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Anyone know any more sums like these?by benorin
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#1
Oct2805, 03:29 AM

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Here are two cool functions defined by power series:
[tex]\sum_{n=1}^{\infty}\frac{z^{n1}}{(1z^{n})(1z^{n+1})}=\left\{\begin{array}{cc}\frac{1}{(1z)^2},&\mbox{ if } z<1 \\\frac{1}{z(1z)^2}, & \mbox{ if } z>1\end{array}\right.[/tex] and [tex]\sum_{n=1}^{\infty}\frac{z^{2^{n1}}}{1z^{2^{n}}}=\left\{\begin{array}{cc}\frac{z}{1z},&\mbox{ if } z<1 \\\frac{1}{1z}, & \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 expectedand somewhat encouraged. But please, post more nifty sums like these. 


#2
Nov1105, 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}{1z^{n}}  \frac{1}{1+z^{n1}}\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? 


#3
Nov1105, 02:07 PM

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



#4
Nov1105, 03:15 PM

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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. 


#5
Nov1505, 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}{1z}+ \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^{n1}}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]. 


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