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| -rw-r--r-- | Fourier Series.page | 19 |
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diff --git a/Fourier Series.page b/Fourier Series.page index 751b1c3..984f3e6 100644 --- a/Fourier Series.page +++ b/Fourier Series.page @@ -24,11 +24,22 @@ $\qquad\sin(2x) = 2\sin(x)\cos(x)$ Rearranging, $$\begin{array}{ccl} \sin(2x).\cos(x) & = & [2\sin(x)\cos(x)].\cos(x)\\ - & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\ - & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\ - & = & \cos y+i\sin y\end{array}$$ + & = & 2 \sin(x) [ 1 - \sin^2(x)]\\ + & = & 2\sin(x) - 2\sin^3(x)\\ +\end{array}$$ + + +Based on the double angle formula, + +$\qquad\sin(3x) = 3\sin(x) - 4\sin^3(x)$ + +Rearranging, + +$\qquad\sin^3(x) = \frac{3\sin(x)-\sin(3x)}{4}$ + +Substituting back in the former equation, we get +$\sin(2x).\cos(x) = 2 \sin(x) - 2 [\frac{3\sin(x)-\sin(3x)}{4}] $ -$\qquad\sin^2(x) = \frac{1-\cos(2x)}{2}$ ##What is the Fourier series actually?</b> ##Why is Fourier series useful? </b> |