Yahoo Answers: Answers and Comments for Can anyone explain Cartwrite's theorem in layman terms? [Mathematics]
Copyright © Yahoo! Inc. All rights reserved.
https://answers.yahoo.com/question/index?qid=20080220084830AAN3QP4
From Anonymous
enUS
Wed, 20 Feb 2008 08:48:30 +0000
3
Yahoo Answers: Answers and Comments for Can anyone explain Cartwrite's theorem in layman terms? [Mathematics]
292
38
https://answers.yahoo.com/question/index?qid=20080220084830AAN3QP4
https://s.yimg.com/zz/combo?images/emaillogous.png

From jaz_will: A pretty good explanation can be find at Theor...
https://answers.yahoo.com/question/index?qid=20080220084830AAN3QP4
https://answers.yahoo.com/question/index?qid=20080220084830AAN3QP4
Fri, 22 Feb 2008 19:51:29 +0000
A pretty good explanation can be find at Theorem of the Day [1].
The idea is the following: we look at functions defined on the unit disc, meaning that the domain of the function is the collection of points in a plane at most a distance 1 from the origin. In particular, we are interested in so called pvalent functions. A function is called pvalent it it takes one (or more) value exactly p times and no values more than p times. (For example, the function sin(x) on the interval (0,2pi) is 2valent, while the function cos(3x) is 6valent.)
We want to know how wild such functions can be.
Cartwright's theorem says that if we also require the function be analytic (which means that it can be exactly represented as its Taylor series), then up to a universal constant, the size of a pvalent function is decided completely by its first p derivatives at the origin.
First, why is this surprising: in general, a function could very well have the first p derivatives very small while the (p+1)th derivative is very big, so that a generic function cannot be completely controlled by its first p derivatives.
Second, why this is sort of expected: we are dealing with complex analytic functions here, so the prototypical example where the first p derivatives are small and the (p+1)th derivative is big is z^(p+1): its first p derivatives are 0, while its (p+1)th is (p+1)factorial, a number that grows super exponentially in p. But as it turns out, z^(p+1) is (p+1)valent.
So Cartwright's theorem basically says that if the Taylor coefficients of the z^N term is much bigger than the ones before it, it will force the function to be at least Nvalent.
A more physical explanation is the following: an analytic function can be thought of as a surface with constant tension: the undulation and waviness of the surface is such that each point is tugging at other points equally. Now, suppose that you have a sheet of rubber, and you want to deform it very wavily. But you also want to do it in such a way that you symmetrically create ripples around the border of the sheet of rubber. So what's going to happen? You will create tension in the rubber sheet that tries its best to evenly distribute itself. But you can imagine that the some of the forces cancel out at the center of the sheet because of the symmetric nature of your deformation. So just measuring the net force at the center of the sheet is not enough to tell you how big the ripples are. The taking of derivatives is a way of breaking this symmetry: by measuring higher order shears and tucks, you can eliminate the effect of the force cancellation due to symmetry. The idea of Cartwright's theorem is that you don't have to measure up to infinite order to understand the behaviour of the deformation: if we know a priori the number of ripples we are going to make in the rubber, this puts a upper limit on how many orders of shear we need to measure before we can reconstruct a rough outline of the deformation.
Hope this helps.