1 Crore+ students have signed up on EduRev. Have you? Download the App 
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 f_{1} + f_{2} are the sums of the corresponding Fourier coefficients of f_{1} and f_{2}.
(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 Fourierseries 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) = f_{1} + f_{2} where f1 = x and f_{2} = π.
The Fourier coefficient of f_{2} = π is a_{0} = π, a_{n}= 0 and b_{n} = 0 . The Fourier coefficient of f_{1} = x is
Thus the Fourier series of f (x) is
Halfrange 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(a_{0} and a_{n} ) . This is the even periodic extension f_{1}(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 f_{2} (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 halfrange expansion of the function f (x) as shown in figure below.
Even periodic extension
Let us calculate the integral
and
Hence the first halfrange 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 halfrange 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.
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 nxn 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 Diracdelta function of strength +1 . Find the amplitudes a_{n} in the Fourier expansion of
and Range: [1,1] hence 2L = 2.
Comparing with
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 x_{0}.
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.
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 a_{0} = 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.
18 docs24 tests
