How do we know that a Fourier series expansion does exist for a given function $f(x)$? I mean, if $f(x)=x$ and we suppose that $x=a_1\sin(x)+a_2\sin(2x)$ with $-\pi\leq x \leq \pi$ the Fourier coefficients $a_1$ and $a_2$ still being the same as if we suppose that $x=a_1\sin(x)+a_2\sin(2x)+...$ So, why there are infinitely many sine or cosine terms?
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I think your question has two parts. The first is existence for a given $f$. A Fourier series exists for a given $f(x)$, $x \in [-\pi,\pi)$ when
$$\int_{-\pi}^{\pi} dx \, |f(x)|^2 \lt \infty$$
The second part I think asks how do we know that the FS coefficients do not change when we add more harmonics to the series. This is because the coefficients do not depend on the number of terms in the sum; they simply depend on the particular harmonic and the function, i.e.,
$$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} dx \, f(x) \, \sin{n x}$$
Ron Gordon
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I know. When we integrate, all other harmonics cancel leaving an squared sine or cosine term. But how can we suppose that there are infinitely many terms? – dot dot Jun 14 '13 at 13:53
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Because of the integrability condition, as well as the fact that the coefficients were derived from a minimization of mean square error betwen the FS and the function. For more details, see http://math.stackexchange.com/questions/364304/fourier-analysis/364371#364371 – Ron Gordon Jun 14 '13 at 13:56
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Note that Ron's condition does not imply that the Fourier series converges (pointwise) to $f$. Only that it converges in the mean of order 2. – GEdgar Jun 14 '13 at 14:16
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Yes. To get the coefficients you find the minimum as in linear regression. – dot dot Jun 14 '13 at 14:51