Fourier Series
Intuition $$ \frac{1}{\sqrt{ 2\pi }}, \frac{\cos(kx)}{\sqrt{ \pi }}, \frac{\sin(kx)}{\sqrt{ \pi }}, \dots $$$$ \int_{0}^{2\pi} (\sin (kx))^{2} \, dx = \int_{0}^{2\pi} (\cos(kx))^{2} \, dx = \pi $$$$ \int_{0}^{2\pi}\sin(kx)\sin(hx) \, dx = \int_{0}^{2\pi}\cos(kx)\cos(hx) \, dx = 0 $$$$ \int_{0}^{2\pi}\sin(kx)\cos(hx) \, dx = \int_{0}^{2\pi} \sin(kx) \, dx = \int_{0}^{2\pi}\cos(hx) \, dx = 0 $$Proofs of the relations In this section we quickly prove why the above equations hold. First we all agree that $\int_{0}^{2\pi} \sin(kx) \, dx = \int_{0}^{2\pi} \cos(hx) \, dx = 0$ because their period divides $2\pi$ and the sum of the area of a period is clearly 0. Or we can explicitly find the primitive and solve ...