# Fundamental frequency in the Fourier series of heat equation.

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?

• If you want to expand in functions with a period of $1$, then you want to use $\{ e^{2\pi i nx }\}_{n=-\infty}^{\infty}$. – COVID-20 Sep 15 at 3:04

$$\{ \sin(n\pi x) \}_{n=1}^{\infty}$$ 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.