Fourier Series/x squared over Minus Pi to Pi

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Theorem

For $x \in \closedint {-\pi} \pi$:

$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$


Proof

From Even Power is Even Function, $x^2$ is an even function.

By Fourier Series for Even Function over Symmetric Range, we have:

$\displaystyle x^2 \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$

where:

\(\displaystyle a_n\) \(=\) \(\displaystyle \frac 2 \pi \int_0^\pi x^2 \map \cos {n x} \rd x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 1 2 a_0\) \(=\) \(\displaystyle \frac 2 {2 \pi} \int_0^\pi x^2 \rd x\) Cosine of Zero is One
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 \pi \cdot \frac {\pi^3} 3\) Primitive of Power, Fundamental Theorem of Calculus
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 3\)


Then:

\(\displaystyle \frac 2 \pi \int_0^\pi x^2 \map \cos {n x} \rd x\) \(=\) \(\displaystyle \frac 2 \pi \intlimits {\frac {2 x \cos n x} {n^2} + \paren {\frac {x^2} n - \frac 2 {n^3} } \sin n x} 0 \pi\) Primitive of $x^2 \cos a x$, Fundamental Theorem of Calculus
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \pi \intlimits {\frac {2 x \cos n x} {n^2} } 0 \pi\) Sine of Multiple of Pi
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \pi \cdot \frac {2 \pi \cos n \pi} {n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {4 \cos n \pi} {n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {-1}^n \frac 4 {n^2}\) Cosine of Multiple of Pi


Substituting for $a_n$ in $(1)$:

$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$

as required.

$\blacksquare$