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340 College Mathematics Fourier Series 341
5.3 Fourier Series
Definition : A series of the form
∞∞
a0 nππxnx
5. FOURIER SERIES f(x)=++abcos()sin()
∑∑
nn
2 ll
nn==11
is called a Fourier series of f(x) with period 2l in the interval
( c, c +2l ) where l is any positive real number and a , a , b are
5.1 Introduction given by the formulae called Euler’s Formulae : 0 n n
In various engineering problems it will be necessary to
express a function in a series of sines and cosines which are
periodic functions. Most of the single valued functions which are 1cl+2
a= fx(),dx
used in applied mathematics can be expressed in the form. 0 l ∫
1 c
a+acosx++axcos2 KK 1cl+2 nxπ
012 a= f(x)cos()dx
2 n ∫
ll
+bsinx++bxsin2 KK c
12 1cl+2 nxπ
within a desired range of values of x. Such a series is called a b= f(x)sin()dx
Fourier Series in the name of the French mathematician Jacques n ∫
ll
Foureier (1768 - 1830) c
These coefficients a , a , b are known as Fourier coefficients.
0 n n
5.2 Periodic Functions
Definition : If at equal intervals of the abscissa ‘x’ the value of In particular if l = π , the Fourier series of f(x) with period 2π in
each ordinate f(x) repeats itself then f(x) is called a periodic the interval (c, c+2π ) is given by
a ∞∞
function. i.e., A function f(x) is said to be a periodic function if 0
f(x)=++acosnxbsinnx
α α 2 ∑∑nn
there exists a real number such that f(x + ) = f(x) for all x. nn==11
The number α is called the period of f(x). and the Fourier coefficients are given by
∴ we have f(x) = f(x + α ) = f(x + 2α ) = f(x + 3α ) 1 c+2π
a= f(x),dx
= …………………..= f(x + n α) = …………. 0 π ∫
Ex : (i) sin x = sin (x + 2π ) = sin (x +4π ) = …………… c
.......….= sin (x + 2n π ) = …………… 1 c+2π
a= f()xcosnπ dx
Hence sin x is a periodic function of the period 2 π . n π ∫
(ii) cos x = cos(x + 2π ) = cos (x + 4π ) = ……… c
1 c+2π
……….. = cos (x + 2n π ) = ……………. b= f(x)sinnπ dx
n π ∫
Hence cos x is a periodic function of the period 2 π . c
We define the Fourier series in terms of these two periodic We shall derive the Euler’s formulae’ for which the following
functions. definite integrals are required.
342 College Mathematics Fourier Series 343
cl+2 1cl+2
(i) dxl= 2 ∴=af()xdx . .
∫ 0 l ∫
c c
c++22lcl . (a)
mππxmx
(ii) cosdx==sin0dx mxπ
∫∫
ll To find a, multiply both sides of (1) by cos where m is a
cc n l
cl+2 mππxnx fixed positive integer and integrate w.r.t x from x= c to x= c+ 2l
(iii) ∫ cossin0dx= for all integers m and n cl+2
c ll mxπ
c++22lcl ∴ ∫ f(x)cos dx
mπxnπxmππxnx c l
(iv) coscosdx==sinsin0dx
∫∫ c++22lcl
llll ∞
cc a0mπxmππxnx
=+cosdxacoscos dx
∑ n
(for all integers m and n such that m≠n) ∫∫
2 lll
n=1
c++22lcl cc
mππxmx cl+2
22 ∞ mππxnx
(v) cosdx==lsin dx
∫∫ +bcossin dx
ll ∑ n ∫
cc n=1 ll
c
cl+2
∞∞
5.4 Derivation of Euler’s Formulae a0 mππxnx
=(0)++acoscosdxb(0)
∑∑
nn
∞∞ 2 ∫ ll
a nππxnx nn==11
We have 0 c
f(x)=++abcossin
∑∑
nn [Using the definite integrals (ii) and (iii) above]
2 ll
nn==11
. . .(1) ∞ cl+2 mππxnx
=≠acoscosdx()mn
To find the coefficients a , a and b , we assume that the series ∑ n ∫
0 n n n=1 c ll
(1) can be integrated term by term from x = c to x = c + 2l cl+2 mxπ
To find a , integrate (1) w.r.t x from c to c + 2l. 2
0 +=acosdx()mn
c+++222lclcl m ∫
a ∞ nxπ c l
0
∴f(x)=+1dxacos dx
∑ n ∞
∫∫∫
2 l
n=1
ccc
=+a(0)al()
∞ cl+2 ∑ nm
nxπ n=1
+ bsin()dx
∑ n ∫ l [Using the definite integrals (iv) and (v) above]
n=1 c
a ∞∞
0 = al()
=(2l)++ab(0)(0) m
∑∑
2 nn cl+2
nn==11 1 mxπ
∴=af(x)cos dx
=al()(using the definite integrals (ii) above) m ∫
0 ll
c
Changing m to n we get
344 College Mathematics Fourier Series 34 5
1cl+2 nxπ 1cl+2 nxπ
a= f(x)cos dx …(b) b= f(x)sin dx
m ∫ n ∫
ll ll
c c
mxπ (b)
To find b , multiply both sides of (1) by sin where m
n l Thus the Euler’s formulae (a), (b), (c) are proved.
is a fixed positive integer and integrate w.r.t x from x = c to
x= c+ 2l Cor. 1 : In particular if l = π and c = 0, we get the Fourier series
cl+2 a ∞∞
mxπ 0
f(x)=++acosnxbsinnx
∴ ∫ f(x)sin dx 2 ∑∑nn
c l nn==11
c++22lcl where the Foureir coefficients are given by
∞
a0 mπxmππxnx 12π
=+sindxasincos dx
∑ n
∫∫ a=f(x),dx
2 lll 0 ∫
n=1
cc π0
∞ cl+2 mππxnx 2π
+∑bsinsin dx 1
n ∫ a= f(x)cosnπ dx
n=1 c ll n π ∫
cl+2 0
a ∞∞mππxnx 2π
0 1
=(0)++a(0)bsinsin dx
∑∑
nn
∫ b= f(x)sinnπ dx
2 nn==11ll n ∫
c π 0
[Using the definite integrals (ii) and (iii) above]
∞ cl+2 mππxnx Cor. 2: In the above formulae if l =π and c=−π , we get the
=≠asinsindx()mn
∑ n ∫ ll Fourier series
n=1 c
cl+2 a ∞∞
mππxnx 0
f(x)=++acosnxbsinnx
+=bsinsindx()mn ∑∑nn
m ∫ ll 2 nn==11
c where the Fourier coefficients are given by
cl+2 mxπ
2 1 π
=+0bsin dx
m ∫ a= fx(),dx
c l 0 π ∫
[Using the definite integrals (iv) above] −π
=bl() [using the definite integral (v)] 1 π
m a= f(x)cosnπ dx
n π ∫
1cl+2 mxπ −π
∴=bf(x)sin dx π
m ∫ 1
ll
c b= f(x)sinnπ dx
n π ∫
Changing m to n we get −π
346 College Mathematics Fourier Series 347
5.7 Even and odd functions
5.5 Conditions for a Fourier series expansion A function f(x) is said to be even if f(-x) = f(x) ∀xin the
It should not be mistaken that every function can be given interval (c, c + 2l) and a function f(x) is said to be odd if
expanded as a Fourier series. In the above formulae we have only f(-x) = -f(x) ∀xin the given interval (c, c + 2l)
shown that if f(x) is expressed as a Fourier series, then the Fourier
coefficients are given by Euler’s formula. It is very cumbersome 5.7.1 Tests for even and odd mature of a function
to discuss whether a function can be expressed as a Fourier series If f(x) is defined by one single expression, f(-x) = f(x)
and to discuss the convergence of this series. However the implies f (x) is even and f(-x) = -f(x) implies f(x) is odd. If f(x) is
following condition called Dirichlet’s condition cover all defined by two or more expressions on parts of the given interval
problems. with 0 as the mid point, f(-x) from the function as defined on one
∞∞ side of 0 = f(x) from the corresponding function as defined on the
a0 nππxnx
f(x)=++abcossin
∑∑
nn other side, implies f(x) is even.
2 ll
nn==11 f(-x) from the function as defined on one side of 0 = -f(x)
provided from the corresponding function as defined on the other side,
(i) f(x) is bounded implies f(x) is odd.
(ii) f(x) is periodic, single – valued and finite
(iii) f(x) has a finite number of discontinuities in any one Examples :
period. 2
(iv) f(x) has at the most a finite number of maxima and (1) f(x) = x + 1 in (-1, 1)
2 2
minima. f(-x) = (-x) + 1 = x + 1 = f(x)
∴ f(x) is even.
These conditions are called Dirichlets conditions. In fact
3
expressing a function f(x) as a Fourier series depends on the (2) f(x) = x in (-1, 1)
3 3
evaluation on the definite integrals f(-x) = (-x ) =- x = - f(x)
1 nxπ 1 nxπ ∴ f(x) is odd.
∫ f(x)cos dx and ∫ f(x)sin dx
llπ l x+−1in (π,0)
within the limits c to c + 2l, 0 to 2π or -π to π according as (3) fx()=
π π π x−1in (0,)π
f(x) is defined for all x in (c, c + 2l) (0, 2 ) or (- , )
f(−x)in (0,ππ)=−−xx1(=−+1)=−−fx()in(,0)
5.6 Interval with 0 as mid point ∴ f(-x) = -f(x)
If c = -l then the interval (c, c + 2l) becomes (-l, l) and ∴ f(x) is odd
further if c = -π , the interval becomes (-π ,π ). These intervals
have 0 as the mid point. For functions defined in such intervals, 5.7.2 Fourier coefficients when f(x) is even and odd
we consider the effect of changing x to –x and classify them as
even and odd functions. From definite integrals, we have
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