Half-Range Fourier Sine Series/Cosine over 0 to Pi
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Theorem
On the interval $\openint 0 \pi$:
\(\ds \cos x\) | \(=\) | \(\ds \frac 8 \pi \sum_{m \mathop = 1}^\infty \frac {m \sin 2 m x} {4 m^2 - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 8 \pi \paren {\frac {\sin 2 x} {1 \times 3} + \frac {2 \sin 4 x} {3 \times 5} + \frac {3 \sin 6 x} {5 \times 7} + \cdots}\) |
Proof
Let $\map f x$ be the function defined as:
- $\forall x \in \openint 0 \pi: \map f x = \cos x$
Let $f$ be expressed by a half-range Fourier sine series:
- $\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$
where for all $n \in \Z_{> 0}$:
- $b_n = \ds \frac 2 \lambda \int_0^\lambda \cos x \sin \frac {n \pi x} \lambda \rd x$
In this context, $\lambda = \pi$ and so this can be expressed more simply as:
- $\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$
where for all $n \in \Z_{> 0}$:
- $b_n = \ds \frac 2 \pi \int_0^\pi \cos x \sin n x \rd x$
First the case when $n = 1$:
\(\ds b_1\) | \(=\) | \(\ds \frac 2 \pi \int_0^\pi \cos x \sin x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \intlimits {\frac {\sin^2 x} 2} 0 \pi\) | Primitive of $\cos x \sin x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \paren {\frac {\sin^2 \pi} 2 - \frac {\sin^2 0} 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \paren {0 - 0}\) | Sine of Multiple of Pi | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
When $n \ne 1$:
\(\ds b_n\) | \(=\) | \(\ds \frac 2 \pi \int_0^\pi \cos x \sin n x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \intlimits {\frac {-\cos \paren {n - 1} x} {2 \paren {n - 1} } - \frac {\cos \paren {n + 1} x} {2 \paren {n + 1} } } 0 \pi\) | Primitive of $\cos x \sin n x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \paren {\paren {\frac {-\cos \paren {n - 1} \pi} {2 \paren {n - 1} } - \frac {\cos \paren {n + 1} \pi} {2 \paren {n + 1} } } - \paren {\frac {-\cos 0} {2 \paren {n - 1} } - \frac {\cos 0} {2 \paren {n + 1} } } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \paren {\paren {\frac {-\cos \paren {n - 1} \pi} {2 \paren {n - 1} } - \frac {\cos \paren {n + 1} \pi} {2 \paren {n + 1} } } + \frac 1 {2 \paren {n - 1} } + \frac 1 {2 \paren {n + 1} } }\) | Cosine of Zero is One and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \pi \paren {\frac {1 - \cos \paren {n - 1} \pi} {\paren {n - 1} } + \frac {1 - \cos \paren {n + 1} \pi} {\paren {n + 1} } }\) | Further simplification |
Thus for $n = 2 m$ for $m \in \Z$:
\(\ds b_n\) | \(=\) | \(\ds \frac 1 \pi \paren {\frac {1 - \cos \paren {2 m - 1} \pi} {\paren {2 m - 1} } + \frac {1 - \cos \paren {2 m + 1} \pi} {\paren {2 m + 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \pi \paren {\frac {1 - \paren {-1} } {\paren {2 m - 1} } + \frac {1 - \paren {-1} } {\paren {2 m + 1} } }\) | Cosine of Multiple of Pi | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \pi \paren {\frac 2 {\paren {2 m - 1} } + \frac 2 {\paren {2 m + 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \pi \paren {\frac {2 \paren {2 m + 1} + 2 \paren {2 m - 1} } {\paren {2 m + 1} \paren {2 m - 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 m + 2 + 4 m - 2} {\pi \paren {2 m + 1} \paren {2 m - 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {8 m} {\pi \paren {4 m^2 - 1} }\) |
and for $n = 2 m + 1$ for $m \in \Z$:
\(\ds b_n\) | \(=\) | \(\ds \frac 1 \pi \paren {\frac {1 - \cos \paren {2 m} \pi} {\paren {2 m - 1} } + \frac {1 - \cos \paren {2 m + 2} \pi} {\paren {2 m + 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \pi \paren {\frac {1 - 1} {\paren {2 m - 1} } + \frac {1 - 1} {\paren {2 m + 1} } }\) | Cosine of Multiple of Pi | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
Thus we have:
\(\ds b_{2 m}\) | \(=\) | \(\ds \frac {8 m} {\pi \paren {4 m^2 - 1} }\) | ||||||||||||
\(\ds b_{2 m + 1}\) | \(=\) | \(\ds 0\) |
and so over the given interval:
\(\ds \cos x\) | \(=\) | \(\ds \sum_{m \mathop = 1}^\infty \frac {8 m} {\pi \paren {4 m^2 - 1} } \sin 2 m x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 8 \pi \sum_{m \mathop = 1}^\infty \frac {m \sin 2 m x} {\paren {4 m^2 - 1} }\) |
$\blacksquare$
Sources
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Chapter One: $\S 5$. Half-Range Sine Series: Example $4$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 23$: Special Fourier Series and their Graphs: $23.13$