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] and [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.
Here's another sum [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?
what's so oddball about those sums? is it that the functions are so similar-looking on different intervals?
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 [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.
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. [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 n^{th} partial sum (including the term outside the series) is given by [tex]-\frac{1}{{z^{2^{n-1}}-1}}[/tex] 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].