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authorsiveshs <siveshs@gmail.com>2010-07-02 03:44:58 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-02 03:44:58 +0000
commit8e66273b1701fe12a436f6e08c41273e0676af89 (patch)
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downloadafterklein-wiki-8e66273b1701fe12a436f6e08c41273e0676af89.tar.gz
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still testing
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@@ -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?</b>
##Why is Fourier series useful? </b>