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Anyone know any more sums like these?

by benorin
Tags: sums
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Oct28-05, 03:29 AM
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Here are two cool functions defined by power series:

[tex]\sum_{n=1}^{\infty}\frac{z^{n-1}}{(1-z^{n})(1-z^{n+1})}=\left\{\begin{array}{cc}\frac{1}{(1-z)^2},&\mbox{ if }
|z|<1 \\\frac{1}{z(1-z)^2}, & \mbox{ if } |z|>1\end{array}\right.[/tex]


[tex]\sum_{n=1}^{\infty}\frac{z^{2^{n-1}}}{1-z^{2^{n}}}=\left\{\begin{array}{cc}\frac{z}{1-z},&\mbox{ if }
|z|<1 \\\frac{1}{1-z}, & \mbox{ if } |z|>1\end{array}\right.[/tex]

The first sum is from (pg. 59, #1) A Course of Modern Analysis by E.T. Whittaker & G.N. Watson, and the second sum (I checked this one rigorously, but not the first) is from (pg. 267, #100b) Theory and Applications of Infinite Series by K. Knopp.

So, any other functions defined by power series that converge to one function for |z|< r and to another function for |z|>r ?

A discussion of the analytic continuation of functions (and, perhaps, the natural boundaries thereof) is nearly expected--and somewhat encouraged. But please, post more nifty sums like these.
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Nov11-05, 10:21 AM
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[tex]\frac{1}{2}\left( z+\frac{1}{z}\right) + \sum_{n=1}^{\infty} \left( z+\frac{1}{z}\right) \left(\frac{1}{1-z^{n}} - \frac{1}{1+z^{n-1}}\right) =\left\{\begin{array}{cc}z,&\mbox{ if }|z|<1 \\\frac{1}{z}, & \mbox{ if } |z|>1\end{array}\right.[/tex]

This sum is also from (pg. 99) A Course of Modern Analysis by E.T. Whittaker & G.N. Watson.

I'm looking for a general theory of such oddball sums, anybody know?
fourier jr
Nov11-05, 02:07 PM
P: 948
what's so oddball about those sums? is it that the functions are so similar-looking on different intervals?

Nov11-05, 03:15 PM
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P: 1,025
Post Anyone know any more sums like these?

No, it's that they converge to different functions inside and outside the unit circle. How do you construct such a power series? That is, given two functions [itex]F_{1}(z)\mbox{ and }F_{2}(z)[/itex], construct a power series

[tex]F(z)=\sum_{k=0}^{\infty}f_{k}(z) = \left\{\begin{array}{cc}F_{1}(z),&\mbox{ if }|z|<R \\F_{2}(z), & \mbox{ if } |z|>R\end{array}\right.[/tex]

for some fixed value of R.
Nov15-05, 07:58 PM
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Here's the general solution to the above problem, this solution is due to J. Tannery circa 1886.

[tex]\theta\left( z\right) =\frac{1}{1-z}+ \sum_{n=1}^{\infty} \frac{z^{2^{n}}}{z^{2^{n+1}}-1} = \left\{\begin{array}{cc}1,&\mbox{ if }\left| z\right| <1 \\0, & \mbox{ if } \left| z\right| >1\end{array}\right.[/tex]

For this one, the nth partial sum (including the term outside the series) is given by


the limits of which are ovbious for the interior and exterior of the unit circle.

Let [tex]F_{1}(z) \mbox{ and } F_{2}(z)[/tex] be defined on the interior and exterior of the circle [tex]\left| z\right| =R[/tex], repectively. Define

[tex]F(z)= \theta\left( \frac{z}{R} \right) F_{1}(z) + \left[ 1 - \theta\left( \frac{z}{R} \right) \right] F_{2}(z)[/tex]

then [tex]F(z)= \left\{\begin{array}{cc} F_{1}(z) ,&\mbox{ if }\left| z\right| <R \\ F_{2}(z) , & \mbox{ if } \left| z\right| >R\end{array}\right.[/tex].

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