Half-Range Fourier Sine Series/Sine of Half x over 0 to Pi, Minus Sine of Half x over Pi to 2 Pi
Jump to navigation
Jump to search
Theorem
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
- $\map f x = \begin {cases}
\sin \dfrac x 2 & : 0 \le x < \pi \\ -\sin \dfrac x 2 & : \pi < x \le 2 \pi \end {cases}$
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
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text I$: $2$.