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

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Theorem

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

$\map f x$ and its $7$th approximation
$\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:

\(\ds a_n\) \(=\) \(\ds \int_{-1}^1 \map f x \cos n \pi x \rd x\)
\(\ds b_n\) \(=\) \(\ds \int_{-1}^1 \map f x \sin n \pi x \rd x\)

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


Thus:

\(\ds a_0\) \(=\) \(\ds \int_{-1}^1 \map f x \rd x\) Cosine of Zero is One
\(\ds \) \(=\) \(\ds \int_{-1}^0 1 \rd x + \int_0^1 \map \cos {\pi x} \rd x\)
\(\ds \) \(=\) \(\ds \bigintlimits x {-1} 0 + \intlimits {\frac {\map \sin {\pi x} } \pi} 0 1\) Primitive of Power, Primitive of $\cos \pi x$
\(\ds \) \(=\) \(\ds \paren {0 - \paren {-1} } - \paren {\frac {\sin \pi} \pi - \frac {\sin 0} \pi}\)
\(\ds \) \(=\) \(\ds 1 - 0\) Sine of Multiple of Pi
\(\ds \) \(=\) \(\ds 1\)

$\Box$


For $n > 0$:

\(\ds a_n\) \(=\) \(\ds \int_{-1}^1 \map f x \cos n \pi x \rd x\)
\(\ds \) \(=\) \(\ds \int_{-1}^0 \cos n \pi x \rd x + \int_0^1 \cos \pi x \cos n \pi x \rd x\) Definition of $f$


Splitting this up into bits:

\(\ds \) \(\) \(\ds \int_{-1}^0 \cos n \pi x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {\sin n \pi x} {n \pi} } {-1} 0\) Primitive of $\cos n \pi x$
\(\ds \) \(=\) \(\ds \paren {\frac {\sin 0} {n \pi} } - \paren {\frac {\map \sin {-n \pi} } {n \pi} }\)
\(\ds \) \(=\) \(\ds 0\) Sine of Multiple of Pi


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:

\(\ds \) \(\) \(\ds \int_0^1 \cos \pi x \cos n \pi x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {\map \sin {\pi - n \pi} x} {2 \paren {\pi - n \pi} } + \frac {\map \sin {\pi + n \pi} x} {2 \paren {\pi + n \pi} } } 0 1\) Primitive of $\cos \pi x \cos n \pi x$
\(\ds \) \(=\) \(\ds \intlimits {\frac {\map \sin {1 - n} \pi x} {2 \paren {1 - n} \pi} + \frac {\map \sin {1 + n} \pi x} {2 \paren {\paren {1 + n} \pi} } } 0 1\)
\(\ds \) \(=\) \(\ds \paren {\frac {\map \sin {1 - n} \pi} {2 \paren {1 - n} \pi} + \frac {\map \sin {1 + n} \pi} {2 \paren {\paren {1 + n} \pi} } } - \paren {\frac {\sin 0} {2 \paren {1 - n} \pi} + \frac {\sin 0} {2 \paren {\paren {1 + n} \pi} } }\)
\(\ds \) \(=\) \(\ds 0 - 0\) Sine of Multiple of Pi
\(\ds \) \(=\) \(\ds 0\)


Now let $n = 1$.

We have:

\(\ds \) \(\) \(\ds \int_0^1 \cos \pi x \cos \pi x \rd x\)
\(\ds \) \(=\) \(\ds \int_0^1 \cos^2 \pi x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac x 2 + \frac {\sin 2 \pi x} {4 \pi} } 0 1\) Primitive of $\cos^2 \pi x$
\(\ds \) \(=\) \(\ds \paren {\frac 1 2 + \frac {\sin 2 \pi} {4 \pi} } - \paren {\frac 0 2 + \frac {\sin 0} {4 \pi} }\)
\(\ds \) \(=\) \(\ds \paren {\frac 1 2 + 0} - \paren {0 + 0}\) Sine of Multiple of Pi
\(\ds \) \(=\) \(\ds \frac 1 2\)


There is only one non-vanishing term:

$a_1 = \dfrac 1 2$

$\Box$


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

\(\ds b_n\) \(=\) \(\ds \int_{-1}^1 \map f x \sin n \pi x \rd x\)
\(\ds \) \(=\) \(\ds \int_{-1}^0 \sin n \pi x \rd x + \int_0^1 \cos \pi x \sin n \pi x \rd x\)


For $-1 < x < 0$:

\(\ds \) \(\) \(\ds \int_{-1}^0 \sin n \pi x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {-\cos n \pi x} {n \pi} } {-1} 0\) Primitive of $\sin n \pi x$
\(\ds \) \(=\) \(\ds \paren {\frac {-\cos 0} {n \pi} } - \paren {\frac {-\map \cos {-n \pi} } {n \pi} }\)
\(\ds \) \(=\) \(\ds \frac {\map \cos {-n \pi} - 1} {n \pi}\) Cosine of Zero is One and rearrangement
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^n - 1} {n \pi}\) Cosine of Multiple of Pi
\(\ds \) \(=\) \(\ds \begin {cases} 0 & : \text {$n$ even} \\ \dfrac {-2} {n \pi} & : \text {$n$ odd} \end {cases}\)
\(\ds \leadsto \ \ \) \(\ds b_{2 r + 1}\) \(=\) \(\ds \dfrac {-2} {\pi \paren {2 r + 1} }\) substituting $n = 2 r + 1$ and simplifying


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

First let $n > 1$.

\(\ds \) \(\) \(\ds \int_0^1 \cos \pi x \sin n \pi x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {-\map \cos {n \pi - \pi} x} {2 \paren {n \pi - \pi} } - \frac {\map \cos {n \pi + \pi} x} {2 \paren {n \pi + \pi} } } 0 1\) Primitive of $\cos \pi x \sin n \pi x$
\(\ds \) \(=\) \(\ds \intlimits {\frac {-\map \cos {n - 1} \pi x} {2 \paren {n - 1} \pi} - \frac {\map \cos {n + 1} \pi x} {2 \paren {n + 1} \pi} } 0 1\)
\(\ds \) \(=\) \(\ds \paren {\frac {-\map \cos {n - 1} \pi} {2 \paren {n - 1} \pi} - \frac {\map \cos {n + 1} \pi} {2 \paren {n + 1} \pi} } - \paren {\frac {-\cos 0} {2 \paren {n - 1} \pi} - \frac {\cos 0} {2 \paren {n + 1} \pi} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 {2 \pi} \paren {\frac {1 - \map \cos {n - 1} \pi} {n - 1} + \frac {1 - \map \cos {n + 1} \pi} {n + 1} }\) Cosine of Zero is One and simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {2 \pi} \paren {\frac {1 - \paren {-1}^{n - 1} } {n - 1} + \frac {1 - \paren {-1}^{n + 1} } {n + 1} }\) Cosine of Multiple of Pi
\(\ds \) \(=\) \(\ds \dfrac 1 {2 \pi} \paren {\begin {cases} \paren {\dfrac 2 {n - 1} + \dfrac 2 {n + 1} } & : \text {$n$ even} \\ 0 & : \text {$n$ odd} \end {cases} }\)
\(\ds \leadsto \ \ \) \(\ds b_{2 r}\) \(=\) \(\ds \dfrac 1 \pi \paren {\dfrac 1 {2 r - 1} + \dfrac 1 {2 r + 1} }\) substituting $n = 2 r$ and simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 \pi \dfrac {\paren {2 r + 1} + \paren {2 r - 1} } {\paren {2 r + 1} \paren {2 r - 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {4 r} {\pi \paren {2 r + 1} \paren {2 r - 1} }\)


Now let $n = 1$.

\(\ds \) \(\) \(\ds \int_0^1 \cos \pi x \sin \pi x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {\sin^2 \pi x} {2 \pi} } 0 1\) Primitive of $\cos \pi x \sin \pi x$
\(\ds \) \(=\) \(\ds \paren {\frac {\sin^2 \pi} {2 \pi} } - \paren {\frac {\sin^2 \pi 0} {2 \pi} }\)
\(\ds \) \(=\) \(\ds 0 - 0\) Sine of Multiple of Pi
\(\ds \) \(=\) \(\ds 0\)


In summary:

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

$\Box$


Finally:

\(\ds \map f x\) \(\sim\) \(\ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}\)
\(\ds \) \(=\) \(\ds \frac {1 + \cos \pi x} 2 + \sum_{r \mathop = 1}^\infty \paren {\dfrac {4 r \sin 2 r \pi x} {\pi \paren {2 r + 1} \paren {2 r - 1} } - \dfrac {2 \map \sin {2 r + 1} \pi x } {\pi \paren {2 r + 1} } }\) substituting for $a_0$, $a_n$ and $b_n$ from above

$\blacksquare$


Sources

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