WEBVTT
Kind: captions
Language: en
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so welcome to the lecture series on mathematical
methods and their applications so in the last
00:00:24.420 --> 00:00:29.990
lecture we have seen that what fourier series
are and how we can expand the function f x
00:00:29.990 --> 00:00:38.260
in the form of cos and sine terms ok so
now we will see that when the fourier series
00:00:38.260 --> 00:00:45.940
is convergent so let us see first convergence
of fourier series for continuous functions
00:00:45.940 --> 00:00:52.130
if a periodic function f x with period two
l is continuous in interval minus l to plus
00:00:52.130 --> 00:00:59.410
l and has continuous first and second derivatives
at each point in that interval then the fourier
00:00:59.410 --> 00:01:04.820
series which we have already discussed a naught
by two plus summation a n cos n pi x by l
00:01:04.820 --> 00:01:14.799
plus summation b n sin n pi x by l of f x
is convergent ok so whats the proof so proof
00:01:14.799 --> 00:01:20.430
we can see
now in this series in this series a naught
00:01:20.430 --> 00:01:33.350
by two plus summation n from one to infinity
a n cos n pi x by l plus summation n from
00:01:33.350 --> 00:01:45.310
one to infinity b n sin n pi x by l in this
series a naught we already know is given
00:01:45.310 --> 00:01:55.920
by one upon l minus one minus l to l integral
minus l to l f x d x a n is given by one by
00:01:55.920 --> 00:02:07.280
l integral minus l to plus l f x cos n pi
x by l into d x and b n is given by one by
00:02:07.280 --> 00:02:19.470
l integral minus l to plus l f x sin n pi
x by l into d x so this this we already said
00:02:19.470 --> 00:02:25.510
in the last lecture that these are the this
is given by eulers formula a naught a n and
00:02:25.510 --> 00:02:35.450
b n now if we if we see a n ok a n is nothing
but what a n is given by one by l minus integral
00:02:35.450 --> 00:02:45.989
minus l to plus l f x cos n pi x by l into
d x now you integrate it by parts ok you take
00:02:45.989 --> 00:02:50.760
it first function this second function this
integrate by parts eta it is nothing but equals
00:02:50.760 --> 00:02:57.609
to one by l now first as it is integration
of second it is nothing but sin n pi x by
00:02:57.609 --> 00:03:09.569
l upon n pi by l from minus l to plus l minus
integral derivative of first and integral
00:03:09.569 --> 00:03:27.359
of second now when x is l eta it is sin n
pi which is zero and when x is minus l which
00:03:27.359 --> 00:03:33.390
is again sin minus n pi which is again zero
this first term is zero eta it is nothing
00:03:33.390 --> 00:03:41.080
but minus it is l l cancels out it is minus
one upon n pi and it is integral minus l to
00:03:41.080 --> 00:03:54.980
plus l f dash sin n pi x by l into d x
so this we obtain now again you apply by
00:03:54.980 --> 00:04:03.480
parts so it is first and second so it is minus
one by n pi now first as it is integration
00:04:03.480 --> 00:04:12.879
of second integral second is nothing but minus
of cos n pi x upon l divided by n pi by l
00:04:12.879 --> 00:04:21.269
and it is from minus l to plus l minus integration
derivative of first integration of second
00:04:21.269 --> 00:04:34.050
that is plus cos cos n pi x by l upon n pi
by l into d x now it is minus one upon n pi
00:04:34.050 --> 00:04:44.569
ok now since f dash is continuous and f is
periodic so f dash l is same as f dash minus
00:04:44.569 --> 00:04:53.520
l ok since f x is continuous and it is periodic
so f dash l will be same as f dash minus l
00:04:53.520 --> 00:04:58.909
and at x equal to l and x equal to minus l
this value is same which is minus one k to
00:04:58.909 --> 00:05:10.050
the power n so the first term will be zero
ok the plus it is l upon n pi integral minus
00:05:10.050 --> 00:05:24.169
l to plus l f double dash x into cos n pi
x by l into d x ok this is this is zero because
00:05:24.169 --> 00:05:33.139
because function is periodic that is f of
x plus two l is same as f of x and derivative
00:05:33.139 --> 00:05:40.889
is continuous so f dash of x plus two l is
also equal to f dash x for all x and when
00:05:40.889 --> 00:05:47.690
you substitute x as minus l so f dash l will
be same as f dash minus l ok so these two
00:05:47.690 --> 00:05:54.139
term are same f dash l is same as f dash minus
l and upper and lower limit the value of this
00:05:54.139 --> 00:05:58.639
is same this term is zero and we left with
this expression
00:05:58.639 --> 00:06:04.750
now we have to show that anyhow that this
is convergent so to show that this is convergent
00:06:04.750 --> 00:06:12.759
again since f double dash x is continuous
and in the interval minus l to l so we can
00:06:12.759 --> 00:06:21.960
say that mod of f double dash x will be less
than m for some m because it is continuous
00:06:21.960 --> 00:06:26.939
f double dash is continuous in the statement
as given in the statement itself that f double
00:06:26.939 --> 00:06:32.060
dash is continuous so f double there will
exist some m such that f double dash will
00:06:32.060 --> 00:06:40.120
be less than capital m so if we apply this
over here eta mod of a n and mod of cos of
00:06:40.120 --> 00:06:48.729
n is always less than equal to one mod of
cos of n x upon l is always less than equals
00:06:48.729 --> 00:06:56.449
to one so mod of a n will be less than strictly
less than one upon n pi integral minus l to
00:06:56.449 --> 00:07:08.250
plus l we can take mod inside mod of f double
dash x into cos n pi x by l into d x which
00:07:08.250 --> 00:07:18.919
is equal to one upon n pi it is less than
or it is less than mod of this it is less
00:07:18.919 --> 00:07:24.879
than mod of this not the entire expression
now when you take this expression inside when
00:07:24.879 --> 00:07:33.120
you take this inside and applying these inequalities
so this is nothing but less than minus l to
00:07:33.120 --> 00:07:40.330
plus l m into one d x which is nothing but
ok
00:07:40.330 --> 00:07:48.060
now this this into this is also there so it
is l upon sorry it is l upon n square pi square
00:07:48.060 --> 00:07:58.180
l upon n square pi square so it is nothing
but two m l upon n square pi square ok so
00:07:58.180 --> 00:08:04.419
a n will be less than this now similarly we
can obtain that mod of b n will be less than
00:08:04.419 --> 00:08:11.569
two similarly it will be less than this this
we can similarly obtain so so what about
00:08:11.569 --> 00:08:19.229
this series so what about this series a naught
by two plus summation n from one to infinity
00:08:19.229 --> 00:08:32.500
a n cos n pi x by l plus summation n from
one to infinity b n sin n pi x by l this mod
00:08:32.500 --> 00:08:45.060
will be less than mod a naught upon two plus
now this mod mod of a n in each for each n
00:08:45.060 --> 00:08:51.560
mod of cos n x is less than equal to one mod
of b n which is less than this and mod of
00:08:51.560 --> 00:08:59.540
sin n x by l which is less than equal to one
so what we will obtain finally this is
00:08:59.540 --> 00:09:08.350
four m by l upon pi square will come outside
two from here two from here this is four
00:09:08.350 --> 00:09:14.100
m l upon pi square and it is summation one
by n square that is one by one square plus
00:09:14.100 --> 00:09:25.660
one by two square plus one by three square
and so on ok because when you take the mod
00:09:25.660 --> 00:09:33.590
and when mod come inside in each in each term
then using these inequalities we will obtain
00:09:33.590 --> 00:09:40.400
this expression because it is mod of a n a
naught by two plus mod of a n is less than
00:09:40.400 --> 00:09:47.560
two m l by n square pi square and mod of this
is less than equal to one again this is less
00:09:47.560 --> 00:09:52.980
than equal to one we know that summation one
by n to power p is convergent when p is greater
00:09:52.980 --> 00:09:58.530
than one eta here p is two it is convergent
as this series is convergent and this series
00:09:58.530 --> 00:10:05.180
is convergent means inside the series also
convergent ok
00:10:05.180 --> 00:10:14.030
this series is convergent means because because
f x is always less than equal to mod of f
00:10:14.030 --> 00:10:25.970
x ok and if f x is convergent then this f
x is also convergent ok so this eta this series
00:10:25.970 --> 00:10:34.160
will be convergent so hence we can say that
if function is a periodic and first and second
00:10:34.160 --> 00:10:40.470
order partial first and second order derivatives
are continuous then this series is convergent
00:10:40.470 --> 00:10:45.660
this series of f x is convergent so here is
the this is the same proof which i have discussed
00:10:45.660 --> 00:10:49.621
here ok
now piecewise continuous function we have
00:10:49.621 --> 00:10:55.100
already discussed this thing in laplace transforms
also that piecewise continuous means that
00:10:55.100 --> 00:11:02.330
you divide interval a comma b into sub into
finite number of subintervals and if in each
00:11:02.330 --> 00:11:07.000
subinterval function is continuous we say
the function is piecewise continuous ok the
00:11:07.000 --> 00:11:11.990
same same thing we will define here suppose
the function is said to be piecewise continuous
00:11:11.990 --> 00:11:19.010
in minus l to plus l if function is defined
and continuous for all l except at finite
00:11:19.010 --> 00:11:26.510
number of points ok if a function at a point
say x naught is not continuous then left hand
00:11:26.510 --> 00:11:32.700
limit and right hand limit at that point exists
and finite ok that means discontinuity are
00:11:32.700 --> 00:11:40.060
of jump type ok now at end point of interval
that is at minus l and plus l the if it is
00:11:40.060 --> 00:11:45.760
minus l the right hand limit if it is plus
l then the left hand limit exist and they
00:11:45.760 --> 00:11:50.720
are finite so then we say the function is
if these three properties hold then we say
00:11:50.720 --> 00:11:56.860
that function is piecewise continuous in minus
l to plus l or in other words we can same
00:11:56.860 --> 00:12:02.520
revise the same definition that to divide
interval into finite number of subintervals
00:12:02.520 --> 00:12:07.270
and in each subinterval function is continuous
and if there is a discontinuity that must
00:12:07.270 --> 00:12:11.500
be of jump type ok
now if a function is piecewise continuous
00:12:11.500 --> 00:12:16.860
what is the convergence of condition for fourier
series a function f x can be expressed as
00:12:16.860 --> 00:12:22.630
fourier series this in the interval minus
l to plus l where a naught a n b n are constants
00:12:22.630 --> 00:12:29.120
provided function is periodic with period
two l single valued and finite function is
00:12:29.120 --> 00:12:33.710
piecewise continuous in minus l to plus l
piecewise continuous we already discussed
00:12:33.710 --> 00:12:39.990
what it is and f x has left hand derivative
and right hand derivative at each point in
00:12:39.990 --> 00:12:47.190
that interval so this is if function is piecewise
continuous if these properties holds then
00:12:47.190 --> 00:12:55.140
it is converges to this expression then
function f x will converge to this expression
00:12:55.140 --> 00:13:02.240
ok can be expressed in this way ok now we
have a theorem that if f x and f dash x that
00:13:02.240 --> 00:13:07.160
is the first derivative be piecewise continuous
function on the interval minus l to plus l
00:13:07.160 --> 00:13:13.130
then the fourier series converges fourier
series f x converges to f x at the point of
00:13:13.130 --> 00:13:18.490
continuity if there is if it is a continuous
point then f x will converge to suppose x
00:13:18.490 --> 00:13:24.130
naught is the point of conti x naught is the
point in minus l to plus l where it is continuous
00:13:24.130 --> 00:13:32.060
then f x will converge to f x naught ok but
if it is a point of discontinuity say x naught
00:13:32.060 --> 00:13:38.830
belongs to minus l to plus l then fourier
series will converge to the average of left
00:13:38.830 --> 00:13:43.960
hand limit and right hand limit at that point
will converge to the half of f x naught plus
00:13:43.960 --> 00:13:48.900
plus f x naught minus
now at both point both the end points of the
00:13:48.900 --> 00:13:53.160
interval minus l to plus l if you take minus
what is the value of the function at minus
00:13:53.160 --> 00:13:59.950
l what is the value of function at l then
the fourier series will converge to half of
00:13:59.950 --> 00:14:07.300
f of minus l plus that is the right hand deriv
right hand limit at x equal to minus l and
00:14:07.300 --> 00:14:12.770
f l minus that is the left hand limit at x
equal to l the average of these two will be
00:14:12.770 --> 00:14:18.440
the value of the function at the end points
of the interval minus l to plus l ok i am
00:14:18.440 --> 00:14:23.790
not discussing the proofs here ok proof of
this theorem so the proof can be obtained
00:14:23.790 --> 00:14:28.070
so these results we will used while solving
some problems
00:14:28.070 --> 00:14:33.890
now let us try this problem find the fourier
series expansion of the function this and
00:14:33.890 --> 00:14:40.040
deduce this expression ok so we will solve
the problem in the usual way as we did in
00:14:40.040 --> 00:14:47.720
earlier to express to find out the fourier
series expansion of this function so function
00:14:47.720 --> 00:14:52.800
is periodic ok we assume here function is
periodic of period two pi function is defined
00:14:52.800 --> 00:15:01.100
from minus pi to plus pi but function is periodic
that is period is two pi ok so let us find
00:15:01.100 --> 00:15:06.770
a naught first eta what will be a naught a
naught will be one by pi minus l to minus
00:15:06.770 --> 00:15:15.810
pi to plus pi f x d x by the definition so
it is one by pi now from minus pi to zero
00:15:15.810 --> 00:15:27.450
it is nothing but minus pi and from zero to
pi it is nothing but x so using this it is
00:15:27.450 --> 00:15:35.300
nothing but one by pi and it is minus pi zero
minus minus plus pi and it is plus pi square
00:15:35.300 --> 00:15:43.320
by two eta that will be nothing but minus
pi by two so that is a naught now if you find
00:15:43.320 --> 00:15:53.820
a n a n is nothing but one by pi integral
minus pi to plus pi f x cos n x d x because
00:15:53.820 --> 00:16:02.510
here l is pi here l is pi so instead of cos
n pi x by l if you take l equal to pi eta
00:16:02.510 --> 00:16:08.870
it will be nothing but cos n x d x ok now
it can be expressed as written as one upon
00:16:08.870 --> 00:16:17.630
pi integral minus pi to pi it is minus pi
to zero it is nothing but minus pi it is nothing
00:16:17.630 --> 00:16:29.530
but minus pi cos n x d x plus zero to pi it
is nothing but x x cos n x d x
00:16:29.530 --> 00:16:39.040
now it is one by pi minus pi will come out
it is sin n x upon n from minus pi to zero
00:16:39.040 --> 00:16:49.570
plus it is x sin n x upon n minus one into
integral of this that is minus minus plus
00:16:49.570 --> 00:17:01.040
cos n x upon n square from zero to pi so it
is nothing but one by pi it is minus pi now
00:17:01.040 --> 00:17:07.159
it is zero when x is zero it is pi it is zero
again when x is minus pi this term is zero
00:17:07.159 --> 00:17:14.290
plus again this term when x is pi and zero
it is zero eta it is zero now here it is one
00:17:14.290 --> 00:17:20.039
by n square times it is minus one k to the
power n minus one this is nothing but one
00:17:20.039 --> 00:17:28.980
upon pi square n minus one k to the power
n minus one so this is this will be a n ok
00:17:28.980 --> 00:17:36.420
now what is b n b n will be nothing but again
one by pi i break it here minus pi to zero
00:17:36.420 --> 00:17:49.390
it is minus pi sin n x d x plus zero to pi
it is nothing but x sin n x d x now when we
00:17:49.390 --> 00:17:57.770
take it it is one by pi it is minus pi it
is minus cos n x upon n you integrate it ok
00:17:57.770 --> 00:18:04.190
from minus pi to pi minus pi to zero plus
you apply by parts here eta we will obtain
00:18:04.190 --> 00:18:16.180
it is x minus cos n x upon n minus one into
minus sin n x upon n square so it is zero
00:18:16.180 --> 00:18:25.440
to pi so when you simplify this it is one
by pi minus pi minus minus plus cos zero is
00:18:25.440 --> 00:18:34.570
one minus it is minus one k to the power n
one by n will come out ok
00:18:34.570 --> 00:18:43.530
now here plus this term is zero when x is
pi and zero because of sin n x and this term
00:18:43.530 --> 00:18:51.211
is zero when x is zero only one term remaining
that is minus of pi by n into minus one k
00:18:51.211 --> 00:19:05.370
to the power n ok now this is minus this is
minus and now what we will obtain from here
00:19:05.370 --> 00:19:21.490
it is it is nothing but it is minus and here
it is also minus it is nothing but it is
00:19:21.490 --> 00:19:29.390
ok let us see it once again it is minus pi
it is minus cos n x upon n minus pi to zero
00:19:29.390 --> 00:19:37.160
eta when zero it is one when minus pi it is
minus one k to the power n it is plus ok now
00:19:37.160 --> 00:19:45.290
x is the integration of this is minus cos
n x upon n ok minus derivative of this
00:19:45.290 --> 00:19:51.490
and integral of this so that is minus sin
n x upon n square from zero to pi now when
00:19:51.490 --> 00:19:59.791
it is zero eta it is one it is this and when
it is pi it is minus pi upon n minus one k
00:19:59.791 --> 00:20:06.550
to the power n ok so it is nothing but
something like one upon n will come out pi
00:20:06.550 --> 00:20:15.710
will cancel and it is one minus it is one
it is minus of this and minus of this will
00:20:15.710 --> 00:20:25.900
add up eta it is two times so it will be minus
two times minus one k to the power n ok
00:20:25.900 --> 00:20:34.620
so this will come here so now we have a
naught a n and b n in our hand so what
00:20:34.620 --> 00:20:41.670
will be the fourier series now a naught is
minus pi by two so fourier series of f x will
00:20:41.670 --> 00:20:48.430
be nothing but a naught is minus pi by two
eta a naught by two plus summation n from
00:20:48.430 --> 00:20:59.330
one to infinity it is one upon n square
pi minus one k to the power n minus one plus
00:20:59.330 --> 00:21:13.040
into into cos n x plus summation n from one
to infinity one upon n into one minus two
00:21:13.040 --> 00:21:20.720
into minus one k to the power n into sin n
x so this is the fourier series of f x which
00:21:20.720 --> 00:21:29.110
we will obtain from here after simplifying
now this this will converge to this ok because
00:21:29.110 --> 00:21:35.370
function is we have seen that the conditions
of convergence of fourier series is piecewise
00:21:35.370 --> 00:21:44.800
continuous holds here ok now we want to deduce
this expression so it is one by one square
00:21:44.800 --> 00:21:51.240
now n square is only here so that means this
term is here but we dont want this term so
00:21:51.240 --> 00:21:58.400
we can put x equal to zero ok we put x equal
to zero so it is nothing but f zero is equals
00:21:58.400 --> 00:22:06.530
to minus pi by four plus summation n from
one to infinity one upon n square pi and minus
00:22:06.530 --> 00:22:14.220
one to power n minus one and cos zero is one
ok now what what will be f zero because it
00:22:14.220 --> 00:22:20.640
is not continuous at x equal to zero left
hand limit is minus pi and right hand limit
00:22:20.640 --> 00:22:32.130
is zero ok because what is f zero minus f
zero minus is nothing but minus pi and f zero
00:22:32.130 --> 00:22:39.940
plus is nothing but zero by the definition
of function so what will be f zero is the
00:22:39.940 --> 00:22:46.490
average of two one by two minus pi plus zero
which is nothing but minus pi by two
00:22:46.490 --> 00:22:55.640
so this is minus pi by two is equals to minus
pi by four plus one upon pi come outside when
00:22:55.640 --> 00:23:04.220
n is one eta it is minus one minus one minus
two minus two upon one square now when n is
00:23:04.220 --> 00:23:10.880
two it is zero when n is three it is again
minus two upon three square minus two upon
00:23:10.880 --> 00:23:17.550
five square and so on so when you simplify
this it is minus pi by two plus pi by four
00:23:17.550 --> 00:23:24.330
is equals to minus two upon pi one upon one
square plus one upon three square plus one
00:23:24.330 --> 00:23:31.170
upon five square and so on so it is minus
pi by two minus pi by four which is minus
00:23:31.170 --> 00:23:40.670
two by pi into this expression
so hence the value of this expression is nothing
00:23:40.670 --> 00:23:47.440
but pi square by eight pi square by eight
so hence we obtain the required result so
00:23:47.440 --> 00:23:52.630
in this way if the function is discontinuous
at a point what i want to illustrate in this
00:23:52.630 --> 00:23:57.120
example i want to restate that if you want
to find out the value of function at a point
00:23:57.120 --> 00:24:02.360
where it is discontinuous then to find out
the value of that point is nothing but the
00:24:02.360 --> 00:24:08.059
average of the left hand limit at that point
and right hand limit at that point ok you
00:24:08.059 --> 00:24:11.900
find out left hand limit and right hand limit
and the average of these two will be the value
00:24:11.900 --> 00:24:19.410
of function at that point eta all other things
are as usual we did earlier but the important
00:24:19.410 --> 00:24:24.000
point to note here is what how can you find
the value of function at the point of discontinuity
00:24:24.000 --> 00:24:29.420
ok if function is discontinuous
now it is the same same type of example eta
00:24:29.420 --> 00:24:35.370
we can easily solve this example also again
suppose you want to find out a naught so a
00:24:35.370 --> 00:24:43.580
naught is nothing but now it is from minus
pi to plus pi it is one by pi integral minus
00:24:43.580 --> 00:24:53.140
pi to plus pi f x d x so it is one by pi now
you can break it from minus pi to minus pi
00:24:53.140 --> 00:25:06.530
by two it is it is minus one from minus pi
by two pi by two it is zero and from pi by
00:25:06.530 --> 00:25:16.050
two to pi it is it is one so you can simplify
this it is a very simple thing you can very
00:25:16.050 --> 00:25:24.510
easily simplify it is minus pi by two minus
plus ok negative will come outside actually
00:25:24.510 --> 00:25:31.980
negative of minus pi by two minus minus plus
pi and it is zero it is pi minus pi by two
00:25:31.980 --> 00:25:41.850
and which is nothing but when you simplify
eta it is zero ok now what will be a n again
00:25:41.850 --> 00:25:47.380
a n you can compute a n will be one upon pi
again you will break the interval into three
00:25:47.380 --> 00:25:56.100
ok from minus pi by from minus pi to minus
pi by two it is minus one into cos n x from
00:25:56.100 --> 00:26:01.970
minus pi by two to pi by two it is zero no
need to calculate that this is zero plus integral
00:26:01.970 --> 00:26:12.200
pi by two to pi it is one into cos n x d x
now it is nothing but one upon pi negative
00:26:12.200 --> 00:26:25.770
of sin n x upon n from minus pi to minus pi
by two and plus sin n x upon n again from
00:26:25.770 --> 00:26:35.470
pi by two to pi so it is nothing but one upon
pi minus of sin minus one by n comes out and
00:26:35.470 --> 00:26:45.270
it is sin ok it is negative of negative of
sin n pi by two and sin pi is zero
00:26:45.270 --> 00:26:53.490
again here sin n pi is zero and it is nothing
but minus of one by n sin n pi by two which
00:26:53.490 --> 00:27:01.960
is nothing but again it is zero so it is zero
ok so that you can easily simplify similarly
00:27:01.960 --> 00:27:09.260
when you find b n what will be b n it is one
upon pi again minus pi to plus pi minus pi
00:27:09.260 --> 00:27:23.480
by two it is minus of sin n x d x and plus
pi by two to pi it is sin n x d x so when
00:27:23.480 --> 00:27:31.100
you simplify it is one upon pi it is minus
minus plus cos n x upon n from minus pi to
00:27:31.100 --> 00:27:41.230
minus pi by two and it is minus of cos n x
upon n from pi by two to pi so this value
00:27:41.230 --> 00:27:53.570
is nothing but one upon pi it is cos n pi
by two upon n one by n one by n comes out
00:27:53.570 --> 00:28:00.140
and minus minus one k to the power n minus
again minus one k to the power n and minus
00:28:00.140 --> 00:28:14.299
cos n pi by two is it ok one upon n pi will
come out and it is sin cos n pi by two minus
00:28:14.299 --> 00:28:21.770
of cos n pi minus one k to the power n negative
is here so it is nothing but minus one k to
00:28:21.770 --> 00:28:32.220
the power minus minus one k to the power n
and it is minus of cos n pi by two ok eta
00:28:32.220 --> 00:28:37.620
it is ok minus is outside so minus is outside
eta minus will be here something like this
00:28:37.620 --> 00:28:44.910
ok minus minus is outside eta it is nothing
but one upon n pi so it is two times two will
00:28:44.910 --> 00:28:54.960
come outside two times cos n pi by two minus
minus one k to the power n ok
00:28:54.960 --> 00:29:00.890
so only only we have in this expression a
naught a n are zero so for this function f
00:29:00.890 --> 00:29:07.950
x we have only the sine terms on the right
hand side because a naught and a n are zero
00:29:07.950 --> 00:29:13.260
ok eta we have on the right hand side only
sine terms so what will be f x f x will be
00:29:13.260 --> 00:29:20.240
nothing but summation n from one to infinity
two upon pi can come outside it is one
00:29:20.240 --> 00:29:31.920
upon n and it is cos n pi by two minus minus
one k to the power n into sin n x so this
00:29:31.920 --> 00:29:39.679
will be the expression of this function so
f at pi by two will be nothing but one by
00:29:39.679 --> 00:29:50.190
two f of pi by two minus plus f of pi by two
plus so it is nothing but one by two f of
00:29:50.190 --> 00:29:57.980
pi by two minus is zero and f of pi by two
plus is one eta it is nothing but one by two
00:29:57.980 --> 00:30:02.610
so we substitute x equal pi by two both the
sides on the left hand side we have one by
00:30:02.610 --> 00:30:08.370
two and the on the right hand side we have
two upon pi one minus one by three plus one
00:30:08.370 --> 00:30:13.440
by five and so on so value of this is nothing
but one by two ok
00:30:13.440 --> 00:30:20.640
so in this way if you want to find out fourier
series we first have to find out a naught
00:30:20.640 --> 00:30:29.260
a n and b n and if we want a series expression
value of the series expression and we have
00:30:29.260 --> 00:30:35.549
to substitute x as a point where function
is discontinuous then you first find f x naught
00:30:35.549 --> 00:30:41.130
minus f x naught plus and the average of these
two will give the value of function at that
00:30:41.130 --> 00:30:44.140
point so
thank you very much for this lecture