Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am looking for a continuous function to be used in fourier series graph that have the same value at both $-\pi$ and $\pi$ but has a very poor differentiability at a point.

I have one: $\sqrt{(\pi\vert x\vert) - x^{2}}$ through trial and error and it indeed have poor differentiability at $x=0$. Now I have no issue computing its fourier coefficient for $a_{0}$ and $b_{n}$ but its another story for $a_{n}$. Hence I could not graph it. I have also consulted various books but to no avail.

Hence I would like to know if anyone here have a such function in mind so that I can graph its fourier's partial sums and compare it with its function.

share|improve this question
    
If you just want to take a look, you can compute coefficient numerically. Or even ask, say, Mathematic to plot truncated Fourier series for this function - there is such routine. –  Ilya Jan 10 '13 at 13:07
1  
@Ilya - Yes I did using Mathematica. When plotting the $a_{n}$, even Mathematica produced an error saying exceed limit, etc and couldn't continue. This equation took about 40 mins to be process by Mathematica on my computer. –  Sandra Jan 10 '13 at 13:17
1  
This is the $a_{n}$ computed by Mathematica => $\frac{(\pi BesselJ[1, (\frac{n\pi}{2}] \cos \frac{n \pi}{2}}{n}$. When plotting it as a graph, Mathematica can't continue. –  Sandra Jan 10 '13 at 13:21
    
Why not use $f(x) = |x|$? You can find the corresponding Fourier series here. –  Ayman Hourieh Jan 10 '13 at 13:26
    
@AymanHourieh - Yes I had done this. Thank you. I am looking for one other example. –  Sandra Jan 10 '13 at 13:30

1 Answer 1

If you are looking for something simple and not differentiable at $x=0$, try the following:

$$ |x| = \frac{\pi}{2} - \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2 n-1)^2} \cos{(2 n-1) x}$$

Plot the partial sums in Mathematica against $|x|$. What happens at $x=0$?

share|improve this answer
    
Yes I had plotted $\vert x \vert$ as one of my example. I am looking for another one with the same value at $(-\pi,\pi)$ and had more worst differentiability at a point than $\vert x \vert$. The one I gave in my first post had a very bad differentiability at $x=0$ but too bad even Mathematica can't draw out its partial sums. –  Sandra Jan 10 '13 at 13:33
    
Another one to try might be $|x|^{-1/2}$. Its Fourier transform has interesting properties at least. –  Ron Gordon Jan 10 '13 at 13:34
1  
From the graph of this function, I can't explain about the one point I need. I am trying to explain about the convergence and hence differentiability at one point just like $\vert x \vert$. I have even look through Handbooks of Mathematics & Tables and integrals reference books but to no avail. –  Sandra Jan 10 '13 at 13:47
    
Be careful re convergence of Fourier series at a point. Fourier series are designed to converge to a function over an interval according to an $L^2$ metric (i.e., the integral of the square of the error vanishes as the number of terms in the partial sum increases). At a particular point, however, you may never get convergence in the sense you hope. An example is the Fourier series for $\sin{x}/x$, which is known for its Gibb's phenomenon. –  Ron Gordon Jan 10 '13 at 13:55
1  
That is true. I am trying to show Paul du Bois-Reymond work from 1876 that says continuous function whose Fourier series diverge at a point. At present even reviewing all the available journal, there is no such function stated anywhere except proofs given by various authors that based on partial sums if I remember correctly. –  Sandra Jan 10 '13 at 14:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.