# Thread: Finding a Laurent series

1. ## Finding a Laurent series

The exercise states : Find the first 4 terms different from $0$ of the Laurent series of $g(z)=e^z \cos z$, around $z_0=0$.

My attempt : $g(z)=\left ( 1+z+\frac{z^2}{2}+ \frac{z^3}{3!}+ \frac{z^4}{4!}+... \right ) \left ( 1- \frac{z^2}{2}+\frac{z^4}{4!}-\frac{z^6}{6!} +... \right )=1+...$ ( I don't really know the other terms but I think they are all $0$ when $z=0$.)

So what can I answer? The only term seems to be $1$, and if there are others they are all worth $0$ in $z=0$. ( I realize it could not be so, but I don't think so).

2. Originally Posted by arbolis
The exercise states : Find the first 4 terms different from $0$ of the Laurent series of $g(z)=e^z \cos z$, around $z_0=0$.

My attempt : $g(z)=\left ( 1+z+\frac{z^2}{2}+ \frac{z^3}{3!}+ \frac{z^4}{4!}+... \right ) \left ( 1- \frac{z^2}{2}+\frac{z^4}{4!}-\frac{z^6}{6!} +... \right )=1+...$ ( I don't really know the other terms but I think they are all $0$ when $z=0$.)

So what can I answer? The only term seems to be $1$, and if there are others they are all worth $0$ in $z=0$. ( I realize it could not be so, but I don't think so).
I think you're confusing the value of the function at $z$ with the point at which the power series is centered. Since both the series for $e^z$ and $\cos z$ are centered at $0$, by using the product of the power series you're already expanding around $0$. Also notice that the Laurent series is just the Taylor expansion since it's a product of entire functions.

Now using the Cauchy product formula we have:

Let $e^z= \sum_{n=0} ^{ \infty } \frac{z^n}{n!} = \sum_{n=0} ^{ \infty} a_nz^n$ and $\cos z = \sum_{n=0} ^{ \infty } \frac{(-1)^nz^n}{(2n+1)!} = \sum_{n=0} ^{ \infty} b_nz^n$ then $e^z\cos z = \sum_{n=0} ^{ \infty} c_nz^n$ where $c_n = \sum_{k=0} ^{n} a_kb_{n-k}$

So to compute the second term in the expansion we have $a_0b_2+a_1b_1+a_2b_0= \frac{1}{5!} + \frac{-1}{3!} + \frac{1}{2!} = \frac{41}{5!}$ which would be the second coefficient in the series (if I made no mistake), now just compute two more which are non-zero.

3. Sorry, I messed up the power series for $\cos z$, it should be $\sum_{n=0} ^{ \infty} \frac{(-1)^nz^{2n}}{2n!} =\sum_{n=0} ^{ \infty} b_nz^{2n}$. The rest is the same although with the exception that $c_n= \sum_{k+2l=n} a_kb_l$ so $c_2$ should be $0$.