Anyone know any more sums like these?

[benorin]

Here are two cool functions defined by power series:

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

and

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

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.

 

[benorin]

Here's another sum

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

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]

what's so oddball about those sums? is it that the functions are so similar-looking on different intervals?

 

[benorin]

How do you construct a series like this?

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 $F_{1}(z)\mbox{ and }F_{2}(z)$, construct a power series

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

for some fixed value of R.

 

[benorin]

This is sweet... Here's the general solution to the above problem

Here's the general solution to the above problem, this solution is due to J. Tannery circa 1886.

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

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

$$-\frac{1}{{z^{2^{n-1}}-1}}$$

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

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

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

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