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Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point. $$ z\mapsto\frac{1}{e^z - 1} $$ The point is $z_0=0$ (four terms of laurent series).

I have wrote $e^z -1$ as $z+z^2/2!+z^3/3!$....

Now i don't know how to proceed with this further.

Please answer in detail I am very weak with this.

Thank you.

E. Joseph
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rohit
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3 Answers3

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The other answer has its benefits, but if you want to check your answer or simply bash out the "important" coefficients and ignore all that insignicantly small stuff, then here goes. Just note that we need to be careful about convergence.

We have that $\frac{1}{e^z-1} = \frac{1}{z(1+(z/2!+z^2/3!+...))}$ where we let $(z/2!+z^2/3!+...) = P(z)$. Then:

$$\frac{1}{e^z-1} = 1/z - P(z)/z + P(z)^2/z - P(z)^3/z + ... \\ = 1/z - 1/2 - z/3! + z/(2!)^2 - z^2/4! + 2z^2/(2!\cdot 3!) - z^2/(2!)^3 + O(z^3) \\ = 1/z - 1/2 + z/12 + z^2\cdot (1/6-1/24-1/8) + O(z^3) \\ = 1/z - 1/2 + z/12 + O(z^3).$$

Christopher K
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    I think it misses one important point: To use the series for $1 \over (1+z)$, you need to prove that $|z| < 1$, otherwise, the series does not converge. So here, you need to prove that $|P(z)| < 1$ before using the series, which I don't think it's possible in the domain $0 < |z| < 2 \pi$ – le duc quang Apr 23 '16 at 10:20
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    @leducquang late reply, but I think this method is valid: for all small enough $z, |P(z)| < 1$ as $P(z) \rightarrow 0$ as $ z \rightarrow 0$. Let's assume we are working in some annulus of sufficiently small radius s.t. this holds. Then we apply Chris Ks argument and generate terms of the Laurent series. But we know the Laurent series is unique on the whole of the annulus $0 < |z| < 2 \pi$ so it must extend. – Evgeny T Apr 16 '18 at 15:40
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Let $f(z) = (e^z - 1)^{-1}$ which has a simple pole at $z = 0$ (easy enough to see). Consider $$ h(z) = \frac{e^z - 1}{z} = \sum_{n = 0}^{\infty}\frac{z^n}{(n + 1)!} $$ $h$ is an entire function (prove it to yourself). Now let $g(z) = \frac{1}{h(z)}$ which is analytic over some area (I leave where as an exercise). Now $$ f(z) = \frac{1}{zh(z)} = \frac{g(z)}{z} = \sum_{n = -\infty}^{\infty}a_nz^n $$ Furthermore, $a_n$ can be found \begin{align} a_n &= \frac{1}{2\pi i}\int_{|z| = R}\frac{f(z)}{z^{n+1}}dz\\ &= \frac{1}{2\pi i}\int_{|z| = R}\frac{g(z)}{z^{n+2}}dz \end{align} where $0<R<2\pi$. $g$ is analytic on the inside of $|z| = R$. By Cauchy's Theorem, $a_n = 0$ for $n\leq -2$. $$ f(z) = \sum_{k = -1}^{\infty}a_nz^n $$ Now compute the first few $a_n$. To find the derivative of $g$, we should first find the derivative of $h$. $$ h^{(k)}(z) = \sum_{n=k}^{\infty}\frac{n!z^{n-k}}{(n-k)!(n+1)!} $$ Therefore, $$ h^{(k)}(0) = \frac{1}{k+1} $$ for all $k\geq 0$ We can easily see that $1 = g(z)h(z)$ so $0=g'h+gh'$. In general, $$ 0 = (gh)^{(k)}(z) = \sum_{i = 0}^k\binom{k}{i}h^{k-i}(z)g^i(z) $$ At $z = 0$, $h^{k-i}(0) = \frac{1}{k - i + 1}$; therefore, $$ 0 = (gh)^{(k)}(0) = \sum_{i = 0}^k\binom{k + 1}{i}g^i(0) $$ Going back to the coefficient $a_n$, we have $$ a_n = \frac{1}{2\pi i}\int_{|z| = R}\frac{g(z)}{z^{n+2}}dz = \frac{g^{(n+1)}(0)}{(n+1)!} $$ for all $n\geq -1$. $$ 0 = \sum_{j = 0}^k\frac{a_{j-1}}{(k-(j-1))!} $$ So $a_{-1} = 1$, $a_0 = -1/2$, all positive even terms are zero.... Let $B_k := (-1)^{k-1}(2k)!a_{2k-1}$ be Bernoulli numbers. Note that $F(z) = \frac{1}{e^z - 1} -\frac{1}{z} + \frac{1}{2}$ is an odd function. Therefore, $$ f(z) = \frac{1}{z} -\frac{1}{2} +\sum_{k=1}^{\infty}a_{2k-1}z^{2k-1} = \frac{1}{z} -\frac{1}{2} +\sum_{k=1}^{\infty}(-1)^{k-1}\frac{B_k}{(2k)!}z^{2k-1} $$

Mathman
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dustin
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    This is many years later but could you possibly expound on some of this? Particularly, I'm not sure what you mean by $h^{(k)} = 1/k+1$. Do you mean $h^{(k)}(0) = 1/k+1$? Also, how are you getting that zero equals the sum of something in terms of derivatives of $h$ and $g$? – inkievoyd Feb 16 '18 at 00:47
  • @inkievoyd I don't remember any more and I haven't done any math in awhile. – dustin Feb 16 '18 at 03:59
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(This is a rewrite of dustin's answer for my own reference.)

Clearly $f(z)=\frac{1}{e^z - 1}$ has a pole at $z=0$. Since $\lim_{z\to0}zf(z)=1$, the pole is simple. Thus $f(z)$ has a Laurent series expansion $\sum_{n\ge-1}a_nz^n$ about zero with $a_{-1}=1$. Now, as both $$g(z)=\frac{z}{e^z - 1}=\sum_{n\ge0}a_{n-1}z^n$$ and $$\frac{1}{g(z)}=\frac{e^z - 1}{z}=\sum_{n\ge0}\frac{z^n}{(n+1)!}$$ are analytic at zero, we have $$\left(\sum_{n\ge0}a_{n-1}z^n\right)\left(\sum_{n\ge0}\frac{z^n}{(n+1)!}\right)=1.$$ By comparing coefficients on both sides, we see that $a_0=-\frac12$ and $\{a_n\}_{n\ge-1}$ is given by the recurrence relation $a_{-1}=1$ and $$\sum_{k=0}^n\frac{a_{k-1}}{(n-k+1)!}=0\quad(n\ge1).$$ In particular, $a_0=-\frac12$. Finally, in a deleted neighbourhood of zero, it is straightforward to verify that $$\sum_{n\ge1}a_nz^n=f(z)-\frac{a_{-1}}{z}-a_0 =\frac{1}{e^z-1}-\frac{1}{z}+\frac{1}{2}$$ is an odd function. Therefore, all positive even terms $a_2,a_4,\ldots$ are actually equal to zero and the previous recurrence relation can be rewritten as $a_{-1}=1,\ a_0=-\frac12$ and $$\frac{1}{(2n+1)!}-\frac{1}{2(2n)!}+\sum_{k=1}^n\frac{a_{2k-1}}{(2n-2k+1)!}=0\quad(n\ge1).$$

Remark. The coefficients $B_n$ of the Taylor expansion $$\frac{z}{e^z-1}=\sum_{n\ge0}\frac{B_nz^n}{n!}$$ are known as Bernoulli numbers. Thus $a_n=\frac{B_{n+1}}{(n+1)!}$ for every $n\ge-1$.

Ramen Nii-chan
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