Fourier Cosine Coefficients for Even Function over Symmetric Range

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

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


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

$a_n = \dfrac 2 \lambda \displaystyle \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$


Proof

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

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

By definition of Fourier series:

$a_n = \dfrac 1 \lambda \displaystyle \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x$


From Cosine Function is Even:

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

for all $a$.

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


Thus:

\(\displaystyle a_n\) \(=\) \(\displaystyle \frac 1 \lambda \int_{-\lambda}^{-\lambda + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 \lambda \int_{-\lambda}^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \lambda \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x\) Definite Integral of Odd Function: Corollary

$\blacksquare$


Also see


Sources