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# Engineering Tables/Fourier Transform Table 2

Signal Fourier transform
unitary, angular frequency
Fourier transform
unitary, ordinary frequency
Remarks
$g(t)\!\equiv\!$

$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\!\!G(\omega) e^{i \omega t} d \omega \,$
$G(\omega)\!\equiv\!$

$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\!\!g(t) e^{-i \omega t} dt \,$
$G(f)\!\equiv$

$\int_{-\infty}^{\infty}\!\!g(t) e^{-i 2\pi f t} dt \,$
10 $\mathrm{rect}(a t) \,$ $\frac{1}{\sqrt{2 \pi a^2}}\cdot \mathrm{sinc}\left(\frac{\omega}{2\pi a}\right)$ $\frac{1}{|a|}\cdot \mathrm{sinc}\left(\frac{f}{a}\right)$ The rectangular pulse and the normalized sinc function
11 $\mathrm{sinc}(a t)\,$ $\frac{1}{\sqrt{2\pi a^2}}\cdot \mathrm{rect}\left(\frac{\omega}{2 \pi a}\right)$ $\frac{1}{|a|}\cdot \mathrm{rect}\left(\frac{f}{a} \right)\,$ Dual of rule 10. The rectangular function is an idealized low-pass filter, and the sinc function is the non-causal impulse response of such a filter.
12 $\mathrm{sinc}^2 (a t) \,$ $\frac{1}{\sqrt{2\pi a^2}}\cdot \mathrm{tri} \left( \frac{\omega}{2\pi a} \right)$ $\frac{1}{|a|}\cdot \mathrm{tri} \left( \frac{f}{a} \right)$ tri is the triangular function
13 $\mathrm{tri} (a t) \,$ $\frac{1}{\sqrt{2\pi a^2}} \cdot \mathrm{sinc}^2 \left( \frac{\omega}{2\pi a} \right)$ $\frac{1}{|a|}\cdot \mathrm{sinc}^2 \left( \frac{f}{a} \right) \,$ Dual of rule 12.
14 $e^{-\alpha t^2}\,$ $\frac{1}{\sqrt{2 \alpha}}\cdot e^{-\frac{\omega^2}{4 \alpha}}$ $\sqrt{\frac{\pi}{\alpha}}\cdot e^{-\frac{(\pi f)^2}{\alpha}}$ Shows that the Gaussian function $\exp(-\alpha t^2)$ is its own Fourier transform. For this to be integrable we must have $\mathrm{Re}(\alpha)>0$.
$e^{i a t^2} = \left. e^{-\alpha t^2}\right|_{\alpha = -i a} \,$ $\frac{1}{\sqrt{2 a}} \cdot e^{-i \left(\frac{\omega^2}{4 a} -\frac{\pi}{4}\right)}$ $\sqrt{\frac{\pi}{a}} \cdot e^{-i \left(\frac{\pi^2 f^2}{a} -\frac{\pi}{4}\right)}$ common in optics
$\cos ( a t^2 ) \,$ $\frac{1}{\sqrt{2 a}} \cos \left( \frac{\omega^2}{4 a} - \frac{\pi}{4} \right)$ $\sqrt{\frac{\pi}{a}} \cos \left( \frac{\pi^2 f^2}{a} - \frac{\pi}{4} \right)$
$\sin ( a t^2 ) \,$ $\frac{-1}{\sqrt{2 a}} \sin \left( \frac{\omega^2}{4 a} - \frac{\pi}{4} \right)$ $- \sqrt{\frac{\pi}{a}} \sin \left( \frac{\pi^2 f^2}{a} - \frac{\pi}{4} \right)$
$e^{-a|t|} \,$ $\sqrt{\frac{2}{\pi}} \cdot \frac{a}{a^2 + \omega^2}$ $\frac{2 a}{a^2 + 4 \pi^2 f^2}$ a>0
$\frac{1}{\sqrt{|t|}} \,$ $\frac{1}{\sqrt{|\omega|}}$ $\frac{1}{\sqrt{|f|}}$ the transform is the function itself
$J_0 (t)\,$ $\sqrt{\frac{2}{\pi}} \cdot \frac{\mathrm{rect} \left( \frac{\omega}{2} \right)}{\sqrt{1 - \omega^2}}$ $\frac{2\cdot \mathrm{rect} (\pi f)}{\sqrt{1 - 4 \pi^2 f^2}}$ J0(t) is the Bessel function of first kind of order 0, rect is the rectangular function
$J_n (t) \,$ $\sqrt{\frac{2}{\pi}} \frac{ (-i)^n T_n (\omega) \mathrm{rect} \left( \frac{\omega}{2} \right)}{\sqrt{1 - \omega^2}}$ $\frac{2 (-i)^n T_n (2 \pi f) \mathrm{rect} (\pi f)}{\sqrt{1 - 4 \pi^2 f^2}}$ it's the generalization of the previous transform; Tn (t) is the Chebyshev polynomial of the first kind.
$\frac{J_n (t)}{t} \,$ $\sqrt{\frac{2}{\pi}} \frac{i}{n} (-i)^n \cdot U_{n-1} (\omega)\,$

$\cdot \ \sqrt{1 - \omega^2} \mathrm{rect} \left( \frac{\omega}{2} \right)$

$\frac{2 i}{n} (-i)^n \cdot U_{n-1} (2 \pi f)\,$

$\cdot \ \sqrt{1 - 4 \pi^2 f^2} \mathrm{rect} ( \pi f )$

Un (t) is the Chebyshev polynomial of the second kind


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# Riemann-Roch theorem for singular curves

It might be a naive question, but I just realized I had not thought about this before. If $C$ is a smooth curve, for any line bundle $D$ we have the Riemann-Roch formula: $$\chi(D)=\deg D+1-g(C).$$ Does this nicely extend also to singular curves as $\chi(D)=\deg D+1-p_a(C)$ ?

(if yes, is it an easy consequence of the smooth case?)

• One source is in the exercises to Hartshorne IV.1. Mar 3 '14 at 19:01
• @Andrew: You could turn this into an answer by adding the content of the exercise. Mar 3 '14 at 22:16

format: plain


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# Most important asssumption in linear models ... and the second most

library(tidyverse)
library(mosaic)

We are following here the advise of Gelman and Hill (2007, p. 46-47).

# Validity

Quite obviously, the right predictors must be included in the model in order to learn something from the model. The “right” predictors means: avoiding the wrong ones, and including the correct ones. Easier said than done, particularly with a look to the causal inference aspects. Let’s turn to the next most important assumption.

# Additivity and linearity as the second most important assumptions in linear models

We assume that $$y$$ is a linear function of the predictors. If y is not a linear function of the predictors, we cannot expect the model to deliver correct insights (predictions, causal coefficients). Let’s check an example.

## Simulate an exponentially distributed assocation

We will only use positive values, let’s add some small value to the y’s in order they are all positive.

len <- 42  # 42 x values
set.seed(42)
x <- rep(runif(len), 30)  # each x value repeated 30 times
y <- dexp(x, rate = 5) + rnorm(length(x), mean = 0, sd = .1)  # add some noise

y <- y + 1
gf_point(y ~ x)

## Linear model

### Compute the Model as linear model (not transformed)

lm1 <- lm(y ~ x)
coef(lm1)
#> (Intercept)           x
#>    3.797730   -3.458985

### Plot it

gf_point(y ~ x) %>%
gf_abline(slope = coef(lm1)[2], intercept = coef(lm1)[1], color = "red")

Doesn’t fit well.

### Check the linearity assumption

Actually the plot above vividly shows us that the linear model is not doing a good job. But anyhow, let’s put that in a form more typically used for checking model assumptions.

gf_point(lm1\$residuals ~ fitted(lm1)) %>%
gf_smooth() %>%
gf_lims(y = c(-3, 3))

If we had a linear trend, we should see no strong pattern whatsoever. The mean residual should jump around zero at each value of the x-axis, with the variance remaining (more or less) constant. This is not the case here.

## Second model: log-tranformed in y

gf_point(log(y) ~ x)

lm2 <- lm(log(y) ~ x)
coef(lm2)
#> (Intercept)           x
#>    1.331698   -1.556101
gf_point(log(y) ~ x) %>%
gf_abline(slope = coef(lm2)[2], intercept = coef(lm2)[1], color = "red")

Quite ok. Note that the discrepancies stem completely from the fact that we added some Gaussian noise.

gf_point(resid(lm2) ~ fitted(lm2)) %>%
gf_smooth() %>%
gf_lims(y = c(-3, 3))

Much better than the first model.


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If I can solve this simple urn probability problem, what do I need to solve this more complex urn problem?

I can solve this problem. An urn contains $4$ white and $5$ black balls; a second urn contains $5$ white and $4$ black ones. One ball is transferred from the first to the second urn. What is the probability that a ball then drawn from the second urn will be white?

Let $T_W$ represent the transferring of a white ball to the second urn while $T_b$ transfers a black one. Let $W$ represent the extraction of a white ball. Then the law of total probability yields

\begin{align*} P(B) &= P(B|T_B) P(T_B) + P(B|T_P) P(T_P)\\ &= \frac{6}{10} \frac{4}{9} + \frac{5}{10} \frac{5}{9}\\ &= \frac{24}{90} + \frac{25}{90}\\ &= \frac{49}{90}. \end{align*}

This problem is very clear to me. How can I adjust my knowledge to be able to solve the following one?

An urn contains $3$ white and $4$ black balls; a second urn contains $4$ white and $3$ black ones. (In extractions, the ball always goes back to the same urn where it came from --- id est, with replacement.) Extract a ball from a randomly chosen urn. If the ball is white, choose the next from urn $1$. If it's black, choose the next from urn $2$. Given that the first ball came from urn $1$, what's the probability that the second one is white?

• Use exactly the same approach. what are the ways to get a white ball in the second draw? (i) the first draw is black and the second draw is white, or (ii) the first draw is white and the second draw is also white. What happens in the two cases? – onetimething Apr 15 '18 at 0:34
• @Onetimething Here's what I do, but I'm totally not sure. To begin there are two possibilities: I could draw from first urn or second urn. However, the problem tells me the first ball came from first urn, so I can completely ignore the part of the sample space that describes first draw from second urn. Then I get a new problem: what's the probability we can draw a white ball? Now I have two possibilities: if the first ball was white, I'll use the first urn; otherwise, second urn. Based on that, I set up the following. – Joep Awinita Apr 15 '18 at 15:50
• Let $B_1$ mean black ball was the first drawn; let $W_1$ mean white ball was the first drawn. Then $P(W) = P(W| W_1) P(W_1) + P(W|B_1)P(B_1) = \frac{3}{7}\frac{3}{7} + \frac{4}{7}\frac{4}{7} = \frac{25}{49}$. – Joep Awinita Apr 15 '18 at 15:50
• @Onetimething If this is right, there's a next problem for me. Same problem but I'd want the probability that the third ball is a white given the first ball came from urn 1. I have not seen a way to solve it yet. It must be a very similar reasoning, but I haven't spotted it yet. Any directions will be appreciated. – Joep Awinita Apr 15 '18 at 19:02
• Yes, this seems perfectly reasonable to me. For the case of three draws too as you mentioned use the same approach. But there will be a lot more terms to add/multiply! – onetimething Apr 15 '18 at 23:58


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# Math Help - Is this non-linear equation solvable?

1. ## Is this non-linear equation solvable?

Hi,I need to solve this equationf''(x) + a*f'(x)^2 +b* f'(x)=0.Any clue?Thanks a lot,Isatis

2. ## Re: Is this non-linear equation solvable?

Hey isatis55.

Consider u = f'(x), then:

du/dx + a*u^2 + b*u = 0 or
du/dx = -a*u^2 - b*u which is separable.

3. ## Re: Is this non-linear equation solvable?

Originally Posted by isatis55
Hi,I need to solve this equationf''(x) + a*f'(x)^2 +b* f'(x)=0.Any clue?Thanks a lot,Isatis
First to simplify the notation I will use $f(x)=y$

While harder for this problem, in general, any 2nd order ODE can be reduced to a first order ODE by the substitution $u=y'$

By the chain rule we get

$\frac{d}{dx}y'=\frac{du}{dy}\underbrace{\frac{dy}{ dx}}_{y'=u} \iff y''=u\frac{du}{dy}$

This will change $y$ from the dependent variable to the independent variable of a new function u.

This gives

$u\frac{du}{dy}+au^2+bu=0 \iff \frac{du}{dy}+au=-b$

This can be solved by an integrating factor.


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# Determine whether the series is convergent or divergent.$\frac {1}{5} + \frac {1}{7} + \frac {1}{9} + \frac {1}{11} + \frac {1}{13} + \cdot \cdot \cdot$

## DivergesHint: Use Integral test

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### Video Transcript

were given a series and were asked to determine whether this series is convergent or divergent. So the series is 1/5 plus 1/7 plus 1/9 plus 1 11 plus 1 13 and so on. When I say and so on, what does this mean? That is, how would you continue find the next term in this some? Well, we see that we'd simply find the next odd number after 13, just 15 and take the reciprocal. So, nurse, we can write this as the sum from and equals, Let's say, zero to infinity of one over five plus to end course. This could also be written as these some from n equals one to infinity of one over two n Plus three doesn't really affect our method either way. Now, let's take ffx be the function 1/2 X plus three. Now we know that F of X is non negative on the interval from one to infinity. Indeed, it's non negative for all positive values of X. We also know that F is mon atomically decreasing on the interval from one to infinity. This is because it's an inverse function. In fact, it's decreasing for all X such that the denominator is non zero and therefore it follows that from the integral test follows that our series converges if and only if the integral from one to infinity of one over two. X plus three. The X converges. The question is, does this interval converge well and to go from one to infinity of 1/2 x plus three d x. This is the same as the limit as T approaches infinity of the integral from one duty of 1/2 x plus three d x, and this is equal to taking anti derivatives limit as T approaches infinity of one half times the natural log of two X plus three from X equals one t and plugging in. This is the limit as T approaches infinity of one half times the natural laws of two t plus three minus one half times the natural log of Just see, that's two times one is two plus three is five. Of course you know that limit This T purchase infinity of the natural log of two T plus three is again infinity. This is infinity minus one half natural log of five. Which of course is just infinity and therefore the integral diverges. And so it follows that these series also diverges by the integral test.

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Draw number lines and locate the points on them:(a) $\frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{4}{4}$(b) $\frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{7}{8}$(c) $\frac{2}{5}, \frac{3}{5}, \frac{8}{5}, \frac{4}{5}$

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To do :

We have to locate the given points on the number line.

Solution :

(a) In order to represent a fraction on the number line, we need to divide the line segment between two whole numbers into 'n' equal parts where n represents the denominator of the fraction.

Therefore,

If we have to represent the fractions $\frac{1}{2}, \frac{1}{2}(=\frac{2}{4}), \frac{3}{4}$ and $\frac{4}{4}(=1)$ on the number line, we need to divide the line segment between 0 and 1 into four equal parts.

In the above figure,

Point A represents $\frac{1}{4}$, point B represents $\frac{1}{2}$, point C represents $\frac{3}{4}$ and point D represents $\frac{4}{4}$.

(b) In order to represent a fraction on the number line, we need to divide the line segment between two whole numbers into 'n' equal parts where n represents the denominator of the fraction.

Therefore,

If we have to represent the fractions $\frac{1}{8}, \frac{2}{8}, \frac{3}{8}$ and $\frac{7}{8}$ on the number line, we need to divide the line segment between 0 and 1 into eight equal parts.

In the above figure,

Point A represents $\frac{1}{8}$, point B represents $\frac{2}{8}$, point C represents $\frac{3}{8}$ and point D represents $\frac{7}{8}$.

(c) In order to represent a fraction on the number line, we need to divide the line segment between two whole numbers into 'n' equal parts where n represents the denominator of the fraction.

Therefore,

If we have to represent the fractions $\frac{2}{5}, \frac{3}{5}, \frac{8}{5}$ and $\frac{4}{5}$ on the number line, we need to divide the line segment between 0 and 1  and also between 1 and 2 into five equal parts each.

In the above figure,

Point A represents $\frac{2}{5}$, point B represents $\frac{3}{5}$, point C represents $1\frac{3}{5}=\frac{8}{5}$ and point D represents $\frac{4}{5}$.

Updated on 10-Oct-2022 13:32:51


text:
Thanks:  0

# Thread: Definite integral on elliptic integral where modulus is function of variable

1. How to prove:

$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}}$

where $k(\theta)=\sqrt\frac{4Rh \tan \theta}{Z^2+(R+h \tan \theta)^2}$

and $K[k(\theta)]$ is the complete elliptic integral of the first kind, defined by

$K[k(\theta)]= \int_0^{\frac{\pi }{2}}\frac{\,d\phi}{\sqrt{1-k^2(\theta)\sin^2 \phi}}$

and h, R and Z $\gt 0$

2. Can you show us what you have tried and where you are stuck? This will give our helpers a better idea how to provide help without perhaps offering suggestions that you may already be trying.

Originally Posted by bshoor
How to prove:

$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}}$

where $k(\theta)=\sqrt\frac{4Rh \tan \theta}{Z^2+(R+h \tan \theta)^2}$

and $K[k(\theta)]$ is the complete elliptic integral of the first kind, defined by

$K[k(\theta)]= \int_0^{\frac{\pi }{2}}\frac{\,d\phi}{\sqrt{1-k^2(\theta)\sin^2 \phi}}$

and h, R and Z $\gt 0$
Please make correction of the post:
$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}d\theta=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}}$

Originally Posted by MarkFL
Can you show us what you have tried and where you are stuck? This will give our helpers a better idea how to provide help without perhaps offering suggestions that you may already be trying.
I have tried in different ways. But the integral becomes more and more complicated. Anyway the identity can be modified to :

$\int_{0}^{\frac{\pi }{2}}{\sqrt {\sin \theta cos\theta}k(\theta) K[k(\theta)]}d\theta=\pi \sqrt{ \frac{Rh}{R^2 + (h+Z)^2}}$

by multiplying both side by $2 \sqrt{Rh}$

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text:
# Centers of quotients of Lie Groups

Exercise 7.11 in Fulton's Representation Theory asks to prove that:

(a) Show that any discrete normal subgroup of a connected Lie group $G$ is in the center $Z(G)$

(b) If $Z(G)$ is discrete, show that $G/Z(G)$ has trivial center.

I was able to solve (a) relatively easily (for $y \in N$, with $N$ a normal discrete subgroup of $G$) by considering the continuous mapping $\phi_y : G \to G$ for $g \mapsto gyg^{-1}y^{-1}$. Since the map is continuous and $G$ is connected, its image must be connected. But since the image is contained in $N$ which is discrete, then the image must be trivial, implying that $gy = yg$ so $y \in Z(G)$ so $N \subset Z(G)$.

However, I am unsure of how to proceed in (b). I know that $G/Z(G)$ is the same as the group of inner automorphisms of $G$, but I don't know if that's how I should proceed or if I should take a different tactic.

Thanks for any help with this problem.

Later Edit:

Since I have to turn this in shortly, I want to say that I eventually ended up using a result from Stillwell's "Naive Lie Theory" that states that $Z(G)$ discrete implies that there are no nondiscrete normal subgroups. Thus, it is pretty easy to show that $G/Z(G)$ is simple, implying that its center must be trivial. However, Stillwell's result uses machinery that won't be introduced in Fulton's text for quite a while, so I'm still unsatisfied with the result. I eventually decided there were at least three possible approaches to the problem:

1. What I ultimately did, except actually using the machinery available up to this point in Fulton to prove Stillwell's result and proceeding from there
2. Possibly something involving $G/Z(G) \cong Inn(G)$, though I still don't know what
3. Proving that $G=[G,G]$ (the commutator of $G$), then using Grun's Lemma to immediately show that the center of $G/Z(G)$ is trivial, though I'm not sure that having $Z(G)$ discrete even implies this fact

I still would like to know a decent solution, so any help/suggestions/proofs would still be more than welcome.

-
Are you sure about the result of Stillwell's? Perhas it requires knowing $G$ is simple already? Because if you pick your favorite Lie group $G$ with trivial center, say, $SO(3)$. Then $Z(G\times G) = \{(e,e)\}$ so is discrete but $G\times\{e\}$ is a nondiscrete normal subgroup. –  Jason DeVito Oct 12 '11 at 11:33
Hmm, if you look at the preimage of the center of the quotient, is there some way to conclude that it must be discrete, given that Z(G) is discrete? –  Tobias Kildetoft Oct 12 '11 at 12:35

Let $Z=Z(G)$. Let $a \in G$ such that $aZ$ is in the center of $G/Z$. Then for all $b \in G$ we have $abZ=baZ$ so that $ba=abz_b$ ($=az_bb$) for some $z_b \in Z$. Thus $bab^{-1}=az_b$. Consider the normal subgroup, $N$, generated by $a$ and $Z$. In particular, $N = \{a^nz \,|\, n \in \mathbb{Z};\;z \in Z\}$. If you can show this is discrete [$N = \cup_{n\in\mathbb{Z}} a^nZ$], you'll have $N \subseteq Z$ (actually $N=Z$) by part (a) and so $a\in Z$ and so $aZ=Z$.


text:
QUESTION

# BUS 661 Assignment British Airways Swipe Card Debacle

This document of BUS 661 Assignment British Airways Swipe Card Debacle consists of:

a. From each change perspective, what are the key issues to understanding the wildcat strike?b. Assume that you have been retained as a change consultant by BA management to advise them on how to avoid such a situation in the future. What lessons emerge from each perspective and what recommendations would you draw from each in constructing your advice to BA management? If appropriate, role-play the presentation of this advice to senior management of BA.c. Is there one change perspective, or a combination of change perspectives, that provides the best way of understanding the swipe card issue? Why?d. What broad conclusions emerge from this analysis?

• @
• 2 orders completed
Tutor has posted answer for $14.29. See answer's preview$14.29


text:
## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises: 34

#### Answer

$a_{n}=3*(\frac{1}{10})^{n-1}$ Or: $a_{n}=3*(10)^{1-n}$

#### Work Step by Step

We are given: $3, 0.3, 0.03, 0.003$ ... We see that each term is the previous term divided by $10$, starting with $3$. We find the pattern: $a_1=3*10^{0}$ $a_2=3*10^{-1}$ $a_3=3*10^{-2}$ $a_4=3*10^{-3}$ Therefore: $a_{n}=3*(\frac{1}{10})^{n-1}$ Or: $a_{n}=3*(10)^{1-n}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
1. The problem statement, all variables and given/known data

A metal sphere with radius a is supported on an insulating stand at the center of a hollow, metal spherical shell with radius b. There is charge +Q on the inner sphere and charge -Q on the outer shell. Take the potential V to be zero at infinite separation.

Calculate the potential V(r) for:

i) r < a
ii) a < r < b
iii) r > b

2. Relevant equations

1) $$E = -\frac{\partial V}{\partial r}$$

2) $$V = \int E d\ell$$ (along radial direction)

3. The attempt at a solution

(iii) I already know; V is obviously 0 outside both shells (since E is also zero outside both, as a Gaussian sphere drawn beyond r(b) will enclose Q-Q = 0 charge).

I derived (ii) as a problem in lecture last week.

This leaves only (i) as the problem. The attempt was to use the second of the two equations up top, first finding the electric field so it can be applied:

Applying Gauss’ law, we can find that the Electric field inside the inner sphere is:

$$\frac{q’}{q} = \frac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi a^{3}}$$

$$q’ = q\frac{r^{3}}{a^{3}}$$

From there, using a Gaussian sphere with a radius r < a:

$$E(4\pi r^{2}) = \frac{q\frac{r^{3}}{a^{3}}}{\epsilon_{0}}$$

$$E = \frac{1}{4\pi \epsilon_{0}} \frac{qr}{a^{3}}$$

And then applying the equation for potential difference:

$$\int E d\ell = -\Delta V$$

Where dl = dr, choosing a radial path from infinity to r.

$$\int \frac{1}{4\pi \epsilon_{0}} \frac{qr}{a^{3}} dr = -\Delta V$$

$$\frac{1}{4\pi \epsilon_{0}} \frac{qr^{2}}{2a^{3}} = \Delta V = V_{r} – V_{\infty} = V_{r}$$

I’m suspicious of this solution though, as it doesn’t reduce to
$$\frac{1}{4\pi \epsilon_{0}} \frac{q}{a}$$
For the case where r = a.

Did the mistake lie in how I calculated the electric field, the technique I used to determine potential, or something else altogether?

http://ift.tt/1enSHWu


text:
# Probability one hitter is better

1. Nov 7, 2015

### KenNKC

Let's say baseball player A gets 90 hits in 300 at bats, and player B gets 25 hits in 100 at bats. The true ability of both players is unknown. What is the probability that player A has a higher true ability than player B? Would the Student's t-test be used to calculate this? Thanks.

2. Nov 7, 2015

### Hornbein

Student's t test is used on sample sizes of less than thirty.

Other than that, hits of baseball players follow which distribution? Once you know that, you can look up the test in a statistics text.

3. Nov 7, 2015

4. Nov 8, 2015

### Ronie Bayron

Don't you think that's obvious enough Player A hits (3/10) player B (2.5/10), so player A wins B. (0.5/10) difference B~A.
Even so that probability of hitting a ball per pitch is 50% and every event is independent with one over another. There's no way B would surpass A.

5. Nov 8, 2015

### Staff: Mentor

If player B is batting .250 in the major leagues and player A is batting .300 in their high school league, then player B is still probably better. The true ability required to hit depends also on the true ability of the pitcher.

Even within the same league, the difficulty of the pitchers must be accounted for. I use a similar approach for my pinewood derby.

6. Nov 8, 2015

### Ronie Bayron

Say, if we have to defeat the variability of the true ability of the pitcher for instance, and use one pitching machine for the two (A&B) instead, would that counts? I think B is defeated most probably.

7. Nov 8, 2015

### Staff: Mentor

Sure. If you are performing a controlled experiment then you could control for that. The OP wasn't clear what conditions the data were acquired under. But they sound like regular game play rather than experimentation. In regular play the skill of the competition matters greatly.

8. Nov 8, 2015

### KenNKC

Thanks for the replies. To clarify, I'm assuming both players faced equal quality pitching. Assuming a normal distribution, the true talent of player A with a 95% certainty is between .248 and .352. For player B the range is from .165 to .335. So player A's true talent could be .270 for example, and he just had a lucky 300 at bats. And player B's true talent could be .320 and he had an unlucky 100 at bats. So although Player A is more likely to be better than B based on the sample data, Player B could also be the better one. The part I am struggling with is coming up with a formula that gives this probability.

9. Nov 8, 2015

### Staff: Mentor

So the normal approximation to the binomial distribution is $\mu= np$ and $\sigma=np(1-p)$. You can take the PDF for two such normal variables and simply multiply them together to get a joint PDF.

Once you have the joint PDF, then you just integrate in the region $X>Y$ to get the probability.

I think.

10. Nov 8, 2015

### WWGD

Isn't this a difference of means test?

11. Nov 8, 2015

### Staff: Mentor

You could do it that way, but it would probably be better to do a one sided test for equal proportions.

Either way would be much simpler than my last suggestion.

12. Nov 8, 2015

### insightful

Also, remember that saying, "With 95% certainty we cannot reject the hypothesis that A is better than B," is not the same as saying, "There is a 95% probability that A is better than B."

13. Nov 8, 2015

### KenNKC

The test for equal proportions did the trick. I have a book with results of problems of the same type as my example, but the method to derive them isn't included. The equal proportions test got the right answers. For my example, the probability is about 83%. Thanks for everyone's input.

14. Nov 8, 2015

### Staff: Mentor

15. Nov 8, 2015

### Staff: Mentor

I think you are alluding to Bayesian vs frequentist methods? A frequentist approach would give the probability of the data given the hypothesis. The Bayesian approach would give the probability of the hypothesis given the data.


text:
$\displaystyle \sum_{n \le x} y^{\Omega(n)}=O\left( \frac{x(\log x)^{y-1}}{2-y} \right)$ for $1 \le y<2$

Primary tabs

Major Section:
Reference
Type of Math Object:
Theorem

Mathematics Subject Classification

Would someone mind proofreading this and see if the range on y can be extended further than $1 \le y < 2$? (The maximum is $0 \le y <2$, which seems to be possible.) I would greatly appreciate it. Thanks.


text:
Fri 19 December 2014

Monte Carlo integration is an extremely powerful method for evaluating high-dimensional integrals. It's a method that, I find, is not nearly as well known as it should be. This post is an attempt at fixing this.

The post was originally written as an IPython notebook which you can download here.

In []:
from __future__ import print_function

import matplotlib.pyplot as plt
import matplotlib.cm as cm
from matplotlib import rcParams
import numpy as np
#import seaborn as sns

In []:
tableau20 = [(31, 119, 180), (174, 199, 232), (255, 127, 14), (255, 187, 120),
(44, 160, 44), (152, 223, 138), (214, 39, 40), (255, 152, 150),
(148, 103, 189), (197, 176, 213), (140, 86, 75), (196, 156, 148),
(227, 119, 194), (247, 182, 210), (127, 127, 127), (199, 199, 199),
(188, 189, 34), (219, 219, 141), (23, 190, 207), (158, 218, 229)]
tableau20 = [ (r/255.,g/255.,b/255.) for (r,g,b) in tableau20 ]

rcParams["lines.linewidth"] = 2
#rcParams["lines.color"] = cm.Blues(1.0)
rcParams["text.usetex"] = True
rcParams["font.family"] = "sans-serif"
rcParams["font.sans-serif"] = "Computer Modern sans serif"
rcParams["mathtext.fontset"] = "custom"
rcParams["mathtext.default"] = "regular"
#rcParams["font]
rcParams["font.size"] = 18

#sns.set_style("whitegrid")
#sns.set_context("notebook", font_scale=1.5)

We want to integrate a function $f(x) = x^2$ in the domain $0 < x \le 3$. Intuitively, we can model the function as lots of small rectangles stacked vertically next to each other, and sum the area of each rectangle. If we decrease the width of each rectangle, we will eventually tend to the correct answer.

In []:
plt.figure(figsize=(7,5))
x = np.linspace(0.,3.,200)
plt.plot(x, x**2, color=tableau20[0])

rectangle_width = 0.3
nrectangles = int(round(3./rectangle_width))

for irectangle in range(nrectangles):
left = irectangle*rectangle_width
right = (irectangle+1)*rectangle_width
mid_point = (left + right)/2.
plt.fill_between([left, right], [mid_point**2, mid_point**2], facecolor=tableau20[1],edgecolor=(1.,1.,1.))

plt.xlabel("$x$")
plt.ylabel("$f(x)$")
plt.gca().spines["top"].set_visible(False)
plt.gca().spines["right"].set_visible(False)
plt.gca().get_xaxis().tick_bottom()
plt.gca().get_yaxis().tick_left()

This is the basic premise behind most traditional methods of integration. Setting this out mathematically, we approximate the integral $\int_a^b f(x) \; \mathrm{d}x$ as the sum of the area of $N$ rectangles. If the area of the $i$-th rectangle from the left (from $a$) is $A_i$, we can then write this approximation as,

$$\int_a^b f(x) \; \mathrm{d}x \simeq \sum_i^N A_i \; .$$

If we denote the mid-point of each rectangle as $x_i$, then the height of the $i$-th rectangle will be $f(x_i)$. The width of all the rectangles is identical: $\frac{b-a}{N}$. The area $A_i$ is therefore

$$A_i = f(x_i) \times \frac{b-a}{N} \; .$$

In []:
f = plt.figure(figsize=(7,5))
x = np.linspace(0.,3.,200)
ax0 = f.gca()
ax0.plot(x, x**2, color=tableau20[0])

rectangle_width = 0.3
nrectangles = int(round(3./rectangle_width))

key_rect = nrectangles-2

for irectangle in range(nrectangles):
if irectangle != key_rect:
left = irectangle*rectangle_width
right = (irectangle+1)*rectangle_width
mid_point = (left + right)/2.
ax0.fill_between([left, right], [mid_point**2, mid_point**2], facecolor=tableau20[1],edgecolor=(1.,1.,1.),alpha=0.3)

# Plotting key rectangle
left = key_rect*rectangle_width
right = (key_rect+1)*rectangle_width
mid_point = (left + right)/2.
ax0.fill_between([left, right], [mid_point**2, mid_point**2], facecolor=tableau20[1], edgecolor=(1.,1.,1.))

ax0.annotate("$x_i$",xy=(mid_point,0.), xytext=(mid_point, 1.), ha="center",arrowprops={"arrowstyle":"->","linewidth":0.5})

arrow_kwargs = {'arrowprops':dict(arrowstyle="-",
ec="k",
shrinkA=5, shrinkB=5,
linewidth=0.5
)}

ax0.annotate('', xy=(right, 0.05),
xytext=(right, mid_point**2),
**arrow_kwargs)
ax0.annotate("$f(x_i)$", xy=(right+0.15, 0.5*mid_point**2))

arrow_kwargs = {'arrowprops':dict(arrowstyle="-",
ec="k",
shrinkA=5, shrinkB=5,
linewidth=0.5
)}
ax0.annotate('', xy=(left, mid_point**2), xytext=(right, mid_point**2),**arrow_kwargs)
ax0.annotate("$(b-a)/N$", xy=(mid_point, mid_point**2+0.8),ha="center",backgroundcolor=(1.,1.,1.,0.7))

ax0.annotate("$A_i$", xy=(mid_point,5.),ha="center")

ax0.set_xlabel("$x$")
ax0.set_ylabel("$f(x)$")
ax0.spines["top"].set_visible(False)
ax0.spines["right"].set_visible(False)
ax0.get_xaxis().tick_bottom()
ax0.get_yaxis().tick_left()

Let's code this up in Python for the function $f(x) = x^2$ between 0 and 3. We will use 100 rectangles.

In []:
import numpy as np

nrectangles = 100 # number of x-values to use to calculate the average.

# Integration domain
xlow = 0.
xhigh = 3.

# Integrand
f = lambda x: x**2

xvalues = np.linspace(xlow, xhigh, nrectangles)
fvalues = f(xvalues) # f(x_i) for each rectangle
areas = fvalues * (xhigh - xlow)/nrectangles # A_i for each rectangle
integral = sum(areas)

print("Integral: ", integral)

Integral:  9.04545454545

These values are a little off the correct ones (the correct value is 9). Increasing the number of points would lead to an improved estimate.

Monte Carlo integration¶

Monte Carlo integration works in a similar manner, but, instead of distributing the rectangles uniformly in the integration domain, we distribute them randomly. Let's see this in action.

In []:
plt.figure(figsize=(7,5))
xlow, xhigh = 0., 3.
npoints = 10
plt.plot(x, x**2, color=tableau20[0])
xvalues = xlow + (xhigh-xlow)*np.random.random(npoints)
rectangle_width = 0.3
nrectangles = int(round((xhigh-xlow)/rectangle_width))

for irectangle in range(nrectangles):
mid_point = xvalues[irectangle]
left = mid_point-rectangle_width/2.
right = mid_point+rectangle_width/2.
plt.fill_between([left, right], [mid_point**2, mid_point**2], facecolor=(174/255., 199/255., 232/255.,0.5),edgecolor=(1.,1.,1.))

plt.xlabel("$x$")
plt.ylabel("$f(x)$")
plt.gca().spines["top"].set_visible(False)
plt.gca().spines["right"].set_visible(False)
plt.gca().get_xaxis().tick_bottom()
plt.gca().get_yaxis().tick_left()

As you can see, some of the rectangles overlap, and parts of the data range are not covered by rectangles. It's not even terribly clear that this gives the right answer, but you can show, with a bit of maths, that the more rectangles you use, the closer you get to the right answer. Let's code this in Python. As before, we will calculate $\int_0^3 x^2 \; \mathrm{d}x$.

In []:
# Number of rectangles
nrectangles = 100

# Integrand
f = lambda x: x**2

# Integration domain
xlow = 0.
xhigh = 3.

# randomly distributed rectangle mid-points
xvalues = xlow + (xhigh-xlow)*np.random.random(nrectangles)
fvalues = f(xvalues) # f(x_i) for each rectangle
areas = fvalues * (xhigh-xlow)/nrectangles # Area for each rectangle

integral = sum(areas)
print("Integral: ", integral)

Integral:  8.93542385496

Note that if you are running this yourself, you might get very different answers (we are using random numbers, after all). Increasing the number of points leads to estimates of the integral that converge to 9, as expected.

Monte Carlo integration in Python¶

There is a Python module for Monte Carlo integration, called scikit-monaco. There are instructions on the website on how to install it if you don't have it yet. For example, to calulate the integral of $f(x) = x^2$ over the domain $0 < x \le 3$:

In []:
from skmonaco import mcquad
nrectangles = 1e5
value, error = mcquad(lambda x: float(x**2), nrectangles, [xlow], [xhigh])
print("{} +/- {}".format(value, error))

8.99218977184 +/- 0.0254341400804

Why Monte Carlo integration?¶

If you have read this far, you might have noticed that using Monte Carlo integration gives a less accurate result than grid-based method, for the same number of points. So why is Monte Carlo integration useful?

For a given computational cost, traditional methods perform very badly if you increase the number of dimensions. So far, we have only been integrating a function of one variable. Let's now imagine that we have a function of $N$ variables, where $N$ is an integer. I will call these variables $x_1, x_2, \ldots x_N$. We now want to integrate the function $f(x_1, x_2, \ldots x_N) = x_1^2 + x_2^2 + \ldots + x_N^2$ over the domain $0< x_1 \le 1,\; 0 < x_2 \le 1,\; \ldots 0 < x_N \le 1$.

Let's look at an example with $N = 3$. The function is

$$f(x_1, x_2, x_3) = x_1^2 + x_2^2 + x_3^2 = \sum_{i=1}^3 x_i^2$$

We can represent this function in Python:

In []:
f = lambda xs : sum(xs**2)

# Let's evaluate f at x_1 = 0.2, x_2 = 0.3, x_3 = 0.4:
f(np.array([0.2, 0.3, 0.4]))

We want to compare how grid-based methods compare to Monte Carlo methods as we increase the number of dimensions. To have a consistent metric, we will assume we have a budget of $2^{18} = 262144$ function evalutations (I chose this number because it divides nicely into different powers of two, which makes generating a grid quite a lot easier). We will compare the accuracy of the prediction after this many evaluations. Download the attached IPython notebook to see the code used for generating the grid.

In []:
import itertools

xlow = 0.
xhigh = 1.
npoints = 2**18

def integrate_analytical(ndimensions):
return ndimensions * (1./3.)

def integrate_grid(ndimensions):
"""
Integrate f by calculating the average value of a set of
points distributed on an ndimensional grid in the
integration domain.
"""
assert ndimensions in (1,2,3,6,9,18),\
"Number of dimensions must be a divisor of 18."
npoints_per_dimension = 2**(18//ndimensions)
dx = (xhigh-xlow)/float(npoints_per_dimension) # spacing between points along each dimension
coordinates = np.linspace(xlow+dx/2., xhigh-dx/2., npoints_per_dimension)
# array of coordinates along each direction. Eg. if "npoints_per_dimension" is 4,
# coordinates is array([ 0.375,  1.125,  1.875,  2.625])

# "points" is a list of points. For instance, if "npoints_per_dimension" is 4
# and "ndimensions" is 9, "points" is:
# [ [ 0.375, 0.375, 0.375, 0.375, 0.375, 0.375, 0.375, 0.375, 0.375],
#   [ 0.375, 0.375, 0.375, 0.375, 0.375, 0.375, 0.375, 0.375, 1.125],
#   [ 0.375, 0.375, 0.375, 0.375, 0.375, 0.375, 0.375, 0.375, 1.875],
#   ...
#   [ 2.625,  2.625,  2.625,  2.625,  2.625,  2.625,  2.625,  2.625]]
points = np.array(list(itertools.product(coordinates, repeat=ndimensions)))

# Evaluate "f" at each point.
fvalues = [ f(point) for point in points ]

averagef = sum(fvalues) / npoints # Average value of "f" in the domain of integration

# Calculate the integral from the average by multiplying it by the integration volume
integration_volume = (xhigh-xlow)**ndimensions
integral = averagef * integration_volume
return integral

def relative_error_grid(ndimensions):
estimate = integrate_grid(ndimensions)
true_value = integrate_analytical(ndimensions)
return abs(estimate - true_value) / true_value

def relative_error_mc(ndimensions):
estimate, absolute_error = mcquad(f, npoints, [xlow]*ndimensions, [xhigh]*ndimensions)
return absolute_error/estimate

In []:
ndimensions = [ 1, 2, 3, 6, 9, 18 ]
grid_errors = [ relative_error_grid(n) for n in ndimensions ]
mc_errors = [ relative_error_mc(n) for n in ndimensions ]

In []:
plt.figure(figsize=(7,5))
plt.yscale("log")
plt.plot(ndimensions, grid_errors, color=tableau20[0])
plt.plot(ndimensions, mc_errors, color=tableau20[2])
plt.ylim(1e-5,1e-1)
plt.xticks([1,2,3,6,9,18])

plt.annotate("Monte Carlo", xy=(1.03,0.38), xycoords="axes fraction", color=tableau20[2])
plt.annotate("Grid", xy=(1.03,0.93), xycoords="axes fraction", color=tableau20[0])

plt.xlabel("number of dimensions")
plt.ylabel("relative error")
plt.gca().spines["top"].set_visible(False)
plt.gca().spines["right"].set_visible(False)
plt.gca().get_xaxis().tick_bottom()
plt.gca().get_yaxis().tick_left()

As the plot shows, the relative error in a Monte Carlo integration remains approximately the same as you increase the number of dimensions (it even decreases a little for this particular function). In contrast, the relative error of the grid-based method starts low, but increases exponentially with the number of dimensions.

For this particular problem, the two methods cross over around $d = 4.5$. So for a problem with more than about five dimensions, Monte Carlo integration gives significantly better results for the same computational effort. Using a better grid-based integration method would improve the scaling slightly, but the exponential decrease in accuracy remains.

Conclusion¶

Monte Carlo integration performs much better for high-dimensional integrals. It's not uncommon to see Monte Carlo integration used on problems with several hundreds of dimensions. Monte Carlo integration is used, in various forms, as part of Markov Chain Monte Carlo, a practical method of doing Bayesian statistical analysis, or quantum Monte Carlo, a set of methods for integrating the Schrödinger equation to predict the behaviour of quantum particles.


text:
# Maths

The 2nd, 5th, and 14th term Of an AP are consecutive terms of a GP. Given that the sum of the three terms of the AP is 9, find the
A, first term and the common difference of the AP.
B common ratio of the GP
C sum of the first three terms of the GP

1. 👍 0
2. 👎 0
3. 👁 581
1. (a+4d)/(a+d) = (a+13d)/(a+4d)
a+d + a+4d + a+13d = 9

AP: a=3/13, d=6/13
T2 = 9/13
T5 = 27/13
T14 = 81/13

Clearly the have a common ratio of 3

1. 👍 0
2. 👎 1

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3. ### math

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4. ### mathematics

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1. ### Math

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2. ### mathematics

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3. ### geometric progression

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text:
# Spin question

1. Nov 23, 2007

### lion8172

1. The problem statement, all variables and given/known data

I have a question relating to Griffith's treatment of spin. Griffiths shows, using the commutation relations for angular momentum, that L^2 has eigenvalues l(l+1), where l=0, 1/2, 1, 3/2, .... He also shows that the operator
$$-\hbar^2 \left( \frac{1}{\sin(\theta)} \frac{\partial}{\partial \theta} \left( \sin(\theta) \frac{\partial}{\partial \theta} \right) + \frac{1}{\sin^2(\theta)} \frac{\partial^2}{\partial \phi^2} \right)$$
only admits eigenvalues with integral values of l, by directly solving the differential equation. At one point, he states that the above expression is equal to L^2. My question is as follows: Is it correct to state that the half-integral eigenvalues are automatically excluded when L^2 is parametrized in terms of angles, and is it true to state that L^2 is not, in general, equal to the expression above? Spin and angular momentum satisfy the same commutation relations, so would it be correct to state that "angular momentum" is a kind of "spin" which is generated by an operator of the form $$(\hbar/i) (\vec{r} \times \nabla)$$?

2. Relevant equations

3. The attempt at a solution

2. Nov 23, 2007

### nrqed

Yes, what you say is basically correct. The language people use is slightly different, however. What you call angular momentum (represented as a differential operator acting in space) should be called "orbital angular momentum". What you call spin for the more general concept is usually called "angular momentum".

SO spin and orbital angular momenta are two types of "angular momentum". Spin is also called "intrinsic" angular momentum and is represented by matrices acting in an internal space. Orbital angular momentum is represented by differential operators acting in ordinary space.


text:
# Math Help - Closed set

1. ## Closed set

I am working on this problem, and I am not 100% sure with my argument. Could someone help me verify it?

Let (X,d) be a metric space. Suppose that $$ is a sequence in X such that it has no convergent subsequence. Claim: $A=\{x_1,x_2,x_3,...\}$ is closed in X.

By assumption, for every sequence $$ in A, $$ does not converge. Otherwise, $$ has a convergent subsequence. This implies that A is closed since I'll show that a set A is closed in (X,d) iff for every sequence $$ in A if $$ converges then $lim p_n \in A$.

First direction: Suppose A is closed. Let $$ be a convergent sequence in A. Assume $lim p_n=p \notin A$, then $lim p_n \in X-A$. But X-A is open and $lim p_n \in X-A$ implies that there exists N such that if n>N, then $p_n \in X-A$. Contradicts to our assumption that $$ is a sequence in A.

Second direction: I'll show X-A is open. Let $$ be a sequence in X converging to $p \in X-A$. It suffices to show that there exists N such that if n>N, then $p_n \in X-A$. Assume $p_n \in X-A$ for only finitely many times, then $$ is a sequence in A. By hypothesis, $lim p_n \in A$. But $lim p_n=p \in X-A$. Contradiction. So, X-A is open, ie A is closed.

2. I think there is a technicality that you need to be careful of here.

Let $$ be the sequence $$. Then $$ is a sequence in $A$, but it is not a subsequence of $$.

Thus your statement following "by assumption" is not in fact by assumption, but must be proved.

3. THEOREM If $A$ is a subset of a metric space $(M,d)$ and $p$ is a limit point of $A$ then there is sequence of distinct points from $A$ that converges to $p$.
If you can prove that theorem the you have done this question.
i.e. The set has no limits points.

4. Originally Posted by DrSteve
I think there is a technicality that you need to be careful of here.

Let $$ be the sequence $$. Then $$ is a sequence in $A$, but it is not a subsequence of $$.

Thus your statement following "by assumption" is not in fact by assumption, but must be proved.
Thanks a lot for your correction, DrSteve. I should have seen that.
Thanks Plato. I think I can prove that statement. Could you verify my argument for me? Since $p$ is a limit point of $A$, every neighborhood of $p$ has nonempty intersection with $A-\{p\}$. So, there exists $p_n \in B(p,1/n) \bigcap A-\{p\}$. The sequence $$ converges to p. But how can I be sure that this is a sequence of distinct points. Also, I don't see how proving this means I'm done with the original question. Could you elaborate on that please?

5. Originally Posted by jackie
how can I be sure that this is a sequence of distinct points.
For each $n$ make sure that $\delta_n=\min\left\{\frac{1}{n},d(p,x_{n-1}\right)\}$.
Pick $p_n\in\mathcal{B}(p;\delta_{n})\setminus\{p\}$

6. From a PM.
You know that no subsequence of $$ converges.


text:
# Why do we have constraints on the baryon asymmetry of the universe and not the lepton asymmetry?

What is it about baryons that means that CMB observations can determine their asymmetry without also including a charged leptonic component?


text:
# Change in velocity of a circular orbit?

For a body in a circular orbit will any (i.e. however small) decrease in velocity mean it falls to earth or will it go into an elliptical orbit? I originally thought that it was the first (i.e. falls to earth) but this the answer by User58220 in this question What may be effect of air friction to the velocity of satellite? seems to take the latter point of view. Here are my reasoning,

1. The force is central and therefore never does work on the tangential component of velocity.
2. This means that it will never gain the tangential velocity it has lost and therefore never be able to return to a stable orbit and will continually gain a radial velocity towards the earth until it crash.

So which is it? Could you also give the relevant equations, thanks

If the velocity decreases by a little bit, it goes into an elliptical orbit with the apogee the same as the original orbit. The orbit will be stable if no other changes happen. Only when the velocity decreases to the point where the elliptical orbit intersects the Earth's atmosphere will the object crash into the Earth.

• Typed faster than me! -_- Nov 8 '14 at 16:40
• @Cheeku No, I was just finished before you. :)
– LDC3
Nov 8 '14 at 16:41
• @LDC3 Would this orbit not need a negative eccentricity, given by $$\sqrt{1+\frac{2EL^2}{m\alpha^2}}$$ since in the case for a circular orbit it is already 0 and E is decreasing more?
– user43487
Nov 8 '14 at 17:01
• @Joseph The eccentricity is defined as a ratio of 2 positive numbers; how can it be negative? The only reason you think it is negative is that you have the distances reversed.
– LDC3
Nov 8 '14 at 18:35
• @LDC3 Sorry I meant imaginary, simply from the formula I have given above
– user43487
Nov 8 '14 at 18:58


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• CommentRowNumber1.
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• CommentTimeOct 16th 2013

Created partial map classifier, but itex won’t make harpoons even though this page claims it can.

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text:
## Tuesday, July 3, 2012

### Stefan-Boltzmann and the Green House Effect

I was really going to hold off on this until I had a little better punchline, but since I have already tossed out the ridiculously simple, Ein=Eout+dS-dW I may as well show some of my cards.  Correction: It is S, Entropy and W, Work, not the changes.

The Stefan-Boltzman law is a brilliant derivation of the energy of a two dimensional flat plate showing that the energy radiated is proportional to the absolute temperature of the flat plate object by J=alpha*sigma*T^4, where the sigma is equal to 5.67e-8 or the Stefan-Boltzmann constant.  The alpha is the curious adjustment required to make objects of varying thermodynamic properties proportional to true black bodies or that flat plate in the derivation.  Since any geometric shape can be approximated as a collection of flat plates, the S-B law, applied to nearly all object with this little caveat, "the object cannot re-absorb its own emitted energy."

From the Thermodynamic derivative, "It follows from classical electrodynamics that the radiation pressure $P$ is related to the internal energy density:" P=u/3, where u is the internal energy.

This little caveat has interesting implications.  The internal energy of water is not linear.  That energy is normally expressed as enthalpy.  So for the S-B equation to perfectly apply to water it would need to be based on the absolute enthalpy not the absolute temperature of the water.

For example: the specific heat of water at 4C is 4.205 Joules per gram.  At 32C the specific heat is 4.178 Joules per gram.  Very small difference but 4.178/4.205=0.9935 is the ratio of the change in specific heat.  So instead of 1 which would be a perfect black body we have 0.9935, close enough for most engineering work.  So just looking at the surface of water at 32C we would need a small correction for the emissivity which is better than the standard 0.924 approximation used for more exact estimates of distant objects with the S-B equation.

Where absolute enthalpy comes into play is that the internal energy of the object has to be transferred to the surface of the object, if not the value 0.9935 has to change.  Using the radiation pressure analogy, the enthalpic pressure would be the sum of the specific heats from the average internal energy or enthaply density to the surface.  That would be 121.3 Joules per gram from 4C degrees to 32C degrees.  The 4C temperature of water is pretty special.  It is the temperature of maximum density of pure water and its S-B equivalent energy is approximately equal to the energy of fusion, 334.5 Wm-2 versus 334 Joules/gram.  This would appear to indicate that the surface emissivity is indeed proportional to the internal Enthaply of the system.

Indications and verification are two different critters, but that is the direction I am moving in at present, defining the Absolute Enthaply in a form comparable to radiative pressure.  Kinda fun!


text:
# The Indian tent

The Indian tent is cone-shaped. Its height is 3.5 m. The diameter of the base is 2.5 m. How much canvas is needed to make a tire?

Correct result:

S =  14.595 m2

#### Solution:

$h=3.5 \ \text{m} \ \\ D=2.5 \ \text{m} \ \\ \ \\ r=D/2=2.5/2=\dfrac{ 5 }{ 4 }=1.25 \ \text{m} \ \\ s=\sqrt{ h^2+r^2 }=\sqrt{ 3.5^2+1.25^2 } \doteq 3.7165 \ \text{m} \ \\ \ \\ S=\pi \cdot \ r \cdot \ s=3.1416 \cdot \ 1.25 \cdot \ 3.7165=14.595 \ \text{m}^2$

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text:
# Showing a Function on Groups is One-to-One

• April 9th 2011, 03:42 PM
Relmiw
Showing a Function on Groups is One-to-One
I've got a problem and i'm not sure how to start this. Could somebody give me a hand?

Let G be a group of odd order.
Show that the function f: G --> G defined by f(x) = x^2 is one-to-one.
• April 9th 2011, 08:27 PM
roninpro
To show that a function is one-to-one, you need to show that the only element that maps to the identity is the identity. In other words, $\ker f=\{e\}$. You can try to combine this with Lagrange's theorem: the order of a subgroup divides the order of the group.

Can you see what to do?
• April 9th 2011, 11:28 PM
Deveno
if x^2 = e (trivially, e^2 = e in any group) and x is not e, what can we say about the order of G?
• April 10th 2011, 12:42 AM
tonio
Quote:

Originally Posted by roninpro
To show that a function is one-to-one, you need to show that the only element that maps to the identity is the identity.

This is false in general: the function $f(x)=x^2$ is not 1-1 on $\mathbb{R}^*$ , yet

$f(1)=1$.

It is true if we talk about a group (unitary ring) homomorphism, though.

Tonio

In other words, $\ker f=\{e\}$. You can try to combine this with Lagrange's theorem: the order of a subgroup divides the order of the group.

Can you see what to do?

.
• April 10th 2011, 12:43 AM
tonio
Quote:

Originally Posted by Deveno
if x^2 = e (trivially, e^2 = e in any group) and x is not e, what can we say about the order of G?

I think this won't help if the given function isn't a group homomorphism...

Tonio
• April 10th 2011, 10:29 AM
Deveno
Quote:

Originally Posted by tonio
I think this won't help if the given function isn't a group homomorphism...

Tonio

it's a hint, not the answer. my intent was to get the asker to think about the implications of G having odd order.

if x is any element in such a group, x has odd order, because....

continuing, this means that if x^2 = y^2, and if the order of x is k = 2m+1, x = xe = x(x^k) = x(x^(2m+1)) = x^(2m+2) = (x^2)^(m+1)

= (y^2)^(m+1) = y^(2m+2) so x is in <y>, so <x> < <y>. but by the same token, if the order of y is r = 2n+1, y = ye = y(y^(2n+1))

= y^(2n+2) = (y^2)^(n+1) = (x^2)^(n+1).

so y is in <x>, hence <x> = <y>, so k = r, and m = n. hence x = y^(2m+2) = y^(2m+1)y = ey = y, so g-->g^2 is injective, homomorphism or not.
• April 10th 2011, 10:48 AM
Relmiw
Quote:

Originally Posted by Deveno
it's a hint, not the answer. my intent was to get the asker to think about the implications of G having odd order.

if x is any element in such a group, x has odd order, because....

continuing, this means that if x^2 = y^2, and if the order of x is k = 2m+1, x = xe = x(x^k) = x(x^(2m+1)) = x^(2m+2) = (x^2)^(m+1)

= (y^2)^(m+1) = y^(2m+2) so x is in <y>, so <x> < <y>. but by the same token, if the order of y is r = 2n+1, y = ye = y(y^(2n+1))

= y^(2n+2) = (y^2)^(n+1) = (x^2)^(n+1).

so y is in <x>, hence <x> = <y>, so k = r, and m = n. hence x = y^(2m+2) = y^(2m+1)y = ey = y, so g-->g^2 is injective, homomorphism or not.

That is interesting that you were able to show this without using Lagrange's theorem at all. I understand how this proof works, but i don't think i ever would've thought of it. Thank you for your help
• April 10th 2011, 01:01 PM
Swlabr
Quote:

Originally Posted by tonio
This is false in general: the function is not 1-1 on , yet

.

It is true if we talk about a group (unitary ring) homomorphism, though.

Tonio

What you said isn't true - the function you give maps two elements to the identity, f(-1)=1=f(1), so what you've written doesn't form a counter-example. Specifically, it doesn't form a counter-example because your map is a homomorphism (as your group is abelian)...

I am not sure what your point is - are you perhaps getting confused with 1-1, which means injection, and a "there exists a 1-1 correspondence between the sets", which means there exists a bijection?

Alternatively, counter-example to the comment: A function is an injection if the only element which maps to the identity is the identity' would be $\phi: \mathbb{Z}_3\rightarrow \mathbb{Z}_2, 0\mapsto 0, 1\mapsto 1, 2\mapsto 1$. This isn't a homomorphism, and so your statement works. In general, the kernel of a homomorphism is trivial if and only if the function is an injection. This is easy to prove, and was what Roninpro was trying to say, and indeed did say when you realise he said ker(f)=e...normal functions don't have kernels...
• April 10th 2011, 01:46 PM
Deveno
i think the point tonio was trying to make was, there is a distinction between:

a) a homomorphism f is injective iff ker(f) = {e}, and

b) a function f is injective only if f^-1(e) = {*}, a singleton set (note the "only if").

the thing is, if f is only a function, and not a homomorphism, it could well be that the pre-image of the identity is a singleton set, without f being 1-1. tonio's example

is not the best one, since for his map f^-1(1) = {-1,1}. swlabr's example is a good counter-example for a function, but not for a function of the form x-->x^2.

in my orginal reply, i hinted at the fact that any group of even order must have an element of order 2 (the map x-->x^-1 on G-{e} must have a fixed point),

so there aren't any such counter-examples for groups of even order, x-->x^2 is never injective. as he (rightfully) pointed out, this is not the same as saying for

groups of odd order, x-->x^2 is always injective (~A --> ~B is not the same as A-->B, we have to rule out A --> ~B. here "A" is: G is of odd order,

and "B" is: the squaring map is injective).

judge's decision: tonio gets 5 points for pointing out the logical subtleties of the problem, and loses 2 for improper rebuttal.
• April 10th 2011, 06:19 PM
tonio
Quote:

Originally Posted by Deveno
i think the point tonio was trying to make was, there is a distinction between:

a) a homomorphism f is injective iff ker(f) = {e}, and

b) a function f is injective only if f^-1(e) = {*}, a singleton set (note the "only if").

the thing is, if f is only a function, and not a homomorphism, it could well be that the pre-image of the identity is a singleton set, without f being 1-1. tonio's example

is not the best one, since for his map f^-1(1) = {-1,1}. swlabr's example is a good counter-example for a function, but not for a function of the form x-->x^2.

in my orginal reply, i hinted at the fact that any group of even order must have an element of order 2 (the map x-->x^-1 on G-{e} must have a fixed point),

so there aren't any such counter-examples for groups of even order, x-->x^2 is never injective. as he (rightfully) pointed out, this is not the same as saying for

groups of odd order, x-->x^2 is always injective (~A --> ~B is not the same as A-->B, we have to rule out A --> ~B. here "A" is: G is of odd order,

and "B" is: the squaring map is injective).

judge's decision: tonio gets 5 points for pointing out the logical subtleties of the problem, and loses 2 for improper rebuttal.

I think the rebuttal was not only proper but necessary and important since what was being

rebutted was incorrect, so I should lose only 1 point for giving an incorrect counterexample.

Tonio

Pd. You know what? Take away an other point from me for being a pest and rebutting again...
• April 10th 2011, 06:25 PM
tonio
Quote:

Originally Posted by Swlabr
What you said isn't true -

No, what I said is true, but the example I gave was flawed. Nevertheless it's easy to

fix it: $f:\mathbb{R}\rightarrow \mathbb{R}\,,\,f(x)=|x|$ .

This is not a 1-1 function yet $\ker f=\{0\}$ ...

the function you give maps two elements to the identity, f(-1)=1=f(1), so what you've written doesn't form a counter-example. Specifically, it doesn't form a counter-example because your map is a homomorphism (as your group is abelian)...

I am not sure what your point is - are you perhaps getting confused with 1-1, which means injection, and a "there exists a 1-1 correspondence between the sets", which means there exists a bijection?

No. My point was that to show in general that a function is 1-1 you must show

that $f(x)=f(y)\Longrightarrow x=y$ , and it won't help showing that the unit, in case the

set has an algebraic structure that admits one, is the only elt. mapped to the unit.

The famous $"\mbox{f is 1-1 }\iff \ker f=\{1\}"$ works in general only for homomorphisms, not

for general functions.

Tonio

Alternatively, counter-example to the comment: A function is an injection if the only element which maps to the identity is the identity' would be $\phi: \mathbb{Z}_3\rightarrow \mathbb{Z}_2, 0\mapsto 0, 1\mapsto 1, 2\mapsto 1$. This isn't a homomorphism, and so your statement works. In general, the kernel of a homomorphism is trivial if and only if the function is an injection. This is easy to prove, and was what Roninpro was trying to say, and indeed did say when you realise he said ker(f)=e...normal functions don't have kernels...

.
• April 10th 2011, 11:21 PM
Swlabr
Quote:

Originally Posted by tonio
.

A counter-example where the function $x\mapsto x^2$ maps only the identity to the identity but isn't 1-1 is group with presentation,

$\langle a, b; a^2=b^2\rangle$.

It is torsion free (a group with one-relator is has torsion if and only if the relator is a true power) and so the only element which maps to the identity is the identity. If a and b are non-equal elements then the function is not 1-1 since a and b are mapped to the same element. However, a and b are non-equal elements as the group is not infinite cyclic. This is because it isomorphic to the group with relator $abab^{-1}$ (replace $a$ with $ab$ via a Tietze transformation) and so the group has abelianisation $\mathbb{Z} \times C_2$, and so cannot be infinite cyclic.

Also - my statement `what you said isn't true' was a remenant of a previous edit, when I was still trying to work out what you were saying. It was late. I was tired...
• April 11th 2011, 07:46 AM
Deveno
Quote:

Originally Posted by tonio
I think the rebuttal was not only proper but necessary and important since what was being

rebutted was incorrect, so I should lose only 1 point for giving an incorrect counterexample.

Tonio

Pd. You know what? Take away an other point from me for being a pest and rebutting again...

duly noted, and extra point deducted. but +1 for humor.


text:
Thread: Intel Xeon PHI? View Single Post
2021-07-20, 15:19   #207
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

578010 Posts

Quote:
 Originally Posted by paulunderwood I don't get to do stage 2 P-1.
The benchmarks you show (presumably on Linux) are showing significantly different tradeoffs than what I've observed and documented on Windows 10, Xeon Phi 7250.

Ram does not need to be permanently restricted to available-amount/number-of-workers, or equally allocated between workers. Let prime95 / mprime dynamically allocate what's useful, where useful, leaving the rest for a possible second (, improbable third, etc.) coincident stage 2 to use. When stage 2 completes on a given worker, the additional ram is returned to the available pool for some worker to use when needed. Assuming a Xeon Phi system contains only the 16GB MCDRAM, allow up to somewhere between 2-12GB per whatever worker needs it at the moment for P-1 stage 2. Even if you start out with almost equal exponents at t0, needing P-1 before PRP iteration zero, synchronized, all requiring P-1, allowing some to get allocated enough ram to run stage 2, some won't, and P-1 overall run times will differ, so gradually the workers will desynchronize and reduce collision between P-1 memory needs of the workers. Consider several cases.

Case 1: most workers get PRP with P-1 already performed. One worker is assigned to do P-1 continuously. That one worker should get much of the available RAM allowed to be allocated to it by prime95/mprime.
In local.txt
Code:
Memory=12288 during 7:30-23:30 else 12288

[Worker #1]

[Worker #2]

[Worker #3]

[Worker #4]

...
Case 2: With several workers running PRP, some assignments of which will need P-1 at the beginning, most of the time only one worker will be in P-1 stage 2 at a time, since P-1 overall takes ~1/30 as long as a PRP, and about 1/2 of that 1/30 is P-1 stage 2. On the occasions where one P-1 stage 2 is about to launch while another is already using a lot of memory, prime95 will restrict the newly launching stage 2 to what is left of allowed stage 2 memory global usage. I would chance 12GB as a global limit for prime95 with 4-8 workers, somewhat less for higher worker count, and see what mprime does in that case. With similar memory limit settings as for case 1. (Check free ram with all PRP no P-1 running for your chosen worker count. Allow up to ~90% of that in stage 2. Chancing a LITTLE paging/swapping of other processes isn't the end of the world. A 1-worker instance running 57M LL DC uses only ~47MB ram. A wavefront PRP appears to only need ~100-200MB ram per worker.)

Case 3: A more conservative stance for several workers running PRP and occasional needed P-1. Set both global and per-worker limits, so that with high worker counts there is a guarantee that at least 2 workers running P-1 stage 2 simultaneously each get significant memory for stage 2.
In local.txt
Code:
Memory=12288 during 7:30-23:30 else 12288

[Worker #1]
Memory=6144 during 7:30-23:30 else 6144

[Worker #2]
Memory=6144 during 7:30-23:30 else 6144

[Worker #3]
Memory=6144 during 7:30-23:30 else 6144

[Worker #4]
Memory=6144 during 7:30-23:30 else 6144

...
Case 4: (Not recommended) Restrict all workers to prime95 default 0.3GB ram limit for P-1/ECM stage 2, or available-ram/number-of-workers which might be as low as 12/64= 0.1875 GiB. Stage 2 might not run on any worker, so on average P-1 will be less effective than preceding cases. This I consider a de-optimization. The difference in benchmark PRP performance between 64 and 32 workers on Xeon Phi 7210 is roughly comparable to the loss of P-1 factor probability performance by skipping stage 2. Prime95 readme.txt indicates 0.2GB minimum for 100M stage 2 P-1, and proportional to exponent. Latencies are also very long with high worker counts.

Even a quite large exponent's P-1 stage 2 does not use all 12GiB allowed, leaving ~1 GiB for another worker to run P-1 stage 2 simultaneously.
Extrapolating the readme's guidance for desirable, setting per-worker caps ~1.2GiB could allow up to 10 simultaneous ~110M stage 2's!
With 3GB set on a single-worker 6GB ram system, that's configured to also support dual GPUs' Gpuowl system ram needs & Win10, gives ~4.2% P-1 factoring probability:

Code:
[Sat Jul 17 23:53:29 2021]
UID: Kriesel/test, M105169069 completed P-1, B1=789000, B2=28852000, ...
[Sun Jul 18 16:23:34 2021]
UID: Kriesel/test, M105175207 completed P-1, B1=789000, B2=28854000, ...
[Mon Jul 19 09:27:09 2021]
UID: Kriesel/test, M105186817 completed P-1, B1=789000, B2=28857000, ...
[Tue Jul 20 02:04:44 2021]
UID: Kriesel/test, M105190549 completed P-1, B1=789000, B2=28857000, ...
Compared to ~1.66% for B1=910000 only.
Attached Thumbnails


text:
# zbMATH — the first resource for mathematics

Completely integrable Hamiltonian systems connected with semisimple Lie algebras. (English) Zbl 0342.58017

##### MSC:
 37J35 Completely integrable finite-dimensional Hamiltonian systems, integration methods, integrability tests 17B80 Applications of Lie algebras and superalgebras to integrable systems 70H05 Hamilton’s equations 70F10 $$n$$-body problems
Full Text:
##### References:
 [1] Arnold, V.I.: Mathematical methods of classical mechanisms. Moscow, 1974 (in Russian) [2] Moser, J.: Three integrable Hamiltonian systems connected with isospectral deformations. Adv. Math.16, 197 (1975). · Zbl 0303.34019 [3] Calogero, F.: Solution of the one-dimensionaln-body problems with quadratic and/or inversly quadratic pair potentials. J. Math. Phys.12, 419-436 (1971) [4] Sutherland, B.: Exact results for a quantum many-body problem in one dimension. II. Phys. Rev.A5, 1372-1376 (1972) [5] Lax, P. D.: Integrals of nonlinear equations of evolution and solitary waves. Comm. Pure Appl. Math.21, 467-490 (1968) · Zbl 0162.41103 [6] Calogero, F., Marchioro C., Ragnisco, O.: Exact solution of the classical and quantal one-dimensional many-body problems with the two-body potentialV ?(x)=g 2 a 2/sinh2 a x. Lett. Nuovo Cim.13, 383-387 (1975) [7] Calogero, F.: Exactly solvable one-dimensional many-body problems. Lett. Nuovo Cim.13, 411-416 (1975) [8] Bourbaki, N.: Groupes et algebras de Lie, Chapitres 4 5 et 6. Paris: Hermann 1969 · Zbl 0205.06001 [9] Olshanetsky, M. A., Perelomov, A. M.: Completely integrable classical systems connected with semisimple Lie algebras. I. Lett. Math. Phys. (1976) · Zbl 0343.70010 [10] Perelomov, A. M.: Completely integrable classical systems connected with semisimple Lie algebras. II. Preprint ITEP No. 27 (1976) · Zbl 0343.70010 [11] Adler, M.: A New Integrable System and a Conjecture by Calogero. Preprint (1975) [12] Bateman, H., Erdélyi, A.: Higher transcendental functions, v. 3, N. Y. 1955 · Zbl 0143.29202
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# How to use procedural variables

Copy the text below and it will demonstrate the use of procedural variables, this is a fully working program. You don't even need to understand how it works the syntax is pretty simple.

program Test;

{$mode objfpc}{$H+}
uses
{$IFDEF UNIX}{$IFDEF UseCThreads}
{$ENDIF}{$ENDIF}
Classes;

// Make the Types the type corresponds to a function signature
type
TFuncNoArgsString = function(): String;
TFuncOneArgsString = function(x: string): string;

// Example functions
function Hello: String;
begin
Result := 'Hello There';
end;

function Woah(G: String): String;
begin
Result := 'Woah ' + G;
end;

// Overloaded function takes the two types of function
// pointers created above
begin
WriteLn(f());
end;

begin
WriteLn(f('there!!!'));
end;

var
ptr: Pointer;
List: TList;
begin
// the "@" symbol turns the variable into a pointer.
// This must be done in order pass a function as a
// parameter.  This also demonstrates that pascal

Take(@Hello);
Take(@Woah);

// Now put a function in an untyped pointer
ptr := @Hello;
// Type the pointer and call it all at the same time
WriteLn(TFuncNoArgsString(ptr));
// A TList Example
List := TList.Create;
WriteLn(TFuncNoArgsString(List[0]));
end.
With {$modeswitch classicprocvars on} the @-address-operator is not necessary to refer to methods. Also, if you are using the @-address-operator usage of {$typedaddress on} is advised in order to prevent programming mistakes.


text:
Evaluation symbol...
07-14-2015, 08:24 PM
Post: #1
 salvomic Senior Member Posts: 1,394 Joined: Jan 2015
Evaluation symbol...
hi,
please, help to get evaluation symbol (|) in a program...
I would get this:
Code:
 EXPORT lagrangeValue(lista_x, lista_y, xValue) BEGIN   LOCAL y, temp;   y:= lagrange(lista_x, lista_y) | X=xValue;   RETURN y; END;
but the | symbol isn't evaluated...
What's the correct syntax?

thank you,
Salvo

∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C - DM42, DM41X - WP34s Prime Soft. Lib
07-14-2015, 09:57 PM
Post: #2
 Arno K Senior Member Posts: 448 Joined: Mar 2015
RE: Evaluation symbol...
you perhaps should use: y:=unapply(lagrange(lstA,lstB),x), and then y(xVal),
this at least works for the command line from cas, from home that is another question, the output then given by y:=unapply(lagrange(listA,listB),X) is some kind of weird: program(X,[0],X+1) and y(6) delivers Error: Invalid input, but in a cas program the solution above should work.
I know, this is not what you want to do, but with the prime one often has to use, well, let me call it, workarounds to get something done. I would say ist a bug in the where function, as in simple instructions like x+1|x=6 it works correctly in cas (lower case x) and home (upper case x).
Arno
07-14-2015, 10:06 PM
Post: #3
 salvomic Senior Member Posts: 1,394 Joined: Jan 2015
RE: Evaluation symbol...
(07-14-2015 09:57 PM)Arno K Wrote:  you perhaps should use: y:=unapply(lagrange(lstA,lstB),x), and then y(xVal),
this at least works for the command line from cas, from home that is another question, the output then given by y:=unapply(lagrange(listA,listB),X) is some kind of weird: program(X,[0],X+1) and y(6) delivers Error: Invalid input, but in a cas program the solution above should work.
I know, this is not what you want to do, but with the prime one often has to use, well, let me call it, workarounds to get something done. I would say ist a bug in the where function, as in simple instructions like x+1|x=6 it works correctly in cas (lower case x) and home (upper case x).
Arno

hi Arno, thanks for input.
I'm trying, but very weird results, yes!
I put y:= unapply(lagrange(lista_x, lista_y), xValue); // xValue is 30 now
I get also "program(30,[0],5.12234233945E-5": I should get only about 0.5

Is that as you told me?

Salvo

∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C - DM42, DM41X - WP34s Prime Soft. Lib
07-14-2015, 10:32 PM
Post: #4
 salvomic Senior Member Posts: 1,394 Joined: Jan 2015
RE: Evaluation symbol...
ok,
the best way could be this:
Code:
 EXPORT lagrangeValue(lista_x, lista_y, xValue) BEGIN   LOCAL y, temp;   // y:= unapply(lagrange(lista_x, lista_y), X);   temp:= lagrange(lista_x, lista_y);   y:= EVAL(EXPR(temp+"|X="+xValue));   RETURN y; END;

my input is:
lagrangeValue({29.43,30.97,27.69,28.11,31.58,33.05},{0.4913598528,0.5145891926, 0.4646875083, 0.4711658342, 0.5236885653, 0.5453707057}, 30)
(list of value about 30, list of sines, xValue to get=30)
I get 0.500000000018 instead of 0.5, but it's almost ok.

Salvo

∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C - DM42, DM41X - WP34s Prime Soft. Lib
07-15-2015, 07:12 AM
Post: #5
 Arno K Senior Member Posts: 448 Joined: Mar 2015
RE: Evaluation symbol...
So I had given up too early when I tried to use "|" at some Point. But eval(expr(..))is a godd idea.
I finally decided only to write programs using cas until (if ever) things will be fixed.
Arno
07-15-2015, 07:33 AM
Post: #6
 salvomic Senior Member Posts: 1,394 Joined: Jan 2015
RE: Evaluation symbol...
(07-15-2015 07:12 AM)Arno K Wrote:  So I had given up too early when I tried to use "|" at some Point. But eval(expr(..))is a godd idea.
I finally decided only to write programs using cas until (if ever) things will be fixed.
Arno

I use also CAS programs mostly for analysis and algebra facts but in many cases (like the astronomy program) I write Home program, to be usable both in CAS and in Home... If this is possible, and sometime, with | or solve or ... is a great endeavor, hi.

Salvo

∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C - DM42, DM41X - WP34s Prime Soft. Lib
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