determination of Fourier coefficients
DeterminationOfFourierCoefficients
2009-04-18 09:05:40
2009-04-18 09:05:40
Algorithm
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Suppose that the real function $f$ 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 formulas of the trigonometric identities, 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 $f$.