Fourier series and convergence

[Niles]

Homework Statement

Hi all.

There is an example in my book, where we have the following Fourier series, and the author writes it as:

<br /> f(x) = \sum\limits_{n = 1}^\infty {\left( {\frac{{( - 1)^n }}{{n^2 }}\cos \left( {\frac{{n\pi x}}{p}} \right) + \frac{1}{n}\sin \left( {\frac{{n\pi x}}{p}} \right)} \right)} = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{n^2 }}\cos \left( {\frac{{n\pi x}}{p}} \right)} + \sum\limits_{n = 1}^\infty {\frac{1}{n}\sin \left( {\frac{{n\pi x}}{p}} \right)} <br />

First of all, we are only allowed to split the sum up if, and only if each part in the summation converges, but the author does not check if they do or don't. Is he making a mistake or am I missing something?

 

[marcusl]

These expressions are identically equal because addition is associative. That is, it doesn't matter what in what order you add numbers.

 

[Niles]

Hmm, according to my analysis-book, both series have to converge in order for us to divide the sum up.

 

[statdad]

You are write - if the two series converge absolutely then the original sum equals the sum of the two terms as shown. But I have a question.

You've given this:

The following code was used to generate this LaTeX image:


<br /> <br /> f(x) = \sum\limits_{n = 1}^\infty {\left( {\frac{{( - 1)^n }}{{n^2 }}\cos \left( {\frac{{n\pi x}}{p}} \right) + \frac{1}{n}\sin \left( {\frac{{n\pi x}}{p}} \right)} \right)} = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{n^2 }}\cos \left( {\frac{{n\pi x}}{p}} \right)} + \sum\limits_{n = 1}^\infty {\frac{1}{n}\sin \left( {\frac{{n\pi x}}{p}} \right)} <br />

If this is written as a way of saying "the function f can be formally represented by this series, then convergence isn't an issue (although usually the notation f(x) \sim is used rather than equality)

If the statement is that the first series actually converges to f, then there may be a theorem that says "given these conditions on a function, the Fourier series actually converges and ...", and the work is justified by that. A final possibility is that the reader is supposed to supply the details about why this is valid.

(There is always the possibility of a typographical error in the text, but one of this magnitude would be, I think, rare)

 

[marcusl]

The first sum is convergent, but the second requires thought. Break the second sum into a sum over odd integers plus a sum over even integers. The sum over odds describes a square wave of period 2p, so it is convergent. Every term in the even sum is smaller than the corresponding term in the odd sum, so it is convergent also by comparison.