0
$\begingroup$

I'm stucking trying to solve the following problem:

Given a function $f$ given as $f(z)=\frac{z\cdot \cos z}{z-2}$ I am asked to show that the Taylor series for $f$ around $0$ is given as $$f(z)=\sum_{k=0}^{\infty}a_kz^k $$ where $a_0=0$ and $$a_{2k}=a_{2_k-1}/2 $$ $$ a_{2k+1}=\frac{\left(a_{2k}+\frac{(-1)^{k+1}}{(2k)!}\right)}{2}$$

My work so far:

Given $$f(z)=\frac{z\cdot \cos z}{z-2}=\sum_{k=0}^{\infty}a_kz^k$$ it must be the case that \begin{alignat*}{2} z\cdot \cos z&=\sum_{k=0}^{\infty}a_kz^k(z-2)\\ &= \sum_{k=0}^{\infty}-2a_kz^k+a_kz^{k+1} \\ &= \sum_{k=0}^{\infty}-2a_kz^k+a_{k-1}z^{k} \\ &= \sum_{k=0}^{\infty}(-2a_k+a_{k-1})z^k \\ &= \sum_{k=0}^{\infty}(-a_k+\frac{a_{k-1}}{2})z^k \end{alignat*}

From here I'm not sure how to proceed. I know that $z\cdot \cos z=z\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}z^{2k}=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}z^{2k+1}$, but I'm not sure how this connects with what I've done so far.

Any tips are appreciated.

$\endgroup$

2 Answers 2

2
$\begingroup$

Observe that the function has a simple pole at $z=2$, so the series will converge for $|z| < 2$, in which region we have

$$\frac{1}{z-2} = -\frac{1}{2} \frac{1}{1-\tfrac{z}{2}}=- \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$$

We also, as you point out, have

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

So the Taylor series will be the product

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

Now, from this point, you might be able to find an expression for the $n$-th term in this series using the Cauchy product formula, or you could just write out the first few terms is that is satisfactory.

$\endgroup$
1
  • $\begingroup$ Thanks! I haven't really come across the Cauchy product formula in my course before, so I will have to look into that. $\endgroup$
    – SebaM
    Commented Dec 11, 2020 at 11:10
1
$\begingroup$

Hint: $$ \frac{1}{z-2} = \sum_{n=0}^\infty \frac{-1}{2^{n+1}}z^n \quad\text{for } |z|<2 $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .