1
$\begingroup$

How do we solve

$\sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} f_k = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n (n + 1)} \right)^k f_k$

for $W_{m,k}$ ?

Here is what I have attempted so far:

$\begin{array}{l} \left\{ W_{m, k} : \sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \frac{f^{(k)} (0)}{k!} \left( \frac{n + x}{n (n + 1)} \right)^k \right\}\\ \left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{n = 1}^{\infty} \sum_{k = 0}^{\infty} \frac{\frac{f^{(k)} (0)}{k!} \left( \frac{n + x}{n (n + 1)} \right)^k}{n (n + 1)} \right\}\\ \left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{m = 1}^{\infty} \sum_{k = 0}^{\infty} \frac{\left( \frac{m + x}{m (m + 1)} \right)^k}{m (m + 1)} \frac{f^{(k)} (0)}{k!} \right\}\\ \left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} \frac{\left( \frac{m + 1 + x}{(m + 1) (m + 2)} \right)^k}{(m + 1) (m + 2)} \frac{f^{(k)} (0)}{k!} \right\}\\ \left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} \frac{(m + 1 + x)^k}{((m + 1) (m + 2))^k (m + 1) (m + 2)} \frac{f^{(k)} (0)}{k!} \right\} \end{array}$

$\endgroup$
2
$\begingroup$

$\sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} f_k = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n (n + 1)} \right)^k f_k $

I'll naively expand the right side, not worrying about convergence.

$\begin{array}\\ \sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n(n+1)} \right)^k f_k &=\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k(n+x)^k\\ &=\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k\sum_{m=0}^k \binom{k}{m}x^mn^{k-m}\\ &=\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{m=0}^{\infty}\sum_{k=m}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k \binom{k}{m}x^mn^{k-m}\\ &=\sum_{m=0}^{\infty}x^m\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k=m}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k \binom{k}{m}n^{k-m}\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \left( \frac{1}{n(n+1)} \right)^k \binom{k}{m}n^{k-m}\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\binom{k}{m}\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \left( \frac{n}{n(n+1)} \right)^k n^{-m}\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\binom{k}{m}\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \left( \frac{1}{n+1} \right)^k\frac1{n^{m}}\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\binom{k}{m}\sum_{n=1}^{\infty} \left( \frac{1}{n+1} \right)^{k+1}\frac1{n^{m+1}}\\ \end{array} $

so it looks like $ W_{m, k} =\binom{k}{m}\sum_{n=1}^{\infty} \left( \frac{1}{n+1} \right)^{k+1}\frac1{n^{m+1}} $ and $ W_{m, k} =0$ for $k < m$.

$\endgroup$
3
  • $\begingroup$ Very nice, I verified that gives the right terms. Thank you, I will study this.. looks like I need to learn some more about using the binomial series $\endgroup$ – crow May 17 '18 at 16:52
  • $\begingroup$ I have explicit formulas for some of the $W_{m, n}$ and am posting them as a question. $\endgroup$ – marty cohen May 18 '18 at 2:24
  • $\begingroup$ Here is the question: math.stackexchange.com/questions/2785802/… $\endgroup$ – marty cohen May 18 '18 at 2:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.