Fourier Cosine Coefficients for Odd Function over Symmetric Range

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Theorem

Let $\map f x$ be an odd real function defined on the interval $\openint {-\lambda} \lambda$.

Let the Fourier series of $\map f x$ be expressed as:

$\ds \map f x \sim \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$


Then for all $n \in \Z_{\ge 0}$:

$a_n = 0$


Proof

As suggested, let the Fourier series of $\map f x$ be expressed as:

$\ds \map f x \sim \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$

By definition of Fourier series:

\(\ds a_n\) \(=\) \(\ds \frac 1 \lambda \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x\)


From Cosine Function is Even:

$\cos a = \map \cos {-a}$

for all $a$.

By Odd Function Times Even Function is Odd, $\map f x \cos \dfrac {n \pi x} \lambda$ is odd.

The result follows from the corollary to Definite Integral of Odd Function:

$\ds \dfrac 1 \lambda \int_{-\lambda}^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x = 0$

$\blacksquare$


Also see


Sources

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