TOPIC: FOURIERTOPIC: FOURIER
SERIES ANDSERIES AND
FOURIER INTEGRALFOURIER INTEGRAL
Guided By: Prof. Jaydev PatelGuided By: Prof. Jaydev Patel
Introduction
Jean Baptiste Joseph Fourier
(Mar21st 1768 –May16th
1830) French mathematician,
physicist
Main Work:
(The Analytic Theory of Heat)
•Any function of a variable, whether continuous or
discontinuous, can be expanded in a series of sines
or cosines multiples of the variable
•The concept of dimensional homogeneity in
equations
•Proposal of his partial differential equation for
conductive diffusion of heat
Discovery of the "greenhouse effect“
•Fourier series is very useful in solving ordinary
and partial differential equation.
Even, Odd, and Periodic FunctionsEven, Odd, and Periodic Functions
Fourier Series of a Periodic FunctionFourier Series of a Periodic Function
Definition : A Fourier series may be defined as an expansion of a
function in a series of sines and cosines such as ,
0<x<2π
The coefficients are related to the periodic function f(x)
by definite integrals:
Henceforth we assume f satisfies the following (Dirichlet)
conditions:
(1) f(x) is a periodic function;
(2) f(x) has only a finite number of finite discontinuities;
(3) f(x) has only a finite number of extrem values, maxima and
minima in the interval [0,2p].
The formula for a Fourier series is:
We have formulae for the coefficients (for the
derivations see the course notes):
One very important property of sines and cosines is their
orthogonality, expressed by:
∑
∞=
=












+





+=
n
n
nn
T
xn
b
T
xn
aaxf
1
0
2
sin
2
cos)(
ππ
∫
−
=
2
2
0 )(
1
T
T
dxxf
T
a ∫
−






=
2
2
2
cos)(
2
T
T
n dx
T
xn
xf
T
a
π
∫
−






=
2
2
2
sin)(
2
T
T
n dx
T
xn
xf
T
b
π





=
≠
=











∫
− mn
T
mn
dx
T
xm
T
xn
T
T
2
0
2
sin
2
sin
2
2
ππ
nmdx
T
xm
T
xn
T
T
,allfor0
2
sin
2
cos
2
2
=











∫
−
ππ
These formulae are used in the derivation of the formulae for
Example – Find the coefficients for the Fourier series of:
 To Find
)()2(
0
0
)(
xfxf
xx
xx
xf
=+



≤≤
≤≤−−
=
π
π
π
0a
nn ba ,
Find ,0a
∫
−
=
2
2
0 )(
1
T
T
dxxf
T
a ∫−
=⇒
π
π
π
dxxfa )(
2
1
0
∫=⇒
π
π 0
0
1
xdxa
2
0
π
=⇒ a
f (x) is an even function so:
∫−
=
π
π
π
dxxfa )(
2
1
0
∫=⇒
π
π 0
0 )(
1
dxxfa
π
π 0
2
0
2
1






=⇒
x
a
Find na
∫
−






=
2
2
2
cos)(
2
T
T
n dx
T
xn
xf
T
a
π
∫−






=⇒
π
π
π
π
π
dx
xn
xfan
2
2
cos)(
1
Since both functions are even their product is even:
( )∫−
=
π
π
π
dxnxxfan cos)(
1 ( )∫=⇒
π
π 0
cos
2
dxnxxan
nb
∫
−






=
2
2
2
sin)(
2
T
T
n dx
T
xn
xf
T
b
π
∫−






=⇒
π
π
π
π
π
dx
xn
xfbn
2
2
sin)(
1
( )∫−
=
π
π
π
dxnxxfbn sin)(
1
0=⇒ nb
So we can put the coefficients back into the Fourier series
formula:
∑
∞=
=












+





+=
n
n
nn
T
xn
b
T
xn
aaxf
1
0
2
sin
2
cos)(
ππ
( )( ) ( )∑
∞=
=






−−+=⇒
n
n
n
nx
n
xf
1
2
cos11
2
2
)(
π
π
( ) ( ) +−+−=⇒ xxxf 3cos
9
4
0cos
4
2
)(
ππ
π
Summary of findingSummary of finding
coefficientscoefficients
function
even
function
odd
function
neither
0a
na
nb
0)(
1 2
2
0 == ∫
−
T
T
dxxf
T
a
∫
−






=
2
2
2
cos)(
2
T
T
n dx
T
xn
xf
T
a
π
Though maybe easy to find
using geometry
∫
−






=
2
2
2
sin)(
2
T
T
n dx
T
xn
xf
T
b
π
∫
−






=
2
2
2
sin)(
2
T
T
n dx
T
xn
xf
T
b
π
0
0
0)(
1 2
2
0 == ∫
−
T
T
dxxf
T
a
∫
−






=
2
2
2
cos)(
2
T
T
n dx
T
xn
xf
T
a
π
Though maybe easy to find
using geometry
0
HALF RANGE EXPANSIONSHALF RANGE EXPANSIONS
It often happens in applications, especially when we solve
partial differential equations by the method of separation
of variables, that we need to expand a given function f in
a Fourier series, where f is defined only on a finite interval.
We define an “extended function”, say fext, so that fext is
periodic in the domain of -∞< x < ∞, and fext=f(x) on the
original interval 0<x<L. There can be infinite number of
such extensions.
Fourier extensions: half- and quarter- range cosine and sine
extensions, which are based on symmetry or anti-symmetry
about the endpoints x=0 and x=L.
HRC (HALF RANGE COSINES)HRC (HALF RANGE COSINES)
fext is symmetric about x=0 and
also about x=L. Because of its
symmetry about x=0, fext is an
even function, and its Fourier
series will contain only cosines, no sines. Further,
its period is 2L, so L is half the period.
HRS (HALF RANGE SINE SERIES)HRS (HALF RANGE SINE SERIES)
Fourier IntegralFourier Integral
If f(x) and f’(x) are piecewise continuous in every finite interval, and
f(x) is absolutely integrable on R, i.e.
converges, then
Remark: the above conditions are sufficient, but not necessary.
16
∫ ∫
∞
∞−
∞
∞−
−








=++− dwdttfeexfxf iwtiwx
)(
2
1
)]()([
2
1
π
DIFFERENT FORMS OF FOURIERDIFFERENT FORMS OF FOURIER
INTEGRAL THEOREMINTEGRAL THEOREM
Complex or exponential formComplex or exponential form
THANK YOUTHANK YOU
PREPARED BY :-PREPARED BY :-
• ALAGIYA KEVALKUMAR .A (160280102001)
• BACHAV AHEMEDRAZA .S (160280102002)
• BHADARKA APABHAI GIGABHAI (160280102004)
• BUMBADIYA ASHISH SHARAD (160280102006)
• CHAUDHARY SWAPNILKUMAR (160280102007)
• CHAUHAN DIGVIJAYSINH (160280102008)
• CHAUHAN KARANKUMAR (160280102009)
• DOMADIYA PARIMAL (160280102010)

160280102001 c1 aem

  • 2.
    TOPIC: FOURIERTOPIC: FOURIER SERIESANDSERIES AND FOURIER INTEGRALFOURIER INTEGRAL Guided By: Prof. Jaydev PatelGuided By: Prof. Jaydev Patel
  • 3.
    Introduction Jean Baptiste JosephFourier (Mar21st 1768 –May16th 1830) French mathematician, physicist Main Work: (The Analytic Theory of Heat) •Any function of a variable, whether continuous or discontinuous, can be expanded in a series of sines or cosines multiples of the variable •The concept of dimensional homogeneity in equations •Proposal of his partial differential equation for conductive diffusion of heat Discovery of the "greenhouse effect“ •Fourier series is very useful in solving ordinary and partial differential equation.
  • 4.
    Even, Odd, andPeriodic FunctionsEven, Odd, and Periodic Functions
  • 5.
    Fourier Series ofa Periodic FunctionFourier Series of a Periodic Function Definition : A Fourier series may be defined as an expansion of a function in a series of sines and cosines such as , 0<x<2π The coefficients are related to the periodic function f(x) by definite integrals: Henceforth we assume f satisfies the following (Dirichlet) conditions: (1) f(x) is a periodic function; (2) f(x) has only a finite number of finite discontinuities; (3) f(x) has only a finite number of extrem values, maxima and minima in the interval [0,2p].
  • 6.
    The formula fora Fourier series is: We have formulae for the coefficients (for the derivations see the course notes): One very important property of sines and cosines is their orthogonality, expressed by: ∑ ∞= =             +      += n n nn T xn b T xn aaxf 1 0 2 sin 2 cos)( ππ ∫ − = 2 2 0 )( 1 T T dxxf T a ∫ −       = 2 2 2 cos)( 2 T T n dx T xn xf T a π ∫ −       = 2 2 2 sin)( 2 T T n dx T xn xf T b π      = ≠ =            ∫ − mn T mn dx T xm T xn T T 2 0 2 sin 2 sin 2 2 ππ
  • 7.
    nmdx T xm T xn T T ,allfor0 2 sin 2 cos 2 2 =            ∫ − ππ These formulae areused in the derivation of the formulae for Example – Find the coefficients for the Fourier series of:  To Find )()2( 0 0 )( xfxf xx xx xf =+    ≤≤ ≤≤−− = π π π 0a nn ba ,
  • 8.
    Find ,0a ∫ − = 2 2 0 )( 1 T T dxxf T a∫− =⇒ π π π dxxfa )( 2 1 0 ∫=⇒ π π 0 0 1 xdxa 2 0 π =⇒ a f (x) is an even function so: ∫− = π π π dxxfa )( 2 1 0 ∫=⇒ π π 0 0 )( 1 dxxfa π π 0 2 0 2 1       =⇒ x a
  • 9.
    Find na ∫ −       = 2 2 2 cos)( 2 T T n dx T xn xf T a π ∫−       =⇒ π π π π π dx xn xfan 2 2 cos)( 1 Sinceboth functions are even their product is even: ( )∫− = π π π dxnxxfan cos)( 1 ( )∫=⇒ π π 0 cos 2 dxnxxan nb ∫ −       = 2 2 2 sin)( 2 T T n dx T xn xf T b π ∫−       =⇒ π π π π π dx xn xfbn 2 2 sin)( 1 ( )∫− = π π π dxnxxfbn sin)( 1 0=⇒ nb
  • 10.
    So we canput the coefficients back into the Fourier series formula: ∑ ∞= =             +      += n n nn T xn b T xn aaxf 1 0 2 sin 2 cos)( ππ ( )( ) ( )∑ ∞= =       −−+=⇒ n n n nx n xf 1 2 cos11 2 2 )( π π ( ) ( ) +−+−=⇒ xxxf 3cos 9 4 0cos 4 2 )( ππ π
  • 11.
    Summary of findingSummaryof finding coefficientscoefficients function even function odd function neither 0a na nb 0)( 1 2 2 0 == ∫ − T T dxxf T a ∫ −       = 2 2 2 cos)( 2 T T n dx T xn xf T a π Though maybe easy to find using geometry ∫ −       = 2 2 2 sin)( 2 T T n dx T xn xf T b π ∫ −       = 2 2 2 sin)( 2 T T n dx T xn xf T b π 0 0 0)( 1 2 2 0 == ∫ − T T dxxf T a ∫ −       = 2 2 2 cos)( 2 T T n dx T xn xf T a π Though maybe easy to find using geometry 0
  • 12.
    HALF RANGE EXPANSIONSHALFRANGE EXPANSIONS It often happens in applications, especially when we solve partial differential equations by the method of separation of variables, that we need to expand a given function f in a Fourier series, where f is defined only on a finite interval. We define an “extended function”, say fext, so that fext is periodic in the domain of -∞< x < ∞, and fext=f(x) on the original interval 0<x<L. There can be infinite number of such extensions. Fourier extensions: half- and quarter- range cosine and sine extensions, which are based on symmetry or anti-symmetry about the endpoints x=0 and x=L.
  • 13.
    HRC (HALF RANGECOSINES)HRC (HALF RANGE COSINES) fext is symmetric about x=0 and also about x=L. Because of its symmetry about x=0, fext is an even function, and its Fourier series will contain only cosines, no sines. Further, its period is 2L, so L is half the period.
  • 15.
    HRS (HALF RANGESINE SERIES)HRS (HALF RANGE SINE SERIES)
  • 16.
    Fourier IntegralFourier Integral Iff(x) and f’(x) are piecewise continuous in every finite interval, and f(x) is absolutely integrable on R, i.e. converges, then Remark: the above conditions are sufficient, but not necessary. 16 ∫ ∫ ∞ ∞− ∞ ∞− −         =++− dwdttfeexfxf iwtiwx )( 2 1 )]()([ 2 1 π
  • 17.
    DIFFERENT FORMS OFFOURIERDIFFERENT FORMS OF FOURIER INTEGRAL THEOREMINTEGRAL THEOREM
  • 18.
    Complex or exponentialformComplex or exponential form
  • 19.
  • 20.
    PREPARED BY :-PREPAREDBY :- • ALAGIYA KEVALKUMAR .A (160280102001) • BACHAV AHEMEDRAZA .S (160280102002) • BHADARKA APABHAI GIGABHAI (160280102004) • BUMBADIYA ASHISH SHARAD (160280102006) • CHAUDHARY SWAPNILKUMAR (160280102007) • CHAUHAN DIGVIJAYSINH (160280102008) • CHAUHAN KARANKUMAR (160280102009) • DOMADIYA PARIMAL (160280102010)

Editor's Notes

  • #17 Outline: Central Scientific Problem – Artificial Intelligence Machine Learning: Definition Specifics Requirements Existing Solutions and their limitations Multiresolution Approximation: Limitation Our Approach. Results. Binarization. Plans.