Wendy Gutierrez

2022-01-26

calculating 2 constants in a function

$M=\{f\in C[0,2\pi ],{\int}_{0}^{2\pi}f\left(x\right)\mathrm{sin}x,dx=\pi ,{\int}_{0}^{2\pi}f\left(x\right)\mathrm{sin}2x,dx=2\pi \}$

$a,b\in \mathbb{R},g\in M,g\left(x\right)=a\mathrm{sin}x+b\mathrm{sin}2x,x\in [0,2\pi ]$

Ive

Ive

basgrwthej

Beginner2022-01-27Added 13 answers

The overall pattern is

$\int}_{0}^{2\pi}\mathrm{sin}\left(mx\right)\mathrm{sin}\left(nx\right)dx=\pi {\delta}_{mn$

where$\delta}_{mn$ is 1 when m=n and 0 otherwise.

The average value of$\mathrm{sin}}^{2}\text{}\text{is}\text{}\frac{1}{2$ and sines with different periods are orthogonal over a common period. You can prove the the first from $\mathrm{sin}}^{2}x=\frac{12}{-}\frac{12}{\mathrm{cos}\left(2x\right)$ The $\mathrm{cos}\left(2x\right)$ integrates away. You can prove the first by using the function product identities, which gets you a sum of sine waves that integrate to zero as well. The same thing works for cosines.

This works because the sines and cosines form an orthogonal basis for the$2\pi$ periodic functions and you are effectively expanding a function in that basis.

where

The average value of

This works because the sines and cosines form an orthogonal basis for the