Wikipedia:Reference desk/Archives/Mathematics/2009 August 10

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August 10

TeX help

Looking for TeX code for integral sign having a horizontal dash midway (Dixmier trace related symbol). This would be similar to $\oint$ but with a horizontal line instead of a closed line. Tried a stike (∫) but prefer better. Ref: Noncommutative integral. If TeX doesn't have such, then question: can custom symbols be created in TeX? How?, please point. Thank you. Henry Delforn (talk) 07:54, 10 August 2009 (UTC)

The MnSymbol library might have what you need. Check here and look on page 29. Otherwise search that PDF for "integral" and you might find some others. Maelin (Talk | Contribs) 12:04, 10 August 2009 (UTC)

You can try

 \def\dashint{\mathop{\mathpalette\dodashint{}}\!\int}
 \def\dodashint#1{\setbox0\hbox{$#1\int$}\setbox0\hbox to \wd0{\hss$#1-$\hss}\wd0=0pt\box0}

It won't work in Wikipedia's texvc, if that's what you're asking for. — Emil J. 13:08, 10 August 2009 (UTC)

No didn't work. Came up with: ${\text{∫}}$ which is better than ∫ but still bad. Henry Delforn (talk) 14:59, 10 August 2009 (UTC)

You may be interested in the unicode equivalent, ⨍, U+2A0D, or the slanty version ⨏, U+2A0F. JackSchmidt (talk) 15:12, 10 August 2009 (UTC)

Here's a crude approximation which is accepted by texvc: \,\,{-}\;\;\!\!\!\!\!\!\!\!\!\int $\,\,{-}\;\;\!\!\!\!\!\!\!\!\!\int$ (the two initial spaces are only needed if something precedes the integral). — Emil J. 15:40, 10 August 2009 (UTC)

I don't care what they may say, I don’t care what they may do, you guys are just alright with me, oh yeah. Henry Delforn (talk) 20:14, 10 August 2009 (UTC)

Time Simulation in Matlab

Hi,

I want to produce a plot of two parametrised functions in real time, so i can see the path of the particle being drawn out as time increases; in Matlab. I'm currently studying fluid flows so i want to be able to see the path of a particle given the x and y equations in time. My tutor produces simulations with Maple quite easily but apparently it is possible to produce them with Matlab and since I have a copy of Matlab i'd love to know how.

I've done a bit of searching and have come across Simulink models, but they seem very complicated for what i want to be doing (I have a novice level of understanding on Matlab, I can write scripts with ease but models are beyond me).

Thanks for any help.

Pete 124.184.72.1 (talk) 10:45, 10 August 2009 (UTC)

You can try something like this:

figure;
hold on;
x = 0;
y = 0;
plot(10, 10, 'b-');
for i=1:10
    x1 = x;
    y1 = y;
    x = x + 1;
    y = y + 1;
    plot([x1, x], [y1, y], 'b-');
    drawnow;
    pause(0.1);
end

Hope that helps... --Martynas Patasius (talk) 23:04, 10 August 2009 (UTC)

Complex integration

Hi, it's me again. I'm trying to evaluate $\int _{0}^{\infty }f(x)\,dx$ with $f(x)={\frac {\cos(x)-1}{x^{2}}}$. I know that the integral converges (comparison with x^-2, and well behaved at origin), and numerical estimation yields $\int _{0}^{\infty }f(x)\,dx\approx -{\frac {\pi }{2}}$.

So... I define $g(z)={\frac {e^{iz}-1}{z^{2}}}$. It's got one singularity, a simple pole at the origin, and we have residue equal to i. Thus, any contour $\gamma$ with winding number n around the origin has $-2\pi n$ for its integral. It makes sense to me to take as a contour something like the upper half of $Re^{i}t$, the lower half of $re^{i}t$ (with R large and r small), and the lines that join them. Oriented in the usual way, this should lead to an integral that contains the one we're looking for, with other parts that can be dealt with separately, we hope.

I have no problem with the integral on the large circle going to 0. However, it seems to blow up on the smaller semicircle, and I can't see what I'm doing wrong. Should I be using a different contour? Thanks in advance for any tips. -GTBacchus^(talk) 22:04, 10 August 2009 (UTC)

Heh, I just came here to ask about the same exact problem and even the exact same part of it, the inner circle. The answer is exactly $-\pi /2$ by the way. I was doing it slightly differently, though, the upper half of $(1/R)e^{it}$. I have a solution from someone else and I am just trying to understand what they do for this. Call this contour $\gamma$. Then

$Integral_{\gamma }=\int _{\gamma }{\frac {e^{iz}-iz-1}{z^{2}}}\,dz+\int _{\gamma }{\frac {iz}{z^{2}}}\,dz.$.

The solution I am looking at makes the claim "We then see that the first integrand has a removable singularity at z = 0 and so as $R\to \infty$, this integral goes to 0." I don't get it. StatisticsMan (talk) 22:27, 10 August 2009 (UTC)

That makes some sense to me. Since ${\frac {e^{iz}-1}{z^{2}}}$ has a simple pole with residue i, then subtracting ${\frac {i}{z}}={\frac {iz}{z^{2}}}$ takes away that pole. As long as there are no other negative powers of z in the Laurent expansion, the singularity becomes removable.

I mean, I think that makes sense. Must scribble madly on whiteboard now... -GTBacchus^(talk) 22:34, 10 August 2009 (UTC)

Yea, I get that there is a removable singularity, but why does that make the integral 0? StatisticsMan (talk) 22:36, 10 August 2009 (UTC)

Wait, which part of the contour are we talking about? If it says "as $R\rightarrow \infty$", are we talking about the big circle? -GTBacchus^(talk) 22:40, 10 August 2009 (UTC)

If the singularity is removable, then the residue is zero, so that makes an integral around it equal zero. -GTBacchus^(talk) 23:03, 10 August 2009 (UTC)

First, I didn't use R and r as you did. Instead, I used R and 1/R, so sorry I confused you. So, as R goes to infinity, the radius of the inner circle goes to 0. Second, if the singularity is removable, then the residue is zero, so that makes the integral around the whole inner circle 0 but we have half of the inner circle whether you do the upper half (what I was doing) or lower half (what you were doing). We don't have the whole circle. StatisticsMan (talk) 23:25, 10 August 2009 (UTC)

Let's not use "the residue theorem" directly. By writing out the function as a Laurent series, one sees that residues close to, but outside contours can contribute to integrals. In the present case: if $C_{\epsilon }$ is a upper semi-circle of radius $\epsilon$ centered at 0, we have: $\int _{C_{\epsilon }}{\frac {e^{iz}-1}{z^{2}}}\,\mathrm {d} z=\int _{C_{\epsilon }}{\frac {i}{z}}+h\,\mathrm {d} z$, where $h$ is analytic. Now if you parameterize $C_{\epsilon }$ by $z=e^{i\theta }$ and choose clockwise orientation (which is what we need for our computation here) $\int _{C_{\epsilon }}{\frac {i}{z}}\,\mathrm {d} z=\pi$. Needless to say, this holds in the limit $\epsilon \rightarrow 0$, where the contribution of the bounded term $h$ to the integral vanishes. To wrap up the computation of the real integral, note that the integral along a large circle decays as the radius grows, by Jordan's lemma. Phils 02:18, 11 August 2009 (UTC)

I figured out how to finish it up. The point is, if it is a removable singularity, it means there exists some r such that the function is bounded on a disk of radius r about that point. In other words as long as you make r (or 1/R) small enough, the function must be bounded. So, you use the ML inequality to say the integral is less than or equal to whatever the bound is times the length which is $\pi /R$ (in my case). And, $M\pi /R$ goes to 0 as R goes to infinity. StatisticsMan (talk) 03:05, 11 August 2009 (UTC)

I get it. Thank you both very much! -GTBacchus^(talk) 17:48, 12 August 2009 (UTC)