Odd way to define a Fourier series

I found a solution to a differential equation as a Fourier series written as

$\displaystyle X = a_0 + k \sum_{n \neq 0} \frac{1}{n} a_n e^{-in \tau}$

I've been having some trouble understanding the 1/n in the summand. Is this simply a way to underscore that n can't be 0 in the sum? I really don't know why the summand is written that way.

Thanks!

-Dan

Re: Odd way to define a Fourier series

Well it would definitely make sense that $\displaystyle \displaystyle \begin{align*} n \neq 0 \end{align*}$ because of the $\displaystyle \displaystyle \begin{align*} \frac{1}{n} \end{align*}$. I would make a guess to say that your counter is over all integer values of n except 0, including negatives, otherwise they would just write $\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty} \end{align*}$ would they not?

Re: Odd way to define a Fourier series

Quote:

Originally Posted by

**Prove It** Well it would definitely make sense that $\displaystyle \displaystyle \begin{align*} n \neq 0 \end{align*}$ because of the $\displaystyle \displaystyle \begin{align*} \frac{1}{n} \end{align*}$. I would make a guess to say that your counter is over all integer values of n except 0, including negatives, otherwise they would just write $\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty} \end{align*}$ would they not?

Yes, the sum is over all integers except 0. (In the end it's only over all positive n. There's a relationship due to the boundary conditions that gives a_{-n} in terms of a_n.) My question is basically "Why don't they just absorb the 1/n into the a_n?" Reading the text doesn't give an answer. Could the reason possibly be to force the sum to converge? (What's that called? Regularization or something?)

-Dan