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 We first begin with a few basic identities on the size of sets. Then, we will show that the set of possible functions representing sets is not larger than the set of available functions. This at best indicates that the Fourier series is not altogether impossible.  
   
 ## To show that $(0,1) \sim \mathbb R$
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 ## Cantor's proof for $\mathbb R > \mathbb N$
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 ## 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 $\mathbb R^{\mathbb R}$.
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 #<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:  
@@ -195,11 +196,11 @@ 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.
   
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 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:
   
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 #<b>Why the Fourier decomposition is useful? </b>
 Applications will be covered on Monday July 5, 2010. See you all soon!  
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