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@@ -24,14 +24,15 @@ A set is called countable if it has the same cardinality as the natural numbers,
 
 Cardinality holds some surprises, though. For example, there are as many pairs of natural numbers as there are natural numbers; the set $\mathbb{N}\times\mathbb{N}$ is countable. They can be enumerated by following the zig-zag path through the cartesian grid of pairs below.
 
-![](/orderedpairs.0.png)
+<center>![](/orderedpairs.0.png)</center>
 
-1-->0.
-## Proof that no. of available functions is greater than number of functions required to define the periodic function
-Consider any arbitrary periodic function in the interval $[-\pi,\pi]$. This can be represented as a series of values at various points in the interval. For example,  
-$\qquad f(0) = ... , f(0.1) = ..., f(0.2) = ...$ and so on. At each point, we can assign any real number (i.e. $\in \mathbb R$). So, the number of possible periodic functions in an interval is of the order of $\mathbb R^{\mathbb R}$.
-  
-*--> don't quite remember how this goes.*  
+What is the cardinality of the set of periodic functions on $[-\pi,\pi]$? If we look at the set of all functions, where we demand no continuity or regularity or relationship between function values at nearby points, then there are rather many. The cardinality is $\mathbb R^{\mathbb R}$, which turns out to have the same size as $2^{\mathbb R}$, the set of subsets of the real numbers. On the homework, we will show that the set of lists of real numbers (ie, possible fourier series) has the same cardinality as $\mathbb R$, and that $2^{\mathbb R}$ has a larger cardinality. There just aren't enough fourier series that every arbitrary function could have its own. Luckily, we're not interested in most of those ugly beasts. We'd by happy knowing:
+
+What is the cardinality of the set of *continuous* periodic functions on $[-\pi,\pi]$? To describe a continuous function $f:[-\pi,\pi]\rightarrow \mathbb R$, make a list of its values at the dyadic points $f(n\pi/2^m)$ as follows:
+
+$$(f(0),f(-\pi),f(\pi),f(-\frac{\pi}{2}),f(\frac{\pi}{2}),f(-\frac{3\pi}{4}),f(-\frac{\pi}{4}),f(\frac{\pi}{4}),f(\frac{3\pi}{4}),\dots)$$
+
+When you start graphing $f$ from the list above, drawing dots at each of the recorded values, a pointillist picture begins to emerge. Since $f$ is continuous, dense pointillism is a faithful representation of the function (more precisely, to recover $f(x)$, take a limit of dyadic numbers which approach $f$, and consider the limit of their images). Thus a continuous function on the interval can be described by a list of real numbers, just like a fourier series can be. Hence there are as many fourier series as there are continuous functions, and there is a chance that each continuous function could have its own. Cardinality is no obstruction, then: it is not a priori impossible that every continuous function could have its own fourier series.
 
 #<b>Why Fourier decomposition 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:  
@@ -59,7 +60,6 @@ $$\begin{array}{ccl}
  & = & 2\sin(x) - 2\sin^3(x)\\
 \end{array}$$
   
-
 Based on the triple angle formula,  
   
 $$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$  
@@ -128,7 +128,7 @@ We now proceed to define certain operations on these functions in Hilbert space.
   
 $$
 \begin{array}{ccl}
-inner product, (f,g) & = & \int_0^{2\pi} f \, g \,dx\\
+(f,g) & = & \int_0^{2\pi} f \, g \,dx\\
 \mid f \mid ^2  = (f, f) & = & \int_0^{2\pi} f^2 \,dx\\
 \end{array}
 $$
@@ -136,20 +136,20 @@ $$
 This is the inner product of 2 real-number functions. For a function on complex numbers, the above definition must be altered as follows:    
 $$
 \begin{array}{ccl}
-inner product, (f,g) & = & \int_0^{2\pi} f \,\bar g \,dx\\
+(f,g) & = & \int_0^{2\pi} f \,\bar g \,dx\\
 \mid f \mid ^2  = (f, f) & = & \int_0^{2\pi} f^2 \,dx\\
 \end{array}
 $$  
   
-*Note: These are purely definitions, and we are  defining the inner product to ensure that the inner product of f and f is a real number.*
+*Note: These are purely definitions, and we are  defining the inner product to ensure that the inner product of $f$ and $f$, called the *norm* of $f$, 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*)  
+Any basis vectors could conceivably have been used, so long as they are orthonormal. (*Note: These particular basis vectors are chosen to prove that Fourier series exists*)  
   
-In order to prove orthonormality of the basis vectors:  
+To prove the orthonormality of the basis vectors, we compute: 
   
 $$
 \begin{array}{ccl}
@@ -209,14 +209,19 @@ $$ a_m = \int_0^{2\pi} \, f \, \frac{1}{\sqrt{2\pi}} \, e^ {-inx} \, dx $$
   
 This is the common definition for the terms of the Fourier series.
 
-##Proving that this function is does indeed completely represent $f$
-It is important to note at this point that we have simply expressed the periodic function $f$ in terms of a sum of arbitrary orthonormal vectors $f_n$. We haven't quite shown yet that the sum of orthonormal vectors actually completely represents $f$. Put in another way, there could be some components of $f$ that are not described by the vectors $f_n$. It is necessary to prove first that no other such components exist.
-  
-*--> don't quite remember this part*  
-  
-Now, we know that the entire function space can be described by the defined basis vectors. We can now prove that, the sum of terms that we computed in the previous section does indeed converge to $f$ as follows:
-  
-*--> don't quite remember this part*    
+##The Fourier series reproduces $f$
+
+It is important to note at this point that we have simply expressed the periodic function $f$ in terms of a sum of arbitrary orthonormal vectors $f_n$. We haven't quite shown yet that the sum of orthonormal vectors actually completely represent $f$. Put in another way, there could be some components of $f$ that are not described by the vectors $f_n$.
+
+**Theorem** Let $f:I\rightarrow \mathbb C$ be a continuous, periodic function with Fourier coefficients $a_n$, and suppose that the sum
+
+$$\sum_{n=-\infty}^\infty a_n e^{inx}$$ converges for all $x$. Then 
+
+$$f(x) = \sum_{n=-\infty}^\infty a_n e^{inx}.$$
+*Proof.* 
+Consider the difference $g:=f-\sum_{n=-\infty}^\infty a_n e^{inx}$.
+****inner product with delta functions
+
   
 #<b>Why the Fourier decomposition is useful? </b>
 Applications will be covered on Monday July 5, 2010. 
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