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I was asked to find the first three terms in the taylor series of $\psi (z)$ around $z=0$ where $(e^z-1)^2=z^2 \psi(z)$

and I'm having a few difficulties.

My original idea was to say $\psi (z)=\frac{(e^z-1)^2}{z^2}$ and then find the taylor series of $(e^z-1)^2$ at $z=0$, find the taylor series of $\frac{1}{z^2}$ at $z=0$ and multiply. The problem is that there is no such thing as taylor series of $\frac{1}{z^2}$ at $z=0$ since it is not analytic, continuous, or even defined at $z=0$. So that was a bad idea.

My second idea "worked", but I want to check my answer.

The idea was to find the taylor series of $f(z)=(e^z-1)^2$. It's easy to see that $f(0)=f'(0)=0$ so the first 2 terms are $0$, and for any $n \geq 2$: $f^{(n)}(z)=2e^z$ and so $f^{(n)}(0)=2$

So the taylor series of $f$ around $z=0$ is $f(z)=\frac{2}{2!}z^2+\frac{2}{3!}z^3+\frac{2}{4!}z^4+...$

So we have

$\frac{2}{2!}z^2+\frac{2}{3!}+\frac{2}{4!}z^4+...=z^2 \psi(z)$, and so $\psi(z)=\frac{2}{2!}+\frac{2}{3!}z+\frac{2}{4!}z^2+...$ and we specified the first three terms.

Is this result correct? I'm questioning myself because it defies logic that it's not possible in the previous way (of finding the taylor of $\frac{1}{z^2}$) but i'm getting a correct result here.

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  • $\begingroup$ Your $n^{\text{th}}$ derivatives are wrong. The first idea can work by finding the Laurent series. Regarding the very last paragraph, consider $z\mapsto \frac z z$, same issue. $\endgroup$
    – Git Gud
    Commented Jan 31, 2015 at 12:51
  • $\begingroup$ That is correct. I made a mistake finding the derivative. $\endgroup$ Commented Jan 31, 2015 at 12:52

3 Answers 3

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The method is correct. But you made an error in the calculation of the derivative. The first derivative is $2(e^z -1)e^z = 2(e^{2z}-e^z)$, the second is then $2(2e^{2z}-e^z)$ and so on. So you have $ (2^{n} - 2)$; not what you claimed.

The reason this works but it seems the former does not work is similar to the fact that you can compute $12 : 4 = 3$ in the integers but $12 \ \frac{1}{4}$ makes no sense in the integers. Of course it makes sense in the rationals.

The analog of the rationals here, would be Laurent series.

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We have

$$\psi(z)=\left(\frac{e^z-1}{z}\right)^2=\left(1+\frac {z}2+\frac{z^2}{6}+o(z^2)\right)^2=1+z+\frac{7}{12}z^2+\frac14z^3+o(z^3)$$

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You've made an error in your second approach. Your derivatives of $f$ are wrong.

If $f(z)=(e^z-1)^2$, then $f^{(n)}(z) = 2^ne^{2z}-2e^z$, so $f^{(n)}(0)=2^n-2$.

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