Fourier Series for Logarithm of Sine of x over 0 to Pi
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Theorem
- $\ds \map \ln {\sin x} = -\ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n$
where $0 < x < \pi$.
Proof 1
We find the Half-Range Fourier Cosine Series over $\openint 0 {\dfrac \pi 2}$ for $\map \ln {\sin x}$.
By definition:
- $\ds \map \ln {\sin x} \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos 2 n x$
where for all $n \in \Z_{\ge 0}$:
- $\ds a_n = \frac 4 \pi \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \ \d x$
By Definite Integral from 0 to Half Pi of Logarithm of Sine x:
- $a_0 = \dfrac 4 \pi \paren {-\dfrac \pi 2 \ln 2} = -2 \ln 2$
By Definite Integral from 0 to Half Pi of Logarithm of Sine x by Cosine of 2nx:
- $a_n = \dfrac 4 \pi \paren {-\dfrac \pi {4 n} } = -\dfrac 1 n$
Therefore:
| \(\ds \map \ln {\sin x}\) | \(\sim\) | \(\ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos 2 n x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n\) |
$\blacksquare$
Proof 2
| \(\ds \sum_{n \mathop = 1}^\infty \dfrac {\cos 2 n x} n\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \dfrac {\map \exp {2 i n x} + \map \exp {-2 i n x} } {2 n}\) | Euler's Cosine Identity: $\cos z = \dfrac {\map \exp {i z} + \map \exp {-i z} } 2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 \sum_{n \mathop = 1}^\infty \paren {\frac {\paren {\map \exp {2 i x} }^n} n + \frac {\paren {\map \exp {-2 i x} }^n} n}\) | Power of Power | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac 1 2 \paren {\map \ln {1 - \map \exp {2 i x} } + \map \ln {1 - \map \exp {-2 i x} } }\) | Power Series Expansion for Logarithm of 1 + x: Corollary: $-\map \ln {1 - x} = \ds \sum_{n \mathop = 1}^\infty \dfrac {x^n} n$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac 1 2 \paren {\map \ln {1 - \map \exp {-2 i x} - \map \exp {2 i x} + 1} }\) | Sum of Logarithms | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac 1 2 \paren {\map \ln {2 - 2 \paren {\frac {\map \exp {-2 i x} + \map \exp {2 i x} } 2} } }\) | simplifying and multiplying top and bottom by $2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac 1 2 \paren {\map \ln {2 - 2 \map \cos {2 x} } }\) | Euler's Cosine Identity: $\cos z = \dfrac {\map \exp {i z} + \map \exp {-i z} } 2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac 1 2 \paren {\map \ln {4 \sin^2 x} }\) | Double Angle Formula for Cosine: Corollary $2$: $1 - \cos 2 \theta = 2 \sin^2 \theta$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {\map \ln {4 \sin^2 x }^{\frac 1 2} }\) | Logarithm of Power | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {\map \ln {2 \sin x} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {\ln 2 + \map \ln {\sin x} }\) | Sum of Logarithms | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sum_{n \mathop = 1}^\infty \dfrac {\cos 2 n x} n\) | \(=\) | \(\ds -\ln 2 - \map \ln {\sin x}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \ln {\sin x}\) | \(=\) | \(\ds -\ln 2 - \sum_{n \mathop = 1}^\infty \dfrac {\cos 2 n x} n\) | rearranging |
$\blacksquare$