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Let $f\colon G\rightarrow\mathbb{C}$ be a complex valued function given by $f(z)=\exp(\frac{z}{1-z})$.

Prove that the Taylor series' coefficients of $f$ at $0$ are $$a_0=1 \qquad \qquad a_n=\sum_{s=1}^n \frac{1}{s!} \binom{n-1}{s-1}$$

Thoughts: My idea was to do induction in n. I found $a_0=1$ pretty easy as $\exp(0)=1$. It's the induction step that is causing me some trouble and I need some help with! Also is there an easier way than induction to prove this? Thank you!

CruZ
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  • What you want to show by induction is that $f^{(s)}(0) = \binom{n-1}{s-1}$. Perhaps you can try to guess a pattern for the $s$th derivative of $f$? You can't use $f^{(s-1)}(0)$ to find $f^{(s)}(0)$, but you can use $f^{(s-1)}(z)$ to find $f^{(s)}(z)$... – Luke Collins Feb 24 '20 at 21:59
  • Hmm I can't quite see the pattern. I have differentiated 3 times and it seems $f^{(s)}(z)=\frac{exp(\frac{z}{1-z})g(z)}{(1-z)^{2s}}$ where g(z) is the part I can't see the pattern of... – CruZ Feb 24 '20 at 22:13
  • Actually, you're right, upon inspection the derivatives do seem to complicate themselves. How about a more direct approach: have you tried simplifying $\sum_{n=0}^\infty a_n z^n$ by substituting those $a_n$ directly? I'm sure that the exponential function should pop up somehow, maybe write the binomial as $\frac{(n-1)!}{(n-s)!(s-1)!}$ – Luke Collins Feb 24 '20 at 22:35
  • I can't really get exp to pop up... Am I missing a common way to express it? – CruZ Feb 24 '20 at 22:44
  • What is $G$ here? – YiFan Tey Feb 24 '20 at 23:18
  • See the answer here: https://math.stackexchange.com/questions/3558912/radius-of-convergence-when-coefficient-is-a-sum – Kavi Rama Murthy Feb 24 '20 at 23:29

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Notice that \begin{align*} \exp\Big(\frac z{z-1}\Big) &= \sum_{k=0}^\infty \frac1{k!}\Big(\frac{z}{1-z}\Big)^k\\[3pt] &= \sum_{k=0}^\infty \frac{z^k}{k!}(1-z)^{-k}\\[3pt] &= 1+ \sum_{k=1}^\infty \frac{1}{k!}\Big(\sum_{n=0}^\infty\frac{(k+n-1)!}{(k-1)!\,n!}\, z^{n+k}\Big)\\[3pt] &= 1+\sum_{k=1}^\infty \frac{1}{k!}\Big(\sum_{n=k}^\infty\frac{(n-1)!}{(k-1)!\,(n-k)!}\, z^n\Big) \quad[n\leftarrow n-k]\\[3pt] &= 1+\sum_{k=1}^\infty \frac{1}{k!}\Big(\sum_{n=k}^\infty\binom{n-1}{k-1} z^n\Big)\\[3pt] &=1+\sum_{k=1}^\infty \sum_{n=k}^\infty\frac{1}{k!}\binom{n-1}{k-1} z^n\\[3pt] &=1+\sum_{n=1}^\infty \Big(\sum_{s=1}^n\frac{1}{s!}\binom{n-1}{s-1}\Big) z^n, \end{align*} where the third step applies the binomial theorem to $(1-z)^{-k}$, and the last step follows by swapping the order of summation.

Luke Collins
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  • Sorry, why can you change the order of summation? Don't you have to know the series is absolute convergent? Also how do you get the s to appear? I see that because of the binomial coefficient the terms of the inner sum are zero from a certain point so that parts makes sense with it ending up as a finite sum – CruZ Feb 25 '20 at 16:17
  • Yes, the series converges absolutely within some radius, as long as $z\neq 1$. I relabelled $k$ to $s$ so that things match up with what you asked in the question. – Luke Collins Feb 25 '20 at 16:37
  • But how do you get the sum to start at s=1? I'm a bit rusty on index changes and such in sums – CruZ Feb 25 '20 at 17:05
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    Let's use an "indicator function". For a logical proposition $\phi$, let $\mathbf 1_\phi = 1$ if $\phi$ is true, and $0$ otherwise. Then we have $$\sum_{k=1}^\infty \sum_{n=k}^\infty a_{n,k} = \sum_{k=1}^\infty \sum_{n=1}^\infty \mathbf 1_{n\geqslant k},a_{n,k} = \sum_{n=1}^\infty\sum_{k=1}^\infty \mathbf 1_{n\geqslant k},a_{n,k}= \sum_{n=1}^\infty \sum_{k=1}^na_{n,k}= \sum_{n=1}^\infty \sum_{s=1}^na_{n,s}.$$ – Luke Collins Feb 25 '20 at 17:15