Half-Range Fourier Sine Series/x by Pi minus x over 0 to Pi
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Theorem
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
- $\map f x = x \paren {\pi - x}$
Then its half-range Fourier sine series can be expressed as:
- $\ds \map f x \sim \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 r + 1} x} {\paren {2 r + 1}^3}$
Proof
By definition of half-range Fourier sine series:
- $\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$
where for all $n \in \Z_{> 0}$:
- $b_n = \ds \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x$
Thus by definition of $f$:
\(\ds b_n\) | \(=\) | \(\ds \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \int_0^\pi x \paren {\pi - x} \sin n x \rd x\) | Definition of $f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \int_0^\pi x \sin n x \rd x - \frac 2 \pi \int_0^\pi x^2 \sin n x \rd x\) | Linear Combination of Definite Integrals and simplifying |
Splitting it up into two:
\(\ds 2 \int_0^\pi x \sin n x \rd x\) | \(=\) | \(\ds 2 \intlimits {\frac {\sin n x} {n^2} - \frac {x \cos n x} n} 0 \pi\) | Primitive of $x \sin n x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \paren {\paren {\frac {\sin n \pi} {n^2} - \frac {\pi \cos n \pi} n} - \paren {\frac {\sin 0} {n^2} - \frac {0 \cos 0} n} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \frac {\pi \cos n \pi} n\) | Sine of Multiple of Pi and removing vanishing terms |
and:
\(\ds \frac 2 \pi \int_0^\pi x^2 \sin n x \rd x\) | \(=\) | \(\ds 2 \intlimits {\frac {2 x \sin n x} {n^2} + \paren {\frac 2 {n^3} - \frac {x^2} n} \cos n x} 0 \pi\) | Primitive of $x^2 \sin n x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \paren {\paren {\frac {2 \pi \sin n \pi} {n^2} + \paren {\frac 2 {n^3} - \frac {\pi^2} n} \cos n \pi} - \paren {\frac {0 \sin 0} {n^2} + \paren {\frac 2 {n^3} - \frac {0^2} n} \cos 0} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \paren {\paren {\frac 2 {n^3} - \frac {\pi^2} n} \cos n \pi - \frac 2 {n^3} \cos 0}\) | Sine of Multiple of Pi and removing vanishing terms | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \paren {\paren {\frac 2 {n^3} - \frac {\pi^2} n} \cos n \pi - \frac 2 {n^3} }\) | Cosine of Zero is One |
Thus:
\(\ds b_n\) | \(=\) | \(\ds -2 \frac {\pi \cos n \pi} n - \frac 2 \pi \paren {\paren {\frac 2 {n^3} - \frac {\pi^2} n} \cos n \pi - \frac 2 {n^3} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \paren {1 - \cos n \pi} } {\pi n^3} - 2 \frac {\pi \cos n \pi} n + \frac {2 \pi \cos n \pi} n\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \paren {1 - \paren {-1}^n} } {\pi n^3}\) | Cosine of Multiple of Pi and simplifying |
When $n$ is even, $\paren {-1}^n = 1$ and so $\dfrac {4 \paren {1 - \paren {-1}^n} } {\pi n^3} = 0$.
When $n$ is odd, $n$ can be expressed as $n = 2 r + 1$ for $r \ge 0$.
Thus:
\(\ds b_n\) | \(=\) | \(\ds \frac {4 \paren {1 - \paren {-1}^{2 r + 1} } } {\pi \paren {2 r + 1}^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \paren {1 + 1} } {\pi \paren {2 r + 1}^3}\) | as $\paren {-1}^{2 r + 1} = -1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 8 {\pi \paren {2 r + 1}^3}\) |
Hence:
- $\ds \map f x \sim \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 r + 1} x} {\paren {2 r + 1}^3}$
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 23$: Special Fourier Series and their Graphs: $23.18$