Fourier Series/Cosine of x over Minus Pi to Zero, Minus Cosine of x over Zero 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 $7$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

$\blacksquare$


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