Tuesday, November 24, 2020

Looking At Fourier Series

 Fourier series and their applications have great utility in many areas of  science, from celestial mechanics in astronomy to heat -radiation transfer and solutions of related differential equations in physics.  By way of reference, in previous posts we saw the use of the Taylor series expansion of a function f(x) about a point x = a:  

f(x= a) = f(a) + f ’(a) × (x – a) + f ” (a) (x – a)2 / 2!  + f ”’ (a) (x – a)3 / 2!               +   f n (a) (x– a)n / n!  +  ….


Here we have, in effect, an approximation to a given function f(x) by means of polynomials.  This approximation is based o getting as close a fit as possible to some particular point, say x = a.   The series usually converges in an interval about a, im fact having a as its center.  The problem is that the fit to the function f(x) in general is only expected realistic in a restricted neighborhood of a.   So the Taylor series does a decent job locally.  But over a fairly wide interval the Fourier series is a much better choice.

Whereas power series use powers of x as the fundamental elements Fourier series use sines and cosines (trigonometric functions) as their basic components, e.g.

f(x) =  o  / 2    +   å¥ n = 1    a n  cos nx   +   b n  sin nx

 

If  f is continuous on [-p,  p  of period 2p , i.e.  f (p) =f (-p)  then f can be represented as a uniformly convergent Fourier series.  

Further, we can compute the n and  n  Fourier coefficients  by integrating both sides of the Fourier series as follows:

ò p -p   f(x) dx  =  ò p -p  [ o  / 2    +   å¥ n = 1    a n  cos nx +   b n  sin nx ] dx  

To determine  n  for example multiply both sides of the above by cos mx and integrate, further keeping in mind the following  orthogonality conditions

1) ò p -p   cos mx cos nx dx = 0 (if m   n and  p  if m = n)  

2) ò p -p   sin mx sin nx dx = 0 (if m   n and  p  if m = n) 

3) ò p -p   sin mx cos nx dx = 0 (m, n assume any +ve integer) 

Then we obtain:  n  = 1/ p  ò p -p   f(x) cos mx dx   

Similarly:  b n  = 1/ p  ò p - f(x) sin mx dx

o  = 1/ p  ò p - f(x)  dx    

Example Problem:  If the series:

A   +   å¥ n = 1   { a n  cos npx/ L   +   b n  sin npx/ L}  

converges uniformly to (-L, L), show that for n = 1,2, 3....

n  = 1/  ò L-L   f(x) (cos npx/ L) dx     

And:   A   =    o  / 2

Solution:  Multiplying cos mpx/ L  by:

f(x)  =  å¥ n = 1   { a n  cos npx/ L   +   b n  sin npx/ L}  

And integrating  from -L to L we get:

ò L-L   f(x) (cos mpx/ L) dx   =  A ò L-L   (cos mpx/ L) dx   +

奠n = 1   [ a n  ò L-L   cos mpx/ L)( cos npx/ L) dx +  

n  ò L-L   (cos mpx/ L)( sin npx/ L)dx ]

=   a m  L    Or:   m  = 1/  ò L-L   f(x) (cos mpx/ L)  dx  

If m = 1, 2, 3...  

And:   n  = 1/  ò L-L   f(x) (cos npx/ L) dx       for n = 1, 2, 3...

To find A take:

ò L-L   f(x)  dx    =  2AL,    Or: A =   1/ 2L ò L-L   f(x)  dx  

Then let m = 0 in the first part of problem, so that:

o  = ò L-L   f(x)  dx   Or:   A =   o   /2

It should be noted that the set of Fourier coefficients is defined even if f(x) is not continuous. They depend only on the existence of the integral.  For example consider a function with a jump discontinuity:   Let  - p   <    h   <  p  .

f(x)  =   {0   for - p   <  x  <  h  

         =   {1  for    h <  x  <    p 

o  = 1/ p  ò p -p   f(x)  dx  1/ p ò p  h  (1) dx 

1/ (p  -  h)

n  = 1/ p ò p  h   cos nx  dx = 1/  [sin nx/n]p  h

 n  =  1/  sin nh /n

Lastly: 

n  = 1/ p ò p  h   sin nx  dx = 1/  [-cos nx/n]p  h

1/ p [- cos np/n   +   cos nh/n ]

=  1/ p [ cos n h/n   -   1/n]      (n even)

=  1/ p [ cos n h/n   +   1/n]      (n odd)


Problems for the Math whiz:

1)   From the example problem, show that:

 b m  = 1/  ò L-L   f(x) (sin mpx/ L) dx  

2) For the discontinuity example, consider what happens when: x = h = 0

Compute the new Fourier coefficients at the 'jump point'   and also give the general and specific Fourier series.

3(a) Find the Fourier coefficients corresponding to the function

f(x) =   { 0   for  -5 < x <  0

              {3   for    0  < x  <  5

Over a period P  = 10

b) Hence, write the corresponding Fourier series



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