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As part ofa voluntary homework assignment, I want to find the Taylor series around 0 for the complex valued function \begin{equation} f(z):=\frac{1}{1-\exp(z-1)} \end{equation}

My approach so far is to express $\exp(z-1)$ and $\frac{1}{1-z}$ as power series. Since $\exp(z)= \sum \frac{z^k}{k!}$ and $\frac{1}{1-z} = \sum{z^k}$, this leads me to \begin{equation} \frac{1}{1-\exp(z-1)} = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^m}{m!} \right)^k = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^{mk}}{(m!)^k} \right) \end{equation} To obtain a power series around 0, I use the binomial series: \begin{equation} \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^{mk}}{(m!)^k} \right) = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \sum_{n=0}^{mk} \frac{1}{(m!)^k} \binom{mk}{n} z^n \cdot (-1)^{mk-n} \end{equation} However, I am not sure how to proceed at this point. Can the last expression be simplified?

Of course, one could find the solution by using derivatives, but I should find the coefficients of the Taylor series based on the geometric and exponential power series. Am I overlooking something?

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  • $\begingroup$ Did you try by separating out $e^{-1}$ from $e^{z-1}$? treating $e^{-1}$ as a fixed constant may help write the power series in powers of $z$. Also try to write few terms explicitly. Sometimes too many summations may be more confusing. $\endgroup$
    – skylark
    Commented Jun 3, 2018 at 16:32

3 Answers 3

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HINT: An easier place to start would be $$ \frac{1}{1-e^{z-1}} = \sum_{n=0}^{\infty}\frac{e^{nz}}{e^n} $$ Then expand the exponential $e^{nz}$ in the variable $nz$, instead of raising a power series to an exponent.

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Following Himadri's suggestion:

\begin{equation} f(z):=\frac{1}{1-e^{-1}\exp(z)} = \sum_{k=0}^{\infty} e^{-k}e^{zk} = \sum_{k=0}^{\infty} e^{-k} \sum_{m=0}^{\infty} \frac{(zk)^{m}}{m!} \end{equation} Now exchange the summation orders: \begin{equation} f(z) =\sum_{m=0}^{\infty} \frac{z^m}{m!}\; \sum_{k=0}^{\infty} e^{-k} k^m =\sum_{m=0}^{\infty} \frac{z^m}{m!}\; c_m \end{equation} So the difficulty arises in evaluating the coefficients $c_m = \sum_{k=0}^{\infty} e^{-k} k^m$. Write $$ \sum_{k=0}^{\infty} e^{-k} k^m = \lim_{a\to -1}\sum_{k=0}^{\infty} e^{ak} k^m \\ = \lim_{a\to -1}\sum_{k=0}^{\infty} \frac{\partial^m}{\partial a^m}e^{ ak} = \\ = \lim_{a\to -1} \frac{\partial^m}{\partial a^m}\sum_{k=0}^{\infty} e^{ak} = \\ = \lim_{a\to -1} \frac{\partial^m}{\partial a^m}\frac{1}{1-e^{a}} $$ and proceed from there. A few coefficients are (note they do not include the $1/m!$): $$ \begin{align} c_0 &= \frac{e}{e-1} &\\ c_1 &= \frac{e}{(e-1)^2} &\\ c_2 &= \frac{e(1+e)}{(e-1)^3} &\\ c_3 &= \frac{e(1 + 4e + e^2)}{(e-1)^4} &\\ c_4 &= \frac{e (1 + 11 e + 11 e^2 + e^3)}{(e-1)^5} & \end{align} $$ They can be written as a finite sum $c_m = e\sum_{j=0}^{m}{m\brace j}\frac{j!}{\left(e-1\right)^{j+1}}$, see Markus Scheuer's answer.

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  • $\begingroup$ Are those $\frac{\delta}{\delta a}$ expressions meant to be functional derivatives or normal partial derivatives? $\endgroup$ Commented Jun 3, 2018 at 17:51
  • $\begingroup$ @probably_someone Partial derivatives. I had used a wrong "partial" - fixed now. $\endgroup$
    – Andreas
    Commented Jun 3, 2018 at 17:54
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I think it is more convenient to start with \begin{align*} f(z)=\frac{1}{1-e^{z-1}}&=\sum_{n=0}^\infty e^{(z-1)n} \end{align*} and calculate the Taylor series via \begin{align*} f(z)=\sum_{n=0}^\infty f^{(n)}(0)\frac{z^n}{n!} \end{align*}

We obtain \begin{align*} \color{blue}{\left.\left(\frac{d^k}{dz^k}f(z)\right)\right|_{z=0}}&=\left.\sum_{n=0}^\infty n^ke^{(z-1)n}\right|_{z=0}\tag{1}\\ &=\sum_{n=0}^\infty n^k\left(\frac{1}{e}\right)^n\tag{2}\\ &=\sum_{n=0}^\infty \sum_{j=0}^k{k\brace j}n^{\underline{j}}\left(\frac{1}{e}\right)^n\tag{3}\\ &=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\sum_{n=0}^\infty n^{\underline{j}}\left(\frac{1}{e}\right)^{n-j}\tag{4}\\ &=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\left.\left(\frac{d^j}{dz^j}\frac{1}{1-z}\right)\right|_{z=\frac{1}{e}}\tag{5}\\ &=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\frac{j!}{\left(1-\frac{1}{e}\right)^{j+1}}\tag{6}\\ &\,\,\color{blue}{=e\sum_{j=0}^{k}{k\brace j}\frac{j!}{\left(e-1\right)^{j+1}}}\\ \end{align*}

Comment:

  • In (1) we differentiate $k$ times.

  • In (2) we evaluate the expression at $z=0$.

  • In (3) we represent $n^k$ with the help of the Stirling numbers of the second kind as sum of falling factorials $n^{\underline{j}}=n(n-1)\cdots(n-j+1)$.

  • In (4) we exchange the sums.

  • In (5) we use the geometric series expansion.

  • In (6) we differentiate $j$ times and evaluate at $z=\frac{1}{e}$.

We conclude \begin{align*} \color{blue}{\frac{1}{1-e^{z-1}}=e\sum_{n=0}^\infty \left(\sum_{j=0}^{n}{n\brace j}\frac{j!}{\left(e-1\right)^{j+1}}\right)\frac{z^n}{n!}} \end{align*}

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