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For a function $f(x)$ where $x\in[0,1]$, the Fourier series has a fundamental frequency of $2\pi$. But I noticed that in the Fourier series expansion of the solution of heat equation (with same domain and homogeneous boundary conditions), the fundamental frequency was $\pi$. But why is that the case? Won't the set of functions having $2\pi$ as the fundamental frequency already form a complete set of basis? So, shouldn't the extra frequencies and the corresponding terms ($\sin \pi x, \sin 3\pi x, \sin 5\pi x$....) be redundant?

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  • $\begingroup$ If you want to expand in functions with a period of $1$, then you want to use $\{ e^{2\pi i nx }\}_{n=-\infty}^{\infty}$. $\endgroup$ – COVID-20 Sep 15 at 3:04
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$\{ \sin(n\pi x) \}_{n=1}^{\infty}$ is a complete orthogonal basis of $L^2[0,1]$. And $\{ e^{2\pi i nx} \}_{n=-\infty}^{\infty}$ is a complete orthonormal basis of $L^2[0,1]$. You can expand $\sin(\pi x)$ in $L^2[0,1]$ using a series $\sum_{n=-\infty}^{\infty}a_n e^{2\pi inx}$. That might seem a little strange, but no more strange than being able to expand $\cos(\pi x)$ in an $L^2$ convergent series of functions $\{ \sin(n\pi x) \}_{n=0}^{\infty}$. You'll get $L^2$ convergence, but obviously that won't translate to pointwise convergence at every point of $[0,1]$, and it doesn't have to in order to get $L^2[0,1]$ convergence.

In the same way, you can expand $\cos(\pi x/19)$ in a series of $\{ \sin(n\pi x) \}_{n=1}^{\infty}$, and the series will converge in $L^2[0,1]$. It all seems unlikely at first glance, but it's all part of the Mathemagic.

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