# Derivative of Fourier series

1. Sep 25, 2014

### etf

Let's say I have Fourier series of some function, $$f(t)$$, $$f(t)=\frac{a0}{2}+\sum_{n=1}^{\infty}(an\cos{\frac{2n\pi t}{b-a}}+bn\sin{\frac{2n\pi t}{b-a}})$$, where $$a$$ and $$b$$ are lower and upper boundary of function, $$a0=\frac{2}{b-a}\int_{a}^{b}f(t)dt$$, $$an=\frac{2}{b-a}\int_{a}^{b}f(t)cos\frac{2n\pi t}{b-a}dt$$, bn=$$\frac{2}{b-a}\int_{a}^{b}f(t)sin\frac{2n\pi t}{b-a}dt$$. My question is, can I find derivative of Fourier series on this way:
$$\frac{d}{dt}\sum_{n=1}^{\infty}(an\cos{\frac{2n\pi t}{b-a}}+bn\sin{\frac{2n\pi t}{b-a}})=\frac{d}{dt}(\sum_{n=1}^{\infty}an\cos{\frac{2n\pi t}{b-a}}+\sum_{n=1}^{\infty}bn\sin{\frac{2n\pi t}{b-a}})=$$

Last edited: Sep 25, 2014
2. Sep 25, 2014

### etf

$$=\sum_{n=1}^{\infty}\frac{d}{dt}(an\cos{\frac{2n\pi t}{b-a}})+\sum_{n=1}^{\infty}\frac{d}{dt}(bn\sin{\frac{2n\pi t}{b-a}})$$
?
Are there some limitations when I can do this and when I can't do this?
PS. I had to write this in second message since for some reason my latex code can't fit in one message :)

3. Sep 25, 2014

### mathman

I think I know what you meant - your equation is an identity (I don't think you meant it). I presume your question is can you differentiate term by term to get the derivative of a Fourier series. The answer is no different from the same question for any infinite series.

4. Sep 25, 2014

### HallsofIvy

Staff Emeritus
As long as a sum of functions is uniformly convergent, then you can differentiate the sum "term by term". Fourier series (and Taylor's series) are always uniformly convergent.

5. Sep 25, 2014

### The_Duck

I think the Fourier series of a discontinuous function is only pointwise convergent, not uniformly convergent.

6. Sep 25, 2014

### jbunniii

Indeed, the partial sums of a Fourier series are continuous, and the uniform limit of a sequence of continuous functions is continuous. So no discontinuous function can have a uniformly convergent Fourier series.

Moreover, uniform convergence of a sequence of differentiable functions $f_n$ to some function $f$ is not sufficient to ensure that $f$ is differentiable. If the sequence of derivatives $f_n'$ converges uniformly, then it is sufficient, and $\lim f_n' = f'$.

Last edited: Sep 25, 2014