The fourier series dilemma

Question

According to fourier's theorem,
$$f\left(x\right)=a_0+a_1\cos x+a_2\cos2x...+b_{1\ }\sin x+b_2\sin2x...$$
For some coefficients $a_0, a_1...b_1,b_2...$,
Now if we consider this then each term is periodic and hence $f\left(x\right)$ must also be periodic with period $2\pi$,ie.
$$f\left(x+2\pi\right)=a_0+a_1\cos\left(x+2\pi\right)...b_1\sin\left(x+2\pi\right)...=a_0+a_1\cos x...+b_1\sin x...=f\left(x\right)$$
But this is not true as we know that every function is not periodic.
So my question is this a violation of fourier theorem or there are some conditions in the theorem which can explain this dilemma. Please Help $:)$

Answer 1

The idea is that you're making a expansion of the function $f(x)$ that you assume to be $2\pi$-periodic, hence your result $f(x + 2\pi) = f(x)$. If you wanted the expansion to have period $L$ then the series should look something like

$$ f\left(x\right)=a_0+a_1\cos \left(\color{blue}{\frac{2\pi x}{L}}\right)+a_2\left(2\color{blue}{\frac{2\pi x}{L}}\right)...+b_{1\ }\sin \left(\color{blue}{\frac{2\pi x}{L}}\right)+b_2\sin\left(2\color{blue}{\frac{2\pi x}{L}}\right)... $$

which then satisfies

$$ f(x + L) = f(x) $$

If the function is not periodic at all, then you take the limit $L\to \infty$, the sum becomes and integral (intuitively speaking) and you end up with a Fourier transform