1. ## power series

Determine the power series expansion of $\displaystyle f(z)$ at $\displaystyle z_0$ without computing derivatives
(i) $\displaystyle f(z) = sinz ,$ $\displaystyle z_0 = i;$

(ii)$\displaystyle f(z )= \frac{1}{z^2-3z+2} ,$ $\displaystyle z_0 = 0;$

(iii)$\displaystyle f(z) = Log z,$ $\displaystyle z_0 = 1;$

So how the heck would I do this? If someone could show me step by step how to do one of them I'm pretty sure I could do the rest, but I don't know how to do it without derivatives as I would have just taken the taylor series

2. You basically have to know what the power series are for $\displaystyle \sin x, \frac{1}{x}, \ln x$. Then you can just use substitution.

3. To elaborate, for number 2, you would use the power series centered at x = 2 for $\displaystyle f(x) =\frac{1}{x}$, which is:

$\displaystyle f(2) + \frac{(x - 2)f'(2)}{1!} + \frac{(x-2)^2f''(2)}{2!} + ...$

$\displaystyle = \frac{1}{2} - \frac{(x-2)}{2^2} + \frac{(x-2)^2}{2^3} - \frac{(x-2)^3}{2^4} + ... + \frac{(x-2)^n(-1)^n}{2^{n+1}} + ...$

Then you substitute $\displaystyle x = z^2 - 3z + 2$.

4. Here's what I did for(1):

$\displaystyle \sin(z-i)=\sum_{n=0}^{\infty}\frac{(-1)^n(z-i)^{2n+1}}{(2n+1)!}$

Now, how do we express this in terms of $\displaystyle sin(z)$? Well:

$\displaystyle \sin(z-i)=\sin(z)\cos(i)-\sin(i)\cos(z)$
$\displaystyle \cos(z-i)=\sin(z)\sin(i)+\cos(z)\cos(i)$

Now, I'll leave it to the reader to show:

$\displaystyle \sin(z)=\cosh(1)\sum_{n=0}^{\infty}\frac{(-1)^n(z-i)^{2n+1}}{(2n+1)!}+i\sinh(1)\sum_{n=0}^{\infty}\f rac{(-1)^n(z-i)^{2n}}{(2n)!}$

Guess that's what you meant Iceman? Or no?