Coefficients of Cosine Terms in Convergent Trigonometric Series

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\map S x$ be a trigonometric series which converges to $\map f x$ on the interval $\openint \alpha {\alpha + 2 \pi}$:

$\map f x = \dfrac {a_0} 2 + \ds \sum_{m \mathop = 1}^\infty \paren {a_m \cos m x + b_m \sin m x}$


Then:

$\forall n \in \Z_{\ge 0}: a_n = \dfrac 1 \pi \ds \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x$


Proof

\(\ds \map f x\) \(=\) \(\ds \dfrac {a_0} 2 + \sum_{m \mathop = 1}^\infty \paren {a_m \cos m x + b_m \sin m x}\)
\(\ds \leadsto \ \ \) \(\ds \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x\) \(=\) \(\ds \int_\alpha^{\alpha + 2 \pi} \paren {\dfrac {a_0} 2 + \sum_{m \mathop = 1}^\infty \paren {a_m \cos m x + b_m \sin m x} } \cos n x \rd x\)
\(\ds \) \(=\) \(\ds \int_\alpha^{\alpha + 2 \pi} \dfrac {a_0} 2 \cos n x \rd x\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} \paren {a_m \cos m x + b_m \sin m x} \cos n x \rd x}\)
\(\ds \) \(=\) \(\ds \frac {a_0} 2 2 \pi \delta_{n 0} + \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} \paren {a_m \cos m x + b_m \sin m x} \cos n x \rd x}\) $\ds \int_\alpha^{\alpha + 2 \pi} \cos n x \rd x = \begin {cases} 0 & : n \ne 0 \\ 2 \pi & : n = 0 \end {cases}$
\(\ds \) \(=\) \(\ds a_0 \pi \delta_{n 0} + \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} a_m \cos m x \cos n x \rd x}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} b_m \sin m x \cos n x \rd x}\)
\(\ds \) \(=\) \(\ds a_0 \pi \delta_{n 0} + \sum_{m \mathop = 1}^\infty \paren {\int_\alpha^{\alpha + 2 \pi} a_m \cos m x \cos n x \rd x}\) $\ds \int_\alpha^{\alpha + 2 \pi} \sin m x \cos n x \rd x = 0$
\(\ds \) \(=\) \(\ds a_0 \pi \delta_{n 0} + \sum_{m \mathop = 1}^\infty a_m \pi \delta_{m n}\) $\ds \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x = \begin {cases} 0 & : m \ne n \\ \pi & : m = n \end {cases}$


Thus when $n = 0$ we have:

$\ds \int_\alpha^{\alpha + 2 \pi} \map f x \rd x = \int_\alpha^{\alpha + 2 \pi} \map f x \cos 0 x \rd x = a_0 \pi$

and when $n \ne 0$ we have:

$\ds \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x = a_n \pi$

Hence the result.

$\blacksquare$


Also see


Sources