Fourier series convergence - holder continuity and differentiability

hlin818
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Homework Statement



Given each of the functions f below, describe the set of points at which the Fourier
series converges to f.

b) f(x) = abs(sqrt(x)) for x on [-pi, pi] with f(x+2pi)=f(x)

Homework Equations



Theorem: If f(x) is absolutely integrable, then its Fourier series converges to f at the points where f is also holder continuous or differentiable.

The Attempt at a Solution



I managed to prove that f(x) is holder continuous on [-pi,pi] so by the theorem above since f(x) is absolutely integrable its Fourier series converges to f(x) at every point on this interval. But f(x) is not differentiable at x=0. What did I analyze incorrectly? Or in this context are holder continuity and differentiability not equivalent conditions (i.e. one can fail and the other can hold)?
 
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I may be wrong about the holder condition, but it looks to me like f(x) is holder continuous as long as the exponent in the condition is equal to or less than 1/2.
 
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Sorry about the bump, but this question is killing me. I don't feel like the book explained this convergence criteria well at all.
 
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final bump
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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