differentiation and integration of Fourier series.

If I have the fourier series of $|x|$ for $-l < x < l$ and I make it periodic with period $2l$ I get a cos series: $$\frac{l}{2} -\frac{4l}{\pi^2}\sum_{m=0}^\infty\cos\left((2m+1)\pi\frac{x}{l}\right)\frac{1}{(2m+1)^2}.$$

If I take the term wise integral and derivatives I can see the the integral should converge more quickly and its terms go as $1/m^3$ and the derivative should converge more slowly going as $1/m$. However is it possible to show that they converge to a particular value in order to see if its the same as the original function?

derivative: $-\frac{4l}{\pi}\sum_{m=0}^\infty\sin\left((2m+1)\pi\frac{x}{l}\right)\frac{1}{2m+1}.$

integral: $\frac{lx}{2} -\frac{4l^2}{\pi^3}\sum_{m=0}^\infty\sin\left((2m+1)\pi\frac{x}{l}\right)\frac{1}{(2m+1)^3}$

• Do I understand your question correctly: Given those two Fourier series, how do we find the function they converge to (if any)? If so, you have a hint by differentiating $|x|$ where possible and to calculate the Fourier series of the result. The same for the integral, but integrating, of course. – mickep Apr 17 '15 at 7:26
• @mickep my question was saying how do i know if the that the derivative of the series converges to the derivative of the function – Goods Apr 17 '15 at 7:42
• OK, then I guess you are happy with the answer of Benjamin. – mickep Apr 17 '15 at 8:56

Your series converges uniformly to the continuous function f(x) = |x| in the interval of periodicity. Consequently the integrated series as you stated above converges also uniformly to the integral of f , i.e.,$\int_0^x {f(t)dt}$, the integral of |x|. The derived series converges everywhere and uniformly away from 0 in [-l, l]. Therefore, it converges to the derivative of f and to 0 at 0,i.e., 1 for x > 0 and -1 for x < 0 and 0 for x = 0. I do not quite understand by "to show that they converge to a particular value in order to see if its the same as the original function?" If you mean how to sum the derived series and the integrated series other than by what I just said, then it is a different question altogether.