## Calculating the area of a sectorApril 5, 2011

Posted by Phi. Isett in Uncategorized.
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Here’s a nice way to calculate the area of a sector.

sectorArea

Take your sector of a disk (it could be the whole disk but for now it’s better to picture only a partial sector with 2 radial “edges”) of radius R.  Now, foliate it into circular arcs having radii $0 < r < R$, starting each arc at the same edge of the sector.  Now, straighten each of those circular arcs so that they are perpendicular to the edge from which they begin.  The circular arc of radius $r$ has length $r \theta$ where $\theta$ is the angle of the sector.

After straightening, you end up with a triangle of base $R$ and of height $R \theta$ (this is because $R \theta$ is the length of the outermost circular arc).

Punch-line (not obvious): the map preserves area.  So you have the formula:

$A = \frac{R^2 \theta}{2}$

So why does bending the circles preserve area?

This part requires much more detail, but is still interesting.  Imagine chopping the triangle up into extremely small squares with sides perpendicular to the edges of the triangle.  I claim that the area of each of those little squares will be preserved.  If you look at one of those little (“infinitesimal”) squares $\partial_x \wedge \partial_y$ with (bottom) horizontal edge $\partial_x$ and (left) vertical edge $\partial_y$ inside of the triangle and try to map it back to the sector, you don’t end up with a rectangle but rather you get a parallelogram $\partial_x^\# \wedge \partial_y^\#$ with an oblique angle.  (At least, if your square is small enough, back in the sector it really does look like a parallelogram.  At the end of the post, I’ll say how to make this heuristic argument fully rigorous.)

When viewed in the sector, the edge which used to be vertical, $\partial_y^{\#}$, now points in the angular direction and has the same length $|\partial_y|$ (this is because vertical line segments in the triangle correspond to circular arcs in the sector).  But the horizontal edge, upon being mapped back to the sector, does not point in the radial direction, but instead makes an oblique angle with $\partial_y^\#$.  Fortunately, we never change the area of a parallelogram by shifting one edge in a direction parallel to the other.  So after shifting $\partial_x^\#$ in the direction of $\partial_y^\#$, we see that it is only the component of $\partial_x^\#$  perpendicular to $\partial_y^\#$ (i.e. the “radial” component rather than the “angular” component) that matters.  To see what this radius is, look at where $\partial_x$ lives in the triangle.  It has edges on two different vertical lines $x = x_0$ and $x = x_0 + |\partial_x|$.  Those lines, remember, used to be circles of radius $x = x_0$ and $x = x_0 + |\partial_x|$ respectively, so we conclude that the radial component of $\partial_x^\#$ has the same length $|\partial_x|$ as the original.

That is essentially why the map preserves volumes.  It’s more complicated than polar coordinates, where going back from a rectangle to a sector takes squares to rectangles.  In that case you don’t need to parallel shift, but volume is not preserved.

A formalization of this intuition involves a bit of differential geometry and the notion of “exterior powers of tangent spaces” to make the parallelograms into something mathematical.  This topic is discussed somewhat in this previous post on the Cauchy Schwartz inequality.

## Uniformly continuous function on a bounded setMarch 3, 2011

Posted by Phi. Isett in Calculus.
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Here’s a problem which is a little odd and kind of interesting; I’ll write it down before I forget about it.

Every function $f : (a,b) \to {\mathbb R}$ which is uniformly continuous on a bounded interval is bounded.  You can prove this by bounding f(x) – f(y) by breaking the total change into a bunch of small changes (using uniform continuity) just as in one common proof of the Fundamental Theorem of Calculus (this technique also gets used in the proof of Sard’s theorem, Harnack’s inequality, and some other things).

Now replace the bounded interval $(a,b)$ by a bounded subset $S \subseteq {\mathbb R}$.  Exercise: are uniformly continuous functions still bounded?

## LaTeX2WP, Princeton grad student seminar, and characteristic polynomial coefficientsNovember 15, 2009

Posted by Phi. Isett in Uncategorized.

(It’ll be pretty cool if it catches on! But for me it probably means I will help to run the blog, and for this purpose I’ll definitely need the converter.)

I need to at least try to write some kind of math, so I’ll explain something which I think is cute: how to express the coefficients of a characteristic polynomial of a matrix in terms of sums of determinants of other matrices constructed from its entries.  Actually, I’ll first give an example which contains all the ideas. Consider the ${3 \times 3}$ matrix ${A}$ whose entries are… Let’s say

$\displaystyle A = \left( \begin{array}{ccc} 1 & 2 & 5 \\ 3 & 4 & 7 \\ 6 & 8 & 9 \end{array} \right)$

The characteristic polynomial ${\chi(x)}$ is the determinant of the matrix ${xI - A}$ where $I$ is the ${3 \times 3}$ identity matrix. ${\chi(x)}$ is a degree 3 polynomial with a leading coefficient 1. In terms of the ${\Lambda^3({\mathbb R}^3)}$, we have

$\displaystyle \begin{array}{c} x-1 \\ -3 \\ -6 \end{array} \wedge \begin{array}{c} -2 \\ x-4 \\ -8 \end{array} \wedge \begin{array}{c} -5 \\ -7 \\ x-9 \end{array} = \chi(x)\cdot e_1 \wedge e_2 \wedge e_3 . \ \ \ \ \ (1)$

I discussed the one-dimensional vector space ${\Lambda^3({\mathbb R}^3)}$ and its geometric meaning in a previous post about the Cauchy-Schwartz inequality and integration. We know that ${\chi(x) = x^3 + c_2 x^2 + c_1 x + c_0}$, where ${c_0 = \chi(0)}$, and plugging in ${0}$ into the above expression, we see that ${c_0 = (-1)^3 \det(A) = \det(-A)}$. The point of this entry is that we can calculate the other derivatives by differentiating, and use the multi-linearity of the wedge product to differentiate easily.  Below the fold I will give an example of how this computation works out, I will state what nice, general formula is proven by this method, and I will discuss the geometric meaning of this computation.  (And at the end I will ask a question about this LaTeX2WP/Python business which is still troubling me)

## A Big Delay for Me and a Fourier TransformSeptember 20, 2009

Posted by Phi. Isett in Uncategorized.
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I’m not exactly sure if anybody reads this blog (as nobody has commented yet).  Just in case anybody does, I should have said a long time ago: I probably won’t post here again until the end of October.  I am still preparing for my general exam and that sort of monopolizes my time.  After that I have maybe 10-13 entries planned, but right now I am resisting the urge to actually write them.

So that the entry is not completely lame, a way to compute the Fourier transform of $f(x) = e^{-|x|}$ on the real line:

Differentiating in the sense of distributions, we have $f'' - f = -2 \delta$ where $\delta$ is the delta-function (the density function corresponding to a point mass at the origin).  By taking the Fourier transform of both sides, we conclude (depending on “where you put the $2 \pi$“)

$\hat{f}(\xi) = \frac{2}{1 + (2 \pi \xi)^2}$

(In particular, we’ve actually computed the integral of $f$ to be $2$ corresponding to $\xi = 0$)

—-  It should be noted, of course, that there are more elementary ways to compute this Fourier transform………  Also note that the Fourier transform has a meromorphic continuation into the complex plane whose poles can be anticipated from the physical space representation.

## A Proof of Liouville’s Theorem in complex analysisAugust 31, 2009

Posted by Phi. Isett in Uncategorized.
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Liouville’s theorem in complex analysis says that the only bounded holomorphic functions are constant.  This is my favorite

Proof:

Pretend you had a non-constant, bounded holomorphic function $f$ on all of $\mathbb{C}$.  Since $f$ is bounded at $\infty$, Riemann’s theorem on removable singularities implies that $f$ extends to a holomorphic (and hence continuous) function on the Riemann sphere $\mathbb{C} \cup \{ \infty \}$, which is compact.  If $f$ were not constant, the open mapping theorem would apply to this extension, and the image of the Riemann sphere would be an open subset of $\mathbb{C}$.  But this cannot be the case, because $\mathbb{C}$ has no nonempty subsets which are both open and compact.  $\Box$

There are some downsides to this proof.  It does not rely on the theory of Riemann surfaces (not a downside).  It does, however, rely on some relatively (though not truly) heavy machinery, and in order to be a correct proof, one isn’t allowed to use Liouville’s theorem to develop this machinery (but it is possible).  Liouville’s theorem follows rather immediately from Cauchy’s integral formulae, and I don’t personally know how to establish things like analyticity, Riemann’s theorem and the open mapping theorem without this tool (although I would be interested if anyone else does!).  I also don’t think it really generalizes very well to other PDE for which a Liouville theorem holds.

## On the geometric meaning of the Cauchy Schwarz inequality, an intro to exterior powers, and surface integralsAugust 19, 2009

Posted by Phi. Isett in Uncategorized.
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This is a brief remark on the Cauchy Schwarz inequality and one way of understanding its geometric meaning (at least in the context of a real inner product space).  In a real inner product space $V$, the inner product $<\cdot, \cdot>$ allows for a generalization of intuitive geometric notions of “length”, “angle”, and “perpendicular” for vectors in $V$.  For two elements $u, v \in V$, I will write $u \wedge v$ to indicate the parallelogram formed by taking $u$ and $v$ as edges.  The reader may as well simply imagine $V = {\mathbb{R}}^n$ for his favorite $n$ and that the inner product is the usual dot product.

The Cauchy Schwarz inequality says that the area of a parallelogram $u \wedge v$ is positive unless u and v are co-linear (it is also equivalent to the triangle inequality, but I will be talking about this formulation instead).  If $u$ and $v$ are co-linear (they point in the same direction or perhaps opposite directions; maybe $v = - u$) the parallelogram one forms with these two vectors is degenerate and has zero area; otherwise, you wind up with a parallelogram which has positive area.  In fact, the volume of $u \wedge v$ is $|| u || \cdot || v ||$ when the two vectors $u$ and $v$ are perpendicular.  If $u$ and $v$ fail to be perpendicular, then we can observe that shifting the edge $u$ by any amount in the direction of the other edge $v$ does not change the area of the parallelogram; thus the area of $A(u \wedge v)$ is the same as the area $A [ (u + \alpha v) \wedge v ]$ for any $\alpha \in \mathbb{R}$.  By choosing $\alpha$ to minimize the length of the first edge (pick $\alpha = - \frac{}{\|v\|^2}$), we can make both edges perpendicular.  The area of the resulting parallelogram must be non-negative (and must be positive when $u$ and $v$ are linearly independent), giving the inequality $\| (u + \alpha v) \|^2 \| v \|^2 = \|u\|^2\|v\|^2 - | |^2 \geq 0$.  (This is actually a self-contained proof of the Cauchy-Schwartz inequality, and it’s the usual proof, but with some motivation regarding how to choose $\alpha$.)

Of course, the Cauchy-Schwartz inequality is also equivalent to the triangle inequality, and the relationship between these two geometric interpretations can be seen by inspecting the parallelogram one forms by drawing “$u + v = v + u$“.

I hope to write on how formalizing this area-of-parallelogram concept works (only on an intuitive level, for now).  For those already familiar with exterior powers of vector spaces and exterior algebras (and their geometric meaning in terms of parallelograms), I can cut to the chase and say that there’s an induced inner product on the exterior powers and the Cauchy Schwartz inequality is what you get by writing $< u \wedge v , u \wedge v > \geq 0$.

I will also discuss the relation to surface integrals and determinants.

## Proposal: Mathematical ExpositionJuly 28, 2009

Posted by Phi. Isett in Essays, polymath proposals.
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Disclaimer:  My last entry discussed how the use of personal profiles, a bookmarking system and selective provision of statistics can, in combination, help polymath deal with many of the challenges it faces, including the issues of how to provide leadership and promote overall efficiency.  But when I wrote it, I considered only the problem of producing research mathematics en masse, so I do not claim that the same principles necessarily apply to the collaborative production of mathematical exposition.

First of all, I am not suggesting we discontinue working on problems — rather I’m suggesting a (necessarily) smaller side-project.  Polymath should try to write a textbook… or something expository.

## What Polymath Needs is Wasted TimeJuly 26, 2009

Posted by Phi. Isett in Essays, Uncategorized.
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The following is an extensive reply to a post on Terence Tao’s blog.

My point of view is based in part on my experience as a moderator of a webforum of up to a dozen active members which devoted a few years to the collaborative production of a complex storyline with several, deeply interwoven subplots.  Time will tell how well large collaborations can produce mathematics –they are certainly an amazing tool for story-writing, and some comparisons can be made upon abstraction so the experience may be relevant.  The Google Groups format had tremendous advantages and shortcomings, but an inability to harness people’s free time ultimately lead to our story’s stagnation.

——————————————————————–

At the moment, polymath seems to function in many ways analogously to various forms of entertainment and “time-wasting” (reading blogs and webcomics, participating in forums, watching movies, watching YouTube, etc.) – indeed, this “wasted time” is in some sense exactly the incredible resource which polymath must compete to harness, although within a more restricted audience and for the noble purpose of serving mathematics.  I am sort of joking, but this is only my interpretation of Prof. Tao’s original request that the participation in the latest polymath problem solving experiment be casual.

## Hello World, the Lagrange Multiplier Theorem, and Kuhn-Tucker conditionsJuly 26, 2009

Posted by Phi. Isett in Calculus, economics, Uncategorized.
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This is a blog which was started on a whim in order to host a very extensive reply to a blog entry of Terence Tao.  This entry is not the reply — the next entry is.

I may continue this blog to contain pieces of mathematics that I find pretty, important and/or not so commonly known.  In any case, I’d feel silly devoting an entire blog to a single reply.  So here.. I’ll take this opportunity to include my favorite proof of the Lagrange multiplier theorem, and later on maybe I’ll post a nice proof of Stirling’s formula, a group algebraic proof of the Riemann-Lebesgue lemma, or something…

The Lagrange Multiplier Theorem helps us solve constrained maximization / minimization problems, making it (among other things) extremely important in economics.  A (weak form of) the Lagrange Multiplier theorem can be stated as follows:

Let $f : \mathbb{R}^n \to \mathbb{R}$ be a real-valued, continuously differentiable function (called the “objective function” — the thing we want to maximize or minimize), and let $g : \mathbb{R}^n \to \mathbb{R}^m$ be continuously differentiable (called the “constraint function” for reasons which will soon be clear).

If $x_0$ is an extremizer of the restriction of $f$ to the (codimension $m$) “constraint manifold” given by $M \equiv \{ x : g(x) = 0 \}$, then there is a unique linear functional $\lambda : \mathbb{R}^m \to \mathbb{R}$ satisfying the equality of linear maps $Df(x_0) = \lambda \circ Dg(x_0)$.

We assume $m \leq n$ — things tend to be more complicated when the number of constraints exceeds the number of degrees of freedom.

Proof:

We know that $f$ must satisfy some first order condition: namely (by further restricting $f$ to trajectories on $M$) that $Df(x_0)v = 0$ for any $v \in \mathbb{R}^n$ which can be realized as a velocity at $x_0$ by a path within the constraint manifold. An exercise in the implicit function theorem shows that these velocities include all of $\mbox{ Ker } Dg(x_0)$ (these are all such velocities — they are the directions in which one can move without changing $g$, and together they are called the “tangent space” of $M$ at $x_0$ under generic circumstances).

Finally, now that $\mbox{ Ker } Dg(x_0) \subseteq \mbox{ Ker } Df(x_0)$, it is a trivial exercise in linear algebra to show that there is a unique linear functional $\lambda : \mathbb{R}^m \to \mathbb{R}$ defined by the identity $Df(x_0) = \lambda \circ Dg(x_0)$.  Basically, we have shown that the zero set of the linear map $Dg(x_0)$ is contained in the zero set of $Df(x_0)$, but because the two maps are linear, their level sets are all translations of the zero level set, and therefore we conclude that every level set of $Dg(x_0)$ is contained in a level set of $Df(x_0)$.

Does anyone know how to turn off the annoying preview thing for hyperlinks..? Never mind, got it.

Edit (29 Aug, 2009):

The same ideas can be used to give a proof of the Kuhn-Tucker conditions for a constrained maximization problem where  we replace the system of equalities $\{ g(x) = 0 \}$ with the system of inequalities $\{ g_i(x) \leq 0 | i =1, \ldots, m \}$.  We simply replace the role of trajectories within the constraint manifold by trajectories going into the constraint set.