# Fourier Series/Cosine of x over Minus Pi to Zero, Minus Cosine of x over Zero to Pi

## Theorem

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

$\map f x$ and its $7$th approximation
$\map f x = \begin {cases} \cos x & : -\pi < x < 0 \\ -\cos x & : 0 < x < \pi \end {cases}$

Then its Fourier series can be expressed as:

 $\displaystyle \map f x$ $\sim$ $\displaystyle -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r x} {4 r^2 - 1}$ $\displaystyle$ $=$ $\displaystyle -\frac 8 \pi \paren {\frac {\sin 2 x} {1 \times 3} + \frac {\sin 4 x} {3 \times 5} + \frac {\sin 6 x} {5 \times 7} + \frac {\sin 8 x} {7 \times 9} + \dotsb}$

## Proof

It is apparent by inspection that $\map f x$ is an odd function over $\openint {-\pi} \pi$.

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

$\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$

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

$b_n = \displaystyle \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x$

for all $n \in \Z_{>0}$.

Thus by definition of $f$:

$\displaystyle b_n = -\frac 2 \pi \int_0^\pi \cos x \sin n x \rd x$

When $n \ne 1$, we have:

 $\displaystyle b_n$ $=$ $\displaystyle -\frac 2 \pi \int_0^\pi \cos x \sin n x \rd x$ $\displaystyle$ $=$ $\displaystyle -\frac 2 \pi \intlimits {\frac {-\cos \paren {n - 1} x} {2 \paren {n - 1} } - \frac {\cos \paren {n + 1} x} {2 \paren {n + 1} } } 0 \pi$ Primitive of $\cos x \sin n x$ $\displaystyle$ $=$ $\displaystyle -\frac 2 \pi \paren {\paren {\frac {-\cos \paren {n - 1} \pi} {2 \paren {n - 1} } - \frac {\cos \paren {n + 1} \pi} {2 \paren {n + 1} } } - \paren {\frac {-\cos \paren {n - 1} 0} {2 \paren {n - 1} } - \frac {\cos \paren {n + 1} 0} {2 \paren {n + 1} } } }$ $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \paren {\paren {\frac {\cos \paren {n - 1} \pi} {n - 1} + \frac {\cos \paren {n + 1} \pi} {n + 1} } - \paren {\frac 1 {n - 1} + \frac 1 {n + 1} } }$ Cosine of Zero is One and simplifying $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \paren {\frac {\cos \paren {n - 1} \pi} {n - 1} - \frac 1 {n - 1} + \frac {\cos \paren {n + 1} \pi} {n + 1} - \frac 1 {n + 1} }$ simplifying

When $n = 1$, we have:

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

Hence:

 $\displaystyle \map f x$ $\sim$ $\displaystyle \sum_{n \mathop = 1}^\infty b_n \sin n x$ $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \sum_{n \mathop = 2}^\infty \paren {\frac {\cos \paren {n - 1} \pi} {n - 1} - \frac 1 {n - 1} + \frac {\cos \paren {n + 1} \pi} {n + 1} - \frac 1 {n + 1} } \sin n x$ substituting for $b_n$ from above and rearranging

When $n$ is odd, we have $n = 2 r + 1$ for $r \ge 1$, and so:

 $\displaystyle$  $\displaystyle \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\cos \paren {\paren {2 r + 1} - 1} \pi} {\paren {2 r + 1} - 1} - \frac 1 {\paren {2 r + 1} - 1} + \frac {\cos \paren {\paren {2 r + 1} + 1} \pi} {\paren {2 r + 1} + 1} - \frac 1 {\paren {2 r + 1} + 1} } \sin \paren {2 r + 1} x$ $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\cos 2 r \pi} {2 r} - \frac 1 {2 r} + \frac {\cos \paren {2 r + 2} \pi} {2 r + 2} - \frac 1 {2 r + 2} } \sin \paren {2 r + 1} x$ $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac 1 {2 r} - \frac 1 {2 r} + \frac 1 {2 r + 2} - \frac 1 {2 r + 2} } \sin \paren {2 r + 1} x$ Cosine of Multiple of Pi $\displaystyle$ $=$ $\displaystyle 0$ everything vanishes

When $n$ is even, we have $n = 2 r$ for $r \ge 1$, and so:

 $\displaystyle$  $\displaystyle \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\cos \paren {2 r - 1} \pi} {2 r - 1} - \frac 1 {2 r - 1} + \frac {\cos \paren {2 r + 1} \pi} {2 r + 1} - \frac 1 {2 r + 1} } \sin 2 r x$ $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {-1} {2 r - 1} - \frac 1 {2 r - 1} + \frac {-1} {2 r + 1} - \frac 1 {2 r + 1} } \sin 2 r x$ Cosine of Multiple of Pi $\displaystyle$ $=$ $\displaystyle -\frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {\frac 1 {2 r - 1} + \frac 1 {2 r + 1} } \sin 2 r x$ $\displaystyle$ $=$ $\displaystyle -\frac 2 \pi \sum_{r \mathop = 1}^\infty \frac {\paren {2 r + 1} + \paren {2 r - 1} } {\paren {2 r - 1} \paren {2 r + 1} } \sin 2 r x$ $\displaystyle$ $=$ $\displaystyle -\frac 2 \pi \sum_{r \mathop = 1}^\infty \frac {4 r} {4 r^2 - 1} \sin 2 r x$ Difference of Two Squares and simplifying $\displaystyle$ $=$ $\displaystyle -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r x} {4 r^2 - 1}$ simplifying

$\blacksquare$