Even, Odd functions & Half-Range Expansion | Mathematical Models - Physics

Even & Odd function

A function g (x) is said to be even if g (-x) = g (x), so that its graph is symmetrical with respect to vertical axis.

A function h (x) is said to be even ifh (-x) = -h (x).

Since the definite integral of a function gives the area under the curve of the function between the limits of integration, we have

Fourier Cosine Series and Fourier Sine Series
Fourier series of an even function of period 2L, is a “Fourier cosine series”

with coefficients( note integration from 0 to L)

Fourier series of an odd function of period 2L, is a “Fourier sine series”

 with coefficients
NOTE:
 (i) For even function f (x);

and
(ii) For odd function f (x);


The Case of Period 2π

If L = π, and f(x) is even function then

with coefficients

If  f(x) is odd function then

with coefficients

Sum and Scalar Multiple

(a) The Fourier coefficients of a sum f₁ + f₂ are the sums of the corresponding Fourier coefficients of f₁ and f₂.

(b) The Fourier coefficients of cf are c times the corresponding Fourier coefficients of f.

Example 4: Find the Fourier series of the periodic function f (x) as shown in figure:

We have already calculated the Fourier series of the periodic function f (x) as shown in figure:

The Fourier series is
The function given in the problem can be obtained by adding k to the above function. Thus the Fourier series of a sum k+ /'(x) are the sums of the corresponding Fourier

series of k and /(x).
The Fourier series is f  

Example 5: Find the Fourier series of the periodic function
f (x) = x + π ; ( - π < x < π) having period 2π

Let f(x) = f₁ + f₂   where f1 = x and f₂ = π.

The Fourier coefficient of f₂ = π is a₀ = π, a_n= 0 and b_n = 0 . The Fourier coefficient of f₁ = x is

Thus the Fourier series of f (x) is

Half-Range Expansion

Half-range expansions are Fourier series. The idea is simple and useful. We could extend f (x) as a function of period L and develop the extended function into a Fourier series. But this series would in general contain both cosine and sine terms.

We can do better and get simpler series. For our given function f (x) we can calculate

Fourier cosine series coefficient(a₀ and a_n ) . This is the even periodic extension f₁(x) of f (x) in figure (b).

For our given function f (x) we can calculate Fourier sine series coefficient (b_n ). This

is the odd periodic extension f₂ (x) off (x) in figure (c).

Both extensions have period 2L . Note that f(x) is given only on half the range, half the

interval of periodicity of length 2L .

Example 6: Find the two half-range expansion of the function f (x) as shown in figure below.

Even periodic extension





Let us calculate the integral
 

and



Hence the first half-range expansion of f (x) is

This Fourier cosine series represents the even periodic extension of the given function/ (x), of period 2 L as shown in figure.

Odd periodic extension




Let us calculate the integral
 




Hence the other half-range expansion of f (x) is

This Fourier sine series represents the odd periodic extension of the given function f (x).
of period 2L as shown in figure.

Complex Fourier series

The Fourier series

can be written in complex form, which sometimes simplifies calculations.

∵ e^inx = cos nx + i sin nx and e^-inx = cos nx -i sin nx

Thus

Lets take

where coefficients c_0, c_n and k_n are given by


We can also write (2) as (take k_n = c_-n)

This is so called complex form of the Fourier series or, complex Fourier series off (x) .
The c_n are called complex Fourier series coefficients of f (x).

For a function of period 2L

Example 7: Find the complex Fourier series of the periodic function
f (x) = e^x; ( - π < x < π), having period 2π

Let Fourier series is f(x) =




Thus Fourier series is
Let us derive the real Fourier series
∵ (1 + in) e^inx = (1 + in) (cos nx + i sin nx) = (cos nx - nsin nx) + i (cos nx + sin nx)
∵ n varies from -∞ to + ∞, equation (1) has corresponding term with -n instead of n .
Thus

∵ (1 -in) e^-inx = (1 -in) (cos nx - i sin nx) = (cos nx-n sin nx) - i (cos nx+sin nx)
Let’s add these two expressions;
(1 + in) e^inx + (1 - in)e^-inx = 2 (cos nx - n sin nx), n = 1,2,3........
For n = 0,    
Thus

Example 8: Consider the periodic function f(t) with time period T as shown in the figure below.

The spikes, located at  where n = 0,±1,±2,..., are Dirac-delta function of strength +1 . Find the amplitudes a_n in the Fourier expansion of

and Range: [-1,1] hence 2L = 2.
Comparing with

Approximation by Trignometric Polynomials

Fourier series have major applications in approximation theory, that is, the approximation of functions by simpler functions.

Let f (x)be a periodic function, of period 2π for simplicity that can be represented by a Fourier series. Then the N^th partial sum of the series is an approximation to f (x) :

We have to see whether (1) is the "best” approximation to f by a trignometric polynomial of degree N, that is , by a function of the form

where "best” means that the "error” of approximation is minimum.
The total square error of F relative to / on the interval -π < x < π is given by

The function F is a good approximation to f but |f - F| is large at a point of discontinuity x₀.

Minimum square error

The total square error of F relative to f on the interval - π < x < n is minimum if and only if the coefficients of f(x) are the Fourier coefficients of f(x). This minimum value E* is given by

From (3) we can see that E* cannot increase as N increases, but may decrease. Hence with increasing N the partial sums of the Fourier series of f yields better and better approximations to f.

Parseval’s Identity

Since E* > 0 and equation (3) holds for every N, we obtain

Now Parseval’s Identity is

Example 9: Compute the total square error of F with N = 3 relative to
f (x) = x + π    (- π < x < π)
on the interval - π < x < π .

Fourier coefficients are a₀ = π, a_n = 0 and b_n =
Its Fourier series is given by

Hence,


Although |f (x) - f(x)| is large at x = ± π . where f is discontinuous, F approximates f quite well on the whole interval.