tag:blogger.com,1999:blog-46280875252175613602017-02-07T20:54:40.131-08:00Mathemagician"The purpose of math is insight, not math" --My adaptation of R.W. Hamming's basic philosophy.Reid Atchesonhttp://www.blogger.com/profile/00136130547038951751noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4628087525217561360.post-39016744902211168982011-03-14T15:17:00.001-07:002011-03-14T15:51:30.912-07:00In Honor of Pi DayIn honor of $\pi$ day here are some cool facts about $\pi$.<div>
<br /></div><div><b>1. </b>Not only is $\pi$ an irrational number, it's what we call a <i>transcendental </i>number. This means that we can't compute <meta equiv="content-type" content="text/html; charset=utf-8">$\pi$ exactly as the root of a polynomial with rational coefficients (we can however get arbitrarily close this way.) It is among the few known numbers which have this property, though we know there are quite a lot of transcendental numbers (uncountably many, in fact), it's a frustratingly difficult thing to prove that a particular number is transcendental.</div><div>
<br /></div><div><b>2.</b>\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi ^2}{6} \]</div><div>
<br /></div><div><b>3. </b> \[ \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \]</div><div>
<br /></div><div><b>4. </b>If a given complex valued function $f$ defined on an open set $U$ containing $0$ of complex numbers is given by a power series of the form</div><meta equiv="content-type" content="text/html; charset=utf-8"><div>\[ f(z) = \sum_{n=0}^{\infty} a_n z^n \]</div><div>then the coefficients $a_n$ are given by the formula</div><div>\[ a_n = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z^n} dz \]</div><div>for any closed curve $\gamma$ bounding $0$ (of course $\gamma$ needs to be in $U$)</div><div>
<br /></div><div>Of further interest on #4 is that the condition that a complex valued function $f$ be given by a power series as above in a given set $U$ is in fact <i>equivalent</i> to the fact that $f$ be only differentiable <i>once</i> in that set. So in complex analysis, having one derivative is equivalent to having all derivatives is equivalent to being analytic. This is a huge departure from real analysis and has to do with a slight strengthening of the derivative obtained when moving from the real line to the complex plane.</div><div>
<br /></div><div><b>5. (Stirling's formula) </b></div><div>\[ \lim_{n\to \infty} \frac{n!}{\sqrt{2\pi n} \left(\frac{n}{e} \right)^n}=1 \]</div><div>This gives a way to approximate the large number $n!$ for large values of $n.$</div><div>
<br /></div><div>.</div><div>.</div><div>.</div><div>.</div><div>.</div><div>.</div><div>
<br /></div><div>
<br /></div><div>$\pi$ appears in a lot of unexpected places in mathematics, but what would be a post about pi without relating it to another (arguably more important) number $e$?</div><div>
<br /></div><div>$\mathbf{n+1}$</div><div>
<br /></div><div>\[ e^{i\pi}+1 = 0 \]</div><div>
<br /></div>Reid Atchesonhttp://www.blogger.com/profile/00136130547038951751noreply@blogger.com0tag:blogger.com,1999:blog-4628087525217561360.post-82567100152212234672010-03-21T21:03:00.000-07:002010-03-21T22:29:55.573-07:00Cool Pictures!Well, moderately cool for me. I finally finished the code, and was able to get some visualization ready for it. I'm going to do some very complicated geometries next, but for now we'll make do with the unit sphere. The great thing about the finite element method is that it can handle complicated geometries with little to no change in the solver code, in my case I won't have to change a single thing, all I have to do is input the mesh.<br /><br />Here is what the mesh generator gave me for the unit sphere, (chopped in half)<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_qHrMyxkjq00/S6bsxNy5dcI/AAAAAAAAAAw/xIYQXbEDybE/s1600-h/cutinhalf.jpg"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 320px; height: 240px;" src="http://2.bp.blogspot.com/_qHrMyxkjq00/S6bsxNy5dcI/AAAAAAAAAAw/xIYQXbEDybE/s320/cutinhalf.jpg" alt="" id="BLOGGER_PHOTO_ID_5451304729239451074" border="0"></a><br /><br />and here is another picture of the same sphere<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_qHrMyxkjq00/S6btAtHF79I/AAAAAAAAAA4/_oeYtmHtP6s/s1600-h/smoothside.jpg"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 320px; height: 240px;" src="http://4.bp.blogspot.com/_qHrMyxkjq00/S6btAtHF79I/AAAAAAAAAA4/_oeYtmHtP6s/s320/smoothside.jpg" alt="" id="BLOGGER_PHOTO_ID_5451304995343691730" border="0"></a><br /><br />The way I visualized the solution was by taking a cross section and moving it through the domain, having the cross section take on colors associated with the functional values. Here's the video for that<br /><br /><br /><object style="height: 344px; width: 425px"><param name="movie" value="http://www.youtube.com/v/jE0FTKsalpc"><param name="allowFullScreen" value="true"><param name="allowScriptAccess" value="always"><embed src="http://www.youtube.com/v/jE0FTKsalpc" type="application/x-shockwave-flash" allowfullscreen="true" allowScriptAccess="always" width="425" height="344"></object></embed>Reid Atchesonhttp://www.blogger.com/profile/00136130547038951751noreply@blogger.com0tag:blogger.com,1999:blog-4628087525217561360.post-24108506233172900602010-03-18T21:07:00.000-07:002010-03-18T21:20:19.687-07:00More cool thingsJust finished some Matlab code which approximates the solution to the 3D Poisson equation with Dirichlet boundaries. Several problems in electrostatics utilize this equation as a simplified model for steady state analysis. The scalar function solution has proven harder to visualize than I expected, but hopefully I will have some neat pictures up soon. I'm taking a break on the code for a day or two though.<br /><br /><br />I can't decide what to make my next post, I really want to write something about the horror genre.Reid Atchesonhttp://www.blogger.com/profile/00136130547038951751noreply@blogger.com0tag:blogger.com,1999:blog-4628087525217561360.post-26518730530197323512010-03-10T07:25:00.000-08:002010-03-21T22:17:59.793-07:00Winding NumberFirst post, so I'll start on a neat topic. There is a distinct difference between real and complex analysis, the difference is not only the existence of the imaginary unit $i,$ but the failure of common inverse functions. The most important function in analysis is the function $\exp$ defined for every complex number $z$ by the equation<br /><br />\[\exp(z) = \sum_{n=0}^{\infty}\frac{z^n}{n!}.\]<br /><br />There are various ways to define this function, but the above equation is the most concise. Common notation for this function is $\exp(z) = e^z.$ A natural question then is whether or not there exists a function $f$ such that for every complex number $z$ we have <br /><br />\[e^{f(z)}=z.\]<br /><br />The answer is no for the general case of the entire complex plane. However, if $z$ happens to be a positive real number, then the answer is yes. In this case the function $f$ is unique, and we denote it by $\ln.$ For every positive number $x$ we have <br /><br /><br />\[\ln(x)= \int_1^x{{\frac{1}{t}dt}.\]<br /><br /><br />Something goes wrong in the complex case which prevents the existence of a universal inverse of $\exp.$ For one thing, the notion of integration of complex valued functions is much different than that of real valued functions. For functions like $\exp$ we have for any closed path $\gamma$ that<br /><br />\[\int_{\gamma}\exp(z)dz = 0 \]<br /><br />(For definitions of these concepts, see Rudin's <span style="font-style:italic;">Real and Complex Analysis</span> page 200-201.) The notion of a path integral is not so difficult, as it is simply the same thing as one might have encountered in a vector calculus course. Functions which satisfy the above equation for any closed path $\gamma$ in a region $\Omega$ are said to be <span style="font-style:italic;">holomorphic</span> in $\Omega.$ This fact can be used to show that the path of integration of holomorphic functions does not matter, and one can simply pick the easiest path to evaluate the integral. <br /><br />This is where the problem occurs, there is a function $\ln$ in complex analysis, but it fails to be holomorphic in any region which includes a portion of the negative real axis. The question then is what exactly happens if we have a given path $\gamma$ and we perform the integration<br /><br /><br />\[\int_{\gamma}{{\frac{1}{z}dz}?\]<br /><br />The answer is (intuitively): It depends on how many times the path "loops" around the origin. For every loop the path makes, the integral picks up another $2\pi i.$ This actually has some useful consequences, as it makes the Cauchy integral formula a possibility, which leads to power series representation of functions holomorphic on a given region. There is a sophisticated theorem which says that for any closed path $\gamma$ the above integral, when divided by $2\pi i$ is always an integer. The proof of this is quite subtle, but I can present the special case of a circle. Furthermore, I will consider the full "winding number," which is actually defined to be<br /><br /><br />\[Ind_{\gamma}(\alpha)\int_{\gamma}{{\frac{1}{z-\alpha}dz}.\]<br /><br />for every complex number $\alpha.$ The main purpose of this is that it translates the origin to another location, letting us consider how many times the path loops around any given point.<br /><br />A circle is the set of points of distance $r>0$ from a given center point $\alpha$. Complex variables gives a very simple way to express this , for every $t\in[0,2\pi]$ we define $\gamma(t)= \alpha + re^{it}$ This path $\gamma$ traces out the path of a circle, it begins and ends on the point $\alpha + r$, and loops exactly once around the center $\alpha$. We will let it loop $n$ times around $\alpha$ by taking instead $t\in[0,2n\pi]$<br /><br />Suppose that $ | z - \alpha | < r $, then<br /><br />$<br />\begin{aligned}<br /><br />\frac{1}{2\pi i}\int_{\gamma}{\frac{1}{w-z}dw}= \frac{1}{2\pi i}\int_{\gamma}{\frac{1}{(w-\alpha)-(z-\alpha)}dw}\\<br /><br />\\= \frac{ir}{2\pi i}\int_0^{2n\pi}{\frac{e^{it}}{re^{it}-(z-\alpha)}}\\<br /><br />\\= \frac{1}{2\pi}\int_0^{2n\pi}{\frac{1}{1- \frac{z-\alpha}{re^{it}}}dt}\\<br /><br />\\= \frac{1}{2\pi}{\sum_{m=0}^{\infty}\int_0^{2n\pi}\left(\frac{z-\alpha}{re^{it}}}\right)^mdt}\\<br /><br /><br />\\= n + 0 + 0 + \ldots\\<br /><br /><br /><br />\end{aligned}<br />$<br /><br />Although a simple fact, it has some nice consequences, from this and path independence one can prove power series representation of holomorphic functions. This gives us eventually Liouville's theorem, which can be used to prove some very deep theorems in the theory of Banach algebras.Reid Atchesonhttp://www.blogger.com/profile/00136130547038951751noreply@blogger.com0