


Definition of Fourier Series and Typical Examples




Baron Jean Baptiste Joseph Fourier (1768 − 1830) introduced the idea that any periodic function can be represented
by a series of sines and cosines which are harmonically related.
To consider this idea in more detail, we should introduce some definitions and common terms.

Basic Definitions
A function f (x) is said to have period P if
f (x + P) = f (x) for all x.
Let the function f (x) has period 2 π.
In this case, it is enough to consider behavior of the function on the interval [−π, π].
 Suppose that the function f (x) with period 2π is absolutely integrable on
[−π, π] so that the following socalled Dirichlet integral
is finite:
 Suppose also that the function f (x) is a single valued, piecewise continuous
(must have a finite number of jump discontinuities), and piecewise monotonic (must have a finite number of maxima and minima).
If the conditions 1 and 2 are satisfied, the Fourier series for the function f (x)
exists and converges to the given function (see also
Convergence of Fourier Series
about convergence conditions.)
At a discontinuity x_{0}, the Fouries Series converges to
The Fourier series of the function f (x) is given by
where the Fourier coefficients a_{0}, a_{n}, and b_{n}
are defined by the integrals
Sometimes alternative forms of the Fourier series are used. Replacing a_{n} and b_{n} by the new variables
d_{n} and φ_{n} or d_{n} and θ_{n}, where
we can write:
Fourier Series of Even and Odd Functions
The Fourier series expansion of an even function f (x)
with the period of 2 π does not involve the terms with sines and has the form:
where the Fourier coefficients are given by the formulas
Accordingly, the Fourier series expansion of an odd 2 πperiodic function f (x)
consists of sine terms only and has the form:
where the coefficients b_{n} are
Below we consider expansions of 2 πperiodic functions into their Fourier series, supposing that these expansions exist and are convergent.

Example 1


Let the function f (x) be 2 πperiodic and suppose that it is presented by the Fourier series:
Calculate the coefficients a_{0}, a_{n}, and b_{n}.
Solution.
To define a_{n}, we integrate the Fourier series on the interval [−π, π]:
For all n > 0,
Therefore, all the terms on the right of the summation sign are zero, so we obtain
In order to find the coefficients a_{n} at m > 0, we multiply both sides of
the Fourier series by cos mx and integrate term by term:
The first term on the right side is zero. Then, using the wellknown trigonometric identities, we have
if m ≠ n.
In case when m = n, we can write:
Thus,
Similarly, multiplying the Fourier series by sin mx and integrating term by term, we obtain the expression
for b_{m}:
Rewriting the formulas for a_{n}, b_{n}, we can write the final expressions for the Fourier coefficients:

Example 2


Find the Fourier series for the square 2 πperiodic wave defined on the interval [−π, π]:
Solution.
First we calculate the constant a_{0}:
Find now the Fourier coefficients for n ≠ 0:
Since , we can write:
Thus, the Fourier series for the square wave is
We can easily find the first few terms of the series. By setting, for example, n = 5, we get
The graph of the function and the Fourier series expansion for n = 10 is shown in Figure 1.

 
Fig.1,
n = 10

 Fig.2,
n = 5,
n = 10


Example 3


Find the Fourier series for the sawtooth wave defined on the interval [−π, π] and
having period 2 π.
Solution.
Calculate the Fourier coefficients for the sawtooth wave.
Since this function is odd (Figure 2), then a_{0} = a_{n} = 0.
Find the coefficients b_{n}:
To calculate the latter integral we use integration by parts formula:
Let . Then so the integral becomes
Substituting and for all integer values of n,
we obtain
Thus, the Fourier series expansion of the sawtooth wave is (Figure 2 above)

Example 4


Let f (x) be a 2 πperiodic function such that
for . Find the Fourier series for the parabolic wave.
Solution.
Since this function is even, the coefficients b_{n} = 0. Then
Apply integration by parts twice to find:
Since and for integer n,
we have
Then the Fourier series expansion for the parabolic wave is (Figure 3)

 
Fig.3,
n = 2,
n = 5

 Fig.4,
n = 1,
n = 2


Example 5


Find the Fourier series for the triangle wave
defined on the interval [−π, π].
Solution.
The constant a_{0} is
Determine the coefficients a_{n}:
Integrating by parts, we can write
Then
The values of sin nx at x = 0 or
x = ± π are zero. Therefore,
When n = 2k, then .
When n = 2k + 1, then
Since the function f (x) is even, the Fourier coefficients b_{n}
are zero. Therefore, the complete Fourier expansion for the triangle wave (see Figure 4 above) is

Example 6


Find the Fourier series for the function
defined on the interval [−π, π].
Solution.
First we find the constant a_{0}:
Now we calculate the coefficients a_{n}:
Notice that
Since cos (n − 1)π = (−1)^{n −1},
we get the following expression for the coefficients a_{n}:
It's seen that a_{n} = 0 for odd n. For even n, when
n = 2k (k = 1,2,3,...), we have
Calculate the coefficients b_{n}. Start with b_{1}:
The other coefficients b_{n} for n > 1 are zero. Indeed,
Thus, the Fourier series of the given function is given by
Graphs of the function and its Fourier expansions for n = 2 and n = 8 are shown
in Figure 5.

 
Fig.5,
n = 2,
n = 8

 Fig.6,
n = 10


Example 7


Find the Fourier series for the function
defined on the interval [−π, π].
Solution.
Compute the coefficients a_{n}:
(These results are obvious since this function is odd.)
Calculate the coefficients b_{n}:
Thus, the Fourier series expansion of the function is given by
The graph of the function and the Fourier series expansion for n = 10 are shown
in Figure 6 above.



