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1st attempt:

The power series expansion for $\cos z$ is: $$f(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}...$$

Dividing by $z^2$ gives:

$$f(z) = \frac{1}{z^2}-\frac{1}{2!}+\frac{z^2}{4!}-\frac{z^4}{6!}+...$$

$$f(z) = \frac{1}{z^2}+\sum_{n=0}^\infty \frac{(-1)^{n+1}z^{2n}}{(2n+2)!}$$

But this is where I get stuck.

2nd attempt:

Using the power series representation for $\cos z$, we can just represent this all as:

$$\frac{\sum_{n=0}^\infty \frac{(-1)^nz^{2n}}{(2n)!}}{z^2}$$

But then how does this relate to finding the Laurent series?

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    $\begingroup$ "But this is where I get stuck" - You have already solved the problem at this point (apart from a small mistake; a missing $(-1)^n$), you have found the Laurent series. $\endgroup$
    – Winther
    Mar 14 '16 at 1:58
  • $\begingroup$ @Winther edited to fix the mistake, thanks. $\endgroup$ Mar 14 '16 at 2:27
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$$\begin{align} \frac1{z^2}\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}&=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n-2}}{(2n)!}\\\\ &=\sum_{n=-1}^{\infty}\frac{(-1)^{n-1}z^{2n}}{(2n+2)!}\\\\ &=\frac1{z^2}+\sum_{n=0}^{\infty}\frac{(-1)^{n-1}z^{2n}}{(2n+2)!} \end{align}$$

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  • $\begingroup$ Are you sure about the sign of $(-1)^n$? $\endgroup$ Mar 14 '16 at 1:58
  • $\begingroup$ @HenryW. Henry, thanks for catching the typo! Much appreciated and edited accordingly. - Mark $\endgroup$
    – Mark Viola
    Mar 14 '16 at 2:00
  • $\begingroup$ Ok so then I don't get what is so special about Laurent series. In my mind, I thought "ok, find the taylor series of both functions evaluated at $z=0$ and then divide one by the other." If my professor were to ask me what I learned about Laurent series from this problem, I wouldn't know what to say. $\endgroup$ Mar 14 '16 at 2:28
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    $\begingroup$ @whatwhatwhat You are correct. The LS is unique inside the annulus. Well done!! You didn't butcher it at all. - Mark $\endgroup$
    – Mark Viola
    Mar 14 '16 at 2:42
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    $\begingroup$ @whatwhatwhat Yes, the answers are the same. $\endgroup$
    – Mark Viola
    Mar 14 '16 at 2:43
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You are only one step away from the correct answer. Your first attempt got the sign wrong. The second one is correct, though. The following is based on your second attempt: $$ \frac{1}{z^2} \sum_{n=0}^\infty \frac{(-1)^nz^{2n}}{(2n)!} = \sum_{n=-1}^\infty \frac{(-1)^{n+1}z^{2n}}{(2n+2)!} $$ It can be broken down to the pricipal part and holomorphic part: $$ = \frac{1}{z^2} + \sum_{n=0}^\infty \frac{(-1)^{n+1}z^{2n}}{(2n+2)!} $$ Don't forget that Laurent series is unique.

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