Half-Range Fourier Sine Series/Sine of Half x over 0 to Pi, Minus Sine of Half x over Pi to 2 Pi

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Theorem

Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:

$\map f x$ and its $7$th approximation
$\map f x = \begin {cases}

\sin \dfrac x 2 & : 0 \le x < \pi \\ -\sin \dfrac x 2 & : \pi < x \le 2 \pi \end {cases}$


Then its Fourier series can be expressed as:

\(\ds \map f x\) \(\sim\) \(\ds \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin n x} {4 n^2 - 1}\)
\(\ds \) \(=\) \(\ds \frac 8 \pi \paren {\frac {\sin x} {1 \times 3} - \frac {2 \sin 2 x} {3 \times 5} + \frac {3 \sin 3 x} {5 \times 7} - \frac {4 \sin 4 x} {7 \times 9} + \frac {5 \sin 5 x} {9 \times 11} - \dotsb}\)


Proof

By definition of half-range Fourier sine series:

$\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \dfrac {n x} 2$

where:

\(\ds b_n\) \(=\) \(\ds \frac 2 {2 \pi} \int_0^{2 \pi} \map f x \sin \frac {n \pi x} {2 \pi} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \int_0^{2 \pi} \map f x \sin \frac {n x} 2 \rd x\)

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


Thus:

\(\ds b_n\) \(=\) \(\ds \frac 1 \pi \int_0^{2 \pi} \map f x \sin \dfrac {n x} 2 \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \int_0^\pi \sin \dfrac x 2 \sin \dfrac {n x} 2 \rd x + \frac 1 \pi \int_\pi^{2 \pi} -\sin \dfrac x 2 \sin \dfrac {n x} 2 \rd x\) Definition of $f$


When $\dfrac {n x} 2 \ne \dfrac x 2$, that is, when $n \ne 1$, we have:

\(\ds \int \sin \dfrac x 2 \sin \dfrac {n x} 2 \rd x\) \(=\) \(\ds \frac {\sin \paren {\dfrac x 2 - \dfrac {n x} 2} } {2 \paren {\dfrac 1 2 - \dfrac n 2} } - \frac {\sin \paren {\dfrac x 2 + \dfrac {n x} 2} } {2 \paren {\dfrac 1 2 + \dfrac n 2} } + C\) Primitive of Sine of $\dfrac x 2$ by Sine of $\dfrac {n x} 2$
\(\ds \) \(=\) \(\ds \frac {\sin \paren {\dfrac {\paren {n - 1} x} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} x} 2} } {n + 1} + C\)


and so for $n \ne 1$:

\(\ds b_n\) \(=\) \(\ds \frac 1 \pi \int_0^{2 \pi} \map f x \sin \dfrac {n x} 2 \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \intlimits {\frac {\sin \paren {\dfrac {\paren {n - 1} x} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} x} 2} } {n + 1} } 0 \pi - \frac 1 \pi \intlimits {\frac {\sin \paren {\dfrac {\paren {n - 1} x} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} x} 2} } {n + 1} } \pi {2 \pi}\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \paren {\paren {\frac {\sin \paren {\dfrac {\paren {n - 1} \pi} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} \pi} 2} } {n + 1} } - \paren {\frac {\sin \paren {\dfrac {\paren {n - 1} 0} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} 0} 2} } {n + 1} } }\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac 1 \pi \paren {\paren {\frac {\sin \paren {\dfrac {\paren {n - 1} 2 \pi} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} 2 \pi} 2} } {n + 1} } - \paren {\frac {\sin \paren {\dfrac {\paren {n - 1} \pi} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} \pi} 2} } {n + 1} } }\)
\(\ds \) \(=\) \(\ds \frac 2 \pi \paren {\frac {\sin \paren {\dfrac {\paren {n - 1} \pi} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} \pi} 2} } {n + 1} }\) Sine of Multiple of Pi and simplifying


When $\dfrac {n x} 2 = \dfrac x 2$, that is, when $n = 1$, we have:

\(\ds \int \sin \dfrac x 2 \sin \dfrac x 2 \rd x\) \(=\) \(\ds \int \sin^2 \dfrac x 2 \rd x + C\)
\(\ds \) \(=\) \(\ds \frac x 2 - \frac {\sin 2 \dfrac x 2} {4 \dfrac 1 2} + C\) Primitive of $\sin^2 \dfrac x 2$
\(\ds \) \(=\) \(\ds \frac x 2 - \frac {\sin x} 2 + C\) simplifying


and so for $n = 1$:

\(\ds b_n\) \(=\) \(\ds \frac 1 \pi \int_0^{2 \pi} \map f x \sin \dfrac x 2 \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \intlimits {\frac x 2 - \frac {\sin x} 2} 0 \pi - \frac 1 \pi \intlimits {\frac x 2 - \frac {\sin x} 2} \pi {2 \pi}\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \paren {\paren {\frac \pi 2 - \frac {\sin \pi} 2} - \paren {\frac 0 2 - \frac {\sin 0} 2} }\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac 1 \pi \paren {\paren {\frac {2 \pi} 2 - \frac {\sin 2 \pi} 2} - \paren {\frac \pi 2 - \frac {\sin \pi} 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \paren {\frac \pi 2 - \frac {2 \pi} 2 + \frac \pi 2}\) Sine of Multiple of Pi
\(\ds \) \(=\) \(\ds 0\) everything vanishes


Hence:

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


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

\(\ds \) \(\) \(\ds \sum_{r \mathop = 1}^\infty \frac 2 \pi \paren {\frac {\sin \paren {\dfrac {\paren {\paren {2 r + 1} - 1} \pi} 2} } {\paren {2 r + 1} - 1} - \frac {\sin \paren {\dfrac {\paren {\paren {2 r + 1} + 1} \pi} 2} } {\paren {2 r + 1} + 1} } \sin \dfrac {\paren {2 r + 1} x} 2\)
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 1}^\infty \frac 2 \pi \paren {\frac {\sin \paren {\dfrac {2 r \pi} 2} } {2 r} - \frac {\sin \paren {\dfrac {\paren {2 r + 2} \pi} 2} } {2 r + 2} } \sin \dfrac {\paren {2 r + 1} x} 2\)
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 1}^\infty \frac 2 \pi \paren {\frac {\sin r \pi} {2 r} - \frac {\sin \paren {r + 1} \pi} {2 r + 2} } \sin \dfrac {\paren {2 r + 1} x} 2\)
\(\ds \) \(=\) \(\ds 0\) Sine of Multiple of Pi


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

\(\ds \) \(\) \(\ds \sum_{r \mathop = 1}^\infty \frac 2 \pi \paren {\frac {\sin \paren {\dfrac {\paren {2 r - 1} \pi} 2} } {2 r - 1} - \frac {\sin \paren {\dfrac {\paren {2 r + 1} \pi} 2} } {2 r + 1} } \sin \dfrac {2 r x} 2\)
\(\ds \) \(=\) \(\ds \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\sin \paren {r - \frac 1 2} \pi} {2 r - 1} - \frac {\sin \paren {r + \frac 1 2} \pi} {2 r + 1} } \sin r x\)
\(\ds \) \(=\) \(\ds \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\paren {-1}^{r - 1} } {2 r - 1} - \frac {\paren {-1}r} {2 r + 1} } \sin r x\) Sine of Half-Integer Multiple of Pi
\(\ds \) \(=\) \(\ds \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\paren {-1}^{r - 1} } {2 r - 1} + \frac {\paren {-1}^{r - 1} } {2 r + 1} } \sin r x\)
\(\ds \) \(=\) \(\ds \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \paren {\frac {2 r + 1 + 2 r - 1} {\paren {2 r - 1} \paren {2 r + 1} } } \sin r x\)
\(\ds \) \(=\) \(\ds \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {4 r} {4 r^2 - 1} \sin r x\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \frac 8 \pi \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r \sin r x} {4 r^2 - 1}\) simplifying

$\blacksquare$


Sources

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