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Let $ f_t(x) := \begin{cases} \frac{xe^{tx}}{e^x-1} & \quad x\neq0\\ 1 & \quad x=0\\ \end{cases} $ where $t\in \mathbb{R}$.

(a) Prove that there is a $\delta$ which doesn't depend on $t$ such that $f_t(x)$ has a power series expansion about 0 on $(-\delta, \delta)$.

(b) Let $f_t(x)= \sum_{n=0}^{\infty} a_n(t)\frac{x^n}{n!}$ for $x\in (-\delta, \delta)$. First, show that the identity $$ \sum_{n=0}^{\infty} a_n(t)\frac{x^n}{n!} = e^{tx} \sum_{n=0}^{\infty} a_n(0)\frac{x^n}{n!} $$ holds and using this identity prove that $a_n(t) = \sum_{m=0}^n \binom{n}{m} a_m (0) t^{n-m}$.

For part (a), I have tried find a rule to be able to express $f_t(x)$ in an equivalent power series form, but I have no idea how to deal with the extra $t$ component.

As for (b), I assume that showing that the identity holds involves some part of using the fact that $e^{tx} = \sum_{k=0}^{\infty} \frac{(tx)^k}{k!}$, but I have not yet figured out how...

What would be the method to tackle both problems? Any tips or help would be most appreciated!

EDIT: I have tried and done (I think?) part (b)! The identity follows from the fact that $f_t(x) = e^{tx} \cdot f_0(x)$, and the proof follows from finding the Cauchy product of $e^{tx}$ and $f_0(x)$. Still no luck with (a), though...

  • If you know a little bit of complex analysis, that would simplify part a) greatly. – Daniel Fischer Sep 07 '15 at 09:32
  • @DanielFischer Which theorem should I be looking into? –  Sep 07 '15 at 09:41
  • Power series representations of holomorphic functions, typically the result about the domain of validity of the Taylor series is done as an application of Cauchy's integral formula. But it's doubtful whether using results from complex analysis would be acceptable for this exercise. If not, you have to go through the "real methods" way and fiddle around with estimates for the derivatives. That, however, can still be simplified considerably by proving that the Cauchy product of the power series of $f$ and $g$ converges to $fg$ at least on the intersection of the intervals of convergence. – Daniel Fischer Sep 07 '15 at 09:48

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