From 8e66273b1701fe12a436f6e08c41273e0676af89 Mon Sep 17 00:00:00 2001 From: siveshs Date: Fri, 2 Jul 2010 03:44:58 +0000 Subject: still testing --- Fourier Series.page | 17 ++++++++++++++++- 1 file changed, 16 insertions(+), 1 deletion(-) (limited to 'Fourier Series.page') diff --git a/Fourier Series.page b/Fourier Series.page index 4cf9014..1298bc9 100644 --- a/Fourier Series.page +++ b/Fourier Series.page @@ -8,12 +8,27 @@ To show that Fourier series is plausible, let us consider some arbitrary trignom $1.\quad\sin^2(x) = ?$ Based on the double angle formula, -$\cos(2x) = 1 - 2 \sin^2(x)$ + +$\qquad\cos(2x) = 1 - 2 \sin^2(x)$ Rearranging, $\qquad\sin^2(x) = \frac{1-\cos(2x)}{2}$ +$2.\quad\sin(2x).\cos(2x) = ?$ + +Based on the double angle formula, + +$\qquad\sin(2x) = 2\sin(x)\cos(x)$ + +Rearranging, +$$\begin{array}{ccl} +\sin(2x).\cos(x) & = & (\2\sin(x)\cos(x))\cos(x)\\ + & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\ + & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\ + & = & \cos y+i\sin y\end{array}$$ + +$\qquad\sin^2(x) = \frac{1-\cos(2x)}{2}$ ##What is the Fourier series actually? ##Why is Fourier series useful? -- cgit v1.2.3