# How many terms in a series expansion

General: If $f \in C^1$ is a periodic function defined over some multi-dimensional space, then it should be possible to express $f$ as a FINITE fourier series.

1. is this true of any periodic basis?
2. is there a way to determine the number of terms in this finite series?

Specific: I have a function that is smooth and continuous and is defined on the unit hypersphere. I want to know:

1. is it possible to represent this function as a FINITE series in the hyperspherical harmonic basis?
2. how many terms will it take?
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This isn't true. Almost every periodic function will have infinitely many nonzero terms in its Fourier series expansion. Requiring that the function be continuous or even smooth doesn't change this, although it does place some limits on the asymptotic behavior of the Fourier coefficients as $|k|\rightarrow\infty$. – mjqxxxx Jan 3 '12 at 5:01
@ mjqxxxx: Under what circumstances is a function guaranteed to have a finite fourier series? – okj Jan 3 '12 at 21:32
@okj : When your function itself is a sinusoid or a linear combination of sinusoids of harmonically related frequencies, then the Fourier series has only finite non zero terms. In other words take any Fourier series and make all coefficients except some finite number of them to zero, then the resulting function is what you want. – Rajesh Dachiraju Jan 4 '12 at 0:21

If $f$ is a $C^1$ function, $2\pi$-periodic, defined on the real line, then $$\widehat{f'}(n)=\frac 1{2\pi}\int_0^{2\pi}f'(x)e^{-2i nx}dx=-\frac 1{2\pi}\int_0^{2\pi}f(x)e^{-2inx}(-2in)dx=\frac{in}{\pi}\widehat f(n),$$ so $\{n\widehat{f}(n)\}\in\ell^2(\mathbb Z)\}$. But it doesn't imply that only finitely many terms are nonzero. For example, let $\{\alpha_n\}_{n=-\infty}^{+\infty}$ a bounded sequence and $f(x)=\sum_{n=-\infty}^{+\infty}\alpha_n e^{-|n|}e^{inx}$. $f$ is $C^1$ and all the terms are not $0$, so there are a lot of function, which are $C^1$ (an in fact $C^{\infty}$ as the example shows), but all the Fourier coefficients are different from $0$.