summaryrefslogtreecommitdiffstats
path: root/Fourier Series.page
blob: 4e5b610face72af952e10af8a41bb3e700161cb0 (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
#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 the time interval(period of the function). Then it can be expressed as : 
  
$$
\begin{array}{ccl}
f & = & \Sigma e^{inx}\\ 
& = & a_0 + \Sigma a_n\cos nx + \Sigma b_n\sin nx\\
\end{array}
$$  
  
#Why is Fourier series useful? </b>
Applications will be covered on Monday July 5, 2010. See you all soon!