Expand $z^4\cos(z-1)$ around $z=1$

Question

Expand $z^4\cos(z-1)$ around $z=1$ to Laurent series

We take $w=z-1$

$$(w+1)^4\cos(w)=(w+1)^4\sum_{n=0}^{\infty}(-1)^n\frac{w^{2n}}{2n!}=(w^4+4w^3+6w^2+4w+1)\sum_{n=0}^{\infty}(-1)^n\frac{w^{2n}}{2n!}$$

How should I continue?

Answer 1

To expand $f (z) =z^4\cos (z-1)$ arround $z=1$, we expand $g (z)=f (z+1) $ arround $z=0$ and we get $f (z)=g (z-1 )$.

$$g (z)=(1+z)^4\cos (z) $$ $$=(1+4z+6z^2+4z^3+z^4)\sum_{n=0}^\infty a_nz^n $$ $$=\sum_{n=0}^\infty(a_ {n-4}+4a_{n-3}+6a_{n-2}+4a_{n-1}+a_n)z^n $$

with $$a_{2p}=(-1)^p\frac {1}{(2p)!} $$ $$a_{2p+1}=0$$ and $$j <0\implies a_j=0$$

replace $z $ by $z-1$ in $g (z) $ to get $f (z) $.