5
$\begingroup$

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?

  • 2
    [In that case...](http://dx.doi.org/10.1098/rspa.1995.0096)2011-11-24

1 Answers 1

7

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 $\sum g_i^{'}$ and at least one point of convergence for $\sum g_i$.

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

  • 0
    $\sum_{i=1}^\infty 1/i^x$ 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 > 1.2011-11-24