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##Why Fourier series possible?</b>
We first begin with a few basic identities on the size of sets. Show that the set of possible functions representing sets is not larger than the set of available functions?
##Why Fourier series is plausible?</b>
To show that Fourier series is plausible, let us consider some arbitrary trignometric functions and see if it is possible to express them as the sum of sines and cosines:
$1.\quad\sin^2(x) = ?$
Based on the double angle formula,
$$\cos(2x) = 1 - 2 \sin^2(x)$$
Rearranging,
$$\sin^2(x) = \frac{1-\cos(2x)}{2}$$
$2.\quad\sin(2x)\cdot\cos(2x) = ?$
Based on the double angle formula,
$$\qquad\sin(2x) = 2\sin(x)\cos(x)$$
Rearranging,
$$\begin{array}{ccl}
\sin(2x)\cdot\cos(x) & = & [2\sin(x)\cos(x)]\cdot\cos(x)\\
& = & 2 \sin(x) [ 1 - \sin^2(x)]\\
& = & 2\sin(x) - 2\sin^3(x)\\
\end{array}$$
Based on the double angle formula,
$$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$
Rearranging,
$$\sin^3(x) = \frac{3\sin(x)-\sin(3x)}{4}$$
Substituting back in the former equation, we get
$$
\begin{array} {ccl}
\sin(2x) & = & 2\sin(x) - 2 [\frac{3\sin(x)-\sin(3x)}{4}]\\
& = & \frac{1}{2}\sin(x) + \frac{1}{2}\sin(3x)\\
\end{array}
$$
Thus, we see that both these functions could be expressed as sums of sines and cosines. It is possible to show that every product of trignometric functions can be expressed as a sum of sines and cosines:
$$
\begin{array}{ccl}
e^{i\theta} & = & \cos \theta + i \sin \theta\\
e^{-i\theta} & = & \cos \theta - i \sin \theta\\
\end{array}{ccl}
$$
Solving for \cos \theta and \sin \theta\\
$$
\begin{array}{ccl}
\cos \theta & = & \frac{1}{2}e^{i\theta} + \frac{1}{2}e^{-i\theta}\\
\sin \theta & = & \frac{1}{2i}e^{i\theta} - \frac{1}{2i}e^{-i\theta}\\
\end{array}
$$
##What is the Fourier series actually?</b>
##Why is Fourier series useful? </b>
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