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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 triple 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 $\sin \theta$
$$
\begin{array}{ccl}
\cos \theta & = & \frac{1}{2}e^{i\theta} + \frac{1}{2}e^{-i\theta}\\
\sin \theta & = & \frac{1}{2i}e^{i\theta} - \frac{1}{2i}e^{-i\theta}\\
\end{array}
$$
It is easy to show that any product of cosines and sines can be expressed as the product of exponentials which will reduce to a sum of sines and cosines.
As a final test to see if the Fourier series really could exist for any periodic function, we consider a periodic function with a sharp peak as shown below
<center>![*Peak Function Image*](/peak_func.gif) </center>
If it is possible to approximate the above function using a sum of sines and cosines, then it can be argued that *any* continuous periodic function can be expressed in a similar way. This is because any function could be expressed as a number of peaks at every position.
It turns out that the above function can be approximated as the difference of two cosines, namely, $\cos^{2n}(x) + cos^{2n+1}(x)$
<center>![$\cos^{2n}(x)$](/cos10x.gif) ![$cos^{2n+1}(x)$](/cos11x.gif) </center>
Summing these two functions we get the following:
<center>![$\cos^{2n}(x) + cos^{2n+1}(x)$](/cos10x-cos11x.gif)</center>
#What is the Fourier series actually?</b>
Now, to begin proving that the Fourier series is a true fact let us begin with the following hypothesis:
Let $f : \mathbb I \rightarrow \mathbb C$ be a continuous, periodic function where $I$ is some time interval(period of the function). Then it can be expressed as :
$$
\begin{array}{ccl}
f & = & \Sigma e^{inx}\\
& = & a_0 + \Sigma a_n\cos nx + \Sigma b_n\sin nx\\
\end{array}
$$
We begin proving this hypothesis by considering that any function on the right-hand side of our hypothesis is uniquely defined by the co-efficients of the terms a_0 through a_n and b_1 through b_n. This can be taken to mean that every function is really a vector in an n-dimensional Hilbert space.
We now proceed to define certain operations on these functions in Hilbert space. One operation that will be particularly useful is that of the inner product of two functions in Hilbert space:
---> define inner product here
This is the definition for a function of real numbers. For a function on complex numbers, the above definition must be altered as follows:
*Note: These are purely definitions, and we are now definining the inner product to ensure that inner product of f and f is a real number.*
#Why is Fourier series useful? </b>
Applications will be covered on Monday July 5, 2010. See you all soon!
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