written 7.8 years ago by |
Given $f(x) = \sqrt{1 - cosx} (0,2\pi)$
Fourier series is given by
$f(x) = \frac{a_{0}}{2} + \sum\limits_{n = 1}^{\infty}a_{n}cosnx + \sum\limits_{n = 1}^{\infty}b_{n}sinnx$
We know $1 - cos2x = 2sin^{2}x$
Similarly,
$1 - cosx = 2sin^2\frac{x}{2}$
$\therefore f(x) = \sqrt{2sin^{2}\frac{x}{2}} = \sqrt{2}sin\frac{x}{2}$
Let us first find,$a_{0}$
$\therefore a_{0} = \frac{1}{\pi}\int\limits_{0}^{2\pi}f(x)dx$
$\therefore a_{0} = \frac{1}{\pi}\int\limits_{0}^{2\pi}\sqrt{2}sin\frac{x}{2}dx$
$\therefore a_{0} = \frac{\sqrt{2}}{\pi}[-2cos\pi + 2]$
$\therefore a_{0} = \frac{4\sqrt{2}}{\pi}$
Now,
$a_{n} = \frac{1}{\pi}\int\limits_{0}^{2\pi}f(x) cosnxdx$
$a_{n} = \frac{1}{\pi} \int\limits_{0}^{2\pi}\sqrt{2}sin\frac{x}{2}cosnxdx$
$a_{n} = \frac{\sqrt{2}}{\pi}\int\limits_{0}^{2\pi}sin\frac{x}{2}cosnxdx$
$a_{n} = \frac{\sqrt{2}}{\pi}\frac{1}{2}\int\limits_{0}^{2\pi}sin(n + 1/2)x - sin(n - 1/2)xdx$
$[\because cosA sinB = \frac{1}{2}[sin(A + B) - sin(A - B)]$
$a_{n} = \frac{\sqrt{2}}{\pi}\frac{1}{2}\int\limits_{0}^{2\pi}sin\bigg(\frac{2n + 1)}{2}\bigg)x - sin\bigg(\frac{2n - 1}{2}\bigg)x dx$
$a_{n} = \frac{\sqrt{2}}{\pi}\bigg[\frac{-cos\bigg(\frac{2n + 1}{2}\bigg)x}{\frac{2n + 1}{2}} - \frac{\bigg(-cos\bigg(\frac{2n - 1}{2}\bigg)x\bigg)}{\frac{2n - 1}{2}}\bigg]_{0}^{2\pi}$
$a_{n} = \frac{\sqrt{2}}{\pi}\bigg[\frac{2}{2n + 1}\bigg(-cos\bigg(\frac{2n +1}{2}\bigg)x\bigg) - \frac{2}{2n - 1}\bigg(cos\bigg(\frac{2n - 1}{2}\bigg)x\bigg)\bigg]_{0}^{2\pi}$
$a_{n} = \frac{\sqrt{2}}{\pi}\bigg[\frac{1}{2n + 1} (-cos(2n\pi + \pi) + \frac{1}{2n - 1}(cos(2n\pi - \pi)) + \frac{1}{2n + 1} - \frac{1}{2n - 1}\bigg]$
$a_{n} = \frac{\sqrt{2}}{\pi}\bigg[\frac{1}{2n + 1}cos2n\pi + \frac{1}{2n - 1}(-cos2n\pi) + \frac{1}{2n + 1} - \frac{1}{2n - 1}\bigg]$
$a_{n} = \frac{\sqrt{2}}{\pi}\bigg[\frac{1}{2n + 1} - \frac{1}{2n - 1} + \frac{1}{2n + 1}-\frac{1}{2n - 1}\bigg]$
$a_{n} = \frac{\sqrt{2}}{\pi}\bigg[\frac{2}{2n + 1} - \frac{2}{2n - 1}\bigg]$
$a_{n} = \frac{\sqrt{2}}{\pi}*2\bigg[\frac{2n - 1 - (2n + 1)}{4n^2 - 1}\bigg]$
$a_{n} = \frac{2\sqrt{2}}{\pi}\frac{(-2)}{4n^2 - 1} = \frac{-4\sqrt{2}}{\pi(4n^2 - 1)}$
Hence,
$a_{n} = \frac{-4\sqrt{2}}{\pi(4n^2 - 1)}$
$b_{n} = \frac{1}{\pi}\int\limits_{0}^{2\pi}cos\bigg(\frac{1}{2} + n \bigg)x - cos\bigg(\frac{1}{2} - n\bigg)x*dx$
$b_{n} = \frac{\sqrt{2}}{\pi}\bigg[\frac{sin\bigg(\frac{1}{2} + n \bigg)x}{\bigg(\frac{1}{2} + n\bigg)} - \frac{sin\bigg(\frac{1}{2} - n\bigg)x}{\bigg(\frac{1}{2} - n\bigg)}\bigg]_{0}^{2\pi}$
$b_{n} = \frac{\sqrt{2}}{\pi}(0) = 0$
Hence putting the value of the coefficients in this series we get
$F(x) = \frac{1}{2}\frac{4\sqrt{2}}{\pi} + \sum\limits_{n = 1}^{\infty} \frac{-4\sqrt{2}}{\pi(4n^2 - 1)}cosnx$
$\sqrt{1 - cosx} = \frac{2\sqrt{2}}{\pi} - \frac{4\sqrt{2}}{\pi}\sum\limits_{n = 1}^{\infty}\frac{1}{(4n^2 - 1)}cosnx$.............(1)
Now to deduce
$\frac{1}{2} = \sum\limits_{n = 1}^{\infty}\frac{1}{(4n^2 - 1)}$
Put x = 0 in equation (1)
$0 = \frac{2\sqrt{2}}{\pi} - \frac{4\sqrt{2}}{\pi}\sum\limits_{n = 1}^{\infty}\frac{1}{(4n^2 - 1)}$
i.e.
$\frac{2\sqrt{2}}{\pi} = \frac{4\sqrt{2}}{\pi}\sum\limits_{n = 1}^{\infty}\frac{1}{(4n^2 - 1)}$
$\frac{1}{2} = \sum\limits_{n = 1}^{\infty}\frac{1}{(4n^2 - 1)}$
Hence proved