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I'm a student of mechanical engineering and I have a problem with computing residues of a complex function. I've read some useful comments in the other threads. Now I've got some ideas about essential singularity and series expansion in computing the residue. However, I still can't find the solution to my problem. I arrived at a complex function in the process of finding a solution to a mechanical problem. Then I had to obtain the residues to proceed to the next steps. The function has the following form:

$$f(z)=\frac{\exp(Az^N+Bz^{-N})}{z}$$

where $A$, $B$ and $N$ are real constants $(N \geq 3)$.

I want to compute the resiude at $z=0$. I wrote the Laurent series of $f$, but got an infinite sum. I do not even know if I am at the right direction. I would be really thankful if someone could give me a hint on this and put me back in the right direction.

t.b.
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Benyamin Gholami
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  • I formatted your post in order to make it more readable. I guess you want $N$ to be a natural number, else your $f$ isn't defined in a neighborhood of zero. – t.b. Jul 08 '11 at 15:24
  • What do you mean by "I got an infinite sum"? The Laurent series is of course going to be an infiite sum, but I don't think that is what you meant... – Mariano Suárez-Álvarez Jul 08 '11 at 15:26
  • @Theo Buehler: Thanks for modification. And as you guessed, N is a natural number (integer). – Benyamin Gholami Jul 11 '11 at 06:27
  • @Mariano Suárez-Alvarez♦: You are right. I meant the infinite sum does not converge. However, with the help of friends here I'm at the right direction now. – Benyamin Gholami Jul 11 '11 at 06:30

2 Answers2

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From each even term in the exponential series, you get one contribution where the positive and negative powers of $z$ cancel out. This is the middle term of the binomial expansion, and to get the coefficient $a_{-1}$ in the Laurent series you need to sum this over all even terms, which leads to

$$a_{-1}=\sum_{k=0}^\infty\frac1{(2k)!}\binom{2k}{k}A^kB^k=\sum_{k=0}^\infty\frac{(AB)^k}{k!^2}\;.$$

joriki
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I'm assuming $N$ is an integer here to avoid issues with branches of the logarithm.

Note that $\exp(Az^N) = \sum_{m=0}^{\infty} {A^mz^{mN} \over m!}$ and $\exp(Bz^{-N}) = \sum_{m=0}^{\infty} {B^mz^{-mN} \over m!}$, so that $\exp(Az^N + Bz^{-N})$ is the product of these two series, and ${\displaystyle{\exp(Az^N + Bz^{-N}) \over z}}$ is the product of these two series divided by $z$.

The residue at $z = 0$ will be the cofficient of the ${1 \over z}$ term in this product divided by $z$, or in other words the constant term in the product. A constant term is obtained in the multiplication when you multiply a term ${A^mz^{mN} \over m!}$ by a term ${B^mz^{-mN} \over m!}$ (with the same $m$). So the constant term in the overall product is the sum of these, that is, $\sum_{m=0}^{\infty} {(AB)^m \over (m!)^2}$ and this will be your residue.

Zarrax
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  • Thank you both for your good answers. Your comments really helped. In the process of my solution I arrived at a contour integral. Then I had to use the residue theorem to solve the integral. Also I arrived at another integral with the following format: $$f(z)=\frac{\exp(Az^N+Bz^{-N})}{z-a}$$
    where a is a constant and can be in/or out of the closed contour. My problem is that how I should obtain the residue in both cases (when a is in and out of the contour). Can I use the Laurent expansion like the previous problem? How the singularity at z=0 affect the solution?     – Benyamin Gholami Jul 11 '11 at 07:43