confusion about power series derivatives

Question

we have a formula to find the derivative of the power seires $f(z):=\sum a_n(z-z_0)^n $

Why isn't $f^{(n)}(z_0)=0$? Because the summand has factor $(z-z_0)^{n-k}$ which becomes $(z_0-z_0)^{n-k}=0$. How should I interpret this formula?

Answer 1

We have to intepret $z^{0}$ as $1$ even when $z=0$ in power series expansions. For example $f(z)= \sum\limits_{k=0}^{\infty} a_k z^{k}$ stands for $a_0+a_1z+a_2z^{2}+\cdots$ so when you calculte $f(0)$ you have to take $a_0 0^{0}$ s $a_0$ and not $0$.

Answer 2

The same way you interpret $f'(2)$: first take the derivative, then evaluate at $2$. If you do it in the other order, you always get zero, which is not interesting. Perhaps slightly better (if more baroque) notation would be $$ \left. f^{(n)}(z) \right|_{z = z_0} \text{.} $$

Notice that evaluating in this order, you get \begin{align*} f^{(n)}(z_0) &= \left. n! a_n(z-z_0)^{n-n} \right|_{z = z_0} \\ &= \left. n! a_n (z-z_0)^0 \right|_{z = z_0} \\ &= \left. n! a_n 1 \right|_{z = z_0} \\ &= n! a_n \text{.} \end{align*}

Answer 3

Yes, most of the $(z-z_0)^\ell$ terms will become $0$ when you insert $z = z_0$. However, exactly one of the terms will have $(z-z_0)^0 = 1$. Inserting $z_0$ into that term will give you $(z_0-z_0)^0 = 1$.

Answer 4

$$ \begin{align} \frac{d}{dz}a_0 + a_1(z-z_0) + a_2(z-z_0)^2 &=& a_1 + a_2 \cdot 2(z-z_0) \implies f^{(1)}(z_0) = a_1\\ \frac{d^2}{dz^2}a_0 + a_1(z-z_0) + a_2(z-z_0)^2 &=& a_2 \cdot 2\implies f^{(2)}(z_0) = 2\cdot a_2 \end{align} $$ a term will always survive that is independent of $(z-z_0)$.