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) = a 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 a n and b n Fourier coefficients by integrating both sides of the Fourier series as follows:
ò p -p f(x) dx = ò p -p [ a o / 2 + å¥ n = 1 a n cos nx + b n sin nx ] dx
To determine a 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: a n = 1/ p ò p -p f(x) cos mx dx
Similarly: b n = 1/ p ò p -p f(x) sin mx dx
a o = 1/ p ò p -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....
a n = 1/L ò L-L f(x) (cos npx/ L) dx
And: A = 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 +
å¥ n = 1 [ a n ò L-L cos mpx/ L)( cos npx/ L) dx +
b n ò L-L (cos mpx/ L)( sin npx/ L)dx ]
= a m L Or: a m = 1/L ò L-L f(x) (cos mpx/ L) dx
If m = 1, 2, 3...
And: a n = 1/L 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:
a o = ò L-L f(x) dx Or: A = 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
a o = 1/ p ò p -p f(x) dx = 1/ p ò p h (1) dx
= 1/ p (p - h)
a n = 1/ p ò p h cos nx dx = 1/ p [sin nx/n]p h
a n = - 1/ p sin nh /n
Lastly:
b n = 1/ p ò p h sin nx dx = 1/ p [-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/L 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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