on the characteristic of the field $K$. This is because the degrees of irreducible representations over a splitting field depend only on the characteristic of the field.
Default case: characteristic zero
By default, when referring to the lcm of degrees of irreducible representations, we refer to the case of characteristic zero, and we can in particular take $K = \mathbb{C}$.
Related facts
What it divides
Any divisibility fact stating that the degree of every irreducible representation over a splitting field must divide some fixed number implies that the lcm also divides that fixed number. Some of these are listed below: | 343 | 1,638 | {"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": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-33 | latest | en | 0.904259 |
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• 0:40 Step 1: Visualize It
• 1:34 Step 2: Define the Problem
• 1:47 Step 3: Write an Equation
• 2:22 Step 4: Find the Min/Max
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Lesson Transcript
Instructor: Sarah Wright
In this lesson, you'll learn how to follow a five-step process to solve complex optimization problems by visualizing, defining, writing an equation, finding the minimum or maximum, and answering the question.
## Five Steps to Solve Optimization Problems
We've seen that we can solve optimization problems by following a five-step process. It is: visualize the problem, define the problem, write an equation for it, find the minimum or maximum for the problem (usually the derivatives or end-points) and answer the question.
Let's try it for a more complex problem. Take a sheet of paper that is 20 cm long by 10 cm wide. Cut a square from each corner, and fold the sheet to make an open box. How large of a square should you cut to get the largest box volume?
## Step 1: Visualize It
The first step is to visualize it. So let's draw out on a sheet of paper a rectangle that is 20 cm long by 10 cm wide. I'm going to cut a square from each corner and fold the sheet to make an open box. At this point, you may want to grab your own sheet of paper, cut the corners and fold it to make a box. I've color-coded the squares that I've cut out and the adjoining edges to create edges on the box. Okay, I've visualized it. I don't know how large each of those squares is going to be. I'm going to call those x cm. So, I've cut an x-cm square from each corner of the box. When I do that, I end up with a box that is 10 - 2x cm deep, 20 - 2x cm wide and x cm tall.
## Step 2: Define the Problem
Our second step is to define the problem. What do we need to know? We want to maximize the volume. Once we have the volume maximized, we need to know how large of a square we've cut out from each corner to get that maximum volume.
## Step 3: Write an Equation
So let's write an equation for it. Here's my box: It's x tall, 10 - 2x deep and 20 - 2x wide. The volume of that box is volume equals length times width times height, or v = (x)(10 - 2x)(20 - 2x). I have one equation, and I have one thing that I can change. I can change x to maximize the volume. So x is my independent variable and my volume is my dependent variable.
## Step 4: Find the Min/Max
I'm going to find the minimum or maximum of this equation. Let's write it out. I'm going to expand this, and I get 4x^3 - 60x^2 + 200x. All I did was multiply x by what was in the parentheses (both sets), because I don't want to deal with any parentheses when I'm taking a derivative. I want it all written out. So now I'm going to take that derivative, dv/dx, differentiate the right-hand side and I get 12x^2 - 120x + 200.
That's not so bad, but I'm trying to find the critical point of this equation. That is, where the first derivative equals 0. Let's set that to 0 and divide both sides by 4. So my critical point is where 0 = 3x^2 - 30x + 50. I don't see an easy way to factor this, so I'm going to use the quadratic formula to find what values of x solve this equation. I get x = (30 +/- the square root of (900-600)) / 6. I can expand that out to get 5 +/- (the square root of 300)/6, or 5 +/- (10 * the square root of 3)/6, which finally simplifies to 5 +/- (5 * the square root of 3)/3. I don't like fractions or square roots, so I'm going to use a calculator to write this out and find the values of x that solve this equation. That is, the values of x that are critical points for my volume equation. Those two values are x=7.89 and x=2.11.
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Did you know… We have over 160 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 1,201 | 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} | 4.65625 | 5 | CC-MAIN-2018-30 | longest | en | 0.907729 |
https://mail.haskell.org/pipermail/haskell-cafe/2009-May/061061.html | 1,503,142,412,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105341.69/warc/CC-MAIN-20170819105009-20170819125009-00242.warc.gz | 799,478,523 | 1,967 | Miguel Mitrofanov miguelimo38 at yandex.ru
Wed May 6 12:06:02 EDT 2009
```On 6 May 2009, at 19:27, Adrian Neumann wrote:
> Hello,
>
> I'm trying to prove the unfold fusion law, as given in the chapter
> "Origami Programming" in "The Fun of Programming". unfold is defined
> like this:
>
> unfold p f g b = if p b then [] else (f b):unfold p f g (g b)
>
> And the law states:
>
> unfold p f g . h = unfold p' f' g'
> with
> p' = p.h
> f' = f.h
> h.g' = g.h
>
> Foremost I don't really see why one would want to fuse h into the
> unfold. h is executed once, at the beginning and is never needed
> again. Can someone give me an example?
>
> So, this is what I got so far:
>
> unfold p f g.h = (\b -> if p b then [] else (f b): unfold p f g (g
> b).h
> = if p (h b) then [] else (f (h b)) : unfold p f g (g (h b))
> = if p' b then [] else f' b: unfold p f g (h (g' b))
= if p' b then [] else f' b : (unfold p f g . h) g' b
= if p' b then [] else f' b : unfold p' f' g' (g' b) -- NB!
= unfoldr p' f' g' b
This "proof" is actually biting itself on the tail; however, you can
make it work, for example, like this:
(A_n) take n ((unfold p f g . h) b) = take n (unfold p' f' g' b)
Now, A_0 is obvious (take 0 whatever = []), and A_n follows from
A_{n-1} by the previous argument. By induction, A_n holds for all n.
``` | 473 | 1,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-34 | longest | en | 0.821493 |
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# hw - 1 Population 1 n = xbar = 25000 SD = 6000 Population 2...
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1. Population 1 : n = ? ; xbar = 25000 ; SD = 6000 Population 2 : n = ? ; xbar = 28000 ; SD = 9000 What is the 95% confidence interval for the difference? I do not have the n (the number of observations), and I need that value in order to calculate the Confidence Interval. The Confidence interval asks that I multiply the 95% coefficient of 1.96 by the Standard Error. To calculate the Standard Error, I must divide the Standard Deviations by the number of observations, respectively. 2. Population 1 : n = 100 ; xbar = .57 ; SD = (1-0)√(.57)(.43) = .495 Population 2 : n = 100 ; xbar = .32 ; SD = (1-0)√(.32)(.68) = .466 What is the 95% Confidence Interval for the difference? Xbar (1-2) = .57 - .32 = .25 SE (1-2) = √[(.245/100) + (.217/100) = .068 CI(95) = .25 +/- 1.96(.068) = .25 +/- .13 3. Population 1 : n = 70 ; xbar = .26 ; SD = .11 Population 2 : n = 148 ; xbar = .23 ; SD = .13 Can I be 95% confident the researcher’s theory is correct? Xbar(1-2) = .26 - .23 = .03 SE (1-2) = √[(.012/70) + (.017/148) = √.00017 + .00011 = .017 CI(95) = .03 +/- 1.96(.017) = .03 +/- .033 No. I cannot be 955 confident that the researcher’s theory is correct because the confidence interval contains 0 within its range, meaning that it is possible that having your name at the top of the ballot has no effect on the candidates’ chances of winning.
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https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-connecting-concepts-through-application/chapter-1-linear-functions-1-5-finding-equations-of-lines-1-5-exercises-page-75/18 | 1,537,734,186,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159744.52/warc/CC-MAIN-20180923193039-20180923213439-00368.warc.gz | 753,516,155 | 12,758 | Intermediate Algebra: Connecting Concepts through Application
Published by Brooks Cole
Chapter 1 - Linear Functions - 1.5 Finding Equations of Lines - 1.5 Exercises - Page 75: 18
y=14x + 150
Work Step by Step
We can treat the quantity of shirts as j and the price of them as y, so our two points will be (15,360) and (25,500). Using $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ to find the slope, we get m= $\frac{500-360}{25-15}$ = $\frac{140}{10}$ = $14$ Now, using y=mx+b: y=14x + b 500=14(25)+b 500=350+b 150=b y=14x + 150
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 216 | 681 | {"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} | 4.25 | 4 | CC-MAIN-2018-39 | longest | en | 0.810519 |
http://s3-us-west-1.amazonaws.com/slickgood/confidence-interval-bootstrap.html | 1,580,166,460,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251728207.68/warc/CC-MAIN-20200127205148-20200127235148-00069.warc.gz | 140,630,394 | 9,232 | ## Confidence interval bootstrap,deepak chopra meditation techniques list,secret videos on netflix canada - You Shoud Know
03.04.2015
For aerospace structures, the material strengths used in stress analysis are often based on the statistical variation of the materials.
The confidence interval is used to describe how much faith we have in these reliability values. The following illustrates what the 95% confidence means for a B-basis allowable based on a normal distribution. The figure shows the population true 10th percentile level at 1.28 standard deviations to the left of the mean. How to Determine the Confidence Interval for a Population Proportion.The TI-89 Titanium graphing calculator is a powerful, hand held calculator.
Confidence interval - Wikipedia, the free encyclopedia.we are then interested in finding a 95 onfidence interval for p1a??p2, the difference in the two population proportions.
Here you have a list of opinions about Confidence interval and you can also give us your opinion about it. You will see other people's opinions about Confidence interval and you will find out what the others say about it. In statistics, a confidence interval (CI) is a type of interval estimate of a population parameter.
Confidence intervals consist of a range of values (interval) that act as good estimates of the unknown population parameter; however, in infrequent cases, none of these values may cover the value of the parameter.
In applied practice, confidence intervals are typically stated at the 95% confidence level. Certain factors may affect the confidence interval size including size of sample, level of confidence, and population variability. Confidence intervals were introduced to statistics by Jerzy Neyman in a paper published in 1937. In the image below, you can see a graph with the evolution of the times that people look for Confidence interval. Thanks to this graph, we can see the interest Confidence interval has and the evolution of its popularity.
You can leave your opinion about Confidence interval here as well as read the comments and opinions from other people about the topic. Every science research starts with a detailed analysis of the field of interest, defining the clear aim and hypothesis, by thorough planning of the research design and the way of data collecting and data analysis, as well as the reporting of the results.
Let us presume, for example, that we want to know what the average cholesterol concentration in the population is. In statistics, for any statistical measure, a confidence interval presents a range of possible values within which, with some certainty, we can find the statistical measure of the population.
As such, confidence interval is a realistic estimate of (in)precision and sample size of certain research (3). Only the studies with a large sample will give a very narrow confidence interval, which points to high estimate accuracy with a high confidence level. First we have to determine the confidence level for estimating the mean of a parameter in a population. The lower margin of a confidence interval is calculated by deducting the previous formula result from the mean. In our sample example, with the mean cholesterol concentration in the population, the confidence interval would be calculated by using the Z value because of the large size of the sample (N=121) that is normally distributed.
Due to the fact that today there are a lot of statistical softwares that calculate and provide confidence intervals for the majority of statistical indicators, we shall rarely calculate a confidence interval manually.
Confidence interval can be calculated for difference or ratio between any two statistical indicators, so we could examine if this difference or ratio is of any statistical significance. How do we define the confidence interval when it is related to a ratio as, for example, in OR? As the level of confidence decreases, the size of the corresponding interval will decrease.
In most practical research, the standard deviation for the population of interest is not known. Confidence interval: definition, Definition of confidence interval, from the stat trek dictionary of statistical terms and concepts. Commonly used reliability values are 99% (referred to as A-basis allowables) and 90% reliability values (referred to as B-basis allowables). A confidence interval of 95% is specified in both MIL-HDBK-5 and MIL-HDBK-17 for calculating A and B basis values.
We expect 90% (for a B-basis value) of the strength values from the population to be greater than this value, or falling on the right side of the 10th percentile line.
If the 95r 99confidence interval does not contain the population proportion (one half), it is.
If the 95r 99confidence interval does not contain the population proportion (one half), it is.Below is the general formula to estimate a population proportion with a 95confidence interval.
We need to derive a formula for the.If the samples size n and population proportion p satisfy the condition that np a?? 5 . The level of confidence of the confidence interval would indicate the probability that the confidence range captures this true population parameter given a distribution of samples.
However, when presented graphically, confidence intervals can be shown at several confidence levels, for example 90%, 95% and 99%.
And below it, you can see how many pieces of news have been created about Confidence interval in the last years.
To answer this question we select a sample for which we believe that represents the population and in this sample we calculate the mean of cholesterol concentration. By random selection we have chosen one sample of 121 individuals and on this sample we have calculated the mean cholesterol concentration. Therefore, we can consider confidence interval also as a measure of a sample and research quality. Depending on the confidence level that we choose, the interval margins of error also change. In other words: if we select randomly hundred times a sample of 121 individuals and calculate the mean cholesterol concentration and the confidence interval of that estimate in the sample, then in ten out of these hundred samples the confidence interval will not include the actual mean of the population. It should be kept in mind that a confidence interval is accurate only for the samples that follow a normal distribution, whereas it is approximately accurate for large samples that are not distributed normally. However, it is important to know the input based on which the confidence interval is calculated, so we could better understand its meaning and interpretation.
They describe the same thing, but in two different ways and are complementary to each other.
Let us go back to our example of cholesterol concentration in the population to see how the confidence interval can be used for estimating statistical significance of the difference between two means. Let us remember what the definition of confidence interval was: it defines margins of error within which we can expect the actual value with 95% confidence.
Let us imagine that we have assessed the degree of carotid artery stenosis for all our individuals from the sample (N = 121) using echo-colour Doppler sonography analysis.
In the last twenty years, every day there are more journals that require reporting of the confidence intervals for each of their key results.
Wallpaper that displayed are from unknown origin, and we do not intend to infringe any legitimate intellectual, artistic rights or copyright. This means that only 1% and 10% of the material is likely to fall below these strength values. How frequently the observed interval contains the parameter is determined by the confidence level or confidence coefficient. Therefore, we can consider confidence interval also as a measure of the sample and research quality. Many journals therefore require providing the key results with respective confidence intervals (4,5).
The most used confidence intervals in the biomedical literature are the 90%, 95%, 99% and not so often 99.9% one.
The larger the confidence interval is, the higher is the possibility that this interval includes also the mean of cholesterol concentration in the population. Although there are some other ways of calculating it, the confidence interval is generally and most frequently calculated using standard error. Most often we decide for the 95% confidence what means that we will allow that only in 5% cases the actual mean of population does not fall into our interval. For small samples (N < 30) the t value should be used instead of the Z value in the formula for confidence interval, with N-1 degrees of freedom (9). The P value describes probability that the observed phenomenon (deviation) occurred by chance, whereas the confidence interval provides margins of error within which it is possible to expect the value of that phenomenon.
Do women in our sample have indeed a lower cholesterol concentration then men, or did the observed difference only occur by chance?
Our confidence interval contains also a zero (0) meaning that it is quite possible that the actual value of the difference will equal zero, namely, that there is no difference between the cholesterol concentration between men and women.
It means that the odds of cholesterol concentration being and not being a risk factor for carotid artery stenosis are even. Reporting of this confidence interval provides additional information about the sample and the results. If you are the legitimate owner of the one of the content we display the wallpaper, and do not want us to show, then please contact us and we will immediately take any action is needed either remove the wallpaper or maybe you can give time to maturity it will limit our wallpaper content view.
More specifically, the meaning of the term "confidence level" is that, if CI are constructed across many separate data analyses of replicated (and possibly different) experiments, the proportion of such intervals that contain the true value of the parameter will match the given confidence level. This value is represented by a percentage, so when we say, "we are 99% confident that the true value of the parameter is in our confidence interval", we express that 99% of the hypothetically observed confidence intervals will hold the true value of the parameter.
By analysing some sample characteristics we actually want to have an overview of the situation in the population.
In this case we use the arithmetic mean as a point estimate of cholesterol concentration in the population.
We repeat sampling for 100 times and each time we calculate a certain arithmetical mean of cholesterol concentration.
Standard error is a standard deviation of sample means, calculated out of hundred random population samples (8). We are interested whether those two groups are different in average cholesterol concentration and is the cholesterol concentration a risk factor for development of carotid artery stenosis. That is, for any cholesterol concentration the odds of that person having or not having carotid artery stenosis are the same.
It is, moreover, above all useful and irreplaceable supplement to a classical hypothesis testing and to the generally accepted P value. All of the content we display the wallpapers are free to download and therefore we do not acquire good financial gains at all or any of the content of each wallpaper.
Whereas two-sided confidence limits form a confidence interval, their one-sided counterparts are referred to as lower or upper confidence bounds. After any particular sample is taken, the population parameter is either in the interval realized or not; it is not a matter of chance. Depending on the confidence level that we choose, the interval margins of error and respective range also change.
The thing we want to know is the answer to the question: can we consider this calculated mean as a good estimate for cholesterol concentration in the population. For small samples the t value is higher than the Z value what logically means that the confidence interval for smaller samples with the same confidence level is larger.
The P value calculated by the t-test is 0.426 proving that the difference of the cholesterol concentration between men and women is not statistically significant. This also confirms the percentage of correctly classified individuals considering the cholesterol concentration (50.41%). It should become a standard of all scientific journals to report key results with respective confidence intervals because it enables better understanding of the data to an interested reader. Therefore, based on what we find out about our sample using the descriptive analysis we make conclusions for the population as a whole (2). Out of a total of hundred related 95%-confidence intervals, 95% of them will contain the actual arithmetical mean of the population (?).
There is a rule for same sized samples: the smaller the confidence level is, the higher is the estimate accuracy. Only half of the individuals are correctly categorised in an adequate group – so the selection does not depend on cholesterol concentration but on a pure chance. If a corresponding hypothesis test is performed, the confidence level is the complement of respective level of significance, i.e. Let us now see how the range of confidence interval and its margins of error change depending on the confidence level in our example of cholesterol concentration estimate in the population (Figure 1).
Many statistical textbooks contain tables with t values for matching confidence level and different degrees of freedom (1). What is hereby understood is the fact that everything we conclude about the sample is reliably applicable to the whole population.
The confidence interval contains the parameter values that, when tested, should not be rejected with the same sample. Greater levels of variance yield larger confidence intervals, and hence less precise estimates of the parameter. Only the studies with a large sample will give a very small confidence interval, which points to high estimate accuracy with a high confidence level. Confidence intervals of difference parameters not containing 0 imply that there is a statistically significant difference between the populations. The P value describes probability that the observed phenomenon (difference) occurred by chance, whereas the confidence interval provides margins of error within which it is possible to expect the value of that phenomenon. In the last twenty years, increasing number of journals require reporting of the confidence intervals for each of their key results.
It is, moreover, very useful and irreplaceable supplement to a classical hypothesis testing and to the generally accepted P value. It should become a standard of all scientific journals to report key results with respective confidence intervals because it enables better understanding to the interested reader. | 2,807 | 14,977 | {"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-2020-05 | latest | en | 0.884152 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=131&t=26395&p=79772 | 1,597,177,152,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738819.78/warc/CC-MAIN-20200811180239-20200811210239-00216.warc.gz | 375,968,544 | 12,214 | ## Entropy of gas vs liquid vs solid
$\Delta S = \frac{q_{rev}}{T}$
Nancy Le - 1F
Posts: 52
Joined: Fri Sep 29, 2017 7:07 am
### Entropy of gas vs liquid vs solid
Why is the entropy of a gas much greater than that of a liquid or solid?
Ava Harvey 2B
Posts: 54
Joined: Fri Sep 29, 2017 7:04 am
### Re: Entropy of gas vs liquid vs solid
The basic idea of why gases have a greater entropy than liquids or solids has to do with the spacing of particles. Gases have a higher entropy because they occupy a greater volume in comparison to liquids and solids. Gas molecules move freely at high speeds and are more spread out, thus they have more room and more positions in which they can be arranged. The more positions or microstates that can be occupied, the greater the entropy. Hope that helps!
Salma Quintanilla 1J
Posts: 29
Joined: Wed Nov 23, 2016 3:02 am
### Re: Entropy of gas vs liquid vs solid
Liquids have more energy and entropy than solids. When there is an increased number of gas molecules, there is also an increase in entropy.
Ya Gao
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time
### Re: Entropy of gas vs liquid vs solid
When the molecules are in gas phase, there are many more possible states they can occupy compared to the possible states molecules can occupy when they are in liquid phase or solid phase. With more possible states, the entropy will be higher.
Brandon Fujii 1K
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am
### Re: Entropy of gas vs liquid vs solid
Whenever a substance is being heated AND is not going through a phase change, the temperature of the substance (average Kinetic Energy of each particle) increases. As the total kinetic energy of the particles of the substance increases, entropy increases. Therefore, because a gas has a higher temperature than that of a solid, the gas has a greater entropy.
Wayland Leung
Posts: 51
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 1 time
### Re: Entropy of gas vs liquid vs solid
What does it mean when there are more possible states a gas can occupy?
RussellChin_3A
Posts: 42
Joined: Sat Jul 22, 2017 3:01 am
### Re: Entropy of gas vs liquid vs solid
By more "possible states" I think you mean more possible positions, as this determines the entropy of a substance. For a gas, it can occupy more possible positions, because like the person above said, gases can occupy a larger volume compared to liquids and solids. They are more able to freely move in a space, therefore can occupy more possible positions.
Justin Folk 3I
Posts: 43
Joined: Wed Sep 21, 2016 2:56 pm
### Re: Entropy of gas vs liquid vs solid
Because gas has more "randomness" by intuition. THere is inherently order in a liquid and solid-- they are all contained and there is only a set place that they can go. Gravity dictates this in a liquid and the lattice of a solid dictates this | 751 | 2,872 | {"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": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.916161 |
http://thetopsites.net/article/53617125.shtml | 1,606,279,650,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181179.12/warc/CC-MAIN-20201125041943-20201125071943-00600.warc.gz | 84,608,880 | 6,730 | ## How to convert a huge list-of-vector to a matrix more efficiently?
list.as.matrix package
convert list to matrix python
r convert list to dataframe
large list to dataframe r
convert list to matrix mathematica
convert to list in r
convert list to numeric matrix r
create matrix from list python
I have a list of length 130,000 where each element is a character vector of length 110. I would like to convert this list to a matrix with dimension 1,430,000*10. How can I do it more efficiently?\ My code is :
```output=NULL
for(i in 1:length(z)) {
output=rbind(output,
matrix(z[[i]],ncol=10,byrow=TRUE))
}
```
This should be equivalent to your current code, only a lot faster:
```output <- matrix(unlist(z), ncol = 10, byrow = TRUE)
```
How to Create a Data Frame from Scratch in R, To convert a huge list of a vector to a matrix, you can use the unlist() function before creating a matrix as follows: output <- matrix(unlist(z), ncol Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Learn more How to convert a huge list-of-vector to a matrix more efficiently?
I think you want
```output <- do.call(rbind,lapply(z,matrix,ncol=10,byrow=TRUE))
```
i.e. combining @BlueMagister's use of `do.call(rbind,...)` with an `lapply` statement to convert the individual list elements into 11*10 matrices ...
Benchmarks (showing @flodel's `unlist` solution is 5x faster than mine, and 230x faster than the original approach ...)
```n <- 1000
z <- replicate(n,matrix(1:110,ncol=10,byrow=TRUE),simplify=FALSE)
library(rbenchmark)
origfn <- function(z) {
output <- NULL
for(i in 1:length(z))
output<- rbind(output,matrix(z[[i]],ncol=10,byrow=TRUE))
}
rbindfn <- function(z) do.call(rbind,lapply(z,matrix,ncol=10,byrow=TRUE))
unlistfn <- function(z) matrix(unlist(z), ncol = 10, byrow = TRUE)
## test replications elapsed relative user.self sys.self
## 1 origfn(z) 100 36.467 230.804 34.834 1.540
## 2 rbindfn(z) 100 0.713 4.513 0.708 0.012
## 3 unlistfn(z) 100 0.158 1.000 0.144 0.008
```
If this scales appropriately (i.e. you don't run into memory problems), the full problem would take about 130*0.2 seconds = 26 seconds on a comparable machine (I did this on a 2-year-old MacBook Pro).
Slow Matrices? Try Lists., How do I convert a list to a vector in R? I have a list of length 130,000 where each element is a character vector of length 110. I would like to convert this list to a matrix with dimension 1,430,000*10.
It would help to have sample information about your output. Recursively using `rbind` on bigger and bigger things is not recommended. My first guess at something that would help you:
```z <- list(1:3,4:6,7:9)
do.call(rbind,z)
```
See a related question for more efficiency, if needed.
LongVectors: Long Vectors, How do I convert a vector to a Dataframe in R? Sometimes in R, I have a list z of length with each element being a vector of the same length .I want to convert this list into an matrix, where each element of the list corresponds to a column in the matrix.
You can also use,
```output <- as.matrix(as.data.frame(z))
```
The memory usage is very similar to
```output <- matrix(unlist(z), ncol = 10, byrow = TRUE)
```
Which can be verified, with `mem_changed()` from `library(pryr)`.
list.as.matrix: Convert list to matrix in sprof: Profiling, timing and , How do I convert a list into a Dataframe in R? Hi all, I'm looking for a way to automatically convert a list of values into a matrix. For example, A1 A1 1000 A1 A2 998 A1 A3 468 A1 A4 491 A2 A1 998 A2 A2 1000 A2 A3 464 A2 A4 488 A3 A1 468 A3 A2 464 A3 A3 1000 A3 A4 992 A4 A1 491 A4 A2 488 A4 A3 992 A4 A4 1000 should be converted to A1
you can use as.matrix as below:
```output <- as.matrix(z)
```
Using vectors and matrices in R, List data structures can be more efficient than their matrix and array counterparts. But data frames have big performance problems of their own. LIST) will transform any list containing equal length vectors into a matrix, Goal: from a list of vectors of equal length create a matrix where each vector becomes a row. Example: > a <- list() 10 1 2 3 4 5
1. Vectors, Matrices, and Arrays, Vectors of 2^31 or more elements were added in R 3.0.0. Currently all atomic (raw, logical, integer, numeric, complex, character) vectors, lists and expressions can Arrays (including matrices) can be based on long vectors provided each of and even then factors will be more efficient (4 bytes per element rather than 8). I have a data.frame that looks like this. x a 1 . x b 2 . x c 3 . y a 3 . y b 3 . y c 2 . I want this in matrix form so I can feed it to heatmap to make a plot.
Subsetting · Advanced R., a list of numeric vectors. byrow. boolean. Arrange list entries as rows. Default is to use columns. filler. Say you want to convert a matrix to a list, where each element of the list contains one column. list() or as.list() obviously won't work, and until now I use a hack using the behavior of tapply : x <- matrix(1:10,ncol=2)
Chapter 2 Working with vectors, matrices, and arrays, It was mentioned earlier that all the elements of a vector must be of the same mode. as a list that can store different types of objects without having to change their with data of a single mode, a matrix may be more efficient than a data frame. data frame, and is very handy to check to make sure a large data.frame really The built-in matrix function has the nice option to enter data byrow.Combine that with an unlist on your source list will give you a matrix. We also need to specify the number of rows so it can break up the unlisted data.
• Wait- when you say that each entry has 11 characters, you mean that it is a vector with 11 items? I originally thought that each was a string with 11 characters in it. Can you show `z[1:2]` as an example?
• @user1787675: I still don't understand. What is an "entry"? Is it a vector? Can you show `z[1:2]`?
• +1, but I'd recommend setting `USE.NAMES=FALSE` in `unlist` in order to save time and memory.
• It should be `use.names` (i.e. in lowercase).
• Just to clarify, it should be `output <- matrix(unlist(z), ncol = 10, byrow = TRUE, use.names=FALSE)` to be the most efficient. | 1,735 | 6,274 | {"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} | 2.921875 | 3 | CC-MAIN-2020-50 | latest | en | 0.669266 |
https://corplingstats.wordpress.com/2012/11/20/algorithm-snippets/ | 1,591,322,248,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492427.71/warc/CC-MAIN-20200605014501-20200605044501-00459.warc.gz | 295,135,090 | 25,110 | # Binomial algorithm snippets
### Introduction
Elsewhere on this blog I summarise an analysis of the performance of a broad range of different confidence interval calculations and 2 × 2 contingency tests against equivalent ‘exact’ Binomial tests calculated from first principles.
For transparency, it is necessary to show how I went about computing these results.
Many of these algorithms are summarised in mathematical terms in this paper. However, for those who wish to recreate the computation, here is the code in the programming language C.
Warning: colleagues have pointed out that this post is not for the faint hearted!
### Binomial test
Precalculated
• long double array fact[n] contains factorial of n, written n!
A ‘long double’ is a 64-bit floating point number, which can express very large numbers. Since the factorial of n is the product of the series 1…n, it can get very large (see below). However with 64 bits we have a limit of n=1,500, which is more than enough. A mere ‘double’ is a 32 bit floating point number, and a ‘float’ a 16 bit one.
Core functions
• combinatorial function nCr
• single Binomial probability BinP(p, n, x)
• cumulative Binomial probability CumBinP(p, n, x)
• population Binomial probability PopBinP(n, x)
```long double nCr(int n, int r)
{
if (n>LIMIT) return -1;
return fact[n]/(fact[r]*fact[n-r]);
}```
In other words, provided n is not too large, define the combinatorial function nCr (see footnote 5 in the Binomial paper) as follows:
nCr = n!/r!{– r}!
This is a fraction of two very large numbers, so we tell C to use 64 bit arithmetic to be as accurate as possible. Bear in mind that for this paper we need to deal with n = 500:
500! = fact[500] = 1.22013682599111007 × 101134.
Other functions are defined similarly.
Here are the single Binomial probability, equation (5), and equation (6), representing the cumulative Binomial probability. pow is the power function.
```long double BinP(float p, int n, int x)
{
if (p=1) return 0;
return nCr(n, x) * pow(p,x) * pow(1-p,n-x);
}
long double CumBinP(float p, int n, int x)
{
register int r;
register long double c = 0;
for (r = x; r<=n; r++)
c += BinP(p, n, r);
return c;
}```
Although this code can seem forbidding, in fact it is just another way of writing the mathematical formulae in the paper. Instead of using the summation sign in the above, we perform an explicit loop where r increases from x to n, and add up the result. The code ‘r++’ simply means add one to r.
Finally, here is one last function: to compute the population cumulative Binomial. I include this because we need a way of calculating the inverse of equation (6). This simply adds up single Binomial probabilities until it exceeds Pcrit = α/2.
```float popCumBinP(int n, int x)
{
register int r;
register double c = 0;
float p = x/(float)n;
for (r=0; r Pcrit)
return (r-1)/(float)n;
}
return (r-1)/(float)n;
}```
### Find Binomial P
Input
• p = observed o / total n.
This function requires a bit more explanation. It performs a simple search method to find P for an exact tail area Pcrit = α/2. In the paper at a number of places we refer to “a search procedure”. I thought therefore I should give an example of what such a procedure might look like.
The algorithm is accurate to 4 decimal places (it continues while absolute difference d > 0.00001), with a for loop providing a stopping condition.
The way it works is it picks a value for P and tests it. If the result is too high, it tries a lower value of P; if too low, it tries a higher value of P. Provided that the function we are searching is monotonic (always increasing or always decreasing) this type of procedure can be guaranteed to zoom in on the correct value if you keep dividing by 2.
Note that this process is quite computationally expensive, performing the CumBinP function potentially hundreds of times with different values of P until it converges on the value. This is not the optimum approach, as it is possible to use the gradient to improve the estimate of P and increase the rate of convergence, but it is robust, which is more important here.
```float findBinP(int n, int o)
{
float p2 = o/(float)n, p = p2/1.5;
long double a, d, a2, a3, d2, d3, pdiff;
int i;
for (i=0;i0.00001) // accurate to 4 dps
{
if (p+p2>1)
d2 = 1;
else
{
a2 = CumBinP(p+p2, n, o);
d2 = fabsl(a2-Pcrit);
}
if (p<p2)
d3 = 1;
else
{
a3 = CumBinP(p-p2, n, o);
d3 = fabsl(a3-Pcrit);
}
pdiff = p2;
if (d3 < d2)
{
pdiff=-p2;
d2 = d3;
}
if (d2 < d)
p+=pdiff;
else
p2 = p2/2;
}
else return p;
}
return p;
}```
### Fisher sum test
Core functions
• Fisher point probability Fisher(a, b, c, d)
• Fisher sum probability FisherSum(a, b, c, d)
These are equations (13) and (14) in the paper. I have included these because these are two of the trickier calculations to perform.
```long double Fisher(int a, int b, int c, int d)
{
return (fact[a+c]/fact[a]) * // arranged like this
(fact[b+d]/fact[b]) * // to minimise overflow error
(fact[a+b]/fact[c]) *
(fact[c+d]/fact[d]) / fact[a+b+c+d];
}```
The FisherSum function (equation 14) requires some explanation.
Take a matrix [[a, b], [c, d]].
a b c d
The instruction is to sum Fisher probabilities for that pattern and add this to the Fisher probabilities for each more extreme version of the matrix, noting that all the time a+b, c+d, a+c and b+d must be constant. Consider the matrix below, where row and column totals both equal 20.
5 15 15 5
A “more extreme” version of this matrix which satisfies the condition that the row and column totals sum to 20 is the following.
4 16 16 4
This is further away from the flat matrix (all cells equal to 10), but the totals are still locked to 20. So, the way to enumerate all versions of the matrix which are more extreme is to traverse the matrix diagonally, either by increasing a and d and decreasing b and c, or the opposite (as above), until no other combination is possible. We keep adding individual Fisher probabilities until we hit the limit below.
0 20 20 0
To make this clearer, we can define row totals n₁ = a+b and n₂ = c+d.
```float FisherSum(int a, int b, int c, int d)
{
double p = 0;
int n1 = a+b, n2 = c+d;
if (a/(float)n1 > c/(float)n2)
for (;(a=0);a++,c--)
p+=Fisher(a,n1-a,c,n2-c);
else
for (;(a>=0)&&(c<=n2);a--,c++)
p+=Fisher(a,n1-a,c,n2-c);
success = p<=Pcrit;
return p;
}``` | 1,770 | 6,354 | {"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 | 4 | CC-MAIN-2020-24 | latest | en | 0.847802 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/chapter-7-chapter-test-page-554/2 | 1,686,342,594,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00465.warc.gz | 867,404,595 | 11,738 | ## Algebra and Trigonometry 10th Edition
$1$
We know that $\sin^2 x+\cos^2 x=1$ $\csc^2 \beta (1-\csc^2 \beta)=\csc^2 \beta \sin^2 \beta$ or, $=(\dfrac{1}{\sin^2 \beta}) \times \sin^2 \beta$ or, $\csc^2 \beta (1-\csc^2 \beta)=1$ | 109 | 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": 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} | 3.515625 | 4 | CC-MAIN-2023-23 | latest | en | 0.512516 |
https://www.ukessays.com/essays/computer-science/the-principle-of-duality-states-computer-science-essay.php | 1,596,900,901,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737883.59/warc/CC-MAIN-20200808135620-20200808165620-00128.warc.gz | 878,641,120 | 15,214 | # The Principle Of Duality States Computer Science Essay
2246 words (9 pages) Essay
1st Jan 1970 Computer Science Reference this
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To simplify a SOP for a Boolean expression using a K map, first identify all the input combinations that produce an output of logic level 1 and place them in their appropriate K map cell. Consequently, all other cells must contain zero (0). Second, group the adjacent cells that contain 1 in a manner that maximizes the size of the groups but also minimizes the total number of groups. All 1’s in the output must be included in a group even if the group is only one cell. Third, as each SOP term represents an AND expression, each (AND) grouping is written with only the input variables that are common to the group. Finally, the simplified expression is formed by ORing each of the (AND) groups. To illustrate let us consider the function X =A’BC+ABC’+ABC whose truth table and K map are illustrated below:
Truth table specifications for a logic function may not to include all possible combinations of the input binary digits for the input variables, yet they may still be complete specifications of the logic function for the prescribed application. In these situations certain input combinations will not occur due to the nature of the application. When the input combinations are irrelevant or cannot occur, the output states are in the Truth table and the K map are filled with an X and are referred to as don’t care states.
When simplifying K maps with don’t care states, the contents of the undefined cells (1 or 0) are chosen according to preference. The aim is to enlarge group sizes thereby eliminating as many input variables from the simplified expression as possible. Only those X’s that assist in simplifying the function should be included in the groupings. No additional X’s should be added that would result in additional terms in the expression.
## THE TABULATION METHOD
The K-map method is convenient as long as the number of variables does not exceed five or six, but when the number increases it becomes difficult to use this method. The tabulation method overcomes this difficulty, besides it is suitable for computer mechanisation. It was first formulated by Quine and later improved by McCluskey. It is also known as the Quine-McCluskey method.
The tabular method of simplification consits of two parts:
Determination of Prime Implicants.
Selection of Essential Prime Implicants.
The first part is covered in the software provided, the second part is not. Each minterm and its combined cells are included in the Minterm record, and linked lists techniques are used. Linked lists are the best way to manage unknown amount of data at run time instead of using a large array which is a waste of memory. The record used is:
Pminterm = ^Minterm;
Minterm = Record
Number : Word; {This is to hold the decimal equivalent of the minterm}
NumberOfOnes : Byte; {This is to hold the no. of 1s in the field Number}
Selected : Boolean; {used as a tick when this term is selected}
CellStr : String; { a string to hold all minterms that form a cell with this term }
DashPlaceStr : String {a string to hold the place of the dash }
Next : Pminterm { a pointer to the next record to form a linked list}
End;
The software procedure summarise as follows:
The user enters the minterms in decimal equivalent.
The program sorts minterms in the number of ones included in the binary equivalent (SortMinTerms).
Any two minterms that differ from each other by only one variable are combined (done in FirstTabulation procedure), two minterms fit into this category if the number in the lower group is greater than that in the upper one, and the two numbers differ by a power of 2 e.g. ( 2d = 0010b and 10d = 1010b the difference is 8d which is a power of 2). Then the two numbers are copied to the second linked list ( First, Second, Vertex : Pminterm;). This procedure is carried for all the minterms. Matching groups are copied to second linked list while first one deleted.
Then a second tabulation is carried in the SecondTabulation procedure, here each cell contained in the CellStr field of the Minterm record are compared together, a matching is found if the numbers in one cell are greater than the other one and they differs by a power of 2 e.g. (0,2 and 8,10 differs by 8 which is a power of two) . This procedure is carried for all the records in the second linked list and matching cells are copied to the first linked list, and any incomparable cells are copied to the Vertex linked list. Then the second linked list is deleted.
The step 4 is repeated m – 1 time where m is the number of inputs, transferring cells between first and second linked lists.
The last linked list and the vertex linked list prime implicants are printed in the ABC equivalent using the WMterm procedure.
My program architecture can be used to simplify any number of inputs, but practical limitations are in the Number field in the Minterm record which is a word i.e. 16 bits (16 inputs). Bearing in mind as well the simplified expression is contained in a string which is only 255 characters.
Some inputs and corresponding outputs to try on the program:
Boolean Expression
Simplified Boolean Expression
F(A,B,C) = S(0,6,7)
A’B’C’ + AB
F(A,B,C) = S(0,4,6,7)
B’C’ + AB + AC’
F(A,B,C,D,E,F) = S(4,5,6,7,36,37,38,39)
B’C’D
F(A,B,C,D,E,F,G,H) = S(16,17 up to 31,144,145 up to 159)
B’C’D
## VARIABLE ENTERED METHOD (VEM)
The conventional logic minimization is time consuming & easy only for 4-5 variables. It becomes difficult to solve using conventional K-map method when number of variables goes on increasing. In that case Variable Entered Method will be good idea to use. It represents values of functions in terms of its variables called map entered variables.
A new method for obtaining a compact subsumptive general solution of a system of Boolean equations is presented. The method relies on the use of the variable-entered Karnaugh map (VEKM) to achieve successive elimination through successive map folding. It is superior in efficiency and simplicity to methods employing Marquand diagrams or Conventional Karnaugh maps; it requires the construction of significantly smaller maps and produces such maps in a minimization-ready form. Moreover, the method is applicable to general Boolean equations and is not restricted to the two-valued case.
## Combinational Sequential Circuits
Combinational logic circuits implement Boolean functions. Boolean functions are mappings of input bitstrings to output bitstrings. These circuits are functions of input only.
What does that mean? It means that if you feed in an input to a circuit, say, 000, then look at its output, and discover it is, say, 10, then the output will always be 10 for that circuit, if 000 is the input. 000 is mapped to 10.
If that value were not the same every single time, then the output must not completely depend on 000. Something else must be affecting the output. Combinational logic circuits always depend on input.
Another way to define something that is a function of input is to imagine that you are only allowed to use input variables xk-1,…,x0, i.e. data inputs, cm-1,…,c0, i.e., control inputs, to write the function. This function can not depend on global variables or other variables.
Like combinational logic circuits, a sequential logic circuit has inputs (labelled with x with subscripts) and outputs (labelled with z with subscripts).
Unike combinational logic circuits, a sequential logic circuit uses a clock.
Also, there is a box inside the circuit called State.
This box contains flip flops. Assume it has k flip flops. The flip flops basically store a k-bit number representing the current state.
The output z is computed based on the inputs (x with subscripts) and the state coming out of the state box (q with subscripts).
The state may be updated at each positive clock edge. When there’s not a positive clock edge, the state remains unchanged.
The information needed to update to the state (called the next state) comes from the current state (the current value of q) and the input, which is fed through combinational logic, and fed back into the state box, telling the state box how to update itself.
A sequential circuit uses flip flops. Unlike combinational logic, sequential circuits have state, which means basically, sequential circuits have memory.
The main difference between sequential circuits and combinational circuits is that sequential circuits compute their output based on input and state, and that the state is updated based on a clock. Combinational logic circuits implement Boolean functions, so they are functions only of their inputs, and are not based on clocks.
## The S-R Latch
A bistable multivibrator has two stable states, as indicated by the prefix bi in its name. Typically, one state is referred to as set and the other as reset. The simplest bistable device, therefore, is known as a set-reset, or S-R, latch.
To create an S-R latch, we can wire two NOR gates in such a way that the output of one feeds back to the input of another, and vice versa, like this:
The Q and not-Q outputs are supposed to be in opposite states. I say “supposed to” because making both the S and R inputs equal to 1 results in both Q and not-Q being 0. For this reason, having both S and R equal to 1 is called an invalid or illegal state for the S-R multivibrator. Otherwise, making S=1 and R=0 “sets” the multivibrator so that Q=1 and not-Q=0. Conversely, making R=1 and S=0 “resets” the multivibrator in the opposite state. When S and R are both equal to 0, the multivibrator’s outputs “latch” in their prior states.
## The Clocked D-Latch
Since the enable input on a gated S-R latch provides a way to latch the Q and not-Q outputs without regard to the status of S or R, we can eliminate one of those inputs to create a multivibrator latch circuit with no “illegal” input states. Such a circuit is called a D latch, and its internal logic looks like this:
Note that the R input has been replaced with the complement (inversion) of the old S input, and the S input has been renamed to D. As with the gated S-R latch, the D latch will not respond to a signal input if the enable input is 0 — it simply stays latched in its last state. When the enable input is 1, however, the Q output follows the D input.
Since the R input of the S-R circuitry has been done away with, this latch has no “invalid” or “illegal” state. Q and not-Q are always opposite of one another.
## Master-Slave Flip-Flops
A master-slave flip-flop is constructed from two seperate flip-flops. One circuit serves as a master and the other as a slave. The logic diagram of an SR flip-flop is shown in figure below. The master flip-flop is enabled on the positive edge of the clock pulse CP and the slave flip-flop is disabled by the inverter. The information at the external R and S inputs is transmitted to the master flip-flop. When the pulse returns to 0, the master flip-flop is disabled and the slave flip-flop is enabled. The slave flip-flop then goes to the same state as the master flip-flop.
http://wearcam.org/ece385/lectureflipflops/flipflops/fig9.gif
Logic diagram of a master-slave flip-flop
The timing relationship is shown in Figure below and is assumed that the flip-flop is in the clear state prior to the occurrence of the clock pulse. The output state of the master-slave flip-flop occurs on the negative transition of the clock pulse. Some master-slave flip-flops change output state on the positive transition of the clock pulse by having an additional inverter between the CP terminal and the input of the master.
http://wearcam.org/ece385/lectureflipflops/flipflops/fig10.gif
Timing relationship in a master slave flip-flop
## Edge Triggered Devices
Another type of flip-flop that synchronizes the state changes during a clock pulse transition is the edge-triggered flip-flop. When the clock pulse input exceeds a specific threshold level, the inputs are locked out and the flip-flop is not affected by further changes in the inputs until the clock pulse returns to 0 and another pulse occurs. Some edge-triggered flip-flops cause a transition on the positive edge of the clock pulse (positive-edge-triggered), and others on the negative edge of the pulse (negative-edge-triggered). The logic diagram of a D-type positive-edge-triggered flip-flop is shown in figure below:
http://wearcam.org/ece385/lectureflipflops/flipflops/fig11.gif
D-type positive-edge triggered flip-flop
When using different types of flip-flops in the same circuit, one must ensure that all flip-flop outputs make their transitions at the same time, ie., during either the negative edge or the positive edge of the clock pulse.
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250
Direction: In these questions, the relationship between different elements is shown in the statements. These statements are followed by three conclusions. Mark answer.
Q.1
Statements: a ≥ b ≠ c = d ≮ e
Conclusions:
I. c = e
II. c ≥ e
III. a > d
(A) I and II follow
(B) All follow
(C) Either I or II follows.
(D) None follow.
(E) Only II follow.
Q.2
Statements: a ≠ b ≱ c = d ≤ e ≯ f
Conclusions:
I. c ≤ f
II. b < e
III. a > d
(A) I and III follow.
(B) Only II follow.
(C) All follow.
(D) I and II follow
(E) None of these.
Q.3
Statements: a ≮ b ≤ c ≠ d > e ≱ f
Conclusions:
I. a < c
II. d > f
III. c = e
(A) I and II follow.
(B) All follow.
(C) Only I follow.
(D) II and III follow.
(E) None follow.
Q.4
Statements: a = b < c ≯ d ≠ e ≥ f ≮ g
Conclusions:
I. a ≱ d
II. d > f
III. b < f
(A) Only I follow.
(B) None follow.
(C) I and III follow.
(D) Only II follow.
(E) None of these.
Q.5
Statements: a < b ≯ c ≱ d = e ≤ f < g
Conclusions:
I. a < d
II. d ≤ f
III. b ≱ g
(A) II and III follow.
(B) None follow.
(C) All follow.
(D) I and II follow.
(E) None of these
Ans 1. (C) Either I or II follow.
Solution:
a ≥ b > < c = d ≥ e
Ans 2. (D) I and II follow
Solution:
a > < b < c = d ≤ e ≤ f
Ans 3. (E) None follow.
Solution:
a ≥ b ≤ c > < d > e < f
Ans 4. (A) Only I follow.
Solution:
a = b < c ≤ d > < e ≥ f ≥ g
Ans 5. (C) All follow.
Solution:
a < b ≤ c < d = e ≤ f < g | 604 | 1,532 | {"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} | 3.875 | 4 | CC-MAIN-2020-16 | latest | en | 0.64133 |
https://cn.maplesoft.com/support/help/maple/view.aspx?path=componentLibrary%2FsignalBlocks%2Fcontrollers%2FPID | 1,709,336,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00421.warc.gz | 176,509,356 | 23,263 | PID
Description
The PID component models an ideal proportional-integral-derivative (PID) controller. This controller differs from the ones normally used in the industrial setup as the proportional gain, k, is decoupled from the system. This does not affect the integral and the derivative gain performance. For a more practically useful PID-controller, use the Lim PID component.
The equation of this controller is $y=k\left(1+\frac{1}{{T}_{i}s}+{T}_{d}s\right)u$ where the derivative is estimated numerically based on the time step, ${N}_{d}$. The greater the value of ${N}_{d}$, the better the estimate of the derivative.
The Signal Size parameter allows the block to operate on a vector of signals rather than a single signal.
Initialization
The PID block can be initialized in different ways controlled by parameter initType. DoNotUse_InitialIntegratorState is added for backward compatibility reasons.
initType Integrator initType Derivative initType NoInit NoInit NoInit SteadyState SteadyState SteadyState InitialState InitialState InitialState InitialOutput and initial equation: y = y_start NoInit SteadyState DoNotUse_InitialIntegratorState InitialState NoInit
In many cases, the most useful initial condition is SteadyState because initial transients are no longer present. If $\mathrm{initType}=\mathrm{InitPID}\cdot \mathrm{SteadyState}$, then in some cases difficulties might occur. The reason is the equation of the integrator, $\stackrel{.}{y}=ku$. The steady state equation, $\stackrel{.}{x}=0$ leads to the condition that the input to the integrator is 0. If the input $u$ is already (directly or indirectly) defined by another initial condition, the initialization problem is singular (that is, has none or infinitely many solutions). This situation occurs often in mechanical systems, where, for example, $u=\mathrm{desiredSpeed}-\mathrm{measuredSpeed}$. Because speed is both a state and a derivative, it is natural to initialize it with 0. As sketched, this is not possible. The solution is to not initialize ${u}_{m}$ or the variable that is used to compute ${u}_{m}$ by an algebraic equation. If the parameter Limits At Initial is $\mathrm{false}$, the limits at the output of this component are removed from the initialization problem, which leads to a much simpler equation system. After initialization has been performed, it is checked with an assert whether the output is in the defined limits.
Connections
Name Description Modelica ID $u$ Real input vector u $y$ Real output vector y
Parameters
Name Default Units Description Modelica ID Signal Size $1$ Dimension of input and output signals signalSize k $1$ Gain k ${T}_{i}$ $0.5$ $s$ Time Constant of Integrator Ti ${T}_{d}$ $0.1$ $s$ Time Constant of Derivative block Td ${N}_{d}$ $10$ The greater ${N}_{d}$, the more ideal the derivative component Nd Initial Values NoInit Type of initialization; see Initialization section initType ${x}_{{i}_{0}}$ $0$ Initial or guess value for the output of the integrator component xi_start ${x}_{{d}_{0}}$ $0$ Initial or guess value for state of derivative component xd_start ${y}_{0}$ $0$ Initial value of output y_start | 744 | 3,157 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 30, "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-2024-10 | latest | en | 0.822315 |
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```Chapter 8: Quadratic Equations Exercise – 8.13
Question: 1
Find the roots of the equation (x - 4) (x + 2) = 0
Solution:
The given equation is (x - 4)(x + 2) = 0
Either x - 4 = 0 therefore x = 4
Or, x + 2 = 0 therefore x = -2
The roots of the above mentioned quadratic equation are 4 and -2 respectively.
Question: 2
Find the roots of the equation (2x + 3)(3x - 7) = 0
Solution:
The given equation is (2x + 3)(3x - 7) = 0.
Either 2x + 3 = 0, therefore x = (-3)/2
Or, 3x -7 = 0, therefore x = 7/3
The roots of the above mentioned quadratic equation are x = (-3)/2 and x =7/3 respectively.
Question: 3
Find the roots of the quadratic equation 3x2 - 14x - 5 = 0
Solution:
The given equation is 3x2 - 14x - 5 = 0
= 3x2 - 14x - 5 = 0
= 3x2 - 15x + x - 5 = 0
= 3x(x - 5) + 1(x - 5) = 0
= (3x + 1)(x - 5) = 0
Either 3x + 1 = 0 therefore x = (-1)/3
Or, x - 5 = 0 therefore x = 5
The roots of the given quadratic equation are 5 and x = (-1)/3 respectively.
Question: 4
Find the roots of the equation 9x2 - 3x - 2 = 0.
Solution:
The given equation is 9x2 - 3x - 2 = 0.
= 9x2 - 3x - 2 = 0.
= 9x2 - 6x + 3x - 2 = 0
= 3x(3x - 2) + 1(3x - 2) = 0
= (3x - 2)(3x + 1) = 0
Either, 3x - 2 = 0 therefore x = 2/3
Or, 3x + 1 = 0 therefore x = (-1)/3
The roots of the above mentioned quadratic equation are x = 2/3 and x = (-1)/3 respectively.
Question: 5
Find the roots of the quadratic equation
Solution:
The given equation is
Cancelling out the like terms on both the sides of the numerator. We get,
= x2 + 4x - 5 = 7
= x2 + 4x - 12 = 0
= x2 + 6x - 2x - 12 = 0
= x(x + 6) - 2(x - 6) = 0
= (x + 6)(x - 2) = 0
Either x + 6 = 0
Therefore x = -6
Or, x - 2 = 0
Therefore x = 2
The roots of the above mentioned quadratic equation are 2 and – 6 respectively.
Question: 6
Find the roots of the equation 6x2 + 11x + 3 = 0.
Solution:
The given equation is 6x2 + 11x + 3 = 0.
= 6x2 + 11x + 3 = 0.
= 6x2 + 9x + 2x + 3 = 0
= 3x(2x + 3) + 1(2x + 3) = 0
= (2x + 3)(3x + 1) = 0
Either, 2x + 3 = 0 therefore x = (-3)/2
Or, 3x + 1 = 0 therefore x = (-1)/3
The roots of the above mentioned quadratic equation are x = (– 3)/2 and x = (– 1)/2 respectively.
Question: 7
Find the roots of the equation 5x2 - 3x - 2 = 0
Solution:
The given equation is 5x2 - 3x - 2 = 0.
= 5x2 - 3x - 2 = 0.
= 5x2 - 5x + 2x - 2 = 0
= 5x(x - 1) + 2(x - 1) = 0
= (5x + 2)(x - 1) = 0
Either 5x + 2 = 0 therefore x = (- 2)/5
Or, x - 1 = 0 therefore x = 1
The roots of the above mentioned quadratic equation are 1 and x = (- 2)/5 respectively.
Question: 8
Find the roots of the equation 48x2 - 13x - 1 = 0
Solution:
The given equation is 48x2 - 13x - 1 = 0.
= 48x2 - 13x - 1 = 0.
= 48x2 - 16x + 3x - 1 = 0.
= 16x(3x - 1) + 1(3x - 1) = 0
= (16x + 1)(3x - 1) = 0
Either 16x + 1 = 0 therefore (-1)/16
Or, 3x - 1 = 0 therefore x = 1/3
The roots of the above mentioned quadratic equation are x = (-1)/16 and x = 1/3 respectively.
Question: 9
Find the roots of the equation 3x2 = – 11x – 10
Solution:
The given equation is 3x2 = – 11x – 10
= 3x2 = – 11x - 10
= 3x2 + 11x + 10 = 0
= 3x2 + 6x + 5x + 10 = 0
= 3x(x + 2) + 5(x + 2) = 0
= (3x + 2)(x + 2) = 0
Either 3x + 2 = 0 therefore x = (- 2)/3
Or, x + 2 = 0 therefore x = -2
The roots of the above mentioned quadratic equation are (-2)/3 and – 2 respectively.
Question: 10
Find the roots of the equation 25x(x + 1) = – 4
Solution:
The given equation is 25x(x + 1) = 4
= 25x(x + 1) = – 4
= 25x2 + 25x + 4 = 0
= 25x2 + 20x + 5x + 4 = 0
= 5x(5x + 4) + 1(5x + 4) = 0
= (5x + 4)(5x + 1) = 0
Either 5x + 4 = 0 therefore x = (- 4)/5
Or, 5x + 1 = 0 therefore x = (- 1)/5
The roots of the quadratic equation are x = (- 4)/5 and x = (- 1)/5 respectively.
Question: 11
Find the roots of the quadratic equation
Solution:
The given equation is
Cross multiplying both the sides. We get,
= 2 = 3x(x - 2)
= 2 = 3x2 - 6x
= 3x2 - 6x - 2 = 0
= 3x2 - 3x - 3x - 2 = 0
The roots of the above mentioned quadratic equation areandrespectively.
Question: 12
Find the roots of the quadratic equation x − 1/x = 3
Solution:
The given equation is x − 1/x = 3
= x2 - 1 = 3x
= x2 - 1 - 3x = 0
The roots of the above mentioned quadratic equation areandrespectively.
Question: 13
Find the roots of the quadratic equation
Solution:
The given equation is
Cancelling out the like numbers on both the sides of the equation
= x2 - 3x - 28 = -30
= x2 - 3x - 2 = 0
= x2 - 2x - x - 2 = 0
= x(x - 2) - 1(x - 2) = 0
= (x - 2)(x - 1) = 0
Either x - 2 = 0
Therefore x = 2
Or, x - 1 = 0
Therefore x = 1
The roots of the above mentioned quadratic equation are 1 and 2 respectively.
Question: 14
Find the roots of the quadratic equation a2x2 - 3abx + 2b2 = 0
Solution:
The given equation is a2x2 - 3abx + 2b2 = 0
= a2x2 - 3abx + 2b2 = 0
= a2x2 - abx - 2abx + 2b2 = 0
= ax(ax - b) - 2b(ax - b) = 0
= (ax - b)(ax - 2b) = 0
Either ax - b = 0 therefore x = b/a
Or, ax - 2b = 0 therefore x = 2b/a
The roots of the quadratic equation are x = 2b/a and x = b/a respectively.
Question: 15
Find the roots of the 4x2 + 4bx - (a2 - b2) = 0
Solution:
= - 4(a2 - b2) = - 4(a - b)(a + b)
= -2(a - b) * 2(a + b)
= 2(b - a) * 2(b + a)
= 4x2 + (2(b - a) + 2(b + a)) – (a - b)(a + b) = 0
= 4x2 + 2(b - a)x++ 2(b + a)x + (b - a)(a + b) = 0
= 2x(2x + (b - a)) + (a + b)(2x + (b - a)) = 0
= (2x + (b - a))(2x + b + a) = 0
Either, (2x + (b - a)) = 0
Or, (2x + b + a) = 0
The roots of the above mentioned quadratic equation areandrespectively.
Question: 16
Find the roots of the equation ax2 + (4a2 - 3b)x - 12ab = 0
Solution:
The given equation is ax2 + (4a2 - 3b)x - 12ab = 0
= ax2 + (4a2 - 3b)x - 12ab = 0
= ax2 + 4a2x - 3bx -12ab = 0
= ax(x - 4a) – 3b(x - 4a) = 0
= (x - 4a)(ax - 4b) = 0
Either x - 4a = 0
Therefore x = 4a
Or, ax - 4b = 0
Therefore x = 4b/a
The roots of the above mentioned quadratic equation are x = 4b/a and 4a respectively.
Question: 17
Find the roots of
Solution:
The given equation is
= (x + 3)(2x - 3) = (x + 2)(3x - 7)
= 2x2 - 3x + 6x - 9 = 3x2 - x - 14
= 2x2 + 3x - 9 = 3x2 - x - 14
= x2 - 3x - x - 14 + 9 = 0
= x2 - 5x + x - 5 = 0
= x(x - 5) + 1(x - 5) = 0
= (x - 5)(x + l) - 0
Either x - 5 - 0 or x + 1= 0
x = 5 and x = –1
The roots of the above mentioned quadratic equation are 5 and -1 respectively.
Question: 18
Find the roots of the equation
Solution:
The given equation is
= 3(4x2 - 19x + 20) = 25(x2 - 7x + 12)
= 12x2 - 57x + 60 = 25x2 – 175x + 300
= 13x2 - 78x - 40x + 240 = 0
= 13x2 - 118x + 240 = 0
= 13x2 - 78x - 40x + 240 = 0
= 13x(x - 6) - 40(x - 6) = 0
= (x - 6)(13x - 40) = 0
Either x - 6 = 0 therefore x = 6
Or , 13x - 40 = 0 therefore x = 40/13
The roots of the above mentioned quadratic equation are 6 and 40/13 respectively.
Question: 19
Find the roots of the quadratic equation
Solution:
The given equation is
= 4(2x2 + 2) = 17(x2 - 2x)
= 8x2 + 8 = 17x2 - 34x
= 9x2 - 34x - 8 = 0
= 9x2 - 36x + 2x - 8 = 0
= 9x(x - 4) + 2(x - 4) = 0
= (9x + 2)(x - 4) = 0
Either 9x + 2 = 0 therefore x = (-2)/9
Or, x - 4 = 0 therefore x = 4
The roots of the above mentioned quadratic equation are x = (-2)/9 and 4 respectively.
Question: 20
Find the roots of the quadratic equation
Solution:
The equation is
= x(3x - 5) = 6(x2 - 3x + 2)
= 3x2 - 5x = 6x2 - 18x + 12
= 3x2 - 13x + 12 = 0
= 3x2 - 9x - 4x + 12 = 0
= 3x(x - 3) - 4(x - 3) = 0
= (x - 3)(3x - 4) = 0
Either x - 3 = 0 therefore x = 3
Or, 3x - 4 = 0 therefore 4/3
The roots of the above mentioned quadratic equation are 3 and 4/3 respectively.
Question: 21
Find the roots of the quadratic equation
Solution:
The equation is
= 6(4x) = 5(x2 - 1)
= 24x = 5x2 - 5
= 5x2 - 24x - 5 = 0
= 5x2 - 25x + x - 5 = 0
= 5x(x - 5) + 1(x - 5) = 0
= (5x + 1)(x - 5) = 0
Either x - 5 = 0
Therefore x = 5
Or, 5x + 1 = 0
Therefore x = (-1)/5
The roots of the above mentioned quadratic equation are x = (-1)/5 and 5 respectively.
Question: 22
Find the roots of the quadratic equation
Solution:
The equation is
= 2(5x2 + 2x + 2) = 5(2x2 - x - 1)
= 10x2 + 4x + 4 = 10x2 - 5x - 5
Cancelling out the equal terms on both sides of the equation. We get,
= 4x + 5x + 4 + 5 = 0
= 9x + 9 = 0
= 9x = – 9
x = –1
x = – 1 is the only root of the given equation.
Question: 23
Find the roots of the quadratic equation
Solution:
The given equation is
Now we solve the above quadratic equation using factorization method
Therefore,
Now, one of the products must be equal to zero for the whole product to be zero for the whole product to be zero. Hence, we equate both the products to zero in order to find the value of x.
Therefore,
Or,
The roots of the above mentioned quadratic equation are andrespectively.
Question: 24
Find the roots of the quadratic equation
Solution:
The given equation is
= (2x2 - 2x(a + b) + a2 + b2)ab = (a2 + b2)(x2 - (a + b)x + ab)
= (2abx2 - 2abx(a + b) + ab(a2 + b2)) = (a2 + b2)(x2 - (a2 + b2)(a + b)x + (a2 + b2)(ab)
= (a2 + b2 - 2ab)x - (a + b)(a2 + b2 - 2ab)x = 0
= (a - b)2x2 - (a + b)(a + b)2x2 = 0
= x(a - b)2(x - (a + b)) = 0
= x(x - (a + b)) = 0
Either x = 0
Or, (x - (a + b)) = 0
Therefore x = a + b
The roots of the above mentioned quadratic equation are 0 and a + b respectively.
Question: 25
Find the roots of the quadratic equation
Solution:
The given equation is
Cancelling out the like terms on both the sides of numerator and denominator. We get,
= (x - 1)(x - 4) = 18
= x2 - 4x - x + 4 = 18
= x2 - 5x - 14 = 0
= x2 - 7x + 2x - 14 = 0
= x(x - 7) + 2(x - 7) = 0
= (x - 7)(x + 2) = 0
Either x - 7 = 0
Therefore x = 7
Or, x + 2 = 0
Therefore x = – 2
The roots of the above mentioned quadratic equation are 7 and – 2 respectively.
Question: 26
Find the roots of the quadratic equation
Solution:
The given equation is
= (x - c)(ax - 2ab + bx) = 2c(x2 - bx - ax + ab)
= (a + b)x2 - 2abx - (a + b)cx + 2abc = 2cx2 - 2c(a + b)x + 2abc
Question: 27
Find the roots of the Question x2 + 2ab = (2a + b)x
Solution:
The given equation is x2 + 2ab = (2a + b)x
= x2 + 2ab = (2a + b)x
= x2 - (2a + b)x + 2ab = 0
= x2 - 2ax - bx + 2ab = 0
= x(x - 2a) - b(x - 2a) = 0
= (x - 2a)(x - b) = 0
Either x - 2a = 0
Therefore x = 2a
Or, x - b = 0
Therefore x = b
The roots of the above mentioned quadratic equation are 2a and b respectively.
Question: 28
Find the roots of the quadratic equation (a + b)2x2 - 4abx - (a - b)2 =0
Solution:
The given equation is (a + b)2x2 - 4abx - (a - b)2 = 0
= (a + b)2x2 - 4abx - (a - b)2 = 0
= (a + b)2x2 - ((a + b)2 - (a-b)2 )x - (a - b)2 = 0
= (a + b)2x2 - (a + b)2 x + (a - b)2 x - (a - b)2 = 0
= (a + b)2x(x - 1) (a + b)2(x - 1) = 0
= (x - 1)(a + b)2x + (a + b)2) = 0
Either x - 1 = 0
Therefore x = 1
Or, (a + b)2x + (a + b)2) = 0
The roots of the above mentioned quadratic equation areand 1 respectively.
Question: 29
Find the roots of the quadratic equation a(x2 + 1) - x(a2 + 1) = 0
Solution:
= a(x2 + 1) - x(a2 + 1) = 0
= ax2 + a - a2x - x = 0
= ax(x - a) - 1(x - a) = 0
= (x - a)(ax - 1) = 0
Either x - a = 0
Therefore x = a
Or, ax - 1 = 0
Therefore x = 1/a
The roots of the above mentioned quadratic equation are (a and x = 1/a respectively.
Question: 30
Find the roots of the quadratic equation
Solution:
The given equation is
= x2 + (a + 1/a)x + 1 = 0
= x2 + ax + x/a +(a × 1/a) = 0
= x(x + a) +1/a(x + a) = 0
= (x + a)(x + 1/a) = 0
Either x + a = 0
Therefore x = – a
Or , (x + 1/a) = 0
Therefore x = 1/a
The roots of the above mentioned quadratic equation are x = 1/a and a respectively.
Question: 31
Find the roots of the quadratic equation abx2 + (b2 - ac)x - bc = 0
Solution:
The given equation is abx2 + (b2 - ac)x - bc = 0
= abx2 + (b2 - ac)x - bc = 0
= abx2 + b2x - acx - bc = 0
= bx(ax + b) - c(ax + b) = 0
= (ax + b)(bx - c) = 0
Either, ax + b = 0
Therefore x = (-b)/a
Or, bx - c = 0
Therefore x = c/b
The roots of the above mentioned quadratic equation are x = c/b and x = (-b)/a respectively.
Question: 32
Find the roots of the quadratic equation a2b2x2 + b2x - a2x - 1 = 0
Solution:
The given equation is a2b2x2 + b2x - a2x - 1 = 0
= a2b2x2 + b2x-a2x - 1 = 0
= b2x(a2x + 1) - 1(a2x + 1)
= (a2x + 1)( b2x - 1) = 0
Either (a2x + 1) = 0
Therefore x = (-1)/a2
Or, (b2x - 1) = 0
Therefore x = 1/b2
The roots of the above mentioned quadratic equation are x = 1/b2 and x = (-1)/a2 respectively.
```
### Course Features
• 728 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
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By De Simone A., Mundici D.
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Extra info for A Cantor-Bernstein Theorem for Complete MV-Algebras
Example text
What’s the greatest power of 10 that’s a factor of 9,200,000? 197. The number 6,915 is divisible by which numbers from 2 to 6, inclusive? ) 198. ) Using that information, what is the remainder when you divide ? 193. What’s the greatest power of 10 that’s a factor of 30,940,050? 199. You know that the number 612 is divisible 194. The number 78 is divisible by which num- by 9 (because 6 + 1 + 2 = 9). Using that information, what is the remainder when you divide ? bers from 2 to 6, inclusive? ) 200.
Mina took a long walk on the beach each day of her eight-day vacation. On half of the days, she walked 3 miles and on the other half she walked 5 miles. How many miles did she walk altogether? 161. Simon noticed a pair of square numbers that add up to 130. He then noticed that when you subtract one of these square numbers from the other, the result is 32. What is the smaller of these two square numbers? 31 32 Part I: The Questions 162. If Donna took 20 minutes to read 60 pages of 166. Yianni just purchased a house priced at a 288-page graphic novel, how long did she take to read the whole novel, assuming that she read it all at the same rate?
The number 380 is divisible by which numbers from 2 to 6, inclusive? ) by 6 (because it’s an even number whose digits add up to 9, which is divisible by 3). Using that information, what is the remainder when you divide ? 41 42 Part I: The Questions Working with Prime and Composite Numbers 203. Is 151 a prime number? 201–210 201. Which of the following are prime numbers 204. Is 161 a prime number? and which are composite numbers? 205. Is 223 a prime number? 39 206. Is 267 a prime number? 41 207. | 790 | 3,336 | {"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} | 3.46875 | 3 | CC-MAIN-2018-47 | longest | en | 0.896558 |
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We are providing you with the test of the day which will help you in upcoming exams like IDBI Executive 2018. From this, you can practice questions with the timer that will help you to improve speed. IDBI Executive Exam 2018 is one of the biggest opportunity for you guys to grab on, set you own mark and proceed with full confidence. All the very best for your preparations and All the future exams.
Q1. Ratio of cost price to selling price of an article is 5 : 6. If 20% discount is offered on marked price of article then marked price is what percent more than cost price?
100/3%
50%
40%
200/3%
60%
Solution:
Q2. Ramesh has 20% savings with him from his monthly salary. If expenditure on clothing is 25% of overall expenditure and his total expenditure except clothing is 3600 then find his saving.
1000
1500
1600
1200
900
Solution:
Q3. A milkman has 20 L of pure milk. He sells 5 L of it at profit of 20% and he mixes 5 L water in remaining quantity of milk. If he sells whole quantity of mixture at cost price then find his overall percentage profit?
25%
20%
24%
28%
30%
Solution:
Q4. A bus starts off from a station at 9 A.M at a speed of 90 km/hr. After 1 hour reduces its speed by If person on car sets off from the same station at 10 AM on the same day from which bus started and moves in the direction in which bus is moving then what should be the speed of car so that it overtakes bus at 12 Noon.
105 km/hr
90 km/hr
75 km/hr
80 km/hr
120 km/hr
Solution:
Q5. Average age of A, B and C is 10 year less than age of C. If ratio of age of A, B and C is 3 : 4 : 5 then what is the sum of age of A & C.
80 year
70 year
85 year
78 year
72 year
Solution:
Directions (6-10): Study the following information carefully and answer the questions given below.
Ten persons are sitting in twelve seats in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In row 1, A, B, C, D and E are seated and all of them are facing south, and in row 2, P,Q, R, S and T are sitting and all of them are facing north. One seat is vacant in each row. Therefore, in the given seating arrangement each member seated in a row faces another member of the other row. All of them like a different colour i.e. Red, blue, yellow, green, white, black, cream, pink, brown and violet. A sits second to left of person who likes red colour, the one seat which is on the extreme end is adjacent to either A or the person who likes red colour. T sits one of the extreme ends of the row. There are three persons sits between T and S, who likes blue colour. Immediate neighbor of T faces B. One of immediate neighbor of B faces R, who likes yellow colour. There are no vacant seats adjacent to R. C likes green colour and sits second to left of vacant seat. One of immediate neighbor of R is Q and he likes white colour. One of immediate neighbor of Q faces D and he likes black colour. The one who likes cream colour sits immediate left of the person who likes pink colour. The one who likes brown colour sits third to left of person who likes violet colour.
Q6. Who among following sits on the extreme end of the row?
E,Q
B,P
B,S
D,P
D,S
Solution:
Q7. Who among following likes violet colour?
P
R
Q
T
S
Solution:
Q8.Who among following sits second to left of T?
S
P
R
Q
None of these
Solution:
Q9. Who among following sits opposite to C?
P
Q
R
S
T
Solution:
Q10.Who among following likes cream colour?
A
B
C
D
E
Solution:
Directions (11-15): In each of the question given below a/an idiom/phrase is given in BOLD which is then followed by five options which then tries to decipher its meaning as used in the sentence. Choose the option which gives the meaning of the phrase most appropriately in context of the given sentence.
Q11. I’m banking on you to help with the charity event.
pretend
rely
doubt
faith
inhibit
Solution:
The phrasal verb “bank on” means “to base your hopes on something / someone”. Thus, the most appropriate meaning among the given options is “rely” which means “depend on with full trust or confidence”. Hence option (b) is the correct choice.
Q12. His plans to trek through South America fell through when he got sick.
ascend
overturn
slip
fail
deteriorate
Solution:
The phrasal verb “fell through” means “to fail; doesn’t happen”. Thus, the most appropriate meaning among the given options is “fail”. Hence option (d) is the correct choice. Ascend means rise or move up through the air.
Q13. We hope they can iron out their differences and get on with working together.
resolve
question
conclude
facilitate
upset
Solution:
The phrasal verb “iron out” means to remove problems or find solutions, or to resolve by discussion, eliminate differences. Thus, among the given options, the most appropriate meaning is “resolve”. Hence option (a) is the correct choice.
Q14. He finished his cup of tea and got on with the gardening.
yield
retrogress
persist
stay
last
Solution:
The phrasal verb “get on with” means to start or continue doing something, especially work. Thus, among the given options, the most appropriate meaning is “persist” which means “continue to exist; be prolonged”. Hence option (c) is the correct choice. Retrogress means go back to an earlier state, typically a worse one.
Q15. He’s never made an effort to keep up with current events.
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Contradictions. Pile Driving. Y = Driving Speed What are the Knobs?. Driving Speed = f ( Ground Hardness, Pile Diameter, Tip Angle …). Driving Speed is Slow. Ground is Hard. Pile Diameter is Large. Tip Shape is Blunt. Tip Shape is Sharp. Support is Poor. Contradiction. Pile Driving.
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Pile Driving
Y = Driving Speed
What are the Knobs?
Driving Speed = f ( Ground Hardness, Pile Diameter, Tip Angle …)
Driving Speed is Slow
Ground is Hard
Pile Diameter is Large
Tip Shape is Blunt
Tip Shape is Sharp
Support is Poor
Pile Driving
Thin
Thick
Use a knob and a setting
Pile Diameter
Pile Driving
In order to drive fast, the pile must be sharp. In order to support heavy loads, the pile must be blunt
Conflicting Requirements
Supporting
Driving
Sharpness
Compromise Solutions
Time Consuming
Guarantees Risk
Delays the Solution
Idealized Solutions
Conflicting Requirements
Supporting
Driving
Blunt
Sharp
It must be sharp and blunt
Setting the Knob to Both Settings
Rather than compromising, Do both (Resolve the Contradiction)
We will consider four ways:
Separation in Space
Separation by Direction
Separation by Scale
Separation in Time
Blunt
Sharp
Pile Sharpness
Separation in Space
Non-Uniform
Carrier
Completely Separate
Touching
Nesting
Composite or Mixture
Copy
Real
McCoy
Red & Green
Round
inside Square
Allowed to turn while driving
Constrained while supporting
Separation by Direction, (Path or Plane)
Green
View
Flexible
Square
View
Stiff
Round
View
Red
View
Separation by Scale or Between the Parts and the Whole
Thin
Thick
Blunt & Sharp
Blunt & Sharp
Massive and Light Table
Square & Round
Sand Paper:
Counter
Weight
Object
Heavy & Light Object
Smooth Object made from from Partially Rough Objects
Inflexible Particles on a Flexible Carrier
Chain is Flexible
Separation in Time
Shaping
Cap
Material
Pumped Concrete
Concrete
Drive
Blunt
Pile
Second
Pile
Catches
Lip Edge
of First
Pile
Drive
Sharp
Copy
Remove
Sharp
Copy
Drive
First
Pile
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# Floating in the waters of the equatorial pacific, an array
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Floating in the waters of the equatorial pacific, an array [#permalink]
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03 Jul 2009, 02:13
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78.Floating in the waters of the equatorial pacific, an array of buoys collect and transmit data on long term interactions between the ocean and the atmosphere, interactions that affect global climate.
A. an array of buoys collect and transmit data on long term interactions between the ocean and the atmosphere, interactions that affect
B. an array of buoys collects and transmits data on long term interactions between the ocean and the atmosphere, with interactions affecting
C. an array of buoys collect and transmit data on long term interactions between the ocean and the atmosphere that affects
D. an array of buoys collects and transmits data on long term interactions between the ocean and the atmosphere that is affecting
E. an array of buoys collect and transmit data on long term interactions between the ocean and the atmosphere as
Oa is
[Reveal] Spoiler:
a
Pl explain the reference of array here. is it singular or plural?
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Manager
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03 Jul 2009, 06:04
What is the source of OQ?
I think the array should be singular.
If you observe, all the answer choices started with 'an'.
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03 Jul 2009, 06:17
Interesting question. Initially I thought 'an array of buoys' is singular so looked into B and D options ... but these two options don't look right to me.
then I decided 'an array of buoys' is plural .. so A fits here.
Please Post OA and OE for this one.
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Director
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03 Jul 2009, 09:09
I think this question has wrong underlined portion.
http://www.mbacrashcourse.com/sentence- ... ys-sc.html
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03 Jul 2009, 09:36
How can be an array be singular?
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03 Jul 2009, 10:05
Quote:
I think the array should be singular.
If you observe, all the answer choices started with 'an'.
Array is a collective noun.
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Re: another interesting [#permalink] 03 Jul 2009, 10:05
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# Floating in the waters of the equatorial pacific, an array
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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®. | 1,170 | 4,283 | {"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} | 3.09375 | 3 | CC-MAIN-2017-47 | latest | en | 0.874396 |
https://everything2.com/title/Feistel+network | 1,632,606,270,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057775.50/warc/CC-MAIN-20210925202717-20210925232717-00513.warc.gz | 273,269,083 | 6,318 | Feistel networks are a strategy for designing iterated block ciphers developed by Horst Feistel at IBM in 1974, that has been used in the design of many block ciphers, including DES, Blowfish, Twofish, and CAST. A Feistel network works in this way: Take a block of length n bits and divide it into two equal-sized halves, called L and R. A round of the cipher can be calculated from the previous round by:
Li = Ri-1
Ri = Li-1 XOR f(Ri-1, Ki)
where Ki is the subkey used in the ith round and f is an arbitrary round function (usually key mixing and s-box transformations). This is a very popular way of designing block ciphers because the whole construction is guaranteed to be invertible, even if the function f happens not to be, so long as the input to f may be reconstructed. Because the XOR function is used to combine the left half with the output of the round function, we may recompute Li-1 given Ri and Ri-1 (which is incidentally Li by definition), provided we can reconstruct the value of Ei:
Li-1 = Ri XOR f(Li, Ei)
Ri-1 = Li
So with the inverse structure here, we are essentially running the Feistel network backwards!
There is also an alternative structure called an unbalanced Feistel network, used in ciphers like Skipjack and MacGuffin, where the left and right halves are not of equal size. The characteristics of this sort of construction are still quite uncertain, although its use in a cipher such as Skipjack developed by an organization like the NSA is somewhat reassuring. MacGuffin was created to catalyze research in this method of cipher construction.
Log in or register to write something here or to contact authors. | 384 | 1,651 | {"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-2021-39 | latest | en | 0.94047 |
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