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I was thinking about derivative of infinite sum of functions, i.e.

$$f(x) = \sum_{i = 0}^\infty g_i(x)$$

$g(x)$ is continuous in domain of $f$

Because if $(f+g)'(x) = f'(x) + g'(x)$ then $\left(\sum\limits_{i = 0}^{\infty} g_i(x)\right)' = \sum\limits_{i = 0}^{\infty} g_i'(x)$ isn't it?

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Your function $f$ is not well-defined except $g(x) = 0$ for all $x$. I suppose you mean something different. Also remember that having $g$ being continuous does not make it differentiable. –  Matthias Klupsch Nov 24 '11 at 18:33
That doesn't always work. The canonical example is due to Weierstrass... –  J. M. Nov 24 '11 at 18:35
OK, but what if I set $g(x) = \frac{1}{n^x}$. In this question I'm interested especially in $\zeta(z)$ function –  Łukasz Niemier Nov 24 '11 at 18:37
In that case... –  J. M. Nov 24 '11 at 18:43

1 Answer 1

up vote 5 down vote accepted

First I assume you mean $g_i$ instead of $g$, and you have to suppose at least that the $g_i$ are all differentiable (more than just continuous).

Even then this is in general false. One common case where it is true is when you assume uniform convergence of $\Sigma g_i^{'}$ and at least one point of convergence for $\Sigma g_i$.

A counter example under your hypothesis : take $g_i^{'}(x) = cos(i \pi x)/i^2$. then $\Sigma g_i$ converges since it converges normally ($\Sigma \frac{1}{i^2}< \infty$) but $\Sigma g_i^{'}$ diverges at 0 (since $\Sigma \frac{1}{i} = \infty$).

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And what if $g_i(x) = 1/i^x$? That is especially interesting me. –  Łukasz Niemier Nov 24 '11 at 18:46
then it is true for $x>1$ because the convergence is uniform on all compacts of $]1, +\infty[$ –  Glougloubarbaki Nov 24 '11 at 18:53
$\sum_{i=1}^\infty 1/i^x$ is the Riemann zeta function $\zeta(x)$ for $\Re x > 1$. The series of derivatives $\sum_{i=1}^\infty -\ln(i)/i^x$ also converges for $\Re x > 1$, and uniformly on compact sets, so by the "One common case" Glougloubarbaki mentioned the sum is indeed $\zeta'(x)$ for $\Re x > 1$. –  Robert Israel Nov 24 '11 at 18:54

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