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