Difference between revisions of "HWK2MA530Spring2014" - Rhea

Latest revision as of 08:33, 31 January 2014

Homework 2 collaboration area

Here it is again:

$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz. $

This is the place.

Exercise 1

Suppose that $ \varphi(z) $ is a continuous function on the trace of a path γ. Prove that the function

$ f(z)=\int_{\gamma}\frac{\varphi(\zeta)}{\zeta-z}d\zeta $

is analytic on $ \mathbb{C}-\text{tr }\gamma $.

Discussion

--A function is said to be analytic on an open set Ω if it is $ \mathbb{C} $-differentiable at every point in Ω. Since the definition of analytic involves an open set, to complete Exercise 1 we need to have an open set somewhere. Since tr γ is the continuous image of a compact set, it is compact. Since $ \mathbb{C} $ is a Hausdorff space, tr γ is closed. Alternately one can note that $ \mathbb{C} $ is homeomorphic to $ \mathbb{R}^{2} $ where we know that a compact set is closed and bounded. Since the complement of a closed set is open, $ \mathbb{C}-\text{tr }\gamma $ is open. The long and short of this is: since$ \mathbb{C}-\text{tr }\gamma $ is an open set, one just has to show that $ \varphi $ is $ \mathbb{C} $-differentiable at every point in this set in order to complete the exercise.

Exercise 2

If

$ U=\sum_{n=0}^{\infty}u_{n}\text{ and }V=\sum_{n=0}^{\infty}v_{n} $

are given by the sum of absolutely convergent series, show that

$ UV=\sum_{n=0}^{\infty}p_{n}\text{ where } p_{n}=\sum_{k=0}^{n}u_{k}v_{n-k} $

and that this sum converges absolutely.

Discussion

--We can define power series

$ \sum_{n=0}^{\infty}u_{n}z^{n},\text{ }\sum_{n=0}^{\infty}v_{n}z^{n},\text{ and }\sum_{n=0}^{\infty}p_{n}z^{n} $.

That we have absolute convergence of the first and second series with $ z=1 $ tells us that their radius of convergence is $ \geq 1 $. We could then try to use Hadamard's formula to find the radius of convergence of the third series, using information about these first two series. If we can show that this radius of convergence is $ >1 $, this will imply that we have the absolute convergence when $ z=1 $. One issue that might arise is that we really need the radius of convergence of this third series to be strictly bigger than $ 1 $ since we can't say much about convergence for $ z $ with modulus equal to the radius of convergence. It seems we may be hard pressed to obtain this strict inequality since we do not have a strict inequality for the first two series. Is Hadamard's formula the way to go? Or should we drop these power series altogether and just deal with series of complex numbers?

--It seems that Hadamard's formula may not be the way to go. These power series are actually useful (psychologically) in that the powers of $ z $ help put the coefficients in the right place. More precisely, let $ r_{m}, s_{m},\text{ and }t_{m} $ be the $ m $th partial sums for the respective $ u,v,\text{ and }p $ power series defined above. We can then relate the partial sums of these series:

$ r_{m}s_{m}=t_{m}+\left(u_{m}v_{1}+\cdots+u_{1}v_{m}\right)z^{m+1}+\cdots+u_{m}v_{m}z^{m}. $

This relationship can be taken advantage of to relate the series of coefficients: use the power series to obtain a desirable inequality, then let $ z=1 $. Now we certainly could have obtained this inequality just by working with the coefficient series themselves, but the powers of $ z $ naturally group the terms in a natural and useful way; everything amounts to polynomial multiplication.

It only takes a little more work from here to show the absolute convergence of the series we're interested in.

$ \sqrt{\frac{1+\pi}{1+z^2}} $

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