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I have to expand $f(z) = \frac{e^z -1}{z}$ into a power series around $z = 0$.

  1. question: Does this mean that I have to somehow use the Taylor series? I know that a power series is $f(z) = \sum_{n = 0}^{\infty} a_n (z - z_0)^n$.

As far as I understand it, a Taylor series is a special case of a power series. But maybe I'm just overthinking it (?)

  1. I know that the power series of $e^z$ is $e^z = \sum_{n = 0}^{\infty} \frac{z^n}{n!}$. Then, I have $f(z) = \frac{e^z -1}{z} = \frac{e^z}{z} - \frac{1}{z} = \frac{\sum_{n = 0}^{\infty} \frac{z^n}{n!}}{z} - \frac{1}{z} = \sum_{n = 0}^{\infty} \frac{z^n z^{-1}}{n!} - \frac{1}{z} = \sum_{n = 0}^{\infty} \frac{z^{n-1}}{n!} - \frac{1}{z}$.

But here I'm stuck by determing the power series for $\frac{1}{z}$. Any help on how to determine it?

syphracos
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1 Answers1

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Based on the power series for $e^{z}$ around $0$, one gets the desired result as follows:

\begin{align*} e^{z} = 1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + \ldots & \Rightarrow e^{z} - 1 = z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + \ldots \Rightarrow \frac{e^{z} - 1}{z} = 1 + \frac{z}{2!} + \frac{z^{2}}{3!} + \ldots \end{align*}

Hopefully this helps!

Átila Correia
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  • Thanks! Am I then correct with my comment from above? So, in the end I get $f(z) = \sum_{n = 1}^{\infty} \frac{z^{n-1}}{n!}$ ? – syphracos Nov 02 '22 at 21:25
  • @syphracos Yes, you are correct! – Átila Correia Nov 02 '22 at 21:26
  • Great! Thanks again! But still, the question I posted above. If I have problems like this, for finding the power series, do I have a distinction between a power series approach or a Taylor series approach? https://math.stackexchange.com/questions/1733888/expanding-1-z-about-z-1-using-taylor-series-vs-power-series here I found that these are 2 separate approaches. So, I can either 1. algebraically manipulate the function until I (maybe) find an expression which contains a known power series (for example trig functions, exponential, logarithmic etc. ) or 2. try to find the Taylor series ? – syphracos Nov 02 '22 at 21:31
  • @syphracos It depends on the problem. Sometimes it is easier to calculate the coefficients of the power series using the derivatives and other times it is easier to manipulate the function until you get something which you are familiar with. In any case, it is helpful to have some power series by heart so that you can compare with. – Átila Correia Nov 02 '22 at 21:41
  • yeah, I see what You mean.. just one more thing. The next exercise is that I have to show that $g(z) = \frac{z}{e^z - 1}$ can be expanded into a power series around $z = 0$. This is basically the inverse of the previous problem. Here I get $g(z) = \frac{1}{\sum_{n = 1}^{\infty} \frac{z^{n-1}}{n!}}$. So, I have $g(z) = \frac{1}{f(z)}$. How to show here that $g(z)$ can be expanded into a power series? – syphracos Nov 02 '22 at 21:46
  • @syphracos I think this may help you. – Átila Correia Nov 02 '22 at 22:21
  • I don't quite understand how he gets the the expression for $\frac{\omega}{e^{\omega} - 1}$ in the accepted answer – syphracos Nov 08 '22 at 21:13