Wendy Gutierrez

Answered question

2022-01-26

calculating 2 constants in a function
$M=\left\{f\in C\left[0,2\pi \right],{\int }_{0}^{2\pi }f\left(x\right)\mathrm{sin}x,dx=\pi ,{\int }_{0}^{2\pi }f\left(x\right)\mathrm{sin}2x,dx=2\pi \right\}$
$a,b\in \mathbb{R},g\in M,g\left(x\right)=a\mathrm{sin}x+b\mathrm{sin}2x,x\in \left[0,2\pi \right]$
Ive

Answer & Explanation

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 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.

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