summaryrefslogtreecommitdiffstats
path: root/Fourier Series.page
blob: 91bad7ad0bbd21a6b2822de575e9a78eb5efc16c (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
##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,  
  
$\qquad\cos(2x) = 1 - 2 \sin^2(x)$  
  
Rearranging,  
  
$\qquad\sin^2(x) = \frac{1-\cos(2x)}{2}$  
  
$2.\quad\sin(2x).\cos(2x) =  ?$  
   
Based on the double angle formula,  
  
$\qquad\sin(2x) = 2\sin(x)\cos(x)$  
  
Rearranging,
$$\begin{array}{ccl} 
\sin(2x).\cos(x) & = & [2\sin(x)\cos(x)].\cos(x)\\
 & = & 2 \sin(x) [ 1 - \sin^2(x)]\\
 & = & 2\sin(x) - 2\sin^3(x)\\
\end{array}$$
  

Based on the double angle formula,  
  
$\qquad\sin(3x) = 3\sin(x) - 4\sin^3(x)$  
  
Rearranging,  
  
$\qquad\sin^3(x) = \frac{3\sin(x)-\sin(3x)}{4}$  

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:  
  
$$
\begin{array}{ccl}
e^{i\theta} & = & \cos \theta + i \sin \theta\\
\end{array}
$$
##What is the Fourier series actually?</b>

##Why is Fourier series useful? </b>

$(\nearrow)\cdot(\uparrow)=(\nwarrow)$