Cfrgtkky

The question reads as follows:

Let the function $f$ be given by $f(z)=frac{z}{2}+frac{z}{e^z-1}$ if
$zneq0$ and $f(z)=1$ if $z=0$. Show that $f$ is analytic at $z=0$ and
that $f(z)=f(-z)$. Deduce that there is a Taylor series
$f(z)=a_0+a_2x^2+a_4x^4+...$ valid for $|z|<2pi$. Find $a_0,a_2,a_4$.

Clearly the function is analytic at $z=0$, and $f(z)$ is even for all $z$. Now, the singularities occur where $z=2nipi$ where $ninmathbb{Z}$. Given $|z|<2pi$, we have that $f(z)$ is analytic within this disk. Hence a Taylor series exists for $f(z)$ such that $z_0=0$. Thus $f(z)=a_0+a_1z+a_2x^2+a_3z^3+a_4x^4+....$ in $|z|<2pi$ where $a_n=frac{f^{n}(0)}{n!}$ where $ninmathbb{Z^+}$.

Since $f(z)$ is even this implies that only the even powers exist. Thus $f(z)=a_0+a_2x^2+a_4x^4+...$. I am having trouble finding the coefficients, $a_0,a_2,a_4$.

Thank you in advanced for you help.

To find the first terms of the Taylor series of $f$, there are various possibilities. One can simply differentiate the function and use $a_n = frac{1}{n!}f^{(n)}(0)$ to obtain the coefficients, but that soon becomes unwieldy. Still, for the coefficients up to order $4$, it's doable.

Another standard way, when one has such a quotient, is to expand the numerator (trivial here) and the denominator, and divide both by a suitable power of $z$ times a constant, so that the denominator has the form $1 - h(z)$, and expand $frac{1}{1-h(z)}$ in a geometric series. That leads to (ignoring the $frac{z}{2}$ summand)

$$begin{align}
frac{z}{e^z-1} &= frac{z}{z + frac{z^2}{2} + frac{z^3}{6} + dotsb}\
&= frac{1}{1 + left(frac{z}{2} + frac{z^2}{6} + dotsbright)}\
&= 1 - left(frac{z}{2} + frac{z^2}{6} + dotsbright) + left(frac{z}{2} + frac{z^2}{6} + dotsbright)^2 - dotsb\
&= 1 - frac{z}{2} +left(frac14 - frac16right)z^2 + dotsb\
&= 1 - frac{z}{2} + frac{z^2}{12} + dotsb
end{align}$$

The coefficients $a_0$ and $a_2$ are found in this way without much computation, but finding $a_4$ involves some more computation. It's still not too hard.

Yet another way is to multiply the unknown expansion with the known expansion of the denominator to obtain a recurrence for the coefficients:

$$begin{align}
z &= frac{z}{e^z-1}(e^z-1)\
&= left(sum_{k=0}^infty frac{b_k}{k!}z^kright)cdotleft(sum_{m=1}^infty frac{z^m}{m!}right)\
&= sum_{n=1}^infty left(sum_{k=0}^{n-1}frac{b_k}{k!(n-k)!}right)z^n\
&= sum_{n=1}^infty left(sum_{k=0}^{n-1} binom{n}{k}b_kright)frac{z^n}{n!}.
end{align}$$

Equating the coefficients yields $1 = binom{1}{0}b_0$, so $b_0 = 1$, and for $n > 1$, the recurrence

$$sum_{k=0}^{n-1}binom{n}{k}b_k = 0.tag{1}$$

For $n = 2$, we obtain $0 = binom{2}{0}b_0 + binom{2}{1}b_1 = 1 + 2b_1$, so $b_1 = -frac12$. For $n = 3$, it becomes

$$0 = binom{3}{0}b_0 + binom{3}{1}b_1 + binom{3}{2}b_2 = 1 - frac{3}{2} + 3b_2 Rightarrow b_2 = frac16.$$

Note that we have $a_k = frac{b_k}{k!}$, so this agrees with the result above.

Using the fact that $b_{2k+1} = 0$ for $k > 0$ by the evenness of $f$, we need not compute $b_3$ and further odd coefficients, which saves some work. Computing $a_4$ with the recurrence $(1)$ is then not much work.