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

## 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

$\blacksquare$