Fourier Series/1 over -1 to 0, Cosine of Pi x over 0 to 1

Theorem
Let $\map f x$ be the real function defined on $\openint {-1} 1$ as:


 * $\map f x = \begin{cases}

1 & : -1 < x < 0 \\ \map \cos {\pi x} & : 0 < x < 1 \end{cases}$

Then its Fourier series can be expressed as:


 * $\map f x \sim \displaystyle \dfrac 1 2 + \frac {\cos \pi x} 2 + \sum_{r \mathop = 1}^\infty \paren {\dfrac {4 r \sin 2 r \pi x} {\pi \paren {2 r + 1} \paren {2 r - 1} } - \dfrac {2 \map \sin {2 r + 1} \pi x } {\pi \paren {2 r + 1} } }$

Proof
By definition of Fourier series:


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

where:

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

Thus:

For $n > 0$:

Splitting this up into bits:

For $0 < x < 1$, there are two cases that need to be addressed: when $n = 1$ and when $n > 1$.

First let $n > 1$.

We have:

Now let $n = 1$.

We have:

There is only one non-vanishing term:


 * $a_1 = \dfrac 1 2$

Now for the $\sin n \pi x$ terms:

For $-1 < x < 0$:

For $0 < x < 1$, there are two cases that need to be addressed: when $n = 1$ and when $n > 1$.

First let $n > 1$.

Now let $n = 1$.

In summary:


 * $\begin {cases} a_0 = \dfrac 1 2 \\ a_1 = \dfrac 1 2 \\ b_{2 r} = \dfrac {4 r} {\pi \paren {2 r + 1} \paren {2 r - 1} } \\ b_{2 r + 1} = \dfrac {-2} {\pi \paren {2 r + 1} } \end {cases}$

Finally: