I am working on my thesis, one topic therein is Euler's series transformation, would you kindly look over my derivation of it?(adsbygoogle = window.adsbygoogle || []).push({});

Workin with the function f(x) defined by the following power series (convergent for at least [tex]-1<x\leq +1[/tex],)

[tex]f(x)=\sum_{k=1}^{\infty}(-1)^{k-1}a_{k}x^{k} [/tex]

consider the change of variables given by

[tex]x=\frac{y}{1-qy},[/tex]

where q is a constant, and for which

[tex]x^k=\left( \frac{y}{1-qy}\right) ^k = y^k \left( \frac{1}{1-qy}\right) ^k = \frac{y^k}{(k-1)!q^{k-1}}\frac{d^{k-1}}{dy^{k-1}} \left( \frac{1}{1-qy}\right) = \frac{y^k}{(k-1)!q^{k-1}}\frac{d^{k-1}}{dy^{k-1}} \left( \sum_{n=0}^{\infty}q^ny^n \right) [/tex]

[tex]= \frac{y^k}{(k-1)!q^{k-1}} \left( \sum_{n=k-1}^{\infty}\frac{n!}{(k-n+1)!}q^ny^{n-k+1} \right) = \frac{1}{q^{k-1}} \sum_{n=k-1}^{\infty} \left(\begin{array}{c}n\\k-1\end{array}\right) q^ny^{n+1} [/tex]

so that we have (applying the change of variables to f(x),)

[tex] f\left( \frac{y}{1-qy}\right) = \sum_{k=1}^{\infty}(-1)^{k-1}a_{k}\left( \frac{y}{1-qy}\right) ^{k} = \sum_{k=1}^{\infty}(-1)^{k-1}a_{k} \frac{1}{q^{k-1}} \sum_{n=k-1}^{\infty} \left(\begin{array}{c}n\\k-1\end{array}\right) q^ny^{n+1} [/tex]

re-index to start with k=0 to get

[tex] f\left( \frac{y}{1-qy}\right) = \sum_{k=0}^{\infty} \sum_{n=k}^{\infty} (-1)^{k}\frac{a_{k+1}}{q^{k}} \left(\begin{array}{c}n\\k\end{array}\right) q^ny^{n+1} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} (-1)^{k}\frac{a_{k+1}}{q^{k}} \left(\begin{array}{c}n\\k\end{array}\right) q^ny^{n+1} = \sum_{n=0}^{\infty} q^ny^{n+1} \sum_{k=0}^{n} (-1)^{k} \left(\begin{array}{c}n\\k\end{array}\right) \frac{a_{k+1}}{q^{k}}[/tex]

where the order of summation was reversed according to the rule

[tex]\sum_{k=0}^{\infty} \sum_{n=k}^{\infty} b_{k,n}= \sum_{n=0}^{\infty} \sum_{k=0}^{n}b_{k,n} [/tex]

Also note, that (according to the transformation variables given above,) the value [tex]y=\frac{1}{q+1}[/tex] corresponds to [tex]x=1[/tex]; thus, considering f(1) gives

[tex]f(1) = \sum_{k=1}^{\infty}(-1)^{k-1}a_{k} = \frac{1}{q+1}\sum_{n=0}^{\infty} \left( \frac{q}{q+1}\right) ^{n} \sum_{k=0}^{n} (-1)^{k} \left(\begin{array}{c}n\\k\end{array}\right) \frac{a_{k+1}}{q^{k}}[/tex]

which, upon setting q=1, gives Euler's Series transfromation as a special case, namely

[tex]\boxed{ \sum_{k=1}^{\infty}(-1)^{k-1}a_{k} = \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}\sum_{k=0}^{n} (-1)^{k} \left(\begin{array}{c}n\\k\end{array}\right) a_{k+1} }[/tex]

which is useful for series convergence acceleration (sometimes) and for the analytic continuation for functions defined by series. Any thoughts?

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# Euler's Series Transformation

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