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http://www.startribune.com/science/245697811.html | 1,417,232,901,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931012025.85/warc/CC-MAIN-20141125155652-00250-ip-10-235-23-156.ec2.internal.warc.gz | 853,416,166 | 28,501 | ## The problem with infinity is that you can't stop
• Updated: February 15, 2014 - 5:11 PM
You might think that if you simply started adding the natural numbers, 1 plus 2 plus 3 and so on all the way to infinity, you would get a pretty big number.
So it came as a shock to many people when, in a recent video, a pair of physicists purported to prove that this infinite series actually adds up to … minus 1/12.
More than 1.6 million people have viewed this calculation, which plays a key role in modern physics and quantum theory; the answer, as absurd as it sounds, has been verified to many decimal places in lab experiments.
Even the makers of the video — Brady Haran, a journalist, and Ed Copeland and Antonio Padilla, physicists at the University of Nottingham in England — admit there is a certain amount of “hocus-pocus,” or what some mathematicians have called dirty tricks, in their presentation. But there is broad agreement that a more rigorous approach to the problem gives the same result, as shown by a formula in Joseph Polchinski’s two-volume textbook “String Theory.”
So what’s going on with infinity? “This calculation is one of the best-kept secrets in math,” said Edward Frenkel, a mathematics professor at the University of California, Berkeley, and author of “Love and Math: The Heart of Hidden Reality.” “No one on the outside knows about it.”
The first one down this road was the great 18th-century mathematician Leonhard Euler, who was born in Switzerland but did most of his work in Berlin and St. Petersburg, Russia. In 1749, he used a bag of mathematical tricks to solve the problem of adding the natural numbers from 1 to infinity. Clearly, if you stop adding anywhere along the way — at a quintillion (1 with 18 zeros after it), say, or a googolplex (10 to the 100th power zeros) — the sum will be enormous. The problem with infinity is that you can’t stop. You never get there. It’s more of a journey than a destination.
In modern terms, Frenkel explained, the gist of the calculations can be interpreted as saying that the infinite sum has three parts: One blows up when you go to infinity, one goes to zero, and minus 1/12. The infinite term, he said, is just thrown away.
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Interactive: Guide to new restaurants, 2014 | 555 | 2,427 | {"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-2014-49 | longest | en | 0.96063 |
https://rdrr.io/cran/SpiceFP/man/SpiceFP-package.html | 1,639,003,702,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00426.warc.gz | 545,525,732 | 8,453 | # SpiceFP-package: A Sparse and Structured Procedure to Identify Combined... In SpiceFP: Sparse Method to Identify Joint Effects of Functional Predictors
## Description
A set of functions allowing to implement the 'SpiceFP' approach which is iterative. It involves transformation of functional predictors into several candidate explanatory matrices (based on contingency tables), to which relative edge matrices with contiguity constraints are associated. Generalized Fused Lasso regression are performed in order to identify the best candidate matrix, the best class intervals and related coefficients at each iteration. The approach is stopped when the maximal number of iterations is reached or when retained coefficients are zeros. Supplementary functions allow to get coefficients of any candidate matrix or mean of coefficients of many candidates.
## Details
The main function of the package is the spicefp function. It directly performs the three main steps of the SpiceFP approach, by using intermediate functions of the package.
1) At he first step, contingency tables are constructed by defining joint modalities using class intervals or bins. Several candidate partitions are then defined. For each statistical individual i and each candidate partition (denoted u here), the 2 (resp. 3) functional predictors are transformed into frequency bi(resp. tri)-variate histograms (or contingency tables), stored as row vectors. The combination of these row vectors for all individuals enables the construction of a candidate explanatory matrix indexed by u (denoted here X^u). The function candidates is designed to build these candidate matrices.
2) At the second step, for each candidate explanatory matrix, an edge matrix is defined to represent the contiguity constraints between modalities of the contingency table.
3) Finally at the last step, the best class intervals and related regression coefficients are defined by: i) performing a Generalized Fused Lasso using each candidate explanatory matrix. The SpiceFP model is the following
y_i = X_i^u β^u + \varepsilon_i,
where β^u is the coefficient to be estimated on the 2D (resp. 3D) intervals. The estimator of β is obtained as follows:
\hat{β}^{u,γ}(λ) = argmin \frac{1}{2} \|y - X^u β\|_2^2 + λ \|D ^{u,γ} β\|_1,
where λ is a penalty parameter that controls the smoothness of the coefficients, and γ is the ratio between the regularization parameters of parsimony and fusion. ii) choosing the best candidate matrix and selecting its variables using an information criterion and checking the shutdown conditions to stop the approach. Indeed, SpiceFP may be used in an iterative way. It therefore allows to identify up to K best candidate matrices and related coefficients.
## Author(s)
Maintainer: Girault Gnanguenon Guesse girault.gnanguenon@gmail.com
Authors:
Other contributors:
SpiceFP documentation built on Sept. 15, 2021, 9:07 a.m. | 612 | 2,915 | {"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} | 2.84375 | 3 | CC-MAIN-2021-49 | longest | en | 0.877297 |
https://www.askiitians.com/forums/Trigonometry/sina-cosb-1-2-cosa-sinb-2-3-then-find-the-val_153773.htm | 1,723,392,368,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00519.warc.gz | 517,397,241 | 42,859 | # Sina+cosb=1/√2 & cosa+sinb=√2/√3 then find the value of : (1) sin(a+b) (2) tan(a-b/2)
Vikas TU
14149 Points
7 years ago
Sina+cosb=1/√2
Sina = 1/√2 – cosb
squaring both sides,
(sina)^2 = 1/2 + (cosb)^2 – 2cosb/√2
1 – (cosa)^2 = 1/2 + (cosb)^2 – 2cosb/√2
1/2 – (cosa)^2 = (cosb)^2 – 2cosb/√2.................(1)
This is the eqn. 1
Do it same squaring method in cosa+sinb=√2/√3
and bring the final eqn. in cosa and cosb terms
then after solve those eqns. andd get cosa and cosb values. | 225 | 488 | {"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} | 3.796875 | 4 | CC-MAIN-2024-33 | latest | en | 0.467647 |
https://www.physicsforums.com/threads/could-traveling-at-high-speed-pay-off-financially-back-on-earth.622115/ | 1,544,681,759,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824525.29/warc/CC-MAIN-20181213054204-20181213075704-00082.warc.gz | 1,011,055,329 | 18,105 | # Could traveling at high speed pay off financially back on Earth?
1. Jul 19, 2012
### drbanner
Hypothetical Space/time perception concept with financial implications.
If you were to invest a $100,000 in an account with 3% annual interest, how much interest would you have accrued in$ if you were to travel at 80 % the speed of light directly away from earth for 1 year, and then directly back to earth for 1 year at the same speed? How much time would have went by on earth?
2. Jul 19, 2012
### DaveC426913
You age 2 years, Earth ages 3.
http://www.1728.org/reltivty.htm
At the bottom of the page is a calculator. Enter .8 and press c=1 to see the time dilation factor.
3. Jul 19, 2012
### ghwellsjr
At 0.8c, gamma = 1/√(1-0.82) = 1/√(1-0.64) = 1/√(0.36) = 1/0.6 = 1.666 or exactly 5/3. Since you are gone 2 years, 3.3333 years will pass on earth. Assuming that the interest is compounded three times a year at 1%, you will end up with $110462.21. By the way, if you had stayed on earth, with the same compounding, you would end up with$106152.02 so you have \$4310.19 more by taking the trip but I doubt you will get any bank to honor your claim unless the bank goes with you.
Last edited: Jul 19, 2012
4. Jul 19, 2012
### phinds
Which would be WAY less than it would cost to make the trip
5. Jul 19, 2012
### PAllen
Another obvious point is that even if you make the example more extreme:
Put X dollars in a Trust with with 'reliable' management company, travel such that after one year of your time passes, 100 years of earth time passes. You come back and have lot's of dollars - but probably not so much more in spending power than when you left. If you instruct that you want minimal risk of decrease, the best you can hope for over the long haul is a little better than inflation.
6. Jul 20, 2012
### ghwellsjr
Worse yet, I just got a notice from one of my financial institutions warning:
So you better not try this in California.
7. Jul 20, 2012
### PAllen
Many states have such a law. That's why I mentioned a Trust, managed e.g. by a Bank's trust department. It is completely exempt from such laws. FYI: even in states with such a law, if you set up online access to your account and simply check the balance periodically on line, that counts as 'activity' for the purposes of such laws (at least in my state it does; we ran into this issue in one account). The big issue for a long term investment is management that is both competent and free of conflict of interest, and must be a corporate entity with successor policies, else you are at great risk over a very long time window.
8. Jul 20, 2012
### ghwellsjr
Now there's an interesting problem: how often would you have to contact your bank while traveling on a spaceship to make sure they got a call from you every three years (so as not to lose your funds)? For this particular scenario, when would the traveler make a call so that it arrived at the bank after three years?
9. Jul 20, 2012
### harrylin
I don't follow this - the bank simply counts the number of bank years, and it has to honour that obligation. However, not only the trip will cost much more, there's also inflation.-
10. Jul 20, 2012
### Staff: Mentor
11. Jul 20, 2012
### ghwellsjr
You're right, I don't know what I was thinking concerning the need for the bank to take the trip. Thanks for catching this.
But what about my question for the need to contact the bank every 3 years? When does the traveler need to send a message to the bank so that it will arrive 3 years after he leaves?
12. Jul 20, 2012
### PAllen
Just have the traveler set up a trust, problem avoided.
13. Jul 20, 2012
### DaveC426913
Your time is 2 years. Earth time is 3.3333 years. All shorter intervals will have a similar ratio.
14. Jul 20, 2012
### ghwellsjr
Re: The best available discussion of this topic
Very witty, even hilarious, thanks for sharing this.
15. Jul 20, 2012
### ghwellsjr
Too late, the traveler already left and got a communication from his bank, along with the first interest credited, sent four months (1/3 year) after he left that he had to communicate at least every 3 years to satisfy the state. Now the question is, when will he receive the message and when will he need to respond in order for the reply to get to the bank at the three year deadline?
16. Jul 20, 2012
### ghwellsjr
If you are saying that he can send the message 1.8 years after he left, it will be too late, the state will already have his funds by the time he gets back.
17. Jul 20, 2012
### ghwellsjr
Not according to Paul Krugman:
Competition will have driven the travel costs to zero. (You gotta read the paper.)
18. Oct 20, 2012
### ghwellsjr
I'm still waiting for the correct answer to my question in post #15.
To summarize the situation, a guy puts money in a bank and leaves in a spaceship at 0.8c. After four months (earth time) the bank sends him an email message informing him that he needs to respond to the bank at least every 3 years in order to avoid losing his money to the state.
"Now the question is, when will he receive the message and when will he need to respond in order for the reply to get to the bank at the three year deadline?"
19. Oct 20, 2012
### HallsofIvy
Since we have already injected a bit of "reality" into this, all of this is assuming that the bank does not go broke and is able to pay your your money!
20. Oct 20, 2012
### ghwellsjr
A bank offering 3% interest probably will go broke but even if they were only offering 0.1% interest, my question still stands. It has nothing to do with the interest rate, only what will happen to the principle if he doesn't respond within 3 years. Our state is already broke so they would love to collect this winfall. How can we make sure that doesn't happen? | 1,528 | 5,804 | {"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-2018-51 | latest | en | 0.950994 |
https://www.codeproject.com/Articles/653355/Color-Constancy-Gray-World-Algorithm?msg=4662072 | 1,508,405,569,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823260.52/warc/CC-MAIN-20171019084246-20171019104246-00894.warc.gz | 889,499,300 | 22,944 | 13,193,829 members (48,547 online)
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Posted 15 Sep 2013
# Color Constancy: Gray World Algorithm
, 15 Sep 2013
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Color constancy is a mechanism of detection of color independent of light source. The light source many introduce color casts in captured digital images To solve the color constancy problem a standard method is to estimate the color of the prevailing light
## Color Constancy
Color constancy is a mechanism of detection of color independent of light source. The light source many introduce color casts in captured digital images To solve the color constancy problem a standard method is to estimate the color of the prevailing light and then, at the second stage, remove it. Once the color of light in individual channels is obtained the each color pixel is normalized by a scaling factor .
Two of the most commonly used simple techniques for estimating the color of the light are the Grey-World and Max-RGB algorithms. These two methods will work well in practice if the average scene color is gray or the maximum is white.
## Gray world assumption
The Gray World Assumption is a white balance method that assumes that your scene, on average, is a neutral gray. Gray-world assumption hold if we have a good distribution of colors in the scene. Assuming that we have a good distribution of colors in our scene,the average reflected color is assumed to be the color of the light. Therefore, we can estimate the illumination color cast by looking at the average color and comparing it to gray.
Gray world algorithm produces an estimate of illumination by computing the mean of each channel of the image.
One of the methods of normalization is that the mean of the three components is used as illumination estimate of the image. To normalize the image of channel i ,the pixel value is scaled by $s_1 = \frac{avg}{avg_i}$ where $avg_i$ is the channel mean and $avg$\$ is the illumination estimate .
Another method of normalization is normalizing to the maximum channel by scaling by \$s_i\$ $r_i = \frac{max(avg_R,avg_G,avg_B)}{avg_i}$
Another method of normalization is normalizing to the maximum channel by scaling by norm \$m_i\$ $m_i = \sqrt{(avg_r*avg_r+avg_g*avg_g+avg_b*avg_b)}$
$r_i = \frac{max(m_R,m_G,m_B)}{m_i}$
Some of the images are taken from http://research.edm.uhasselt.be/~oancuti/Underwater_CVPR_2012/ image set
## Code
For code refer to site https://github.com/pi19404/m19404/tree/master/ColorConstancy/ The files are color_constancy.cpp and color_constancy.hpp.
The class for performing gray world transformation is gray_world.
`Mat run2(Mat,int p,int m); //Processing in RGB color space`
The norm factor is $p=1$ for gray world algorithm and various normalization techniques can be passed as $m=(1,2,3)$
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https://dubtam.org/amazing-ratio-and-proportion/ | 1,620,350,256,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.18/warc/CC-MAIN-20210506235514-20210507025514-00243.warc.gz | 181,457,530 | 8,289 | # Amazing ratio and proportion Wonderful
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### Two quantities are in direct proportion when they increase or decrease in the same ratio.
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34 68 is an example of a proportion. In order to work with ratio and proportions you first have to completely understand. If we solve this proportional statement we get. Therefore the ratio defines the relation between two quantities such as ab where b is not equal to 0. B and c.
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C are given then a. KS3 Maths Ratio and proportion learning resources for adults children parents and teachers. 14082020 From the wrist to the tip of the fingers is 110 of the human height. Ratio and Proportion PDF. Ratio and proportion Calculate how the amounts of ingredients required will change according to the scale of production The Engineering Process scheme of work provides students with an in-depth understanding of some engineering materials and how they are being developed in industry.
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Source: pinterest.com
Two quantities are in direct proportion when they increase or decrease in the same ratio. It is a statement that two ratios are equal. But they also have more than a sprinkling of richness that leads students merrily along the path towards hypotheses and generalisations. Ratio Venn Diagrams Age Ratios Ratio and Cuisenare Rods Ratio and Polygon Angles Ratio Questions in Life Harder Ratio Questions. 34 68 is an example of a proportion.
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D are equal that is if. 14082020 From the wrist to the tip of the fingers is 110 of the human height. Must Practice 11 Plus 11 Ratio and Proportion Past Paper Questions. 20 x 5 25 x 4. B and b.
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KS3 Maths Ratio and proportion learning resources for adults children parents and teachers. It is a statement that two ratios are equal. In order to work with ratio and proportions you first have to completely understand. B and b. Ratio and Proportion PDF.
Source: pinterest.com
This is called solving the proportion. Ratios are usually written in the form ab and can be used on maps to show the scale in relation to real life. 14082020 From the wrist to the tip of the fingers is 110 of the human height. 16092008 A proportion is an equation with a ratio on each side. B and b.
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http://stats.stackexchange.com/questions/24283/significance-testing-of-slopes-with-replicates/24295 | 1,369,168,292,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700497024/warc/CC-MAIN-20130516103457-00051-ip-10-60-113-184.ec2.internal.warc.gz | 247,921,915 | 12,769 | # Significance testing of slopes with replicates
I've made replicate measurements of a parameter (fluorescence) which is expected to increase with time, and I'm having a hard time understanding how to test for significance of the slope of the parameter vs. time (using a linear least squares model).
Intuitively, it makes sense to me to combine all time-fluorescence pairs for each replicate into a grand dataset, and look at the significance of the slope of that fit. However, the intercept of the parameter is expected to differ between replicates due to an instrumental artifact. It's not clear to me whether it is legitimate to combine the data for each replicate due to the expected difference in intercepts. Many thanks for any help.
to put this in R terms:
t1 <- 1:10
f1 <- rnorm(10)
t2 <- 1:10
f2 <- rnorm(10)
t3 <- 1:10
f3 <- rnorm(10)
Is it legitimate to look at the significance of the slope of
model <- lm(c(f1, f2, f3) ~ c(t1, t2, t3))
Or should I be doing something different?
-
It's not legitimate to just fit a line through the whole thing for two reasons:
• even if you had only one replicate, you will have autocorrelation in the residuals over time. Observation at $t_n$ is related to the observation at $t_{n-1}$ so you can't treat them as independent. There is actually less information in each observation than if they were independent. So you need a model that takes this into account.
• as it happens, you have three replicates. The intercept and the slope are likely to vary for each replicate. You could address this by allowing an interaction of slope and intercept with a fixed factor for replicate, but better is just to treat this as a source of grouped randomness.
The lme() function in Pinheiro and Bates' library(nlme) can solve both of these problems for you (although working out the exact way to treat the residuals can be quite an involved issue).
In terms of how the data should be structured, I think it's best to have three columns - one for which replicate you have, and one each for the time and for the actual measurement. As well as being a good format to fit a model to, this is also a good format for drawing plots easily using ggplot2.
So something like this.
library(ggplot2)
library(nlme)
t1 <- 1:10
f1 <- rnorm(10,13,1) + t1 * rnorm(10,3,1)
t2 <- 1:10
f2 <- rnorm(10,14,2) + t1 * rnorm(10,2,.5)
t3 <- 1:10
f3 <- rnorm(10,14,1) + t1 * rnorm(10,4, 1.5)
fALL <- c(f1, f2, f3)
tALL <- c(t1, t2, t3)
replicate <- rep(c("One", "Two", "Three"), rep(10,3))
fluoro <- data.frame(fALL, tALL, replicate)
rm(f1, f2, f3, t1, t2, t3, replicate) # clean up
fluoro
p <- ggplot(fluoro, aes(x=tALL, y=fALL, color=replicate))
# couple of different versions of the plots
p + geom_line()
p + geom_smooth(method="lm") + geom_point()
# example only - this may be the wrong error structure
model <- lme(fALL ~ tALL, data=fluoro, random=~1|replicate, correlation=corAR1())
-
In the original poster's formulation how is there autocorrelation in the residuals? The dependent variables are 30 iid $N(0,1)$ variables. In the verbal description, the poster seems to describe a situation where there's a deterministic trend in the mean over time, not temporal autocorrelation. – Macro Mar 8 '12 at 4:13 In the OP's code they are indeed iid $N(0,1)$ but in his description they are ten measurements in time on the same variable - although he does not state it himself I am sure there is likely to be autocorrelation in addition to the deterministic trend. Imagine the time between measurements approaching zero. – Peter Ellis Mar 8 '12 at 13:54 I think @PeterEllis is right. The actual measurements are a physical parameter of a sample (fluorescence). The measured fluorescence is probably given by something like Fm = F0 + noise + drift(time), where F0 is the 'true' fluorescence signal, noise is random and drift(time) is some long-wavelength instrumental drift that is a function of time. I guess F0 and drift(time) must both introduce autocorrelation. – Drew Steen Mar 8 '12 at 19:55
If you want to include a different intercept for each of the three sequences of values, you can try
n <- factor(rep(1:3, each=10))
lm( c(f1, f2, f3) ~ c(t1, t2, t3) + n )
-
Or if you also want to allow a different slope you could try lm( c(f1, f2, f3) ~ c(t1, t2, t3) * n ). But this doesn't get around the problem that the residuals are related to eachother over time. – Peter Ellis Mar 8 '12 at 0:09 (also @PeterEllis): These are good points; I hadn't thought about the autocorrelation issue. The slopes are my real 'measurement' (they represent a rate). Perhaps then it is conservative, but not overconservative, to simply do a one-sample t-test on the fitted slopes, like so: t.test(c(slope1, slope2, slope3), mu=0) – Drew Steen Mar 8 '12 at 1:27 | 1,295 | 4,783 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2013-20 | latest | en | 0.951317 |
http://mathhelpforum.com/advanced-algebra/11548-coefficients-vectors-method.html | 1,527,242,048,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867055.20/warc/CC-MAIN-20180525082822-20180525102822-00093.warc.gz | 191,557,382 | 10,623 | 1. ## The coefficients-vectors method
Ok, I understand how to get it this far. But how are they getting this?
2. Hello there!
I'm not sure what you are trying to do, multiplication or dot product (the single dot usually indicates dot product).
But the matrix below is the result of multiplying those two matricies. Is that what you are having a problem with?
3. nm, i just needed to sit back and look at it more.
4. It makes no difference which method you use to multiply matrices, whichever one you understand the best. The important thing is that you know how to multiply them. The one I use is the standard dot product between the rows and coloums.
5. Hello, Grunt!
Someone taught you a very messy way of multiplying matrices.
. . Shame on them!
ThePerfectHacker referred to a "standard dot product".
. . I thought everyone used it.
Let's multiply "a row times a column".
Code:
| 4 |
[ 1 2 3 ] · | 5 | = (1)(4) + (2)(5) + (3)(6)
| 6 |
= 4 + 10 + 18 = 32
I hope you see what pairs are being multipled.
Code:
| 3 1 |
| 1 0 2 | | |
| | · | 2 1 |
| -1 3 1 | | |
| 1 0 |
We will multiply each row in the first matrix
. . by each column in the second matrix.
Here we go . . .
Row-1 times Column-1
Code:
| 3 |
| 1 0 2 | · | 2 | = (1)(3) + (0)(2) + (2)(1) = 5
| 1 |
In Row-1, Column-1 of the product matrix (upper left).
Row-1 times Column-2
Code:
| 1 |
| 1 0 2 | · | 1 | = (1)(1) + (0)(1) + (2)(0) = 1
| 0 |
This goes in Row-1, Column-2 of the product (upper right).
Row-2 times Column-1
Code:
| 3 |
|-1 3 1 | · | 2 | = (-1)(3) + (3)(2) + (1)(1) = 4
| 1 |
This goes in Row-2, Column-1 of the product (lower left).
Row-2 times Column-2
Code:
| 1 |
|-1 3 1 | · | 1 | = (-1)(1) + (3)(1) + (1)(0) = 2
| 0 |
This goes in Row-2, Column-2 of the product (lower right).
| 5 1 |
| 4 2 | | 668 | 1,945 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-22 | latest | en | 0.891846 |
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# Building social network with Neo4j and Python
Social phenomena is coming. We have lot’s of social applications that we are using every day, let’s say Facebook, twitter, Instagram. Lot’s of such kind apps based on social graph and graph theory. I would like to share my knowledge and expertise about how to work with graphs and build large social graph as engine for Social network using python and Graph databases. We'll compare SQL and NoSQL approaches for friends relationships.
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### Building social network with Neo4j and Python
1. 1. Building social network with Neo4j and Python Andrii Soldatenko 3-4 July 2016 @a_soldatenko
2. 2. Agenda: • Who am I? • 101 Graph data structure in Python • neo4j overview • PEP-249 and Neo4j • other python neo4j clients
3. 3. Andrii Soldatenko • Backend Python Developer at • CTO in Persollo.com • Speaker at many PyCons and Python meetups • blogger at https://asoldatenko.com
4. 4. Preface
5. 5. Books
6. 6. (Graphs)-[:ARE]->(Everywhere)
7. 7. 101 Graph data structure graph = {'A': ['B', 'C'], 'B': ['C', 'D'], 'C': ['D'], 'D': ['C'], 'E': ['F'], 'F': ['C']} from https://www.python.org/doc/essays/graphs/
8. 8. BDFL
9. 9. 101 Graph data structure def find_path(g, s, e, path=[]): path = path + [s] if s == e: return path if not s in g: return None for node in g[s]: if node not in path: newpath = find_path(g, node, e, path) if newpath: return newpath return None
10. 10. 101 Graph data structure >>> graph = {'A': ['B', 'C'], 'B': ['C', 'D'], 'C': ['D'], 'D': ['C'], 'E': ['F'], 'F': ['C']} >>> find_path(graph, 'A', 'D') ['A', 'B', 'C', 'D'] >>>
11. 11. Graph database <id=234>
roles=Jummy Dugan <id=12>
born=1971 name=“Halle Berry”
12. 12. SQL vs NoSQL
13. 13. Who is my boss?
14. 14. Consistency
15. 15. Availability
16. 16. Cypher Query Language MATCH (relationship, pattern matches) WHERE (filter condition) RETURN (what to return, nodes, rels, properties)
17. 17. Cypher: Example MATCH (n:Person {name: "Tom Hanks”}) -[:ACTED_IN]->(m) RETURN n,m SELECT n,m FROM persons;
18. 18. Cypher: Example
19. 19. Cypher: Example ╒══════════════════════╕ !tomHanksMovies.title ! ╞══════════════════════╡ !Charlie Wilson's War ! "######################\$ !The Polar Express ! "######################\$ !A League of Their Own ! !Apollo 13 ! "######################\$ !The Green Mile ! "######################\$
20. 20. Advanced Cypher http://neo4j.com/docs/cypher-refcard/current/
21. 21. Pros
22. 22. Contra - unable to handle lots of data - when you want to update all or a subset of entities
23. 23. Python
24. 24. Python 200 OK
25. 25. https://github.com/nigelsmall/py2neo Graph Data Types Graph Databases Batch & Manual Indexing Calendar subgraph neokit – Command Line Toolkit for Neo4 Object-Graph Mapping
26. 26. PEP 249 Python Database API Specification v2.0 https://www.python.org/dev/peps/pep-0249/
27. 27. Python DB API 2.0 for Neo4j https://github.com/jakewins/neo4jdb-python class Connection(object): def __init__(self, db_uri): self.errorhandler = default_error_handler self._host = urlparse(db_uri).netloc self._http = http.HTTPConnection(self._host) self._tx = TX_ENDPOINT self._messages = [] self._cursors = set() self._cursor_ids = 0
28. 28. neo4jdb-python https://github.com/jakewins/neo4jdb-python >>> from neo4j.contextmanager import Neo4jDBConnectionManager >>> >>> >>> neo_url = 'http://localhost:7474' >>> manager = Neo4jDBConnectionManager(neo_url) >>> with manager.write() as w: ... w.execute("CREATE (TheMatrix:Movie {title:'The Matrix'})") ...
29. 29. Object Graph Mapper
30. 30. class FriendRel(StructuredRel): since = DateTimeProperty(default=lambda: datetime.now(pytz.utc)) met = StringProperty() class Person(StructuredNode): name = StringProperty() friends = RelationshipTo('Person', 'FRIEND', model=FriendRel) rel = jim.friend.connect(bob) rel.since # datetime object
31. 31. (n)->[:Questions]
32. 32. Thank You @a_soldatenko andrii.soldatenko@gmail.com | 1,264 | 4,252 | {"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-2021-17 | latest | en | 0.784052 |
https://www.justintools.com/unit-conversion/area.php?k1=pings&k2=deciares | 1,652,912,930,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00799.warc.gz | 971,875,139 | 27,703 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
AREA Units Conversionpings to deciares
1 Pings
= 0.330579 Deciares
Category: area
Conversion: Pings to Deciares
The base unit for area is square meters (Non-SI/Derived Unit)
[Pings] symbol/abbrevation: (ping)
[Deciares] symbol/abbrevation: (da)
How to convert Pings to Deciares (ping to da)?
1 ping = 0.330579 da.
1 x 0.330579 da = 0.330579 Deciares.
Always check the results; rounding errors may occur.
Definition:
The deciare is ten square meters.
In relation to the base unit of [area] => (square meters), 1 Pings (ping) is equal to 3.30579 square-meters, while 1 Deciares (da) = 10 square-meters.
1 Pings to common area units
1 ping = 3.30579 square meters (m2, sq m)
1 ping = 33057.9 square centimeters (cm2, sq cm)
1 ping = 3.30579E-6 square kilometers (km2, sq km)
1 ping = 35.583242737048 square feet (ft2, sq ft)
1 ping = 5123.9847479695 square inches (in2, sq in)
1 ping = 3.9536919351616 square yards (yd2, sq yd)
1 ping = 1.2763726548536E-6 square miles (mi2, sq mi)
1 ping = 5123984747.9695 square mils (sq mil)
1 ping = 0.000330579 hectares (ha)
1 ping = 0.00081687777684427 acres (ac)
Pingsto Deciares (table conversion)
1 ping = 0.330579 da
2 ping = 0.661158 da
3 ping = 0.991737 da
4 ping = 1.322316 da
5 ping = 1.652895 da
6 ping = 1.983474 da
7 ping = 2.314053 da
8 ping = 2.644632 da
9 ping = 2.975211 da
10 ping = 3.30579 da
20 ping = 6.61158 da
30 ping = 9.91737 da
40 ping = 13.22316 da
50 ping = 16.52895 da
60 ping = 19.83474 da
70 ping = 23.14053 da
80 ping = 26.44632 da
90 ping = 29.75211 da
100 ping = 33.0579 da
200 ping = 66.1158 da
300 ping = 99.1737 da
400 ping = 132.2316 da
500 ping = 165.2895 da
600 ping = 198.3474 da
700 ping = 231.4053 da
800 ping = 264.4632 da
900 ping = 297.5211 da
1000 ping = 330.579 da
2000 ping = 661.158 da
4000 ping = 1322.316 da
5000 ping = 1652.895 da
7500 ping = 2479.3425 da
10000 ping = 3305.79 da
25000 ping = 8264.475 da
50000 ping = 16528.95 da
100000 ping = 33057.9 da
1000000 ping = 330579 da
1000000000 ping = 330579000 da
(Pings) to (Deciares) conversions | 875 | 2,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-21 | latest | en | 0.639101 |
https://math.stackexchange.com/questions/4867396/any-copula-to-force-one-variable-larger-than-the-other-in-the-joint | 1,716,830,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00720.warc.gz | 346,229,360 | 34,114 | # Any copula to force one variable larger than the other in the joint?
I want to transform two known marginal distributions into a joint distribution by using copula.
As I understand commonly used copulas cannot make sure that in the joint distribution one variable is always larger than the other. I wonder if there is any that can help. Thanks a lot.
If $$\mu$$ and $$\nu$$ are the two marginal distributions and if there exists a coupling $$(X,Y)$$ of $$\mu$$ and $$\nu$$ such that $$\Pr(X \geqslant Y) = 1$$, then this implies that $$\mu$$ stochastically dominates $$\nu$$.
So this stochastic domination is a necessary condition for the existence of your copula. Now, if this condition holds true, you can take the copula $$C(u, v) = \min(u,v)$$, which is the bivariate cumulative distribution function of a random pair $$(U, U)$$ where $$U \sim \mathcal{U}(0,1)$$. | 223 | 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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.87781 |
https://brainly.in/question/65574 | 1,484,824,759,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280587.1/warc/CC-MAIN-20170116095120-00389-ip-10-171-10-70.ec2.internal.warc.gz | 799,461,576 | 10,388 | The angles of a quadilateral are in AP whose common difference is 10 degree. Find the angels
2
by eshitarai16lk
2014-12-28T11:28:59+05:30
This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Since the angles are in A.P., let the angles are p, p+10, p+20, p+30
sum of angles=360
⇒ p + p+10 + p+20 + p+30 = 360
⇒ 4p + 60 = 360
⇒ 4p = 360-60 = 300
⇒ p = 300/4 = 75
So the angles are
p = 75
p+10 = 75+10 = 85
p+20 = 75+20 = 95
p+30 = 75+30 = 105 | 242 | 695 | {"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} | 4.03125 | 4 | CC-MAIN-2017-04 | latest | en | 0.892214 |
https://it.mathworks.com/matlabcentral/answers/383843-how-to-find-a-velocity-when-you-have-time-and-position-values-in-an-array | 1,716,975,803,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00599.warc.gz | 255,214,756 | 26,263 | # How to find a velocity when you have time and position values in an array?
3 visualizzazioni (ultimi 30 giorni)
Kelly Harmison il 21 Feb 2018
Modificato: Kelly Harmison il 21 Feb 2018
I am required to differentiate by calling values from an array. I am also required to use either a for or a while loop.
##### 2 CommentiMostra NessunoNascondi Nessuno
James Tursa il 21 Feb 2018
What have you done so far? What specific problems are you having with your code? What is the relationship between velocity and position?
Kelly Harmison il 21 Feb 2018
Modificato: Kelly Harmison il 21 Feb 2018
The main idea is to know how to differentiate on Matlab. I have two vectors: time and position. I am supposed to use a loop to call specific values from the position vector and find their slope (velocity).
Accedi per commentare.
### Risposte (1)
Roger Stafford il 21 Feb 2018
Modificato: Roger Stafford il 21 Feb 2018
Let x and y be row vectors of the same length where x gives successive values of the independent variable and y the corresponding dependent variable values - in this case x values are times and y values distances. It is not necessary that x values be equally-spaced. To get a second order approximation of the derivative at each point, do this:
x1 = x([3,1:end-1]); x2 = x([2:end,end-2]);
y1 = y([3,1:end-1]); y2 = y([2:end,end-2]);
dydx = ((y2-y).*(x-x1).^2+(y-y1).*(x2-x).^2)./((x2-x).*(x-x1).*(x2-x1));
The row vector dydx will give approximate values of the derivative of y with respect to x. This is a vectorized solution - no for loops or while loops are necessary.
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Kelly Harmison il 21 Feb 2018
Thanks!
Accedi per commentare.
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Translated by | 524 | 1,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-22 | latest | en | 0.72381 |
http://math.stackexchange.com/questions/470541/4-textth-power-of-a-2-times-2-matrix | 1,469,434,887,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824217.36/warc/CC-MAIN-20160723071024-00245-ip-10-185-27-174.ec2.internal.warc.gz | 157,945,230 | 21,815 | # $4^\text{th}$ power of a $2\times 2$ matrix
$$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$$ is given as a matrix. What is the result of $$ad + bc \text{ if } A^4=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$
Note that $A^4$ is the $4^\text{th}$ power of the matrix $A$.
I tried to use some trigonometric expressions but it gets very complicated and couldn't solve it.
-
Hint: can you describe the geometric effect of multiplying by your matrix as a transformation of the plane? – Mark Bennet Aug 18 '13 at 15:45
This is a high school question of the topic matrix. I dont think that it involves transformation of a plane. – guest Aug 18 '13 at 15:48
Most high schools curricula don't even mention matrices, but it would be an extremely strange one that did so without explaining their relation to geometry. – Henning Makholm Aug 18 '13 at 15:52
Also, are you completely sure that what is being asked for isn't $ad-bc$ rather than $ad+bc$? – Henning Makholm Aug 18 '13 at 15:53
Yes it is a.d+b.c , not a.d-b.c. I think in that case we could first find the determinant of A and then calculate the 4th power of that value. – guest Aug 18 '13 at 15:56
All you need to do here is to recognise that $A$ represents an anti-clockwise rotation about the origin by $x$. If you square the matrix then you repeat the rotation a second time, i.e. rotate by $2x$. Similarly, the cube $A^3$ can be thought of as performing the rotation three times. Hopefully you can see that $A^4$ is an anti-clockwise rotation about the origin by $4x$. As a matrix, this looks like $$A^4 = \left[ \begin{array} 1\cos(4x) & -\sin(4x) \\ \sin(4x) & \cos(4x) \end{array}\right]$$
If you want $ad+bc$ then you have $$\cos(4x)\cos(4x)-\sin(4x)\sin(4x) \equiv \cos(4x+4x) \equiv \cos(8x)$$ Here I used the double angle formula. If, however, you actually wanted the determinant, i.e. $ad-bc$ then you have $$\cos(4x)\cos(4x)+\sin(4x)\sin(4x) \equiv \cos(4x-4x) \equiv 1$$
Note that the last result is obvious. A rotation preserves volume and so its matrix must have determinant one. (If the determinant had been $2$, say, then all areas would double.)
-
Altough your solution is a bit complicated for me,the answer is cos(8x). Thank you for your explanation. – guest Aug 18 '13 at 16:10
@guest You're welcome. What level are you working at? Would you like me to explain anything further? Just let me know and I can edit the post. – Fly by Night Aug 18 '13 at 16:11
Here is a hint:
We have
$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$
Use matrix multiplication to compute (I'll do it for you) ...
$B=A^2 = \left(\begin{array}{cc}\cos^2 x -\sin^2 x& -2\sin x \cos x \\ 2\sin x \cos x & \cos^2 x -\sin^2 x\end{array}\right)$
Now simplify using the trigonometry you know, and compute $B^2=A^4$.
-
I found the result as cos(8x) and it is correct. Thanks to you and to@Fly by Night. – guest Aug 18 '13 at 16:11
Here's my way of looking at this: let $J$ be the $2 \times 2$ matrix given by
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$;
then $J^2 = -I$, where $I$ is the $2 \times 2$ identity matrix:
$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Now observe that
$A(x) = \cos x I + \sin x J$,
and that since $J^2 = -I$, $J$ behaves algebraically exactly like $i = \sqrt {-1} \in \mathbb{C}$, $\mathbb{C}$ being the ordinary complex numbers. In particular we have, for $a, b$ real
$(aI + bJ)^2 = (a^2 - b^2)I + 2abJ$,
and if $c,d$ are real as well,
$(aI + bJ)(cI + dJ) = (ac - bd)I + (ad + bc)J$.
These formulas are easy to verify simply using elementary matrix algebra, one really doesn't even need to look at specific matrix entries, just maneuver expressions as if they were polynomials in $J$, and reduce using $J^2 = -I$. Furthermore, they show that de Moivre's classic formula
$(\cos \theta + i\sin \theta)^n = \cos n \theta + i\sin n \theta$
holds for matrices of the form $\cos \theta I + \sin \theta J$:
$(\cos \theta I + \sin\theta J)^n = \cos n \theta I + \sin n \theta J$.
Proving the above equation merely relies on a simple induction on the exponent $n \in \Bbb{Z}$, $n \ge 0$: if
$(\cos \theta I + \sin\theta J)^k = \cos k \theta I + \sin k \theta J$,
then multiplying through by $\cos \theta I + \sin \theta J$ yields
$(\cos \theta I + \sin\theta J)^{k + 1} = (\cos \theta I + \sin \theta J)(\cos k \theta I + \sin k \theta J)$,
and we have using the above formulas
$(\cos \theta I + \sin \theta J)(\cos k \theta I + \sin k \theta J)$
$= (\cos \theta \cos k \theta - \sin \theta \sin k \theta)I + (\sin \theta \cos k \theta + \cos \theta \sin k \theta)J$
$=\cos(k + 1) \theta I + \sin(k + 1) \theta J$,
this last equality relying on the standard addition formulas for $\sin (x + y)$ and $\cos (x + y)$. For more information on this derivation, see this wikipedia page.
The OP's specific concern is neatly wrapped up by observing that since
$A = \cos x I + \sin x J$,
it follows from what we have seen that
$A^4 = \cos 4x I + \sin 4x J$,
and we have $ad + bc$ (using now our OP guest's $a, b, c, d$) as
$ad + bc = \cos^2 4x - \sin^2 4x = \cos 8x$,
as our friend Fly by Night pointed out in his answer to this question.
What I like about the present approach is, that once we have
$(\cos \theta I + \sin\theta J)^n = \cos n \theta I + \sin n \theta J$,
it is really not much extra work to calculate all sorts of stuff about $A^n$ for any $n \in \Bbb{Z}$, $n \ge 0$; for example
$A^{17} = \cos 17 x I + \sin 17 x J$,
and we have
$ad + bc = \cos 34 x$
in this case.
Hope I haven't been too long winded, or gone too far afield. Cheeer-i-o, Ladies and Gents!
-
Did you notice that the OP said that my "solution is a bit complicated for" him? – Fly by Night Aug 18 '13 at 21:12
@Fly by Night: No, in fact, I did not. I confess that I am none too diligent about reading comments that aren't directed at me or directly relate to one of my posts. Not that I don't want to/enjoy doing so, just busy and somewhat preoccupied with other matters. I can only hope that, if my answer overshot the mark, it is communicative enough that guest or some other SE member finds in it at least some of the satisfaction I attained in it's composition. And if someone learns something from it, even if they can't grasp the whole thing, that is very good by me. Thanks for the feedback! – Robert Lewis Aug 18 '13 at 21:31
\begin{eqnarray*} A & = & \left(% \begin{array}{rr} \cos\left(x\right) & -\sin\left(x\right) \\ \sin\left(x\right) & \cos\left(x\right) \end{array} \right) = \cos\left(x\right) - {\rm i}\sin\left(x\right)\,\sigma_{y} \\ A^{2} & = & \cos^{2}\left(x\right) - \sin^{2}\left(x\right) - 2{\rm i}\sin\left(x\right)\cos\left(x\right)\,\sigma_{y} = \cos\left(2x\right) - {\rm i}\sin\left(2x\right)\,\sigma_{y} \\&&\mbox{Then} \\ A^{4} & = & \cos\left(4x\right) - {\rm i}\sin\left(4x\right)\,\sigma_{y} = \left(% \begin{array}{rr} \cos\left(4x\right) & -\sin\left(4x\right) \\ \sin\left(4x\right) & \cos\left(4x\right) \end{array} \right) \quad\Longrightarrow\quad {\rm det}\,\left(A^{4}\right) = 1 \\[5mm]&&\mbox{} \end{eqnarray*}
$$ad + bc = \cos^{2}\left(4x\right) - \sin^{2}\left(4x\right) = \cos\left(8x\right)$$
$\sigma_{y}$ is a Pauli Matrix: http://en.wikipedia.org/wiki/Pauli_matrices
Indeed, ${\rm det}\,\left(A^{4}\right) =\left({\rm det}\,A\right)^{4} = 1^{4} = 1$. Also,
$$A' = -\sin\left(x\right) - {\rm i}\cos\left(x\right)\,\sigma_{y} = -{\rm i}\left\lbrack\cos\left(x\right) - {\rm i}\sin\left(x\right)\,\sigma_{y}\right\rbrack = -{\rm i}A \ \Longrightarrow\ A = {\rm e}^{-{\rm i}\,x\,\sigma_{y}}$$ since $A'' + A = 0$ with $\left.A\right\vert_{x\ =\ 0} = 1$ and $\left.A'\right\vert_{x\ =\ 0} = -{\rm i}\,\sigma_{y}$. That means $$A^{n} = {\rm e}^{-{\rm i}\,n\,x\,\sigma_{y}} =\left(% \begin{array}{rr} \cos\left(nx\right) & -\sin\left(nx\right) \\ \sin\left(nx\right) & \cos\left(nx\right) \end{array} \right)\,, \quad A\ \mbox{is a Rotation Matrix}$$
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# Arch | Classification of Arches
## Classification of Arch
• An arch may be defined as a mechanical arrangement of wedge-shaped blocks of stones or bricks mutually supporting each other and also supported at the end by piers or abutments.
• The function of an Arch is to carry the weight of the structure above the opening because of their shape the block support each other with mutual of their personal weight also the structure remains in position on the support.
• Intrados- Here is the inner curve from an arch.
• Soffits- It is the inner surface of an arch.
• Extrados- It is the outer curve of an arch.
• Key- It is the curve-shaped unit fixed at the crown of the arch.
• Springing- It is an imaginary line joining the springing points of either end.
• Abutment- That is the end support about an arch.
• Voussoirs- This is a wedge-shaped unit of masonry forming an arch.
#### Precast Arches
• Precast arches remain installed on an in-situ concrete strip foundation and also are placed end-to-end to produce the desired structure length.
• Wherever large expanses are to be crossed, for example, o a flood plain, multiple arches span equally placed side at the side and provide an aesthetically pleasing structure.
• The precast arch system is suitable for highways, rails, mining and hydraulic applications.
#### Type of Arches
• A flat arch when the shape of the arch is usually flat, it is called a flat arch.
• Segment arch When the centre of the arch is situated below the springing line, it is called a segmented arch.
• Semi-circular arch When the shape of the arch is semi-circular and the centre of the arch lies on the springing line it is called as semi-circular.
• Semi-elliptical arch When the shape of an arch is Semi-elliptical it is called a Semi-elliptical arch such an arch has more than one centre.
• Inverted arch Such an arch is used to increase the bearing capacity of the soil.
• Pointed arch That type of arch consists of two curves, which imply meeting at the apex of a triangle.
• Reliving arch When the arch is constructed over a wooden joint or a flat arch it is known as relieving arch.
• Horseshoe arch When the shape of an arch is horse-shoe type it is called a horseshoe arch.
• Stilled arch When the shape of an arch is Semi-circular type and attached at the tops of two vertical portions it is known as a Stilled arch.
• Venetian Arch When the depth at the crown is more than that of the springing line it is known as Venetian Arch.
• Florentine Arch That type of arch is similar to the Venetian arch except that it has a shape from a semi-circular curve.
## Classification of Arches
#### Based on Shape
1. Flat Arch
2. Segmental Arch
3. Semi-circular Arch
4. Semi-Elliptical Arch
5. Inverted Arch
6. Pointed Arch
7. Horse Shoe Arch
8. Venetian Arch
9. Florentine Arch
10. Stilled Arch
#### Based on Centers
1. One Centre arch
2. Two Centre arch
3. Three Centre arch
#### Based on Workmanship
1. Rough Arch
2. Axed or Rough Cut Arch
3. Gauged Arch
Based on Material Arch
1. Stone Arch
2. Brick Arch
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### Borrowing in Subtraction in Various Contexts
```Date: 10/25/2006 at 23:48:36
From: Evelyn
Subject: 12 5/11 - 2 6/11
How can you subtract 12 5/11 - 2 6/11? You can't take 6 apples from 5!
```
```
Date: 10/26/2006 at 10:09:57
From: Doctor Ian
Subject: Re: 12 5/11 - 2 6/11
Hi Evelyn,
You're right, but you CAN take 6 apples from 16. What do I mean by that?
Well, note that
12 5/11
= 12 + 5/11 This is what we MEAN by a mixed number
= 11 + 1 + 5/11 Because 12 = 11 + 1
= 11 + 11/11 + 5/11 Because 1 = 11/11
= 11 + 16/11 Add the fractions
Were you able to follow that? If so, note that your subtraction can
be written this way:
11 16/11
- 2 6/11
----------
Can you do it now?
Note that this is EXACTLY the same thing we do with a "normal"
subtraction, like
43
- 17
----
where we "borrow" or "regroup", trading one group of 10 for 10
"groups" of 1, so we can complete the subtraction in each column:
1
3 3 40 + 3 is the same as 30 + 13
- 1 7
------
2 6
|____ 13 - 7 = 6
It's just the same idea, appearing in a different situation. You can
use the same idea again when dealing with times, e.g.,
4:22 3:82 4 hours + 22 minutes is the
- 1:40 -> - 1:40 same as 3 hours + 82 minutes.
------ ------
2:42
or other units,
4 gallons, 2 quarts 3 gallons, 6 quarts
- 1 gallon, 3 quarts -> - 1 gallon, 3 quarts
--------------------- ---------------------
2 gallons, 3 quarts
So this single idea turns out to be pretty useful! The trick is
remembering that it IS the same idea, and not a bunch of unrelated
techniques.
Does that make sense? Let me know if you need more help.
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Elementary Fractions
Elementary Place Value
Elementary Subtraction
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December 12, 2013
# Posts by tabby
Total # Posts: 220
trig/precalc
bobpursely: Thank you so much! God bless you for taking the time to help!
trig/precalc
a car's flywheel has a timing mark on it's outer edge. the height of the timing mark on the rotating flywheel is given by y=3.55sin[x - (pi/4)]. graph one full cycle of this function. I was not taught how to graph a function like this, only the simple y= sin x and y= c...
Art and Writing
Lauren: All arguments are very good points made by both sides. And yes, it is hard to make a living as an artist or writer. They aren't called starving artists for nothing. On the other side, if art and creativity are your passion, nothing less will ever satisfy you. (And ...
English
Dear Anon - If you want evidence to support your claim, may I suggest you pick an airline and use it fictitiously in your essay. By this I mean, write something like "Your friend Miranda moves to LA, California. You live in Houston, TX. You were very close and enjoyed tal...
English College composition I
Your thesis statement should be the sentance that grabs the reader's attention and gets them interested while at the same time states your viewpoint. Be sure you know where you stand on your topic - you obviously do. I like your first thesis statement, especially the part ...
Math
If sinθ=1/5 and θ is in Quadrant I, find the exact value of sin2θ.
Math/Trig
Given A=56degrees, C=61degrees, and b=10.5, solve triangle ABC. If no triangle exists, explain why. If two solutions exist, write both. Round your answer to the nearest tenth
Math/Trigonometry
1. Solve: 2 cos² x - 3 cos x + 1 = 0 for 0 ≤ x < 2pi. 2. Solve: 2 sin x - 1 = 0 for 0° ≤ x < 360° 3. Solve: sin² x = cos² x for 0° ≤ x < 360° 4. Solve: sin x - 2sin x cos x = 0 for 0 ≤ x < 2pi.
Trig/Math
f(x)=1/3sin(2/3x-π/4)+4 Find the amplitude, period, phase shift, and vertical shift of the function. Amplitude:1/3 Period:?????? Phase Shift:??????? Vertical Shift:+4 Please Help ?
Trig/Math
f(x)=1/3sin(2/3x-π/4)+4 Find the amplitude, period, phase shift, and vertical shift of the function. Amplitude:1/3 Period:?????? Phase Shift:??????? Vertical Shift:+4 Please Help ?
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It is often the case that a number is naturally associated to the outcome of a random experiment: the number of boys in a three-child family, the number of defective light bulbs in a case of 100 bulbs, the length of time until the next customer arrives at the drive-through window at a bank. Such a number varies from trial to trial of the corresponding experiment, and does so in a way that cannot be predicted with certainty; hence, it is called a random variable. In this chapter and the next we study such variables.
• 6.1: Random Variables
A random variable is a number generated by a random experiment. A random variable is called discrete if its possible values form a finite or countable set. A random variable is called continuous if its possible values contain a whole interval of numbers.
• 6.2: Probability Distributions for Discrete Random Variables
The probability distribution of a discrete random variable X is a list of each possible value of X together with the probability that X takes that value in one trial of the experiment. The probabilities in the probability distribution of a random variable X must satisfy the following two conditions: Each probability P(x) must be between 0 and 1 and the sum of all the probabilities is 1 .
• 6.3: The Binomial Distribution
Suppose a random experiment has the following characteristics. There are n identical and independent trials of a common procedure. There are exactly two possible outcomes for each trial, one termed “success” and the other “failure.” The probability of success on any one trial is the same number p. Then the discrete random variable X that counts the number of successes in the n trials is the binomial random variable with parameters n and p. We also say that X has a binomial distribution
• 6.E: Discrete Random Variables (Exercises)
6: Discrete Random Variables is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts. | 2,169 | 6,334 | {"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} | 3.84375 | 4 | CC-MAIN-2024-30 | latest | en | 0.197254 |
https://us.metamath.org/ilegif/df-enq0.html | 1,721,061,369,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00607.warc.gz | 538,404,465 | 5,881 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > df-enq0 Unicode version
Definition df-enq0 7252
Description: Define equivalence relation for nonnegative fractions. This is a "temporary" set used in the construction of complex numbers, and is intended to be used only by the construction. (Contributed by Jim Kingdon, 2-Nov-2019.)
Assertion
Ref Expression
df-enq0 ~Q0
Distinct variable group: ,,,,,
Detailed syntax breakdown of Definition df-enq0
StepHypRef Expression
1 ceq0 7114 . 2 ~Q0
2 vx . . . . . . 7
32cv 1331 . . . . . 6
4 com 4508 . . . . . . 7
5 cnpi 7100 . . . . . . 7
64, 5cxp 4541 . . . . . 6
73, 6wcel 1481 . . . . 5
8 vy . . . . . . 7
98cv 1331 . . . . . 6
109, 6wcel 1481 . . . . 5
117, 10wa 103 . . . 4
12 vz . . . . . . . . . . . . 13
1312cv 1331 . . . . . . . . . . . 12
14 vw . . . . . . . . . . . . 13
1514cv 1331 . . . . . . . . . . . 12
1613, 15cop 3531 . . . . . . . . . . 11
173, 16wceq 1332 . . . . . . . . . 10
18 vv . . . . . . . . . . . . 13
1918cv 1331 . . . . . . . . . . . 12
20 vu . . . . . . . . . . . . 13
2120cv 1331 . . . . . . . . . . . 12
2219, 21cop 3531 . . . . . . . . . . 11
239, 22wceq 1332 . . . . . . . . . 10
2417, 23wa 103 . . . . . . . . 9
25 comu 6315 . . . . . . . . . . 11
2613, 21, 25co 5778 . . . . . . . . . 10
2715, 19, 25co 5778 . . . . . . . . . 10
2826, 27wceq 1332 . . . . . . . . 9
2924, 28wa 103 . . . . . . . 8
3029, 20wex 1469 . . . . . . 7
3130, 18wex 1469 . . . . . 6
3231, 14wex 1469 . . . . 5
3332, 12wex 1469 . . . 4
3411, 33wa 103 . . 3
3534, 2, 8copab 3992 . 2
361, 35wceq 1332 1 ~Q0
Colors of variables: wff set class This definition is referenced by: enq0enq 7259 enq0sym 7260 enq0ref 7261 enq0tr 7262 enq0er 7263 enq0breq 7264 enq0ex 7267
Copyright terms: Public domain W3C validator | 880 | 1,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.11444 |
https://www.physicsforums.com/threads/modular-arithmetic.983141/ | 1,579,613,719,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604397.40/warc/CC-MAIN-20200121132900-20200121161900-00420.warc.gz | 1,034,227,653 | 15,170 | # Modular arithmetic
Gold Member
## Homework Statement:
22x^2 + 13(x + 2y)^3 = 11 has no integer solutions x and y
## Homework Equations:
Module asthmatic.
I am thinking of taking modular of 13 to both sides of equation. So it will become
22x^2 is 11 mod 13.
And try all the values from x equal to zero to x equal to 11.
Is their easier way to solve it
I think it is a good idea, what I would do is to define $z=x^2$ to have the equation $az=c \pmod{m}$ which is very easy to solve without trying an error (which is not a problem for this small numbers, but its always nice to have a method to solve things), and then you only need to solve $z = x^2 \pmod{m}$ which again can be done without trying an error.
Well, as I said, the equations $ax=b \pmod{m}$ and $x^2=a \pmod{m}$ are very common and important and you can find the solutions of both without trying and error (if you have not studied this then it's OK). Of course, that means that, if they don't have a solution you can prove it without having to try all the possibilities. (As I said, since you are working $\pmod{13}$ try and error is not a big deal, but if you were working at $mod{97}$ it would be really long, and a number with 10 digits would be impossible except maybe for a computer.) | 340 | 1,260 | {"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} | 3.5625 | 4 | CC-MAIN-2020-05 | latest | en | 0.97259 |
https://www.explorebesttoday.com/interesting/what-is-cash-reserve-ratio.html | 1,624,350,577,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488512243.88/warc/CC-MAIN-20210622063335-20210622093335-00285.warc.gz | 674,522,175 | 33,530 | # What is cash reserve ratio
## What is mean by cash reserve ratio?
The percentage of cash required to be kept in reserves, vis-a-vis a bank’s total deposits, is called the Cash Reserve Ratio. The cash reserve is either stored in the bank’s vault or is sent to the RBI. Banks do not get any interest on the money that is with the RBI under the CRR requirements.
## What is cash reserve ratio with example?
Cash reserve Ratio (CRR) is the amount of Cash that the banks have to keep with RBI. This Ratio is basically to secure solvency of the bank and to drain out the excessive money from the banks. For example, if you deposit Rs 100 in your bank, then bank can’t use the entire Rs 100 for lending or investment purpose.
## What is the SLR and CRR?
CRR and SLR are the two ratios. CRR is a cash reserve ratio and SLR is statutory liquidity ratio. Under CRR a certain percentage of the total bank deposits has to be kept in the current account with RBI which means banks do not have access to that much amount for any economic activity or commercial activity.
5.00 per cent
## What mean by SLR?
Statutory liquidity ratio
## Why is cash reserve ratio important?
The CRR (4 per cent of NDTL) requires banks to maintain a current account with the RBI with liquid cash. … While ensuring some liquid money against deposits is the primary purpose of CRR, its secondary purpose is to allow the RBI to control liquidity and rates in the economy.
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## How is cash reserve ratio calculated?
There is no cash reserve ratio formula. In technical terms, CRR is calculated as a percentage of net demand and time liabilities (NDTL). NDTL for banking refers to the aggregate savings account, current account and fixed deposit balances held by a bank.
## How does cash reserve ratio work?
Definition: Cash Reserve Ratio (CRR) is a specified minimum fraction of the total deposits of customers, which commercial banks have to hold as reserves either in cash or as deposits with the central bank. … The aim here is to ensure that banks do not run out of cash to meet the payment demands of their depositors.
## What is required reserve ratio formula?
Formula for Required Reserve Ratio
The reserve ratio is simply a fraction of deposits that banks hold in reserves. … For example: If the required reserve ratio is 10%, you can simply convert it to a fraction by dividing this percentage number by 100 to get the ratio of reserves to deposits.
## What is CRR and SLR rate 2020?
9th October 2020 – RBI keeps Repo Rate unchanged at 4%IndicatorCurrent RateCRR3%SLR18.50%Repo Rate4.00%Reverse Repo Rate3.35%
## What is the purpose of CRR and SLR?
Basic differences between CRR and SLR.SLR (Statutory Liquidity Ratio)Cash Reserve Ratio (CRR)This ratio is used by the RBI to control the bank’s leverage for credit expansion.CRR is issued by the central bank to control the liquidity in the market.
## Which banks maintain CRR and SLR?
CRR regulates the flow of money in the economy whereas SLR ensures the solvency of the banks. CRR is maintained by RBI, but RBI does not maintain SLR. The liquidity of the country is regulated by CRR while SLR governs the credit growth of the country.16 мая 2020 г.
You might be interested: When was the federal reserve bank created
## What happens when cash reserve ratio increases?
If the CRR is raised to 6%, then a bank must keep Rs 6 for every Rs 100 deposits. Cash deposit to be maintained with RBI by a bank increases with increase in CRR. When CRR is increased, then the banks would not have more money at their disposal to sanction loans. | 822 | 3,637 | {"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} | 2.734375 | 3 | CC-MAIN-2021-25 | latest | en | 0.951255 |
http://www.johndcook.com/blog/2010/01/31/parameters-from-percentiles/ | 1,481,269,227,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542687.37/warc/CC-MAIN-20161202170902-00452-ip-10-31-129-80.ec2.internal.warc.gz | 532,997,111 | 14,897 | # Parameters and percentiles
The doctor says 10% of patients respond within 30 days of treatment and 80% respond within 90 days of treatment. Now go turn that into a probability distribution. That’s a common task in Bayesian statistics, capturing expert opinion in a mathematical form to create a prior distribution.
Things would be easier if you could ask subject matter experts to express their opinions in statistical terms. You could ask “If you were to represent your belief as a gamma distribution, what would the shape and scale parameters be?” But that’s ridiculous. Even if they understood the question, it’s unlikely they’d give an accurate answer. It’s easier to think in terms of percentiles.
Asking for mean and variance are not much better than asking for shape and scale, especially for a non-symmetric distribution such as a survival curve. Anyone who knows what variance is probably thinks about it in terms of a normal distribution. Asking for mean and variance encourages someone to think about a symmetric distribution.
So once you have specified a couple percentiles, such as the example this post started with, can you find parameters that meet these requirements? If you can’t meet both requirements, how close can you come to satisfying them? Does it depend on how far apart the percentiles are? The answers to these questions depend on the distribution family. Obviously you can’t satisfy two requirements with a one-parameter distribution in general. If you have two requirements and two parameters, at least it’s feasible that both can be satisfied.
If you have a random variable X whose distribution depends on two parameters, when can you find parameter values so that Prob(Xx1) = p1 and Prob(Xx2) = p2? For starters, if x1 is less than x2 then p1 must be less than p2. For example, the probability of a variable being less than 5 cannot be bigger than the probability of being less than 6. For some common distributions, the only requirement is this requirement that the x‘s and p‘s be in a consistent order.
For a location-scale family, such as the normal or Cauchy distributions, you can always find a location and scale parameter to satisfy two percentile conditions. In fact, there’s a simple expression for the parameters. The location parameter is given by
and the scale parameter is given by
where F(x) is the CDF of the distribution representative with location 0 and scale 1.
The shape and scale parameters of a Weibull distribution can also be found in closed form. For a gamma distribution, parameters to satisfy the percentile requirements always exist. The parameters are easy to determine numerically but there is no simple expression for them.
For more details, see Determining distribution parameters from quantiles. See also the ParameterSolver software.
Update: I posted an article on CodeProject with Python code for computing the parameters described here.
Related posts:
## 7 thoughts on “Parameters and percentiles”
1. Gregor Gorjanc
This is cool!
2. Super awesome. In addition to the practical problem you’ve solved (and could there be survey applications? customer response? employee preference? dating sites?) it makes me think of several abstract issues.
For one, what’s the minimum storage size of a distribution? If you manually specified “all” of the percentiles of the exponential distribution, it could take an infinite amount of storage space, right? But you only need one parameter (or should I say three: base, λ, and the function definition) to generate it. On the other hand what’s so special about the functions that we know? A sum of exponentials is simple in some other linear basis but I can’t think of a cheap way to store that in a computer.
For two, every probability distribution is just a 1-generalised function s.t. ∫ƒ is a function while ƒ is not guaranteed to be one. Which is why it makes sense to store the CDF’s (they’re also monotonic which has to be good for making them small in some way). So now chase the diagram backwards: your solution with storing percentiles of CDF’s is applicable to any 1-generalised function. Does that buy us anything for free?
3. Pingback: What distribution can be closely (or precisely) fit to the “5 number summary” statistics? | Q&A System
4. Well, that is a useful post. I shall add that we can elicit the judgements using an alternative way by asking the “expert” to provide e.g. the lower and upper quartiles.
In regard to how to assess the fitted distribution then we may use the feedback stage by reporting to the “expert” properties of the fitted distribution for confirmation.
5. As usual I tried solving it first before reading your solution, and for once I got it right! (I am a stats novice, so that is an accomplishment for me.) My method was to first scale in order to find the standard deviation, then multiply the standard deviation by the standard quantile of p1 to figure out how far q1 is from the mean. I think that is mathematically equivalent to what you showed. Here is my R code using a normal distribution:
` p1 <- .1 p2 <- .8 q1 <- 30 q2 <- 90`
`# scale to find the sd: sd <- (q2 - q1) / (qnorm(p2) - qnorm(p1)) # multiply sd by standard quantile to find the mean: mu 0.1 pnorm(90, mu, sd) # => 0.8 `
6. *Oops, the code above should read:
p1 <- .1
p2 <- .8
q1 <- 30
q2 <- 90
# scale to find the sd:
sd <- (q2 – q1) / (qnorm(p2) – qnorm(p1))
# use a given quantile and sd to find the mean:
mu <- q1 – qnorm(p1) * sd
# check for correctness:
pnorm(30, mu, sd)
pnorm(90, mu, sd)
7. I recently read Morgan2014, Use (and abuse) of expert elicitation in support of decision making for public policy.
In it there are some nice tips on how to ask if you want more info on the shape. It is also good to be aware of the ubiquitous over confidence.
http://www.ncbi.nlm.nih.gov/pubmed/24821779 | 1,334 | 5,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-50 | latest | en | 0.928116 |
https://math.answers.com/questions/What_is_1134_divided_by_6 | 1,660,609,899,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00743.warc.gz | 374,712,159 | 39,499 | 0
# What is 1134 divided by 6?
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2011-11-17 12:03:39
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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1134/6 = 189 | 128 | 327 | {"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-2022-33 | latest | en | 0.743875 |
kosterfjord.se | 1,709,019,688,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474671.63/warc/CC-MAIN-20240227053544-20240227083544-00013.warc.gz | 20,674,907 | 27,007 | # Keno Algorithm: In 5 Easy Steps
Keno Algorithm: In 5 Easy Steps
Randomness exams for big samples It’s This That I Use My Games For Young Women For of Keno:Daniel Corriveau driven immediately following your dog may make several checks in a casino with the keno performance on the ruined computer chip machine. With this she or he figured the latest broken/missing home pc chip shot will probably deliver comparable phone numbers, not likely randomly numbers. A good standard gambler may not take full advantage of a misstep inside RNG application, yet he’ll see, ascertain as well as guess for many details which are often taken out in just a few match sessions.
Methods to Define and additionally Sign up typically the Algorithmic program:Writing (at betting house table keno and / or online) the best way various amounts usually are replicated plus pay attention to the regularity involved with the look of them can offer with a poker player your opportunity improve it’s prospects of winning. Watch typically the extractions while in in one week and see which usually phone numbers are generally to appear sometimes, note these folks and after that enjoy those numbers. It’s commended for you to guarantee the minimum regarding 10 numbers of having far better chances. You will discover practices to spend time playing through, though observing/determining any keno algorithmic rule is among the most best.
Web based keno algorithmic program:These kinds of over the internet algorithmic rule has the ability to guide bettors engaged to find helpful numbers from the keno game. Formulated seeing that a web-based plan and even operating under a keno simulator, this valuable algorithm causes estimations involving quantities, and these can be “extracted” on the game. Every one professional provides two solutions: a single is related to the amount of bargains he or she would like carry out and then the other relates to how many volumes, the ball player really wants to use on a new ticket. In that case, subsequent to showing up in the “calculate” tab, all the applications are preparing it has the estimations and also several possibilities succeeding numbers to solution on.
Keno Protocol meant for Far better Risks of Being successful:Keno may be a alluring online casino pastime, an awesome mix off lady luck, strategy and probabilities. Remaining an exceptional test, perhaps you may improve your chances of profiting merely as it were readily protocol for sets of rules, determined by easy statistical rules. This kind of algorithmic program is tested possibly at home, frequently online (and this can be the most effective way). The operation suggests seeing every extractions conducted in a few days, note them and after that take up accordingly. It’s always a lot easier to determine the criteria found in an existing on line casino, for the reason that the sole thing you want to do is normally to observe the tables. With regard to web based keno, you should spend money on seats for not less than few units, consequently, based on information made to bring about your current observations. Then simply, you will need to can guess on more numbers (minimum 10 numbers) remember which gaming more than usual, might boost up chances. | 652 | 3,254 | {"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-2024-10 | latest | en | 0.949057 |
https://appedreview.com/app/ruler-professional/ | 1,685,781,517,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00047.warc.gz | 131,513,217 | 18,168 | # Ruler Professional
## Review Overview
### 7.1
#### Ruler Professional: A Best App for Measurement
Summary : An Easy-To-Use Ruler App
[asa_item id=”527471088″]
Ruler Professional provides users with an interactive ruler. When launched, Ruler Professional shows users a 6″ ruler with two red lines. Users can adjust the red lines, and Ruler Professional will report the distance between the two lines (up to 6.5″). Users can also change the metric from inches to centimeters by clicking the “Info” button at the top of the screen.
### Instructional Ideas
1. Teachers can have students measure common objects (e.g., pencil, eraser, book) found in the classroom to build their familiarity with measurement. Teachers can require students to log the items they measured and their length.
2. Teachers can create a homework assignment in which they give students a variety of measurements (e.g., 2″, 3.75″, 5.25″) and require students to locate objects in their home of that length. Teachers can add a twist by requiring students to use their iPad’s camera to photograph the objects as well.
3. Teachers can build students’ addition skills with this app. To do so, teachers give students a series of objects to measure, and students are to measure all the objects, add their measurements together, and then determine an overall length.
4. Teachers can have students use this app to help them solve word problems dealing with measurement and length. For example, if Ted has a pen that measures 4.5 inches and Tonya has a pencil that measures 12.3 centimeters, who has the longer writing utensil?
A1. Rigor A2. 21st Century Skills A3. Conn. to Future Learning A4. Value of Errors A5. Feedback to Teacher A6. Level of Material A7. Cooperative Learning A8. Accom. of Individual Diff.
B1. Ability to Save Progress B2. Platform Integration B3. Screen Design B4. Ease of Use B5. Navigation B6. Goal Orientation B7. Information Presentation B8. Media Integration B9. Cultural Sensitivity
C1. Learner Control C2. Interactivity C3. Pace C4. Flexibility C5. Interest C6. Aesthetics C7. Utility | 496 | 2,087 | {"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-2023-23 | latest | en | 0.904189 |
http://mathoverflow.net/feeds/question/104009 | 1,371,665,524,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709006458/warc/CC-MAIN-20130516125646-00018-ip-10-60-113-184.ec2.internal.warc.gz | 161,064,488 | 2,482 | 'Condition number' for Rayleigh-Ritz quotient - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:12:03Z http://mathoverflow.net/feeds/question/104009 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104009/condition-number-for-rayleigh-ritz-quotient 'Condition number' for Rayleigh-Ritz quotient Felix Goldberg 2012-08-05T10:44:01Z 2012-12-07T22:10:22Z <p>Suppose that $A$ is a Hermitian matrix and that $u,v$ are two vectors. Is there some known function $\kappa(A)$ so that $||u-v|| \leq \kappa(A) |\frac{u^{*}Au}{u^{*}u}-\frac{v^{*}Av}{v^{*}v}|$?</p> <p>UPDATE: Andrew T. Baker has shown that the answer is "no" in general (to take an even simpler counter-example, take $A=I$) - so let's add the <strong>assumption</strong> that $u$ is a simple eigenvector of $A$.</p> http://mathoverflow.net/questions/104009/condition-number-for-rayleigh-ritz-quotient/104199#104199 Answer by Andrew T. Barker for 'Condition number' for Rayleigh-Ritz quotient Andrew T. Barker 2012-08-07T14:02:05Z 2012-08-07T16:38:23Z <p>I think the answer for a general Hermitian matrix $A$ is no.</p> <p>Let $u, v$ be distinct eigenvectors of $A$ with the same eigenvalue $\lambda$ and normalized so that $u^\ast u = v^\ast v = 1$. Then $\| u - v \| > 0$ and \begin{equation*} | u^\ast Au - v^\ast A v | = | \lambda u^\ast u - \lambda v^\ast v | = | \lambda - \lambda | = 0. \end{equation*}</p> <p>A concrete counterexample is \begin{equation*} A = \left( \begin{array}{ccc} 2& & \\ & 2& \\ & & 1 \end{array} \right), u = \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right), v = \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right) \end{equation*}</p> <p>ADDED: if we assume all eigenvalues are distinct, then the above argument leads to some kind of bound. Again, $u,v$ are eigenvectors of $A$ with corresponding eigenvalues $\lambda_1, \lambda_2$ and normalized with $u^\ast u = v^\ast v = 1$. Then \begin{equation*} \| u - v \| \leq 2 \end{equation*} and \begin{equation*} | u^\ast Au - v^\ast A v | = | \lambda_1 - \lambda_2 | \end{equation*}</p> <p>so that if we choose \begin{equation*} \kappa(A) \geq \min_{\lambda_i, \lambda_j} \frac{2}{|\lambda_i - \lambda_j|} \end{equation*} we can guarantee your inequality will hold.</p> http://mathoverflow.net/questions/104009/condition-number-for-rayleigh-ritz-quotient/104210#104210 Answer by Sharkos for 'Condition number' for Rayleigh-Ritz quotient Sharkos 2012-08-07T16:31:49Z 2012-08-07T16:31:49Z <p>Another trivial problem, I'm afraid:</p> <p>The two sides scale differently under rescaling of $u, v$. That is, if it holds for some $\kappa, u \neq v$ then simply multiply $u, v$ by some huge number $Z$ - then the left-hand side is now $Z$ times bigger whilst the right-hand side is constant.</p> <p>How many further qualifications are necessary?!</p> http://mathoverflow.net/questions/104009/condition-number-for-rayleigh-ritz-quotient/115736#115736 Answer by RLC for 'Condition number' for Rayleigh-Ritz quotient RLC 2012-12-07T19:10:42Z 2012-12-07T22:10:22Z <p>If u is an eigenvector for an eigenvalue that is in the interior of the spectrum, then there is no such bound becomes there are numerous vectors that generate Rayleigh quotients for points in the convex hull of the spectrum. And as for Sharkos comment, you would want something like the sine of the angle between u and v.</p> <p>An absolute condition number for a Rayleigh quotient would look more like $|x^*Ax-y'Ay|<=\kappa$(x'Ax) ||x-y||\$, where K would be the condition number for x'Ax, where x and y are unit vectors. It shows how changes in the Rayleigh quotient can be bounded by changes in the vector.</p> <p>Your suggested condition number is more like the condition number for the generating vector. </p> | 1,225 | 3,802 | {"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} | 3.796875 | 4 | CC-MAIN-2013-20 | latest | en | 0.639574 |
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The Seven Primes (Posted on 2013-04-26)
Let a, b, c be positive integers such that a, b, c, a+b-c, a+c-b, b+c-a, a+b+c are 7 distinct primes.
The sum of two of a, b, c is 800.
If d be the difference of the largest prime and the least prime among those 7 primes, find the maximum value of d.
No Solution Yet Submitted by Danish Ahmed Khan No Rating
Subject Author Date analytical solution xdog 2013-04-26 21:16:45 computer solution Charlie 2013-04-26 17:47:32
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https://www.jiskha.com/display.cgi?id=1304286270 | 1,511,192,444,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806070.53/warc/CC-MAIN-20171120145722-20171120165722-00504.warc.gz | 797,864,652 | 3,797 | # calculus
posted by .
Express as a single logarithm
ln(a)+1/2ln(b)
2ln4-ln2
• calculus -
ln a + (1/2) ln b
= lan a + ln )b^(1/2))
= ln (a√b)
you try the second, and let me know what you got
• calculus -
2ln4-ln2
=ln(16)-ln2
=ln(16/2)
=ln8?
• calculus -
correct
• calculus -
Thank you =)
## Similar Questions
1. ### algebra 3-4
im in a cal class but were reviewing algebra stuff now. and i have 1 more questions. it says. express the given quantity as a single loragithms 2ln4-ln2 so is it 2ln(2) please tell me if im getting the right answers and if im not then …
2. ### Calculus
Write the logarithm of a single quantity: 2ln(x)+ln(4)-(1/2)ln(9)
3. ### calculus
Express as a single logarithm ln(a)+1/2ln(b) 2ln4-ln2
4. ### Algebra
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5. ### Algebra II
Did I do this correctly? Use log laws to write the following expression as a single logarithm. 3/2 ln25 + 3 ln2 = ln25 3/2 + ln2^3 = ln125 + ln2^3 =ln125 + ln8 =ln(125*8) =ln1000
Did I do this correctly? Use log laws to write the following expression as a single logarithm. 3/2 ln25 + 3 ln2 = ln25 3/2 + ln2^3 = ln125 + ln2^3 =ln125 + ln8 =ln(125*8) =ln1000
7. ### ALGEBRA
Contract the expressions. That is, use the properties of logarithms to write each expression as a single logarithm with a coefficient of 1. text ((a) ) ln\(3\)-2ln\(4\)+ln\(8\) ((b) ln\(3\)-2ln\(4+8\) (c) )ln\(3\)-2(ln\(4\)+ln\(8\))
8. ### ALGEBRA
Contract the expressions. That is, use the properties of logarithms to write each expression as a single logarithm with a coefficient of 1. text((a) ) ln\(3\)-2ln\(4\)+ln\(8\) text((b) )ln\(3\)-2ln\(4+8\) text((c) )ln\(3\)-2(ln\(4\)+ln\(8\))
9. ### Calculus
Write as a single logarithm of a single quantity ln(3)+1/2ln(x+2)-4ln(1+sqrtx)
10. ### PreCalculus - Need help, due tonight
1. Rewrite the expression 1/2ln(x – 6) + 3ln(x +1) as a single logarithm. 2. Rewrite the expression 1/3[ln x – 2ln(1 – x)] as a single logarithm.
More Similar Questions | 736 | 2,028 | {"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} | 4.03125 | 4 | CC-MAIN-2017-47 | latest | en | 0.784607 |
https://crypto.stackexchange.com/questions/72217/canonical-embedding-vs-plaintext-slots-in-ring-lwe | 1,582,960,718,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148671.99/warc/CC-MAIN-20200229053151-20200229083151-00056.warc.gz | 326,668,518 | 32,739 | # Canonical embedding vs. plaintext slots in Ring-LWE
I'm working on the canonical embedding mentioned in [LPR10] and [LPR13]. What confuses me is that the difference and the relationship between the canonical embedding and the concept of ''plaintext slot'' in other works (e.g., [SV11]).
Let's focus on the $$m$$-th cyclotomic number field $$K = \mathbb{Q}(\zeta_m)$$ where $$\zeta_m$$ is an abstract element of order $$m = 2^k$$ for some $$k$$. For canonical embedding $$\sigma$$, it is comprised of $$\left| \mathbb{Z}_m^* \right| = \varphi(m)$$ embeddings $$\sigma_i: K \mapsto \mathbb{C}$$. More precisely, $$\sigma_i(\zeta_m) = \omega_m^i$$ and $$\sigma(a) = (\sigma_i(a))_{i \in \mathbb{Z}_m^*}$$ for the primitive $$m$$-th root of unity $$\omega_m$$ and $$a \in K$$. The canonical embedding maps an element in $$K$$ to the vector space $$H$$ which is defined to be $$\begin{gather} H = \{ \textbf{x} \in \mathbb{C}^{\mathbb{Z}_m^*}: x_i = \overline{x_{m-i}},\ \forall i \in \mathbb{Z}_m^* \}. \end{gather}$$ Notice that $$x_i = \overline{x_{m-i}}$$ holds for all $$i \in \mathbb{Z}_m^*$$. So using canonical embedding, it seems that there are only $$\varphi(m)/2$$ instead of $$\varphi(m)$$ ''slots'' for an vector in $$H$$ to encode complex numbers. If we want to encode a vector $$\mathbf{v}$$ of real numbers into an element in $$K$$ via $$\sigma^{-1}$$, is it true that the number of elements in $$\mathbf{v}$$ should be at most $$\varphi(m)/2$$ instead of $$\varphi(m)$$? (Question 1)
To generate plaintext slots, we can apply Chinese Remainder Theorem (CRT) to the polynomial ring $$R_p = \mathbb{Z}_p[X]/\Phi_m(X)$$ for the $$m$$-th cyclotomic polynomial $$\Phi_m(X)$$. Note that $$\Phi_m(X) = \prod_{i \in \mathbb{Z}_m^*} (X - \zeta_m^i) \mod p$$ where $$(\zeta_m)^m \equiv 1 \mod p$$ and CRT will result in $$\varphi(m)$$ slots for component-wise addition and multiplication. This CRT is indeed a kind of number-theoretic transform (NTT) by evaluating some polynomial $$b \in R_p$$ at $$b(\zeta_m^i)$$ for $$i \in \mathbb{Z}_m^*$$, which is quite similar to the situation where the embedding $$\sigma_i$$ is applied to $$\mathcal{O}_K$$, the ring of integers of $$K$$. Why do we have $$\varphi(m)$$ slots here while there are only $$\varphi(m)/2$$ ''slots'' in canonical embedding? (Question 2)
Many thanks
My understanding (updated on 29/07/19)
It seems that the difference between canonical embedding and plaintext slot comes from the difference between $$\omega_m$$ and $$\zeta_m$$, in which $$\omega_m^m = 1 \in \mathbb{C}$$ and $$\zeta_m^m = 1 \in \mathbb{Z}_p$$. For some $$a \in \mathcal{O}_K$$ and $$i \in \mathbb{Z}_m^*$$, we have $$\begin{gather} \sigma_i(a) = a(\omega_m^i) = a(\overline{\omega_m^{m-i}}) = \overline{a(\omega_m^{m-i})} \in \mathbb{C}, \end{gather}$$ in which the last equality is derived from the property of complex conjugate. The last equality implies that once one half of the $$\sigma_i(a)$$'s are fixed, the rest of them will be fixed as complex conjugate, which results in the factor $$1/2$$. However, note that there is no counterpart of this property of complex conjugate in $$\mathbb{Z}_p$$. That is, for some $$b \in R_p$$, $$\begin{gather} b(\zeta_m^i) = b((\zeta_m^{m-i})^{-1}) \neq (b(\zeta_m^{m-i}))^{-1} \mod p. \end{gather}$$ So it is safe to encode different values to the $$\varphi(m)$$ slots and there must exist a coresponding polynomial in $$R_p$$ using NTT.
Am I right?
• While this isn't a full answer, one thing to note is that $\mathbb{Q}(\zeta_m)$ is a dimension $\varphi(m)$ field extension over $\mathbb{Q}$, but $H$ is a dimension $\varphi(m)/2$ field over $\mathbb{C}$. Using $\mathbb{C}\cong\mathbb{R}^2$, you get that $H$ is actually a $\varphi(m)$ field over $\mathbb{R}$. This is a way of looking at things to get the dimensions to match up, although $\mathbb{Q}$ being countable and $\mathbb{R}$ being uncountable is an obstacle to this being the full argument. – Mark Jul 28 '19 at 23:13
• Of course since the embedding isn't a bijection, there's no need for them to have the same cardinality. Still, computing $\sigma^{-1}$ would require having some good description of its domain, which is left out of the above argument. – Mark Jul 28 '19 at 23:14
• To encode a vector of reals, one may abandon the imaginary part of each entry in $\mathbb{C}_{\mathbb{Z}_m^*}$. This will result in one half of the vector is identical to the other half. The inverse mapping $\sigma^{-1}$ is similar to fast fourier transform (FFT), I think. – X.G. Jul 30 '19 at 14:18
• @Mark, thank you for your helpful comments. It takes me a day to think about the isomorphism $\mathbb{C}\cong\mathbb{R}^2$. Of course we have such unitary matrix B (c.f., Section 2.2 in LPR13 ) that contributes to the isomorphism $H \mapsto \mathbb{R}^{\varphi(m)}$. However, it seems that the conponent-wise addtion and multipilication in $H$ cannot be transfromed into $\mathbb{R}^{\varphi(m)}$ using B. So, although we have dimensions match up, we still have $\varphi(m)/2$ space in $H$ to encode reals. – X.G. Jul 30 '19 at 14:26
• I think you're assuming that if we want to encode an element with real coefficients in $\mathbb{Q}(\zeta_m)$ (so $\varphi(m)$ coefficients), we can do it by encoding an element with real coefficients in $H$ (so $\varphi(m)/2$ real coefficients). This isn't the case, because $\sigma$ doesn't map real numbers to real numbers (note that specifically $\sigma(1) = \zeta_m$ is complex). In general, you're asking about the "canonical" or "minkowski" embedding, of which there are non-cryptographic resources you can consult. It might be useful for you to try some explicit computations with a CAS. – Mark Jul 30 '19 at 19:41 | 1,781 | 5,702 | {"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": 55, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-10 | latest | en | 0.755939 |
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Partial_fractions_in_complex_analysis | 1,632,739,196,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00626.warc.gz | 1,081,316,782 | 11,972 | # Partial fractions in complex analysis
In complex analysis, a partial fraction expansion is a way of writing a meromorphic function f(z) as an infinite sum of rational functions and polynomials. When f(z) is a rational function, this reduces to the usual method of partial fractions.
## Motivation
By using polynomial long division and the partial fraction technique from algebra, any rational function can be written as a sum of terms of the form 1 / (az + b)k + p(z), where a and b are complex, k is an integer, and p(z) is a polynomial. Just as polynomial factorization can be generalized to the Weierstrass factorization theorem, there is an analogy to partial fraction expansions for certain meromorphic functions.
A proper rational function, i.e. one for which the degree of the denominator is greater than the degree of the numerator, has a partial fraction expansion with no polynomial terms. Similarly, a meromorphic function f(z) for which |f(z)| goes to 0 as z goes to infinity at least as quickly as |1/z|, has an expansion with no polynomial terms.
## Calculation
Let f(z) be a function meromorphic in the finite complex plane with poles at λ1, λ2, ..., and let (Γ1, Γ2, ...) be a sequence of simple closed curves such that:
• The origin lies inside each curve Γk
• No curve passes through a pole of f
• Γk lies inside Γk+1 for all k
• ${\displaystyle \lim _{k\rightarrow \infty }d(\Gamma _{k})=\infty }$ , where d(Γk) gives the distance from the curve to the origin
Suppose also that there exists an integer p such that
${\displaystyle \lim _{k\rightarrow \infty }\oint _{\Gamma _{k}}\left|{\frac {f(z)}{z^{p+1}}}\right||dz|<\infty }$
Writing PP(f(z); z = λk) for the principal part of the Laurent expansion of f about the point λk, we have
${\displaystyle f(z)=\sum _{k=0}^{\infty }\operatorname {PP} (f(z);z=\lambda _{k}),}$
if p = -1, and if p > -1,
${\displaystyle f(z)=\sum _{k=0}^{\infty }(\operatorname {PP} (f(z);z=\lambda _{k})+c_{0,k}+c_{1,k}z+\cdots +c_{p,k}z^{p}),}$
where the coefficients cj,k are given by
${\displaystyle c_{j,k}=\operatorname {Res} _{z=\lambda _{k}}{\frac {f(z)}{z^{j+1}}}}$
λ0 should be set to 0, because even if f(z) itself does not have a pole at 0, the residues of f(z)/zj+1 at z = 0 must still be included in the sum.
Note that in the case of λ0 = 0, we can use the Laurent expansion of f(z) about the origin to get
${\displaystyle f(z)={\frac {a_{-m}}{z^{m}}}+{\frac {a_{-m+1}}{z^{m-1}}}+\cdots +a_{0}+a_{1}z+\cdots }$
${\displaystyle c_{j,k}=\operatorname {Res} _{z=0}\left({\frac {a_{-m}}{z^{m+j+1}}}+{\frac {a_{-m+1}}{z^{m+j}}}+\cdots +{\frac {a_{j}}{z}}+\cdots \right)=a_{j},}$
${\displaystyle \sum _{j=0}^{p}c_{j,k}z^{j}=a_{0}+a_{1}z+\cdots +a_{p}z^{p}}$
so that the polynomial terms contributed are exactly the regular part of the Laurent series up to zp.
For the other poles λk where k 1, 1/zj+1 can be pulled out of the residue calculations:
${\displaystyle c_{j,k}={\frac {1}{\lambda _{k}^{j+1}}}\operatorname {Res} _{z=\lambda _{k}}f(z)}$
${\displaystyle \sum _{j=0}^{p}c_{j,k}z^{j}=[\operatorname {Res} _{z=\lambda _{k}}f(z)]\sum _{j=0}^{p}{\frac {1}{\lambda _{k}^{j+1}}}z^{j}}$
To avoid issues with convergence, the poles should be ordered so that if λk is inside Γn, then λj is also inside Γn for all j < k.
## Example
The simplest examples of meromorphic functions with an infinite number of poles are the non-entire trigonometric functions, so take the function tan(z). tan(z) is meromorphic with poles at (n + 1/2)π, n = 0, ±1, ±2, ... The contours Γk will be squares with vertices at ±πk ± πki traversed counterclockwise, k > 1, which are easily seen to satisfy the necessary conditions.
On the horizontal sides of Γk,
${\displaystyle z=t\pm \pi ki,\ \ t\in [-\pi k,\pi k],}$
so
${\displaystyle |\tan(z)|^{2}=\left|{\frac {\sin(t)\cosh(\pi k)\pm i\cos(t)\sinh(\pi k)}{\cos(t)\cosh(\pi k)\pm i\sin(t)\sinh(\pi k)}}\right|^{2}}$
${\displaystyle |\tan(z)|^{2}={\frac {\sin ^{2}(t)\cosh ^{2}(\pi k)+\cos ^{2}(t)\sinh ^{2}(\pi k)}{\cos ^{2}(t)\cosh ^{2}(\pi k)+\sin ^{2}(t)\sinh ^{2}(\pi k)}}}$
sinh(x) < cosh(x) for all real x, which yields
${\displaystyle |\tan(z)|^{2}<{\frac {\cosh ^{2}(\pi k)(\sin ^{2}(t)+\cos ^{2}(t))}{\sinh ^{2}(\pi k)(\cos ^{2}(t)+\sin ^{2}(t))}}=\coth ^{2}(\pi k)}$
For x > 0, coth(x) is continuous, decreasing, and bounded below by 1, so it follows that on the horizontal sides of Γk, |tan(z)| < coth(π). Similarly, it can be shown that |tan(z)| < 1 on the vertical sides of Γk.
With this bound on |tan(z)| we can see that
${\displaystyle \oint _{\Gamma _{k}}\left|{\frac {\tan(z)}{z}}\right|dz\leq \operatorname {length} (\Gamma _{k})\max _{z\in \Gamma _{k}}\left|{\frac {\tan(z)}{z}}\right|<8k\pi {\frac {\coth(\pi )}{k\pi }}=8\coth(\pi )<\infty .}$
(The maximum of |1/z| on Γk occurs at the minimum of |z|, which is kπ).
Therefore p = 0, and the partial fraction expansion of tan(z) looks like
${\displaystyle \tan(z)=\sum _{k=0}^{\infty }(\operatorname {PP} (\tan(z);z=\lambda _{k})+\operatorname {Res} _{z=\lambda _{k}}{\frac {\tan(z)}{z}}).}$
The principal parts and residues are easy enough to calculate, as all the poles of tan(z) are simple and have residue -1:
${\displaystyle \operatorname {PP} (\tan(z);z=(n+{\frac {1}{2}})\pi )={\frac {-1}{z-(n+{\frac {1}{2}})\pi }}}$
${\displaystyle \operatorname {Res} _{z=(n+{\frac {1}{2}})\pi }{\frac {\tan(z)}{z}}={\frac {-1}{(n+{\frac {1}{2}})\pi }}}$
We can ignore λ0 = 0, since both tan(z) and tan(z)/z are analytic at 0, so there is no contribution to the sum, and ordering the poles λk so that λ1 = π/2, λ2 = -π/2, λ3 = 3π/2, etc., gives
${\displaystyle \tan(z)=\sum _{k=0}^{\infty }\left[\left({\frac {-1}{z-(k+{\frac {1}{2}})\pi }}-{\frac {1}{(k+{\frac {1}{2}})\pi }}\right)+\left({\frac {-1}{z+(k+{\frac {1}{2}})\pi }}+{\frac {1}{(k+{\frac {1}{2}})\pi }}\right)\right]}$
${\displaystyle \tan(z)=\sum _{k=0}^{\infty }{\frac {-2z}{z^{2}-(k+{\frac {1}{2}})^{2}\pi ^{2}}}}$
## Applications
### Infinite products
Because the partial fraction expansion often yields sums of 1/(a+bz), it can be useful in finding a way to write a function as an infinite product; integrating both sides gives a sum of logarithms, and exponentiating gives the desired product:
${\displaystyle \tan(z)=-\sum _{k=0}^{\infty }\left({\frac {1}{z-(k+{\frac {1}{2}})\pi }}+{\frac {1}{z+(k+{\frac {1}{2}})\pi }}\right)}$
${\displaystyle \int _{0}^{z}\tan(w)dw=\log \sec z}$
${\displaystyle \int _{0}^{z}{\frac {1}{w\pm (k+{\frac {1}{2}})\pi }}dw=\log \left(1\pm {\frac {z}{(k+{\frac {1}{2}})\pi }}\right)}$
Applying some logarithm rules,
${\displaystyle \log \sec z=-\sum _{k=0}^{\infty }\left(\log \left(1-{\frac {z}{(k+{\frac {1}{2}})\pi }}\right)+\log \left(1+{\frac {z}{(k+{\frac {1}{2}})\pi }}\right)\right)}$
${\displaystyle \log \cos z=\sum _{k=0}^{\infty }\log \left(1-{\frac {z^{2}}{(k+{\frac {1}{2}})^{2}\pi ^{2}}}\right),}$
which finally gives
${\displaystyle \cos z=\prod _{k=0}^{\infty }\left(1-{\frac {z^{2}}{(k+{\frac {1}{2}})^{2}\pi ^{2}}}\right).}$
### Laurent series
The partial fraction expansion for a function can also be used to find a Laurent series for it by simply replacing the rational functions in the sum with their Laurent series, which are often not difficult to write in closed form. This can also lead to interesting identities if a Laurent series is already known.
Recall that
${\displaystyle \tan(z)=\sum _{k=0}^{\infty }{\frac {-2z}{z^{2}-(k+{\frac {1}{2}})^{2}\pi ^{2}}}=\sum _{k=0}^{\infty }{\frac {-8z}{4z^{2}-(2k+1)^{2}\pi ^{2}}}.}$
We can expand the summand using a geometric series:
${\displaystyle {\frac {-8z}{4z^{2}-(2k+1)^{2}\pi ^{2}}}={\frac {8z}{(2k+1)^{2}\pi ^{2}}}{\frac {1}{1-({\frac {2z}{(2k+1)\pi }})^{2}}}={\frac {8}{(2k+1)^{2}\pi ^{2}}}\sum _{n=0}^{\infty }{\frac {2^{2n}}{(2k+1)^{2n}\pi ^{2n}}}z^{2n+1}.}$
Substituting back,
${\displaystyle \tan(z)=2\sum _{k=0}^{\infty }\sum _{n=0}^{\infty }{\frac {2^{2n+2}}{(2k+1)^{2n+2}\pi ^{2n+2}}}z^{2n+1},}$
which shows that the coefficients an in the Laurent (Taylor) series of tan(z) about z = 0 are
${\displaystyle a_{2n+1}={\frac {T_{2n+1}}{(2n+1)!}}={\frac {2^{2n+3}}{\pi ^{2n+2}}}\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2n+2}}}}$
${\displaystyle a_{2n}={\frac {T_{2n}}{(2n)!}}=0,}$
where Tn are the tangent numbers.
Conversely, we can compare this formula to the Taylor expansion for tan(z) about z = 0 to calculate the infinite sums:
${\displaystyle \tan(z)=z+{\frac {1}{3}}z^{3}+{\frac {2}{15}}z^{5}+\cdots }$
${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}={\frac {\pi ^{2}}{2^{3}}}={\frac {\pi ^{2}}{8}}}$
${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{4}}}={\frac {1}{3}}{\frac {\pi ^{4}}{2^{5}}}={\frac {\pi ^{4}}{96}}.}$ | 3,190 | 8,733 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 34, "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} | 3.96875 | 4 | CC-MAIN-2021-39 | latest | en | 0.861464 |
https://www.teacherspayteachers.com/Product/Math-Word-Problems-for-Spring-and-Summer-Addition-and-Subtraction-1-10-NO-PREP-3467870 | 1,539,949,905,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512395.23/warc/CC-MAIN-20181019103957-20181019125457-00125.warc.gz | 1,091,376,851 | 20,799 | # Math Word Problems for Spring and Summer Addition and Subtraction 1-10 NO PREP
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Hands-on, NO PREP, addition and subtraction word problems that are engaging! Perfect for K-2!
This resource is part of a bundle. Check it out by clicking on the link below:
Math Word Problems (BUNDLE): Addition and Subtraction 1-10, NO PREP
In math, we teach addition and subtraction concepts. In literacy, we also teach the concept of beginning, middle and end when reading or writing a story. On a daily basis, students should also be using sight words, basic sentence structures and seasonal vocabulary while reading and writing.
We have combined it ALL in this amazing hands-on NO PREP, student friendly resource. Students will enjoy the fun cut and paste activities!
Try a free sample: Math Word Problems FREEBIE: Addition and Subtraction 1-10, NO PREP
This resource contains:
-explicit instructions with photos to help use this resource in your classroom
-a reference page with a seasonal word list with images, as well as key words to create an addition story
-27 math subtraction stories (3 each for subtraction 2-10)
-9 templates to create your own subtraction story
**Each “create your own” subtraction story has 4 options for differentiation!**
-a reference page with a seasonal word list with images, as well as key words to create a subtraction story
The 4 resources in this addition and subtraction math word problem BUNDLE are:
Math Word Problems for Fall: Addition and Subtraction 1-10, NO PREP(available now)
Math Word Problems for Winter: Addition and Subtraction 1-10, NO PREP (available now)
Math Word Problems for Every Day Addition and Subtraction 1-10 NO PREP (available now)
Math Word Problems for Spring and Summer Addition and Subtraction 1-10 NO PREP (available now)
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I am SO excited to announce that this resource is done in collaboration with Yvette Rossignol, a "Top 100" TPT seller! As a result, a similar resource is also available in French in Yvette’s TPT store. If you are in a school where the same grade level is taught in French and in English, this is a great opportunity for both to use a similar resource. I know from experience that students, teachers and parents love it when they know that students are using similar resources.
Follow our stores to be the first to know about future bilingual resources.
Since Yvette and I are great friends and neighbors, you never know what we will come up with over a walk, a boat ride or a glass of wine!
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 726 | 3,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-43 | latest | en | 0.855705 |
https://codedump.io/share/ZrHbpRLJvjek/1/linq-aggregate-algorithm-explained | 1,477,334,609,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719677.59/warc/CC-MAIN-20161020183839-00069-ip-10-171-6-4.ec2.internal.warc.gz | 830,811,564 | 9,409 | alexanderb - 1 month ago 5x
C# Question
# LINQ Aggregate algorithm explained
This might sound lame, but I have not been able to find a really good explanation of
`Aggregate`
.
Good means short, descriptive, comprehensive with a small and clear example.
The easiest to understand definition of `Aggregate` is that it performs an operation on each element of the list taking into account the operations that have gone before. That is to say it performs the action on the first and second element and carries the result forward. Then it operates on the previous result and the third element and carries forward. etc.
Example 1. Summing numbers
``````var nums = new[]{1,2,3,4};
var sum = nums.Aggregate( (a,b) => a + b);
Console.WriteLine(sum); // output: 10 (1+2+3+4)
``````
This adds `1` and `2` to make `3`. Then adds `3` (result of previous) and `3` (next element in sequence) to make `6`. Then adds `6` and `4` to make `10`.
Example 2. create a csv from an array of strings
``````var chars = new []{"a","b","c", "d"};
var csv = chars.Aggregate( (a,b) => a + ',' + b);
Console.WriteLine(csv); // Output a,b,c,d
``````
This works in much the same way. Concatenate `a` a comma and `b` to make `a,b`. Then concatenates `a,b` with a comma and `c` to make `a,b,c`. and so on.
Example 3. Multiplying numbers using a seed
For completeness, there is an overload of `Aggregate` which takes a seed value.
``````var multipliers = new []{10,20,30,40};
var multiplied = multipliers.Aggregate(5, (a,b) => a * b);
Console.WriteLine(multiplied); //Output 1200000 ((((5*10)*20)*30)*40)
``````
Much like the above examples, this starts with a value of `5` and multiplies it by the first element of the sequence `10` giving a result of `50`. This result is carried forward and multiplied by the next number in the sequence `20` to give a result of `1000`. This continues through the remaining 2 element of the sequence.
Example 2, above, uses string concatenation to create a list of values separated by a comma. This is a simplistic way to explain the use of `Aggregate` which was the intention of this answer. However, if using this technique to actually create a large amount of comma separated data, it would be more appropriate to use a `StringBuilder`, and this is entirely compatible with `Aggregate` using the seeded overload to initiate the `StringBuilder`.
``````var chars = new []{"a","b","c", "d"};
var csv = chars.Aggregate(new StringBuilder(), (a,b) => {
if(a.Length>0)
a.Append(",");
a.Append(b);
return a;
});
Console.WriteLine(csv);
``````
Updated example: http://rextester.com/YZCVXV6464
Source (Stackoverflow) | 692 | 2,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2016-44 | longest | en | 0.813745 |
http://aznakay.info/diagram/original-two-way-light-wiring-diagram | 1,575,695,715,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540495263.57/warc/CC-MAIN-20191207032404-20191207060404-00133.warc.gz | 15,637,322 | 4,955 | # Original Two Way Light Wiring Diagram Light Wiring Diagram New Two Way Switch | Wiring
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Two way light wiring diagram - Right here we a have a schematic (fig 1) which makes it easy to visualize how this circuit works. In this kingdom the lamp is off, converting the location of either transfer will transfer the stay to the lamp turning it on. In case you now alternate the location of the other transfer the circuit is broken over again. Cable d (fig 2) is a 3 middle and earth, that is the ‘3 twine manage’ that links the two light switches together.?com on the primary switch connects to com on the second one transfer, l1 on the first switch connects to l1 on the second, and l2 on the primary transfer connects to l2 on the second one. Here's a way switching answer posted for certainly one of our users who had run the strength feed to one of the switch containers and had no radial circuit to pick out up a impartial at the lamp holder. 2 way switching approach having two or greater switches in specific locations to govern one lamp. They're stressed out so that operation of both transfer will control the mild.?this association is frequently observed in stairways, with one switch upstairs and one transfer downstairs or in long hallways with a transfer at either cease. Here we have a way switching system that utilises two unmarried gang two-way switches and a three cord manipulate, shown within the new harmonised cable colors. It's miles possible to obtain a similar end result the use of a two wire manage which, even though it saves on cable, isn't always recommended. This is the preferred technique.
All earth wires have to connect with the earth terminal within the transfer returned-container and if you are the use of metal switches there ought to be a loop from this earth terminal to the one on the switch plate (see note a on fig 2). Fig 2 under shows how we achieve this configuration.?similar to any loop-in loop-out radial circuit, the switch cable from the ceiling rose incorporates two wires, a everlasting live and a switched stay. That is cable c below, one twine connects to l1 and the opposite to l2 at the pinnacle transfer. Word: the grey wire in cable ‘d’ is a switched stay and the blue twine in cable ‘c’ and black wire in cable ‘d’ are permanent lives and hence need to be marked with brown sheathing at each quit as proven. | 513 | 2,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-51 | latest | en | 0.915229 |
https://ask.cvxr.com/t/how-to-express-this-objective-funtion-in-cvx-or-tfocs/5081 | 1,680,300,446,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949689.58/warc/CC-MAIN-20230331210803-20230401000803-00320.warc.gz | 138,779,842 | 5,060 | # How to express this objective funtion in CVX or TFOCS?
Dear all,
How to express the following objective function with CVX?
\min_{z>0} z*ln(\frac{1}{N}\sum_{k=1}^{N} \exp{(\frac{k}{z})}-\alpha)
It should be convex.
I also tried smooth_logsumexp in TFOCS and It always reports error in the “-” before \alpha.
Here \alpha is a negative constant.
Thank you in advance.
I believe this is {concave,constant,convex} respectively for alpha {positive, zero, negative}.
Yes. Thank you for such a quick reply. Here alpha is negative. I should have stated here. Let me add it.
I have removed the non-convex designation. I don’t see how to enter this into CVX. If someone else does, great, but I am not optimistic.
It may not be the problem of convexity since here the variable is z.
I tried smooth_logsumexp in TFOCS. It works well without the constant term. But when I added the constant term, it reported error in the symbol - or + before the constant term \alpha. I guess it may have something to do with the function handle function. But I don’t know how to use it correctly. Could you help me with it?
Without the constant term, i.e., with alpha = 0, the objective function is just a constant.
I will have to defer to someone else for help with TFOCS. But I will re-classify the thread as TFOCS, and change its title, in order to gain attention of TFOCS experts.
Thank you so much. Yesterday, someone told me this could be done through exponential cone. Could you provide information regarding to exponential cone in CV
I’m not saying it can’t be done with exponential cone, but I didn’t see how, else I would have told you. log_sum_exp does invoke the exponential cone, BTW.
Fro an example of explicit use of exponential cone construct in CVX, look at mcg’s first post in Solve optimization problems of exp function . However, any use in CVX of log, exp, entr, rel_entr, log_sum_exp will invoke the exponential cone “under the hood”, so the someone who told you might be referring to any of this. Anyhow, if the someone really does know how to input this problem into CVX, please have them or you post it here.
Thank you so much for your help and I will ask him in detail. | 540 | 2,184 | {"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-2023-14 | latest | en | 0.924291 |
http://clay6.com/qa/38971/a-particle-moves-with-simple-harmonic-motion-in-a-straight-line-in-first-ta | 1,529,775,587,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865145.53/warc/CC-MAIN-20180623171526-20180623191526-00545.warc.gz | 61,601,851 | 26,456 | # A particle moves with simple harmonic motion in a straight line . In first $\;\tau\;s\;$ after starting from rest it travels a distance a , and in next $\;\tau\;s\;$ it travels 2a , in same direction , then :
(a) amplitude of motion is 3a(b) time period of oscillations is $\;8\tau\;$(c) amplitude of motion is 4a(d) time period of oscillations is $\;6\tau\;$ | 106 | 362 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-26 | latest | en | 0.712237 |
https://blog.csdn.net/Sumujingling/article/details/51539210 | 1,529,894,976,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00069.warc.gz | 564,980,147 | 11,179 | # CUDA之矩阵乘法——复数
### 代码
__device__ float GetReal(const Matrix A, int row, int col) {
return A.real[row * A.stride + col];
}
__device__ float GetImag(const Matrix A, int row, int col) {
return A.imag[row * A.stride + col];
}
__device__ void SetElement(Matrix A, int row, int col, float valueR, float valueI) {
A.real[row * A.stride + col] = valueR;
A.imag[row * A.stride + col] = valueI;
}
__device__ Matrix GetSubMatrix(Matrix A, int row, int col) {
Matrix Asub;
Asub.width = BLOCK_SIZE;
Asub.height = BLOCK_SIZE;
Asub.stride = A.stride;
Asub.real = &A.real[A.stride * BLOCK_SIZE * row+ BLOCK_SIZE * col];
Asub.imag = &A.imag[A.stride * BLOCK_SIZE * row+ BLOCK_SIZE * col];
return Asub;
}
__global__ void CMatMulKernel(Matrix A, Matrix B, Matrix C) {
int blockRow = blockIdx.y;
int blockCol = blockIdx.x;
Matrix Csub = GetSubMatrix(C, blockRow, blockCol);
float CvalueR = 0;
float CvalueI = 0;
for (int m = 0; m < (A.width / BLOCK_SIZE); ++m) {
Matrix Asub = GetSubMatrix(A, blockRow, m);
Matrix Bsub = GetSubMatrix(B, m, blockCol);
__shared__ float AsR[BLOCK_SIZE][BLOCK_SIZE];
__shared__ float AsI[BLOCK_SIZE][BLOCK_SIZE];
__shared__ float BsR[BLOCK_SIZE][BLOCK_SIZE];
__shared__ float BsI[BLOCK_SIZE][BLOCK_SIZE];
AsR[row][col] = GetReal(Asub, row, col);
AsI[row][col] = GetImag(Asub, row, col);
BsR[row][col] = GetReal(Bsub, row, col);
BsI[row][col] = GetImag(Bsub, row, col);
for (int e = 0; e < BLOCK_SIZE; ++e)
{
CvalueR += AsR[row][e] * BsR[e][col]-AsI[row][e]*BsI[e][col];
CvalueI += AsR[row][e] * BsI[e][col]+AsI[row][e]*BsR[e][col];
}
}
SetElement(Csub, row, col, CvalueR,CvalueI);
}
void CMatMul(const Matrix A, const Matrix B, Matrix C) {
Matrix d_A;
d_A.width = d_A.stride = A.width;
d_A.height = A.height;
size_t size = A.width * A.height * sizeof(float);
cudaMalloc((void**)&d_A.real, size);
cudaMalloc((void**)&d_A.imag, size);
cudaMemcpy(d_A.real, A.real, size,
cudaMemcpyHostToDevice);
cudaMemcpy(d_A.imag, A.imag, size,
cudaMemcpyHostToDevice);
Matrix d_B;
d_B.width = d_B.stride = B.width;
d_B.height = B.height;
size = B.width * B.height * sizeof(float);
cudaMalloc((void**)&d_B.real, size);
cudaMalloc((void**)&d_B.imag, size);
cudaMemcpy(d_B.real, B.real, size,
cudaMemcpyHostToDevice);
cudaMemcpy(d_B.imag, B.imag, size,
cudaMemcpyHostToDevice);
Matrix d_C;
d_C.width = d_C.stride = C.width;
d_C.height = C.height;
size = C.width * C.height * sizeof(float);
cudaMalloc((void**)&d_C.real, size);
cudaMalloc((void**)&d_C.imag, size);
dim3 dimBlock(BLOCK_SIZE, BLOCK_SIZE);
dim3 dimGrid(B.width / dimBlock.x, A.height / dimBlock.y);
CMatMulKernel<<<dimGrid, dimBlock>>>(d_A, d_B, d_C);
cudaMemcpy(C.real, d_C.real, size,
cudaMemcpyDeviceToHost);
cudaMemcpy(C.imag, d_C.imag, size,
cudaMemcpyDeviceToHost);
cudaFree(d_A.real);
cudaFree(d_A.imag);
cudaFree(d_B.real);
cudaFree(d_B.imag);
cudaFree(d_C.real);
cudaFree(d_C.imag);
}
## 为什么这么久过去了我依然不能上传图片。
#### CUDA复数类的定义
2015-05-25 11:44:04
#### 实数及复数矩阵加法并行CUDA
2014年08月28日 19.8MB 下载
#### CUDA中的复数定义、开内存空间以及运算
2018-03-19 20:58:39
#### CUDA调用cuFFT后对复数求模
2018-06-07 17:21:26
#### CUDA 中 FFT 的使用
2015-07-02 11:55:56
#### CUDA的初始化
2013-09-30 11:03:40
#### lesson6 复数及复指数
2015-12-20 22:29:56
#### 《GPU高性能编程CUDA实战》学习笔记（四）
2016-09-09 14:24:52
#### CUDA cuFFT使用
2017-02-22 11:23:53
#### CUDA并行算法系列之FFT快速卷积
2016-09-21 11:53:26 | 1,155 | 3,317 | {"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} | 2.75 | 3 | CC-MAIN-2018-26 | latest | en | 0.276501 |
https://physics.stackexchange.com/questions/645460/problems-with-sun-earth-moon-simulations-in-vpython | 1,627,169,699,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00215.warc.gz | 456,709,420 | 38,697 | # Problems with Sun-Earth-Moon simulations in vPython
I'm trying to simulate the Sun-Earth-Moon system using the Verlet algorithm with vpython. I'm using natural units and a cartesian coordinate system, and am having trouble working out the physics.
I tried calculating the initial velocity of the moon by adding Earth's velocity with respect to the sun to the moon's velocity with respect to Earth, but what I'm getting is just an elliptical orbit of the moon around the sun.
I will point out that the simulation seems ok for just Earth, but I don't know how to make the moon orbit it.
I guess I need help with determining the correct initial velocity for the moon and any other remarks on my calculations and explanations on the physics would be greatly appreciated.
The code:
from vpython import *
from math import *
earth_mass = (5.97237 / 1.9855) * 10 ** (-6) # Earth's proportional mass to the sun in natural units
moon_mass = (7.24767306 / 1.9855) * 10 ** (-8) # Moon's proportional mass to the sun in natural units
moon_period = 27.32152778 / 365.25 # Moon's period around Earth in natural units
earth_moon_dist = 0.00256955529 # AU, Moon is ~384400 km from Earth
scene = canvas(fov=0.01, background=color.black)
dt = 0.001 # chosen dt
half_dt = dt / 2
half_dt_squared = (dt ** 2) / 2
M = 1 # Natural unit - 1 Sun mass
G = 4 * (pi ** 2) # Gravitational constant in natural units
def acceleration(body_1, body_2):
# Calculates acceleration of body_1
sun_acc = (G * M / body_1.pos.mag2) * (-body_1.pos.hat) # Force sun exerts on body_1 divided by body_1's mass
body_1_acc = (G * body_2.m / (body_1.pos - body_2.pos).mag2) * (
-(body_1.pos - body_2.pos).hat) # Force body_2 exerts on body_1 divided by body_1's mass
return sun_acc + body_1_acc
# Bodies initialization
# Sun
sun = sphere(color=color.yellow, radius=.05, m=M, texture=r"https://i.imgur.com/lrI62Ot.jpg")
# Earth
earth_r = 1 # Natural unit - 1 AU
earth = sphere(pos=vector(earth_r, 0, 0), radius=.005, texture=textures.earth, make_trail=True, interval=dt,
trail_color=color.blue, m=earth_mass, v=vector(0, 0, 0), a=vector(0, 0, 0))
earth_initial_velocity = sqrt((G * M) * (1 / earth_r))
earth.v = vector(0, earth_initial_velocity, 0)
# Moon
moon_r = earth.pos.x + earth.radius + earth_moon_dist
moon = sphere(pos=vector(moon_r, 0, 0), radius=.001, texture=r"https://i.imgur.com/0lAj5pJ.jpg", make_trail=True,
interval=dt,
trail_color=color.white, m=moon_mass, v=vector(0, 0, 0), a=vector(0, 0, 0))
moon_initial_velocity = earth_initial_velocity + sqrt(G * earth.m / earth_moon_dist)
moon.v = vector(0, moon_initial_velocity, 0)
# Calculate accelerations
earth.a = acceleration(earth, moon)
moon.a = acceleration(moon, earth)
# Simulation
t = 0
while t <= 31557600: # 1 Year in seconds
rate(100)
# Verlet algorithm - update positions
earth.pos += (earth.v * dt + earth.a * half_dt_squared)
moon.pos += (moon.v * dt + moon.a * half_dt_squared)
# Verlet algorithm - partially update velocities
earth.v += (earth.a * half_dt)
moon.v += (moon.a * half_dt)
# Verlet algorithm - update accelerations according to new positions
earth.a = acceleration(earth, moon)
moon.a = acceleration(moon, earth)
# Verlet algorithm - partially update velocities according to new accelerations
earth.v += (earth.a * half_dt)
moon.v += (moon.a * half_dt)
t += dt
EDIT: To emphasize, each time I'm calculating the acceleration of the moon, I use:
$$\vec{a_{M}}=\vec{a_{EM}}+\vec{a_{SM}}=\frac{\vec{F_{EM}}}{m_{M}}+\frac{\vec{F_{SM}}}{m_{M}}=\frac{G*M_{S}}{|\vec{r_{SM}}|^2}\hat{r_{SM}}+\frac{G*M_{E}}{|\vec{r_{EM}}|^2}\hat{r_{EM}}$$,
where $${F_{SM}}$$ is the force the sun exerts on the moon, $$M_{S}$$ is the mass of the sun, $$m_{M}$$ is the mass of the moon, and $$r_{SM}=\vec{S}-\vec{M}$$ with $$\vec{S}$$ the position vector of the sun (in my case the origin) and $$\vec{M}$$ the position vector of the moon.
The same for the force the earth exerts on the moon.
Are there any other forces I should be taking into consideration?
• Related. You may need a more careful visualization to see the Earth’s effect on the Moon’s orbit around the Sun, which is convex everywhere. – rob Jun 14 at 11:08
• I haven't looked at this in any detail, but you might try expanding some "+=". I can't remember the details, but I once hit a case where += did something unexpected. It took a long time to figure out what was going on. Interestingly, it was also in a Vpython program. – garyp Jun 14 at 11:51
• @rob I am pretty certain there is something wrong with my simulation as the moon is supposed to orbit around the earth, I tried looking in the link you sent but I didn't understand how it could help me. Could you please elaborate? – orav94 Jun 14 at 12:02
• @garyp Tried that just now, the code seems to function propely, it's the physics that are the problem. – orav94 Jun 14 at 12:03
• @rob I have added a clarification as to how I'm calculating the acceleration. They do not stay together and act like independent planets orbiting around the sun, the orbits don't even cross. This is why I think there is something wrong with my calculations. – orav94 Jun 14 at 12:51 | 1,506 | 5,138 | {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-31 | latest | en | 0.787797 |
https://askfilo.com/physics-question-answers/a-solid-rubber-ball-of-density-d-and-radius-r-falls-vertically-through-air | 1,718,283,533,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00634.warc.gz | 96,178,244 | 32,494 | World's only instant tutoring platform
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# A solid rubber ball of density 'd' and radius 'R' falls vertically through air. Assume that the air resistance acting on the ball is F KRV where K is constant and V is its velocity. Because of this air resistance the ball attains a constant velocity called terminal velocity after some time.Then is:
A
B
C
D
## Text solutionVerified
Let any time t, the velocity be V.
Acceleration is then
but once it attains terminal velocity ;
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Question Text A solid rubber ball of density 'd' and radius 'R' falls vertically through air. Assume that the air resistance acting on the ball is F KRV where K is constant and V is its velocity. Because of this air resistance the ball attains a constant velocity called terminal velocity after some time.Then is: Updated On May 27, 2022 Topic Mechanical Properties of Fluids Subject Physics Class Class 11 Answer Type Text solution:1 Video solution: 1 Upvotes 256 Avg. Video Duration 11 min | 405 | 1,718 | {"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-2024-26 | latest | en | 0.811512 |
https://codereview.stackexchange.com/questions/156474/find-the-lowest-and-second-lowest-values-in-an-array-of-numbers | 1,713,395,936,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00656.warc.gz | 155,430,608 | 41,400 | # Find the lowest and second lowest values in an array of numbers
Trying to find way to do this with only one pass through the array, so I won't run a sort on it and take arr[0..1]. Not sure how to make it look better:
def LowestSecondLowest(arr)
lowest,second_lowest = nil,nil
arr.each do |n|
if second_lowest.nil?
second_lowest=n
elsif lowest.nil?
if n>second_lowest
lowest=second_lowest
second_lowest=n
elsif n<second_lowest
lowest=n
end
elsif n<lowest
second_lowest=lowest
lowest=n
elsif n>lowest && n<second_lowest
second_lowest=n
end
end
"#{lowest} #{second_lowest}"
end
• It often results in simpler code to start with lowest, second_lowest = VERY_LARGE_NUMBER, VERY_LARGE_NUMBER Feb 27, 2017 at 22:08
• Array#min can return the two lowest values: [3,2,1].min(2) Mar 21, 2017 at 12:10
Because Array lets you put repeated elements of same value, I believe that the two lowest elements in [1, 2, 1] are [1, 1], not [1, 2]. This is the very same behavior one would face if sorting the array and getting the two first elements. Also I believe that in [1] there is a lowest and no 2nd lowest (also, the very same behavior if sorting and getting the first two).
With this in mind, I changed your code to obey these premises (and make it a bit more readable if I may say). Take a look.
def two_lowest arr
# if arr has no elements, there is no answer
# If arr has only one element, this is the lowest
if arr.size < 2 then
return arr.first, nil
end
lowest, second_lowest = nil, nil
arr.each do |n|
if lowest.nil? or n < lowest
# if we have no lowest or we found an element lower than current lowest,
# update our lowest and 2nd lowest
second_lowest = lowest
lowest = n
elsif second_lowest.nil? or n < second_lowest
# if we have no 2nd lowest or we found an element between lowest and 2nd
# lowest, update our 2nd lowest
second_lowest = n
end
end
return lowest, second_lowest
end
If you want to test it
tests = [
[],
[1],
[1, 1],
[1 ,2],
[1, 2, 1],
[1, 2, 3],
[1, 1, 2, 3],
[3, 2, 1, 1],
]
for test in tests
puts "In #{test}:"
puts "#{two_lowest test}"
puts
end
• Thanks very much, Gabriel. I understand the logic behind your revisions. I think my sloppiness came from not considering what would get caught in previous if/elsif statements, and therefore trying to accommodate unnecessary values in subsequent if/elsif statements Feb 28, 2017 at 14:14
• Treating corner cases before main iteration/recursion is a "trick" I usually do to simplify my code. So after all ifs I can forget about corner cases and write just for the general case. I am glad I could help you, it is my first answer in CodeReview :-) Feb 28, 2017 at 17:28
• Would you mind elaborating for me what you mean? I'm still rather new to coding and general strategies like you mention here are not familiar to me. Thanks! Mar 1, 2017 at 19:36
• This strategy is called "early exit". It is discussed here. I hope it helps! Mar 1, 2017 at 22:52 | 838 | 2,923 | {"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} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.822599 |
http://math.stackexchange.com/questions/617625/on-familiarity-or-how-to-avoid-going-down-the-math-rabbit-hole/618217 | 1,394,723,960,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678677514/warc/CC-MAIN-20140313024437-00011-ip-10-183-142-35.ec2.internal.warc.gz | 76,285,290 | 44,012 | # On “familiarity” (or How to avoid “going down the Math Rabbit Hole”?)
Anyone trying to learn mathematics on his/her own has had the experience of "going down the Math Rabbit Hole".
For example, suppose you come across the novel term vector space, and want to learn more about it. You look up various definitions, and they all refer to something called a field. So now you're off to learn what a field is, but it's the same story all over again: all the definitions you find refer to something called a group. Off to learn about what a group is. Ad infinitum. That's what I'm calling here "to go down the Math Rabbit Hole."
Upon first encountering the situation described above one may think: "well, if that's what it takes to learn about vector spaces, then I'll have to toughen up, and do it." I picked this particular example, however, because I'm sure that the course of action it envisions is one that is not just arduous: it is in fact utterly misguided.
I can say so with some confidence, for this particular case, thanks to some serendipitous personal experience. It turns out that, luckily for me, some kind calculus professor in college gave me the tip to take a course in linear algebra (something that I would have never thought of on my own), and therefore I had the luxury of learning about vector spaces without having to venture into the dreaded MRH. I did well in this class, and got a good intuitive grasp of vector spaces, but even after I had studied for my final exams (let alone the first day of class), I couldn't have said what a field was. Therefore, from my experience, and that of pretty much all my fellow students in that class, I know that one does not need to know a whole lot about fields to get the hang of vector spaces. All one needs is a familiarity with some field ($\mathbb{R}$, say).
Now, it's hard to pin down more precisely what this familiarity amounts to. The only thing that I can say about it is that it is a state somewhere between, and quite distinct from, (a) the state right after reading and understanding the definition of whatever it is one wants to learn about (say, "vector spaces"), and (b) the state right after (say) acing a graduate-level pure math course in that topic.
Even harder than defining this familiarity is coming up with an efficient way to attain it...
I'd like to ask all the math autodidacts reading this: how do you avoid falling into the Math Rabbit Hole? And more specifically, how do you efficiently attain enough familiarity with pre-requisite concepts to move on to the topics that you want to learn about?
PS: John von Neumann allegedly once said "Young man, in mathematics you don't understand things. You just get used to them." I think that this "getting used to things" is much of what I'm calling familiarity above. The problem of learning mathematics efficiently then becomes the problem of "getting used to things" quickly.
EDIT: Several answers and comments have suggested to use textbooks rather than, e.g., the Wikipedia, to learn math. But textbooks usually have the same problem. There are exceptions, like Gilbert Strang's books, which generally avoid technicalities and instead focus on the big picture, and they are indeed ideal introductions to a subject, but they are exceedingly rare. For example, as I already mentioned in one comment, I've been looking for an intro book on homotopy theory that focuses on the big picture, to no avail; all the books I've found bristle with technicalities from the get go: Hausdorff this, locally compact that, yadda yadda...
I'm sure that when one mathematician asks another for an introduction to some branch of math, the latter does not start spewing all these formal technicalities, but instead gives a big-picture account, based on simple examples. I wish authors of mathematics books sometimes wrote books in such an informal vein. Note that I'm not talking here about books written for math-phobes (in fact I detest it when a math book adopts a condescending "for-dummies", "let's-not-fry-our-little-brains-now" tone). Informal does not mean "dumbed down". There's a huge gap in the mathematics literature (at least in English), and I can't figure out why.
(BTW, I'm glad that MJD brought up Strang's Linear Algebra book, because it's a concrete example that shows it's not impossible to write a successful math textbook that stays on the big picture, and doesn't fuss over technicalities. It goes without saying that I'm not advocating that all math books be written this way. Attention to such technical details, precision, and rigor are all essential to doing mathematics, but they can easily overwhelm an introductory exposition.)
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Sometimes you just have to swallow the red pill. – Daniel Rust Dec 24 '13 at 22:11
Try reading books that are called an introduction to _____. – user4140 Dec 24 '13 at 22:13
@DanielRust: the choice of linear algebra for my example is better than I thought. – kjo Dec 24 '13 at 22:14
Read a book. Don't learn through Wikipedia. Not that there's anything wrong with Wikipedia (browsing wiki is a great way to learn), but it should be for exploration or reference. Once you're ready to sit down and learn a subject, you find a good book. – Jack M Dec 25 '13 at 11:59
Don't go down the rabbit hole too far. You might end at Bourbaki Vol. 1. – Hagen von Eitzen Dec 26 '13 at 18:56
show 10 more comments
Your example makes me think of graphs.
Imagine some nice, helpful fellow came along, and made a big graph of every math concept ever, where each concept is one node and related concepts are connected by edges. Now you can take a copy of this graph, and color every node green based on whether you "know" that concept (unknowns can be grey).
How to define "know"? In this case, when somebody mentions that concept while talking about something, do you immediately feel confused and get the urge to look the concept up? If no, then you know it (funnily enough, you may be deluding yourself into thinking you know something that you completely misunderstand, and it would be classed as "knowing" based on this rule - but that's fine and I'll explain why in a bit). For purposes of determining whether you "know" it, try to assume that the particular thing the person is talking about isn't some intricate argument that hinges on obscure details of the concept or bizarre interpretations - it's just mentioned matter-of-factly, as a tangential remark.
When you are studying a topic, you are basically picking one grey node and trying to color it green. But you may discover that to do this, you must color some adjacent grey nodes first. So the moment you discover a prerequisite node, you go to color it right away, and put your original topic on hold. But this node also has prerequisites, so you put it on hold, and... What you are doing is known as a depth first search. It's natural for it to feel like a rabbit hole - you are trying to go as deep as possible. The hope is that sooner or later you will run into a wall of greens, which is when your long, arduous search will have born fruit, and you will get to feel that unique rush of climbing back up the stack with your little jewel of recursion terminating return value.
Then you get back to coloring your original node and find out about the other prerequisite, so now you can do it all over again.
DFS is suited for some applications, but it is bad for others. If your goal is to color the whole graph (ie. learn all of math), any strategy will have you visit the same number of nodes, so it doesn't matter as much. But if you are not seriously attempting to learn everything right now, DFS is not the best choice.
So, the solution to your problem is straightforward - use a more appropriate search algorithm!
Immediately obvious is breadth-first search. This means, when reading an article (or page, or book chapter), don't rush off to look up every new term as soon as you see it. Circle it or make a note of it on a separate paper, but force yourself to finish your text even if its completely incomprehensible to you without knowing the new term. You will now have a list of prerequisite nodes, and can deal with them in a more organized manner.
Compared to your DFS, this already makes it much easier to avoid straying too far from your original area of interest. It also has another benefit which is not common in actual graph problems: Often in math, and in general, understanding is cooperative. If you have a concept A which has prerequisite concept B and C, you may find that B is very difficult to understand (it leads down a deep rabbit hole), but only if you don't yet know the very easy topic C, which if you do, make B very easy to "get" because you quickly figure out the salient and relevant points (or it may be turn out that knowing either B or C is sufficient to learn A). In this case, you really don't want to have a learning strategy which will not make sure you do C before B!
BFS not only allows you to exploit cooperativities, but it also allows you to manage your time better. After your first pass, let's say you ended up with a list of 30 topics you need to learn first. They won't all be equally hard. Maybe 10 will take you 5 minutes of skimming wikipedia to figure out. Maybe another 10 are so simple, that the first Google Image diagram explains everything. Then there will be 1 or 2 which will take days or even months of work. You don't want to get tripped up on the big ones while you have the small ones to take care of. After all, it may turn out that the big topic is not essential, but the small topic is. If that's the case, you would feel very silly if you tried to tackle the big topic first! But if the small one proves useless, you haven't really lost much energy or time.
Once you're doing BFS, you might as well benefit from the other, very nice and clever twists on it, such as Dijkstra or A*. When you have the list of topics, can you order them by how promising they seem? Chances are you can, and chances are, your intuition will be right. Another thing to do - since ultimately, your aim is to link up with some green nodes, why not try to prioritize topics which seem like they would be getting closer to things you do know? The beauty of A* is that these heuristics don't even have to be very correct - even "wrong" or "unrealistic" heuristics may end up making your search faster.
-
You mean something like this? xkcd.com/761 – Erel Segal Halevi Dec 25 '13 at 16:06
That is exactly the approach that Khan Academy is taking. I am not saying that KA is the greatest thing ever, but I find, we should all try to re-structure our knowledge in non-linear DAGs, rather than the wanna-be linear crap that it is right now. Look at every advanced math text book and be amazed by how they assume that in chapter 12, you still know what "Corollary 4.32" is. – Domi Dec 26 '13 at 5:43
+1 to the question and the answer. I'm not much of a math person so to speak, but this question and it's answer apply probably to any area of study. I have run into the rabbit hole problem many times trying to learn some new thing about a programming language or how to administer a Linux server. I never have the patience to reach that recursive return you mentioned. I look forward to trying your BFS strategy. – TecBrat Dec 26 '13 at 12:37
An elegantly succinct and informative answer using analogies from computer science! Kudos to you! – Paul Dec 26 '13 at 17:20
There is another way that graphs may be relevant, here. If you think of the egress nodes of the graph (those with no successors) as destinations, nodes where the knowledge becomes actionable, or useful, or just produce a result, then a question is "how to get to them." A top-down approach forces you to learn everything before you can do anything, that is, to visit every predecessor of your destination to pick up every bit of knowledge that your destination may depend on. A "shortest-path" algorithm will minimize some cost function like your pain or time. – Reb.Cabin Dec 26 '13 at 18:48
show 2 more comments
You don't learn what a vector space is by swallowing a definition that says
A vector space $\langle V, S\rangle$ is a set $V$ and a field $S$ that satisfy the following 8 axioms: …
Or at least I don't, and from the sound of things that isn't working for you either. That definition is for someone who not only already knows what a field is, but who also already knows what a vector space is, and for whom the formal statement may illuminate what they already know.
Instead, if you want to learn what a vector space is, you pick up an elementary textbook on linear algebra and you start reading it. I picked up Linear Algebra and its Applications (G. Strang, 1988) from next to the bed just now, and I find that "vector space" isn't even defined. The first page of chapter 2 (“Vector Spaces and Linear Equations”) introduces the idea informally, leaning heavily on the example of $\Bbb R^n$, which was already introduced in Chapter 1, and then emphasizes the crucial property: “We can add any two vectors, and we can multiply vectors by scalars.” The next page reiterates this idea: “a real vector space is a set of ‘vectors’ together with rules for vector addition and multiplication by real numbers.” Then there follow three examples that are different from the $\Bbb R^n$ examples.
A good textbook will do this: it will reduce those 8 axioms to a brief statement of what the axioms are actually about, and provide a set of illuminating examples. In the case of the vector space, the brief statement I quoted, boldface in the original, was it: we can add any two vectors, and we can multiply vectors by scalars.
You don't need to know what a field is to understand any of this, because it's restricted to real vector spaces, rather than to vector spaces over arbitrary fields. But it sets you up to understand the idea in its full generality once you do find out what a field is: “Just like the vector spaces you're used to, except instead of the scalars being real numbers, they can be elements of any field.”
If you find yourself chasing an endless series of definitions, that's because you're trying to learn mathematics from a mathematical encyclopedia. Well, it's worth a try; it worked for Ramanujan. But if you find that you're not Ramanujan, you might try what the rest of us non-Ramanujans do, and try reading a textbook instead. And if the textbook starts off by saying something like:
A vector space $\langle V, S\rangle$ is a set $V$ and a field $S$ that satisfy the following 8 axioms: …
then that means you have mistakenly gotten hold of a textbook that was written for people who already know what a vector space is, and you need to put it aside and get another one. (This is not a joke; there are many such books.)
The Strang book is really good, by the way. I recommend it.
One last note: It's not usually enough to read the book; you have to do a bunch of the exercises also.
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"Just like the vector spaces you're used to, except instead of the scalars being real numbers, they can be elements of any field.” -- replace "real numbers" with "field" and "field" by "ring" and that's the first definition of a module I was given. :) – tomasz Dec 25 '13 at 0:14
I too have had difficulty getting any traction on new areas from sources such as Wikipedia, and moreover am frustrated to see areas I do know explained in ways that will not help a beginner. For an autodidact, sources of free online textbooks are valuable; reddit.com/r/mathbooks is one place to start. – half-integer fan Dec 25 '13 at 3:54
@kjo: Dear kjo, One point about homotopy theory is that it is at a higher level of sophistication than basic linear algebra. The reason is that to formalize the intuitive ideas (of continuity, spaces and shapes, deformations, etc.) takes a bunch of technical machinery (the language of topological spaces, continuous maps between them, product topologies, etc.). It is not true that all (or even most) books on homotopy theory will presume you already know the definitions, but they will all presume a facility at absorbing definitions and joining the dots between formal definitions and ... – Matt E Dec 25 '13 at 3:56
... underlying intuitions. It sounds like you're still at the stage of learning how to do this, and so possibly homotopy theory isn't quite the right thing for you to be reading right now (even if that's your ultimate goal). There are good avenues for learning more foundational material in topology, building (for example) on what you know about linear algebra. Perhaps this would be a good avenue for you to pursue? Regards, – Matt E Dec 25 '13 at 3:57
Amusingly enough, most lecturers do begin with a formal definition, since they know what they're talking about, so this dreadful definition comes very natural to them. Teaching people is much harder than one might think... – Dunno Dec 25 '13 at 19:26
show 3 more comments
A very well known mathematician showed me how he avoids the rabbit hole. I copied his method, and now I can stay out of it most of the time.
I had private weekly seminars with him. Every week, he would research a topic he knew nothing about (that was our deal and that's what was in it for him). I would name the topic (examples: Bloom Filters, Knuth-Bendix Theorem, Linear Logic), and the following week he would give a zero-frills Power-Point presentation of what he found out. The presentations had a uniform pattern:
Motivating Example
Definitions
Lemmas and Theorems
Applications
By beginning with the motivating example, we never got lost in the thicket of technicalities, and the Applications section would circle back and explain the Motivating Example (and maybe some others if time allowed) in terms of the technicalities.
This is how he taught himself a topic without going down the MRH.
Limit your rabbit-hole time (one week)
your presentation must be one hour long
Focus on a Motivating Example
do just enough technicalities to explain the example and optional variations
I have since copied this style. When I teach myself a new topic, I make a slide presentation like that, and then I present it to others in a weekly reading group.
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Wow, thanks for posting this. It's an amazing approach. The reading group in particular; it's a resource I'd not considered at all. Of course, it's not a trivial matter to form such a group and keep it going, but I can see how it would be invaluable. Also, let me say that I'm sooo envious: how I'd love to have a knowledgeable mathematician (even one who was not well-known) as my math tutor, or even just as my math "sensei"... (If I had the I'd hire such a person, of course.) – kjo Dec 26 '13 at 0:17
Yes, I was very lucky to have this opportunity. The reading-group concept works quite well. Just circulate the responsibility for next week's presentation amongst the participants. It helps if you can provide lunch. I'm doing it now with about five of us going through category theory. Forcing the motivating example, no matter how weedy, keeps us out of the rabbit hole. It really works! – Reb.Cabin Dec 26 '13 at 1:12
This approach embodies some basic learning-cycle wisdom. When I was training math/science teachers to function in a developing-country environment for the Peace Corps, we used something called the 4-Mat method to get the basics to subject-matter experts who didn't have (much) teaching experience. 4-Mat has 4 stages that closely mirror yours: aboutlearning.com/what-is-4mat – Faust Dec 27 '13 at 11:48
I think that sometimes, you don't really need to know exactly what every term used means, not right away, anyway. Most of the time, a vague idea is enough to get you started.
Check out the definition (without necessarily understanding it at first -- ploughing through a huge mess of a formal definition is not always helpful at this point, but it helps to see its general structure), then see some examples, tinker a little, see how it works. If I told you everything about horse riding for a month, you probably wouldn't be as good at horse riding as you would be if you had instead tried practised horse riding for a week (and not just because I don't know a thing about horse riding ;) ).
As you get deeper into the subject matter, it might help to understand the details of the definitions, as well as the auxiliary objects. What are they for? What do they really mean? But at first, you shouldn't expect to understand everything, especially when studying more in-depth stuff which (unlike vector spaces) can get you really deep into... rabbit hole.
Familiarity comes with experience. There is no other way.
As a side comment about your vector spaces example: I don't think you can really understand linear algebra if you restrict yourself to reals. They have characteristic zero, are not algebraically closed, they are naturally ordered... this can be very misleading. It's good for starters, but I wouldn't say you understand vector spaces if you just understand real vector spaces.
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+1 for the final paragraph. You don’t understand VS if your (only) intuition is just $k^n$ for some field $k$, and your understanding is extremely limited if you only understand real vector spaces. – Christopher Creutzig Dec 25 '13 at 10:30
Same +1 from me. I actually understood what vector spaces are even for when we were doing various things with matrices (kernel, image, rank etc) – Dunno Dec 25 '13 at 19:30
It is a good idea to learn about vector spaces first in the context of real scalars rather than general fields. But afterward, it is worthwhile to observe that, in most of what you learned (everything short of inner product spaces, in the usual presentations of the subject), you never used the fact that the real numbers come with an ordering; you never needed to consider whether numbers were positive or negative. And for some purposes, like eigenvalues and eigenvectors, it's actually helpful to allow complex numbers into your picture. In fact, all you needed about the real numbers was that you can add, subtract, multiply, and divide them (except of course that you can't divide by $0$) and you can manipulate equations as you learned in elementary algebra. That's why it's safe to allow complex numbers into your picture --- they share all those essential (for linear algebra) properties of the real numbers. And at this point, you know what a field is, even if you've never seen the definition or even the word, because a field is just a collection of things that resemble numbers to the extent that you can add, subtract, multiply, and divide them (except of course that you can't divide by $0$) and you can manipulate equations as you learned in elementary algebra. The formal axioms that define "field" are just the result of the observation that all those algebraic rules you learned are consequences of just a few of the rules; i.e., most of them are redundant. So "field" can be defined by giving just the necessary rules, not all the redundant ones. Of course, that makes it easier to check that something is a field, because you have far fewer rules to verify, and it also makes it easier to write the definition of "field" in a book, because it's shorter than it would otherwise be. But the true idea of "field" remains that all the usual manipulations of equations are valid.
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I have fallen into this vortex with a lot of my studying. The only way I think you can get out of it is to first start off by reading soft math books that don't focus in the details/proofs but try to convey what the general goal of the topic is.
I'm referring here to such books like Garrity's "All the Mathematics You Missed: But Need to Know for Graduate School" which outlines the different disciplines of mathematics, what their roles are and how they relate.
I think anybody who self-studies (and if you're learning math I think you have no choice) has figured out that there are two kinds of math books: those that explain things to you so that you understand them and those that assume you already know the basics and put a new perspective on things.
For me a recent example is combinatorics. A lot of people on here suggested Peter Cameron's book - which is great if you already know a lot of stuff he neglects to mention. If you don't, you're in hell trying to figure out where he is coming up with stuff. And then there is there is Brualdi's combinatorics book which explains things so that you understand them and is a joy to read. Now to properly understand and appreciate combinatorics, I think you need both books (Brualdi first and Cameron second). But I would have saved myself a lot of grief if I had started off with Brualdi first.
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Thanks for the pointer to Garrity's book. I confess that I saw it some time ago on my way out from a bookstore, and snap-judged it (basically "by its cover") right into the "for-dummies" category. Now I see that this quick assessment was quite off-base. I'll give it a more serious look. Also, thanks for the pointer to Brualdi's book. BTW, I see that you are a Don Q fan. Teaching oneself math does sometimes strike as a rather quixotic project. – kjo Dec 25 '13 at 17:05
Garrity's title is misleading, in my opinion. The book is not a math book, but a book about the fields of mathematics. Its role is to build the graph/nodes of topics other people mentioned here. Other books I have found in the same vein are Paul Sally's "Tools of the Trade" and Louis Lyons's "All You Wanted to Know About Mathematics But Were Afraid to Ask". – rocinante Dec 25 '13 at 17:52
Thanks once more for the additional titles. Those two are completely new to me. – kjo Dec 25 '13 at 18:08
Learning in an organized manner (as opposed to "on your own") may help. Plans of what order the topics should be studied in have been worked out by teachers and textbook authors over the years. Trying to start in the middle (with "Vector Space", for example) probably will be difficult.
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Metacademy is a community driven, open source web platform that tries to solve this exact problem. In their own words:
When you try to learn a given concept, Metacademy can show you the full prerequisite structure of the concept and provide a custom learning plan for you to learn the concept as efficiently as possible, complete with curated learning resources as well as discussions on the concept's relationship to other relevant subjects
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Thank you for that link. It's nuggets like this that make me really really glad the Internet exists. – rocinante Dec 31 '13 at 2:59
how do you avoid falling into the Math Rabbit Hole? And more specifically, how do you efficiently attain enough familiarity with pre-requisite concepts to move on to the topics that you want to learn about?
the "rabbit hole" you describe exists in all scientific fields and in many ways is part of the difficulty of its highly specialized nature in the modern age in which it can take many years of study to get to the frontiers of modern research. yet, it is somewhat unacknowledged by experts and considered unavoidable/inevitable. (its great to get some visibility on this prominent issue with your question here.)
in many ways it is unavoidable, yet here are a few ways/strategies around it.
• toy problems. there are sometimes simple problems in an advanced theory that are accessible with a few key definitions in that theory. that doesnt mean they are solvable with those pieces only that they are expressible. this can be psychological leverage to delve deeper in the field.
• textbooks instead of/vs papers. textbooks often are more organized than papers, have a better sense of the overall map of the theory, and are written with beginners in mind sometimes with very careful description and order of introduction of concepts (eg calculus). there can be a lot of variation in coverage/style in textbooks on the same subject. try to find the best ones and then pick a textbook that suits your style.
• "survey papers". these are papers that dont try to prove anything new but "survey" the field. if the field is significant then typically these papers exist. they are not always easy to locate. "insiders" know about them.
• brilliant teacher-writers. often a field has a few writers who are known for their expository rather than research skills, and it pays to focus on their conceptualizations of the field. in some rare cases this can overlap eg Feynman comes to mind. (and conversely there may be few hard-core high-prestige researchers who clearly have little interest in making it all comprehensible/accessible to neophytes or disregard the problem entirely, understanding their primary agenda can help avoid frustration.)
• software/algorithms. approaching mathematical subjects from the pov of algorithms that compute the various entities can be a useful pedagogical approach and has increasing relevance as some key advanced areas of math/TCS are starting to overlap. (eg combinatorics field).
• seek discrete versions of continuous problems. sometimes it seems that math gets most abstract with continuous aspects. discrete versions of the same problem exist and can be simpler to understand/conceptualize. for example the Riemann hypothesis has a lot of very advanced continous math associated with the Zeta function, but there is a discrete version of the conjecture that doesnt even mention the Zeta function.
• possibly approach the field from the problems it addresses/solves (aka "applications") rather than its overarching theories. more generally, there are often equivalent formulations of problems, one simpler to conceptualize for beginners than another.
• look for interesting "bridge theorems" between fields that you know about, and fields that you are interested in. this is a theorem that shows something like an "uncanny correspondence" that seems to have hints it can or will be be developed further.
• wikipedia can be useful but note it often is unnecessarily complex/technical on difficult areas of math/science. at times on some topics it reads as if its written by experts for other experts. raid the references at the end of the article.
• blogs written by experts are increasingly more common and contain remarkably sophisticated expositions, some even surpassing what can be found in textbooks. it will be hard to find specific pages on subjects but once found can be gems. use google and search only on a specific blog, use their topic organization links, etc.
• visual approaches to mathematics. many concepts can be graphed or have visual representations that may be more accessable or intuitive to beginners.
• there are sometimes articles written on "common misconceptions in this field". they can be useful for beginners to avoid common mistakes.
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also—focus on the statements/assertions/definitions of (key) theorems and not so much the technical proofs. math has a pyramidal structure and one can study it top-down looking for "important theorems" 1st. – vzn Dec 27 '13 at 1:50
# In favor of the rabbit hole
## Begin at the beginning, and go on till you come to the end: then stop.
My answer is somewhat contrarian, but I do believe it strongly: You have to fall into the rabbit hole, and you have to go as deep as possible at all times.
Math isn't science. To understand why a frog croaks we might study its respiratory system, to understand this we might get into how cells divide, then the chemical processes in living organisms, organic molecules in general, then the physics of atoms, the sub-atomic particles, etc. By now we have certainly gone too far. The best way of understanding how a frog breathes is to make simplifying assumptions at much higher scales. So we assume that cells are little machines that do a certain thing, or at least that atoms are small balls that bounce off each other and stick together.
To understand vector spaces, there is no point trying to 'gloss over' fields. Best is to know fields inside out. If you can't learn them inside out, you should know the basics as well as possible. The rabbit hole doesn't go on forever - in all cases you will reach either basic definitions, or things that we all agree on intuitively (like the counting numbers).
## Sentence first, verdict afterwards
I have adopted a discipline of studying mathematics, where I never go past a single word that I don't know what it means, and I never skip over a statement that I can't understand, or justify, or understand the justification of. As soon as I have a question of the type "but why wouldn't that work if that condition wasn't true", I stop and think about it until I understand it. If I need to go back earlier in the book or to another book, I do so, even if it means I end up reading books backwards.
I don't always work like this, and there could be places where it is unnecessary, but your example definitely isn't one of them. Suppose that you are trying to learn about vector spaces. Frankly you aren't doing yourself any favors if you believe deep down that all fields are $\mathbb{R}$ and all vector spaces are $\mathbb{R}^n$ for some small natural $n$. Each time you justify something or try to picture something using this model you are storing up problems for yourself when you encounter infinite dimensional spaces, finite fields, or spaces over $\mathbb{C}$ and your mental models don't work anymore. Not to mention the huge problems you would have encountered if you had some misconception about a field (a field is a special case of a group, or a field is an ordered set with some other properties...) and had continued studying vector spaces for a while with this wrong picture in your mind.
Obviously you wouldn't adopt this method to read a newspaper, or a history book. If you don't know what an arquebuse is, but you have an idea that it's some kind of weapon, you might as well assume it's a type of sword. When you find out it's a kind of gun, you can simply slot this knowledge into the understanding of whatever you were reading about arquebuses. Similarly if you think Wisconsin is in Canada. Either it doesn't matter to the story which country it is in, or it does matter, and you find out soon enough where it is, with little effort wasted in reinterpreting other parts of the story.
## Curiouser and curiouser
Now suppose that you do follow my method. You start to research fields. The information about fields you need is this:
• You need to know the axioms that define a field. These easily fit on a sheet of notepaper in large caps written in sharpie. All of them are pretty much self-explanatory. Knowing the names (commutativity etc.) will be invaluable. Arguable any pure mathematician should know them.
• You should know several examples of fields, eg $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, $\mathbb{F_p}$. If the p-adics or other more advanced fields are accessible to you, then all the better, but if not you can live without them. Again, there is nothing here that is not worth knowing to anyone in any branch of pure maths (or in other words, this material is something all undergraduate mathematicians have to learn before they specialize).
• You should know some examples of things that are a bit like fields, but not fields. Eg $\mathbb{N}$, $\mathbb{Z}$, residue sets modulo a composite, reduced residue sets, $\mathbb{R}^n$, boolean rings. There is nothing extra to learn here, you only need to be able to check that various objects you have already heard of don't satisfy the definition of a field. If you haven't heard of some space, you don't need to know that it's not a field. In each case you just figure out one line of the definition that you can point to and say 'This fails, therefore not a field'.
• You should be aware of some theorems about fields. In particular, you should know which properties of 'numbers' apply to all fields. This is again just a case of chasing the definition of a field through some knowledge you already have. Is the quadratic formula applicable in any field? No because square roots are not part of the definition. Is the formula for the solution of a linear equation applicable in any field? Yes, because the multiplicative inverse must exist because...
• You should be aware of some differences between fields. Which fields are finite? Which fields are complete (if you already know what complete means)? What is the characteristic of a field (a one line definition)?
• You should understand how a field has two groups embedded in it.
Now, in all of this there are really only three things which could cause you to get pulled further down the rabbit hole. $\mathbb{C}$, $\mathbb{F}_p$, and the definition of a group. All of these are things that everyone should know. In the case of the first two, all you need to know is the how the field structure works, not any other properties. This can be worked out or taught in an hour (how to add, how to multiply, how to divide, in $\mathbb{C}$ or $\mathbb{F}_p$). As for groups, you need to know the equivalent of the list above for fields. But you would honestly not be wasting your time if you were to read a whole introductory book on groups, studying every proof in detail, even if your goal is vector spaces. A vector space is also a group, as are half of the next thousand spaces/objects you are going learn about. Furthermore, as you learn about groups you are learning:
• how abstract algebra works: you don't assume inverses, commutativity, etc unless the axioms say you can.
• how to do logical proofs, eg how to prove that two sets are equal
• notation in abstract algebra, eg writing binary operations as sums, as products, or in some other way.
• some examples of groups that you can later think about when you are trying to consider vector spaces or fields in generality (for instance, why can you add a multiplicative structure to some additive groups but not others).
All of these are going to help you massively. Even if you NEVER get to vector spaces (you will), it will have been worth you while. Someone who understands groups is a better mathematician than someone who thinks they know what a vector space is, but doesn't understand groups.
Now, instead of thinking of $\mathbb{R}$ everytime you read a fact about a vector space, you can think of vector spaces over all the fields you know. Rather than limiting your imagination, you are challenging it to understand the statements you read in as close to full generality as possible.
Did I load my example? If you are studying something more specialized, then the 'rabbit hole' subjects are not things everyone needs to know, but they are things you need to know. If you are studying Brownian motion, any fact about elementary probability you come across is something you should be able to master.
## Everything's got a moral, if only you can find it
When we speak to professors and other experienced mathematicians about difficult mathematical concepts, they are often able to summarize them with beautiful and compelling generalizations ('Harmonic analysis is about dualities between smooth behavior at small scales and convergent behavior near infinity'). This often makes us worry, because we aren't able to figure these things out ourselves, and they don't appear in our elementary textbooks.
The person who can make that generalization isn't fluent at doing manipulations and solving problems about Fourier transforms because she has understood 'what harmonic analysis is about'. She is fluent, AND she can come out with nice generalizations, because of all the work she has done on the mechanics of harmonic analysis, solving problems, reading slowly through proofs etc. To emulate this person, we shouldn't try to be able to summarize a topic in a neat way. We should try and acquire their detailed knowledge of the mechanics of their subject. Once we've done this, the flashes of insight will come by themselves.
You should also know that some harmonic analysts might think that this 'motivation' for harmonic analysis is wrong, irrelevant or trivial. However all kinds of harmonic analysts would be able to follow each others work by giving the precise definitions they are working with, starting with the things they all agree with.
Seeing the big picture is nice, but you can often afford to miss it. Miss the small picture, and you're not doing mathematics anymore, just reading about it.
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This is all very well in theory but completely inapplicable in practice. There is no "beginning" and "ending" in mathematics - unless you're talking about learning things in the historical/chronological way when they were developed. And if you do that, you've just traded your rabbit hole for all the circles of hell in Dante's Inferno. – rocinante Dec 31 '13 at 2:43
@rocicante: Firstly, as I said, this is how I work 'in practice'. Secondly, of course there is a beginning in mathematics - every definition refers directly or indirectly to undefined terms, every theorem follows directly or indirectly from axioms. – jwg Jan 1 at 23:56
"Seeing the big picture is nice, but you can often afford to miss it. Miss the small picture, and you're not doing mathematics anymore, just reading about it." This is true in my case. Thanks for providing this view. – Gabriel Fair Jan 3 at 19:26
My 2 cents: break the text into chunks you can swallow by allowing yourself multiple passes, where you accept less and less on faith and intuition each time through. I dimly recall Terry Tao giving similar advice on his blog somewhere; hopefully someone can give a link.
For elementary topics, a lot of us are able to "absorb" everything at once, all the definitions, examples, theorems, proofs, unspoken intuitions, etc. At some point it just becomes too much, so do the only thing mathematicians know how to do: break the problem into pieces. Get a high level overview on the first pass, get a sense for what's important on the second pass, figure out what you care to understand on the third pass, and break up those parts you care about iteratively until you reach something manageable.
To take a hypothetical, suppose you're reading a chapter with 50 propositions, lemmas, theorems, and examples. On the first pass you might read the introduction and very quickly skim the rest. After you have a very hazy idea of what the chapter is about, re-skim enough to identify the important theorems and examples. Here I sometimes write a paragraph or two summarizing the chapter. (It'll be incomplete and might be badly wrong, which is fine. It's kind of funny to read my own hazy ideas after I actually understand a topic.)
Now figure out what you actually want to understand; at some point there just isn't time to understand every detail, so you'll have to pick and choose anyway. Perhaps the first half of the chapter proves Theorem M and the second half Theorem Y, which you don't care about yet. Great, no need to slog through Lemmas P, Q, and R, since those are only used in the second half. Maybe Theorem M roughly says Example A is the "only" example; that really says you should get comfortable with Example A, maybe play around with it a little on scratch paper before starting the journey to Theorem M. If it's a long journey, break it up into multiple passes too--maybe start by reading its proof so you know which lemmas are important, and make another rough summary. Perhaps make the path easier by assuming an abstract object is something particular, eg. that a general field is just $\mathbb{R}$. It's fine to read non-linearly.
Eventually you'll have to actually "reach the bottom" and get your hands dirty with technical details, but at least for me it's very helpful to have a high level overview of where things fit into each other before getting down and dirty. I have no hope of retaining much of a complicated topic otherwise. Oh, sometimes I write myself a technical summary, which can be helpful--perhaps rigorous definitions and statements of the ingredients to Theorem M together with a proof of Theorem M in my own words using those ingredients. If I really want to understand something, I add "proof ideas" to a technical summary, where (at the time) I can reconstruct rigorous proofs from my summary. Exercises can also help; sometimes I look for relevant exercises after the first few pages, to get me used to basic definitions that are fundamental to the rest of the chapter.
This works better with some sources than others. If you can't even begin to give a hazy summary (maybe you have no idea what any of the words mean), you're probably reading something written for experts, which doesn't yet include you, so a gentler introduction might be in order. Always be kind to yourself: written math is typically the product of years (or decades, or centuries) of labor by brilliant people. It's no wonder it often takes a long time to absorb.
Finally, I'd like to note that I wrote this post roughly in the style I advocate. The first paragraph is a sketchy overview of the main idea, the second fills in some details, and the third through fifth "get down and dirty" with actionable advice.
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"write a paragraph or two summarizing the chapter. " This is critical into formally learning the topic into a form that can be recalled easier. – Gabriel Fair Jan 3 at 20:17
It's possible to visualise a 2D or 3D spatial object with some structure in your imagination, which is made of abstract stuff and behaves in a certain way. For example, for a vector space, I visualise something like a couple of arrows from a point (the origin), pointing in general direction (in the actual imagined instance, the directions are quite specific), and a part of a plane stretched between and a bit beyond them. And all these are in a state of frozen animation.
As one gets used to vector spaces (for example), this sort of thing helps. The reason is that a lot of questions about all vector spaces could be answered by considering just one non-trivial example that you are familiar with. This speeds up the search through the list of things you know, by using extensive spatial reasoning and memory capabilities of the brain.
In general, the behaviour of, say, a vector space, is infinitely complex. The rabbit hole is infinite! (And splits into exponentially many burrows as a function of depth). That is to say, while the definitions are small and finite, the set of all possible behaviours is infinitely complex, for objects such as groups, fields, vector spaces.
If someone spent a lifetime working with vector spaces, they would probably have in their mind a museum of representative visualised objects, and some efficient ways to choose which to use in order to answer some question or other. Some people, I'm pretty sure, also have at least one other way of thinking that is much faster than this kind of spatial visualisation but consistently yields intuitive answers. Some students in the International Mathematical Olympiad can solve previously unseen problems ten times faster than the average professional mathematician, and I'd like to know how. I've also read that some (most?) chess grandmasters maintain their level of performance when they are asked to continuously solve simple sums while playing chess. So perhaps there is another form of familiarity.
Anyway, to answer the question:
1. It's very useful to explore the nearby reaches of the rabbit hole. With a short Ariadne's thread. It's easy to miss interesting areas even there, so a solid set of exercises from a classic set of lecture notes is invaluable to get good coverage rapidly, and train whatever construct is your understanding.
2. If you have some goal, that's even better, and while some pure mathematicians find their nirvana in a random walk through relatively scary reaches of the rabbit hole (which gets rapidly more horrendous and useless with depth; though I'm sure there are elusive veins of gold and astonishing shortcuts etc.), you can't go wrong with a useful goal from the real world.
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I object to the notion that the rabbit hole is infinite. But perhaps we should start acting like it is :) – Eric Stucky Dec 25 '13 at 16:51
@EricStucky Hmm, Lewis Carroll also didn't seem to think so... – Evgeni Sergeev Dec 25 '13 at 23:08
The real question from the OP:
How do you avoid falling into the Math Rabbit Hole?
MJD's answer hints the the answer to this question: Read a textbook. Textbooks are designed to introduce the material in the correct order, without overburdening or going into too much detail. They are designed to help you progress.
Knowing which textbook to read is a problem unto itself, and for that you should really ask a professional (See: asking a doctor instead of self-diagnosis). To get an idea if the textbook is suitable for you, skim through the first chapter only. If you feel that you grasp or could grasp the content of that chapter, then you're golden. If not, then go find the textbook which will bring you up to speed for that first chapter.
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I think that's a really simple (and potentially hazardous) way of going about things. I think the OP knows to read a textbook instead of Wikipedia; the sticking point is to figure out which textbook is appropriate. The accessibility of the first chapter is often misleading. – rocinante Dec 25 '13 at 15:13
@rocinante: That is why I mention that he should ask which textbook to read. – dotancohen Dec 25 '13 at 16:16
What you are talking about is the process of learning and curiosity.
When researching, you have to keep in mind your goal. Why do you need to study rings when you are just trying to implement an FFT? You don't. So you have to know when to cut off curiosity in exchange for working towards your set out goals.
Now if you don't have defined goal in your study of math, then your goal was to wander about in the rabbit hole. And there's nothing wrong with that, you'll become a stronger analytical thinker the more you study mathematics.
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I read a lot of math on my own as a young person. I would go sit in the stacks, and sift through the books on what I wanted to learn about till I found one that "spoke to me". Part of that test was if I could work some exercises. Once I found such a book, I went through it, working as many exercises as I could. If I couldn't find such a book, I just spun my wheels for a bit till my interests shifted.
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Avoid swallowing definitions and going from there. While they give you the illusion you are learning, you are potentially missing insights and masking problems.
What I'd suggest is reading some university textbooks aimed at first and second year mathematic-students. In my experience (limited) they start from the ground up and go from there.
However, if you don't want to spend your time working through books, most universities offer non-degree programs. You can select the classes you want and learn what you want thoroughly. This has the advantage that you'll have access to cheap summaries, exercises and somebody willing to explain it to you.
For vector spaces specifically, every introductory book in linear algebra should do.
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Here's my short naive answer, not always possible but really good when it is: ask someone who knows and whose pedagogical skills you respect. Interrupt when you need less (or more) technical precision in his/her response.
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I wish more american students were more assertive in the classroom on when to ask for less and more precision. But after learning about student participation in the classroom in Asian and Indian countries, Americans don't seem so bad. – Gabriel Fair Jan 3 at 20:35
I'm not good at algebra but as far i can tell you is that when you need it, you will get it. "deep stuff" exists only as answers to deeper questions, and this kind of learning appears to be systematic in every age. Is tempting to dig on Wikipedia, but previously said is like you here acting like a graph visitor algorithm: visiting adjacent nodes and stacking previous. Note also that Wikipedia is turning a Monster - contributors are more concerned about the completeness than the audience. And this completeness in a mathematical sense is ... formality and verbosity!!! Thats why we need books and exercises. After a lot of exercises your brain will ask "why i'm not organizing or relating those mathematical objects in that in someway"... and then you go to Wikipedia looking for answers to find out that another set of brains did that. And then you acknowledge that you understand 5 or 6 hyperlinks and you say "thanks wikipedia! without you wouldnt be possible" :P
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As one of many "Alices" in the MRH called stackexchange I rather think that studying math with out structure is like waking up in free fall. You see everything around you, you may even come to understand a few things that wiz by your head but you can't shake the feeling that you are falling. More to the point it is like reading a book in a foreign language with a translation book near by. Its a difficult read but doable
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Interesting to point out how much of human information is being assembled and dissected within online forums like SE. I believe that in the next few years we will see crawl bots that parse the internet, fall into the MRH and will invent applicable algorithms on how to escape from the MRH. – Gabriel Fair Jan 3 at 20:47
This question has much broader implication. "Going down the Math Rabbit Hole" simply reveals the nature of being any kind of academic, not just a mathematician. This kind of no ending thing is good for some people who enjoy having things they do not know so that they can keep doing things (whether the things they are doing are worthwhile is a completely different question). This no ending thing, however, really is bad for other people since it indeed is frustrating to know that you can never finish and you can never have a closure. As an actuarial graduate who tries to make up sufficient mathematics, I find it is helpful to take some fundamental subjects like calculus and algebra without which it simply is too difficult to do anything else. Also I suggest a healthy attitude towards study. Nowadays, inter-discipline work is the norm. One could never acquire all the knowledge they need for their work. For my own purpose, I only need to know enough so that I can communicate with my colleagues who are mathematicians, which is the ideal balance in my opinion.
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And this is becoming more common. Academic fields can get so specialized that working together with colleagues to research a subject is all but mandatory. – Gabriel Fair Jan 3 at 20:41
The thing is, you don't have to know what a field is in order to work with vector spaces. The field "stuff" relates vector spaces to a generalization which connects them to other spaces. Skip the rabbit hole, read the rest of the chapter, and do some exercises.
You don't have to understand the word "beverage" in order to swallow the Kool-Aid.
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https://wiki-helper.com/if-i-12-cm-b-10-m-and-volume-1080-m3-then-h-is-kitu-39796905-10/ | 1,716,503,385,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00154.warc.gz | 538,837,389 | 31,417 | # If I = 12 cm, b = 10 m andvolume = 1080 m3, then h is ?
If I = 12 cm, b = 10 m and
volume = 1080 m3, then h is ?
1. ### Correct question:–
• If Length = 12 m, Breadth = 10 m and Volume = 1080 m³, then Height is ?
### Given:–
• Length (l) = 12 m
• Breadth (b) = 10 m
• Volume of Cuboid = 1080 m³
### To Find:–
• Height (h) of the cuboid
### Formula used:–
• Volume of Cuboid = l × b × h
### Solution:–
Volume of Cuboid = l × b × h
[tex]⟹1080 = 12 × 10 × h[/tex]
[tex]⟹1080 = 120 × h[/tex]
[tex]⟹h= \dfrac{1080}{120} [/tex]
[tex]⟹h= \dfrac{108}{12} [/tex]
[tex]{\boxed{⟹h= 9 m}}[/tex]
Hence, the Height of the given Cuboid is 9 m.
━━━━━━━━━━━━━━━━━━
In cuboid :
• lenght : 12 cm
• Volume : 1080 cm³
To find :
• Height of cuboid
Solution :
• Let height of cuboid be h cm
Volume of cuboid : l x b x h
• 1080 = l x b x h
• 1080 = 12 x 10 x h
• 1080 = 120 x h
• 1080/120 = h
• 108/12 = h
• 9 = h | 388 | 921 | {"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} | 4.28125 | 4 | CC-MAIN-2024-22 | latest | en | 0.483813 |
https://de.zxc.wiki/wiki/R%C3%BCckwirkungsabweichung | 1,718,827,156,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00760.warc.gz | 163,807,272 | 11,207 | # Feedback deviation
If a measuring device is built into an apparatus, the original reality changes. The measuring device influences the physical quantity to which the measurement applies. Its effect on the measured value leads to a measurement error that in the metrology basic standard 1319 DIN -1 feedback deviation is called.
## Measurement of temperature
To measure the temperature of a gas flowing through a pipe, z. B. installed a resistance thermometer . The measuring insert and the thermometer protection tube have a different heat dissipation than the original tube wall. In the equilibrium between this increased heat dissipation on the one hand and the supply of heat through the gas on the other hand, a temperature is created in the thermometer that deviates from the gas temperature. The temperature difference to the environment is regularly determined to be too small.
Given the large number of parameters such as installation depth, diameter and flow velocity, the deviation cannot be specified quantitatively. At most, there are empirical values as to when the measurement deviation becomes negligibly small.
## Measurement of electrical quantities
In electrical circuits, one speaks of the fact that a measuring device exerts an influence on the circuit if its insertion changes the quantity to be measured. The influence and reaction-free measurement of the electrical current is only possible with an ideal current source or the electrical voltage with an ideal voltage source or with ideal measuring devices. A measurement deviation must be expected in every real case. The cause is self-consumption through its measuring connections. If parameters are known for electrical measuring devices, the feedback deviation can often be determined mathematically. Such marks come into question
• the internal resistance ${\ displaystyle R_ {i}}$
• the maximum voltage or current consumption:
• in the case of an ammeter, the voltage drop at the end of the measuring range ,${\ displaystyle U_ {I \ mathrm {(MB)}}}$${\ displaystyle I _ {\ mathrm {MB}}}$
• in the case of voltmeter, the current consumption at full scale value or the voltage-related resistance${\ displaystyle I_ {U \ mathrm {(MB)}}}$${\ displaystyle U _ {\ mathrm {MB}}}$ ${\ displaystyle \ varrho}$
• occasionally (most likely with measuring devices for variable quantities) the power consumption at full scale value.
In the ideal case (no voltage drop on the ammeter with ) or (no current consumption of the voltmeter with ). Otherwise the measured value always contains a systematic measurement deviation with a negative sign. How big this turns out is not a property of the measuring device alone, but always the result of its interaction with the circuit. ${\ displaystyle U_ {I \ mathrm {(MB)}} = 0}$${\ displaystyle R_ {i} = 0}$${\ displaystyle I_ {U \ mathrm {(MB)}} = 0}$${\ displaystyle R_ {i} = \ infty}$
The application of the indicators for self-consumption should be shown in examples. - The calculated deviation cannot be verified experimentally by repeating the measurement with the same measuring device.
Circuit with voltmeter
example 1
In the circuit shown here, = 24 V; = 10 kΩ. The control of the transistor should be set so that = becomes. It is assumed that the transistor is independent of , an approximation acceptable above about 2V. The voltmeter gives: = 15 V and = 10 kΩ / V. The question is how far the measured and set voltage is also the voltage that results when the measuring device is removed after the setting. ${\ displaystyle U_ {0}}$${\ displaystyle R_ {a}}$${\ displaystyle U_ {CE}}$${\ displaystyle {\ tfrac {1} {2}} U_ {0}}$${\ displaystyle I_ {C}}$${\ displaystyle U_ {CE}}$
${\ displaystyle U _ {\ mathrm {MB}}}$${\ displaystyle \ varrho}$
At = 15 V, = 0.1 mA flows through the measuring device . With an actually applied voltage of 12 V, the current is lower in the ratio 12:15, i.e. = 0.08 mA. During the measurement the current flows through it . After removing the measuring device, only the unchanged current flows through it , the voltage drop on is = 0.8 V lower. Correspondingly increases to 12.8 V. The absolute feedback deviation is = - 0.8 V, the relative deviation = - 0.8 / 12.8 = - 6%. ${\ displaystyle U _ {\ mathrm {MB}}}$${\ displaystyle I_ {U \ mathrm {(MB)}} = 1 / \ varrho}$${\ displaystyle I_ {U}}$${\ displaystyle R_ {a}}$${\ displaystyle I_ {C} + I_ {U}}$${\ displaystyle R_ {a}}$${\ displaystyle I_ {C}}$${\ displaystyle R_ {a}}$${\ displaystyle R_ {a} \ cdot I_ {U}}$${\ displaystyle U_ {CE}}$${\ displaystyle F}$${\ displaystyle f}$
If the setting were made using a measuring device with 1 kΩ / V, = 0.8 mA and accordingly the voltage change to 8 V when the measuring device was removed. The setting of the transistor control can be omitted in such a measuring device after this preliminary planning. ${\ displaystyle I_ {U}}$${\ displaystyle R_ {a}}$
Circuit with ammeter
Example 2
The voltage source is ideal; the load resistance is ohmic with = 3.0 Ω. ${\ displaystyle R_ {L}}$
The ammeter has the measuring ranges 1 | 3 | 10 | 30… 1000 mA; Self-consumption indicator = 0.6 V in all areas. ${\ displaystyle U_ {I \ mathrm {(MB)}}}$
In the measuring range = 300 mA it just shows full deflection. ${\ displaystyle I _ {\ mathrm {MB}}}$
With an internal resistance = 2.0 Ω one determines = 1.5 V. ${\ displaystyle R_ {i} = U_ {I \ mathrm {(MB)}} / I _ {\ mathrm {MB}}}$${\ displaystyle U_ {0} = I (R_ {L} + R_ {i})}$
If you switch the ammeter to the measuring range 1 A, then = 0.6 Ω, and the current increases to 1.5 V / 3.6 Ω = 0.42 A; that is 40% more than first measured. Such a discrepancy is a sure sign of a defective measuring device (which can be excluded with this consideration) or of a feedback deviation. The current after removing the ammeter is even higher at = 0.50 A. The current measured at the beginning differs by −40% compared to the current without the measuring device. ${\ displaystyle R_ {i}}$${\ displaystyle U_ {0} / R_ {L}}$
Despite the larger systematic measurement deviation in the 300 mA range, you should not switch to the 1 A range. Because in the 1 A range the error limits to be considered due to a class symbol are greater than in the 300 mA range. The deviation due to retroactive effects is easy to calculate and correctable; with error limits, the effort to correct the measured value would be much higher.
Example 3
The circuit and meter are the same as before. Only now = 70 V and = 68 kΩ are given, both with 1% relative error limit. ${\ displaystyle U_ {0}}$${\ displaystyle R_ {L}}$
If the expected current is slightly more than 1 mA, select the 3 mA measuring range, which includes an internal resistance = 0.6 V / 3 mA = 0.2 kΩ. ${\ displaystyle R_ {i}}$
The displayed current is . The right current is . ${\ displaystyle I_ {a} = U_ {0} / (R_ {L} + R_ {i})}$${\ displaystyle I_ {r} = U_ {0} / R_ {L}}$
According to the definition of the relative measurement error is
{\ displaystyle {\ begin {aligned} f & = {\ frac {I_ {a} -I_ {r}} {I_ {r}}} = {\ frac {I_ {a}} {I_ {r}}} - 1 \\ & = {\ frac {R_ {L}} {R_ {L} + R_ {i}}} - 1 = {\ frac {-R_ {i}} {R_ {L} + R_ {i}} } = {\ frac {-0 {,} 2} {68 {,} 2}} = - \; 0 {,} 3 \; \% \ end {aligned}}}
According to the rules of error propagation , the current has an error limit of 2%. This means that the feedback deviation of the current is significantly less than its error limit and cannot be taken into account here.
Example 4
The current from a battery with 1.2 V is measured with a digital ammeter at 90% of the full scale value. If the measuring device drops 0.20 V at the end of the measuring range, the consumer is missing 0.18 V. That is 15% of the battery voltage; the consumer receives 85%. After removing the measuring device, the voltage at the consumer increases in the ratio 100/85 ≈ 1.18, i.e. an additive of 18%. With an ohmic load , the current increases to the same extent; it flows 18% higher than measured. | 2,104 | 7,992 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 45, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-26 | latest | en | 0.897134 |
id3s.ns01.us | 1,607,099,350,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00418.warc.gz | 46,835,626 | 6,538 | # Compiled and Solved Problems in Geometry and Trigonometry.
Free Geometry Problems and Questions writh Solutions. Free geometry tutorials on topics such as reflection, perpendicular bisector, central and inscribed angles, circumcircles, sine law and triangle properties to solve triangle problems. Also geometry problems with detailed solutions on triangles, polygons, parallelograms, trapezoids, pyramids and cones are included.
Problem Solving in Geometry with Proportions Worksheet. PROBLEM SOLVING IN GEOMETRY WITH PROPORTIONS WORKSHEET. Problem 1: . Solution: One way to solve this problem is to set up a proportion that compares the measurements of the Titanic to the measurements of the scale model.
## Problems Geometry Solving Complex Methods of.
WebMath is designed to help you solve your math problems. Composed of forms to fill-in and then returns analysis of a problem and, when possible, provides a step-by-step solution. Covers arithmetic, algebra, geometry, calculus and statistics.Problem Solving in Geometry with Proportions.. Solution: One way to solve this problem is to set up a proportion that compares the measurements of the Titanic to the measurements of the scale model. Problem Solving Strategy: Multiply each side by 11.25.If you find problems that are in the Resources section which are not in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Finally, additions to and improvements on the solutions in the AoPSWiki are always welcome.
Geometry and problem solving 217 Solution: The knowledge of the subject should tell us that if we know the cartesian coordinates A ( x 1; y 1 ), B ( x 2; y 2 ) and C ( x 3; y 3 ) of the.Depending on your school they will go, Applied Geometry (D average or lower), Geometry ( C and above), Problem Solving A (D in geometry), Algebra 2 (C or better in Geometry), Problem solving B (D.
Simulation-Driven Design: Solving the Geometry Problem. Abstract Over the years, computer -aided design (CAD) system geometry has evolved: wireframes to surfaces to solids, and parametric to direct. Simulation, however, remains stuck using the same old beam, shell, or simplified-solids paradigm invented more than 50 years ago.
Free math problem solver answers your algebra homework questions with step-by-step explanations.
Solution: One of the first rules of solving these types of problems involving circles is to carefully note whether we are dealing with the radius or the diameter. In this problem, the circle is described using the diameter, which is 4 inches.
Teacher guide Solving Geometry Problems: Floodlights T-5 SUGGESTED LESSON OUTLINE Individual work (5 minutes) Return students’ solutions to them, and remind them of the Floodlights problem. Recall the work we did (last lesson) on shadows.
Title: Loren C. Larson - Problem-Solving Through Problems - (Problems Books in Mathematics Vol 5) - Springer-Verlag, 1982 - 332p - OK. Author: Igor.
We present a new iterative algorithm for the molecular distance geometry problem with inaccurate and sparse data, which is based on the solution of linear systems, maximum cliques, and a minimization of nonlinear least-squares function. Computational results with real protein structures are presented in order to validate our approach.
Practicing and solving a lot of geometric problems will make you familiar with all these theoremas and their application. Also that can better your creativity, because when you check one problem's solution (even if you don't solve it by yourself), you get an additional idea for a solving in future.
Solution: We can solve this problem by carefully considering the information presented and by applying what we know about solving equations. We know, first of all, that the perimeter P of a rectangle obeys the following formula, where l is the length and w is the width.
Geometry Problem 1464. Quadrilateral, Interior Point, Midpoint of Sides, Equal Sum of Areas. Step-by-step Illustration using GeoGebra. Geometry Problem 1463. Parallelogram, Interior Point, Opposite Triangles with Equal Sum of Areas. Step-by-step Illustration using GeoGebra. Geometry Problem 1462.
We provide the general solution of problems concerning AC star circuits by turning them into geometric problems. We show that one problem is strongly related to the Fermat-point of a triangle. | 935 | 4,421 | {"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} | 3.734375 | 4 | CC-MAIN-2020-50 | latest | en | 0.890479 |
http://fricas.github.io/api/ExtensionField.html | 1,721,504,511,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00534.warc.gz | 13,491,715 | 6,800 | # ExtensionField FΒΆ
ExtensionField F is the category of fields which extend the field `F`
0: %
from AbelianMonoid
1: %
from MagmaWithUnit
*: (%, %) -> %
from LeftModule %
*: (%, F) -> %
from RightModule F
*: (%, Fraction Integer) -> %
*: (F, %) -> %
from LeftModule F
*: (Fraction Integer, %) -> %
*: (Integer, %) -> %
from AbelianGroup
*: (NonNegativeInteger, %) -> %
from AbelianMonoid
*: (PositiveInteger, %) -> %
from AbelianSemiGroup
+: (%, %) -> %
from AbelianSemiGroup
-: % -> %
from AbelianGroup
-: (%, %) -> %
from AbelianGroup
/: (%, %) -> %
from Field
/: (%, F) -> %
`x/y` divides `x` by the scalar `y`.
=: (%, %) -> Boolean
from BasicType
^: (%, Integer) -> %
from DivisionRing
^: (%, NonNegativeInteger) -> %
from MagmaWithUnit
^: (%, PositiveInteger) -> %
from Magma
~=: (%, %) -> Boolean
from BasicType
algebraic?: % -> Boolean
`algebraic?(a)` tests whether an element `a` is algebraic with respect to the ground field `F`.
annihilate?: (%, %) -> Boolean
from Rng
antiCommutator: (%, %) -> %
associates?: (%, %) -> Boolean
from EntireRing
associator: (%, %, %) -> %
characteristic: () -> NonNegativeInteger
charthRoot: % -> Union(%, failed) if F has Finite or F has CharacteristicNonZero
coerce: % -> %
from Algebra %
coerce: % -> OutputForm
coerce: F -> %
from CoercibleFrom F
coerce: Fraction Integer -> %
coerce: Integer -> %
commutator: (%, %) -> %
degree: % -> OnePointCompletion PositiveInteger
`degree(a)` returns the degree of minimal polynomial of an element `a` if `a` is algebraic with respect to the ground field `F`, and `infinity` otherwise.
discreteLog: (%, %) -> Union(NonNegativeInteger, failed) if F has Finite or F has CharacteristicNonZero
divide: (%, %) -> Record(quotient: %, remainder: %)
from EuclideanDomain
euclideanSize: % -> NonNegativeInteger
from EuclideanDomain
expressIdealMember: (List %, %) -> Union(List %, failed)
exquo: (%, %) -> Union(%, failed)
from EntireRing
extendedEuclidean: (%, %) -> Record(coef1: %, coef2: %, generator: %)
from EuclideanDomain
extendedEuclidean: (%, %, %) -> Union(Record(coef1: %, coef2: %), failed)
from EuclideanDomain
extensionDegree: () -> OnePointCompletion PositiveInteger
`extensionDegree()` returns the degree of the field extension if the extension is algebraic, and `infinity` if it is not.
factor: % -> Factored %
Frobenius: % -> % if F has Finite
`Frobenius(a)` returns `a ^ q` where `q` is the `size()\\$F`.
Frobenius: (%, NonNegativeInteger) -> % if F has Finite
`Frobenius(a, s)` returns `a^(q^s)` where `q` is the size()\$`F`.
gcd: (%, %) -> %
from GcdDomain
gcd: List % -> %
from GcdDomain
gcdPolynomial: (SparseUnivariatePolynomial %, SparseUnivariatePolynomial %) -> SparseUnivariatePolynomial %
from GcdDomain
inGroundField?: % -> Boolean
`inGroundField?(a)` tests whether an element `a` is already in the ground field `F`.
inv: % -> %
from DivisionRing
latex: % -> String
from SetCategory
lcm: (%, %) -> %
from GcdDomain
lcm: List % -> %
from GcdDomain
lcmCoef: (%, %) -> Record(llcm_res: %, coeff1: %, coeff2: %)
from LeftOreRing
leftPower: (%, NonNegativeInteger) -> %
from MagmaWithUnit
leftPower: (%, PositiveInteger) -> %
from Magma
leftRecip: % -> Union(%, failed)
from MagmaWithUnit
multiEuclidean: (List %, %) -> Union(List %, failed)
from EuclideanDomain
one?: % -> Boolean
from MagmaWithUnit
opposite?: (%, %) -> Boolean
from AbelianMonoid
order: % -> OnePointCompletion PositiveInteger if F has Finite or F has CharacteristicNonZero
plenaryPower: (%, PositiveInteger) -> %
prime?: % -> Boolean
primeFrobenius: % -> % if F has Finite or F has CharacteristicNonZero
primeFrobenius: (%, NonNegativeInteger) -> % if F has Finite or F has CharacteristicNonZero
principalIdeal: List % -> Record(coef: List %, generator: %)
quo: (%, %) -> %
from EuclideanDomain
recip: % -> Union(%, failed)
from MagmaWithUnit
rem: (%, %) -> %
from EuclideanDomain
retract: % -> F
from RetractableTo F
retractIfCan: % -> Union(F, failed)
from RetractableTo F
rightPower: (%, NonNegativeInteger) -> %
from MagmaWithUnit
rightPower: (%, PositiveInteger) -> %
from Magma
rightRecip: % -> Union(%, failed)
from MagmaWithUnit
sample: %
from AbelianMonoid
sizeLess?: (%, %) -> Boolean
from EuclideanDomain
squareFree: % -> Factored %
squareFreePart: % -> %
subtractIfCan: (%, %) -> Union(%, failed)
transcendenceDegree: () -> NonNegativeInteger
`transcendenceDegree()` returns the transcendence degree of the field extension, 0 if the extension is algebraic.
transcendent?: % -> Boolean
`transcendent?(a)` tests whether an element `a` is transcendent with respect to the ground field `F`.
unit?: % -> Boolean
from EntireRing
unitCanonical: % -> %
from EntireRing
unitNormal: % -> Record(unit: %, canonical: %, associate: %)
from EntireRing
zero?: % -> Boolean
from AbelianMonoid
AbelianGroup
AbelianMonoid
AbelianSemiGroup
BasicType
BiModule(%, %)
BiModule(F, F)
CancellationAbelianMonoid
canonicalsClosed
canonicalUnitNormal
CharacteristicNonZero if F has Finite or F has CharacteristicNonZero
CommutativeRing
CommutativeStar
DivisionRing
EntireRing
EuclideanDomain
Field
FieldOfPrimeCharacteristic if F has Finite or F has CharacteristicNonZero
GcdDomain
IntegralDomain
LeftOreRing
Magma
MagmaWithUnit
Monoid
NonAssociativeRing
NonAssociativeRng
NonAssociativeSemiRing
NonAssociativeSemiRng
noZeroDivisors
PrincipalIdealDomain
Ring
Rng
SemiGroup
SemiRing
SemiRng
SetCategory
TwoSidedRecip
UniqueFactorizationDomain
unitsKnown | 1,625 | 5,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.610465 |
https://www.mathopenref.com/equation.html | 1,712,993,263,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00327.warc.gz | 837,502,824 | 3,735 | Equation
An equation is a mathematical statement that two things are equal. It consists of two expressions, one on each side of an 'equals' sign. For example: This equation states that 12 is equal to the sum of 7 and 5, which is obviously true. In an equation, the left side is always equal to the right side.
Using variables
The most common equations contain one or more variables. If we let x stand for an unknown number, and write the equation We know the left side and right side are equal, so we can see that x must be 12+5 or 17. This is the only value that x can have that makes the equation a true statement. We say that x=17 'satisfies' the equation.
This process of finding the value of the unknowns is called "solving the equation". We often say that we "solve for x" - meaning solve the equation to find the value of the unknown number x.
A common mistake
You will often see things that are called equations when they are not. For example you may see something like this referred to as an equation: It does not have an equals sign in it, and so is not an equation. It is called an 'expression'.
Algebra
The study of algebra is in large part about learning ways to solve various kinds equations. For example, there are ways to solve quadratic equations such as | 290 | 1,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.963084 |
https://www.jiskha.com/questions/209935/im-doing-some-word-problems-on-finding-the-areas-of-rectangles-its-just-like-all | 1,575,998,635,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00055.warc.gz | 763,766,212 | 5,027 | # Math
I'm doing some word problems on finding the areas of rectangles, its just like all complicated. I don't understand how to do them. For example:
A square field had 3 m added to its length and 2m added to its width. The area is 90 m2(squared). Find the length of a side of the original field.
So first you get (x+3)(x+2)=90, right?
then you get x2+5x+6=90
then x2+5x=84
How do you go on? How do you figure out x? I don't get it! I have other problems like those, could you tell me how to figure out a problem with x2 and something x = a number in general? Please help!
1. 👍 0
2. 👎 0
3. 👁 178
x^2 + 5x + 6 = 90
Now:
x^2 + 5x - 84 = 0
Factor by inspection or use the quadratic formula to solve for x.
A hint is what two factors of 84 differ by 5?
1. 👍 0
2. 👎 0
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asked by Liz on November 4, 2013
10. ### Calculus-Aproximate Areas
Estimate the area under the graph of f(x)=sin(pix) from x=0 to x=1 using the areas of 3 rectangles of equal width, with heights of the rectangles determined by the height of the curve a a) left endpoint: b) right endpoint:
asked by Liz on November 4, 2013
More Similar Questions | 964 | 3,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-51 | latest | en | 0.94552 |
http://www.conversion-website.com/energy/horsepower-hour-to-watt-hour.html | 1,695,647,505,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508977.50/warc/CC-MAIN-20230925115505-20230925145505-00229.warc.gz | 54,314,798 | 4,404 | # Horsepower-hour to watt hours (hph to Wh)
## Convert horsepower-hour to watt hours
Horsepower-hour to watt hours conversion calculator shown above calculates how many watt hours are in 'X' horsepower-hour (where 'X' is the number of horsepower-hour to convert to watt hours). In order to convert a value from horsepower-hour to watt hours (from hph to Wh) simply type the number of hph to be converted to Wh and then click on the 'convert' button.
## Horsepower-hour to watt hours conversion factor
1 horsepower-hour is equal to 745.69987158227 watt hours
## Horsepower-hour to watt hours conversion formula
Energy(Wh) = Energy (hph) × 745.69987158227
Example: Consider a energy of 178 horsepower-hour. Below are the steps to convert them to watt hours.
Energy(Wh) = 178 ( hph ) × 745.69987158227 ( Wh / hph )
Energy(Wh) = 132734.57714164 Wh or
178 hph = 132734.57714164 Wh
178 horsepower-hour equals 132734.57714164 watt hours
## Horsepower-hour to watt hours conversion table
horsepower-hour (hph)watt hours (Wh)
2216405.39717481
2417896.796917974
2619388.196661139
2820879.596404304
3022370.996147468
3223862.395890633
3425353.795633797
3626845.195376962
horsepower-hour (hph)watt hours (Wh)
250186424.96789557
300223709.96147468
350260994.95505379
400298279.94863291
450335564.94221202
500372849.93579114
550410134.92937025
600447419.92294936
Versions of the horsepower-hour to watt hours conversion table. To create a horsepower-hour to watt hours conversion table for different values, click on the "Create a customized energy conversion table" button.
## Related energy conversions
Back to horsepower-hour to watt hours conversion
TableFormulaFactorConverterTop | 468 | 1,687 | {"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-2023-40 | latest | en | 0.615354 |
https://mckinneybedandbreakfast.com/tag/live-draw-sgp/ | 1,709,185,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474784.33/warc/CC-MAIN-20240229035411-20240229065411-00359.warc.gz | 369,181,038 | 25,809 | ## How To Increase Your Chances Of Winning A Lottery
result sgp games are a popular way to win big money. They’re easy to play, and many states offer different types of lottery games. The odds of winning vary from game to game, so it’s important to understand them before playing.
#### The History of Lotteries
In ancient times, lotteries were used as a way to raise funds for public projects or social causes. For example, towns in France and Burgundy would set up lotteries to finance defenses and public works, such as bridges or canals. In colonial America, lotteries were also a way to fund fortifications and local militias.
#### The First European Lotteries
In Europe, the earliest known lotteries were held in the Roman Empire, where each guest at a dinner party would be given a ticket and could choose a prize. These prizes were often items of fancy value, such as dinnerware or glassware.
They were also a popular way to attract tourists, as they offered visitors the chance to win money and exotic prizes such as horses. They were also an important means of fundraising for the construction of religious buildings and castles in medieval Europe.
Some states still use lotteries to raise funds for public projects, and in other countries they have become a main source of funding for many charitable organizations. For instance, the state of New South Wales has a lottery that draws more than one million tickets each week and raises millions of dollars for its public schools, roads, libraries, and other projects.
#### The Math Behind Lotteries
The basic mathematical principle of a lottery is that every combination of numbers has an equal probability of being drawn. There is no such thing as “lucky” numbers, but there are certain strategies you can use to increase your chances of winning the jackpot.
Purchasing more tickets increases your odds of hitting a jackpot. However, this can also increase your costs. If you’re buying more than one ticket, you should consider joining a group to pool your funds and spread the cost of ticket purchases out over multiple players.
You should also avoid playing numbers that have sentimental value, such as your birthday or the number of your favorite team. This can reduce your odds of sharing a winning prize with others, because other players will likely choose the same numbers.
Another strategy for increasing your odds of winning is to select random numbers that aren’t close together. This can improve your chances of not wasting money on a large jackpot, but it won’t increase your odds of winning any other prize divisions, according to Lew Lefton, a faculty member at Georgia Tech’s School of Mathematics.
In the United States, most states and the District of Columbia have some form of lottery. These games range from instant-win scratch cards to pricier games with larger prizes.
If you’re looking to win the big bucks, it’s best to avoid the most popular lottery games. You’ll be much more likely to hit the jackpot with a smaller lottery game, such as a state pick-3.
## What You Need to Know About a Sportsbook
A live draw sgp is a place where people can place bets on sporting events. It is legal in many states, and is a popular pastime for sports fans. Whether you’re a novice or an expert, there are plenty of things to know about sports betting.
#### What Is a Sportsbook?
A sportsbook can be a safe haven for gamblers, or it can be a nightmare. The choice of which one to join depends on your goals and personal preferences. In some cases, the best way to find a suitable sportsbook is to do some research and read reviews online.
Sportsbooks are a great way to make money, but you must be careful when choosing the right one. There are several factors to consider, including the type of sports you want to bet on and the odds offered. These are important because they will determine your profits or losses.
The best sportsbooks offer a wide variety of games, including bola resmi, basketball, and volleyball. Some even offer wagering on horse races. The odds are updated in real time and you can bet on any game at any time, even during the game.
Moreover, some sportsbooks allow you to deposit money through cashier’s checks and bank drafts. However, be aware of the minimum deposit amounts and the fees associated with these methods. You should also check the terms and conditions of a sportsbook’s bonus programs before making your decision.
#### Layoff Accounts
Some sportsbooks offer layoff accounts, which are a convenient way to cover the spread on a mirror bet or protect your winnings from a bad streak. These accounts can be used by anyone who’s interested in betting on sports, but they are particularly helpful for players who aren’t able to make a large bet.
#### Pay per Head Sportsbook Software
When it comes to running a profitable sportsbook, it’s important to choose the right software. This is because it will help you manage your business more efficiently and increase your profits. You can also use it to expand your business, as well as offer new services and features.
A good sportsbook will focus on small details to attract more customers. This means it should offer a variety of betting options, a good customer support team, and a positive reputation.
#### Signup Bonuses
The signup bonuses that are offered by many online sportsbooks are a great way to entice new customers. These are usually very generous, and can be worth hundreds of dollars. The terms of these bonuses vary from sportsbook to sportsbook, but you should always look for ones that have favorable roll-over requirements.
#### How to Become a Sportsbook Owner
If you’re interested in starting your own sportsbook, it is essential to understand the basic business model. This will help you set a clear price point, avoid credit card bets, and maximize your profits. You should also hire a professional team to manage your sportsbook.
A sportsbook can be a lucrative business, but it requires a lot of work and investment. It’s also a risky venture, so you need to be careful and follow all the rules. You can also consult an attorney if you’re not sure of your rights.
## How to Play the Lottery Online
LIVE SGP history, lotteries have been used to finance important government projects, assist the poor, and prepare for war. In the Middle Ages, lotteries were used to improve fortifications. And in modern times, governments recognize the importance of lotteries.
Lottery games vary by region. The US, for instance, offers a wide range of different games. Some of the most popular lotteries are the Mega Millions, the Powerball, and the Lotto America. Each of these lotteries has different rules. Some offer higher odds of winning, while others have jackpots that are extremely high. Oftentimes, the winning jackpot is divided among several participants. These types of lottery games are called progressive lotteries.
Ticket sales close about two hours before a draw takes place. This means that if you buy your ticket online, you have a better chance of winning. However, you have to be in a state that permits online ticket sales. In some states, you can also purchase tickets over the phone. And in many cases, you can also play using an app on your smartphone.
In the Powerball game, you choose five numbers from a pool of 69. You can also opt for an extra pool of numbers, allowing you to increase your chances of winning. You can buy a Powerball ticket for \$2. Afterwards, you’ll need to match the number you choose with the number drawn.
The New Hampshire Lottery was created in 1964 and continues to draw crowds with huge jackpots. You can also choose to play Keno and other draw games, as well as the Mega Millions. Currently, the lottery has a jackpot that is expected to reach \$1 billion. This is one of the largest jackpots in the world.
The US lottery offers 177 different games. Every week, there are roughly 1,000 drawings. The lottery is available in 45 US states. The majority of the proceeds are used for public good. The proceeds of the US Virgin Islands lottery are also given to public good causes.
When playing the lottery, you have to choose one or more numbers that haven’t been drawn in a while. Some lottery enthusiasts believe that past draws can affect future draws. This is known as the gambler’s fallacy. This false belief is based on the idea that random events can influence each other. If you have a 50% chance of winning, you’ll need to buy a hundred tickets a day for 265.6 years. The jackpot will then reset to the predetermined minimum.
In addition to the US lottery, players from around the world can participate in other lotteries. There are hundreds of bingo halls in the United States. Most of these halls offer prizes of \$100,000.
Some lottery sites offer a concierge service. These websites enable players to buy tickets from different countries and locations. While these services have proven to be successful in the past, they haven’t yet changed the market. These services typically require a bank transfer, and you may need to fill out a cash-out request.
## Playing Online Casino Games
data sgp hari ini you’re looking to play blackjack, poker, or roulette, there are many options for online players. Many of these games have been updated to include the latest technology and graphics. The newest versions of these games are faster and more responsive than their predecessors.
When playing online, the first step is to create an account with an online casino. You’ll also be required to make a real money deposit. Once you’ve deposited, you’ll get bonus cash to use on the casino. Some online casinos also offer live casino bonuses. These are often in the form of welcome bonuses. They’re a great way to get started, but you’ll want to choose an operator carefully.
You’ll also need to select the right game. Some casinos only offer a few options, while others offer the full range of table games. In addition to the usual casino games, some of these casinos also offer lottery-like games such as bingo. Some of these games are very popular and are played by many people.
Live casino games are a lot of fun. They’re designed to mimic the experience of playing in a brick and mortar casino. This includes a live dealer, live audio, and real-time video. The games are also played against computer simulations.
These games are great for players who enjoy interacting with other people. There’s also a Las Vegas feel to playing live casino games, but you won’t have to go to a real casino to get it. You can play online and interact with real people, and this can make all the difference to the experience.
One of the newest technologies available to online casino players is virtual reality. This technology will allow players to walk around the virtual casino, touch the cards, and watch other players. It’s also the best way to experience the thrill of a live casino, without having to leave the comfort of your own home. Unlike the physical casino, you won’t have to worry about dress codes, opening hours, or gambling space.
Live casino games are also a lot harder to rig than regular casino games. This is because you’re dealing with real people, not computer generated responses. Live casinos also feature some of the newest technology, such as real-time audio, HD video, and a professional dealer. A dealer can spin the roulette wheel in real time, and also place bets.
The best live casino online is Bovada, which offers a range of live casino games. The site’s most popular game is roulette, and the site also offers blackjack, baccarat, and poker. These games are played in a high-quality streaming studio. The site also features professional dealers and offers a variety of bonuses and incentives to make the experience more enjoyable.
The best live casino online also has the best gaming software. This software manages the video feed, and controls the seamless user experience. It also has random number generators (RNGs) and a fair game system. These make for a fun and realistic online gambling experience.
## Lottery jackpot odds
The odds of winning the lottery jackpot are very low, but there are some exceptions. For instance, in the Mega Millions, there’s a one-in-302.5-million chance that someone will win the jackpot. And if more than one person wins the jackpot, the prize will be split between them. The best way to improve your odds of winning is to play regularly.
When you play the Powerball, the odds are one in 292 million. These figures are based on a formula that determines the probability of winning the jackpot. This figure is quite low, especially when you consider that you are compared to the population of the United States. And yet, many regular lottery players prefer smaller jackpots with better odds, despite the smaller jackpots.
## Lottery pools
Having a Live Singapore pool is a great way to spend time with friends or co-workers, but it is important to understand the rules and make sure that it is legal. If lottery pools are not done correctly, it can lead to major problems for participants. This can include cheating your fellow players, and you may end up being sued if your pool is not legal.
First, it is important to ensure that everyone in your office is invited to play in the pool. This will give you a larger group of people to play with and a higher chance of winning the lottery. However, if your pool is small, you don’t need to make it public. Having a lottery pool is a great way to bond with your colleagues and build the culture of your firm.
## Taxes on lottery winnings
Taxes on lottery winnings depend on the state in which the lottery is played. For example, New York City and Yonkers will each withhold up to three and a half percent of a winner’s winnings, respectively. This is in addition to the federal withholding of twenty-four percent. In addition, seven states have no income tax, which means big lottery winners in those states will not owe state income taxes on their prize money. In addition, a few states don’t even have a lottery at all.
Generally, lottery winnings must be reported as ordinary income. Federal tax rules apply to prizes, awards, sweepstakes, and raffles, but the rules for state and local taxes vary. It’s best to consult a financial advisor about the best way to handle your lottery winnings and ensure you’re getting the most out of your prize.
## Scams related to lotteries
Lottery scams are not uncommon and can come in many forms. Many of these scams target older people, who often have large savings that are at risk of being wiped out. Be cautious and report any suspicious activity immediately. Typically, scammers will contact you by email, asking for a large up-front fee.
Lotteries are regulated to protect the public from fraud and money laundering. They also help protect vulnerable groups and protect minors from predatory loans and fees. While lotteries are largely based on chance, there is an element of skill that can improve one’s chances of winning.
## A Brief History of Data SGP Gambling
Lotteries are a type of gambling that involves drawing numbers to win a prize. Some governments outlaw them, while others endorse them and organize state and national lotteries. Regardless of the legality of a lottery, it’s important to note that they are addictive. Here’s a brief history of lottery gambling.
## Lotteries date back to the Han Dynasty
The history of the Data SGP can be traced back to the Han Dynasty in China, which lasted between 205 and 187 BC. These ancient Chinese people used the lottery to finance major government projects. It is also mentioned in the Chinese Book of Songs, where the game is referred to as “drawing wood” or “lots”. Today, lottery games are popular all over the world and serve as a major source of entertainment for a wide variety of people.
Although there is little resemblance between ancient lotteries and the modern versions, the concept behind the lottery is very old. Ancient Chinese ‘baige piao’, which means ‘white pigeon ticket,’ was used for public works and military training instead of tax collections. In addition, Lotteries were used to fund military and public works projects throughout ancient China, including the construction of the Great Wall of China.
## They are a form of gambling
Lotteries are forms of gambling where people purchase lottery tickets and hope to win the jackpot or other prize. These games are based on random drawing and can be addictive. Many countries outlaw them, while others endorse them and regulate them. Most regulations focus on the sale of tickets to minors and the licensing of vendors who sell them. By 1900, most countries had made gambling illegal. After World War II, many countries banned it entirely.
Lotteries first made their appearance in the United States during the early nineteenth century, when British colonists brought the game with them. At the time, Christians viewed lotteries as a sinful practice. Ten states banned lotteries between 1844 and 1859, but they quickly became popular again. Despite this, there are many downsides to lotteries.
## They raise money for governments
Lotteries are a popular way for local and state governments to raise money. This revenue helps fund a variety of government services, including road construction and education programs. Some governments also use lottery funds to support programs for seniors and tourism. While lottery revenue is valuable, many people question the ethicality of such a system.
Lotteries are games of chance that have a long and storied history. In the early days, these games provided crucial funds to the military during the French and Indian War. In addition, colonial governments used the money from these games to build churches and schools.
Many people wonder if lotteries are addictive. The temptation to play a lottery can be strong, and the winnings can be extremely large. Fortunately, lotteries do not require a high level of skill. Because of this, they seem less addictive than other types of gambling. Additionally, the non-instant nature of lotteries prevents the brain from activating reward centers. As a result, many lottery players are considered low-risk gamblers.
While lotteries are not considered the most dangerous type of gambling, they do have the potential to cause psychological distress and addiction. Problem gambling is an epidemic in the United States, and it is more common in young people.
## They are a game of chance
Lotteries are a popular form of gambling where you place a bet and then wait for a random number to be drawn. Winning a lottery prize depends primarily on luck, but there is a certain amount of skill involved. You can increase your chances of winning by playing more than one lottery game.
Lotteries are a game of chance, meaning the outcome is dependent on randomness. These games of chance are often regulated to prevent money laundering, fraudulent activities, and practices contrary to public order. They also protect minors and vulnerable people from the negative effects of excessive participation.
## Lottery Strategies – How to Increase Your Odds of Winning the Live Draw SGP
Lotteries are a form of gambling that allows governments to raise revenue without increasing taxes. They allow people to purchase pre-determined prizes by purchasing tickets, and you can use strategies to increase your chances of winning. This article will explain how lotteries work, as well as provide some strategies for improving your odds of winning.
## Lotteries are a form of gambling
Lotteries are games of chance that depend on luck and random events. While they are a popular form of gambling, they are not without risk. Governments often tax winning wagers and may prohibit lotteries in their states. Many lottery players become addicted. Nevertheless, the potential for a large jackpot makes lotteries a popular way to make money.
Lotteries are often accompanied by a Live Draw SGP prize of goods or cash. These prizes may be a fixed amount, a percentage of the total revenues, or a combination of items. While the traditional lotto format is a 50-50 draw, many recent lotteries allow purchasers to pick their own numbers, increasing the likelihood of multiple winners.
## They allow governments to raise revenue without increasing taxes
While the lottery presents itself as a viable alternative to other forms of taxation, it is one of the most unfair. Lotteries exploit the poor, desperate, and addicted to make money. Governments don’t necessarily need to increase taxes in order to raise revenue with lotteries, so long as they have an effective plan.
Historically, states have used federal grants for many purposes. These funds often go toward building projects, education, and welfare. With the federal government no longer providing much of the funding for these services, many states are turning to lotteries to cover these expenses. These lotteries generate millions of dollars each year, and the governments keep one-third of the proceeds. However, there are still some people who believe that these programs hurt the lower-income population.
## They offer predetermined prizes
Lotteries are games in which a person draws a number and tries to win the prize. The prizes are determined by the amount of tickets sold and are sometimes large or small. Often the prizes are split between the sponsor and the state. Lotteries can also be used for decision-making purposes, such as the allocation of land, or for a sporting event. These games can be highly addictive, which is why some governments ban them, while others endorse them.
## Strategies to increase your odds of winning
There are many strategies to increase your odds of winning the lottery. One of the most common is buying more lottery tickets, which will increase your chances of winning. However, it can also lead to more expenses over time. Moreover, a recent study conducted in Australia showed that buying more lottery tickets does not affect your chances of winning. This means that you need to combine this strategy with other proven winning methods.
Another way to improve your chances is to develop patience. While it might seem counterintuitive to spend your wealth on yourself, you can also choose to give it to others. While you’re not legally required to do so, it is a social responsibility to help others. Besides, it’s an enriching experience for yourself. After all, money does not make us happy, but it can give us the opportunity to experience joy.
## How to Win the Lottery Result SGP
Drawing lots for land ownership and other rights is a practice that dates back to ancient times. It first gained widespread use in Europe during the late fifteenth and sixteenth centuries. The first lottery to be tied to the United States was in 1612 when King James I of England established a lottery to provide funding to the settlement in Jamestown, Virginia. Over the following centuries, lottery funding was used by public and private organizations to raise funds for wars, colleges, and public-works projects.
## Lotteries
Lotteries have become an integral part of many governments, allowing them to raise much-needed funds without relying on taxpayers. In addition to being a source of revenue, lottery play can result in a huge cash prize. For example, the National Basketball Association has a lottery for the fourteen worst teams each year, and the winning team gets to choose its draft picks. This means that the winning team has the opportunity to select the best college talent.
## Pattern of European lotteries
Lotteries are a popular form of taxation that first appeared in the 15th century in the Low Countries. Towns began holding public lotteries to raise funds for defense and to provide relief for the poor. These lotteries were widely popular and were seen as a convenient, inexpensive taxation method. For example, the emperor Augustus of Rome organized a lottery called the Staatloterij to raise money for the city’s walls. The winners of this lottery received articles of unequal value.
## Odds of winning
If you’re looking for a way to increase your chances of winning the Result SGP, it’s important to know how many winners there are. The odds of winning the Mega Millions lottery are one in 176 million. The odds for the California Super Lotto are one in 42 million. Obviously, these numbers are far from zero, but if you’re a math nerd, it’s possible to calculate the odds of winning with the help of the odds calculators available online.
## Prizes
There are a number of steps to claim Lottery prizes, including filling out a claim form. If you win a prize of \$100 or more, you must visit the Lottery office to claim your prize. You must present your winning ticket, with the barcodes clearly visible. You must also remove scratch-off material from scratch-off tickets, and you must have a valid photo ID.
## Retailers
Retailers of lottery tickets are not immune from the scrutiny of the federal government. According to a recent investigative report by the Virginian-Pilot, some retailers have won multiple times over the past few years. However, they are still no richer than the \$1.6 billion lottery jackpot winner.
## Tax implications
Winning the lottery can be extremely lucrative, but it can also have negative tax implications. Governments can tax lottery winnings at rates up to 37%. Depending on the prize amount, you may have to pay the full amount in taxes or split it over a period of time. It is therefore important to understand the tax implications of winning the lottery, and plan accordingly. | 5,215 | 25,675 | {"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-2024-10 | longest | en | 0.976709 |
http://www.mathworks.com/matlabcentral/fileexchange/15285-geodetic-toolbox/content/geodetic/ell2utm.m | 1,430,024,992,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246652631.96/warc/CC-MAIN-20150417045732-00205-ip-10-235-10-82.ec2.internal.warc.gz | 636,030,999 | 9,628 | Code covered by the BSD License
Geodetic Toolbox
Mike Craymer (view profile)
13 Jun 2007 (Updated )
Toolbox for angle, coordinate and date conversions and transformations. Version 2.97.
[N,E,Zone,lcm]=ell2utm(lat,lon,a,e2,lcm)
```function [N,E,Zone,lcm]=ell2utm(lat,lon,a,e2,lcm)
% ELL2UTM Converts ellipsoidal coordinates to UTM coordinates.
% UTM northing and easting coordinates must be in a 6 degree
% system. Zones begin with zone 1 at longitude 180E to 186E
% and increase eastward. Formulae from E.J. Krakiwsky,
% "Conformal Map Projections in Geodesy", Dept. Surveying
% Engineering Lecture Notes No. 37, University of New Brunswick,
% Fredericton, N.B, 1973. Krakwisky meridian arc length (S)
% replaced with Hermert (1880) expansion. Vectorized.
% Version: 2011-11-13
% Useage: [N,E,Zone,lcm]=ell2utm(lat,lon,a,e2,lcm)
% [N,E,Zone,lcm]=ell2utm(lat,lon,a,e2)
% [N,E,Zone,lcm]=ell2utm(lat,lon,lcm)
% [N,E,Zone,lcm]=ell2utm(lat,lon)
% Input: lat - vector of latitudes (rad)
% lon - vector of longitudes (rad)
% a - optional ref. ellipsoid major semi-axis (m);
% default GRS80
% e2 - optional ref. ellipsoid eccentricity squared;
% default GRS80
% lcm - optional vector of non-standard central meridians
% (rad); scalar if same for all; default = UTM def'n
% Output: N - vector of UTM northings (m)
% E - vector of UTM eastings (m)
% Zone- vector of UTM zones (zeros if lcm specified)
% lcm - central meridian(s) used in conversion (rad)
% Copyright (c) 2011, Michael R. Craymer
% All rights reserved.
% Email: mike@craymer.com
if nargin ~= 2 & nargin ~= 3 & nargin ~= 4 & nargin ~= 5
warning('Incorrect number of input arguments');
return
end
if nargin == 3
lcm=a;
end
if nargin == 2 | nargin == 3
[a,b,e2,finv]=refell('grs80');
f=1/finv;
else
f=1-sqrt(1-e2);
end
if nargin == 3 | nargin==5
Zone=zeros(size(lat));
else
Zone=floor((rad2deg(lon)-180)/6)+1;
Zone=Zone+(Zone<0)*60-(Zone>60)*60;
lcm=deg2rad(Zone*6-183);
end
if abs(lat)>deg2rad(80)
warning('Latitude outside 80N/S limit for UTM');
end
ko=0.9996; % Scale factor
No=zeros(size(lat)); % False northing (north)
No(lat<0)=1e7; % False northing (south)
Eo=500000; % False easting
lam=lon-lcm;
lam=lam-(lam>=pi)*(2*pi);
RN=a./(1-e2*sin(lat).^2).^0.5;
RM=a*(1-e2)./(1-e2*sin(lat).^2).^1.5;
h2=e2*cos(lat).^2/(1-e2);
t=tan(lat);
n=f/(2-f);
%----- Hinks (1927) Bessel series expansion used in DMA definition of UTM
%A0=1 - n + n.^2*5/4 - n.^3*5/4 + n.^4*81/64 - n.^5*81/64;
%A2=3/2*( n - n.^2 + n.^3*7/8 - n.^4*7/8 + n.^5*55/64 );
%A4=15/16*( n.^2 - n.^3 + n.^4*3/4 - n.^5*3/4 );
%A6=35/48*( n.^3 - n.^4 + n.^5*11/16 );
%A8=315/512*( n.^4 - n.^5 );
%S=a*(A0*lat-A2*sin(2*lat)+A4*sin(4*lat)-A6*sin(6*lat)+A8*sin(8*lat));
%----- Helmert (1880) expansion & simplification of Bessel series (faster)
A0=1+n^2/4+n^4/64;
A2=3/2*(n-n^3/8);
A4=15/16*(n^2-n^4/4);
A6=35/48*n^3;
A8=315/512*n^4;
S=a/(1+n)*(A0*lat-A2*sin(2*lat)+A4*sin(4*lat)-A6*sin(6*lat)+A8*sin(8*lat));
%----- Krakiwsky (1973) expansion
%A0=1-(e2/4)-(e2^2*3/64)-(e2^3*5/256)-(e2^4*175/16384);
%A2=(3/8)*( e2+(e2^2/4)+(e2^3*15/128)-(e2^4*455/4096) );
%A4=(15/256)*( e2^2+(e2^3*3/4)-(e2^4*77/128) );
%A6=(35/3072)*( e2^3-(e2^4*41/32) );
%A8=-(315/131072)*e2^4;
%S=a*(A0*lat-A2*sin(2*lat)+A4*sin(4*lat)-A6*sin(6*lat)+A8*sin(8*lat));
E1=lam.*cos(lat);
E2=lam.^3.*cos(lat).^3/6*(1-t.^2+h2);
E3=lam.^5.*cos(lat).^5/120.*(5-18*t.^2+t.^4+14*h2-58*t.^2.*h2+ ...
13*h2.^2+4*h2.^3-64*t.^2.*h2.^2-24*t.^2.*h2.^3);
E4=lam.^7.*cos(lat).^7/5040.*(61-479*t.^2+179*t.^4-t.^6);
E=Eo+ko*RN.*(E1+E2+E3+E4);
N1=S./RN;
N2=lam.^2/2.*sin(lat).*cos(lat);
N3=lam.^4/24.*sin(lat).*cos(lat).^3.*(5-t.^2+9*h2+4*h2.^2);
N4=lam.^6/720.*sin(lat).*cos(lat).^5.*(61-58*t.^2+t.^4+ ...
270*h2-330*t.^2.*h2+445*h2.^2+324*h2.^3-680*t.^2.*h2.^2+ ...
88*h2.^4-600*t.^2.*h2.^3-192*t.^2.*h2.^4);
N5=lam.^8/40320.*sin(lat).*cos(lat).^7.*(1385-311*t.^2+543*t.^4-t.^6);
N=No+ko*RN.*(N1+N2+N3+N4+N5);
```
Contact us | 1,696 | 4,091 | {"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-2015-18 | longest | en | 0.52499 |
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Algebra
Pregunta
2. Look at the graph of $$f ( x ) = - ( x + 3 ) ^ { 2 } + 4$$ . What is the average rate of change between $$[ - 5 , - 3 ]$$ ? Draw a line between the two points and explain what the rate of change that found means.
average rate of change = $$\frac{4- 0}{- 3+ 5} = 2$$ | 129 | 371 | {"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} | 3.328125 | 3 | CC-MAIN-2022-21 | latest | en | 0.497238 |
http://primes.utm.edu/curios/page.php?number_id=997 | 1,534,922,572,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219692.98/warc/CC-MAIN-20180822065454-20180822085454-00517.warc.gz | 317,349,220 | 2,615 | 158 (another Prime Pages' Curiosity)
Curios: Curios Search: Participate: The sum of the first nine Mersenne prime exponents. [Russ] The smallest number such that the sum of the number and its reverse is a prime that is not palindromic, i.e., 158 + 851 = 1009. [Gupta] Inserting 158 zeros between the first four consecutive primes results in a prime, 2*10^(3*159)+3*10^(2*159)+5*10^159+7. [Opao] 75*158^75+1 and 76*158^76+1 are both prime. However, neither 75*a^75+1 nor 76*a^76+1 are prime for any smaller a! [Hartley] (There is one curio for this number that has not yet been approved by an editor.) Prime Curios! © 2000-2018 (all rights reserved) privacy statement | 214 | 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} | 2.578125 | 3 | CC-MAIN-2018-34 | latest | en | 0.852853 |
https://www.fatalerrors.org/a/this-paper-analyzes-the-open-source-library-of-formula-tree-in-detail.html | 1,685,963,653,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651815.80/warc/CC-MAIN-20230605085657-20230605115657-00228.warc.gz | 824,635,098 | 12,403 | # This paper analyzes the open source library of formula tree in detail
Absrtact: the function of the formula tree module is to start from the training set X and function_ Random sampling is carried out in set to generate a formula tree, and the methods of subtree variation, crossover, hoist variation and point variation are provided at the same time.
## 1. Formula tree module
The function of the formula tree module is to start from the training set X and function_ Random sampling is carried out in set to generate a formula tree, and the methods of subtree variation, crossover, hoist variation and point variation are provided at the same time.
### 1.3 build_ Principle of tree algorithm
Data structure used:
terminal_stack: storage is a stack of several element operations
symbol_tree: lisp_tree list tree, Lisp list is based on the structure of generalized table, so it is easy to express a list as a tree structure. S-expressions may be known for their use in Lisp family programming languages. Lisp was invented by John McCarthy in 1958 and first implemented by Steve Russell on IBM 704 computer. It is particularly suitable for artificial intelligence schemes because it effectively processes symbolic information.
In prefix notation, operators write before their operands. For example, an expression
`a * ( b + c ) / d`
Written as
`(/ (* a (+ b c) ) d)`
For example, formula:
It can also write:
When you write it as an S expression, it becomes this
Corresponding binary tree
That is, the s expression corresponds to the preorder traversal of the symbol tree
Algorithm input: function_set[‘add’, ‘sub’, ‘mul’] , arities{2:[‘add’, ‘sub’, ‘mul’]}, method: grow , max_ Depth: 3 (a random number in 2-4)
```method: grow
n_features: 10
max_depth:2
function_index:0
terminal_stack:[2]
Cycle part
############LOOP##########################
1st time
depth: 1 # The length of the program, that is, the length of the list table of the symbol tree, is equal to len(terminal_stack)
choice: 13
choice_index: 1 #depth < max_ Depth or choice_ Index < = len (self. function_set) will start from function_ Select one from set. For example, select add here
function_index: 0
1_terminal_stack: [2, 2]
2nd time
depth: 2
choice: 13
choice_index: 11
2_terminal: 10 # Here terminal is 10 and n_features are equal, so a random float value is generated
3_terminal: 0.8395650516882855
2_terminal_stack: [2, 1]# Add a constant and subtract 1 from the second value
3rd time
depth: 2
choice: 13
choice_index: 0
2_terminal: 8 # Here terminal is 10 and n_ If features are not equal, add a feature column
3_terminal: 8
2_terminal_stack: [2, 0]
2_terminal_stack.pop(): 0 # When it is equal to 0, it is dropped by pop, and then stack minus 1
4th time
depth: 1
choice: 13
choice_index: 5
2_terminal: 0
3_terminal: 0
2_terminal_stack: [0]
2_terminal_stack.pop(): 0
Finally, the shape of the tree becomes this
The process of generating the second tree
method: grow
max_depth:4
function_index:0
function Address of:<functions._Function object at 0x000001E538356EB0>
terminal_stack:[2]
#############LOOP#########################
1st time
depth: 1
choice: 13
choice_index: 4
2_terminal: 3
3_terminal: 3
2_terminal_stack: [1]
2nd time
depth: 1
choice: 13
choice_index: 4
2_terminal: 6
3_terminal: 6
2_terminal_stack: [0]
2_terminal_stack.pop(): 0
Finally, the shape of the tree becomes this
Draw a flow chart
### 1.4 get_subtree get random subtree
To symbol_ The elements in the tree are given weights. If it is an operator, it is 0.9. If it is not an operator, it is 0.1
For example, tree is
`tree1: mul(X6, sub(X3, X1))`
Give weight for so long
`probs: [0.9 0.1 0.9 0.1 0.1]`
Then normalization is divided by sum
`_probs: [0.42857143 0.47619048 0.9047619 0.95238095 1. ]`
Here is the roulette method to select the cutting point
step
1) Randomly generate random values within a (0,1), such as
`s_index:0.8299421213898753`
2) Find the position where the random value should be in the probs, where this position is the position of start
`start: 2`
3) Initialize end = start=2
`stack = 1`
If end - start < stack, then
`node = program[end]`
If node is an operator, stack needs to add the element
`stack += node.arity`
End itself is added until the end of the program
Draw a flow chart
### 1.5 principle of crossover algorithm module
The purpose of the crossover is to cross subtrees
The first step is to obtain the random subtree from the symbol tree module
` start, end = self.get_subtree(random_state)`
The second step is to obtain random subtrees from other symbol tree individuals
` donor_start, donor_end = self.get_subtree(random_state, donor)`
The third step is to obtain the crossed symbol tree
```self.symbol_tree[: start] + donor[donor_start : donor_end] + self.symbol_tree[end : ]
start, end: 1, 4
removed:range(1, 4)
donor: [mul, 6, 0.6656811846792283]
donor_start, donor_end:(0, 3)
donor_removed:[]
Combined subtree
Finally, the results are obtained
self.symbol_tree[: start] + donor[donor_start : donor_end] + self.symbol_tree[end : ]```
### 1.6 subtree_mutation subtree variation
By P_ subtree_ The mutation parameter controls. This is a more radical mutation strategy: one subtree of the winner will be replaced by another completely random new subtree.
``` chicken = self.build_tree(random_state) Directly produce a new subtree
self.crossover(chicken, random_state) # Then the crossover algorithm is used to generate subtree directly```
### 1.7 hoist_mutation hoist
hoist mutation is A method to combat the bloating of formula tree: randomly select A subtree A from the winner formula tree, then randomly select A subtree B from A, then promote B to the original position of A and replace A with B. hoist means "raise, lift".
The first step is to obtain A random subtree A
``` start, end = self.get_subtree(random_state)
subtree = self.symbol_tree[start:end]```
The second step is to obtain A subtree B from subtree A
``` # Get random subtree B
sub_start, sub_end = self.get_subtree(random_state, subtree)
hoist = subtree[sub_start:sub_end]```
Step 3 lift B to the original position of A and replace A with B
``` self.symbol_tree[:start] + hoist + self.symbol_tree[end:]
start, end: 0, 7
mutation: sub_start, sub_end: 3, 6
removed: [0, 1, 2, 6]
The second time
tree1: mul(X8, X0)
mutation_start, end: 0, 3
mutation_subtree : [mul, x8, x0]
mutation: sub_start, sub_end: 1, 2
mutation_hoist : [x8]
removed: [0, 2]
new_tree: [8]```
### 1.8 point_mutation point variation
By P_ point_ The replace parameter controls. A random node will be changed. For example, addition can be replaced with division, and variable X0 can be replaced with constant - 2.5. Point variation can rejoin some previously eliminated functions and variables, so as to promote the diversity of formulas.
The first step is to copy the symbol tree and obtain a random point
``` program = copy(self.symbol_tree) # Make a copy yourself
# Randomly generate a symbol tree length of points, and then find the points less than the point mutation probability to form a list
mutate = np.where(random_state.uniform(size = len(program)) < self.p_point_replace)[0] # Get a random point```
The second step is to traverse the node node of mutate. If the node is a Function, replace it. If not, add a constant or feature
``` if isinstance(program[node], _Function):
arity = program[node].arity # Operator element
replacement_len = len(self.arities[arity]) # Find how many operators are the same as arity
replacement_index = random_state.randint(replacement_len) # Choose one at random
replacement = self.arities[arity][replacement_index] # Find the operator corresponding to index
program[node] = replacement # Replace```
If not function
First case
```# If it is not an operator, it is a constant or a constant is added to the endpoint
if self.const_range is not None:
terminal = random_state.randint(self.n_features + 1)
else:
terminal = random_state.randint(self.n_features) # Randomly generate a number terminal in a (0, n_features)
if terminal == self.n_features: # If terminal and n_ If the features are equal, replace them with a number of float in (0,1)
terminal = random_state.uniform(*self.const_range)
if self.const_range is None:
raise ValueError('A constant was produced with const_range=None.')
program[node] = terminal```
## 2. The fitness module obtains the adaptability of the symbol tree
### 2.1 get_ all_ The indexes interface obtains the indexes of all data
Step 1: perform parameter verification
```if self._indices_state is None and random_state is None: # If_ indices_state and random_state is None
raise ValueError('The program has not been evaluated for fitness\n yet, indices not available.')
if n_samples is not None and self._n_samples is None: #If n_samples is not None
self._n_samples = n_samples
if max_samples is not None and self._max_samples is None: # n_samples represents the number of rows in the dataset
self._max_samples = max_samples # max_samples maximum sampling tree
if random_state is not None and self._indices_state is None:
self._indices_state = random_state.get_state()```
The second step is to obtain random seeds, and then obtain the index of out of bag data and in bag data
``` indices_state = check_random_state(None)
indices_state.set_state(self._indices_state) # Get random_state
not_indices = sample_without_replacement(
self._n_samples,
self._n_samples - self._max_samples,
random_state=indices_state) # Data outside the bag is selected from [0,self._n_samples]_ n_ samples - self._ max_ Samples data
sample_counts = np.bincount(not_indices, minlength=self._n_samples) # Find the number of occurrences of each index
indices = np.where(sample_counts == 0)[0] # If the number of occurrences is zero, the index is the data left in the bag```
Other functions
sample_without_replacement(n_population, n_samples, random_state,method): sampling function, randomly obtain out of bag data, and select n from the set [0, n_population]_ Samples data, with samples returned
Parameter introduction
### 2.2 raw_fitness
Interface
`raw_fitness(self, X, y, sample_weight)`
First execute the algorithm of X to get y_pred, and then according to y,y_pred and weight calculation error
``` # Evaluate the applicability of the symbol tree according to x and y, and return fitness
y_pred = self.execute(X)
raw_fitness = self.metric(y, y_pred, sample_weight)```
### 2.3 fitness module
Interface
`fitness(self, parsimony_coefficient=None)`
First execute the algorithm of X to get y_pred, and then according to y,y_pred and weight calculation error
```if parsimony_coefficient is None:
parsimony_coefficient = self.parsimony_coefficient
penalty = parsimony_coefficient * len(self.symbol_tree) * self.metric.sign # Here is the saving coefficient multiplied by the length of the formula tree. If it is larger, the better. sign is 1, otherwise it is - 1
return self.raw_fitness_ - penalty # fitness is reduced. Here, the formula for punishing the excessive expansion of the tree is given```
## 3. Parallel module
### 3.1 parallel_evolve
Interface
`_parallel_evolve(n_programs, parents, X, y, sample_weight, seeds, params)`
Input parameter
Properties:
### 3.2 built in interface_ tournament
Objective: to find the best symbol tree
```contenders = random_state.randint(0, len(parents), tournament_size) # Generate tournaments_ The number of size (0, len(parents)) is equivalent to how many are randomly selected from the parent class
fitness = [parents[p].fitness_ for p in contenders] # Get the scores of these selected operator formula trees
if metric.greater_is_better: # Judge whether the larger the index is, the better or the smaller the index is
parent_index = contenders[np.argmax(fitness)] # np.argmax find the index of the largest value
else:
parent_index = contenders[np.argmin(fitness)] # The smaller the better
return parents[parent_index], parent_index # Returns the largest object and its index```
### 3.3 operation process:
Step 1: n_programs indicates how many trees there are in a group in the population. Here, we need to cycle n_programs times
initialization
```method = random_state.uniform() # method the probability selected from the crossover subtree hoist point mutation
parent, parent_index = _tournament() # Found a well behaved formula tree```
The second step is to select the type of mutation according to the probability of method
```method_probs definition
self._method_probs = np.array([self.p_crossover, self.p_subtree_mutation,
self.p_hoist_mutation, self.p_point_mutation])
self._method_probs = np.cumsum(self._method_probs)
if method < method_probs[0]: # If the method is less than the crossover probability
# Cross over method
donor, donor_index = _tournament() # The second best formula tree is used as a subtree
program, removed, remains = parent.crossover(donor.symbol_tree, random_state) # The two intersect
genome = {'method':'Crossover',
'parent_idx': parent_index,
'parent_nodes':removed,
'donor_idx':donor_index,
'donor_nodes':remains
}
elif method < method_probs[1]:# If the method is less than the crossover probability
# subtree mutation
program, removed, _ = parent.subtree_mutation(random_state)
genome = {'method':'Subtree Mutation',
'parent_idx':parent_index,
'parent_nodes':removed
}
elif method < method_probs[2]:
# hoist mutation
program, removed = parent.hoist_mutation(random_state)
genome = {'method':'Hoist Mutation',
'parent_idx': parent_index,
'parent_node': removed
}
elif method < method_probs[3]:
# point mutation
program, mutated = parent.point_mutation(random_state)
genome = {'mehtod':'Point Mutation',
'parent_idx':parent_index,
'parent_nodes':mutated
}
else:
# Self copy
program = parent.reproduce()
genome = {'mehtod': 'Reproduction',
'parent_idx':parent_index,
'parent_nodes':[]
}```
In the third step, the formula tree is generated according to the parameters and the program obtained in the second step
``` program = _SymbolTree(function_set=function_set,
arities=arities,
init_depth=init_depth,
init_method=init_method,
n_features=n_features,
metric=metric,
const_range=const_range,
p_point_replace=p_point_replace,
parsimony_coefficient=parsimony_coefficient,
feature_names=feature_names,
random_state=random_state,
symbol_tree = program)```
then
` program.parents = genome`
The genome here stores the information deleted in the process of generating a subtree, and assigns it to parents
Step 4: according to sample_ The weight information in weight gives weight to the feature column.
``` if sample_weight is None: # Calculate the weight of data in the bag
curr_sample_weight = np.ones((n_samples,)) # If there is no weight information, all sample weights are 1
else:
curr_sample_weight = sample_weight.copy()
oob_sample_weight = curr_sample_weight.copy() # Out of bag data```
Calculate the fitness of out of bag data
``` indices, not_indices = program.get_all_indices(n_samples, max_samples, random_state) # Get the selected data index outside the bag
curr_sample_weight[not_indices] = 0 # If the original data belongs to data outside the bag, the weight corresponding to its index is set to zero
oob_sample_weight[indices] = 0 # If the data outside the bag is not in the original data, the weight corresponding to the index is set to zero
program.raw_fitness_ = program.raw_fitness(X, y, curr_sample_weight) # Calculate the fitness of the data in the bag```
Calculate the fitness of out of bag data
``` if max_samples < n_samples:
# Calculate the applicability of out of bag data
program.oob_fitness_ = program.raw_fitness(X, y, oob_sample_weight) # Calculate the fitness of the data in the bag```
Finally, after N cycles, n mutated subtree programs are obtained, which has two private attributes raw_fitness_， oob_fitness_ The applicability of data inside and outside the bag is stored respectively
## 4. Symbolic transformer module
### 4.1.2 method
_ verbose_reporter: control log output
### 4.1.3 fit module
Interface
` fit(self, X, y, sample_weight = None)`
Input:
### Step 1: verify the data
Verification: check whether the lengths of X and y are consistent and hall_of_fame,function_set,_ Whether arities is normal and whether metric is Fitness type itself inherits from TransformerMixin abstract class
Then put the probability in the list and add it step by step
```self._method_probs = np.array([self.p_crossover, self.p_subtree_mutation, self.p_hoist_mutation, self.p_point_mutation])
self._method_probs = np.cumsum(self._method_probs)```
Then check_ method_probs,init_method,const_range,init_depth,feature_names for type check and scope check
### Step 2: take out the parameters and assign values to them
``` params = self.get_params()
print(f'params: {params}')
params['_metric'] = self._metric
params['function_set'] = self._function_set
params['arities'] = self._arities
params['method_probs'] = self._method_probs```
If it wasn't warm_start mode is ready_ programs and run_details_ Dictionaries
```if not self.warm_start or not hasattr(self, '_programs'):
# warm_ Restart when start is false to free the allocated memory
self._programs = [] # _ In programs, there is a list of winners of each generation [LIST1, List2... Listn]
self.run_details_ = { 'generation':[],
'average_length':[],
'average_fitness':[],
'best_length':[],
'best_fitness': [],
'best_oob_fitness':[],
'generation_time':[]
}
prior_generations = len(self._programs) # _ In programs, there is a list of winners of each generation [LIST1, List2... Listn], and its length represents the number of generations of iteration
n_more_generations = self.generations - prior_generations # How many more generations need to be iterated```
Then to n_ more_ Generate for verification
```population_size Represents the number of populations in each generation
if self.warm_start: # Discard previously used random seeds
for i in range(len(self._programs)):
_ = random_state.randint(MAX_INT, size = self.population_size)
if self.verbose:
self._verbose_reporter()# Output log```
### Step 3: run the program in parallel
Loop layer (from prior_generation to generation)
1) Record time found parent class
```start_time = time()
if gen == 0: # If it is the first generation, the parents are None
parents = None
else:
parents = self._programs[gen - 1] # _ The last generation in programs```
2) Parallel running program
```n_jobs, n_programs, starts = _partition_estimators(self.population_size, self.n_jobs) # Divide the population into n_job copies n_programs indicates how many data there are in the first few copies. starts records the starting point of each group of data
seeds = random_state.randint(MAX_INT, size = self.population_size) # Generate population_size is a random seed
population = Parallel(n_jobs=n_jobs, verbose=int(self.verbose > 1))(
delayed(_parallel_evolve)(n_programs[i],parents, X, y,sample_weight, seeds[starts[i]:starts[i + 1]],
params) for i in range(n_jobs))```
The data are merged to obtain the next generation of mutated population
```population[<symboltree._SymbolTree object at 0x00000118ABB89E20>, <symboltree._SymbolTree object at 0x00000118ABB89E80>, <symboltree._SymbolTree object at 0x00000118ABB89F10>, <symboltree._SymbolTree object at 0x00000118ABB89FD0>]
population = list(itertools.chain.from_iterable(population))```
The fitness and length of all individuals of the population are a list
```fitness = [program.raw_fitness_ for program in population]
length = [program.length_ for program in population]```
3) The penalty coefficient constrains fitness
```parsimony_coefficient = None
if self.parsimony_coefficient == 'auto':
parsimony_coefficient = (np.cov(length, fitness)[1, 0] / np.var(length)) # Take the second row and column of the covariance matrix and divide by the variance
for program in population:
program.fitness_ = program.fitness(parsimony_coefficient) # Fitness after contraction
self._programs.append(population) #The newly generated information of this generation is put into_ programs```
4) Delete eliminated individuals
```if not self.low_memory:
for old_gen in np.arange(gen, 0, -1): # Arrange the number of generations to gen in reverse order into a list, similar to [5 4 3 2 1]
indices = []
for program in self._programs[old_gen]: # Find each symbol tree of the population of the previous generation
if program is not None:# If it's not None
for idx in program.parents: # Find his parents_idx parents_idx contains its best performing parent class
if 'idx' in idx:# Find the parent_idx
indices.append(program.parents[idx])
indices = set(indices) # To repeat
for idx in range(self.population_size):# Every individual in the population
if idx not in indices: # If the individual is not in the optimal set, set him to None
self._programs[old_gen - 1][idx] = None
elif gen > 0:
self._programs[gen - 1] = None #Or set the previous generation to None```
### Step 4 information for operation
Corresponding code
```# Record operation information
if self._metric.greater_is_better: # If the bigger the better
best_program = population[np.argmax(fitness)]
else:
best_program = population[np.argmin(fitness)]
self.run_details_['generation'].append(gen)
self.run_details_['average_length'].append(np.mean(length))
self.run_details_['average_fitness'].append(np.mean(fitness))
self.run_details_['best_length'].append(best_program.length_)
self.run_details_['best_fitness'].append(best_program.raw_fitness_)
oob_fitness = np.nan
if self.max_samples < 1.0:
oob_fitness = best_program.oob_fitness_
self.run_details_['best_oob_fitness'].append(oob_fitness) # Accuracy of out of bag data
generation_time = time() - start_time
self.run_details_['generation_time'].append(generation_time) # Running time```
Handling early stopping
```if self._metric.greater_is_better:
best_finess = fitness[np.argmax(fitness)]
if best_finess >= self.stopping_criteria: # When the stop criteria are met
break
else:
best_finess = fitness[np.argmix(fitness)]
if best_finess <= self.stopping_criteria: # When the stop criteria are met
break```
Here the cycle ends and all generations are obtained.
### Step 5 if it is a transformation
a) First get hall_of_fame index
```fitness = np.array(fitness) # Find the fitness of this generation of population
if self._metric.greater_is_better: # The bigger the better, the better. Choose in reverse order
hall_of_fame = fitness.argsort()[::-1][:self.hall_of_fame] #Get hall_of_fame is an index of fitness
else:
hall_of_fame = fitness.argsort()[:self.hall_of_fame] # The smaller the better, the positive order selection```
Calculate all individuals in the last generation of the population (including those belonging to hall_of_fame) to obtain the predicted value
`evaluation = np.array([gp.execute(X) for gp in [self._programs[-1][i] for i in hall_of_fame]])`
If the index is spearman coefficient, calculate the ranking value of each group of data for evaluation
```if self.metric == 'spearman':
evaluation = np.apply_along_axis(rankdata, 1, evaluation)
from scipy.stats import rankdata
evaluation = np.array([[1,2,3,4],
[6,5,7,8],
[9,10,11,12]])
print(np.apply_along_axis(rankdata, 1, evaluation))
#output
[[1. 2. 3. 4.]
[2. 1. 3. 4.]
[1. 2. 3. 4.]]```
Then the correlation coefficient matrix is calculated
```with np.errstate(divide = 'ignore', invalid = 'ignore'): # Remove warnings such as invalid values except 0
correlations = np.abs(np.corrcoef(evaluation)) # Get the correlation coefficient matrix. If it is spearman coefficient, here is spearman correlation coefficient
[[1. 2. 3. 4.]
[2. 1. 3. 4.]
[1. 2. 3. 4.]]
[[1. 0.8 1. ]
[0.8 1. 0.8]
[1. 0.8 1. ]]
np.fill_diagonal(correlations, 0.) # Diagonal element fill 0
components = list(range(self.hall_of_fame)) # hall_of_frame 0 to hall_of_frame number
indices = list(range(self.hall_of_fame))
# X_shape(354, 13)
# evaluation:(50, 354)
# Iteratively delete the least relevant element
while len(components) > self.n_components: # If the minimum relevant individual is greater than the number to be left
# correlations_shape(50, 50)
most_correlated = np.unravel_index(np.argmax(correlations), correlations.shape) # The greater the index correlation that gets the maximum value, the worse
# Rank the correlation matrix by fitness to determine the most inappropriate index
worst = max(most_correlated) # worst is the column with the largest index
components.pop(worst)
indices.remove(worst) # Remove from serial number
correlations = correlations[:, indices][indices, :]
indices = list(range(len(components)))
self._best_programs = [self._programs[-1][i] for i in hall_of_fame[components]]```
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``` Number of 9 digits numbers divisible by nine using the digits from 0 to 9 if each digit is used atmost once is K*8!, then k has the value equal to..????
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6 years ago
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``` Hello student,Please find the answer to your question belowGiven dataNumber of 9 digits numbers divisible by nine using the digits from 0 to 9 if each digit is used atmost once is K*8!Number of ways=9!+8*8!=17*8!So the value of k=17
```
2 years ago
# Other Related Questions on Algebra
If alpha is a real root of the equation ax 2 +bx+c and beta is a real root of equation -ax 2 +bx+c. Show that there exists a root gama of the equation (a/2)x 2 +bx+c which lies between alpha...
Ajay 6 months ago
Small Mistake in last para posting again..............................................................................................................
Ajay 6 months ago
We have Similarly, So if P(x) = a/2 x 2 +bx +c, then and are off opposite sign and hence there must exist a root between the two numbers.
mycroft holmes 6 months ago
In the listed image can you tell me how beta*gamma = 2 ….. . . .. ??
The value of gamma is still not correct, either printing mistake or you gave me wrong value. The correct value of gamma is below
Ajay 5 months ago
Thankyou so much............................. …......................................................................!
Anshuman Mohanty 5 months ago
Yes sorry..... . . . .it is not so clear.. ok the values are beta = α + α^2 + α^4 and gamma = α^3 + α^5 + α^7
Anshuman Mohanty 5 months ago
if |z - i| Options: a*) 14 b) 2 c) 28 d) None of the above
If |z-i| = ?? PLs complete the question
Nishant Vora one month ago
Got it! [z + 12 – 6 i ] can be rewritten as [ z – i + 12 – 5 i] => | z – i | and => |12 – 5 i | = sqrt ( 12^2 + 5^2) = 13......................(2) => | z + 12 – 6 i | => | z + 12 – 6 i |...
Divya one month ago
I tried posting the question several times, it kept cutting off the rest of the question. Here: If | z-1| Options: a*) 14 b) 2 c) 28 d) None of the above
Divya one month ago
sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°......15 solve it Urgent
Ajay, the complete qution isSolution is sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°..... upto 15 terms. sin 78°=0 sin 42°+sin 54°+ sin 66°+ + sin 18° sin 6°+ where )=0.5 (your required answer),...
Kumar 3 months ago
Not any people get my answer why. You can no give answer my question I am join this site
Vivek kumar 5 months ago
Hello If you want to get the solution quick you should post your question in clear manner. Its not clear what you wnat us to solve, and what does 15 at the end of question means?
Ajay 5 months ago
Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2- (tan^2 theta + cot ^2 theta)^2
What needs to be solved here ? The question is incomplete....................................................................
Ajay 6 months ago
i don’t know how to do this...............................................................................................
Saravanan 2 months ago
this is the question :: Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2 - (tan^2 theta + cot ^2 theta)
Naveen Shankar 6 months ago
solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
Let the feet of the altitudes on BC, AC, AB, be D,E,F resp. Let the orthocenter be H. The following can be proved easily: 1. HDCE and HFBD are cyclic quadrilaterals. Then chord HE subtends...
mycroft holmes one month ago
Draw which is Isoceles as OB = OC. Now which means . Let D, be the foot of the perp from O on BC ( which is also the midpoint of BC). Then OD = OC sin (OBC) = R cos A. Hence the required...
mycroft holmes one month ago
a cos A = b cos B 2R sin A cos A = 2R sin B cos B sin 2A = sin 2B Either A = B (isoceles or equilateral) or 2A = 180 o – 2B so that A+B = 90 o .(Right-angled)
mycroft holmes one month ago
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# Why Is This So Hard??
I sit here, staring at a blank page, trying to come up with something to write about. I like math, it should be easy to talk about, right? I’m going to be a teacher; it should be easy to connect to real life, right? Why is this so hard? Then I realized, it’s hard because I’m forcing it. Let’s stop forcing it. Let’s stop forcing kids to do math, let’s engage them in math instead. Let’s reinvent math, let’s do it together, and let’s start now!
A Little Story….
So, 3 and a half years ago my husband and I fostered an abused, adorable puppy from a local pound. She is an amazing dog; she is good with C, she stays in the yard, she is a good listener. But nobody is perfect, right? Libby angry poops. Yup, you read that right. When we leave her for too long, she angry poops (or pees) in the house. Our brand new, built in 2013, house. It doesn’t matter if we make sure she goes before we leave her, she saves it as our punishment for leaving. Disgusted, I pulled out her old wire kennel from the corner of the garage and set it up in the living room. It needed to be cleaned after sitting in the garage for 3 years. I took the vacuum to it, but it wasn’t enough. I pulled out one of C’s wipes (if you didn’t already know, the uses for baby wipes are endless) and began cleaning each wire rectangle. It was a long process. There were a lot of wire rectangles. Would I ever finish? When? How many rectangles are there?
Maybe you already get where I’m going here. This is the perfect situation to talk to students about multiplication. Here I have a multiplication table all set up. Better than that, we can talk about perimeter and surface area, we can cover it all! If you really wanted to get into, I could take a video of me washing each individual rectangle. We could watch the painfully slow, boring process until students beg to know when it’s going to be over. That’s when you hit them with it, “I don’t know, why don’t you figure out how long it will take?” I got this idea from ___ who took a video of a bucket slowly being filled and made his students watch it until they begged to know when it would be over, genius! Check out his TED Talk below.
Let’s Estimate
Estimation can be difficult process. It’s takes a deep understanding of operations, but it’s also an incredibly useful skill. Over the weekend I visited Americano Central, a Mexican market in Minneapolis. If you haven’t been there, I highly recommend it, the food is amazing. On the opposite side of the street is a Mexican grocery store with a beautiful mosaic wall. Take a look!
This would be a great estimation challenge for students. You could use their already amazing multiplication skills from the dog kennel story to challenge them. How could you divide up the wall? About how many pieces fit in each of your created segments? Ask each student to come up with an answer (and a valid justification) and submit their answers. Take a class poll and see what they think. In the end, we don’t know EXACTLY how many mosaic pieces are in the wall, welcome to real life, kids!
Those are just 2, really quick examples of how every day things use some serious math. So, don’t force students to use math, help them to realize just how useful math is!
# Standard Curriculum for the Standard Child
WARNING: This post is far more serious that my previous posts, but hang in there!
Last week in my math course we were asked to look into the Common Core standards. Before starting my research I had the following assumptions about Common Core and state standards
1. Standards were put in place to ensure every child learns what they will need
2. Standards are created by the education community
3. Common Core standards, especially the math methods, are confusing
4. Teachers dislike standards because they limit the depth of the content while expanding the breadth
Assumption 1: Standards were put into place for the children
I still believe this is true, to an extent. The idealistic standards were put in place to ensure that our nation’s children were learning what they needed to grow into successful human beings. After all, we want our nation to continue to be successful, right? Idealistically, that’s true. But where what’s going on in my community…When the baby boomers were in school they were taken care of by their parents. Their parents were more than happy to pass referendums; they would pay more taxes if it meant that their children would have a better life than they did. They build new state of the art schools with swimming pools and sports fields. Now, as adults, those baby boomers who were supported by their parents don’t find it necessary to give their hard earned money to the school. Referendum after referendum fails, teachers are let go, technology is dated, swimming pools are shutdown. Baby boomers are skeptical of the educational system, yet they are the ones designing the standards. Are they capable of looking beyond their pocket books? That brings me to assumption 2
Assumption 2: Standards are created by the education community
I believed that standards were created by the people that worked with children, or at least by those who have researched the way children learn. After some searching, I learned that standards are heavily influenced by corporate American, or more specifically, by the fortune 500 companies. Big companies have big bucks, and a big need for effective employees. The success of a company (which can also be read as “the profit for the executives”) is dependent on the skills of the employees. When you look at it that way, it’s easy to see why big companies are invested (literally) in the educational system. These corporate investments are, in a big way, thanks to the federally produced document “A Nation at Risk.” Check out the links at the bottom if you want to know more. Long story short, I was wrong, standards are always created or pushed for by the education community.
Assumption 3: Common Core standards, especially the math methods, are confusing
There are many MANY memes out there about Common Core math standards. Common Core moves away from the teaching styles that many of us are used to. It seems confusing and complicated to us, but the standards were based on the best standards from the states across the nation and modeled after successful international programs. They stress deep understanding of key ideas and are organized based on research around how students learn. It might be different than the way we were taught, but take one look at our nation’s international math standing and you’ll see that something has got to give.
Assumption 4: Teachers dislike standards because they limit the depth of the content while expanding the breadth
I still believe this is true. Anytime standards are brought up a tension rises in the room. Standards are rigid; they take away some of the freedom teachers once had. But those of you who are just starting out, like I am, standards are what we will know. The standards are well organized, and while they might limit what we teach, they give us the freedom to focus on how to teach it, rather than what to teach.
Moral of the story, assumptions, while sometimes hold true, lots of times let us down. Take the time to analyze your assumptions, I can guarantee you will learn something new.
Some information for you, as promised
Fortune’s article on Fortune 500 companies involvement in schools
New York Times article on corporate involvement in the classroom
A link to A Nation At Risk excerpt
Teaching Styles
# Keeping It Fresh
Hey, teachers…let’s keep it fresh! Yep, I’m bringing it way back in this post! Ok, so it might not be a phrase you want to use in your 6th grade classroom, but don’t throw the baby out with the bath water. Math teachers need to keep things fresh in their classrooms. There is so much research out there about the importance of keeping our classrooms moving, literally. Like most theories, it sounds simple enough, but putting it into practice can be tough. Since day one of returning for my teaching license I’ve been told that we have to be very purposeful about our teaching. We teach the way we are taught. The way I was taught is a far cry from the way I want to teach. Let’s take a peek at 2 very different teaching styles.
A little Ferris, anyone?
Ferris style teaching
Traditional math teaching, or as I like to think of it, Ferris style, is where the teacher stands in the front of the classroom and spews information at students, who are expected to soak it up and cough it out on the next test. We can all get a giggle out of the Ferris clip because we have all been there, but you can bet that we weren’t laughing in the moment. Rote memorization of math facts might have their place in the classroom, but if students do not grasp the concept, you can be sure the information won’t stick. The video emphasizes the lack of student engagement, I couldn’t even focus for the 90 second clip.
Finland style teaching
Finland is known for their educational system. Teachers are well payed and respected, students are top performers, school days are shorter and class sizes are smaller. Add that with their keen ability to get students working in teams and sharing their learning and you hold the key to success, well, part of it anyway. Did you notice how the students were doing real work? They were up, moving through the school, collecting data and analyzing it. They prepared their results and presented it to their colleagues. The team based, cooperative learning style got students engaged in real life math activities; activities that took the math off their papers and into their communities.
Let’s take a lesson or 2 from Finland. Let’s get students moving. Let’s get students engaged. Let’s keep things fresh.
A giggle, if you have time…
# Why, “I’ll never use this in real life” Won’t Fly in My Classroom
I couldn’t have asked for a better weekend! My husband and I celebrated our 5th wedding anniversary on the beautiful shores of Tenmile Lake in Hackensack, Minnesota with our son, my mom, my brother, and my sister-in-law. I pictured myself relaxing on the beach with my feet in the sand, away from the stresses of math homework, but it seems that math in everyday life is inescapable. As a child, I would complain about the uselessness of math, “I’m NEVER going to use this in real life” and my mom would always reply with some scenario where I would, in real life, need to know how to figure out the math content I was studying. That was the downfall to living with an elementary school principal.
Math is inescapable; we take for granted the math that comes to us to easily. Take, for example, my chore of setting the patio table for dinner by the lake on Friday night. There were 5 adults that would need a place setting (C had already eaten and was happily playing in the sand). Without a thought, I arranged the plates in a near perfect pentagon, all equally spaced – I guess geometry does come in handy. My mom was overly generous with dinner and bought 6 steaks, thinking that Cooper would eat at least some. Well, being a 2-year-old, he wanted nothing to do with his portion – that just meant more for the rest of us. We each finished our serving of steak and decided to split up C’s. 5 adults, 5 pieces, should be easy. Here comes the challenge, my mom wanted a small piece, the rest of us were willing to share equally, and the steak is not a perfect rectangle. Once I had an idea of what 1/5th of the steak would look like, I cut off one piece that was just a bit smaller and served it to my mom. From there I had to divide the rest evenly into 4 pieces, so I divided the steak in half, and then each half in half. The steak was perfectly done (medium rare) and thanks to the math skills I swore I would never use, it was also perfectly divided.
The Nitty Gritty
My weekends plans were a fun example of math used in an everyday activity. The United States is lagging in our math and science skills compared to many other countries. This drop in ability may come as a surprise to people not in our profession, but those of us in the classroom know the test reports all too well. Falling test scores leads to public panic leads to more pressure on administration, teachers, and eventually students to improve test scores. Teaching styles are pushed away from teaching in a manner that is proven to reach students and back towards the traditional “I do…you do” approach. Students need to be engaged in their math content, they need to fully understand the real-world application, and they need to figure it out (mostly) on their own. Elementary math IS real world math. It’s up to us as teachers to make that connection clear, to bring interest and engagement back to the content, and to allow our students the time and space to explore math content in a way that is meaningful to them. Bring real world to the classroom; challenge them with real world applications and sit back and watch the productive struggle ensue.
Productive struggle is one of my favorite math terms, and something that I believe is lacking in our mathematics classroom. If you’re interested in learning more about productive struggle in the classroom check out this quick IGNITE video by Robert Kaplinksy by click on his picture below.
Uncategorized
# Hello!
Hey everyone!
I’m here to share my thoughts and research on teaching and learning math. I hope that we can take a fresh look on how students learn so that we can reinvent how we teach math. Let’s throw out rote memorization and focus on real, engaging, meaningful methods!
If you want to know more about me, check out my About Me section! | 2,964 | 13,705 | {"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} | 3.515625 | 4 | CC-MAIN-2018-09 | longest | en | 0.966296 |
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### Beauty bare -- Part 1
If there is a sacred scripture of science it is surely Euclid's Elements. The book (actually 13 "books") has been continuously "in print" since it was written in about 300 B.C., in Alexandria, Egypt. For over two thousand years it was the standard text from which students of mathematics and the exact sciences learned the ropes. Even today it is the basis for every high-school geometry text.
I can't say that I have read all 13 books. At one time I did work my way step-by-step through Book One, using Thomas Heath's profusely annotated edition. It was an exhilarating experience, and a masterful illustration of what rigorous thinking is all about.
Euclid begins with definitions of a point, a line, and so on.
Then he offers 5 Common Notions and 5 Postulates, all of which he assumes the reader will accept as self-evidently true.
For example:
Common Notion 1: Things equal to the same thing are equal to each other.
Postulate 1: One may draw a straight line from any point to any other point.
Postulate 3: One may describe a circle with any center and any radius.
Basic stuff, really. Who's to quibble?
Now the fun begins, as Euclid deduces Propositions from his first principles.
For example, his first Proposition is to construct an equilateral triangle on a given straight line, for which he evokes Common Notion 1 and Postulates 1 and 3.
And on he goes, building a magnificent compilation of Propositions on what has gone before. Proposition 9: To bisect a given triangle. Proposition 37: Triangles on the same base and in the same parallels have equal areas. And so on.
As I worked my way through the book, I kept track of the logic. Here is a diagram I drew at the time showing the pathway to Proposition 47, the Pythagorean Theorem (the square on the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides). Click to enlarge.
The Pythagorean Theorem is by no means self-evidently true, but its truth is hidden in the self-evidently true Common Notions and Postulates, to be revealed by logical thinking.
If the Elements is the sacred scriptures of science, it is not because of its particular contents of propositions, but as a model for a way of thinking -- a way of thinking that has guided us into the universe of the DNA and the galaxies, and out of the gabble and hiss of parochial prejudice. In effect, Euclid is saying: "Here is a way, a truth, and a light."
Tomorrow: Another avenue to the Pythagorean Theorem and some thoughts on truth and beauty.
My daughter Mo and daughter-in-law Patty were recently in Alexandria, walking the streets walked by Euclid, and were honored with a personal tour through the new Alexandrian Library, once the greatest library in the world. You can see some of her photos here. | 649 | 2,836 | {"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} | 3.5 | 4 | CC-MAIN-2016-30 | longest | en | 0.955271 |
http://www.physicsforums.com/showthread.php?t=220586 | 1,371,714,219,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368710963930/warc/CC-MAIN-20130516132923-00075-ip-10-60-113-184.ec2.internal.warc.gz | 621,472,374 | 7,571 | ## Projectile Motion Sail Boat
1. The problem statement, all variables and given/known data
A sailboat is traveling east at 5m/s. A sudden gust of wind gives the boat an acceleration 0.8m/s^2 (40 degrees north of east).
1. What is the boat's speed 6 seconds later when the gust subsides?
2. What is the boat's direction 6 seconds later when the gust subsides?
2. Relevant equations
v = v0 + at
A^2 + B^2 = C^2
3. The attempt at a solution
I'm so stuck, please tell me what I did wrong.
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Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Hi Kster! Your v_0 shouldn't be 0 (it would only be 0 if the wind was North or South). Try again!
Quote by tiny-tim Your v_0 shouldn't be 0 (it would only be 0 if the wind was North or South).
Thank you for replying, ok so I fixed it:
The Opposite Side:
Vo + at = V
(5m/s?) + (0.8 m/s^2 * 6 sec) = 9.8 m/s
My question is, is 9.8m/s my answer to the boat's speed 6 seconds later? or Do I have to use the Pythagorean Theorem to figure out the Hypoteneuse side and that is my answer?
Blog Entries: 27
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Homework Help
## Projectile Motion Sail Boat
Quote by Kster (5m/s?) + (0.8 m/s^2 * 6 sec) = 9.8 m/s
No.
You must do everything in the same direction.
The acceleration is at 40º, so you must use the component of initial velocity along that direction too (using cos).
You then use Pythagoras to combine that result with the perpendicular component of initial velocity (which will be unaffected).
Similar discussions for: Projectile Motion Sail Boat Thread Forum Replies Introductory Physics Homework 3 Introductory Physics Homework 1 Introductory Physics Homework 4 General Physics 9 Introductory Physics Homework 1 | 514 | 1,926 | {"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} | 3.6875 | 4 | CC-MAIN-2013-20 | latest | en | 0.885288 |
https://www.ultimate-guitar.com/forum/showthread.php?t=999032 | 1,511,123,228,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805761.46/warc/CC-MAIN-20171119191646-20171119211646-00738.warc.gz | 910,557,747 | 15,362 | sorry for all these questions but ive recently started IB maths higher without doing Add maths for GCSE so im in need of a bit of help.
ok ive got this question:
-2+bi is a solution to z^2+az+(3+a)=0 find a and b given that they are real.
so i already know that -2-bi is another solution.... NOW WHAT!?!
"Like a midget at a urinal, I was going to have to stay on my toes"
"Like a blind man at an orgy, I was going to have to feel my way through"
substitute z = -2 +bi into the equation z^2 +az + (3+a) = 0. Then substitute z = -2 - bi. Then you have two equations, and two unknowns (a, and b). Expand and solve.
you make something like this (-2+bi)(-2-bi) and you put them together through multiplication and distribution
Quote by horny_cactus
Who's Rick Roll? Sorry for my ignorance I just joined this forum so I don't yet Know that member.
i honestly dont know thats just what i would do
sorry for double post
Quote by horny_cactus
Who's Rick Roll? Sorry for my ignorance I just joined this forum so I don't yet Know that member.
i is the square root of negative 1
and i think the first guy got it...im doing it atm.
"Like a midget at a urinal, I was going to have to stay on my toes"
"Like a blind man at an orgy, I was going to have to feel my way through"
Meh math noob time...
Is it a quadratic, having a a and c... You could use the discriminant and then factorise...Or just bash it with the full quadratic formula...Idk.
Ok. So you got the other solution. Now you see that real part of both solutions is -2. Let's say that formula is (-y +- sqrt(y^2 - 4xz)/2x
x = 1
y = a
z = 3+a
Well, real part of both solutions is -2. it is equal to -y / 2x. So y = 4 x.
x = 1 therefore y = 4.
y = a so a = 4 too.
z = 3+a = 3+4=7
edit: I fotgot b XD
bi = sqrt (y^- 4 xz) / 2x
bi = sqrt (-3)
b = sqrt (3)
Just ask if you don't understand my way of solving...
Last edited by -Rodion- at Nov 13, 2008,
yeah, I get a = 4, and b = sqrt(3)
I reeeaaally should start paying attention on math lessons :S
Quote by metacarpi
I get called a slut all the time, and I'm a dude.
meh imaginary numbers are so pointless
Deacon of Zeppelinism PM TheHeartbreaker to join
speed demon of the UG Jeepers
Member of the Neutral Milk Hotel club PM Hamish5178 to join~
bi - 2 = lonely man
how did u get b= sqrt3
i got (sqrt5)/2
"Like a midget at a urinal, I was going to have to stay on my toes"
"Like a blind man at an orgy, I was going to have to feel my way through"
+- bi = +- sqrt (y^2 - 4xz) / 2x
y^2 - 4xz = 16 - 4*7 = -12
+- bi = sqrt (-12) / 2 = sqrt ( -12/4) = sqrt (-3)
So when does this kind of stuff fit into your daily life ? | 804 | 2,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-47 | longest | en | 0.952673 |
http://planetmath.org/IntegratingtanXOver0fracpi2 | 1,521,762,272,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648103.60/warc/CC-MAIN-20180322225408-20180323005408-00044.warc.gz | 232,662,533 | 3,148 | # integrating $\mathop{tan}\nolimits x$ over $[0,\frac{\pi}{2}]$
Note that what is meant by $\displaystyle\int\limits_{0}^{\frac{\pi}{2}}\tan x\,dx$ is actually $\displaystyle\lim_{t\to\frac{\pi}{2}^{-}}\int\limits_{0}^{t}\tan x\,dx$, since $\tan x$ is defined on $[0,\frac{\pi}{2})$ but not at $\frac{\pi}{2}$.
$\begin{array}[]{ll}\displaystyle\int\limits_{0}^{\frac{\pi}{2}}\tan x\,dx&% \displaystyle=\lim_{t\to\frac{\pi}{2}^{-}}\int\limits_{0}^{t}\tan x\,dx\\ &\\ &\displaystyle=\lim_{t\to\frac{\pi}{2}^{-}}\ln|\sec x|\bigg{|}_{0}^{t}\\ &\\ &\displaystyle=\lim_{t\to\frac{\pi}{2}^{-}}\ln|\sec t|-\ln|\sec 0|\\ &\\ &\displaystyle=\lim_{t\to\frac{\pi}{2}^{-}}\ln|\sec t|\\ &\\ &\displaystyle=\infty.\end{array}$
Title integrating $\mathop{tan}\nolimits x$ over $[0,\frac{\pi}{2}]$ IntegratingtanXOver0fracpi2 2013-03-22 15:57:47 2013-03-22 15:57:47 Wkbj79 (1863) Wkbj79 (1863) 14 Wkbj79 (1863) Example msc 26A42 msc 45-01 ImproperLimits OneSidedLimit | 410 | 954 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 10, "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-2018-13 | latest | en | 0.328232 |
http://www.computerforums.org/forums/social-lounge-off-topic/quick-calc-question-173217.html | 1,527,496,399,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794872114.89/warc/CC-MAIN-20180528072218-20180528092218-00533.warc.gz | 358,038,053 | 15,489 | Computer Forums Quick Calc Question
01-28-2008, 09:25 PM #1 Daemon Poster Join Date: Nov 2004 Posts: 1,246 Quick Calc Question Derivative of (x)*((4+x)^1/2) I keep getting (1/2x+[(4+x)^1/2])/[(4+x)^1/2], but thats wrong. __________________ __________________ Karma/rep is always appreciated 01010010011001010110010001100010011001010110000101 11001001100100 010011110111010101110100 There are only 10 kinds of people in this world. Those who can read binary, and those who can't.
01-29-2008, 01:36 AM #2 Fully Optimized Join Date: Oct 2007 Location: USA Posts: 1,540 Re: Quick Calc Question I dont know man. My math skills are as good as an 8th graders at best. __________________ __________________
01-29-2008, 03:07 PM #3 Fully Optimized Join Date: Dec 2006 Posts: 2,135 Re: Quick Calc Question umm ^ means exponent? if so then thats just x* square root of 4+x __________________ amd athlon x2 4800+ 2.5ghz / 4gb ddr2 800mhz mushkin/ ZOTAC 8600gt 256mb / 320gb HDD 7200RPM / 250 watt psu
01-29-2008, 03:10 PM #4 Daemon Poster Join Date: Oct 2007 Location: United States Posts: 625 Re: Quick Calc Question all I take is QBA I and I think that's hard! __________________ Intel Quad 6600, Asus P5N-E SLI, GeForce GTX 560 1GB, 6GB RAM, 500GB HDD, Zalman 9700NT, X-Fi Xtreme Gamer, 650 watt PSU
01-29-2008, 03:16 PM #5 In Runtime Join Date: Dec 2007 Posts: 391 Re: Quick Calc Question (4+x)^1/2 + x(1/2(4+x)^(-1/2)) say you have an equation u*v for the derivative you say div of the first times the second plus div of second times first u'v + v'u where ' means first derivative __________________ My setup: AMD Athlon 64 X2 Dual-Core 4000+, 2Gb A-data ddr800, MSI K9A2 CF 790X, MSI 2600XT T2D512EZ, Samsung DVD-RW, Ultra 500W PS Vseries, Maxtor 80Gb, Aspire Mid-tower case, Windows xp Home, Westinghouse 19" LCD | 606 | 1,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-22 | latest | en | 0.717939 |
http://mathhelpforum.com/calculus/2173-calculus-help.html | 1,481,064,603,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542002.53/warc/CC-MAIN-20161202170902-00151-ip-10-31-129-80.ec2.internal.warc.gz | 169,824,134 | 11,567 | 1. ## Calculus HELP!!!!
2. Originally Posted by FrozenFlames
Is there time R'(t)=0
Yes, notice that $R(t)$ is a differenciable function, and $R(0)=R(24)$ thus, by Rolle's Theorem (or Mean-Value theorem) there exists a real number, $0 such as,
$R'(t)=0$
This is my 6th Post!!
3. Originally Posted by FrozenFlames
Find the average rate.
First find the integral,
$\int^{24}_0\frac{1}{79}(768+23t-t^2)dt$
which is $\approx 258.8354$
Now divide it by the length which is 24,
thus the answer is $\approx 10.78$
4. Originally Posted by frozenflames
Mid-point Riemann Sum, avoiding too-technical stuff, is about computing the area between a curve and the horizontal axis by subdividing the interval [a,b] into n number of equal sub-divisions, and then by multiplying the midpoint height or depth of a subdivision by the length or width of the subdivision, thereby assuming the area of a subdivision is a rectangle, and then summing all the individual areas of the n number of subdivisions.
Zeez, if you understand that, then it is not my fault.
I am just avoiding using the summation symbol and all those a(i), x(i), a(i-1), etc. With LaTex there'd be no problem showing those, but I'd shun LaTex as long as I can express myself good enough with the old reliable ASCI. I am a simple guy. :-)
So, the interval is [0,24] in hours.
The n is 4.
Hence, length or width of a subdivision is (24-0)/4 = 6 hrs.
It so happens that in the given data, the midpoint heights of the 4 subdivisions are already shown:
Subdiv. [0,6]....midpoint height = reading at t=3, = 10.4
Subdiv. [6,12]....midpoint height = reading at t=9, = 11.2
Subdiv. [12,18]....midpoint height = reading at t=15 = 11.3
Subdiv. [18,24]....midpoint height = reading at t=21, = 10.2
Hence,
INT.(0-->24)[R(t)]dt
= 6*10.4 +6*11.2 +6*11.3 +6*10.2
= 6(10.4 +11.2 +11.3 +10.2)
= 6(43.1)
= 258.6 gallons per 24hrs, or per day, or gpd.
Or,
= 260 gpd. --------answer. Approximate, remember.
Explain that answer in terms of waterflow?
Easy. The waterflow of the pipe is about 260 gpd. Or, the pipe dislodges water at the rate of about 260 gallon per day.
------------------------
Is there a time between 0 and 24 hours where R'(t)=0?
Where the 1st derivative of R(t) is zero?
Yes. Per the given data, the slope of the rate of flow curve is increasing t=from t=0 to t=12, and then decreasing from t=15 to t=24.
That means somewhere between t=12 and t=15, the slope of the rate of flow curve became horizontal---where R'(t) is zero.
------------------------
Q(t) = (1/79)(768 +23t -t^2) will approximate the average rate of waterflow.
Solve for R(t) by using Q(t).
This portion is not clear to me.
Are R(t) and Q(t) not one and the same?
If not, then R refers to rate and Q refers to quantity or volume?
Then Q = R*A
where A refers to the effective cross-sectional area of the pipe.
We do not know the size of the pipe, so let me pass on this portion of your question. | 877 | 2,920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-50 | longest | en | 0.872948 |
http://theconstructor.org/structural-engg/tuned-mass-dampers/1198/ | 1,487,858,209,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171171.24/warc/CC-MAIN-20170219104611-00196-ip-10-171-10-108.ec2.internal.warc.gz | 247,183,706 | 17,288 | ## Register Now
A tuned mass damper is a device mounted in structures to prevent discomfort, damage, or outright structural failure caused by vibration. They are used in high rise buildings to prevent failure of buildings during earthquakes. They are also known as an active mass damper (AMD) or harmonic absorber,
## How Tuned Mass Dampers Work?
A tuned mass damper (TMD) consists of a mass (m), a spring (k), and a damping device (c), which dissipates the energy created by the motion of the mass (usually in a form of heat). In this figure, M is the structure to which the damper would be attached.
From the laws of physics, we know that F = ma and a = F/m. This means that when an external force is applied to a system, such as wind pushing on a skyscraper, there has to be acceleration. Consequently, the people in the skyscraper would feel this acceleration. In order to make the occupants of the building feel more comfortable, tuned mass dampers are placed in structures where the horizontal deflections from the wind’s force are felt the greatest, effectively making the building stand relatively still.
When the building begins to oscillate or sway, it sets the TMD into motion by means of the spring and, when the building is forced right, the TMD simultaneously forces it to the left.
Ideally, the frequencies and amplitudes of the TMD and the structure should nearly match so that EVERY time the wind pushes the building, the TMD creates an equal and opposite push on the building, keeping its horizontal displacement at or near zero. If their frequencies were significantly different, the TMD would create pushes that were out of sync with the pushes from the wind, and the building’s motion would still be uncomfortable for the occupants. If their amplitudes were significantly different, the TMD would, for example, create pushes that were in sync with the pushes from the wind but not quite the same size and the building would still experience too much motion.
The effectiveness of a TMD is dependent on the mass ratio (of the TMD to the structure itself), the ratio of the frequency of the TMD to the frequency of the structure (which is ideally equal to one), and the damping ratio of the TMD (how well the damping device dissipates energy).
Wide span structures (bridges, spectator stands, large stairs, stadium roofs) as well as slender tall structures (chimneys, high rises) tend to be easily excited to high vibration amplitudes in one of their basic mode shapes, for example by wind or marching and jumping people. Low natural frequencies are typical for this type of structures, due to their dimensions, as is their low damping. With GERB Tuned Mass Dampers (TMD), these vibrations can be reduced very effectively.
### The Tuned Mass Damper may consist of:
• Spring
• Oscillating Mass
• Viscodamper
Fig. Tuned Mass Damper
as main components, or may be designed as a pendulum, also in combination with a Viscodamper.
Each TMD is tuned exactly to the structure and a certain natural frequency of it. Such TMD have been designed and built with an oscillating mass of 40 to 10.000 kg (90 to 22.000 lbs) and natural frequencies from 0.3 to 30 Hz. Vertical TMD are typically a combination of coil springs and Viscodampers, while in case of horizontal and torsional excitation in the corresponding horizontal TMD the coil springs are replaced by leaf springs or pendulum suspensions.
Fig. Amplitude – Frequency response of a low damped system without (blue) and with (yellow) tuned mass damper
## Applications of Tuned Mass Dampers:
Tuned mass dampers are mainly used in the following applications:
• Tall and slender free-standing structures (bridges, pylons of bridges, chimneys, TV towers) which tend to be excited dangerously in one of their mode shapes by wind,
• Stairs, spectator stands, pedestrian bridges excited by marching or jumping people. These vibrations are usually not dangerous for the structure itself, but may become very unpleasant for the people,
• Steel structures like factory floors excited in one of their natural frequencies by machines , such as screens, centrifuges, fans etc.,
• Ships exited in one of their natural frequencies by the main engines or even by ship motion.
Tuned Mass Dampers may be already part of the structure’s original design or may be designed and installed later.
### Example of Tuned Mass Damper Application:
Taipei structure has TMD of weight 730 tonnes which is show in below figure: | 973 | 4,476 | {"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-2017-09 | latest | en | 0.972678 |
https://www.dualnoise.com/2012/07/alternative-optimal-solutions-to-coin.html | 1,623,866,242,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00625.warc.gz | 679,223,320 | 20,590 | ## Tuesday, July 17, 2012
### Alternative Optimal Solutions to the Coin Mix Problem
The solution to the 'optimal change' problem analyzed in the previous post is further examined from a practice perspective. Again, we use CPLEX to do this since it is quite convenient for such analyses.
Problem 1 is an integer knapsack problem that is known to be theoretically NP-Hard but fairly easy to solve in practice using Dynamic Programming. For this specific coin instance, enumeration with simple pruning rules that exploit the problem structure would work too.
Globally Robust versus Scenario-Optimal
Problem 2 is relatively trickier. It attempts to minimize the maximum weight across the feasible solutions to each of the change scenarios. The value of the 10-coin min-weight solution in the previous post:
Pennies:Quarters:Dimes:Nickels::4:3:2:1 is slightly more than a dollar (1.04\$). This coin mix will be able to satisfy any random change request in the range 1-100c. Note: If we only cared about change between 1-99c, an optimal solution (35.412g, 99c) is: PQDN::4241.
However, while an optimal solution to Problem 2 is robust in terms of this ability to meet any request, it is not necessarily optimal in terms of meeting specific coin requests. For example, if we only ever use change for a 3pm \$1.68 cup of Starbucks in the cafeteria, then the 4-3-2-1 solution may not be the best. However, if by chance, we choose to downsize to a smaller \$1.29 cup one day, then the 68c solution may not work.
First Among Equals
We only used a single goal within the optimization formulations: either min-sum, min-count, or min-weight. As mentioned in a recent post, practical optimization models almost never go into production carrying just a single goal. Consider these alternative optimal optimal solutions for 96c:
v # wt p n d q
96 9 24.046 1 0 7 1
96 6 24.046 1 0 2 3
The second solution has the same weight but fewer quarters + dimes, and does the same job a little more efficiently. On the other hand, the first solution perhaps allows greater flexibility in terms of meeting other change requests (e.g. 30c, 40c).
Continuing along this path,
a) if we set the lower bound on the number of dimes in Problem 2 = 7, we obtain the following 13-coin alternative to the 4-3-2-1 answer, having the same weight and value (36.546g, \$1.04) PQDN::4171
b) If we want a least-value solution to Problem 2, then an optimal objective function value whose corresponding weight is only 1.5g more than the min-weight is 1.0\$ (37.912g)
PQDN::5241 or PQDN::5091
(If you ever change a dollar bill in the future, do so in a robust manner via this mnemonic: PQDN 5241 :)
Interestingly, these 1\$ solutions include five pennies (a feasible 4-penny solution has a value of 1.04\$ or more).
Depending on the context, one of these solutions is more preferable than the others, and this preference can be quantified by suitably incorporating secondary and tertiary goals within the objective function. In contrast, a pure constraint satisfaction model will solely focus on generating feasible solutions ignoring any preference. These principles are the cornerstone of many mission-critical optimal planning systems and operational decision support applications deployed across industries. | 791 | 3,299 | {"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} | 3.203125 | 3 | CC-MAIN-2021-25 | longest | en | 0.880397 |
http://hypertextbook.com/facts/2003/PerkhaAhmed.shtml | 1,511,271,546,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806388.64/warc/CC-MAIN-20171121132158-20171121152158-00083.warc.gz | 131,640,088 | 5,965 | The Physics
Factbook
An encyclopedia of scientific essays
# Height of the Great Pyramid
Bibliographic Entry Result
(w/surrounding text)
Standardized
Result
Johnson, Charles William. The Great Pyramid: Measurements, Earth Matrix, 2003. "Its height once rose to 146.59 m or 487 ft." 146.59 m
Pyramid, Chamber's Encyclopedia Volume 6, 1884 (gizapyramid.com rare book archives). "Its height was 480 feet and 9 inches, and its base 764 feet square; in other words, it was higher that St. Paul's Cathedral, on an area the size of Lincoln's Inn Field." 145.45 m
Pyramids (Egypt), Microsoft Encarta Online Encyclopedia, 2003. "When newly completed, the Great Pyramid rose 146.7 m (481.4 ft)--nearly 50 stories high."" 146.7 m
Measurements of the Great Pyramid, Rostau. "Height 280 (146.64), now approximately 262 (137.2)" 146.64 m
(when built)
137.2 m
(today)
Ashmawy, Alaa K. The Seven Wonders: The Great Pyramid of Giza, University of South Florida, 2000. "When it was built, the Great pyramid was 145.75 m (481 ft) high. Over the years, it lost 10 m (30 ft) off its top." 147.75 m
Arab proverb: "Man fears time, yet time fears the pyramids"
The ancient Greeks didn't title the Great Pyramid of Giza as the first wonder of the world without just cause. It was the tallest monument ever built for 4,000 years and today the pyramid is the only wonder still remaining of the ancient seven wonders. But what has exactly left people so enthralled with this ancient work for centuries? The Great Pyramid, standing with a base of 55,000 m and each side greater that 20,000 m in area, can be seen from orbit. Scientist and historians are amazed at the accuracy at which the ancient Egyptians designed the pyramid. The pyramid is a precisely aligned structure and looks north with only 3/60th a degree of error. Among other aspects, the design of the pyramid is alleged by some to have a geometric relationship with celestial bodies such as the orbit of the earth around the sun. This is widely disputed and the question remains open whether or not these relationships were done intentionally or accidentally.
When the Great Pyramid of Giza was first built, its height scaled to a fascinating 146 M. However in spite of the pyramid's ability to defeat the tests of time, it has withered slightly due to the harsh weather of the desert, civil war, and just old age in general. Today the pyramid still stand tall but nearly 10 meters shorter then its original height, at 137 M. Despite the allure pertaining to the pyramid's structure and design, the Great Pyramid of Giza can simply be appreciated for the effort with which the ancient Egyptians built this beautiful artwork.
Perkha Ahmed -- 2003 | 664 | 2,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-47 | latest | en | 0.873999 |
http://html.rhhz.net/jckxjsgw/html/70194.htm | 1,721,125,162,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00864.warc.gz | 16,960,170 | 17,110 | 大地线航法在智能船舶上的应用
舰船科学技术 2023, Vol. 45 Issue (1): 180-185 DOI: 10.3404/j.issn.1672-7649.2023.01.033 PDF
Application of geodesic sailing on intelligent ship
WU Zu-xin, ZHENG Zhong-yi
Navigation College, Dalian Maritime University, Dalian 116026, China
Abstract: Sailing along the shortest path can help build an efficient and clean shipping. Geodesic is the shortest path between two points on the surface of space. Due to geodesic's complicated calculation, varying course, and only shorten 5% voyage by average compared with rhumb line, the geodesic sailing (GS) is almost impossible in traditional nautical practice. The intelligent ship can maximize the advantage of the shortest geodetic path by using its calculation, data acquisition, course control, etc, In this paper, a fast and high-precision algorithm is presented to solve GS (direct/inverse) problems firstly. Then, apply the GS to the unmanned ship by calculating the course dynamically based on the current position and destination, and the new order is issued to the autopilot. Finally, alter course restriction mechanism (ACRM) with different strategies are proposed to reduce the autopilot’s work load by avoiding frequent course alternation. The simulation results show that ACRM can help to obtain a balance between the shortest path and the minimum number of alter course.
Key words: geodesic sailing (GS) rhumb line sailing (RLS) intelligent ship alter course restriction mechanism (ACRM)
0 引 言
1 大地线航法
图 1 大地线航法及其辅助球体示意图 Fig. 1 Geodesic sailing diagram and its auxiliary sphere
$\frac{s_{0i}}{b} = \int_0^{{\sigma _{0i}}} {\sqrt {1 + {k^2}{{\sin }^2}\sigma}{\rm{ d}}\sigma } ，$ (1)
${\lambda _{0i}} = {\omega _{oi}} - f\sin {\alpha _0}\int_0^{{\sigma _{0i}}} {\frac{{2 - f}}{{1 + (1 - f)\sqrt {1 + {k^2}{{\sin }^2}\sigma } }}}{\rm{ d}}\sigma 。$ (2)
1.1 正解 φ1, λ1, α1, s12 ${\boldsymbol{\Rightarrow}}$ φ2, λ2, α2
${\beta _1} = \arctan ((1 - f)\tan {\varphi _1}) ，$ (3)
${\alpha _0} = \arctan (\frac{{\sin {\alpha _1}\cos {\beta _1}}}{{\sqrt {{{\cos }^2}{\alpha _1} + {{\sin }^2}{\alpha _1}{{\sin }^2}{\beta _1}} }}) ，$ (4)
${\sigma _{01}} = \arctan (\tan {\beta _1}/\cos {\alpha _1})，$ (5)
${\omega _{01}} = \arctan (\sin {\alpha _0}\tan {\sigma _{01}})，$ (6)
${s_{01}} = b{A_1}({\sigma _{01}} + \sum\limits_{l = 1}^\infty {{C_{1l}}} \sin 2l{\sigma _{01}}) ，$ (7)
${s_{02}} = {s_{01}} + {s_{12}}，$ (8)
${\tau _2} = {s_{02}}/(b{A_1})，$ (9)
${\sigma _{02}} = {\tau _2} + \sum\limits_{l = 1}^\infty {C_{1l}'} \sin 2l{\tau _2} 。$ (10)
$\varepsilon = \dfrac{{\sqrt {1 + {k^2}} - 1}}{{\sqrt {1 + {k^2}} + 1}}，\quad {A_1} = \dfrac{{\left(1 + \dfrac{1}{4}{\varepsilon ^2} + \dfrac{1}{{64}}{\varepsilon ^4} + \dfrac{1}{{256}}{\varepsilon ^6} + \cdots \right)}}{{\left( {1 - \varepsilon } \right)}}，$
$\begin{split}& {C_{11}} = - \frac{1}{2}\varepsilon + \frac{3}{{16}}{\varepsilon ^3} - \frac{1}{{32}}{\varepsilon ^5} + \cdots ，\\ & {C_{12}} = - \frac{1}{{16}}{\varepsilon ^2} + \frac{1}{{32}}{\varepsilon ^4} -\frac{9}{{2048}}{\varepsilon ^6} + \cdots ，\end{split}$
${C_{13}} = - \frac{1}{{48}}{\varepsilon ^3} + \frac{3}{{256}}{\varepsilon ^5} - \cdots ，\quad {C_{14}} = - \frac{5}{{512}}{\varepsilon ^4} + \frac{3}{{512}}{\varepsilon ^6} - \cdots，$
${C_{15}} = - \frac{7}{{1280}}{\varepsilon ^5} + \cdots ，\quad \quad {C_{16}} = - \frac{7}{{2048}}{\varepsilon ^6} + \cdots，$
$\begin{split}& C_{11}' = \frac{1}{2}\varepsilon - \frac{9}{{32}} {\varepsilon ^3} + \frac{{205}}{{1536}}{\varepsilon ^5} + \cdots，\\ & C_{12}' = \frac{5}{{16}}{\varepsilon ^2} - \frac{{37}}{{96}}{\varepsilon ^4} +\frac{{1335}}{{4096}}{\varepsilon ^6} + \cdots，\end{split}$
$C_{13}' = \frac{{29}}{{96}}{\varepsilon ^3} - \frac{{75}}{{128}}{\varepsilon ^5} + \cdots ，\quad C_{14}' = \frac{{539}}{{1536}}{\varepsilon ^4} - \frac{{2391}}{{2560}}{\varepsilon ^6} + \cdots，$
$C_{15}' = \frac{{3760}}{{7680}}{\varepsilon ^5} + \cdots ，\quad \quad C_{16}' = \frac{{38081}}{{61440}}{\varepsilon ^6} + \cdots。$
${\beta _2} = \arctan \left(\frac{{\cos {\alpha _0}\sin {\sigma _{02}}}}{{\sqrt {{{\cos }^2}{\sigma _{02}} + {{\sin }^2}{\alpha _0}{{\sin }^2}{\sigma _{02}}} }}\right)，$ (11)
${\alpha _2} = \arctan (\tan {\alpha _0}/\cos {\sigma _2})，$ (12)
${\omega _{02}} = \arctan (\sin {\alpha _0}\tan {\sigma _{02}})，$ (13)
${\lambda }_{01}={\omega }_{01}-f\mathrm{sin}{\alpha }_{0}{A}_{3}\left({\sigma }_{01}+{\displaystyle \sum _{l=1}^{\infty }{C}_{3l}\mathrm{sin}2l{\sigma }_{01}}\right)，$ (14)
${\lambda }_{02}={\omega }_{02}-f\mathrm{sin}{\alpha }_{0}{A}_{3}\left({\sigma }_{02}+{\displaystyle \sum _{l=1}^{\infty }{C}_{3l}\mathrm{sin}2l{\sigma }_{02}}\right)，$ (15)
${\lambda _{12}} = {\lambda _{02}} - {\lambda _{01}}，$ (16)
${\lambda _2} = {\lambda _1} + {\lambda _{12}}。$ (17)
\begin{aligned}{A}_{3}= &1-\Biggr(\frac{1}{2}-\frac{n}{2}\Biggr)\varepsilon -\Biggr(\frac{1}{4}+\frac{n}{8}-\frac{3{n}^{2}}{8}\Biggr){\varepsilon }^{2}-\Biggr(\frac{1}{16}+\frac{3n}{16}+\frac{{n}^{2}}{16}\Biggr){\varepsilon }^{3}-\\ &\Biggr(\frac{3}{64}+\frac{n}{32}\Biggr){\varepsilon }^{4}-\frac{3}{128}{\varepsilon }^{5}+\cdots ，\end{aligned}
\begin{aligned}{C}_{31}= &\Biggr(\frac{1}{4}-\frac{n}{4}\Biggr)\varepsilon +\Biggr(\frac{1}{8}-\frac{{n}^{2}}{8}\Biggr){\varepsilon }^{2}+\Biggr(\frac{3}{64}+\frac{3n}{64}-\frac{{n}^{2}}{64}\Biggr){\varepsilon }^{3}+\\ &\Biggr(\frac{5}{128}+\frac{n}{64}\Biggr){\varepsilon }^{4}+\frac{3}{128}{\varepsilon }^{5}+\cdots，\end{aligned}
\begin{aligned}{C}_{32}= &\Biggr(\frac{1}{16}-\frac{3n}{32}+\frac{{n}^{2}}{32}\Biggr){\varepsilon }^{2}+\Biggr(\frac{3}{64}-\frac{n}{32}-\frac{3{n}^{2}}{64}\Biggr){\varepsilon }^{3}+\\ &\Biggr(\frac{3}{128}+\frac{n}{128}\Biggr){\varepsilon }^{4}+ \frac{5}{256}{\varepsilon }^{5}+\cdots，\end{aligned}
${C}_{33}=\Biggr(\frac{5}{192}-\frac{3n}{64}+\frac{5{n}^{2}}{192}\Biggr){\varepsilon }^{3}+\Biggr(\frac{3}{128}-\frac{5n}{192}\Biggr){\varepsilon }^{4}+\frac{7}{512}{\varepsilon }^{5}+\cdots ，$
${C}_{34}=\Biggr(\frac{7}{52}-\frac{7n}{256}\Biggr){\varepsilon }^{4}+\frac{7}{512}{\varepsilon }^{5}+\cdots，$
${C_{35}} = \frac{{21}}{{2560}}{\varepsilon ^5} + \cdots。$
1.2 反解 φ1, λ1, φ2, λ2, ${\boldsymbol{\Rightarrow}}$ α1, s12, α2
1.2.1 初始航向预估
${\alpha _1} = \arctan \Biggr( - \frac{{x/(1 + v)}}{{y/v}}\Biggr) 。$ (18)
$x = \frac{{{\lambda _{12}} - \text{π} }}{{f \text{π} \cos {\beta _1}}},y = \frac{{{\beta _1} + {\beta _2}}}{{f \text{π} {{\cos }^2}{\beta _1}}},\frac{{{x^2}}}{{1 + {v^2}}} + \frac{{{y^2}}}{{{v^2}}} = 1。$ (19)
1.2.2 混合解
${\varphi _1} \leqslant 0,{\varphi _1} \leqslant {\varphi _2} \leqslant - {\varphi _1},0 \leqslant {\lambda _{12}} \leqslant \text{π}。$ (20)
图 2 混合问题中起始纬度交换示意图 Fig. 2 Demonstration of swapping points in hybrid geodesic problem
$\cos {\alpha _2} = \frac{{\sqrt {{{\cos }^2}{\alpha _1}{{\cos }^2}{\beta _1} + {{\cos }^2}{\beta _2} - {{\cos }^2}{\beta _1}} }}{{\cos {\beta _2}}}，$ (21)
${\sigma _{02}} = \arctan (\tan {\beta _2}/\cos {\alpha _2}) ，$ (22)
${\omega _{02}} = \arctan (\sin {\alpha _0}\tan {\sigma _{02}})。$ (23)
$\delta {\lambda _{12}} = {\lambda _{12}} - \lambda _{12}^{(0)}，$ (24)
$\delta {\alpha _1} = \frac{{ - \delta {\lambda _{12}}a\cos {\alpha _2}\cos {\beta _2}}}{{{\eta _{12}}b}} ，$ (25)
\begin{aligned} {\eta _{12}} = &\cos {\sigma _{01}}\sin {\sigma _{02}}\sqrt {1 + {k^2}{{\sin }^2}{\sigma _{02}}} - \\ &\cos {\sigma _{02}}\sin {\sigma _{01}}\sqrt {1 + {k^2}{{\sin }^2}{\sigma _{01}}} - \\ & \cos {\sigma _{01}} \cos {\sigma _{02}}((J({\sigma _{02}}) - J({\sigma _{01}}))\theta ，\end{aligned} (26)
$\begin{split}J({\sigma _{0i}}) = &{A_1}\left({\sigma _{0i}} + \sum\limits_{l = 1}^\infty {{C_{1l}}} \sin 1l{\sigma _{0i}}\right) -\\ &{A_2}\left({\sigma _{0i}} + \sum\limits_{l = 1}^\infty {{C_{2l}}} \sin 2l{\sigma _{0i}}\right)。\end{split}$ (27)
${A_2} = \dfrac{{\left(1 - \dfrac{3}{4}{\varepsilon ^2} - \dfrac{7}{{64}}{\varepsilon ^4} - \dfrac{{11}}{{256}}{\varepsilon ^6} + \cdots \right)}}{{(1 + \varepsilon )}} ，$
$\begin{split}& {C_{21}} = \frac{1}{2}\varepsilon + \frac{1}{{16}}{\varepsilon ^3} - \frac{1}{{32}}{\varepsilon ^5} + \cdots，\\ &{C_{22}} = \frac{3}{{16}}{\varepsilon ^2} + \frac{1}{{32}}{\varepsilon ^4} + \frac{{35}}{{2048}}{\varepsilon ^6} + \cdots，\end{split}$
${C_{23}} = \frac{5}{{48}}{\varepsilon ^3} + \frac{5}{{256}}{\varepsilon ^5} + \cdots ，\qquad {C_{24}} = \frac{{35}}{{512}}{\varepsilon ^4} + \frac{7}{{512}}{\varepsilon ^6} + \cdots，$
${C_{25}} = \frac{{63}}{{1280}}{\varepsilon ^5} + \cdots ，\qquad {C_{26}} = \frac{{77}}{{2048}}{\varepsilon ^6} + \cdots。$
α1 = α1 + δα1作为新的初始航向，并重复上述迭代过程，直到δλ12趋于0。通常经过几次迭代后，经差误差便小于10−15
${s_{12}} = {s_{02}} - {s_{01}}。$ (28)
2 航向变更限制机制
ACRM旨在限制频繁的转向，新的舵令只有在满足一定条件后才会被触发，否则保持原航向。ACRM最常见的策略有但不限于：按一定角度/时间/距离转向。在模拟仿真中，GPS位置不是直接获得的，而是通过恒向线算法计算而得，且航向交替过程中没有时间损失和延迟。因此，仿真结果会与实际情况存在些许差异。仿真中，假定GPS刷新率为10 Hz，即船舶每0.1 s步长为1 m。
2.1 按一定角度转向
图 3 ACRM执行流程图：按一定角度转向 Fig. 3 Diagram of ACRM process: alter course at a certain angle
2.2 按一定时间转向
2.3 按一定距离转向
图 4 不同ACRM策略下的航线示意图 Fig. 4 Diagram of voyage under different ACRM strategies
3 结 语
[1] 史国友, 朱公志, 王玉梅, 等. 恒向线主题直接正反解的高精度算法[J]. 大连海事大学学报:自然科学版, 2009, 35(2): 5-9. [2] TSENG W K, EARLE M A, GUO J L. Direct and inverse solutions with geodetic latitude in terms of longitude for rhumb line sailing[J]. Journal of Navigation, 2012, 65(3): 549-559. DOI:10.1017/S0373463312000148 [3] PETROVIC M. Orthodrome-loxodrome correlation by the middle latitude rule[J]. Journal of Navigation, 2013, 67(3): 539-543. [4] PETROVIC M . Middle rules and rhumb-line sailing[J]. Polish Maritime Research, 2017, 24(2). [5] CHEN H , DI W U , HOUPU L I , et al. Mercator rhumb line track calculation formula and its improvement[J]. Hydrographic Surveying and Charting, 2018. [6] CHEN C L, HSU T P, CHANG J R. A novel approach to great circle sailings: the great circle equation[J]. Journal of Navigation, 2004, 57(2): 311-319. DOI:10.1017/S0373463304002644 [7] EARLE, MICHAEL A. Vector solutions for great circle navigation[J]. Journal of Navigation, 2005, 58(3): 451-457. DOI:10.1017/S0373463305003358 [8] TSENG W K, LEE H S. The vector function for distance travelled in great circle navigation[J]. Journal of Navigation, 2007, 60(1): 158-164. DOI:10.1017/S0373463307214122 [9] NASTRO V, TANCREDI U. Great circle navigation with vectorial methods[J]. Journal of Navigation, 2010, 63(3): 557-563. DOI:10.1017/S0373463310000044 [10] HSU T P, CHEN C L, HSIEH T H. A graphical method for great circle routes[J]. Polish Maritime Research, 2017, 24(1): 12-21. DOI:10.1515/pomr-2017-0002 [11] VINCENTY T. Direct and inverse solutions of geodesics on the ellipsoid with application of nested equations[J]. Survey Review, 1975, 23(176). [12] VINCENTY T, Geodetic inverse solution between antipodal points. [EB/OL] http://geographiclib.sf.net/geodesicpapers/vincenty75b.pdf. [13] KARNEY C . Geodesics on an ellipsoid of revolution[J]. Eprint Arxiv, 2011. [14] KARNEY C. Algorithms for geodesics[J]. Journal of Geodesy, 2013, 87(1): 43-55. DOI:10.1007/s00190-012-0578-z [15] TSENG W K, GUO J L, LIU C P, A comparison of great circle, great ellipse, and geodesic sailing[J]. Journal of Marine Science and Technology, 2013, 21(3): 287–299. [16] GAO M, SHI G Y. Ship-collision avoidance decision-making learning of unmanned surface vehicles with automatic identification system data based on encoder—decoder automatic-response neural networks[J]. Journal of Marine Science and Engineering, 2020, 8(10): 754. [17] WEN H Z, ZHANG G. The path planning for unmanned ship based on the prioritized experience replay of deep q-networks[J]. Basic Clinical & Pharmacology & Toxicology, 2020, 126: 128–129. [18] LANG Y, YUAN B. Algorithm application based on the infrared image in unmanned ship target image recognition[J]. Microprocessors and Microsystems, 2021, 80: 103554. DOI:10.1016/j.micpro.2020.103554 [19] RAPP R H. Geometric geodesy part II[J]. Ohio State University Department of Geodetic Science Surveying, 1993. [20] Olver F W, Lozier D W, Boisvert R F, et al. Nist handbook of mathematical functions[M]. Combridge University Press, 2010. | 4,960 | 12,058 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | latest | en | 0.790622 |
https://www.kodytools.com/units/inertia/from/ctyd2/to/ctft2 | 1,726,796,484,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00567.warc.gz | 788,681,788 | 17,813 | # Carat Square Yard to Carat Square Foot Converter
1 Carat Square Yard = 9 Carat Square Feet
## One Carat Square Yard is Equal to How Many Carat Square Feet?
The answer is one Carat Square Yard is equal to 9 Carat Square Feet and that means we can also write it as 1 Carat Square Yard = 9 Carat Square Feet. Feel free to use our online unit conversion calculator to convert the unit from Carat Square Yard to Carat Square Foot. Just simply enter value 1 in Carat Square Yard and see the result in Carat Square Foot.
Manually converting Carat Square Yard to Carat Square Foot can be time-consuming,especially when you don’t have enough knowledge about Moment of Inertia units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Carat Square Yard to Carat Square Foot converter tool to get the job done as soon as possible.
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## How to Convert Carat Square Yard to Carat Square Foot (ct·yd2 to ct·ft2)
By using our Carat Square Yard to Carat Square Foot conversion tool, you know that one Carat Square Yard is equivalent to 9 Carat Square Foot. Hence, to convert Carat Square Yard to Carat Square Foot, we just need to multiply the number by 9. We are going to use very simple Carat Square Yard to Carat Square Foot conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Carat Square Yard} = 1 \times 9 = \text{9 Carat Square Feet}$$
## What Unit of Measure is Carat Square Yard?
Carat square yard is a unit of measurement for moment of inertia. It represents moment of inertia of a single particle rotating at one yard distance from the rotation axis and having a mass of one carat.
## What is the Symbol of Carat Square Yard?
The symbol of Carat Square Yard is ct·yd2. This means you can also write one Carat Square Yard as 1 ct·yd2.
## What Unit of Measure is Carat Square Foot?
Carat square foot is a unit of measurement for moment of inertia. It represents moment of inertia of a single particle rotating at one foot distance from the rotation axis and having a mass of one carat.
## What is the Symbol of Carat Square Foot?
The symbol of Carat Square Foot is ct·ft2. This means you can also write one Carat Square Foot as 1 ct·ft2.
## How to Use Carat Square Yard to Carat Square Foot Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Carat Square Yard and in the first input field, enter a value.
• From the second dropdown, select Carat Square Foot.
• Instantly, the tool will convert the value from Carat Square Yard to Carat Square Foot and display the result in the second input field.
## Example of Carat Square Yard to Carat Square Foot Converter Tool
Carat Square Yard
1
Carat Square Foot
9
# Carat Square Yard to Carat Square Foot Conversion Table
Carat Square Yard [ct·yd2]Carat Square Foot [ct·ft2]Description
1 Carat Square Yard9 Carat Square Foot1 Carat Square Yard = 9 Carat Square Foot
2 Carat Square Yard18 Carat Square Foot2 Carat Square Yard = 18 Carat Square Foot
3 Carat Square Yard27 Carat Square Foot3 Carat Square Yard = 27 Carat Square Foot
4 Carat Square Yard36 Carat Square Foot4 Carat Square Yard = 36 Carat Square Foot
5 Carat Square Yard45 Carat Square Foot5 Carat Square Yard = 45 Carat Square Foot
6 Carat Square Yard54 Carat Square Foot6 Carat Square Yard = 54 Carat Square Foot
7 Carat Square Yard63 Carat Square Foot7 Carat Square Yard = 63 Carat Square Foot
8 Carat Square Yard72 Carat Square Foot8 Carat Square Yard = 72 Carat Square Foot
9 Carat Square Yard81 Carat Square Foot9 Carat Square Yard = 81 Carat Square Foot
10 Carat Square Yard90 Carat Square Foot10 Carat Square Yard = 90 Carat Square Foot
100 Carat Square Yard900 Carat Square Foot100 Carat Square Yard = 900 Carat Square Foot
1000 Carat Square Yard9000 Carat Square Foot1000 Carat Square Yard = 9000 Carat Square Foot
# Carat Square Yard to Other Units Conversion Table
ConversionDescription
1 Carat Square Yard = 0.167225472 Gram Square Meter1 Carat Square Yard in Gram Square Meter is equal to 0.167225472
1 Carat Square Yard = 16.72 Gram Square Decimeter1 Carat Square Yard in Gram Square Decimeter is equal to 16.72
1 Carat Square Yard = 1672.25 Gram Square Centimeter1 Carat Square Yard in Gram Square Centimeter is equal to 1672.25
1 Carat Square Yard = 167225.47 Gram Square Millimeter1 Carat Square Yard in Gram Square Millimeter is equal to 167225.47
1 Carat Square Yard = 167225472000 Gram Square Micrometer1 Carat Square Yard in Gram Square Micrometer is equal to 167225472000
1 Carat Square Yard = 1.67225472e-7 Gram Square Kilometer1 Carat Square Yard in Gram Square Kilometer is equal to 1.67225472e-7
1 Carat Square Yard = 167225472000000000 Gram Square Nanometer1 Carat Square Yard in Gram Square Nanometer is equal to 167225472000000000
1 Carat Square Yard = 0.2 Gram Square Yard1 Carat Square Yard in Gram Square Yard is equal to 0.2
1 Carat Square Yard = 259.2 Gram Square Inch1 Carat Square Yard in Gram Square Inch is equal to 259.2
1 Carat Square Yard = 1.8 Gram Square Foot1 Carat Square Yard in Gram Square Foot is equal to 1.8
1 Carat Square Yard = 6.4566115702479e-8 Gram Square Mile1 Carat Square Yard in Gram Square Mile is equal to 6.4566115702479e-8
1 Carat Square Yard = 0.000167225472 Kilogram Square Meter1 Carat Square Yard in Kilogram Square Meter is equal to 0.000167225472
1 Carat Square Yard = 0.0167225472 Kilogram Square Decimeter1 Carat Square Yard in Kilogram Square Decimeter is equal to 0.0167225472
1 Carat Square Yard = 1.67 Kilogram Square Centimeter1 Carat Square Yard in Kilogram Square Centimeter is equal to 1.67
1 Carat Square Yard = 167.23 Kilogram Square Millimeter1 Carat Square Yard in Kilogram Square Millimeter is equal to 167.23
1 Carat Square Yard = 167225472 Kilogram Square Micrometer1 Carat Square Yard in Kilogram Square Micrometer is equal to 167225472
1 Carat Square Yard = 1.67225472e-10 Kilogram Square Kilometer1 Carat Square Yard in Kilogram Square Kilometer is equal to 1.67225472e-10
1 Carat Square Yard = 167225472000000 Kilogram Square Nanometer1 Carat Square Yard in Kilogram Square Nanometer is equal to 167225472000000
1 Carat Square Yard = 0.0002 Kilogram Square Yard1 Carat Square Yard in Kilogram Square Yard is equal to 0.0002
1 Carat Square Yard = 0.2592 Kilogram Square Inch1 Carat Square Yard in Kilogram Square Inch is equal to 0.2592
1 Carat Square Yard = 0.0018 Kilogram Square Foot1 Carat Square Yard in Kilogram Square Foot is equal to 0.0018
1 Carat Square Yard = 6.4566115702479e-11 Kilogram Square Mile1 Carat Square Yard in Kilogram Square Mile is equal to 6.4566115702479e-11
1 Carat Square Yard = 167.23 Milligram Square Meter1 Carat Square Yard in Milligram Square Meter is equal to 167.23
1 Carat Square Yard = 16722.55 Milligram Square Decimeter1 Carat Square Yard in Milligram Square Decimeter is equal to 16722.55
1 Carat Square Yard = 1672254.72 Milligram Square Centimeter1 Carat Square Yard in Milligram Square Centimeter is equal to 1672254.72
1 Carat Square Yard = 167225472 Milligram Square Millimeter1 Carat Square Yard in Milligram Square Millimeter is equal to 167225472
1 Carat Square Yard = 167225472000000 Milligram Square Micrometer1 Carat Square Yard in Milligram Square Micrometer is equal to 167225472000000
1 Carat Square Yard = 0.000167225472 Milligram Square Kilometer1 Carat Square Yard in Milligram Square Kilometer is equal to 0.000167225472
1 Carat Square Yard = 167225472000000000000 Milligram Square Nanometer1 Carat Square Yard in Milligram Square Nanometer is equal to 167225472000000000000
1 Carat Square Yard = 200 Milligram Square Yard1 Carat Square Yard in Milligram Square Yard is equal to 200
1 Carat Square Yard = 259200 Milligram Square Inch1 Carat Square Yard in Milligram Square Inch is equal to 259200
1 Carat Square Yard = 1800 Milligram Square Foot1 Carat Square Yard in Milligram Square Foot is equal to 1800
1 Carat Square Yard = 0.000064566115702479 Milligram Square Mile1 Carat Square Yard in Milligram Square Mile is equal to 0.000064566115702479
1 Carat Square Yard = 167225.47 Microgram Square Meter1 Carat Square Yard in Microgram Square Meter is equal to 167225.47
1 Carat Square Yard = 16722547.2 Microgram Square Decimeter1 Carat Square Yard in Microgram Square Decimeter is equal to 16722547.2
1 Carat Square Yard = 1672254720 Microgram Square Centimeter1 Carat Square Yard in Microgram Square Centimeter is equal to 1672254720
1 Carat Square Yard = 167225472000 Microgram Square Millimeter1 Carat Square Yard in Microgram Square Millimeter is equal to 167225472000
1 Carat Square Yard = 167225472000000000 Microgram Square Micrometer1 Carat Square Yard in Microgram Square Micrometer is equal to 167225472000000000
1 Carat Square Yard = 0.167225472 Microgram Square Kilometer1 Carat Square Yard in Microgram Square Kilometer is equal to 0.167225472
1 Carat Square Yard = 1.67225472e+23 Microgram Square Nanometer1 Carat Square Yard in Microgram Square Nanometer is equal to 1.67225472e+23
1 Carat Square Yard = 200000 Microgram Square Yard1 Carat Square Yard in Microgram Square Yard is equal to 200000
1 Carat Square Yard = 259200000 Microgram Square Inch1 Carat Square Yard in Microgram Square Inch is equal to 259200000
1 Carat Square Yard = 1800000 Microgram Square Foot1 Carat Square Yard in Microgram Square Foot is equal to 1800000
1 Carat Square Yard = 0.064566115702479 Microgram Square Mile1 Carat Square Yard in Microgram Square Mile is equal to 0.064566115702479
1 Carat Square Yard = 1.67225472e-7 Ton Square Meter1 Carat Square Yard in Ton Square Meter is equal to 1.67225472e-7
1 Carat Square Yard = 0.0000167225472 Ton Square Decimeter1 Carat Square Yard in Ton Square Decimeter is equal to 0.0000167225472
1 Carat Square Yard = 0.00167225472 Ton Square Centimeter1 Carat Square Yard in Ton Square Centimeter is equal to 0.00167225472
1 Carat Square Yard = 0.167225472 Ton Square Millimeter1 Carat Square Yard in Ton Square Millimeter is equal to 0.167225472
1 Carat Square Yard = 167225.47 Ton Square Micrometer1 Carat Square Yard in Ton Square Micrometer is equal to 167225.47
1 Carat Square Yard = 1.67225472e-13 Ton Square Kilometer1 Carat Square Yard in Ton Square Kilometer is equal to 1.67225472e-13
1 Carat Square Yard = 167225472000 Ton Square Nanometer1 Carat Square Yard in Ton Square Nanometer is equal to 167225472000
1 Carat Square Yard = 2e-7 Ton Square Yard1 Carat Square Yard in Ton Square Yard is equal to 2e-7
1 Carat Square Yard = 0.0002592 Ton Square Inch1 Carat Square Yard in Ton Square Inch is equal to 0.0002592
1 Carat Square Yard = 0.0000018 Ton Square Foot1 Carat Square Yard in Ton Square Foot is equal to 0.0000018
1 Carat Square Yard = 6.4566115702479e-14 Ton Square Mile1 Carat Square Yard in Ton Square Mile is equal to 6.4566115702479e-14
1 Carat Square Yard = 0.83612736 Carat Square Meter1 Carat Square Yard in Carat Square Meter is equal to 0.83612736
1 Carat Square Yard = 83.61 Carat Square Decimeter1 Carat Square Yard in Carat Square Decimeter is equal to 83.61
1 Carat Square Yard = 8361.27 Carat Square Centimeter1 Carat Square Yard in Carat Square Centimeter is equal to 8361.27
1 Carat Square Yard = 836127.36 Carat Square Millimeter1 Carat Square Yard in Carat Square Millimeter is equal to 836127.36
1 Carat Square Yard = 836127360000 Carat Square Micrometer1 Carat Square Yard in Carat Square Micrometer is equal to 836127360000
1 Carat Square Yard = 8.3612736e-7 Carat Square Kilometer1 Carat Square Yard in Carat Square Kilometer is equal to 8.3612736e-7
1 Carat Square Yard = 836127360000000000 Carat Square Nanometer1 Carat Square Yard in Carat Square Nanometer is equal to 836127360000000000
1 Carat Square Yard = 1296 Carat Square Inch1 Carat Square Yard in Carat Square Inch is equal to 1296
1 Carat Square Yard = 9 Carat Square Foot1 Carat Square Yard in Carat Square Foot is equal to 9
1 Carat Square Yard = 3.228305785124e-7 Carat Square Mile1 Carat Square Yard in Carat Square Mile is equal to 3.228305785124e-7
1 Carat Square Yard = 0.0058987049363286 Ounce Square Meter1 Carat Square Yard in Ounce Square Meter is equal to 0.0058987049363286
1 Carat Square Yard = 0.58987049363286 Ounce Square Decimeter1 Carat Square Yard in Ounce Square Decimeter is equal to 0.58987049363286
1 Carat Square Yard = 58.99 Ounce Square Centimeter1 Carat Square Yard in Ounce Square Centimeter is equal to 58.99
1 Carat Square Yard = 5898.7 Ounce Square Millimeter1 Carat Square Yard in Ounce Square Millimeter is equal to 5898.7
1 Carat Square Yard = 5898704936.33 Ounce Square Micrometer1 Carat Square Yard in Ounce Square Micrometer is equal to 5898704936.33
1 Carat Square Yard = 5.8987049363286e-9 Ounce Square Kilometer1 Carat Square Yard in Ounce Square Kilometer is equal to 5.8987049363286e-9
1 Carat Square Yard = 5898704936328600 Ounce Square Nanometer1 Carat Square Yard in Ounce Square Nanometer is equal to 5898704936328600
1 Carat Square Yard = 0.0070547923899161 Ounce Square Yard1 Carat Square Yard in Ounce Square Yard is equal to 0.0070547923899161
1 Carat Square Yard = 9.14 Ounce Square Inch1 Carat Square Yard in Ounce Square Inch is equal to 9.14
1 Carat Square Yard = 0.063493131509245 Ounce Square Foot1 Carat Square Yard in Ounce Square Foot is equal to 0.063493131509245
1 Carat Square Yard = 2.2775027085215e-9 Ounce Square Mile1 Carat Square Yard in Ounce Square Mile is equal to 2.2775027085215e-9
1 Carat Square Yard = 3.6866905852054e-7 Kilopound Square Meter1 Carat Square Yard in Kilopound Square Meter is equal to 3.6866905852054e-7
1 Carat Square Yard = 0.000036866905852054 Kilopound Square Decimeter1 Carat Square Yard in Kilopound Square Decimeter is equal to 0.000036866905852054
1 Carat Square Yard = 0.0036866905852054 Kilopound Square Centimeter1 Carat Square Yard in Kilopound Square Centimeter is equal to 0.0036866905852054
1 Carat Square Yard = 0.36866905852054 Kilopound Square Millimeter1 Carat Square Yard in Kilopound Square Millimeter is equal to 0.36866905852054
1 Carat Square Yard = 368669.06 Kilopound Square Micrometer1 Carat Square Yard in Kilopound Square Micrometer is equal to 368669.06
1 Carat Square Yard = 3.6866905852054e-13 Kilopound Square Kilometer1 Carat Square Yard in Kilopound Square Kilometer is equal to 3.6866905852054e-13
1 Carat Square Yard = 368669058520.54 Kilopound Square Nanometer1 Carat Square Yard in Kilopound Square Nanometer is equal to 368669058520.54
1 Carat Square Yard = 4.4092452436976e-7 Kilopound Square Yard1 Carat Square Yard in Kilopound Square Yard is equal to 4.4092452436976e-7
1 Carat Square Yard = 0.0005714381835832 Kilopound Square Inch1 Carat Square Yard in Kilopound Square Inch is equal to 0.0005714381835832
1 Carat Square Yard = 0.0000039683207193278 Kilopound Square Foot1 Carat Square Yard in Kilopound Square Foot is equal to 0.0000039683207193278
1 Carat Square Yard = 1.4234391928259e-13 Kilopound Square Mile1 Carat Square Yard in Kilopound Square Mile is equal to 1.4234391928259e-13
1 Carat Square Yard = 0.00036866905852054 Pound Square Meter1 Carat Square Yard in Pound Square Meter is equal to 0.00036866905852054
1 Carat Square Yard = 0.036866905852054 Pound Square Decimeter1 Carat Square Yard in Pound Square Decimeter is equal to 0.036866905852054
1 Carat Square Yard = 3.69 Pound Square Centimeter1 Carat Square Yard in Pound Square Centimeter is equal to 3.69
1 Carat Square Yard = 368.67 Pound Square Millimeter1 Carat Square Yard in Pound Square Millimeter is equal to 368.67
1 Carat Square Yard = 368669058.52 Pound Square Micrometer1 Carat Square Yard in Pound Square Micrometer is equal to 368669058.52
1 Carat Square Yard = 3.6866905852054e-10 Pound Square Kilometer1 Carat Square Yard in Pound Square Kilometer is equal to 3.6866905852054e-10
1 Carat Square Yard = 368669058520540 Pound Square Nanometer1 Carat Square Yard in Pound Square Nanometer is equal to 368669058520540
1 Carat Square Yard = 0.00044092452436976 Pound Square Yard1 Carat Square Yard in Pound Square Yard is equal to 0.00044092452436976
1 Carat Square Yard = 0.5714381835832 Pound Square Inch1 Carat Square Yard in Pound Square Inch is equal to 0.5714381835832
1 Carat Square Yard = 0.0039683207193278 Pound Square Foot1 Carat Square Yard in Pound Square Foot is equal to 0.0039683207193278
1 Carat Square Yard = 1.4234391928259e-10 Pound Square Mile1 Carat Square Yard in Pound Square Mile is equal to 1.4234391928259e-10
1 Carat Square Yard = 0.011871143684545 Poundal Square Meter1 Carat Square Yard in Poundal Square Meter is equal to 0.011871143684545
1 Carat Square Yard = 1.19 Poundal Square Decimeter1 Carat Square Yard in Poundal Square Decimeter is equal to 1.19
1 Carat Square Yard = 118.71 Poundal Square Centimeter1 Carat Square Yard in Poundal Square Centimeter is equal to 118.71
1 Carat Square Yard = 11871.14 Poundal Square Millimeter1 Carat Square Yard in Poundal Square Millimeter is equal to 11871.14
1 Carat Square Yard = 11871143684.55 Poundal Square Micrometer1 Carat Square Yard in Poundal Square Micrometer is equal to 11871143684.55
1 Carat Square Yard = 1.1871143684545e-8 Poundal Square Kilometer1 Carat Square Yard in Poundal Square Kilometer is equal to 1.1871143684545e-8
1 Carat Square Yard = 11871143684545000 Poundal Square Nanometer1 Carat Square Yard in Poundal Square Nanometer is equal to 11871143684545000
1 Carat Square Yard = 0.014197769684925 Poundal Square Yard1 Carat Square Yard in Poundal Square Yard is equal to 0.014197769684925
1 Carat Square Yard = 18.4 Poundal Square Inch1 Carat Square Yard in Poundal Square Inch is equal to 18.4
1 Carat Square Yard = 0.12777992716433 Poundal Square Foot1 Carat Square Yard in Poundal Square Foot is equal to 0.12777992716433
1 Carat Square Yard = 4.5834742009702e-9 Poundal Square Mile1 Carat Square Yard in Poundal Square Mile is equal to 4.5834742009702e-9
1 Carat Square Yard = 2.58 Grain Square Meter1 Carat Square Yard in Grain Square Meter is equal to 2.58
1 Carat Square Yard = 258.07 Grain Square Decimeter1 Carat Square Yard in Grain Square Decimeter is equal to 258.07
1 Carat Square Yard = 25806.83 Grain Square Centimeter1 Carat Square Yard in Grain Square Centimeter is equal to 25806.83
1 Carat Square Yard = 2580683.41 Grain Square Millimeter1 Carat Square Yard in Grain Square Millimeter is equal to 2580683.41
1 Carat Square Yard = 2580683409643.8 Grain Square Micrometer1 Carat Square Yard in Grain Square Micrometer is equal to 2580683409643.8
1 Carat Square Yard = 0.0000025806834096438 Grain Square Kilometer1 Carat Square Yard in Grain Square Kilometer is equal to 0.0000025806834096438
1 Carat Square Yard = 2580683409643800000 Grain Square Nanometer1 Carat Square Yard in Grain Square Nanometer is equal to 2580683409643800000
1 Carat Square Yard = 3.09 Grain Square Yard1 Carat Square Yard in Grain Square Yard is equal to 3.09
1 Carat Square Yard = 4000.07 Grain Square Inch1 Carat Square Yard in Grain Square Inch is equal to 4000.07
1 Carat Square Yard = 27.78 Grain Square Foot1 Carat Square Yard in Grain Square Foot is equal to 27.78
1 Carat Square Yard = 9.9640743497814e-7 Grain Square Mile1 Carat Square Yard in Grain Square Mile is equal to 9.9640743497814e-7
1 Carat Square Yard = 0.00001145858694386 Slug Square Meter1 Carat Square Yard in Slug Square Meter is equal to 0.00001145858694386
1 Carat Square Yard = 0.001145858694386 Slug Square Decimeter1 Carat Square Yard in Slug Square Decimeter is equal to 0.001145858694386
1 Carat Square Yard = 0.1145858694386 Slug Square Centimeter1 Carat Square Yard in Slug Square Centimeter is equal to 0.1145858694386
1 Carat Square Yard = 11.46 Slug Square Millimeter1 Carat Square Yard in Slug Square Millimeter is equal to 11.46
1 Carat Square Yard = 11458586.94 Slug Square Micrometer1 Carat Square Yard in Slug Square Micrometer is equal to 11458586.94
1 Carat Square Yard = 1.145858694386e-11 Slug Square Kilometer1 Carat Square Yard in Slug Square Kilometer is equal to 1.145858694386e-11
1 Carat Square Yard = 11458586943860 Slug Square Nanometer1 Carat Square Yard in Slug Square Nanometer is equal to 11458586943860
1 Carat Square Yard = 0.000013704355929532 Slug Square Yard1 Carat Square Yard in Slug Square Yard is equal to 0.000013704355929532
1 Carat Square Yard = 0.017760845284674 Slug Square Inch1 Carat Square Yard in Slug Square Inch is equal to 0.017760845284674
1 Carat Square Yard = 0.00012333920336579 Slug Square Foot1 Carat Square Yard in Slug Square Foot is equal to 0.00012333920336579
1 Carat Square Yard = 4.4241851528707e-12 Slug Square Mile1 Carat Square Yard in Slug Square Mile is equal to 4.4241851528707e-12
1 Carat Square Yard = 167225.47 Gamma Square Meter1 Carat Square Yard in Gamma Square Meter is equal to 167225.47
1 Carat Square Yard = 16722547.2 Gamma Square Decimeter1 Carat Square Yard in Gamma Square Decimeter is equal to 16722547.2
1 Carat Square Yard = 1672254720 Gamma Square Centimeter1 Carat Square Yard in Gamma Square Centimeter is equal to 1672254720
1 Carat Square Yard = 167225472000 Gamma Square Millimeter1 Carat Square Yard in Gamma Square Millimeter is equal to 167225472000
1 Carat Square Yard = 167225472000000000 Gamma Square Micrometer1 Carat Square Yard in Gamma Square Micrometer is equal to 167225472000000000
1 Carat Square Yard = 0.167225472 Gamma Square Kilometer1 Carat Square Yard in Gamma Square Kilometer is equal to 0.167225472
1 Carat Square Yard = 1.67225472e+23 Gamma Square Nanometer1 Carat Square Yard in Gamma Square Nanometer is equal to 1.67225472e+23
1 Carat Square Yard = 200000 Gamma Square Yard1 Carat Square Yard in Gamma Square Yard is equal to 200000
1 Carat Square Yard = 259200000 Gamma Square Inch1 Carat Square Yard in Gamma Square Inch is equal to 259200000
1 Carat Square Yard = 1800000 Gamma Square Foot1 Carat Square Yard in Gamma Square Foot is equal to 1800000
1 Carat Square Yard = 0.064566115702479 Gamma Square Mile1 Carat Square Yard in Gamma Square Mile is equal to 0.064566115702479 | 6,412 | 22,312 | {"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} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.88776 |
https://ltwork.net/our-fate-is-predetermined-we-cannot-alter-our-own-destiny--582966 | 1,675,122,617,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00607.warc.gz | 388,265,301 | 10,506 | # Our fate is predetermined; we cannot alter our own destiny. do you agree or disagree explain why?
###### Question:
Our fate is predetermined; we cannot alter our own destiny.
do you agree or disagree
explain why?
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Vtrac
Topic closed. 1 reply. Last post 6 years ago by calabs.
Page 1 of 1
Miami Garden Fl
United States
Member #103959
January 7, 2011
77 Posts
Offline
Posted: May 7, 2011, 11:46 pm - IP Logged
i'm learning Vtrac
and i have a few questions
i live in FL need to know the States that Mirror FL?
Do i record the Actual # from the 1st of the Month to the 30th
or from the 5th to the 24th?
i understand how to find the Vtech #, What is the Vtech exchange ?
waltoy
United States
Member #27050
November 26, 2005
40272 Posts
Offline
Posted: May 8, 2011, 1:15 am - IP Logged
i'm learning Vtrac
and i have a few questions
i live in FL need to know the States that Mirror FL?
Do i record the Actual # from the 1st of the Month to the 30th
or from the 5th to the 24th?
i understand how to find the Vtech #, What is the Vtech exchange ?
waltoy
Hi waltoy....best of luck in learning vtracs. One of the many tools to use here on LP.
A vtrac exchange is when you use the "other" vtrac numbers for a given draw.
For example, if the drawn number is 536, the vtrac for that number is v142. Now using the vtrac for the number, replace the numbers with the "other" vtrac numbers to form it's exchange number.
Example: vtrac 1 = 0,5 vtrac 4 = 3,8 vtrac 2 = 1,6
Now since the drawn number was 536, replace the numbers 5,3,6 with the other vtrac numbers 0,8,1.
Make sense? that's what a vtrac exchange is.
I believe the mirror states for Florida are DE-ID-TS-NJ-IL
Good luck!!
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http://www.numbersaplenty.com/4087968 | 1,596,951,134,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00221.warc.gz | 167,769,856 | 3,963 | Search a number
4087968 = 25397439
BaseRepresentation
bin1111100110000010100000
321200200122020
433212002200
52021303333
6223341440
746514163
oct17460240
97620566
104087968
112342395
121451880
13b01921
1555b3b3
hex3e60a0
4087968 has 48 divisors (see below), whose sum is σ = 10866240. Its totient is φ = 1345536.
The previous prime is 4087957. The next prime is 4087969. The reversal of 4087968 is 8697804.
It can be divided in two parts, 408796 and 8, that added together give a palindrome (408804).
It is a happy number.
It is a tau number, because it is divible by the number of its divisors (48).
It is a super-2 number, since 2×40879682 = 33422964738048, which contains 22 as substring.
It is a partition number, being equal to the number of ways a set of 70 identical objects can be partitioned into subset.
It is not an unprimeable number, because it can be changed into a prime (4087969) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 9093 + ... + 9531.
It is an arithmetic number, because the mean of its divisors is an integer number (226380).
Almost surely, 24087968 is an apocalyptic number.
4087968 is a gapful number since it is divisible by the number (48) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 4087968, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (5433120).
4087968 is an abundant number, since it is smaller than the sum of its proper divisors (6778272).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
4087968 is a wasteful number, since it uses less digits than its factorization.
4087968 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 549 (or 541 counting only the distinct ones).
The product of its (nonzero) digits is 96768, while the sum is 42.
The square root of 4087968 is about 2021.8723995346. The cubic root of 4087968 is about 159.8953482316.
The spelling of 4087968 in words is "four million, eighty-seven thousand, nine hundred sixty-eight". | 631 | 2,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-34 | latest | en | 0.907574 |
https://mathlesstraveled.com/category/complex-numbers/ | 1,686,323,904,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656737.96/warc/CC-MAIN-20230609132648-20230609162648-00609.warc.gz | 422,201,037 | 20,075 | # Category Archives: complex numbers
## The Möbius function proof, part 2 (the subset parity lemma)
Continuing from my previous post, we are in the middle of proving that satisfies the same equation as , that is, and that therefore for all , that is, is the sum of all the th primitive roots of unity. … Continue reading
Posted in arithmetic, combinatorics, complex numbers, primes, proof | Tagged , , , , , , , , , | 3 Comments
## Mystery curve, animated
As a follow-on to my previous post, here’s an animation (17MB) showing how the “mystery curve” arises as a sum of circular motions: Recall that the equation for the curve is . The big blue circle corresponds to the term—it … Continue reading
Posted in complex numbers, geometry, programming | | 6 Comments
## Random cyclic curves
Princeton Press just sent me a review copy of a new book by Frank Farris called Creating Symmetry: The Artful Mathematics of Wallpaper Patterns. It looks amazing and I’m super excited to read it. Apparently John Cook has been reading … Continue reading
Posted in complex numbers, geometry, programming | Tagged , , , , , | 24 Comments
This week’s Monday Math Madness is a nice little problem involving complex exponentiation. Go check it out, and maybe win a prize!
Posted in challenges, complex numbers, links | Tagged , , , , | Comments Off on Monday Math Madness #31
## Video: Möbius transformations revealed
For your viewing pleasure, a fantastically beautiful video about Möbius transformations, which are functions of the form where z, a, b, c, and d are complex numbers, and . For example, is a Möbius transformation with b=2, c=1, and a=d=0. … Continue reading
Posted in complex numbers, geometry, video | Comments Off on Video: Möbius transformations revealed
## Nuclear Pennies Game: Analysis
And now, for the promised analysis of the Nuclear Pennies Game! First, recall the rules of the game: there is a semi-infinite (i.e. with a beginning but no end) strip of squares, each of which can contain a stack of … Continue reading
Posted in algebra, complex numbers, games, proof | 3 Comments | 491 | 2,102 | {"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-2023-23 | latest | en | 0.866091 |
http://mtapreviewer.com/2016/04/13/2005-grade-5-mtap-elimination-questions-2/ | 1,534,782,570,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216718.53/warc/CC-MAIN-20180820160510-20180820180510-00434.warc.gz | 279,417,106 | 11,230 | # 2005 Grade 5 MTAP Elimination Questions Part 2
Below are the 2005 Grade 5 Metrobank-MTAP-DepEd Math Challenge Elimination questions 26-30 from with answers. Questions 1-25 can be read here. If you see any error, kindly comment below
26.) Pat had 12 apples. He gave each of his friends ¾ of an apple and kept ¾ for himself. If there was no apple left, how many friends were with him?
27.) There are 680 Grades 4-6 students in a school. If 35% of them are in Grade 5, how many Grade 5 students are there?
28.) Rose bought a blouse at a discount of 18% for ₱147.60. What was the marked price of the blouse?
29.) There are 240 scouts in a school. If 90 of them are cubs, what percent of the scouts in the school are cubs?
30.) Ann can make one uniform with 2.4 m of cloth. How many uniforms can she make from 20 m of cloth?
31.) Which of the following numbers can be written as the sum of two primes?
a. 25 b. 35 c. 51 d. 57
32.) Find N if N : 84 = 5 : 6. 16m
Q33-34 no figure.
35.) If 8 men can do a piece of work in 6 days, how many men can do it in 4 days at the same rate?
36.) A restaurant charges 4% service charge. If your order amounted to ₱450, how much did you pay?
37.) A closed box for a raffle is 5 dm long, 3.5 dm wide and 7 dm high. If it is to be completely covered with colored paper, how much surface is to be covered?
38.) An aquarium is 6 dm long, 4 dm wide and 4 dm high. How many liters of water does it contain when it is6/5full?
39.) A base angle of an isosceles triangle is 52o. What is the vertex angle?
40.) How many cubic meters of soil will be needed to raise a 4 m long and 2.5 m wide flower plot by 15 cm?
41.) If your step is about 70 cm and you can make 60 steps a minute, how far can you walk in 10 minutes?
42.) A park is 85 m long and 70 m wide. If you jog around it 6 times, how many meters will that be?
43.) The ratio of two numbers is 3 : 5 and their difference is 34. Find the bigger number?
44.) The average of two fractions is7/8and one fraction is3/2. What is the other?
45.) The scale of a map is 1cm : 15 km. If two towns are 4.5 cm apart on a map, how many kilometers apart are they?
46.) A school basketball team won ¾ of its games. If it won 14 more games than it lost, how many games did it play?
47.) The parallel sides of a trapezoidal field are 52 m and 68 m. If they are 48 m apart, what is its area? | 702 | 2,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-34 | longest | en | 0.955001 |
https://math.stackexchange.com/questions/3592341/poker-probability-with-hands-of-6-cards | 1,713,120,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00667.warc.gz | 361,415,899 | 35,497 | # Poker Probability with hands of 6 cards
I am trying to answer Question 1(d). A valid hand would be Ace hearts, 10 hearts, 2 clubs, 7 clubs, Ace spades, J spades.
My attempts have given me 2 different answers that I am unsure of.
What I did was I had to choose 3 suits from 4, then choose 2 cards from each of those suits. When choosing the 2 cards from each suit, do I have to do it 3 times? Would it be 4C3 x (13C2)^3 or 4C3 x 13C2
Thank you.
• Yes you have to do it three times, once for each of the three suits Mar 23, 2020 at 23:42
You are counting ways to select: two from thirteen kinds for each of three from four suits.$$({^{13}\mathrm C_2})^3\cdot{^4\mathrm C_3}$$
Note: $${^{13}\mathrm C_2}\cdot{^4\mathrm C_3}$$ would count ways to select the same two from thirteen kinds in three from four suits.
• @fatimahfatcakes Since there must be a suit with at least two kinds in the hand, we should assume that is asking for a suit with exactly two kinds . Then it becomes as easy as PIE (ie, use the Principle of Inclusion and Exclusion).$$\def\C#1#2{{^{#1}\mathrm C_{#2}}}\C 41\C{13}2\C{39}4-\C 42(\C{13}{2})^2\,\C{26}2+\C43(\C{13}{2})^3$$ Mar 24, 2020 at 0:33 | 385 | 1,173 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-18 | latest | en | 0.930661 |
https://petapixel.com/2014/05/05/mtf-charts-english-translation/ | 1,620,336,821,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988763.83/warc/CC-MAIN-20210506205251-20210506235251-00479.warc.gz | 475,951,734 | 20,934 | # MTF Charts: The English Translation
This post contains absolutely no mathematics. Explaining MTF without math is sort of like doing a high-wire act without a net. It’s dangerous, but for any number of reasons is more likely to keep the audience interested.
### Why Am I Doing This Again?
I wrote an article on reading MTF charts several years ago. It focused on deciphering the MTF maps that many lensmakers publish when they release a lens, like the one below. But I get a lot of emails asking me asking me how to compare the MTF graphs we use in testing at LensRentals to the manufacturer’s MTF charts. Or asking me to show an MTF frequency chart (and if I show it, then lots of emails asking what it means). So I thought I’d write a quick post about the different types of MTF data and charts.
So first, an overview of the common types of MTF charts.
MTF Maps
I use the term “map” because the horizontal axis maps the lens from the center (‘0’mm) to the edge of the frame (24mm). The colored lines show how certain MTF values change as we move from the center to the edge of the field of view.
Except for the ones put out by Zeiss and Leica, these MTF maps are generated from computer models. Looking at the computer-model MTF map is like looking at a retouched photo of a swimsuit model — the real-life the curves are never quite as good as the image suggests.
A real MTF chart, put out for an actual lens, looks like the one below. Notice it has “0″ (the center) in the middle and goes to both sides, showing one dirty little secret: very few copies of any lens have exactly the same MTF curve on both sides.
What About Those MTF50 Numbers in Lens Reviews?
These days more people seem to pay attention to MTF50 tables generated by computerized target analysis, which appear on most review and testing sites.
Those are useful, of course, but now we’ve perhaps gone too far. People who don’t actually know what the MTF50 means think it means sharpest or even best lens, which isn’t absolutely true. Just like Mount Everest is a good, but not perfect, definition of “thew world’s tallest mountain,” the highest MTF50 is a good, but not perfect definition of “sharpest lens.” (Depending on your definition of ‘tallest’, Mauna Kea or Chimboraza might actually be tallest.)
It gets even more confusing. The MTF maps the camera makers put out, like the one at the top, gives MTF results in 10 and 30 lines/mm. Some people think MTF50 means MTF at 50 lines/mm, but it doesn’t. It actually gives you a completely different set of information.
MTF Frequency Graphs
Finally, you sometimes see a frequency MTF graph for a lens, that looks something line this. It kind of looks like the MTF map I showed at first, but actually they have very little in common. The Frequency graph shows you the MTF50, but adds a lot of additional data.
If you bear with me for a few minutes, I’ll explain what these various MTF numbers and graphs are showing us. If you like lab testing, this will let you understand the tests you more clearly. If you hate lab testing, it will still be worthwhile because you’ll be able to say things like, “The MTF50 numbers in that post don’t show anything about the fine detail resolution capabilities of the lens” in online discussions.
### So What Are These Different MTF Numbers of Which You Speak?
MTF
They aren’t actually different MTFs, they are different ways of using MTF (Modulation Transfer Function) to show different information. Let’s start with a simple definition of the Modulation Transfer Function (MTF). We’ll do it without math, which will be better for everyone. (For 95% of people it will be better because they can understand what they’re interested in. For the other 5% it will be better because they can break out various formulae and make lengthy posts about why this definition is incomplete.)
When a lens makes an image, it’s never a perfect image. For example, let’s start with some thick, black and white bars like this:
If I focus a lens on them, the image the lens makes looks almost, but not quite, the same, like this:
If you look closely, you’ll notice the second image is just a tiny bit blurry. If you measure it very carefully, you’ll find the black isn’t quite as black as the original, and the white not quite as white. You might not notice it on a computer screen, but the difference is there.
Now I’m going to show you the MTF formula without any math: MTF = contrast. We can make it just a bit more complex and more accurate without getting mathematical: MTF = blackest-whitest/blackest+whitest.
You can handle that even with some numbers, right? Let’s say pure black is 1 and pure white is 0. Since the original object is pure black and white we have (1-0/1+0)=1. It’s perfection.
The image a lens makes isn’t quite perfect. In my example, the black is 98% as black as the original image, the white is 98% as white (or put another way, the white is now 2% black). If we plug that into the formula (you don’t have to, just showing you) it would be .98-.02/.98+.02=.96. So the MTF of this lens for these thick black bars is .96.
What if we use smaller bars put closer together, like the images below?
Lenses have more trouble as the bars get smaller. Pretty soon, the grayish blur from the edge of one bar starts to touch the edge of the next bar. The image looks like this.
Now the blackest bars in the image are 66% as black as the original, while the white areas are actually 34% black in the image. I won’t bore you with the formula, but the MTF for our close lines is .32. The difference between an MTF of .96 and .32 is pretty obvious just looking at the pictures, but the numbers make things more comparable. Saying “the MTF drops from .96 to .32” tells someone else the results more accurately than if you said, “It goes from barely blurry to pretty blurry, but I can still count the bars.”.
What About Those Line Pair Things
If we want to be all scientific, we can’t just use some arbitrary ‘thick lines’ and ‘thin lines’ like I did above. We need to quantify them, and the quantity we use is lines or line pairs per mm. It’s the same thing, basically; a line pair is a black and white line, a line would just count black lines.
The ‘per mm’ part is mm at the image plane. (The camera sensor is at the image plane). If everything were perfect, the test target would show X number of lines across each mm of the image plane. Thick lines, like the first example, might be 10 lines per mm. Thinner lines, like the second example, might be 40 lines per mm.
The lines per mm count is usually referred to as the “Frequency” or “Spatial Frequency.” So now we have two things we’re looking at: the MTF, which is a measure of contrast, and the Frequency, which is how small the lines were that we used to measure the contrast.
At a single point on the lens (right at the center, for example) I can measure the MTF at different frequencies (line pairs) and make a nice graph of the results. Remember, this isn’t across the entire front of the lens like the very first graph. This is measuring one point on the lens at different frequencies (smaller and smaller lines). Every lens, like the graph below, has lower MTF as the lines get smaller (the frequency gets higher).
If you look at the left side of the graph, the MTF (contrast) measurement, it’s a simple matter to find the MTF50 (or MTF 10 or MTF90 for that matter). It’s simply the frequency (lines per mm) at which the image retain 50% of the test target’s contrast. (Or 10% or 90% of original contrast form MTF10 or MTF90).
In this example, the MTF50 is about 40 LP/mm. When I (and most current reviewers) report the MTF 50, this is what we’re showing you. Most of us use LP/Image Height rather than per mm, so we just multiply LP/mm by 24 (since a full-frame sensor is 24mm high). So in one of my standard reports for this lens I’d say the MTF50 was 960 LP/IH, which is quite good. But my reports wouldn’t tell you the MTF10 is about 77 LP/mm, or the MTF90 (which doesn’t have a gray line in the image above) is about 22 LP/mm.
Also, remember the Frequency graph is for just one point on the camera lens. When we post a set of bar graphs showing center, average, and corner MTF50, we’re showing you the MTF50 at different locations.
So is the MTF50 the Most Important Number?
Why do we lens testers give you the MTF50 numbers? Well, the first reason is it’s the default reading in Imatest software, so it’s simple and easy. There’s also the fact that most of the lens testers use it and people like to be able to compare results from different testers, so there’s kind of an MTF50 gentleman’s agreement going on. Plus, 617 different graphs showing MTF everything at lens locations Everywhere just cause an article to be confusing and chaotic.
But the truth is that MTF50 is probably the most important overall MTF number when evaluating a lens. MTF50 has been shown in numerous studies to be the point where humans perceive an image to be “sharp” rather than blurry. That makes sense – it’s basically where the contrast is greater than 50%.
But MTF50 is not the only important number. Those other frequencies give us different, but important, information. Lower frequencies, like MTF 80 to 90, show how “contrasty” an image is. If you photograph large, bold structures, this area of the frequency curve may be more important to you than the MTF50.
Numbers like the MTF10 or MTF5 are the absolute resolution limit of the lens. They show what the smallest detail is that the lens can possibly resolve. Anything smaller is just smooth gray blur. Trained human observers and image enhancement programs can actually make out some details at MTF5 in a photograph, but most of us need MTF10. Landscape and macro photographers trying to get the most detail in their large prints might consider MTF10 to be more important than the MTF50.
For example, here is a frequency graph of another lens that has exactly the same MTF50 as the lens above, 40 LP /mm. However, the MTF10 is lower (about 69 LP/mm compared to 77) and the MTF90 is higher (30 LP/mm compared to 22). So this lens should be more contrasty but not resolve tiny detail as well as our first lens.
Some lenses with very high resolution have rather poor contrast. Other lenses have very good contrast, but their resolution is poor. Stated another way, some lense have very high MTF10, but poor MTF80. Other lenses have great MTF50 and MTF30, but low MTF10.
One Point
It’s really important to emphasize that the MTF10 in these frequency charts is completely different than the MTF at 10 lines/mm in the MTF map of a lens I show at the very top of this post. The MTF at 10 lines/mm (in the MTF map at the top of the article) is showing you how much contrast thick lines have at various positions as you go from the center of the lens to the edge. The MTF50 in the frequency graph is showing you how small the lines can be that still retain 50% of their original contrast.
### So Why Am I Bringing This Up Now?
Because it’s time to improve. Like everyone else we’ve been showing MTF50 results from Imatest shot at fairly close range charts. That gives us a number for the whole system (camera and lens) in a map across the front of the lens.
At LensRentals, we’ll soon be adding optical bench results to our Imatest results when we test lenses. This will allow us to look at several other things: the performance of just the lens rather than the lens-camera sensor combination, the performance at infinity focusing distances rather than close up, and MTF at different frequencies among them. We’ll even be able to create front element MTF maps of the lenses we test, and compare those to the computer generated MTF map of the manufacturer. That should be fun and interesting. (Well, it will be fun and interesting for us. I’m guessing maybe some manufacturers won’t be so crazy about it.)
When we test lenses we’ll continue to show you the same MTF50 data that we’ve been showing and that everyone else shows. It’s important data and we already have a huge database of MTF50 data for lots of lenses so it’s great for comparison. We’ll also be adding Frequency MTF charts for the center of the lens, and probably for one or two points off-center. We don’t have a big database of this information yet, but I’ll be able to give you some comparisons to one or two similar lenses.
Since we’ll be showing things besides simply the MTF50 map of the front element, I wanted to show what those other values meant and why that is worthwhile information. The bottom line is we’ll be able to give a more thorough evaluation of lenses than we’ve ever been able to do. | 2,956 | 12,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-21 | latest | en | 0.943129 |
https://couponsanddiscouts.com/what-is-an-appropriate-discount-rate-today/ | 1,632,433,062,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00657.warc.gz | 221,461,734 | 8,196 | # What Is An Appropriate Discount Rate Today
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### Discount Rate Definition - Investopedia
Discount Rate: The discount rate is the interest rate charged to commercial banks and other depository institutions for loans received from the Federal Reserve's discount window.
https://www.investopedia.com/terms/d/discountrate.asp
### What is the Discount Rate and Why Does It Matter? - …
Jan 13, 2021 · Businesses considering investments will use the cost of borrowing today to figure out the discount rate, For example, \$200 invested against a 15% interest rate will grow to \$230. Working backwards, \$230 of future value discounted by 15% is worth \$200 today. This is helpful if you want to invest today, but need a certain amount later.
https://smartasset.com/investing/discount-rate
### Choosing a Discount Rate - Impact DataSource
Sep 12, 2011 · The discount rate represents the decision maker’s patience – the lower the discount rate the more patient one is, the higher the discount rate the more impatient. We recently evaluated energy efficiency investments which included significant upfront costs and incremental savings each year during the 20-year life of the efficiency measure.
https://impactdatasource.com/choosing-a-discount-rate/
### What Discount Rate To Use To Determine The Present Value
Oct 28, 2015 · You have to discount the future money by an appropriate value to translate it into today's value. How much you discount it by can vary. You could, for example, use a "risk-free" rate of return ...
https://seekingalpha.com/article/3610726-what-discount-rate-to-use-to-determine-present-value-of-stock
### What You Should Know About the Discount Rate
Sep 02, 2014 · The discount rate is the rate of return used in a discounted cash flow analysis to determine the present value of future cash flows. In a discounted cash flow analysis, the sum of all future cash flows (C) over some holding period (N), is discounted back to the present using a rate of return (r). This rate of return (r) in the above formula is ...
https://propertymetrics.com/blog/npv-discount-rate/
### A Quick Guide to the Risk-Adjusted Discount Rate
May 31, 2021 · For this reason, the discount rate is adjusted to 8%, meaning that the company believes a project with a similar risk profile will yield an 8% return. The present value interest factor is now ( (1 ...
### Discount Rates For Social Security Or Pension Decisions
Jul 19, 2017 · Determining An Appropriate Discount Rate Of Interest For Financial Planning Strategies. While it may seem an abstract exercise, in reality determining an appropriate “discount rate” is actually highly relevant when evaluating many financial planning strategies, particularly ones that compare traditional investment opportunities with fixed cash flows over time, such as whether to take a ...
https://www.kitces.com/blog/net-present-value-discount-rate-formula-retirement-plan-pension-lump-sum-or-social-security-breakeven/
### Discount and capitalization rates in business valuations
The discount rate of 25% is the required rate of return. The terminal value is calculated by using the constant-growth model to capitalize year six income. Income is expected to grow indefinitely at 5% after year five. In valuation theory, a discount rate represents the required rate of return in an asset-valuation model.
http://archives.cpajournal.com/old/16373958.htm
### Considering an alimony buyout? Here's what you need to know.
The periodic payments are discounted back to the present value in today's dollars. To determine the present value, both parties must agree on the appropriate discount rate. Also, keep in mind that periodic payments could be modifiable if there is a significant change in income and/or end if …
### Discounting 101 - Resources for the Future
Jan 16, 2020 · For policy analysis, the discount rate is chosen to reflect the rate at which society is willing to trade off immediate and future values. Further, a number of specific questions and issues arise when determining the appropriate discount rate to use when valuing the benefits of …
https://www.rff.org/publications/explainers/discounting-101/
### Discount Rate Formula: Calculating Discount Rate [WACC/APV]
Aug 16, 2019 · The definition of a discount rate depends the context, it's either defined as the interest rate used to calculate net present value or the interest rate charged by the Federal Reserve Bank. There are two discount rate formulas you can use to calculate discount rate, WACC (weighted average cost of capital) and APV (adjusted present value).
https://www.profitwell.com/recur/all/discount-rate-formula
### FIN 3104 Flashcards | Quizlet
Assume that the first payment is received today. Use a discount rate of 12%, and round your answer to the nearest \$10. What is the present value of \$150 received at the beginning of each year for 16 years? The first payment is received today. Use a discount rate of 9%, and round your answer to the nearest \$10.
https://quizlet.com/530512852/fin-3104-flash-cards/
### Solved: As A Winner Of A Lottery You Are Going To Receive
As a winner of a lottery you are going to receive \$10,000 every year forever, starting one year from today. If the appropriate discount rate is 10%, what is the present value of the award cash flows? \$40,000 \$20,000 80,000 \$100,000 \$125,000 QUESTION 60 In the above question, suppose your prize is growing at a constant rate … | 1,205 | 5,484 | {"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 | 0.886782 |
https://bigwiggames.com/hmdjtu0/exponential-reliability-function-1636ac | 1,619,002,870,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039536858.83/warc/CC-MAIN-20210421100029-20210421130029-00057.warc.gz | 237,974,041 | 13,338 | ## exponential reliability function
As an example, the first term learned by most people when they are introduced to reliability is MTBF (mean time between failures). 2.8). With the ELS approach to inside dependencies of parallel systems with homogeneous components, we find analytical solutions of their reliability characteristics.Proposition 3.2.1If, in a homogeneous and aging parallel system following the ELS dependency rule, the components have piecewise exponential reliability functions with the coordinates (3.1.55), then the system’s reliability function is(3.2.4)RELSt⋅=1RELSt1…RELStz,t≥0, If, in a homogeneous and aging parallel system following the ELS dependency rule, the components have piecewise exponential reliability functions with the coordinates (3.1.55), then the system’s reliability function is, If, in a homogeneous and aging parallel system following the ELS dependency rule, the components have piecewise exponential reliability functions with the coordinates (3.1.55), then the system’s lifetime in the reliability state subset {u, u + 1, …, z}, u = 1, 2, …, z, exhibits an Erlang distribution with the shape parameter n and the intensity parameter nλ(u)/c(u) and the system’s distribution function is given by the vector. The distribution is called "memoryless," meaning that the calculated reliability for say, a 10 hour mission, is the same for a subsequent 10 hour mission, given that the system is working properly at the start of each mission. It has a fairly simple mathematical form, which makes it fairly easy to manipulate. The subsystem S1 reliability structure. The exponential hazard function is shown in figure the figure below. Therefore, now we can formulate the following theorem. Discrete versus continuous reliability analysis. The graphs of the exemplary system S reliability function coordinates. The components of subsystem S3 have the identical, by the assumption, piecewise exponential reliability functions, Then, the reliability function of subsystem S3, according to (2.3.17)–(2.3.18), is given by, Taking into account the reliability structure of the system S, presented in Fig. For the multistate exponential reliability function, From Theorem 12.2 it follows that the probability distributions for the random variables Tk, k = 1, 2, …, n-1 are a mixture of discrete and absolutely continuous distributions, From the above-mentioned theorem, it follows that, This means that a sequence of state changes (n,n−1,….,1,0) with waiting times (Tn>0,Tn−1=0,….,T1=0) is possible. In this section, we analyze a multistate series-“m out of k” system, described in Section 3.1.4, as an “m out of k” system composed of k series subsystems with dependent according to the LLS rule components. 3. The exponential distribution is the only distribution to have a constant failure rate. we get following values of the mean lifetimes of this system in the reliability state subsets {1,2}, {2}: Similarly, applying (2.1.18) and using (2.4.19)–(2.4.20), and considering the formula. Next, the results are presented in the form of tables containing exact algorithms of the procedure while evaluating reliability characteristics of these systems’ reliability in order to provide the reliability practitioners with a simple and convenient tool for everyday practice. The exponential-logarithmic distribution has applications in reliability theory in the context of devices or organisms that improve with age, due to hardening or immunity. Here we look at the exponential distribution only, as this is the simplest and the most widely applicable. In some cases, parameter position (γ) may represent a guaranteed time during which no equipment failures are expected; in other words, 100% reliability until time t = γ. Whenever the exponential reliability function is applied to calculate equipment, product, service, or event reliability, the main assumption is that events occur randomly over time; otherwise it … The subsystem S2 reliability structure. Remembering ‘e to the negative lambda t’ or ‘e to the negative t over theta’ will save you time during the exam. The exponential conditional reliability function is: which says that the reliability for a mission of duration undertaken after the component or equipment has already accumulated hours of operation from age zero is only a function of the mission duration, and not a function of the age at the beginning of the mission. 1.13 shows the exponential PDF (λ = 1.68; γ = 0.46), which represents a failure in the temperature alarm. However, in some cases, electrical and electronic equipment does not have random failure occurrences over time. The mean time to failure (MTTF = θ, for this case) … The parameters a and b correspond to K and l . The exponential reliability function depends only on the failure rate parameter, therefore the equation is simple. The Reliability Function for the Exponential Distribution. Reliability deals with the amount of time a product or value lasts. Fig. 5) The Hazard Function The hazard function of Exponential Power model is given by h x; , ( , )>0 and x exp x , x 0 1 (9) and the allied R function hexp.power( ) given in … ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. URL: https://www.sciencedirect.com/science/article/pii/B9780128005187000120, URL: https://www.sciencedirect.com/science/article/pii/B9780080999494000076, URL: https://www.sciencedirect.com/science/article/pii/B9780128212608000038, URL: https://www.sciencedirect.com/science/article/pii/B978012821260800004X, URL: https://www.sciencedirect.com/science/article/pii/B9780128054277000014, URL: https://www.sciencedirect.com/science/article/pii/B9780128212608000026, Semi-Markov model of system component damage, Semi-Markov Processes: Applications in System Reliability and Maintenance, Reliability of Large Multi-State Exponential Systems, Reliability of Large and Complex Systems (Second Edition), ’, series–parallel and parallel–series systems composed of components having, Reliability of aging multistate dependent systems, Multistate System Reliability with Dependencies, If, in a homogeneous and aging parallel system following the ELS dependency rule, the components have piecewise, Availability analysis of aging-dependent systems under imperfect repair, Gas and Oil Reliability Engineering (Second Edition), (2.4.3)–(2.4.4), (2.4.9)–(2.4.10), and (2.4.13)–(2.4.14). Notice that in the figure the curve begins with a range at 0.46. Fuzzy Probability Function and its Reliability This section introduce the probability density function of exponential distribution which is used commonly in reliability engineering and is used to model the behavior of units that have a constant failure rate (or units that do not degrade with time or wear out). The failure rate is constant if events occur randomly over time. Another important index is failure rate, which is obtained by dividing the PDF and reliability functions to define the failure rate, as follows: The failure rate is constant over time, as shown in Fig. The components Ei(2), i = 1,2,3,4,5, by the assumption, have piecewise exponential reliability functions, Then, applying (2.3.2)–(2.3.3), the reliability function of subsystem S2 is. The exponential distribution can be used to determine the probability that it will take a given number of trials to arrive at the first success in a Poisson distribution; i.e. Therefore, the NHPP model is a straight application of the exponential model. This means that before parameter position value (γ), equipment has 100% reliability. For example, it would not be appropriate to use the exponential distribution to model the reliability of an automobile. The probability of these sequences of events is, Finally, in this case a value of n-level time to failure is, Krzysztof Kołowrocki, in Reliability of Large and Complex Systems (Second Edition), 2014. gamma distribution. {\displaystyle S (t)=P (\ {T>t\})=\int _ {t}^ {\infty }f (u)\,du=1-F (t).} Agnieszka Blokus, in Multistate System Reliability with Dependencies, 2020. The above equation indicates that the reliability R(t) of a product under a constant rate of failure, λ, is an exponential function of time in which product reliability decreases exponentially with … The exponential distribution is actually a special case of the Weibull distribution with ß = 1. Under these assumptions, using the reliability function of an aging series-“m out of k” system with the coordinates given by (3.1.111)–(3.1.112) or by (3.1.113)–(3.1.114) in Proposition 3.1.13, the system’s mean lifetime in the state subsets {u, u + 1, …, z}, u = 1,2, …, z, is given (4.4.19), similarly as for a series-parallel system in Section 4.4.2. We care about your privacy and will not share, leak, loan or sell your personal information. The constant failure rate of the exponential distribution would require the assumption that t… In doing so it is possible to see the range of time without value, which represents the position parameter (γ = 0.46). For example, a gas compressor with many components (eg, electric motor, bearing, valve, and seal) with a compressor failure rate is comprised of different component failure rates and will result in an increased compressor failure rate and not a constant failure rate shape, as shown in Fig. In other words, reliability of a system will be high at its initial state of operation and gradually reduce to its lowest magnitude over time. Table 2.3. Whenever the exponential reliability function is applied to calculate equipment, product, service, or event reliability, the main assumption is that events occur randomly over time; otherwise it makes no sense to use it. The exponential distribution is a simple distribution with only one parameter and is commonly used to model reliability data. In wear-out failure phases, the MTTF is lower than the previous phase, and if it has been considered constant, failure will likely occur before the time expected. Exponential Distribution. Reliability math and the exponential distribution 1. By continuing, you consent to the use of cookies. So the Probability Distribution function of Exponential Distribution is reliability universe is given by R ( t) = e − λ t = e − t ╱ θ. Fig. Despite the inadequacy of the exponential distribution to accurately model the behavior of most products in the real world, it is still widely used in today’s reliability practices, standards and methods. The probability density function (pdf) of an exponential distribution is {\displaystyle f (x;\lambda)= {\begin {cases}\lambda e^ {- (\lambda x)}&x\geq 0,\\0&x<0.\end {cases}}} Here λ > 0 is the parameter of the distribution, often called the rate parameter. Chet Haibel ©2013 Hobbs Engineering Corporation Reliability Math and the Exponential Distribution 0 0 2. Figure 1.15. Limit reliability functions of multi-state series, parallel, ‘m out of n’, series–parallel and parallel–series systems composed of components having exponential reliability functions are fixed. The values of the risk function r(t) of exemplary system. In this article, a new four-parameter lifetime distribution, namely, Weibull-Linear exponential distribution is defined and studied. The components Ei(1), i = 1,2,3,4, have the identical piecewise exponential reliability functions, Then, the reliability function of subsystem S1, according to (2.3.10)–(2.3.12), is. it describes the inter-arrival times in a Poisson process.It is the continuous counterpart to the geometric distribution, and it too is memoryless.. In fact, this does not always happen, because depending on the life cycle time assessed, it will have different PDF configurations for the system's equipment. 1.15. Many specialists consider the system PDF as exponential because they believe that by regarding different PDFs for each component and equipment, the system PDF shape will be exponential. The functions for this distribution are shown in the table below. The exponential hazard function is. During this correct operation, no repair is required or performed, and the system adequately follows the defined performance specifications. R ( t) = e − λ t. Given a failure rate, lambda, we can calculate the probability of success over time, t. Cool. The exponential-logarithmic distribution arises when the rate parameter of the exponential distribution is randomized by the logarithmic distribution. View our, Using The Exponential Distribution Reliability Function, Probability and Statistics for Reliability, Discrete and continuous probability distributions. Applications The distribution is used to model events with a constant failure rate. Copyright © 2021 Elsevier B.V. or its licensors or contributors. This fact influences decisions because the MTTF cannot be constant over time if failure is not represented by the exponential PDF, which means failures are not random. 2.9. The reliability function coordinates of the exemplary system S are illustrated in Fig. We assume its components Eij, i = 1,2, …, k, j = 1,2, …, li, have piecewise exponential reliability functions given by (4.4.17)–(4.4.18). Cookies Policy, Rooted in Reliability: The Plant Performance Podcast, Product Development and Process Improvement, Musings on Reliability and Maintenance Topics, Equipment Risk and Reliability in Downhole Applications, Innovative Thinking in Reliability and Durability, 14 Ways to Acquire Reliability Engineering Knowledge, Reliability Analysis Methods online course, Reliability Centered Maintenance (RCM) Online Course, Root Cause Analysis and the 8D Corrective Action Process course, 5-day Reliability Green Belt ® Live Course, 5-day Reliability Black Belt ® Live Course, This site uses cookies to give you a better experience, analyze site traffic, and gain insight to products or offers that may interest you. The method of using the algorithms is illustrated by several examples. where the reliability function coordinates are given by (2.4.19)–(2.4.20). Definition 5.2 A continuous random variable X with probability density function f(x)=λe−λxx >0 for some real constant λ >0 is an exponential(λ)random variable. In exponential distribution, the reliability function can be calculated by differentiating the cumulative distribution function. 2.8. The exponential PDF represents a random occurrence over time and best represents electronic, electrical, or random events. Franciszek Grabski, in Semi-Markov Processes: Applications in System Reliability and Maintenance, 2015, We suppose that on y the state changes from k to k − 1, k = 1, 2, …, n, are possib e with the positive probabilities. This function gives the probability of an item operating for a certain amount of time without failure. The exponential reliability function is. Reliability is the probability that a system performs correctly during a specific time duration. Also known as the probability density function (pdf), this function is integrated to obtain the probability that the failure time takes a value in a given time interval. Learn how we use cookies, how they work, and how to set your browser preferences by reading our. The subsystem S3 is a homogeneous “3 out of 5” system consisting of five components (Fig. The scheme of the exemplary series system reliability structure. 2.7. In a gas compressor there are components with increased failure rates, such as the seal and bearing, constant failure rates, such as the electric motor, and decreased failure rates, such as the gas valve. The reliability function is defined as the probability of failure-free operation until time . We use cookies to help provide and enhance our service and tailor content and ads. Let $$F^c = 1 - F$$ denote the denote the right-tail distribution function of $$X$$ (also known as the reliability function), so that $$F^c(t) = \P(X \gt t)$$ for $$t \ge 0$$. The following section describes the normal PDF, which is used in many cases by maintenance and reliability specialists. As seen, m ( t ) and l ( t ) are the cumulative distribution function [ F ( t) ] and the probability density function [ f ( t) ], respectively, of the exponential function discussed in the preceding section. The most frequently used function in life data analysis and reliability engineering is the reliability function. 2.12. Also, another name for the exponential mean is the Mean Time To Failor MTTFand we have MTTF = $$1/\lambda$$. The failure rate was calculated based on the PDF and reliability function of Fig. A CDF of a waiting time in state k for the kernel (12.40) is equal to a function Qkk-1 (t), for k = 1,2,…, n. Applying the results (12.42) and (12.43), we get. The primary trait of the exponential distribution is that it is used for modeling the behavior of items with a constant failure rate. Gas compressor and component failure rates. The cumulative hazard function for the exponential is just the integral of After understanding the exponential PDF it is necessary to define the reliability function, the cumulative density function (CDF), and then the failure rate and MTTF as follows: The exponential reliability function depends only on the failure rate parameter, therefore the equation is simple. Your email address will not be published. The exponential PDF equation is: Fig. Its reliability structure is given in Fig. 1.15. The subsystem S2 is a series of nonhomogeneous systems composed of five components (Fig. In the article Random failure and the MTTF we showed that the equation describing the exponential decay of Reliability (random failure) is: . The exponential probability distribution function is widely used in the field of reliability. Reliability follows an exponential failure law, which means that it reduces as the time duration considered for reliability calculations elapses. To further explain reliability engineering concepts we will begin with the exponential PDF because of its simple mathematics compared to other PDFs. The multistate reliability function is called exponential if all its components (except of Rn[0] (t)) are exponential functions [52]. for t > 0, where λ is the hazard (failure) rate, and the reliability function is. The R function sexp.power( ) given in SoftreliaR package computes the reliability/ survival function. Fig. 2.11. Like all distributions, the exponential has probability density, cumulative density, reliability and hazard functions. 2.7. A mathematical model that describes the probability of failures occurring over time. The risk function of the system S is illustrated in Fig. As such, the reliability function is a function of time, in that every reliability value has an associated time value. 2.9). From (12.17), we obtain the Laplace transforms of the multistate reliability function components. Example: A resistor has a constant failure rate of 0.04 per hour. When there is a position parameter, it is represented in the PDF equation by: This means that failure occurs randomly after a period of time and that it is observed in some electrical equipment. 2.11. Uses of the exponential distribution to model reliability data. of a semi-Markov kernel for which the system of equations (12.41) is fulfilled. 2.7, and applying (2.3.2)–(2.3.3), the reliability function of system S is given by, And consequently, using the results (2.4.3)–(2.4.4), (2.4.9)–(2.4.10), and (2.4.13)–(2.4.14), it takes following form. The graph of the risk function r(t) of exemplary system S. To find the moment of exceeding an acceptable level, for instance δ = 0.05, we determine the values of the system risk function, given in Table 2.3. The exponential distribution provides a good model for the phase of a product or item's life when it is just as likely to fail at … 2.10. Thus, if the random variable (rv) denotes the lifetime of an item, then . Many reliability and maintenance professionals incorrectly consider the MTTF the inverse of the failure rate when the PDF is not exponential. Fig. The tables are composed of three parts, containing reliability data of the evaluated system, necessary calculations and results of the system reliability evaluation. In reliability, since we deal with failure times, and times are non-negative values, the lower bound of our functions starts with 0 rather than -∞. What is the resistor's reliability at 100 hours? A common formula that you should pretty much just know by heart, for the exam is the exponential distribution’s reliability function. The distribution is supported on the interval [0, ∞). To calculate the MTTF applying the following equation, it is possible to see that the MTTF is the inverse of the failure rate in the exponential PDF case: This happens only for the exponential PDF. Using the result of Corollary 3.2.1, we determine the reliability characteristics, the mean values and standard deviations of the system lifetimes in the reliability state subsets, for a parallel system following the ELS dependency rule.Corollary 3.2.2If, in a homogeneous and aging parallel system following the ELS dependency rule, the components have piecewise exponential reliability functions with the coordinates (3.1.55), then the system’s mean lifetime in the reliability state subset {u, u + 1, …, z}, u = 1, 2, …, z, is given by(3.2.8)μELSu=cuλu,u=1,2,…,z, If, in a homogeneous and aging parallel system following the ELS dependency rule, the components have piecewise exponential reliability functions with the coordinates (3.1.55), then the system’s mean lifetime in the reliability state subset {u, u + 1, …, z}, u = 1, 2, …, z, is given by, and the standard deviation of the system lifetime is. Fig. Reliability Prediction Using the Exponential Distribution The exponential distribution applies when the failure rate is constant - the graph … Its survival function or reliability function is: S ( t ) = P ( { T > t } ) = ∫ t ∞ f ( u ) d u = 1 − F ( t ) . The Exponential is a life distribution used in reliability engineering for the analysis of events with a constant failure rate. And then, substituting (2.4.23)–(2.4.24) and (2.4.26)–(2.4.27) into (2.1.17), we determine the standard deviations of the exemplary system S lifetimes: The mean values of the exemplary system lifetimes in the particular reliability states 1,2, by (2.1.20) and using (2.4.23)–(2.4.24), are: Assuming that the critical reliability state is r = 1 and applying (2.1.21), we obtain the risk function of the system S. where the reliability function coordinate R(t, 1) is given by (2.4.19). 2.12. We consider a 3-state (z = 2) series system composed of three subsystems S1, S2, S3. The subsystem S3 reliability structure. The failure density function is. This means the position parameter (γ) represents how long one piece of equipment operates without failure; in other words, how long one piece of equipment has 100% reliability. for any time. 2.10). The case where μ = 0 and β = 1 is called the standard exponential distribution. Then, we find that the risk exceeds a permitted level δ = 0.05 for t = 1.516, and by (2.1.22) we conclude that. Another measure of reliability under stress-strength setup is the probability , which represents the reliability of an … Abstract: This paper considers a class of an efficient 'two-stage shrinkage testimator' (TSST) of 'reliability function' of 'exponential distribution', and the class uses additional information which can be obtained from the past practices, and in the form of past initial … The table below example, it would not be appropriate to use the exponential distribution is for. 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Subsystem S1 is a function of time without failure and parameter conditions are true 2021 exponential reliability function B.V. or licensors. And l, cumulative density, cumulative density, cumulative density, reliability and hazard functions example it! ” system consisting of five components ( Fig exponential reliability function coordinates Corporation reliability and... Coordinates of the exemplary series system reliability structure S3 exponential reliability function a series of nonhomogeneous systems composed of five components Fig. 1.13 shows the exponential distribution is often used to model events with a constant failure rate when PDF! Electrical, or random events a homogeneous “ 3 out of 5 ” system consisting of four components (.... Electronic equipment does not have random failure occurrences over time of Fig every value. 100 % reliability random failure occurrences over time = 1 resistor has a constant failure rate when the failure when! 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## SUM TOTAL by expression values
Hello,
is it possible to sum up values by the output of another expression using the TOTAL qualifier? Example:
Column C is an calculated expression based on column B. I would like to aggregate the amount (column D) by the output of column C. The desired output is displayed in column E. In Qlikview, I am trying to achieve this result using the function TOTAL:
However, I receive the following error: "Total specifier must be a valid field name" .
Is there another way to solve this problem? Please find attached the qvw containing the example.
Please also note that this is an simplified example and in reality the currency calculation is very complex containing several IF-statements so it is not possible to use the whole calculation within the total function (total<original expression>).
Many thanks,
Paul
1 Solution
Accepted Solutions
Luminary Alumni
Hi Paul,
First, I would assume that TOTAL qualified (like set analysis) is not evaluated on row level, but on chart level, so I guess you're not able to make it dynamic using conditions. Second, it expects field name as a input. Your variable returns currency code (which is not a field name). I remember trying to use table column in TOTAL without success and I rewrote the expression using creating separate Sum(TOTAL) for each scenario, summing them up by RangeSum() (I might have used Aggr() as well, don't rememnber, it was some time ago).
So, my recomendation is to calculate an extra field in script, which would contain proper currency for this aggregation, and you don't need to do this kind of gymnastics.
Hope this helps.
Juraj
7 Replies
MVP
Try this
If(Match(Customer,'A','B','C'), Sum(TOTAL <Currency> {<Customer = {'A', 'B', 'C'}>}Amount))
Try this?
If(Match(Customer,'A','B','C'), Sum(TOTAL <Currency> {<Customer = {'A', 'B', 'C'}>}Amount))
Best Anil, When applicable please mark the correct/appropriate replies as "solution" (you can mark up to 3 "solutions". Please LIKE threads if the provided solution is helpful
Anonymous
Not applicable
Author
Thank you!
My currency calculation is more complex, so I have updated the example. Outcome should be as follows (green column):
Luminary Alumni
Hi Paul,
First, I would assume that TOTAL qualified (like set analysis) is not evaluated on row level, but on chart level, so I guess you're not able to make it dynamic using conditions. Second, it expects field name as a input. Your variable returns currency code (which is not a field name). I remember trying to use table column in TOTAL without success and I rewrote the expression using creating separate Sum(TOTAL) for each scenario, summing them up by RangeSum() (I might have used Aggr() as well, don't rememnber, it was some time ago).
So, my recomendation is to calculate an extra field in script, which would contain proper currency for this aggregation, and you don't need to do this kind of gymnastics.
Hope this helps.
Juraj
MVP
I agree with juraj.misina, if this is something you can do in the script (creating currency_cal in the script), then you will save yourself from a very complicated expression (not even sure this is possible to do yet).
Anonymous
Not applicable
Author
Hi Juray,
thanks for your input - I have been afraid it's to complicated for an expression. However, it's a little odd that it's not possible to use calculated values for aggregation, since the result is available on row level.
Cheers,
Paul
Contributor III
Hi,
I have a chart with customers and I want to know the distinct number of customers they have gross amount.
For the selected applicable_date I tried following formula
Sum(TOTAL<customer>
{<
APPLICABLE_DATE -={"'-'"},
NUM_APPLICABLE_DATE = {'>=\$(=Floor(Num(Account_Date_From)))<=\$(=Floor(Num(Account_Date_To)))'}
>}
GROSS_AMOUNT)
Please note that Account_Date_From and Account_Date_To are the variables
Anyway, here is the chart
Here's an example of one of customer data. according to this Gross Amount is 270.
Customer Name Application Date Gross Amount NEWFUNDABC 27/02/2032 50 NEWFUNDABC 27/02/2032 20 NEWFUNDABC 27/02/2032 40 NEWFUNDABC 27/02/2032 30 NEWFUNDABC 04/03/2032 20 NEWFUNDABC 04/03/2032 100 NEWFUNDABC 04/03/2032 10 | 1,051 | 4,393 | {"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} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.909724 |
http://mathhelpforum.com/calculus/113662-tangency.html | 1,529,722,871,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864919.43/warc/CC-MAIN-20180623015758-20180623035758-00374.warc.gz | 197,185,696 | 9,622 | 1. ## Tangency
Another problem I'm stuck on.
"If k >/= 1 (greater than or equal to 1), the graphs of $\displaystyle y=sin(x)$ and y=ke^-x intersect for x >/= 0. Find the smallest value of k for which the graphs are tangent. What are the coordinates of the point of tangency?"
2. Hello, Rumor!
I think I have a solution.
Please check my reasoning and my work.
If $\displaystyle k \geq 1$, the graphs of: $\displaystyle y\:=\:\sin x$ and $\displaystyle y\:=\:ke^{-x}$ intersect for $\displaystyle x \geq 0.$
Find the smallest value of $\displaystyle k$ for which the graphs are tangent.
What are the coordinates of the point of tangency?
We have: .$\displaystyle \begin{array}{ccc} f(x) &=& \sin x \\ g(x) &=& ke^{-x} \end{array}$
The graphs intersect at the point of tangency.
. . Hence: .$\displaystyle f(x) \:=\:g(x) \quad\Rightarrow\quad \sin x \:=\:ke^{-x} \quad\Rightarrow\quad e^x\sin x \:=\:k$ [1]
Their slopes are equal at the point of tangency.
. . Hence: .$\displaystyle f'(x) = g'(x) \quad\Rightarrow\quad \cos x \:=\:-ke^{-x} \quad\Rightarrow\quad e^x\cos x \:=\:-k$ [2]
Divide [1] by [2]: . $\displaystyle \frac{e^x\sin x}{e^x\cos x} \:=\:\frac{k}{\text{-}k} \quad\Rightarrow\quad \tan x \:=\:-1 \quad\Rightarrow \quad x \:=\:\frac{3\pi}{4}$
We have: .$\displaystyle \begin{array}{ccccccc}f\left(\frac{3\pi}{4}\right) &=& \sin\frac{3\pi}{4} &=& \frac{1}{\sqrt{2}} \\ \\[-3mm] g\left(\frac{3\pi}{4}\right) &=& ke^{-\frac{3\pi}{4}}\end{array}$
Since $\displaystyle g\left(\tfrac{3\pi}{4}\right) \,=\,f\left(\tfrac{3\pi} {4}\right)$, we have: .$\displaystyle ke^{-\frac{3\pi}{4}} \:=\:\frac{1}{\sqrt{2}}$
Therefore: .$\displaystyle \boxed{k \:=\:\frac{e^{\frac{3\pi}{4}}}{\sqrt{2}}}$
And the point of tangency is: .$\displaystyle \left(\frac{3\pi}{4},\:\frac{1}{\sqrt{2}}\right)$
3. Ah, of course!
Your work looks sound to me. I don't know why I couldn't remember to set them equal. Thank you very much! | 693 | 1,926 | {"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-2018-26 | latest | en | 0.631553 |
https://it.overleaf.com/articles/tagged/math/page/44 | 1,561,033,136,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999210.22/warc/CC-MAIN-20190620105329-20190620131329-00465.warc.gz | 495,461,781 | 11,564 | # Articles — Math
Articles tagged Math
Show all Articles
Solving most cost-effective loan problem
A simple algorithm paper for the sake of practice.
Rodion Efemov
Calculating the Probability for Winning a 649 Lottery
Calculating the Probability for Winning a 649 Lottery using Probability.
Geometria de superfícies isentrópicas
A dinâmica topológica de inversões geométricas foi estudada em [6]. O espa ̧co de parâmetros das medidas de Markov com suporte no atrator do sistema é um aberto de R3 folheado por superfícies de nível compactas definidas pela entropia métrica: superfícies isentrópicas [7]. Neste artigo abordaremos o aspecto geométrico dessas superfícies. Em particular, classificaremos suas geodésicas e pontos umbílicos.
Deyvisson Ribeiro
Brachistochrone Problem Solved
Solving Brachistochrone Problem
Deriving the summation formula of any Converged Arithmetic Series
I have shown here how to derive the summation of a convergent Arithmetic series and get two results as answers
Edge-Disjoint Tree Realization of Tree Degree Matrices that avoid routine induction
Identifying whether a degree matrix has an edge-disjoint realization is an NP-hard problem. In comparison, identifying whether a tree degree matrix has an edge-disjoint realization is easier, but the task is still challenging. In 1975, a sufficient condition for the tree degree matrices with three rows has been found, but the condition has not been improved since. This paper contains an essential part of the proof which improves the sufficient condition.
Ian Seong
Women and mathematics at the Universities in Prague
This study is focused on lives of twelve women who prepared their doctorates in mathematics at the Faculty of Philosophy of the German University in Prague in the years 1882–1945, respectively at the Faculty of Science of the Czech University in Prague in the years 1882–1920 and 1921–1945 (known as Charles University in Prague in the latter period). In the first part, a short description of the historical background about women's studies at the universities in the Czech lands and a statistical overview of all PhD degrees in mathematics awarded at both universities in Prague is given for a better understanding of the situation with women's doctoral procedures. In the second part, a description of the successful doctoral procedures in mathematics of three women at the German University in Prague and of eight women at Charles University in Prague, as well as one unsuccessful doctoral procedure, are presented.
Martina Bečvářová
Hecke groups, linear recurrences and Kepler limits (update 2)
Computations with the the objects mentioned in the title.
Barry Brent
Tobo | 590 | 2,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-26 | longest | en | 0.7816 |
https://81018.com/questions/ | 1,679,660,798,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00144.warc.gz | 106,828,882 | 38,271 | # Is this the beginning of the universe?
###### by Bruce CamberRelated Homepages: Base-2 Model, Integrated Structure, Lemaître-to-today, Subjects-Objects, Time
The nature of Planck Time (tP) is the root question. First, tP is directly correlated with length (a Janus-face of the most-basic length) and with light. It works in much the same way Einstein’s well-known equation, e0=mc2 dynamically relates mass, energy and light.
The four Planck base units are natural units and each is defined by the most fundamental universal constants. Though difficult to envisage, it is hypothesized that each is a Janus-face of the other.
This simple math works. Our first and the most basic formula, the one that Max Planck defined in 1899, is: “Planck Length divided by Planck Time is equal to the speed of light.” Throughout all 202 notations, the calculation is always within .1% of the speed of light in a vacuum (299,792,458 meters/second). In 1899, the best guess for the speed of light was Leon Foucault 1862 calculation using rotating mirrors. It was 298,000 and the Rosa-Dorsey work in 1907 rendered 299,710,000 m/s. Planck’s base units render 2998,792,422.8 m/s. A closer figure would not be rendered until 1972.
Could these four base units (and light) be the very first moment of time? Within this model the answer is, “Yes.” Still quite open are answers to questions about how and when the other dimensionless constants come into play, especially those attributed to the essential natural of the Standard Model of Particle Physics. Given the values of the four Planck units and light, there can be no singularity per se. Yet, the dynamics describe a fundamental transformation nexus between what we understand to be finite and that which appears to be infinite (perhaps best understood as continuity, symmetry, and harmony).
Could this universe start cold? It appears to have been Lemaître’s position and de facto it became our position as well. In 1999 the limitations of the hot big bang were acknowledged by its prime movers at a conference at Cambridge University. “Come up with alternatives!” Now, twenty years later, they’ve tried, but failed to determine the first expression, the first facet, and the fundamentals that define our reality. How does it all come to be? What is the first look-and-feel of that manifestation?
Could it be a sphere? John Archibald Wheeler imagined quantum foam. Others are also suggesting a sphere and these spheres have been called planckspheres. Add light to this concrescence of Planck Time, Planck Length, Planck Mass, and Planck Charge and the rate by which these spheres are created might readily be described as an endless string.
Can cubic close packing of equal spheres (ccp) be applied at this scale? The thrust within our universe, at least within this model, has been going on right from the start. The concept is of a universe filling within infinitesimal spheres that tile-and-tessellate the universe creating multiple grids that are initially described by ccp and the Fourier transform.
Could this stacking amount to a doubling and then a series of doublings? If this simple logic and simple math is on the right path, within 202 doublings (base-2 notations), these Planck base units always encapsulate: 1) the age of the universe, 2) the size of the universe, 3) the total mass of the universe, and 4) the total energy of the universe. It is all happening right now. The universe is expanding. Exploring such a simple model has been our effort since December 2011: https://81018.com/home/
Please explore our little, horizontally-scrolled chart of numbers to get a sense of this emergence and natural inflation. The first 64 notations are the key. The URL is: https://81018.com/chart/ It is too simple, so simple it seems perhaps a bit of silliness. But if you look at the numbers, there is a sweet logic that prevails. https://81018.com/calculations
I’d be pleased to hear from you. Yes, challenge us, coach us. We need all the help we can get.
# More questions.
1. How is big bang cosmology (and base-2 natural inflation) consistent with General Relativity?
2. How does big bang cosmology (and base-2) explain the Hubble expansion of the Universe?
3. How can big bang cosmology (and base-2 natural inflation) explain the abundances of the light elements?
4. How does big bang cosmology (and base-2 natural inflation) explain CMBR, the existence and properties of the cosmic microwave background radiation?
5. How does big bang cosmology (and base-2 natural inflation) account for what is called the horizon problem whereby photons are roughly the same temperature — 2.725 degrees Kelvin — wherever you look in the universe?
6. How does big bang cosmology (and base-2 natural inflation) account for what is called Flatness Problem? Almost all the evidence collected by cosmologists indicates that the Universe is flat. Like a sheet of paper on a desk, spacetime shows almost no curvature whatsoever.
7. What about the magnetic monopole? All magnets have two poles, a north and a south. Even when a magnet is snapped in half the two poles remain. But this particle would effectively be a magnet with only one pole: a magnetic monopole!
8. Why does the universe have more matter than antimatter?[98]
9. Why is the triangular lattice universally optimal? One needs only to look at the simple ccp dynamic gif, the Fourier transform and the first group of foundational notations as understood within a model that begins with Planck’s base units:
Personal history:
Ruminations:
Us
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,265 | 5,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-14 | latest | en | 0.92678 |
https://examfear.com/notes/Class-12/Maths/Application-of-Integrals/3512/Area-between-Two-Curves.htm | 1,620,991,683,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00608.warc.gz | 257,266,598 | 7,059 | Class 12 Maths Application of Integrals Area between Two Curves
Area between Two Curves
The most general class of problems in calculating the area of bounded regions is the one involving the regions between two curves. Let we are given two curves represented by y = f(x), y = g(x), where f(x) ≥ g(x) in [a, b] as shown in figure.
Here the points of intersection of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two curves.
Elementary strip has height f(x) – g(x) and width dx so that the elementary area
dA = [f(x) – g(x)] dx
Total area A can be calculated as A = ab [f(x) – g(x)]
Alternatively,
A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]
– [area bounded by y = g (x), x-axis and the lines x = a, x = b]
A = ab f(x) = ab g(x) = ab [f(x) – g(x)], where f(x) ≥ g(x) in [a, b]
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the figure, then the area of the regions bounded by curves can be calculated as
Total Area = Area of the region ACBDA + Area of the region BPRQB.
= ac [f(x) – g(x)] + cb [g(x) – f(x)]
Example:
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y.
Solution:
The required area is represented by the shaded area OBCDO.
Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we get the point of intersection as B(√2, 1/2) and D(-√2, 1/2).
It can be observed that the required area is symmetrical about y-axis.
So, Area OBCDO = 2 * Area OBCO
Now, draw BM perpendicular to OA.
Therefore, the coordinates of M is (√2, 0).
Therefore, Area OBCO = Area OMBCO – Area OMBO
= ʃ0√2 {√(9 – 4x2)/4} dx - ʃ0√2 x2/4 dx
= (1/2)ʃ0√2 √(9 – 4x2) dx – (1/4)ʃ0√2 x2 dx
= (1/4)[x√(9 – 4x2) + (9/2) * sin-1 2x/3]0√2 - (1/4)[x3/3]0√2
= (1/4)[√2 * √(9 – 8) + (9/2) * sin-1 2√2/3] - (1/12)(√2)3
= (√2/4) + (9/8) * sin-1 2√2/3 - √2/6
= (√2/12) + (9/8) * sin-1 2√2/3
= (1/2)[√2/6 + (9/4) * sin-1 2√2/3]
Therefore, the required area OBCDO = 2 * (1/2)[√2/6 + (9/4) * sin-1 2√2/3]
= [√2/6 + (9/4) * sin-1 2√2/3] units
Example:
Find the area between the curves y = x and y = x2
Solution:
The required area is represented by the shaded area OBAO as
The points of intersection of the curves, y = x and y = x2 is A (1, 1).
Now, draw AC perpendicular to x-axis.
So, Area (OBAO) = Area (∆OCA) – Area (OCABO)
= ʃ01 x dx - ʃ01 x2 dx
= [x2/2]01 – [x3/3]01
= 1/2 - 1/3
= 1/6 units
. | 983 | 2,483 | {"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} | 4.875 | 5 | CC-MAIN-2021-21 | latest | en | 0.905061 |
https://calculatorshub.net/mechanical-calculators/compressed-air-dryer-sizing-calculation/ | 1,723,285,095,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00013.warc.gz | 121,911,909 | 26,895 | Home » Simplify your calculations with ease. » Mechanical Calculators » Compressed Air Dryer Sizing Calculator Online
# Compressed Air Dryer Sizing Calculator Online
## Introduction
When it comes to the realm of pneumatic systems, sizing a compressed air dryer correctly is crucial. This process ensures the efficiency of system operations and prevents undue expenditures. In this article, we delve into the intricacies of compressed air dryer sizing calculations to enable optimization of your systems for peak performance.
## Definition
A compressed air dryer is a device engineered to eradicate water vapor present in compressed air systems. The sizing calculation, on the other hand, pertains to determining the precise capacity of the dryer necessary to handle a particular amount of airflow, typically quantified in cubic feet per minute (CFM). The sizing of the air dryer ensures its ability to effectively remove moisture, even under varied conditions.
## Working Explanation
The compressed air dryer sizing calculator employs a straightforward formula to ascertain the minimum dryer rating essential for a specific compressor output. By multiplying the compressor output (in CFM) by a factor of 1.2, we derive the required rating for the dryer. This 1.2 multiplier is generally accepted as a standard due to the fluctuation in operational conditions such as temperature, pressure, and specific humidity levels.
## Formula and Variable Description
Here's the formula used:
Dryer Rating (in CFM) = Compressor Output (in CFM) * 1.2
Where:
• Dryer Rating is the minimum capacity the dryer should have, measured in CFM.
• Compressor Output is the quantity of compressed air generated by the compressor, measured in CFM.
## Example
For instance, if your air compressor has an output of 500 CFM, the calculation for the minimum dryer rating would be:
Dryer Rating = 500 CFM * 1.2 = 600 CFM
In this case, you would need a dryer rated at a minimum of 600 CFM to handle the output of your air compressor effectively.
## Conclusion
Understanding the compressed air dryer sizing calculation is crucial for maintaining the efficiency of your compressed air system. By using the right dryer size, you can ensure optimal performance and prevent potential problems associated with moisture in compressed air. Always remember to account for changes in operating conditions and consult with experts if necessary to ensure you have the right equipment for your specific needs. | 487 | 2,484 | {"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} | 3.859375 | 4 | CC-MAIN-2024-33 | latest | en | 0.879398 |
https://www.splashmath.com/math-vocabulary/number-sense/number-system | 1,571,449,926,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986688674.52/warc/CC-MAIN-20191019013909-20191019041409-00328.warc.gz | 1,097,010,682 | 30,590 | # Number System - Definition with Examples
The Complete K-5 Math Learning Program Built for Your Child
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## Number System
### Decimal Number System:
The decimal number system consists of 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 and is the most commonly used number system. We use the combination of these 10 digits to form all other numbers. The value of a digit in a number depends upon its position in the number. The place value table for the decimal number system is as:
Each place to the left is ten times greater than the place to its right, that is, as we move from the right to the left, the place value increases ten times with each place.
• A decimal number system is also called the Base 10 system.
• A number 49,365 is read as Forty-nine thousand three hundred sixty-five, where the value of 4 is forty thousand, 9 is nine thousand, 3 is three hundred, 6 is sixty and 5 is five.
### Binary Number System
In the binary number system, we only use two digits 0 and 1. It means a 2 number system.
Example of binary numbers: 1011; 101010; 1101101
Each digit in a binary number is called a bit. So, a binary number 101 has 3 bits. 499787080
Computers and other digital devices use the binary system. The binary number system uses Base 2.
The word hexadecimal comes from Hexa meaning 6, and decimal meaning 10. So, in a hexadecimal number system, there are 16 digits. It consists of digits 0 to 9 and then has first 5 letters of the alphabet as:
The table below shows numbers 1 to 20 using decimal, binary and hexadecimal numbers.
Decimal Binary Hexadecimal 0 0 0 1 1 1 2 10 2 3 11 3 4 100 4 5 101 5 6 110 6 7 111 7 8 1000 8 9 1001 9 10 1010 10 11 1011 A 12 1100 B 13 1101 C 14 1110 D 15 1111 E 16 10000 F 17 10001 11 18 10010 12 19 10011 13 20 10100 14
Fun Facts The decimal number system is also called the the Hindu–Arabic numeral system. Anthropologists hypothesize that the decimal system was the most commonly used number system, due to humans having five fingers on each hand, and ten in both.
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# M251Hfinal_mockup01(sp09) - 5(15 points Solve the heat...
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MATH251H Practice Exam I Spring 2008 120 minutes 1. (15 points) Consider the linear system, x = p 0 2 - 8 0 P x . a. Find the general real-valued solutions. b. Classify the type and stability of the critical point (0,0). 1
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2. (15 points) Consider the linear system, x = p - 1 2 3 4 P x . If x (0) = p 2 a P , and, lim t + x ( t ) = 0 , what is the value of a ?
3. (20 points) Consider the following nonlinear system: b x = x 2 + y 2 - 10 y = 2 x - 6 y a. Find all the critical points. b. For each critical point, linearize the system. What conclusions can you draw about the type and stability of the critical points of the nonlinear system?
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4. (10 points) Use the method of separation of variables to reduce the following partial diFerential equation to two ordinary diFerential equations, u xx - u tx + 5 x 3 u t = 0 . Do not solve the ordinary diFerential equations.
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Unformatted text preview: 5. (15 points) Solve the heat conduction equation, u t = 4 u xx , ≤ x ≤ 3 , u (0 , t ) = 0 u (3 , t ) = 0 u ( x, 0) = sin( 2 π 3 x )-sin( πx ) + 7 sin( 5 π 3 x ) 6. (10 points) Solve the eigenvalue problem, X ′′ = λX, ≤ x ≤ π, with boundary conditions, X ′ (0) = 0 , X ′ ( π ) = 0 . Find all eigenvalues and eigenfunctions. 7. (15 points) Solve the wave equation, u tt = 9 u xx , ≤ x ≤ 4 , u (0 , t ) = 0 u (3 , t ) = 0 u ( x, 0) = 0 u t ( x, t ) = g ( x ) , where g ( x ) is given by, g ( x ) = b 1 , 2 ≤ x ≤ 3 , , otherwise ....
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M251Hfinal_mockup01(sp09) - 5(15 points Solve the heat...
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Ask a homework question - tutors are online | 693 | 2,225 | {"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} | 3.484375 | 3 | CC-MAIN-2017-51 | latest | en | 0.758119 |
http://www.mpiwg-berlin.mpg.de/Galileo_Prototype/OVERVIEW/OVERVIE7.HTM | 1,508,601,354,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824820.28/warc/CC-MAIN-20171021152723-20171021172723-00718.warc.gz | 519,537,822 | 3,063 | Folios 151 to 170
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Folio 151 Folio page 151r Written by Galileo; contains drawings, text, tables. Relation to the Discorsi: elaboration of 2/06-th-06. Folio page 151v Contains drawings, calculation. Folio page 151r Folio page 151v Folio 152 Folio page 152r Written by Galileo; contains drawings, tables, short texts. Relation to the Discorsi: work on 2/02-th-02; related to 2/00-th-00-dialog2; related to 2/00-th-00-dialog1. Folio page 152v Contains drawings. Folio page 152r Folio page 152v Folio 153 Folio page 153r Contains calculations, minimal text. Folio page 153v Contains drawings, calculation. Folio page 153r Folio page 153v Folio 154 Folio page 154r Contains drawings, calculations, minimal texts, tides. Relation to the Discorsi: related to pf. Folio page 154v Contains calculations. Folio page 154r Folio page 154v Folio 155 Folio page 155r Contains drawing, calculations. Folio page 155v Empty page. Folio page 155r Folio page 155v Folio 156The folio is part of a larger sheet of paper combining the folios 156 and 157 Folio page 156r Contains calculations, drawings, minimal text. Folio page 156v Contains drawings, table, minimal text. Folio page 157r Contains drawings. Folio page 157v Contains drawings, short texts, table. Relation to the Discorsi: work on 2/36-th-22. Folio page 157v and Folio page 156r Folio page 156v and Folio page 157r Folio 157See folio 156 --- --- Folio 158The folio is part of a larger sheet of paper combining the folios 158 and 159 (?). Folio page 158r Contains drawings, short text. Folio page 158v The folio is part of a larger sheet of paper combining the folios 158 and 159 (?). Empty page. Folio page 159r The folio is part of a larger sheet of paper combining the folios 158 and 159 (?). Empty page. Folio page 159v The folio is part of a larger sheet of paper combining the folios 158 and 159 (?). Contains drawings, table, calculations. Folio page 159v and Folio page 158r Folio page 158v and Folio page 159r Folio 159See folio 158 --- --- Folio 160 Folio page 160r Written by Galileo; contains text, drawing. Relation to the Discorsi: incomplete draft (proof) of 2/06-th-06. Folio page 160v Empty page. Folio page 160r Folio page 160v Folio 161 Folio page 161r Written by Galileo; contains text, drawings, table, calculations. Relation to the Discorsi: elaboration of 2/19-pr-06. Folio page 161v Written by Galileo; contains text, drawing. Relation to the Discorsi: draft of 2/33-pr-12. Folio page 161r Folio page 161v Folio 162 Folio page 162r Written by Galileo; contains text, drawings. Relation to the Discorsi: draft of 2/38-pr-16. Folio page 162v Empty page. Folio page 162r Folio page 162v Folio 163 Folio page 163r Written by Galileo; contains texts, drawings. Relation to the Discorsi: draft of 2/35-pr-14-lemma-3; incomplete draft (proof) of 2/36-th-22. Folio page 163v Written by Galileo; contains texts, drawings. Relation to the Discorsi: statement and elaboration of 2/03-th-03; draft of 2/25-th-16; elaboration of 2/23-pr-09-schol1. Folio page 163r Folio page 163v Folio 164 Folio page 164r Written by Galileo; contains texts, drawings. Relation to the Discorsi: elaboration of 2/11-th-11; elaboration of 2/10-th-10. Folio page 164v Written by Galileo; contains texts, drawings. Relation to the Discorsi: elaboration of 2/08-th-08; elaboration of mirandum fragment; statement of Veloc-total-space-squared-prop. Folio page 164r Folio page 164v Folio 165 Folio page 165r Empty page. Folio page 165v Contains drawings, calculation. Folio page 165r Folio page 165v Folio 166 Folio page 166r Written by Galileo; contains drawing, short texts, table. Folio page 166v Empty page. Folio page 166r Folio page 166v Folio 167 Folio page 167r Empty page. Folio page 167v Contains drawing, table, calculation. Folio page 167r Folio page 167v Folio 168The folio is part of a larger sheet of paper combining the folios 168 and 169 Folio page 168r Written by Galileo; contains texts, drawings. Relation to the Discorsi: draft of 2/32-th-21; statement and elaboration of 2/31-th-20-lemma. Folio page 168v Empty page. Folio page 169r Empty page. Folio page 169v Written by Galileo; contains text, drawings, calculations, tables. Relation to the Discorsi: elaboration of 2/12-th-12. Folio page 169v and Folio page 168r Folio page 168v and Folio page 169r Folio 169See folio 168 --- --- Folio 170 Folio page 170r Written by Galileo; contains text, drawings. Relation to the Discorsi: incomplete draft of 2/26-pr-10. Folio page 170v Empty page. Folio page 170r Folio page 170v
Folios 151 to 170 | 1,387 | 4,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-43 | latest | en | 0.671081 |
https://stats.stackexchange.com/questions/381791/is-there-any-statistical-test-for-the-difference-between-two-percentiles | 1,726,407,439,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00702.warc.gz | 504,637,419 | 41,606 | # Is there any statistical test for the difference between two percentiles?
I know the R function t.test to perform the statistical test for the difference bewteen two means, is there any test for the difference between, say, the 95th percentile?
• If you treat the percentile as a quotient $x/(x+y)$ you can use a binomial test. Commented Dec 13, 2018 at 7:37
• You can use prop.test to carry out a test of proportions? Or are you after something that compares a specific percentile of a known distribution? Commented Dec 13, 2018 at 20:42
• @André.B I have two list of real numbers and I would like to know wether there is a statistical significative difference between the 95th percentiles of the two lists. Commented Dec 14, 2018 at 7:15
We can use quantile linear regression to test for the quantile difference between two groups.
To show this, let's generate two groups of data: 500 samples of normal data with mean=0, and 500 samples with mean=10. Now we know that our true effect size is 10, and all t-test assumptions are valid.
# Libraries
library(data.table)
library(quantreg)
# Simulate data
n=1000L
rm(df)
set.seed(27703)
df=data.table(value=rnorm(n,0,1),Group=0L)
df[101:200,Group:=1]
dim(df) # 1000 xx
# Set "Delta" (effect size)
df[Group==1,value:=value+10]
# Explore
plot(value~Group,data=df)
We start by noting that a 2-sample t.test:
# Compare means using a 2-sample t-test
fit=t.test(value~Group,data=df,paired=F,var.equal=T)
fit
# t = -100.64, df = 998, p-value < 2.2e-16
fit$$estimate[2]-fit$$estimate[1] # 10.18341 - Close to true Delta
can be reframed as a linear model:
# Compare means using a LM
fit=lm(value~1+Group,data=df)
summary(fit)
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.003999 0.031998 0.125 0.901
# Group 10.183413 0.101188 100.638 <2e-16 ***
We achieve the exact same estimate, t-value, and p-value.
Least squares linear regression models the mean. We can extend this to quantile regression to model a given quantile. Let's compare medians:
# Compare medians using quantile linear regression
fit=rq(value~1+Group,data=df,tau=0.5,method='fn')
summary(fit,se='iid')
# Value Std. Error t value Pr(>|t|)
# (Intercept) 0.00643 0.03515 0.18292 0.85490
# Group 10.24291 0.11116 92.14371 0.00000
# Compare against simple statistics
median(df[Group==0,value]) # 0.005802855 - Very close to estimated intercept
median(df[Group==1,value]) # 10.25729
10.25729 - 0.005802855 # 10.25149 - Very close to estimated Delta
The model results compare well against a simple check of univariate statistics.
Likewise, we can compare extreme quantiles like 0.90:
# Compare 90th percentile
tau=0.9
fit=rq(value~1+Group,data=df,tau=tau,method='fn')
summary(fit,se='iid')
# Value Std. Error t value Pr(>|t|)
# (Intercept) 1.28681 0.04413 29.15848 0.00000
# Group 9.99919 0.13956 71.65002 0.00000
# Compare against simple statistics
quantile(df[Group==0,value],tau) # 1.286451 - Very close to estimated intercept
quantile(df[Group==1,value],tau) # 11.2839
11.2839 - 1.286451 # 9.997449 - Very close to estimated Delta
Some important notes:
Please explore the different se options for summary.rq(). The reported p-values can vary widely depending on the option selected. The iid option is less conservative than the default option.
This example worked well with a large sample size (n=1000). I repeated this same exercise on the sleep dataset (n=20) and achieved less clear results, especially for extreme quantiles (tau>0.9 | tau<0.1). This isn't surprising, given that extreme quantiles have a smaller breakdown point compared to central tendency statistics.
Finally, with rq() it is very common to receive a warning message of "Solution may be nonunique". This is due to calculating medians from even sample sizes. One solution is to use method='fn'. This may also help with smaller sample sizes, since method fn interpolates and the default method br does not. See this discussion from the package author. | 1,161 | 4,054 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-38 | latest | en | 0.783743 |
https://pdfcoffee.com/moment-influence-line-report-pdf-free.html | 1,669,721,453,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00603.warc.gz | 503,577,041 | 11,442 | # Moment Influence Line Report
##### Citation preview
MOMENT INFLUENCE LINE 1.0 OBJECTIVE Part 1 : To plot monet influence line. Part 2 : To apply the use of moment influence on a simply support beam.
2.0 LEARNING OUTCOME 1. Application the engineering knowledge in practical application. 2. To enhance technical competency in structural civil engineering through laboratory application. 3. Communicate effectively in group. 4. To identify problem, solving finding out appropriate solution through laboratory application.
3.0 INTRODUCTION Influence line are important in the design of structures that resist large live loads. We learned that shear and moment diagrams are importantin determining the maximum internal force in a structure. An influence line represents the variation of the reaction, shear, moment, or deflection at a specific point in a member as a concenstrated force moves over the member. Moving loads on beams are common features of design. Many road bridges are constructed from beam and as such have to be designed to carry a knife edge load or a string of wheel loads, or a uniformly distributed load, or perhaps the worst combination of all three. To find the critical moment at a section, influence lines is used.
4.0 THEORY Definition :Influence line is defined as a line representing the changes in either moment, shear force, reaction or displacement at a section of a beam when a unit load moves on the beam. Part 1 :This experiment examines how momentvaries at a cut section as a unit load moves from one end to another ( see diagram 1). From the diagram, moment influence line equation can be written. For 0 ≤ x ≤a shear line is given by : Sy = -x/ L………(1) For a ≤ x ≤b shear line is given by : Sy = 1-x/ L……(2) For a unit load between0 ≤ x ≤a,
MX= (L-x)a L
1(a-x)………….(1)
For unit load between a ≤ 𝒙 ≤ b MX = xb/L- (x-a)……….……(2
1 ( Unit Load ) x
‘Cut’
MX
x
RA = (1- X/L)
RB=(X/L)
MX
a
b
LK
Figure 1 5.0 Part 2 : If the beam are loaded as shown below, the shear force at the “ cut” can be calculated using the influence line. (See diagram 2). Moment at ‘cut ’section = F 1Y1 + F2Y2 + F3Y3 …(3) (Y1, Y2 and Y3 are ordinates derived from the influence line in term of X1, X2, X3, a,b and L)
6.0 APPARATUS
1. Bending moment in a beam apparatus
7.0 PROCUDER
Part 1 1. Check the Digital Force Meter reads Zero with no load. 2. Place hanger with any mass between 150g to 300g at the first grooved hanger support at the left support and record the Digital Force reading in Table 1. 3. Repeat the procedure to the next grooved hanger until to the last grooved hanger at the right hand support. 4. Complete the calculation in Table 1.
Part 2 1. Place three load hangers with any load 50g to 400g on it and place it at any position between the supports. Record the positions and the Digital Force Display reading in Table 2. 2. Repeat the procedure with three other locations. 3. Complete the calculation in Table .
8.0 RESULT Part 1 Location of load
Digital Force
Moment force
Experimental
Theory influence
from left hand
Display
at cut section
Influence line value
line value (Nm)
support (m)
(N)
(Nm)
0.04
0.3
0.0375
0.025
0.013
0.06
0.4
0.05
0.034
0.019
0.08
0.5
0.0625
0.042
0.025
0.10
0.6
0.075
0.051
0.032
0.12
0.7
0.0875
0.059
0.038
0.14
0.9
0.1125
0.076
0.045
0.16
1.0
0.125
0.085
0.051
0.18
1.1
0.1375
0.093
0.057
0.20
1.2
0.15
0.102
0.064
0.22
1.3
0.1625
0.11
0.07
0.24
1.4
0.175
0.119
0.076
0.26
1.6
0.2
0.136
0.083
0.34
-0.6
-0.075
-0.051
0.068
0.36
-0.5
-0.0625
-0.042
0.054
0.38
-0.4
-0.05
-0.034
0.041
0.40
-0.2
-0.025
-0.017
0.027
Notes : 1. Moment at cut section = Digital force x 0.125 2. Experimental Influence line values =
𝑚𝑜𝑚𝑒𝑛𝑡 (𝑁) 𝐿𝑜𝑎𝑑 (𝑁)
3. Calculate the theoretical value using the equation 1 for load position 40 to 260mm and equation 2 for load position 320mm to 380mm
Part 2 : Location
1 2 3 4
Position of hanger from left hand support (m)
100g
200g
300g
0.04 0.08 0.36 0.26
0.1 0.16 0.34 0.4
0.2 0.26 0.08 0.6
Experimental Shear Nm
Theoretical Shear Nm
2.3 2.7 2.1 1.6
0.288 0.334 0.263 0.200
0.261 0.366 0.260 0.190
Notes : 1. Experiment Moment = Digital force Reading × 0.125 2. Theoretical Moment is calculated Using equation (3)
Calculation Part 1 Moment at cut section = 0.2×0.125 =0.025 Experimental influence line values = =
𝑚𝑜𝑚𝑒𝑛𝑡 (𝑁𝑚) 𝑙𝑜𝑎𝑑 (𝑁)
0.0375 1.962
=0.025 m
Theoretical influence lines value Equation 1 for load position 40 to 260 mm
Mx = (0.44 – 0.04) (0.3) – 1(0.3 – 0.04) 0.44 = 0.013 Nm Equation 2 for load position 320mm to 400
When x = 0.34 m Mx = (0.34) (0.14) – (0.34 – 0.3) 0.44 =0.068 Nm Part 2 F1 = =
100g 100 x 9.81 1000
= F2 = =
0.981N 200g 200 x 9.81 1000
= F3 = =
1.962N 300g 300 x 9.81 1000
=
2.943N
*For location 1 Experimental moment at cut section (Nm) = Digital force reading × 0.125 = 2.3 × 0.125 = 0.288 Nm
Moment at cut : ∑Mx = 0 Mx = 1(0.3)-
x
(0.3) – 1 (0.3-x)
0.44 = 0.3 - 0.3x – 0.3 + x 0.44 Mx = 0.318
When x = 0.3 Mx = 0.318x =0.318(0.3) =0.095 Nm
use interpolation to get y1, y2, y3 y1,
0.095 = 0.3
y1 0.04
0.3y = 0.0038 y1 = 0.013 m
y2,
0.095 = y2 0.3
0.1
y2 = 0.032 m
y3,
0.095 = 0.3
y3 0.2
y3 = 0.063 m
Theoritical moment at cut section (Nm) =
F1y1 + F2y2 + F3y3
=
0.981 (0.013) + 1.962 (0.032) + 2.943 (0.063)
=
0.261 Nm
*for location 2 Experimental moment (Nm) = 0.363 Nm When y1 = 0.025 m , y2 = 0.051 m , y3 = 0.082m Theoritical moment (Nm) = 0.366 Nm
*for location 3 Experimental moment (Nm) = 0.263 Nm When y1 = 0.054m , y2 = 0.068m , y3 = 0.025m Theoritical moment (Nm) = 0.260 Nm
*For location 4, Experimental moment (Nm) = 0.4125 Nm When y1 = 0.082m , y2 = 0.027m , y3 = 0.019m Theoritical moment (Nm) = 0.190 Nm
9.0 DISCUSSION
1. Derive equation 1 and 2
ΣFx = 0 ΣFy = RA + RB – 1 =0 RA + RB = 1 RA( L ) – 1( L – x ) = 0 RAL = 1(L- x)
RB
RA = 1( L – x ) L =1- x L = 1 – (1 – x) = x L L
Equation 1 ; 0 ≤ x ≤ a -Mx + RA(a) – 1(a - x) = 0 Mx = (1 – x/L)a – 1(a - x) = (L – x)a – 1(a - x) L
Equation 2 ; a≤ x ≤ b Mx – RB(b) + 1(x - a) = 0 Mx = RB (b) – 1(x - a) = x/L (b) – 1(x -a) = xb/L – 1(x -a)
2. On the same graph paper, plot the theoretical and experimental values against distance from left hand support.
GRAPH EXPERIMENT VALUE (Nm) VERSUS THEORETICAL VALUE (Nm) VERSUS DISTANCE 0.25 0.2
value (Nm)
0.15 0.1 theoretical value
0.05 0 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.34 0.36 0.38 0.4 -0.05 -0.1
experimental value
distance (m)
From the graph, a peak shaped graph can be obtained. The peak is the weakest point of the beam where there is a hinge in the beam. As load is being moved on the beam, the influence line which was constructed can be used to obtain the value of the moment. As load is moved across near to it, the moment will increase. So does the other way round when load is moving further than the hinge, the value of moment will decrease as the load is moving towards the support at the end. As the load is moving along towards the hinge from both side of support, it will come to a peak where the value of moment is the same. 3. Comment on the experimental results and compare it to the theoretical results.
The experimental results that we obtained are quite accurate and compare to the theoretical results, the experimental results are only slightly different with theoretical results. When we were conducted the experiment, we tried to minimize the error by ensuring the Digital Force Meter reads zero with no load before we place the hangers.
PART 2 1.
Calculate the percentage difference between experimental and theoretical results in table 2. Comment on why the results differ.
Experimental Moment (Nm)
Theoretical moment (Nm)
Percentage Different (%)
0.263
0.261
0.363
0.366
0.263
0.26
0.2
0.19
0.77 0.82 1.15 5.26
The experimental results are slightly different from theoretical results are due to human error and instrument sensitivity as the reading of the instrument keep changing when we conducted the experiment.
10 CONCLUSION As a conclusion, both objectives were achieved. Moment influence line could be plot and the influence line can be use to determine the moment. We were able to identify the reaction and behaviour of a beam in terms of its moment reaction value. This method is useful to check every cross section for a particular beam. | 2,940 | 8,422 | {"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-2022-49 | latest | en | 0.878855 |
https://www.physicsforums.com/threads/linear-transformations-polynomials-matrices.198833/ | 1,475,045,635,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661326.74/warc/CC-MAIN-20160924173741-00184-ip-10-143-35-109.ec2.internal.warc.gz | 1,004,675,565 | 15,574 | # Linear Transformations (polynomials/matrices)
1. Nov 17, 2007
### jesuslovesu
[SOLVED] Linear Transformations (polynomials/matrices)
Never mind, I can see it now, thanks
1. The problem statement, all variables and given/known data
Let S be the linear transformation on P2 into P3 over R. S(p(x)) = xp(x)
Let T be the linear transformation on P3 over R into R2x2 defined by T(a0 + a1x + a2x^2 + a3x^3) = [ a0 a1; a2 a3]
Find a formula for TS(p(x)).
3. The attempt at a solution
The first thing I do is find the S(A) where A is the standard basis of P2 and I place that into a transition matrix from the basis B (std. basis of P3).
B,A = [0,0,0;1,0,0;0,1,0;0,0,1]
Then I do the similar steps for fining [T]C,B where C = E2x2
[T]C,B = I4 (identity matrix of a 4x4)
Multiplying the matrix yields: [T]* = [0,0,0;1,0,0;0,1,0;0,0,1]
I am fairly positive that the math up to this point is accurate. (I get the correct range of T).
My question is how do I specifically find the formula for TS(p(x)) using the last matrix that I found? I know it's a mapping of P2 into R2x2, but I don't quite see how they get [0, a0; a1, a2] as the matrix. I know it lines up with TS, but still
Last edited: Nov 17, 2007 | 405 | 1,206 | {"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} | 3.515625 | 4 | CC-MAIN-2016-40 | longest | en | 0.886702 |
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# MTH603 Final Term paper 2017 Share your current paper here.
mujhay ju yaad hain woh bata raha hon aaj k paper me ju aya thay
mcqs me zaiada tar calculation thi us ki base pr mcqs kiay
1. Simpson 1/3
2. Simpson 3/8
4. Euler
5. Language's interpolation formula
6. Gaussian Elimination Method
7. Divided Differences
8. Richardson's Extrapolation
9. Forward difference Formula
..How to Join Subject Study Groups & Get Helping Material?..
Views: 2161
### Replies to This Discussion
concept ho ga to mcq's easy lagein ge..
Simpson 1/3 Simpson 3/8 method kaafi tha.. long mein mein dono thay or mcq's b thay is trhan k..
2 marks ka aik difference operator ka question tha..
backward or forward manipulations thien..
Construct a backward difference table from the following values of x and y
1 question was from Euler`s method
1 question was from tapezoidal rule
Solve by Gauss Seidel iterative method to 3 decimal place up to Two iterations
Langrange formula mein se b tha kuch.. Forgot..
Subjective file se tha almost sara.. do practice must..
& Wish u All the best..
Ayesha Nazir subjective papr kon c file se tha ???
Really helped a lot.. i had prepared the methods.. the questions included in this file..probably from last lectures..
Attachments:
Thanks Ayesha
MTH 603 Current Final Term Papers In ONE thread 25-feb-2017 to 8 march 2017
mcqs me zaiada tar calculation thi us ki base pr mcqs kiay
1. Simpson 1/3
2. Simpson 3/8
4. Euler
5. Language's interpolation formula
6. Gaussian Elimination Method
7. Divided Differences
8. Richardson's Extrapolation
9. Forward difference Formula
9
Note:-
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Mth603 Paper
Attachments:
My MTH603 paper:
MCQS were mostly form Handout and conceptual.
Subjective were from this File:
GOOD LUCK For all
Attachments:
Any bdy have MOAaZ PAST Papers???
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3 | 894 | 3,371 | {"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-2019-22 | longest | en | 0.735358 |
http://www.slidesearchengine.com/slide/elasticity-of-demand | 1,500,576,927,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423320.19/warc/CC-MAIN-20170720181829-20170720201829-00212.warc.gz | 549,110,799 | 8,059 | ELASTICITY OF DEMAND
33 %
67 %
Education
Published on September 23, 2010
Author: shrvnb
Source: authorstream.com
ELASTICITY OF DEMAND : ELASTICITY OF DEMAND What is Elasticity of demand? : The concept of “elasticity of demand” refers to the responsiveness of demand for a commodity to changes in its determinants. This concept was given by the famous Economist Prof. Alfred Marshall. What is Elasticity of demand? Slide 3: DEFINITION PROF. ALFRED MARSHALL: “ The elasticity of demand in a market is great or small , depending on whether the amount demanded increases much or little for a given fall in the price and diminishes much or little for a given rise in price.” Slide 4: FORMULA Proportionate change in the quantity demanded Proportionate change in the determinants of demand Elasticity of demand = Slide 5: TYPES OF ELASTICITY OF DEMAND Slide 6: TYPES OF ELASTICITY OF DEMAND PRICE ELASTICITY OF DEMAND INCOME ELASTICITY OF DEMAND CROSS-ELASTICITY OF DEMAND PROMOTIONAL ELASTICITY OF DEMAND PRICE ELASTICITY OF DEMAND : The extent of response of demand for a commodity to a given change in price, other demand determinants remaining constant, is termed as the price elasticity of demand. Its co-efficient can be measured as: Proportionate change in quantity demanded Proportionate change in price PRICE ELASTICITY OF DEMAND = Ed Slide 8: The formula for calculating price elasticity can be stated as: Ed = Q/Q P/P Alternatively Ed = Q Q X P P Where, Q = the original demand P = the original price Q = the change in demand P = the change in price Slide 9: Depending on the magnitude and proportional change involved in the data on demand and prices, various numerical values of co-efficient of price elasticity can be obtained. Slide 10: TYPES OF PRICE ELASTICITY OF DEMAND Unitary Elastic Demand : The demand is said to be unitary elastic when the proportionate change in the quantity demanded is the same as the proportionate change in price. Here, Example: If the price rises by 10% and demand fallss by 10%, the elasticity is unitary (Ed =1) Unitary Elastic Demand Q = Q P P Slide 12: QUANTITY DEMANDED Q1 Q D D P1 P O X Y (Ed=1) PRICE The demand curve in this case is a rectangular hyperbola curve. In the above figure, when the price rises from PP1 , the demand falls to QQ1. Therefore, Ed =1. Relatively Elastic Demand : The demand is said to be relatively elastic when the proportionate change in the quantity demanded is the greater than the proportionate change in price. Here, Example: If the price rises by 10% and demand falls by 30%, the elasticity is relatively elastic, i.e., (Ed >1) Relatively Elastic Demand Q > Q P P Slide 14: QUANTITY DEMANDED Q1 Q D D P1 P O X Y (Ed >1) PRICE The demand curve in this case has a gradually declining slope (flat curve). In the above figure, when the price rises from PP1 , the demand falls to QQ1. Therefore, Ed >1. Relatively Inelastic Demand : The demand is said to be relatively inelastic when the proportionate change in the quantity demanded is the less than the proportionate change in price. Here, Example: If the price rises by 30% and demand falls by 10%, then the elasticity is relatively inelastic, i.e., (Ed <1). Relatively Inelastic Demand Q < Q P P Slide 16: QUANTITY DEMANDED Q1 Q D D P1 P O X Y (Ed <1) PRICE The demand curve in this case has a sharply declining slope (steep curve). In the above figure, when the price rises from PP1 , the demand falls to QQ1. Therefore, Ed<1. Perfectly Elastic Demand : The demand is said to be perfectly elastic when at a given price or with a slightest fall in the price, there is an infinite extension of demand. Here, Q is any number and P is zero, and therefore, Ed = Perfectly Elastic Demand any number 0 = ∞ Slide 18: QUANTITY DEMANDED D D P O X Y (Ed = ∞) PRICE The demand curve in this case is horizontal to X-axis. It suggest that the demand is ever-rising at the given price. Perfectly Inelastic Demand : Whatever be the change in the price, if the demand does not respond at all (the quantity demanded remains unchanged), the demand is said to be perfectly inelastic . Here, Q is zero and P is any number, and therefore, Ed = Perfectly Inelastic Demand any number 0 = 0 Slide 20: QUANTITY DEMANDED D D Q O X Y (Ed =0) PRICE The demand curve in this case is vertical to Y-axis. It suggest that the demand remains unchanged at any price. INCOME ELASTICITY OF DEMAND : The extent of response of demand for a commodity to a given change in income, other demand determinants remaining constant, is termed as the income elasticity of demand. Its co-efficient can be measured as: Proportionate change in quantity demanded Proportionate change in income INCOME ELASTICITY OF DEMAND Em = Slide 22: The formula for calculating income elasticity can be stated as: Em Q/Q M/M = Alternatively Em = Q Q X M M Where, Q = the original demand P = the original price Q = the change in demand P = the change in price Slide 23: The income elasticity helps us in classifying the commodities. The following points are noted in this regard: When income elasticity is positive (Em), the commodity is of a normal type. When income elasticity is negative (Em), the commodity is inferior. Example: Cereals like jowar, bajra, etc. When income elasticity is positive and greater than one (Em>1), the commodity is a luxury. Example: T.V. sets, cars, etc. When income elasticity is positive but less than one (Em<1), the commodity is an essential one. Example: Foodgrains, medicines, etc. CROSS-ELASTICITY OF DEMAND : The extent of response of demand for a commodity to a given change in the price of some other related commodity, other demand determinants remaining constant, is termed as the cross-elasticity of demand. Its co-efficient can be measured as: CROSS-ELASTICITY OF DEMAND Exy = Proportionate change in demand for X commodity Proportionate change in the price of Y commodity Slide 25: The formula for calculating price elasticity can be stated as: = Qx Qx Py Py Alternatively Exy Exy = Qx Qx X Py Py Where, Q = the original demand P = the original price Q = the change in demand P = the change in price Slide 26: The demand for two complementary goods is negative. Example: The demand for car and petrol. When the price of petrol increases, the demand for cars will decline. The demand for substitute goods is positive. Example: The demand for tea and coffee. When the price of tea increases, the demand for coffee will increase. PROMOTIONAL ELASTICITY OF DEMAND : The extent of response of demand for a commodity to a given change in advertisement expenditure, other demand determinants remaining constant, is termed as the promotional elasticity of demand. Its co-efficient can be measured as: PROMOTIONAL ELASTICITY OF DEMAND Ex = Proportionate change in demand for X commodity Proportionate change in the advertisement expenditure Slide 28: The formula for calculating price elasticity can be stated as: = Qx Qx A A Alternatively Ex Ex = Qx Qx X A A Where, Q = the original demand A = the original advertisement expenditure Q = the change in demand A = the change in advertisement expenditure Slide 29: This elasticity of demand is very useful to business firms to find out the impact of their advertisement expenditure on the demand for their commodity. Future expansions, innovations and modifications are planned by business firms with the help of promotional elasticity of demand.
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Cross elasticity of demand; Elasticity of substitution; Frisch elasticity of labor supply; Income elasticity of demand; Output elasticity; | 2,126 | 9,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-30 | latest | en | 0.847551 |
https://www.theengineeringstreet.com/2018/12/comparison-between-magnetic-and-electric-circuit.html | 1,624,294,771,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488286726.71/warc/CC-MAIN-20210621151134-20210621181134-00566.warc.gz | 934,733,519 | 28,930 | # Comparison Between Magnetic And Electrical Circuit !!!
## Here the Difference Between Magnetic And Electrical Circuit !!!
### A) Analogy Between Magnetic And Electric Circuit
• 1)
Magnetic Circuit :
The Closed Path Followed by The Magnetic Flux Is Called Magnetic Circuit.
Electric Circuit :
The Path In Which Electric Current Can Follow Is Known As Electric Circuit.
• 2)
Magnetic Circuit :
Flux = MMF / Reluctance
Electric Circuit :
Current = EMF / Resistance
• 3)
Magnetic Circuit :
MMF (AT)
This Is the cause of linking of flux in the circuit.
Electric Circuit :
EMF ( Volts)
This is the cause of flow of electric current in the circuit.
• 4)
Magnetic Circuit :
Flux Density = Flux / Area
Electric Circuit :
Current Density = Current / Area
• 5)
Magnetic Circuit :
Reluctance = This opposes the linking of magnetic flux.
Electric Circuit :
Resistance = This Opposes the flow of electric current.
• 6)
Magnetic Circuit :
Permeance = 1/Reluctance
Electric Circuit :
Conductance = 1/Resistance
• 7)
Magnetic Circuit :
Permeability = 1/ Reluctivity
It is the property of the magnetic path to pass more flux.
Electric Circuit :
Conductivity = 1/ Resistivity
It is the property of the conducting material to pass more current.
• 8)
Magnetic Circuit :
S = S1 + S2 + S3 + ............... + Sn
In series Magnetic Circuit.
Electric Circuit :
R = R1 + R2 + R3 + .................+ Rn
In series electric circuit.
• 9)
Magnetic Circuit :
1/S = 1/S1 + 1/S2 + ............. + 1/Sn
In parallel Magnetic Circuit.
Electric Circuit :
1/R = 1/ R1 + 1/R2 + .............. + 1/Rn
In parallel Electric Circuit.
### B) Differences Between Magnetic Circuit And Electric Circuit
The Magnetic Circuit differs from the electric circuit in the following respects :
(1) Resistance is normally independent of current in an electric circuit, whereas reluctance depends on the Flux Density in the magnetic circuit.
For This reason magnetization (B-H) curves of magnetic materials are used for determining necessary excitation.
(2) Flux actually links in the magnetic circuit, whereas current actually flows in the circuit.
(3) In magnetic circuit, energy is needed to create the magnetic flux only at starting but in an electric circuit, energy is consumed till the current flows.
(4) The magnetic circuit has the property of retentivity .
i.e. a small amount of lux called residual flux persists after the removal of the MMF .
Whereas in an electric circuit, the current reduces to zero after the removal of the applied EMF or when circuit is broken.
#### 1 comment:
1. You have shared such a great post. I got some valuable information from this post. Thanks for sharing such a post. Keep Post. electrical enclosures | 645 | 2,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-25 | latest | en | 0.670094 |
https://www.jiskha.com/display.cgi?id=1287367320 | 1,516,559,512,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890823.81/warc/CC-MAIN-20180121175418-20180121195418-00212.warc.gz | 924,960,436 | 3,533 | # Physics
posted by .
A football is thrown with an initial upward velocity component of 15.0 m/s and a horizontal velocity component of 18.0 m/s
The time required for the football to reach the highest point in its trajectory is 1.53s.
It gets 11.5m above the ground.
How much time after it is thrown does it take to return to its original height?
How far has the football traveled horizontally from its original position?
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A football is thrown with an initial upward velocity component of 15.0m/s and a horizontal velocity component of 18.0m/s . How high does it get above the ground?
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More Similar Questions | 643 | 2,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-05 | latest | en | 0.895679 |
http://car-stars.ru/subtraction_of_fractions_(5)/(16)-(1)/(20) | 1,527,102,436,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865702.43/warc/CC-MAIN-20180523180641-20180523200641-00058.warc.gz | 53,107,415 | 7,053 | # (5)/(16)-(1)/(20) - subtraction of fractions
## (5)/(16)-(1)/(20) - step by step solution for the given fractions. Subtraction of fractions, full explanation.
If it's not what You are looking for just enter simple or very complicated fractions into the fields and get free step by step solution. Remember to put brackets in correct places to get proper solution.
### Solution for the given fractions
$\frac{5}{16 }-\frac{ 1}{20 }=?$
The common denominator of the two fractions is: 80
$\frac{5}{16 }= \frac{(5*5)}{(5*16)} =\frac{ 25}{80}$
$\frac{1}{20 }= \frac{(1*4)}{(4*20)} =\frac{ 4}{80}$
Fractions adjusted to a common denominator
$\frac{5}{16 }-\frac{ 1}{20 }=\frac{ 25}{80 }-\frac{ 4}{80}$
$\frac{25}{80 }-\frac{ 4}{80 }= \frac{(25-4)}{80}$
$\frac{(25-4)}{80 }=\frac{ 21}{80}$
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https://www.examrace.com/GMAT/GMAT-MCQs/Mathematics-Questions/Mathematics-Objective-Questions-Part-7.html | 1,623,594,462,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608856.6/warc/CC-MAIN-20210613131257-20210613161257-00452.warc.gz | 690,367,395 | 7,127 | # Mathematics Objective Questions for GMAT Paper 25
Get top class preparation for SAT right from your home: fully solved questions with step-by-step explanation- practice your way to success.
Q-1. If are the roots of , then are the roots of
(a)
(b)
(c)
(d)
Q-2. The number of real roots of the equation is
(a) 1
(b) 2
(c) 3
(d) None of these
Q-3. If S is the set containing values of x satisfying where [x] denotes GIF, then S contains
(a) (2,4)
(b) (2,4]
(c)
(d)
Q-4. Seven people are seated in a circle, how many relative arrangements are possible?
(a) 7!
(b) 6!
(c)
(d)
Q-5. In how many ways can 4 people be seated on a square table, one on each side?
(a) 4!
(b) 3!
(c) 1
(d) None of these
Q-6. Four different items have to be placed in three different boxes. In how many ways can it be done such that any box can have any number of items?
(a)
(b)
(c)
(d)
Q-7. What is the probability that, if a number is randomly chosen from any 31 consecutive natural numbers. it is divisible by 5?
(a)
(b)
(c)
(d) None of these
Q-8. The mean of a binomial distribution is 5, and then its variance has to be
(a) > 5
(b) = 5
(c) < 5
(d) = 25
Q-9. If a is the single A. M. between two numbers a and b and S is the sum of n A. M. between them, then depends upon
(a) Depends upon
(b) n, a, b
(c) n, a
(d) n, b
Q-10. up to equal to
(a) 1
(b) 2
(c)
(d)
Q-11. The odds in favor of India winning any cricket match is 2: 3. What is the probability that if India plays 5 matches, it wins exactly 3 of them?
(a)
(b)
(c)
(d)
Q-12. For an A. P. , The value of is equall to
(a) 4
(b) 6
(c) 8
(d) 10
Q-13.1 + sin x + then
(a)
(b)
(c)
(d)
Q-14.
(a)
(b) X
(c)
(d)
Q-15. The ends of a line segment are P (1,3) and Q (1,1) . R is a point on the line segment PQ such that If R is an interior point of the parabola then
(a)
(b)
(c)
(d) None of these
Q-16. Set of values of which is true is
(a)
(b)
(c)
(d)
Q-17. The distance between the lines and is
(a)
(b)
(c)
(d) None of these
Q-18. Let and is a variable point such that PA + PB is the minimum. Then k is
(a)
(b) 0
(c)
(d) None of these
Q-19. The length of the y-intercept made by the circle is
(a) 6
(b)
(c)
(d) 3
Q-20. If x + y = k is normal to then k =
(a) 3
(b) 6
(c) 9
(d) None of these
Q-21. T he number of values of c such that the straight line touches the curve is
(a) 0
(b) 1
(c) 2
(d) infinite
Q-22. =
(a) 1
(b)
(c)
(d)
Q-23. Locus of the point z satisfying Re is a non – zero real number, is
(a) a straight line
(b) a circle
(c) an ellipse
(d) a hyperbola
Q-24. The points of z satisfying arg lies on
(a) An arc of a circle
(b) A parabola
(c) An ellipse
(d) A straight line
Q-25. The coefficients of the th term and the th term in the expansion are equal, then
(a)
(b)
(c)
(d) None of these
Q-26.
(a) 2e
(b) e
(c) e-1
(d) 3e
Q-27. If in , then
(a)
(b)
(c)
(d)
Q-28.
(a)
(b)
(c)
(d)
Q-29. Two of straight lines have the equations and and common among them if.
(a)
(b)
(c)
(d) Both (a) and (b)
Q-30. If circle passes through the point (3,4) and cutes orthogonally, then the locus of its centre is . Then
(a) 11
(b) 13
(c) 17
(d) 23
Q-31. For what value of x, the matrix A is singular
(a)
(b)
(c)
(d)
Q-32. If 7 and 2 are two roots of the following equation = 0, then its third root will be
(a) -9
(b) 14
(c)
(d) None of these
Q-33. Period of
(a)
(b)
(c)
(d) None of these
Q-34. The range of log n (sin x)
(a)
(b)
(c) (]
(d)
Q-35. is equal to
(a)
(b)
(c)
(d)
Q-36. let The value of is
(a) 0
(b) 1
(c)
(d)
Q-37. For the curve tangent is parallel to x- axis where
(a)
(b)
(c)
(d)
Q-38. is strictly decreasing for
(a)
(b)
(c)
(d)
Q-39. The value of b for which the function is a strictly decreasing function is
(a)
(b)
(c)
(d) [)
Q-40. Maximum value of the expression is
(a)
(b)
(c)
(d) None of these
Q-41. If then the value of tan
(a) 3
(b) 2
(c) 1
(d) 0
Q-42. then is equal to
(a) 10
(b)
(c) 1
(d) -1
Q-43. If
(a)
(b)
(c)
(d) None of these
Q-44. Length of the sub tangent to the curve y is
(a)
(b) a
(c)
(d) None of these
Q-45. The value of c of mean value theorem when in is
(a)
(b)
(c)
(d)
Developed by: | 1,462 | 4,261 | {"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} | 3.78125 | 4 | CC-MAIN-2021-25 | latest | en | 0.821591 |
http://www.physicsforums.com/showthread.php?s=24f5bd63f75ec729f7a8aeb6536b14e5&p=4612550 | 1,394,329,388,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999670048/warc/CC-MAIN-20140305060750-00044-ip-10-183-142-35.ec2.internal.warc.gz | 481,096,943 | 8,509 | # Power at resonance frequency
by Bromio
Tags: coil, frequency, power, resonance, resonance frequency, rlc circuit
P: 61 Hi. When using spectrum analyzer to measure the response of a coil (a RLC circuit), I see that there is a peak at one frequency (resonance frequency). This is logical because we can model a coil as a RLC circuit. If I change the frequency of the sinusoidal source, the peak reduces its value. My question is: why is there a higher peak at resonance frequency? If what I measure is the power dissipated (by the resistor, of course), why is there a dependence with frequency? If complex power is written as $P = P_{loss}+2j\omega\left(W_m-W_e\right)$, where $P_{loss} = \frac{1}{2}\left|I\right|^2R$ is the dissipated power, $W_e$ is the electric energy stored in the capacitor and $W_m$ is the magnetic energy stored in the inductor, why the analyzer doesn't show a constant peak value (given by $P_{loss}$) whatever the frequency of the sinusoidal source? Thank you.
Mentor P: 10,511 I would guess that the driver power reaches its maximum at the resonance frequency, but it is hard to tell if you don't show the setup and explain where you measure what.
P: 61 I rolled two wires around the coil: one coming from the sinusoidal signal generator and another one going to the spectrum analyzer, which also showed "power" (dissipated power, I suppose). So, I was measuring the voltage drop in the coil (which can be modeled as a RLC circuit). Sweeping frequencies I noted there was a higher peak at resonance frequency. Thanks.
Engineering
Thanks
P: 6,038
## Power at resonance frequency
Quote by mfb I would guess that the driver power reaches its maximum at the resonance frequency.
That's the reason, but I agree that what happens can seem like a paradox at first sight.
If the driving frequency is constant, then once the starting transient has decayed the power dissipated in the resistor is (obviously) the same as the power supplied by the signal generator.
The phase angle between the the generator voltage and current changes as you go through the resonant frequency. That is why the power (in watts) from the generator is not necessarily the same as the volt-amps it is producing.
Maybe this is easier to see with a mechanical oscillator like a mass on a spring. If you apply a force an a frequency much lower than resonance, you can ignore the inertia of the mass. The applied force is in phase with the motion, so the positive work you do compressing the spring is equal to the negative work done by the spring when you release it.
At a frequency much higher than resonance, you can ignore the force in the spring compared with the inertia force needed to accelerate the mass. The displacement is now 180 degrees out of phase with the force, but the effect is the same: the work you do on the mass averages out to zero, over a complete cycle.
Away from resonance, a plot of work done against time may have large positive and negative amplitudes in each cycle (analogous to measuring volt-amps) even though the average work done over the cycle is low (analogous to measuring watts).
But when you are at the resonant frequency, the displacements are 90 degree out of phase with the force, and you do work on the system over the complete cycle. And of course as the amplitude increases, a constant force does even more work, because the product of force x distance increases.
P: 578
Quote by Bromio Hi. When using spectrum analyzer to measure the response of a coil (a RLC circuit)
I think you need to supply more detail on your measurement setup. Are you using spectrum analyzer tracking gen for source? If not what is the source, 50 ohms? How is inductor connected to 50 ohm input to spectrum analyzer and the source, coaxial cable? Are you doing anything to match your inductor to 50 ohms (i.e. broadband attenuator)?
To give you an idea where I am going with this, if you are using coax to extend source & spectrum analyzer down to your inductor, you will see periodic dips in response due the the cable lengths.
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July 1, 2010
By
(This article was first published on Statistical Graphics and more » R, and kindly contributed to R-bloggers)
Peter Huber referred to “the rawness of raw data”, a kind of data we would not expect to find in a textbook. The book of Fahrmeir and Tutz on multivariate modelling refers to the visual impairment data from Liang et al., 1992 in table 3.12:
Visual Impairment Data from Liang et al. as found in Fahrmeir and Tutz
Nothing wrong here at first sight; but how would you tell? There are some people who are actually able to look at non-trivial table data and spot “the round peg in the square hole”, but that just won’t work for the rest of us.
As you might guess, I am going to make a case for graphics here.
Let’s start with what the mainstream would do: plot the data in a dotplot like thing using the trellis paradigm of conditioning. I used ggplot2 to make sure to trellis state-of-the-art. A simple
` qplot(count, side, data=visual2, colour=impaired) + facet_grid(age ~ race)`
gives me:
(I still have a hard time to find that syntax intuitive …) Surprisingly this plot already is sufficient to spot the “problem” in the data, although some important properties of the data can’t be seen here.
A mosaic plot makes the whole thing even easier:
(impairment cases highlighted, left and right is left and right)
The left and right cases are (what a surprise) always of the same size, except for the 70+, black – hard to believe that in this group 110 cyclops show up not having a right eye.
In the mosaic plot the higher proportion of the impaired right eyes for 70+ blacks jumps immediately to ones eyes, but what reveals the error is the missing independence between race and side for 70+. That implies that we have too few cases here, and what is ’226′ in the table should actually be ’336′.
Here is the (corrected) data.
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## AutoCAD 20.0 Crack X64 [Updated-2022]
by carodah
How to install the crack 1) put Autocad.crack to autocad folder and click on this file, it will show crack dialog 2) click OK 3) install Autocad 4) open Autocad and run 5) Use the crack If you have a problem Please contact autocad-support@autodesk.com By « autocad-support@autodesk.com » I mean autocad-support@autodesk.com. Please don’t send other emails. Q: What’s the difference between a power curve and a power/frequency curve? From my understanding, a power curve is an extension of a power/frequency curve. Is that correct? My understanding is that a power/frequency curve is used to model the output power (in watts) of a system as a function of the output frequency. Is that correct? A: Mostly, yes. There is a subtle difference. A power curve is a measure of the power of a device, or some kind of output, as a function of some input. A power/frequency curve is a function of the frequency at which some device operates. In electronics, the output power of a device is usually specified in terms of power per frequency: $$P = 10^3.8 \times f^3 \times f$$ where $f$ is frequency and $P$ is power (units watts). A curve is a plot of some kind of variable against another, so these should really be referred to as power/frequency curves, but they’re often called power curves. A power curve in this context is a plot of some power (units watts) against some frequency. Power is a scalar, so it has no units. Frequency is a vector, so frequency is units of $1/time$. Q: Is it possible to get the original node image and replace it with the uploaded image in the node image field? I’ve been tasked with restoring a site from another server that is offline. The site is an older Drupal 6 site. When a node is being created, it automatically creates an image that says « this is my node » and uploads the node into an image field. Is it possible to use the original node image from the database, and replace it with the uploaded image? I need | 492 | 2,014 | {"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} | 2.65625 | 3 | CC-MAIN-2022-33 | latest | en | 0.915589 |
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lec04_ddn
# lec04_ddn - Continuing Issues Phys 221 Lab questions Dr...
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Continuing Issues Phys 221 Lab questions – Dr. Lewicki, Room 142 I-clicker Problems – Dr. Saxena, Room 176 8/30/11 1
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8/30/11 3 From Mechanics (PHYS 220) • Energy Conservative Forces: Kinetic Energy: associated with the state of motion Potential Energy: associated with the configuration of the system Work done by a conservative force is independent of path.
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8/30/11 4 Gravitational Potential Energy Electric Potential Energy 2 2 1 r m m G F g = 2 2 1 r q q k F e = U=0 at r = Potential energy : can be positive or negative Field forces push object into area with lower potential energy: Potential energy decreases , work is done on object – for example, kinetic energy of the object may increase (speed) Total energy is conserved Potential Energy PE = G m 1 m 2 r PE = k q 1 q 2 r
8/30/11 5 Potential Energy of Two Points PE elec = k q 1 q 2 r r m m G U g 2 1 = Gravity: always attraction. U is more negative (“lower”) when objects are closer q 1 q 2 < 0 : attraction.
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Ask a homework question - tutors are online | 459 | 1,790 | {"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-2018-13 | latest | en | 0.795405 |
http://www.cfd-online.com/Forums/cfx/114497-transient-simulations-how-tell-its-converged-ive-read-faq-user-guides-print.html | 1,475,283,675,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662430.97/warc/CC-MAIN-20160924173742-00214-ip-10-143-35-109.ec2.internal.warc.gz | 385,088,643 | 6,095 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- CFX (http://www.cfd-online.com/Forums/cfx/)
- - Transient simulations: how to tell its converged (I've read the FAQ & user guides!) (http://www.cfd-online.com/Forums/cfx/114497-transient-simulations-how-tell-its-converged-ive-read-faq-user-guides.html)
JuPa March 12, 2013 05:59
Transient simulations: how to tell its converged (I've read the FAQ & user guides!)
3 Attachment(s)
Hello,
I have a transient problem. In my problem a plate is heated from below, which heats up the liquid column above it (please see the picture). The large temperature difference between T1 and T2 causes unsteady, buoyant natural convection.
In steady state I know what parameters to look out for to ensure convergence (correct me if I am wrong):
- You must give enough iterations to allow the residuals must fall acceptably low (for single precision no more than 1e-6). I understand you can get effects such as psuedo-steady state.
- Imbalances must be less than 1%, and ideally less than 0.01%.
- You must set monitor points inside your domain to monitor points of interest such as temperature and velocity. If these level out it suggests a steady state.
- You must do a mesh sensitivity study and tailor your mesh.
I know my problem is transient as:
- I've run the simulation in steady state with a specified physical timestep (the timestep was first estimated, and then reduced or increased depending on residual behaviour).
- Steady state imbalances are not acceptable.
- Monitoring key variables such as temperature and velocity in my domain suggest the problem is transient.
- Logically I expect the physical problem to be transient.
I am homing in to the solution in the following way:
Run 1. Do a steady state pure conduction case first (as in go into expert parameters and switch fluid modelling off).
Run 2. Using the results from Run 1. I initialise and do a steady state buoyant simulation using an upwind advection scheme.
Run 3. Using the the results from Run 2. I initialise and do another steady state simulation using a high resolution advection scheme. The residuals and imbalances do not fall.
Run 4. Using the results from Run 3. I initialise and do a transient simulation using a high resolution advection scheme and a second order transient scheme (backward Euler). The residuals do fall and the imbalances do reduce acceptably low.
A picture of the transient residuals are attached. A picture of some velocity vectors are also attached. Hopefully you can get a feel for the problem.
After reading the FAQ, Ansys User Guides, and searching the forum it is still not clear what criteria must be met to ensure a transient simulation has converged! I know there is something called "Transient Statistics" in CFX, however I don't know how these are used to determine convergence.
Any input will be appreciated!
A few further notes:
- A mesh sensitivity study has NOT been conducted yet - it will be the next step (to compare transient results on Mesh 1 vs Mesh 2 vs Mesh 3 etc).
- I've set the problem to be 2D (extruded one cell thick). In reality the domain is a 3D cylinder (and that will be the next step). I want to understand the problem in 2 dimensions before I move onto 3 dimensions.
Thank you
flotus1 March 12, 2013 06:12
To me, your results look like a turbulent flow simulated with a DNS-solver at insufficient spatial and temporal resolution.
Did you try to simulate your case with a turbulence model activated?
JuPa March 12, 2013 06:16
I am doing a turbulent simulation using a 2 equation model (K Omega SST) as I type this reply. This is obviously cheaper than doing a DNS simulation. However DNS may be done in the future.
But I would still like to know how to tell if a transient simulation has converged. Do you know what is meant by transient statistics?
flotus1 March 12, 2013 07:05
The transient statistics can be used for data sampling, e.g. time-averaging of variables.
JuPa March 12, 2013 07:08
Hi flotus1,
How valid are time averaged results of a problem which is very transient? Also how do you view time averaged results?
ghorrocks March 12, 2013 07:13
What Rayleigh number is the flow? That is what will tell you whether the flow is turbulent or not.
This flow is almost certainly 3D transient. Your 2D model is of little use other than to check the setup works, it will have little similarity to the 3D case and you cannot infer any 3D results from it.
The answer to your question (how do I tell it is converged?) is a sensitivity study. Do a baseline simulation, then change the parameter of interest (convergence tolerance, imbalance, time step size, mesh size) and see if the change is less than a tolerance you are happy to live with in parameters you care about. You will find these sensitivity studies are a little iterative - the mesh size affects the time step size affects the convergence - so you might need to do a few goes at it as you home in on a set of parameters which give an accurate solution.
flotus1 March 12, 2013 07:23
The data sampling has to be activated before starting the simulation.
I actually dont know how this is done in CFX, but the manual surely does.
From my experience with data sampling in Fluent, I anticipate this feature to be quite buggy, so dont get disappointed too fast.
Quote:
How valid are time averaged results of a problem which is very transient
Since there are several types of unsteadyness, the question cannot be answered in general.
For transient results that are statistically steady (time-constant mean value), time-averaging makes sense.
In your case, since you are using time-averaged equations (RANS) in an unsteady manner, time-averaging as a post-processing step is, lets say disputable.
Someone correct me if I'm wrong, but the concept of using URANS-methods for solving turbulent flow with steady boundary conditions is of questionable validity.
ghorrocks March 12, 2013 07:27
Well, lets determine whether the simulation is laminar or turbulent before you go too far. What is the Rayleigh number? Dos the literature say this Ra number should be laminar or turbulent for this configuration?
JuPa March 12, 2013 07:34
Quote:
Originally Posted by flotus1 (Post 413378) The data sampling has to be activated before starting the simulation. I actually dont know how this is done in CFX, but the manual surely does.
If I am not mistaken in CFX output control you just need to select transient statistics, and select which variables you want to monitor.
Quote:
Originally Posted by ghorrocks (Post 413383) Well, lets determine whether the simulation is laminar or turbulent before you go too far. What is the Rayleigh number? Dos the literature say this Ra number should be laminar or turbulent for this configuration?
Thanks for your reply. Based on a quick "back of the envelope" calculation the Rayleigh number is in the order of 9 E +7 (90000000).
Incropera and Dewitt suggest a laminar regime between 10^4 and 10^7. And a turbulent regime between 10^7 and 10^11.
The reason I'm doing a 2D simulation is purely because I have done some heat transfer calculations estimating the surface temperatures which are based on some empirical correlations (which agree to within 10% of data). I don't expect CFD and my calculations to agree bang on, but I don't expect them to disagree by a huge amount either.
You mentioned a "baseline simulation". I'm sorry it's the first time I've heard this term - what is it?
I appreciate the mesh resolution will have an effect on the Courant number and hence the timestep.
oj.bulmer March 12, 2013 09:47
If I may ask, why choose a 2D case over axisymemtric; given the latter is more relevant to the cylindrical tank? You should be careful here since buoyant convection flows may be unphysically sensitive to turbulence in case of 2D/axisymmetric cases. Some literature search may help here. Though, given Ra no., is it fair to say the solution is partly transitional? Finally, if you have nicely converged mesh independent 2D solution, just later to realize that 3D would yield completely different results, the maneuver has little relevance with your final objective.
Your plots don't indicate anything about the imbalances. Plot of 3 timesteps (including coefficient loop values) should shed some light on convergence of each timestep more clearly, rather than 100 upwards timesteps as you show now. Convergence of every timestep is essential and will bring more credibility to your time-averaged results.
Baseline simulation would be your first simulation that you find reasonable with best judgement. From there, sensetivity analysis will hep you understand effect of mesh size and timestep on your solution. How did you estimate timestep here? The ideal timestep is a reasonable one that achieves the trade off between convergence target and the running time. You may start with convergence target of 1e-4 and conservation target of 0.01 and then estimate the ideal timestep by sensetivity analysis. Howevever, adaptive timestepping is easiest way to do this. But you must be careful with the convergence targets since too tight values will result in too conservative timesteps in adaptive timestepping. But the final convergence targets will depend on the hardware and time available to you for simulation.
Convergence criteria would be decided by whether your solution is steady at the end, periodic or chaotic. I presume the upper wall disposes off the heat from liquid. it is tricky though, since initially the temperture gradients will be huge, and as fluid keeps circulating, the temperatures are less non-uniform. You can monitor the total heat integrated over the volume. When this value flattens, or shows oscillations with small magnitude around a mean value, there is your approximate indication of convergence, and equilibrium of heat going in and out and its effect on convection etc. You then can take time average values of portion of time for which this heat is flat/definitively periodic.
But obviously, more experienced folks may suggest a better way to judge the convergence.
OJ
JuPa March 12, 2013 10:02
Quote:
Originally Posted by oj.bulmer (Post 413436) If I may ask, why choose a 2D case over axisymemtric; given the latter is more relevant to the cylindrical tank? You should be careful here since buoyant convection flows may be unphysically sensitive to turbulence in case of 2D/axisymmetric cases. Some literature search may help here. Though, given Ra no., is it fair to say the solution is partly transitional?
Hi OJ! The solution is physically non-symmetrical. At this stage I want to see what's going on in the planer dimensions. I.e. 2D axisymmetric will give a worse prediction compared to a regular 2D simulation (assuming one is interested in the physics in the planer direction).
Quote:
Originally Posted by oj.bulmer (Post 413436) Your plots don't indicate anything about the imbalances. Plot of 3 timesteps (including coefficient loop values) should shed some light on convergence of each timestep more clearly, rather than 100 upwards timesteps as you show now. Convergence of every timestep is essential and will bring more credibility to your time-averaged results.
Sure thing - in many respects CFX should have the "Monitor coefficient loop convergence" set to "on" by default when doing transient simulations.
Quote: | 2,556 | 11,420 | {"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} | 2.828125 | 3 | CC-MAIN-2016-40 | latest | en | 0.898502 |
https://books.google.gr/books?id=ujcDAAAAQAAJ&vq=%22the+squares+on+the+whole+line+and+on+one+of+the+parts%22&dq=editions:UOMDLPabq7928_0001_001&lr=&hl=el&output=html&source=gbs_navlinks_s | 1,638,420,515,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00465.warc.gz | 232,849,230 | 12,928 | # Elements of geometry, containing books i. to vi.and portions of books xi. and xii. of Euclid, with exercises and notes, by J.H. Smith
Rivingtons, 1872 - 349 σελίδες
### Τι λένε οι χρήστες -Σύνταξη κριτικής
Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες.
### Περιεχόμενα
INTRODUCTORY REMARKS ON SOLIDS SURFACES LINES 1 and 5 TO XXVI 7 EUCLIDS PROPOSITIONS I To IV 14 EUCLIDS PROPOSITIONS IX TO XII 24 EUCLIDS PROPOSITIONS XVI AND XVII 30 EUCLIDS PROPOSITIONS XVIII TO XXIII 37 C 44
EUCLIDS PROPOSITIONS XVI To XX 143 PROPOSITION C 150 APPENDIX TO BOOKS I 198 II 211 Eucl V 224 EUCL VI 225 CONTAINING THE PROPOSITIONS OCCASIONALLY 231 EUCLIDS PROPOSITION XIII 233
### Δημοφιλή αποσπάσματα
Σελίδα 52 - If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles.
Σελίδα 17 - If two triangles have two angles of the one equal to two angles of the other, each to each ; and one side equal to one side, viz.
Σελίδα 167 - If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.
Σελίδα 69 - The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another. Let ABCD be a parallelogram, of which the diameter is AC...
Σελίδα 106 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line ; it is required to draw a straight line through the point A, parallel to the straight hue BC.
Σελίδα 88 - If a straight line be bisected, and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected; and of the square on the line made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to the point D. Then the squares on AD, DB, shall be double of the squares on AC, CD.
Σελίδα 78 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Σελίδα 91 - In every triangle, the square on the side subtending either of the acute angles, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the acute angle and the perpendicular let fall upon it from the opposite angle, Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular AD from the opposite angle.
Σελίδα 5 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.
Σελίδα 5 - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. | 832 | 3,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-49 | latest | en | 0.700149 |
https://artofproblemsolving.com/wiki/index.php/AoPS_Wiki_talk:Problem_of_the_Day/July_14,_2011 | 1,653,168,082,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00098.warc.gz | 161,813,429 | 9,367 | # AoPS Wiki talk:Problem of the Day/July 14, 2011
## Solution
We begin by factoring the given expression, $\prod_{n=1}^{39}\frac{n^2+6n+9}{n^2+6n+8}$, to $\prod_{n=1}^{39}\frac{(n+3)^2}{(n+2)(n+4)}$. Then, writing this as multiple products and shifting the indices for clarity, we get $\frac{(\prod_{n=4}^{42}n)^2}{(\prod_{n=3}^{41}n)(\prod_{n=5}^{43}n)}$. Clearly, this equals $\frac{(\frac{42!}{6})^2}{(\frac{41!}{2})(\frac{43!}{24})}$. At this point, all that is left is arithmetic. The expression equals $\frac{(42!)^2*2*24}{41!*43!*6^2}=\frac{42!}{41!}*\frac{42!}{43!}*\frac{48}{36}=42*\frac{1}{43}*\frac{4}{3}$. This trivially simplifies to $\frac{168}{129}=\boxed{\frac{56}{43}}$. | 283 | 689 | {"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": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-21 | latest | en | 0.699748 |
https://chess.stackexchange.com/questions/39460/what-is-the-name-of-this-mating-motif | 1,723,686,905,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00409.warc.gz | 128,014,048 | 39,881 | # What is the name of this mating motif?
Anyone know the name of this motif? I have seen it several times, but never knew the name. Source Polgar 5334 problems.
Can one classify this as a Legal's trap motif due to the famous Queen sacrifice(Bxd1)?
• 1.Nf6 gxf6 2. Bxf7? You might get more answers if the reader doesn't have to solve the puzzle Commented Mar 16, 2022 at 1:46
• @Edward, I am sure if a user can't solve this puzzle they will not know the answer :). Commented Mar 16, 2022 at 5:44
• looks like a variant of legal :D Commented Mar 16, 2022 at 6:16
• This position can arise after 1 e4 e5 2 Nf3 Nc6 3 Bb5 a6 4 Ba4 b5 5 Bb3 d6 6 Nc3 Bg4 7 Nd5 Na5 8 c3 Ne7 9 Nxe5 Bxd1 It is indeed a bit like Légall's mate, but it's come about in a strange way. More usually, Légall's mate involves Bxf7+, but here, so long as White has wNd5, White does not threaten Bxf7+. White's attack works only because bNe7 blocks bKe8, so it could be seen as White punishing Ne7? (instead of Nf6). So it was handy for White that they'd earlier played Nd5, which would otherwise have been a bad move. Commented Mar 16, 2022 at 7:05
• @Akavall Perhaps, but if a user knows the answer that doesn't mean they'll care to solve the puzzle. Commented Mar 16, 2022 at 22:46
The move Nxe5 echoes the famous game by Legal, because if black captures the queen, checkmate is then unavoidable. However, I would classify this motif as a `discovered attack`(which includes checkmate itself). There is some more useful information here | 454 | 1,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.960292 |
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