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Currently I am learning Complex Analysis online when the instructor stated that we need to prove that the derivative of a power series is a power series. But isn't that straight forward? The derivative of $$\sum_{k\ge0}a_k(z-z_0)^k$$ Is:$$\sum_{k\ge1} a_k*k(z-z_0)^{k-1}$$ Let $a_k*k=b_k$ (new coefficients), then re-index and you get a power series. Is this a valid proof?

Kamal Saleh
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    No. The fact that these series are uniformly convergent is important – FShrike Dec 08 '22 at 19:35
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    Suppose that $f$ is complex differentiable on some domain. How is the derivative of $f$ actually defined? How do you use that to obtain a power series for $f$? How do you know that the derivative of $f$ is equal to the formal derivative of the power series? All of these things require some argumentation. – Xander Henderson Dec 08 '22 at 19:39
  • This link may be interesting: https://math.stackexchange.com/questions/147869/interchanging-the-order-of-differentiation-and-summation – Little Jonny Dec 08 '22 at 21:38

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If you have a sum of functions $f(z) + g(z)$ then you know the derivative of the sum is simply $f'(z) + g'(z)$. You learned this a long time ago and it feels natural to rely on this fact without thinking. And that's good; this is always true.

If you want to differentiate a sum of several functions, like $f(z) + g(z) + h(z)$ or even $\sum_{k=0}^N f_k(z)$, then you can use the same fact from above but applied multiple times. For example, you could write $\frac{d}{dz} \left[ f(z) + g(z) + h(z) \right] = \frac{d}{dz} \left[ f(z) + g(z)\right] + \frac{d}{dz} h(z) = f'(z) + g'(z) + h'(z)$. It would take more steps, but you could eventually show $\frac{d}{dz} \sum_{k=0}^N f_k(z) = \sum_{k=0}^N f'_k(z)$.

Finally, you have all this intuition built up and you're trying to apply it for $\sum_{k=0}^\infty a_k z^k$. But the method above only works when you're dealing with a finite sum of functions, because you could split them apart one at a time. With the power series version, no finite sequence of "break the sum into two piles" can ever get us down to one term per pile.

As you've noticed, the situation is extra confusing because the faulty conclusion actually turns out to be (basically) true in this case! The derivative really does exist (inside the radius of convergence) and it equals exactly the expression you expect. More complicated methods are required to prove these claims though.

David Clyde
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