Talk:PlanetPhysics/Determination of Fourier Coefficients

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: determination of Fourier coefficients %%% Primary Category Code: 02. %%% Filename: DeterminationOfFourierCoefficients.tex %%% Version: 3 %%% Owner: pahio %%% Author(s): bci1, pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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Suppose that the real \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} $f$ may be presented as sum of the Fourier series: \begin{align} f(x) \;=\; \frac{a_0}{2}+\sum_{m=0}^\infty(a_m\cos{mx}+b_m\sin{mx}) \end{align} 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{align} \int_{-\pi}^\pi\!f(x)\cos{nx}\,dx \,=\, \frac{a_0}{2}\!\int_{-\pi}^\pi\!\cos{nx}\,dx +\!\sum_{m=0}^\infty\!\left(a_m\!\int_{-\pi}^\pi\!\cos{mx}\cos{nx}\,dx+b_m\!\int_{-\pi}^\pi\!\sin{mx}\cos{nx}\,dx\right)\!. \end{align} When\, $n = 0$,\, the equation (2) reads \begin{align} \int_{-\pi}^\pi f(x)\,dx = \frac{a_0}{2}\cdot2\pi = \pi a_0, \end{align} 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 \htmladdnormallink{formulas}{http://planetphysics.us/encyclopedia/Formula.html} of the trigonometric \htmladdnormallink{identities}{http://planetphysics.us/encyclopedia/Cod.html}, getting $$\int_{-\pi}^\pi\cos{mx}\cos{nx}\,dx = \frac{1}{2}\int_{-\pi}^\pi[\cos(m-n)x+\cos(m+n)x]\,dx,$$ $$\int_{-\pi}^\pi\sin{mx}\cos{nx}\,dx = \frac{1}{2}\int_{-\pi}^\pi[\sin(m-n)x+\sin(m+n)x]\,dx.$$ The latter expression vanishes always, since the sine is an odd function.\, If\, $m \neq n$,\, 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{align} \int_{-\pi}^\pi f(x)\cos{nx}\,dx = \pi a_n \end{align} 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{align} \int_{-\pi}^\pi f(x)\sin{nx}\,dx = \pi b_n. \end{align} The equations (4) and (5) imply the formulas $$a_n \;=\; \frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos{nx}\,dx \quad (n = 0,\,1,\,2,\,\ldots)$$ and $$b_n \;=\; \frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin{nx}\,dx \quad (n = 1,\,2,\,3,\,\ldots)$$ for finding the values of the Fourier coefficients of $f$.

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