3
$\begingroup$

Let $f(z)=\cfrac{e^{-3z}}{z^2(z-2)^2}$, find the Laurent series about $z=0$.

On the region $0<|z|<2$, I get

$\cfrac{1}{(z-2)^2}=\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-1}}{2^{n+1}}$,

then $\cfrac{1}{z^2(z-2)^2}=\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-3}}{2^{n+1}}$

And I get confused here, is it ok if I let

$f(z)=\cfrac{e^{-3z}}{z^2(z-2)^2}=e^{-3z}\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-3}}{2^{n+1}}$

Or I have to express $e^{-3z}$ as its Taylor series and the multiplicate?

Thank you

$\endgroup$
  • $\begingroup$ That is exactly it. I think you have a typo, though. between the third and fourth lines, the left hand side changes, the right hand side does not. $\endgroup$ – Alex S Jun 27 '15 at 4:53
  • $\begingroup$ Thank you I've edited it. So, the answer will be $f(z)=e^{-3z}\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-3}}{2^{n+1}}$ ? $\endgroup$ – Frank Jun 27 '15 at 5:11
  • $\begingroup$ It's not OK to leave it like that, as that isn't a Laurent series. Whatever the answer is, it will be the product of the Taylor and Laurent series, but I can't help but feel there must be a better way than just multiplying them! $\endgroup$ – Theo Bendit Jun 27 '15 at 5:18
  • $\begingroup$ I had a play around with multiplying the series manually, and I found that the existence of a closed form for the coefficients is equivalent to the existence of a closed form for a sum of the form $\sum_{k=0}^n \frac{x^k}{k!}$. I'm sceptical that such a closed form exists, but maybe someone else could shine a light here? $\endgroup$ – Theo Bendit Jun 27 '15 at 5:48
  • $\begingroup$ The expansion is doable for sure but the problem is to find the explicit form for the coefficient. $\endgroup$ – Claude Leibovici Jun 27 '15 at 5:55
1
$\begingroup$

You have $$f(z)=\cfrac{e^{-3z}}{z^2(z-2)^2}=\Big(\sum_{m=0}^\infty \frac{(-3)^m }{m!}z^m\Big)\times \Big(\sum_{n=0}^\infty \frac{n}{2^{n+1}}z^{n-3}\Big)=\sum_{i=-2}^\infty c_i z^i$$ and you look for the general expression of the $c_i$'s.

For better legibility, we shall write $$\sum_{i=-2}^\infty c_i z^i=\Big(\sum_{m=0}^\infty a_m z^m\Big)\times \Big(\sum_{n=0}^\infty b_n z^{n-3}\Big)$$

In order to have a degree $(k-3)$, you must add several terms the product of whicb making $z^k$ and then you should arrive to something like $$c_{k+3}=\sum_{m=0}^k a_m \,b_{k-m}$$

$\endgroup$
  • $\begingroup$ Thank you very much, I had the same idea but i wanted to know if ther was another way to do so. $\endgroup$ – Frank Jun 29 '15 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.