Fourier Series/Pi Squared minus x Squared over Minus Pi to Pi

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Theorem

Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as:

$\map f x$ and its $4$th approximation
$\map f x = \pi^2 - x^2$


$f$ can be expressed as a half-range Fourier cosine series thus:

\(\displaystyle \map f x\) \(\sim\) \(\displaystyle \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos n x} {n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 4 \paren {\cos x - \frac 1 4 \cos 2 x + \frac 1 9 \cos 3 x - \cdots}\)


Proof

We have that:

$\pi^2 - \paren {-x}^2 = \pi^2 - x^2$

and so $\map f x$ is even on $\openint {-\pi} \pi$.

It follows from Fourier Series for Even Function over Symmetric Range:

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


where for all $n \in \Z_{> 0}$:

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


Thus by definition of $f$:

\(\displaystyle a_0\) \(=\) \(\displaystyle \frac 2 \pi \int_0^\pi \map f x \rd x\) Cosine of Zero is One
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \pi \int_0^\pi \paren {\pi^2 - x^2} \rd x\) Definition of $f$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \pi \intlimits {\pi^2 x - \frac {x^3} 3} 0 \pi\) Primitive of Power
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \pi \paren {\paren {\pi^3 - \frac {\pi^3} 3} - \paren {0 - \frac 0 3} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \pi \paren {\frac {2 \pi^3} 3}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {4 \pi^2} 3\)

$\Box$


Then for $n > 0$:

\(\displaystyle a_n\) \(=\) \(\displaystyle \frac 2 \pi \int_0^\pi \map f x \cos n x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \pi \int_0^\pi \paren {\pi^2 - x^2} \cos n x \rd x\) Definition of $f$
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi \int_0^\pi \cos n x \rd x - \frac 2 \pi \int_0^\pi x^2 \cos n x \rd x\) Linear Combination of Definite Integrals


Splitting this up into two:

\(\displaystyle \) \(\) \(\displaystyle 2 \pi \int_0^\pi \cos n x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi \intlimits {\frac {\sin n x} n} 0 \pi\) Primitive of $\cos a x$
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi \paren {\frac {\sin n \pi} n - \frac {\sin 0} n}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi \paren {0 - 0}\) Sine of Multiple of Pi
\(\displaystyle \) \(=\) \(\displaystyle 0\)


\(\displaystyle \) \(\) \(\displaystyle -\frac 2 \pi \int_0^\pi x^2 \cos n x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac 2 \pi \intlimits {\frac {2 x \cos n x} {n^2} + \paren {\frac {x^2} n - \frac 2 {n^3} } \sin n x} 0 \pi\) Primitive of $x^2 \cos n x$
\(\displaystyle \) \(=\) \(\displaystyle -\frac 2 \pi \paren {\paren {\frac {2 \pi \cos n \pi} {n^2} + \paren {\frac {\pi^2} n - \frac 2 {n^3} } \sin n \pi} - \paren {\frac {0 \cos 0} {n^2} + \paren {\frac {0^2} n - \frac 2 {n^3} } \sin 0} }\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac 2 \pi \paren {\frac {2 \pi \cos n \pi} {n^2} }\) Sine of Multiple of Pi and removing vanishing terms
\(\displaystyle \) \(=\) \(\displaystyle -\frac {4 \cos n \pi} {n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac {4 \paren {-1}^n} {n^2}\) Cosine of Multiple of Pi
\(\displaystyle \) \(=\) \(\displaystyle \paren {-1}^{n - 1} \frac 4 {n^2}\) simplification

$\Box$


Finally:

\(\displaystyle \map f x\) \(\sim\) \(\displaystyle \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \frac {4 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac 4 {n^2} \cos n x\) substituting for $a_0$ and $a_n$ from above
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos n x} {n^2}\) simplifying

$\blacksquare$


Sources