WEBVTT
Kind: captions
Language: en
00:00:18.850 --> 00:00:24.730
so welcome to the lecture series on mathematical
methods and its applications so we have already
00:00:24.730 --> 00:00:29.310
discussed fourier series we have seen what
fourier series are and how we can solve some
00:00:29.310 --> 00:00:34.230
problems based on that what are properties
what are convergence properties of the fourier
00:00:34.230 --> 00:00:40.780
series that also we have seen and also the
complex of fourier series now for a integrals
00:00:40.780 --> 00:00:48.300
what for a integrals are and why they are
important let us see so if you have a periodic
00:00:48.300 --> 00:00:55.900
function f x which is defined on a real line
r then f x can be represented it by a fourier
00:00:55.900 --> 00:01:01.170
series that we have already seen ok which
is given by a naught by two plus summation
00:01:01.170 --> 00:01:07.700
n cos n pi x pi l plus summation b n sin n
pi x by l that we have already seen
00:01:07.700 --> 00:01:13.670
if f x is not periodic suppose f x is not
a periodic function then it cannot be represented
00:01:13.670 --> 00:01:20.080
it by a fourier series over entire real line
is one two express a non periodic function
00:01:20.080 --> 00:01:27.100
by a fourier series then it is not possible
however f x may be represented it in an integral
00:01:27.100 --> 00:01:37.590
form ok how let us see now first of all suppose
f x have the following three properties f
00:01:37.590 --> 00:01:46.320
x piecewise continuous on every interval
from minus l to plus l f x is absolutely integrable
00:01:46.320 --> 00:01:52.700
on x axis that means integral from minus infinity
to plus infinity mod of f x converges that
00:01:52.700 --> 00:02:00.670
means this value is finite integral minus
to plus infinity mod f x d x value is finite
00:02:00.670 --> 00:02:07.090
and for every x belongs to r f x has left
and right hand derivatives
00:02:07.090 --> 00:02:14.670
so first of all let us suppose the function
f x is periodic ok and satisfy these three
00:02:14.670 --> 00:02:21.810
properties ok again suppose f x is a periodic
function with period two will define interval
00:02:21.810 --> 00:02:26.930
minus l two plus l satisfy the properties
p one to p three which we have discussed then
00:02:26.930 --> 00:02:33.770
it can be specified fourier series in this
way ok basically we derive the fourier integral
00:02:33.770 --> 00:02:39.220
from the fourier series itself ok we have
the fourier series like this we already know
00:02:39.220 --> 00:02:53.250
this f x suppose to a naught upon two plus
summation a n cos n pi x upon l plus b n sin
00:02:53.250 --> 00:03:01.890
n pi x by l so this is the fourier series
expression and it is varying from one to infinity
00:03:01.890 --> 00:03:12.810
or function f x ok we are a naught is given
by a naught is given by one upon l integral
00:03:12.810 --> 00:03:25.190
minus l two plus l f t d t a n is given by
one by l integral minus l two l f t cos n
00:03:25.190 --> 00:03:37.140
pi t by l into d t and b n is given by one
by l integral minus l to plus l f t sin n
00:03:37.140 --> 00:03:51.900
pi t by l into d t
so these are expressions ok now now let
00:03:51.900 --> 00:03:58.280
us substitute these expressions over here
ok so what will obtain f x will be equals
00:03:58.280 --> 00:04:08.300
to one by two l integral minus l to plus l
f t d t plus summation n varying from one
00:04:08.300 --> 00:04:19.900
to infinity a n is nothing but one by l
integral minus l to plus l f t cos n pi t
00:04:19.900 --> 00:04:35.470
by l into d t into cos n pi x upon l ok plus
summation n varying from one to infinity again
00:04:35.470 --> 00:04:48.810
one by l integral minus l to plus l f t sin
n pi t by l into d t and whole multiplied
00:04:48.810 --> 00:05:04.340
by sin n pi x by l ok now let us suppose let
omega n is n pi by l ok so what will be delta
00:05:04.340 --> 00:05:13.740
omega delta omega is omega n minus omega n
minus one which is nothing but n pi by l minus
00:05:13.740 --> 00:05:22.750
n minus one pi by l which is pi by l
so let us substitute these terms in this expression
00:05:22.750 --> 00:05:31.340
ok so what will be effects what effects we
will obtain so f x will be nothing but is
00:05:31.340 --> 00:05:45.250
equal to now now this this one by l one
by l is nothing but delta omega by pi so it
00:05:45.250 --> 00:05:58.130
is delta omega by two pi integral minus l
to plus l f t d t ok from this from this expression
00:05:58.130 --> 00:06:06.230
now plus again this one by l will be nothing
but pi upon delta omega upon pi from this
00:06:06.230 --> 00:06:13.780
expression so one upon pi will be outside
one upon pi will be outside summation n varying
00:06:13.780 --> 00:06:25.110
from one to infinity minus l to plus l f t
cos and pi n pi l by l is nothing but omega
00:06:25.110 --> 00:06:42.600
n omega n t ok into d t and whole multiplied
by cos omega n into x into delta omega ok
00:06:42.600 --> 00:06:48.010
delta omega by pi is here instead of this
so delta omega is here and similarly plus
00:06:48.010 --> 00:06:57.830
one by pi summation n varying from one to
infinity integral minus l to plus l f t sin
00:06:57.830 --> 00:07:12.070
omega n t into d t and whole multiplied by
sin omega n x into delta omega because pi
00:07:12.070 --> 00:07:20.160
n pi l by l is nothing but omega n so it is
omega n x into delta omega ok
00:07:20.160 --> 00:07:31.449
now let l now let let l tend to infinity
because you on to expend the function over
00:07:31.449 --> 00:07:38.800
the entire real line ok we want to expend
the function over the entire real line thats
00:07:38.800 --> 00:07:47.110
why we are we are tending l to infinity so
as l tends to infinity of course minus l to
00:07:47.110 --> 00:07:51.830
plus l interval will tend to minus into plus
infinity that is over the entire real line
00:07:51.830 --> 00:08:00.010
and delta omega will tend to zero delta omega
is nothing but pi upon l and as l tend to
00:08:00.010 --> 00:08:07.680
infinity this will tends to zero ok so and
function f x is absolutely convergent that
00:08:07.680 --> 00:08:15.620
we are already assumed ok so this term so
this integral when l tends to infinity is
00:08:15.620 --> 00:08:24.000
finite and delta omega tend into zero the
first term is tend to zero because delta omega
00:08:24.000 --> 00:08:30.650
tend to zero and as l tend to infinity so
this value since is absolutely convergent
00:08:30.650 --> 00:08:36.760
absolutely convergence so this value is finite
finite into zero will be zero so first term
00:08:36.760 --> 00:08:48.340
is zero and this will sum up as remaining
sum because because l tends to infinity
00:08:48.340 --> 00:08:55.749
l tends to infinity and this delta omega which
tend zero so this sum up as remaining sum
00:08:55.749 --> 00:09:06.160
so this will be nothing but one upon pi integral
zero to infinity ok zero to infinity integral
00:09:06.160 --> 00:09:19.100
minus infinity plus infinity f t cos omega
t into d t whole with cos omega x into d omega
00:09:19.100 --> 00:09:28.760
plus plus integral zero to infinity integral
minus to plus infinity f t it is sin omega
00:09:28.760 --> 00:09:45.550
t into d t and whole multiplied by sin omega
x into d omega ok because l is tend to infinity
00:09:45.550 --> 00:09:52.779
ok and delta omega tend to zero so this will
be sum up as remaining sum so which is given
00:09:52.779 --> 00:09:59.369
by one upon pi integral zero to infinity minus
to plus infinity this term plus this term
00:09:59.369 --> 00:10:06.160
so thats what we have given the p p t also
that this presumed as a remaining sum of the
00:10:06.160 --> 00:10:10.309
different integral as l tend to infinity so
this will give this value
00:10:10.309 --> 00:10:16.189
now here if you take if you take this as a
omega because this is a function of omega
00:10:16.189 --> 00:10:25.329
and this as b omega so what will be f x f
x will be nothing but one by pi times integral
00:10:25.329 --> 00:10:41.230
zero to infinity a omega cos omega x t omega
plus plus integral zero to infinity it
00:10:41.230 --> 00:10:57.279
is b omega sin omega x into d omega so that
will be that will be this expression where
00:10:57.279 --> 00:11:08.550
a omega is nothing but minus infinity to plus
infinity f t cos omega t d t and b omega is
00:11:08.550 --> 00:11:19.930
nothing but minus will plus infinity f t sin
omega t into d t so this is what the fourier
00:11:19.930 --> 00:11:28.079
integral is i mean integral representation
of a function f x ok for a integral representation
00:11:28.079 --> 00:11:35.199
of a function f x given by this where a omega
is a this and b omega is this ok so we we
00:11:35.199 --> 00:11:43.360
expend the function over the entire real
line by assuming that l is tend to infinity
00:11:43.360 --> 00:11:51.800
ok so this is how we can obtain this thing
now this is called fourier integrals of
00:11:51.800 --> 00:12:00.730
f x now suppose f x satisfy the properties
p one to p three ok which we have discussed
00:12:00.730 --> 00:12:05.170
then the fourier integral of f x converges
to f x the point of continuity the same as
00:12:05.170 --> 00:12:10.629
in fourier series this we have also discussed
fourier series also the same follows in fourier
00:12:10.629 --> 00:12:16.999
integrals also that if if satisfy the some
properties p one to p three then the fourier
00:12:16.999 --> 00:12:22.820
integral of f x converges f x at point of
continuity and for discontinuity it converges
00:12:22.820 --> 00:12:28.930
to the average of left hand limit and the
right hand limit at that point ok now let
00:12:28.930 --> 00:12:32.749
us try this problem now
we want fourier integral representation of
00:12:32.749 --> 00:12:36.989
this function so we know that the fourier
integral representation of a function is given
00:12:36.989 --> 00:12:42.569
by this expression f x is given by this expression
a omega is given by this b omega is given
00:12:42.569 --> 00:12:54.380
by this expression now their function is what
function is defined function is one when mod
00:12:54.380 --> 00:13:01.869
x is less than one and zero when mod x is
greater than one ok so this is the function
00:13:01.869 --> 00:13:07.120
now we want fourier integral representation
of this function so this function is defined
00:13:07.120 --> 00:13:11.240
over the entire real line so we want the fourier
integral representation of this function how
00:13:11.240 --> 00:13:17.639
you can find it so we first find a omega and
b omega we substitute the values of a omega
00:13:17.639 --> 00:13:23.399
b omega over here so that will give the fourier
integral representation of this function so
00:13:23.399 --> 00:13:29.739
what is a omega let us find a omega first
what is a omega from here a omega is minus
00:13:29.739 --> 00:13:40.249
plus infinity f t cos omega t d t
now it is defined from minus one to plus
00:13:40.249 --> 00:13:48.149
one ok is one so minus one to plus one is
one otherwise it is zero so it is cos omega
00:13:48.149 --> 00:13:54.550
t d t and cos omega t is an even function
so it is nothing but two times zero to one
00:13:54.550 --> 00:14:04.899
cos omega t d t and it is equals to two times
sin omega t upon omega from zero to one which
00:14:04.899 --> 00:14:18.610
is nothing but two sin omega upon omega because
sin zero is zero ok now what is b omega b
00:14:18.610 --> 00:14:26.329
omega will be nothing but again minus infinity
plus infinity by this definition f t sin omega
00:14:26.329 --> 00:14:35.939
t into d t which is equals to minus one to
plus one sin omega t into d t because from
00:14:35.939 --> 00:14:42.239
minus one to plus one it is one and it is
an odd function it is zero so we have computed
00:14:42.239 --> 00:14:48.799
both a omega and b omega and the fourier integral
representation of a function f x is given
00:14:48.799 --> 00:14:55.449
by this expression as you already know so
this will be nothing but one upon pi integral
00:14:55.449 --> 00:15:04.439
zero to infinity
now what is a omega is two sin omega cos omega
00:15:04.439 --> 00:15:14.660
x upon omega into d omega d omega is zero
so that will be the fourier integral representation
00:15:14.660 --> 00:15:22.220
of this function f x ok now we can also
write this function as zero to infinity
00:15:22.220 --> 00:15:34.299
two sin omega cos omega x upon omega d omega
which will be equals to pi times f x and which
00:15:34.299 --> 00:15:42.759
is equals to pi you multiply simply that by
pi when mod x less than one that is zero when
00:15:42.759 --> 00:15:56.369
mod x greater than one ok this two you can
also bring here pi by two so it is pi by two
00:15:56.369 --> 00:16:06.170
and what happens when x is one when x is one
so it is it is discontinues so it converges
00:16:06.170 --> 00:16:11.259
to average of left hand limit and right hand
limit which is nothing but what will be f
00:16:11.259 --> 00:16:22.600
one f one will nothing but half of f one minus
plus f one plus so that will be nothing but
00:16:22.600 --> 00:16:29.440
one by two one plus zero that is one by two
and you have to multiply pi by two also so
00:16:29.440 --> 00:16:44.100
that will be pi by four ok so in this way
you can find out the value
00:16:44.100 --> 00:16:52.070
the same you can find out when x equals to
minus one also when x equal to minus one we
00:16:52.070 --> 00:17:01.299
are the same value it is x equal to x equal
plus or minus one ok so so this integral
00:17:01.299 --> 00:17:09.559
is equal to basically this integral now
we have to find out integral zero to infinity
00:17:09.559 --> 00:17:16.839
sin omega upon omega d omega so that will
be nothing but you substitute x as zero ok
00:17:16.839 --> 00:17:22.870
its a function of omega so you can arbitrary
choose any value of x so when x is zero so
00:17:22.870 --> 00:17:30.139
when x is zero so this is nothing
but so you can substitute both side as
00:17:30.139 --> 00:17:40.370
zero so it is zero to infinity sin omega upon
omega d omega is equals to pi by two f zero
00:17:40.370 --> 00:17:51.340
and where it is zero when f is zero f is one
it is pi by two ok you can directly see from
00:17:51.340 --> 00:17:56.380
here when x is zero value is pi by two value
of this expression is pi by two so you can
00:17:56.380 --> 00:18:08.990
easily see that this value when x is zero
ok in fact when you put x equal to say half
00:18:08.990 --> 00:18:14.899
when you put x equal to half to when x equal
to half satisfy this expression when x equal
00:18:14.899 --> 00:18:21.370
to half the value of that will also be pi
by two you can easily find out under different
00:18:21.370 --> 00:18:31.120
values of x the value of this expression nothing
but this expression ok now let us try to prove
00:18:31.120 --> 00:18:37.460
this that this is equal to this expression
so how we can prove this this is nothing but
00:18:37.460 --> 00:18:42.889
fourier integral representation of function
f x so what is the fourier integral representation
00:18:42.889 --> 00:18:53.450
of function so what is the expression it is
zero to i mean infinity it is cos it is
00:18:53.450 --> 00:19:05.210
cos pi omega by two into cos omega x upon
one minus omega square d omega is equals to
00:19:05.210 --> 00:19:14.809
it is given to you pi by two cos x when mod
x is less than pi by two and zero when mod
00:19:14.809 --> 00:19:23.679
x is so we have to prove this ok we have to
prove this so what is the fourier integral
00:19:23.679 --> 00:19:31.710
representation of function f x is nothing
but one upon pi times integral integral
00:19:31.710 --> 00:19:47.320
zero to infinity a omega cos omega x t x cos
omega x d omega plus integral zero to infinity
00:19:47.320 --> 00:20:03.260
d omega sin omega x into d omega ok is it
so fourier integral representation is into
00:20:03.260 --> 00:20:13.320
d omega let us same thing and here it is d
t ok now what is what is a omega a omega is
00:20:13.320 --> 00:20:23.470
minus minus will plus infinity it is f t cos
omega t d t and d omega is given by minus
00:20:23.470 --> 00:20:34.630
will plus infinity f t sin omega t d t that
you already know now if you compare this
00:20:34.630 --> 00:20:46.429
with this so what will be a omega so it is
pi f x which is equals to if you compare
00:20:46.429 --> 00:20:52.340
this by this so cos omega x coefficient here
is simply a omega
00:20:52.340 --> 00:21:01.830
so we can simply say that a omega is nothing
but cos pi omega by two upon one minus omega
00:21:01.830 --> 00:21:10.350
square and d omega is zero because there is
no term of sin omega x ok and this integral
00:21:10.350 --> 00:21:21.429
zero to infinity a omega cos omega x d x is
nothing but pi times f x which is equals to
00:21:21.429 --> 00:21:30.389
which is equals to pi by two cos x when mod
x is less than pi by two and zero when x is
00:21:30.389 --> 00:21:41.860
mod x is greater than pi by two so this implies
f x is cos x by two when mod x is less than
00:21:41.860 --> 00:21:54.500
pi by two and zero when mod x is greater than
pi by two ok so you take this f x you take
00:21:54.500 --> 00:22:02.460
this f x and try to obtain the value of a
omega if you obtain a omega as this that means
00:22:02.460 --> 00:22:13.139
we have proved ok any how any how we have
assume that this is true ok to assuming this
00:22:13.139 --> 00:22:19.169
to be true we have find out the coefficients
a omega and d omega and we have find f
00:22:19.169 --> 00:22:28.200
x for this f x if a omega comes out to be
this that means we have proved ok so what
00:22:28.200 --> 00:22:36.700
will be a omega for this f x and b omega a
omega should be this b omega should be this
00:22:36.700 --> 00:22:47.580
so what will be a omega a omega will be given
by minus infinity plus infinity f t cos omega
00:22:47.580 --> 00:22:57.549
t d t now it is given by f x is defined like
this from minus pi by two to plus pi by two
00:22:57.549 --> 00:23:07.580
it is cos t upon two into cos omega t d t
which is nothing but it is even function it
00:23:07.580 --> 00:23:17.740
is two times by two zero to pi by two cos
t into cos omega t d t which is equals to
00:23:17.740 --> 00:23:24.210
now two two cancels out multiply divided by
two again zero to pi by two it is two cos
00:23:24.210 --> 00:23:38.520
a cos b which is nothing but sin a plus b
plus sin a minus b it is equals to one by
00:23:38.520 --> 00:23:46.960
two integral of sin is minus cos so put minus
outside put minus on outside it is cos of
00:23:46.960 --> 00:23:54.830
one plus omega t upon one plus ok both will
be cos sorry cos a to cos a cos b is nothing
00:23:54.830 --> 00:24:02.659
but cos a plus b plus cos a minus b so it
will be cos it will be cos and minus is not
00:24:02.659 --> 00:24:12.429
so it will be sin and plus sin one minus omega
t upon one minus omega and t varying from
00:24:12.429 --> 00:24:22.130
zero to pi by two so this will be nothing
but one by two it is sin one plus omega into
00:24:22.130 --> 00:24:30.659
pi by two upon one plus omega plus sin one
minus omega into pi by two upon one minus
00:24:30.659 --> 00:24:37.149
omega ok now we were simplify this is nothing
but one by two
00:24:37.149 --> 00:24:45.750
now sin ninety plus theta is sin theta to
that is sin pi by two omega upon one plus
00:24:45.750 --> 00:24:52.840
omega again sin ninety minus theta no sin
ninety plus theta is cos theta sorry its cos
00:24:52.840 --> 00:25:06.679
theta so it will be cos theta again here it
is cos theta so this will comes out this cos
00:25:06.679 --> 00:25:15.909
pi omega by two will come out and it will
be nothing but two upon one minus omega square
00:25:15.909 --> 00:25:26.250
so two two cancels out so this will comes
equal to this cos pi omega by two into one
00:25:26.250 --> 00:25:35.049
upon one minus omega square now when you compute
b omega b omega must be zero when you compute
00:25:35.049 --> 00:25:45.029
b omega for the same problem so b omega will
be nothing but what minus plus infinity f
00:25:45.029 --> 00:25:57.250
t sin omega t d t now it is nothing but
minus pi two plus pi by two it is cos t by
00:25:57.250 --> 00:26:05.899
two sin omega t d t its an odd function so
it is zero so hence we have proved because
00:26:05.899 --> 00:26:11.500
we have obtained that a omega is this and
b omega is zero for this function that means
00:26:11.500 --> 00:26:19.510
this results hold ok so we have shown the
result so this how we can find out fourier
00:26:19.510 --> 00:26:25.320
integral representation of a function f x
ok so so
00:26:25.320 --> 00:26:25.820
thank you