INTRODUCTION
The study of a special type of series whose terms are trigonometric functions of a variable was started in the 18th century when J.B.J. Fourier (1768-1830) was successful in his attempt to prove that an arbitrary function given in the interval can be expressed under certain conditions in a trigonometric series of the form
and subsequently, such type of special series is given the name ‘Fourier Series’. It has great importance in the study of many branches of science including social sciences. The knowledge of Fourier series provides a powerful technique for solving problems of a periodic nature which occur in situations like theory of conduction of heat, electrical and mechanical vibrations, propagation of electromagnetic waves, acoustics, optics, etc. Fourier was the first person to initiate an elaborate study of trigonometric series and afterwards Dirichlet, Parseval, Bessel and many other mathematician made major contributions to the study of conditions of a function to be represented in a trigonometric series, convergence, integration, and differentiation of the Fourier series. In this article we are going to study Fourier series and their properties.
Before starting the article of Fourier series, it is essential to study the behaviour of periodic functions, piecewise monotone and piecewise continuous functions and some properties related to them.
PERIODIC, PIECEWISE MONOTONE AND PIECEWISE CONTINUOUS FUNCTIONS
Definition: Let , a function is said to be periodic if there exists a positive constant such that for all ; the least number possessing this property (if such a number exists) is called the period of the function. is called periodic of period or fundamental period . is periodic of period implies also for all .
The trigonometric functions , are periodic of period and , are periodic of period . is periodic of period Every constant function is trivially periodic with no definite Fundamental period.
The following observations are quite clear from the definitions of periodic functions.
- If is defined on an interval of length then can be extended to the whole of by making periodic of period .
- The graph of a periodic function over will be replicas of a portion over an interval of length if is period of the function.
Hence if a function is defined on it can be made to be a periodic function on of period . In fact, sum of two periodic function of same period is also a periodic function of that period.
If is periodic of period then the following results are quite easy to prove:
Definition: A function is said to be piecewise monotone if there exists a partition of into open sub-intervals in each of which is monotone.
Every monotone function is piecewise monotone but the converse is not true.
For example
is piecewise monotone on [0, 1], but is not monotone on [0, 1].
Definition: A function is said to be piecewise continuous if there exists a partition of into open sub-intervals in each of which is continuous.
Every continuous function is piecewise continuous but the converse is not true.
For example, the function defined on by
is piecewise continuous, since it is continuous on , , , but is not continuous on ; and are points of discontinuities of on .
Theorem 2.1.
Definition (Fourier Series): Let be an integrable function on or if is unbounded on , let the improper integral be absolutely convergent.
Then the trigonometric series
is called the Fourier series in corresponding to the function , where , called Fourier Coefficients, are given by
Definition: A sequence of Riemann integrable functions defined on is called orthogonal if
and it is said to be orthonormal if
From Theorem 2.1, it easily follows that are all orthonormal on . The function is orthonormal on
FOURIER SERIES EXPANSION
Not every function can be expanded into Fourier series. There are certain conditions which are to be satisfied by a function for its expansion into Fourier series. The most familiar one is Dirichlet’s conditions.
Dirichlet’s Conditions
Let be a given function. is said to satisfy the Dirichlet’s conditions in if it satisfies one of the following two conditions:
- is bounded on and the interval can be broken into a finite number of open sunintervals of , in each of which is monotone i.e., in other words is bounded and piecewise monotone on ;
- has a finite number of points of infinite discontinuities on but when arbitrarily small neighbourhoods of these points of discontinuities are excluded from , is bounded and piecewise monotone in the remainder of the interval and also the improper integral is absolutely convergent.
Remark: Dirichlet’s conditions can also be put in a slightly different form:
is said to satisfy Dirichlet’s condition on if is a function of bounded variation on or when has a finite number of points of finite discontinuities on and when arbitrarily small neighbourhoods of these points are excluded from , is a function of bounded variation on the reminder of and is absolutely convergent on .
Theorem 3.1.
If the series converges uniformly to on then it is the Fourier series for in .
Proof. Since the series converges uniformly to on , is the sum function of the series and we can write
(1)
By integrating term-by-term on , we get using Theorem 2.1,
Multiplying (1) by on both sides and integrating term-by-term (which is valid, since is a bounded function and after multiplication the series is still uniformly convergent on , we have
Multiplying (1) by on both sides and integrating term-by-term, we have
Hence, from definition the series (1) represents the Fourier series for on .
Note: Fourier series exists for every Riemann integrable function. But that series may or may not converge to .
Theorem 3.2:
If be bounded, integrable and monotone on then
Proof. (i) Since is monotone on , applying second mean value theorem (Weierstrass form) of integral calculus, there exists a such that
Since is convergent, by Cauchy’s condition of convergence
Since is
Hence
for .
Proof: – (ii) For the integra .
We write .
Since is bounded and monotone on , is also bounded and monotone on for and .
By the second mean value theorem,
Since is convergent, there exists such that for all .
Therefore .
Since , for any positive there exists a positive such that
whenever .
Hence whenever
by
,
Hence of , there exists such that whenever
This implies
\ben*}
\left\vert\int_{0}^{a}\varphi(x)\dfrac{\sin nx}{x}dx\right\vert & \leq\left\vert\int_{0}^{h}\varphi(x)\dfrac{\sin nx}{x}dx\right\vert+\left\vert\int_{h}^{a}\varphi(x)\dfrac{\sin nx}{x}dx\right\vert\\
& <\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon\text{ when and .}
\end{align*}
Thus
Corollary 1. If is bounded integrable and monotone on where
,
then
Proof. where
is bounded and integrable on . is monotone increasing on and hence on where .
Then by Theorem 3.2
Corollary 2. If be bounded and integrable and monotone on for then
Proof. From (ii) of Theorem 3.2,
Since we have
Remark: the integrals in Theorem 3.2 are known as Dirichlet’s Integrals.
Theorem 3.3
If be bounded, integrable and monotone on and (not necessarily in the same sense), where , then
where stand for Fourier coefficients of .
Proof. From definition of we have
Theorem 3.4 (Fourier-Dirichlet):
If is bounded, integrable, periodic pf period and it is piecewise monotone on then the Fourier series for in is equal to
at in , when is continuous at ,
at in , when and exist,
at , when and exist.
Proof. The Fourier series for (since it satisfies Dirichlet’s condition in ) in is
where
and
Let,
Then
In the first integral, we will put and in the second integral, we are going to put .
Then
Since satisfies Dirichlet’s conditions in, if we consider as functions of then they satisfy Dirichlet’s conditions in and in respectively. If exist then and have the limits when tends to zero.
Hence, if , we have using Dirichlet’s integrals,
Hence (ii) proved.
If is continuous at where then .
Hence,
which proves (i).
At ,
If and exist, using Dirichlet’s integrals,
We have,
Similarly, it can be shown that
Remark: The Theorem 3.4 establishes the point-wise convergence of the Fourier series for to the function and not the uniform convergence. The Fourier series expression of a function is unique.
Theorem 3.5. (Parseval’s Identity):
If the Fourier series for in converges uniformly to on then
Where are the Fourier coefficients for in .
Proof. The Fourier series of in is
Where
Since the series (1) converges uniformly to on , then we can write (1) as
Multiplying both sides of (2) by and then integrating term-by-term (which is valid, since is bounded on or if unbounded, is absolutely convergent) on we have
Theorem 3.6. (Bessel’s Inequality)
If is bounded and integrable on or if is unbounded on but is convergent, then
Where are the Fourier coefficients for in .
Proof. The Fourier series expression for on is
Where are given by
for all .
Let
Since , we have
From (1)
Again from (1),
Hence using (2), (3) and (4) we have the inequality
Note: Taking in Bessel’s inequality we obtain
Since and hence is integrable on , the series
SOME EXAMPLES ON FOURIER SERIES REPRESENTATION OF FUNCTIONS IN
Example 4.1. We are going to solve the Fourier series representation of in where
and hence we will deduce that .
Solution: is continuous on and hence it is bounded and integrable on . Since for , is monotone increasing on . Thus satisfies Dirichlet’s conditions on so that can be represented by a Fourier series
Hence the Fourier series expansion for in is
Since is continuous on , so we can write
Example 4.2. We are going to find the Fourier series expansion of the function
Hence we will deduce that
Solution: Since is an even function, its derivative is odd and hence is symmetric about origin. in and . Hence is piecewise monotone in . Hence satisfies Dirichlet’s conditions in . Therefore can be represented by the Fourier series
For since is an odd function.
Hence, Fourier series corresponding to in is
Since is continuous on ,
Example 4.3. Obtain the Fourier series expansion of in , where
Then we are going to show that the sum of the series
Solution: is piecewise monotone on . is continuous on and hence is integrable on . Thus satisfies Dirichlet’s conditions in .
So, can be represented by the Fourier series
where
For
Thus, the Fourier series expansion of in is
i.e.,
Since is continuous on ,
Therefore,
Observation. If a function is not defined at then can be defined arbitrarily at those points. However, it is preferable to define at those points in such a way that makes continuous on .
Example 4.4. We will find Fourier series of the periodic function with period defined as
Then we will calculate the sum of the series at .
And we will prove that .
Solution: For the sake of continuity, we define
is piecewise monotone in . It is bounded and periodic with period . is continuous on except at . Hence is integrable on . Thus satisfies Dirichlet’s conditions on . Hence can be represented by Fourier series, which is
Where
Thus, Fourier series for in is
At , the Fourier series for converges to
Hence sum of the series at is .
Since is continuous at , we have
Example 4.5. If on , we will prove that the Fourier series of is given by
Then we will deduce that
and
Solution: is piecewise monotone on . is continuous on and hence is integrable on . Therefore satisfies Dirichlet’s conditions on . Hence can be represented by Fourier series in which is
where
Since is odd function, for
Hence the Fourier series for in is
Since is continuous on , we can write
Hence therefore,
Since
and
hence by Weierstrass M-test,
The 5series
[Remark: We can use comparison test for the test of convergence]
This implies the infinite series
converges uniformly to in . Then by Parseval’s Identity (Theorem 3.5)
Example 4.6. Find the Fourier series of the periodic function with period defined by
The we will find the sum of the series at and then we are going to show that
Solution: We define at and at as , to make continuous at
is piecewise monotone on . It is continuous on except at and and is bounded on . Hence is a bounded periodic and integrable function on which validates the expansion of into Fourier series, which is
Where
Thus, the Fourier series for in is
i.e.
Since is periodic function of period , so . Since is not continuous at , the Fourier series for converges at .
Hence sum of the series at is equal to .
At , the Fourier series for converges to
Therefore
Example 4.7. Let be defined as follows:
We are going to obtain the Fourier’s coefficients and the Fourier series for the function and hence we will find that sum of the series can be represented as
Solution: is piecewise monotone in . It is periodic with period . It is bounded on . is continuous in except at . Hence it is integrable on . Thus satisfies Dirichlet’s conditions on which validates Fourier series expansion of in .
The Fourier coefficients are
For ,
when is odd, . When is even,
For ,
Thus, we have
and
Hence the Fourier series for in is
Since is continuous at , the sum of the series at is
Therefore,
FOURIER SERIES FOR EVEN AND ODD FUNCTIONS
When is an even function on
Then Fourier coefficients becomes zero, since is then odd and hence
Thus the Fourier series for becomes a cosine series.
If is odd i.e.,
then the Fourier coefficients
and
Thus, the Fourier series for will be a sine series.
Example 5.1. We are going to consider the even function on and we will see that it has a cosine series in Fourier’s form which is
Also the series converges to in and by using this series we will show that,
Solution:
is piecewise monotone on . is bounded on . is continuous on and hence is integrable on . Thus satisfies Dirichlet’s conditions on and it can be represented by Fourier series on , which is
Where
For
Hence
Since is odd function, .
Thus, Fourier series for is in form of the cosine series
Since is continuous on , so the series converges to for all and
We have
At , the sum of the series is
Therefore,
Example: 5.2. We will discuss for the odd function
has a sine series in Fourier’s form as
Then we will show that
[NOTE:- sinhx, coshx are hyperbolic functions.]
Solution: is an odd function on and is monotone increasing on ; it is bounded and integrable on . Thus satisfies Dirichlet’s conditions for expansion in Fourier series on which will be a sine series as
Where
Hence Fourier series, which is a sine series, for on is
Since is continuous on , the series converges to on and we have for
At , the sum of the series is
Hence, we have
FOURIER EXPANSION ON .
We have seen that satisfies Dirichlet’s conditions then can be expressed either as a cosine series or as a sine series or as a combination of sine series and cosine series. If is defined on and it is bounded, integrable and piecewise monotone in , then can be extended over to as a bounded, integrable and piecewise monotone even or odd function as desired and hence this function can be expressed as Fourier sine or cosine as desired.
Theorem 6.1.
If is periodic, bounded, integrable and piecewise monotone on , then
Fourier cosine series of on is
where
and
if
Fourier sine series of on is
Where
And
Proof. (i) We define a function on by
Then for all and so is an even function on . Since satisfies Dirichlet’s conditions on , also satisfies Dirichlet’s conditions on , hence we have a Fourier cosine series (by Fourier-Dirichlet theorem)
Where
And
Also
(ii) We define a function on by
Then i.e., is an odd function on . Since satisfies Dirichlet’s conditions on , also satisfies Dirichlet’s conditions on .
Then we have a Fourier sine series
where
and
Hence the theorem proved.
Let we discuss some examples regarding with the theorem 6.1
Example 6.1. We are going to show that for
Then we will deduce that
Solution:
We are to represent in Fourier sine series in . Since satisfies Dirichlet’s conditions on , the Fourier sine series (by Theorem 6.1.) for on is
where
being continuous on , we have
The series
as
and
is a convergent series, hence by Weierstrass M-test, the series
is uniformly convergent and this series is the derivative of the series
Since
Hence using term-by-term differentiation (which is permissible) of the sine series in , we have
Example 6.2. Using the expansions of the functions and in the interval in cosines of multiple of , we are going to prove the equality
Solution:
satisfies Dirichlet’s conditions in . The cosine series for on is
Where
As is continuous on , we have
Let, satisfies Dirichlet’s conditions in . The cosine series for on is
Where
And
As is continuous on , we have
From (1),
From (2),
Therefore, adding these two convergent series in we have
Example 6.3. Show that the following function satisfies Dirichlet’s conditions of convergence of Fourier series in :
,
Find a series of cosines of multiples of which is the Fourier expansion of in . Deduce that
Solution: is piecewise monotone on , it is bounded on . Except at and , is continuous on , and hence is Riemann integrable on . Hence satisfies Dirichlet’s conditions of convergence of Fourier series in .
Fourier cosine series of in is
where
which is non-zero when is odd but not multiple of 3.
Thus, cosine series for in is
i.e.,
i.e.,
Since this series converges to at
So
Example 6.4. We are going to expand the function in a Fourier cosine series, where is not an integer.
Solution: Let
is continuous and piecewise monotone on . So, satisfies Dirichlet’s conditions on . Then Fourier cosine series of on is
Where,
And
Hence the Fourier cosine series of is
Example 6.5. We are going to express the function
in a Fourier series in .Then we will show that
Solution: is continuous, integrable, monotone on . Thus, satisfies Dirichlet’s conditions on and Fourier series for on is
Where
For ,
For
Thus, Fourier series for on is
i.e.,
Since this series converges to at , hence
Hence we got our desired result.
SOME PROBLEMS FOR THE READERS
- Find the Fourier series for the function
- Find the Fourier series for the function . And calculate the sum at .
BIBLIOGRAPHY:
- Fourier Series, Transforms, and Boundary Value Problems
Book by J. Ray Hanna and John H. Rowland - Fourier series
Book by G. P. Tolstov - Fourier Series
Book by Rajendra Bhatia (Best book for Beginners) - Discourse on Fourier series
Book by Cornelius Lanczos
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