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| + | '''Exercise 1''' | ||
| + | For what values of <math>z</math> is the series | ||
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| + | <math>\sum_{n=0}^{\infty}\left(\frac{z}{1+z}\right)^{n}</math> | ||
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| + | convergent? Same question for | ||
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| + | <math>\sum_{n=0}^{\infty}\frac{z^{n}}{1+z^{2n}}.</math> | ||
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| + | '''Discussion''' | ||
| + | |||
| + | -The first series yields readily to an application of Hadamard's formula and a subsequent analysis of those values <math>z</math> for which the limsup is equal to <math>1</math>. The second series seems a little more difficult. Neither Hadamard nor the ratio test seem to be very easy to work out. Can we recognize these terms as the derivative of another series? We know that the derivative of a power series will have the same radius of convergence as the original series. Should we try to decompose the series somehow? For example, we can write | ||
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| + | |||
| + | <math>\sum_{n=0}^{\infty}\frac{z^{n}}{1+z^{2n}}=\sum_{n=0}^{\infty}\left(\frac{\frac{1}{2}z^{n}}{z^{n}-i}+\frac{\frac{1}{2}z^{n}}{z^{n}+i}\right).</math> | ||
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| + | The trouble with this idea is that we know that if the two series in this decompostion converge, then their sum converges. The converse isn't true though... | ||
[[2014 Spring MA 530 Bell|Back to MA530, Spring 2014]] | [[2014 Spring MA 530 Bell|Back to MA530, Spring 2014]] | ||
[[Category:MA5530Spring2014Bell]] [[Category:MA530]] [[Category:Math]] [[Category:Homework]] | [[Category:MA5530Spring2014Bell]] [[Category:MA530]] [[Category:Math]] [[Category:Homework]] | ||
Revision as of 18:04, 15 February 2014
Homework 4 collaboration area
This is the place.
Exercise 1
For what values of $ z $ is the series
$ \sum_{n=0}^{\infty}\left(\frac{z}{1+z}\right)^{n} $
convergent? Same question for
$ \sum_{n=0}^{\infty}\frac{z^{n}}{1+z^{2n}}. $
Discussion
-The first series yields readily to an application of Hadamard's formula and a subsequent analysis of those values $ z $ for which the limsup is equal to $ 1 $. The second series seems a little more difficult. Neither Hadamard nor the ratio test seem to be very easy to work out. Can we recognize these terms as the derivative of another series? We know that the derivative of a power series will have the same radius of convergence as the original series. Should we try to decompose the series somehow? For example, we can write
$ \sum_{n=0}^{\infty}\frac{z^{n}}{1+z^{2n}}=\sum_{n=0}^{\infty}\left(\frac{\frac{1}{2}z^{n}}{z^{n}-i}+\frac{\frac{1}{2}z^{n}}{z^{n}+i}\right). $
The trouble with this idea is that we know that if the two series in this decompostion converge, then their sum converges. The converse isn't true though...
