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#<b>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?
#<b>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> </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>  </center>
Summing these two functions we get the following:
<center></center>
#<b>What is the Fourier Series actually?</b>
##Initial Hypothesis
Now, to prove that the Fourier series is indeed true, we 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 :
$$
\begin{array}{ccl}
f & = & \Sigma e^{inx}\\
& = & a_0 + \Sigma a_n\cos nx + \Sigma b_n\sin nx\\
\end{array}
$$
##Definition of Inner Product of Functions
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 a $2n+1$-dimensional 'Hilbert space'.(*perhaps someone can clarify this?*)
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:
$$
\begin{array}{ccl}
inner product, (f,g) & = & \int_0^{2\pi} f g dx\\
\mid f \mid ^2 = (f, f) & = & \int_0^{2\pi} f^2 dx\\
\end{array}
$$
This is the inner product of 2 real-number functions. For a function on complex numbers, the above definition must be altered as follows:
--> altered function here
*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.*
##Basis Vectors of the Hilbert Space
The basis vectors of this Hilbert space are taken as follows:
basis vectors, $f_n = \frac{1}{\sqrt{2\pi}}e^{inx}$
Any basis vectors could conceivable have been assumed on the condition that the basis vectors are orthonormal. (*Note: These particular basis vectors are chosen to prove that Fourier series exists*)
In order to prove orthonormality of the basis vectors:
--> orthonormal proof goes here
##Determining Coefficients of the Basis vectors
In any vector space, the inner product of a vector and its basis vector gives the coefficient. For example, consider a 2-dimensional vector as shown below:
--> image of a 2d vector
--> demo of coefficient being true
Extending this principle to the case of an n-dimensional vector:
--> compute inner product here and then continue to show what the coefficient formula is
##Proving that this function is does indeed completely represent $f$
--> don't quite remember this part
#<b>Why is Fourier series useful? </b>
Applications will be covered on Monday July 5, 2010. See you all soon!
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