PlanetPhysics/Determination of Fourier Coefficients

Suppose that the real [[../Bijective/|function]] $f$ may be presented as sum of the Fourier series:

$\begin{array}{c}f(x)=\frac{a_0}{2}+\sum _{m=0}^{\mathrm{\infty }}(a_m\cos mx+b_m\sin mx)\end{array}$

Therefore, $f$ is periodic with period $2\pi$.\, For expressing the Fourier coefficients $a_m$ and $b_m$ with the function itself, we first multiply the series (1) by $\cos nx$ ($n\in \mathbb{Z}$) and integrate from $-\pi$ to $\pi$.\, Supposing that we can integrate termwise, we may write

$\begin{array}{c}\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos nxdx=\frac{a_0}{2}\underset{-\pi }{\overset{\pi }{\int }}\cos nxdx+\sum _{m=0}^{\mathrm{\infty }}(a_m\underset{-\pi }{\overset{\pi }{\int }}\cos mx\cos nxdx+b_m\underset{-\pi }{\overset{\pi }{\int }}\sin mx\cos nxdx).\end{array}$

When\, $n=0$,\, the equation (2) reads

$\begin{array}{c}\underset{-\pi }{\overset{\pi }{\int }}f(x)dx=\frac{a_0}{2}\cdot 2\pi =\pi a_0,\end{array}$

since in the sum of the right hand side, only the first addend is distinct from zero.

When $n$ is a positive integer, we use the product [[../Formula/|formulas]] of the trigonometric [[../Cod/|identities]], getting $\underset{-\pi }{\overset{\pi }{\int }}\cos mx\cos nxdx=\frac{1}{2}\underset{-\pi }{\overset{\pi }{\int }}[\cos (m-n)x+\cos (m+n)x]dx,$ $\underset{-\pi }{\overset{\pi }{\int }}\sin mx\cos nxdx=\frac{1}{2}\underset{-\pi }{\overset{\pi }{\int }}[\sin (m-n)x+\sin (m+n)x]dx.$ The latter expression vanishes always, since the sine is an odd function.\, If\, $m\ne n$,\, the former equals zero because the antiderivative consists of sine terms which vanish at multiples of $\pi$; only in the case\, $m=n$\, we obtain from it a non-zero result $\pi$.\, Then (2) reads

$\begin{array}{c}\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos nxdx=\pi a_n\end{array}$

to which we can include as a special case the equation (3).

By multiplying (1) by $\sin nx$ and integrating termwise, one obtains similarly

$\begin{array}{c}\underset{-\pi }{\overset{\pi }{\int }}f(x)\sin nxdx=\pi b_n.\end{array}$

The equations (4) and (5) imply the formulas $a_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos nxdx(n=0,1,2,\dots )$ and $b_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\sin nxdx(n=1,2,3,\dots )$ for finding the values of the Fourier coefficients of $f$.