Welcome to the lecture series on differential
equations for under graduate students, today’s
lecture is on Convergence and sum of the Fourier
Series. Till now in Fourier series, we had
learn that for a periodic function, we can
write a Fourier series for that function,
we had also learn if the function is defined
only on a finite interval on the real line,
we can still extend this and write the either
cosine series or the sine series and we called
it the half range expansions.
In the plots of those Fourier series and the
functions, we had seen that, as with the partial
sums we are including more and more terms,
we are getting more closer to the function
with the series. These things are, how we
are seeing is that is why they are matching
and why they are not matching, here is that
now we are going to discuss this mathematical
property of this Fourier series, which we
are calling convergence and the sum of the
Fourier series.
For the convergence, as it is very easy to
find out the Fourier series, given any periodic
function, we can find out the coefficients
using the Euler’s formula or we can extend
it to the even extension and, or odd extension
and get the sine and cosine series, again
using the Euler’s formula I found easy.
But, proving that the convergence or that
why it is matching that is a little bit more
difficult and beyond the scope of this first
level course. So, I will just quote here one
result about this convergence of the series,
which is due to ((Refer Time: 02:13)) this
result says...
Let, f x be a 2 pi periodic function, which
is twice differentiable with f x, f dash x
and f double dash x are piecewise continuous
on the interval, minus pi to plus pi. Then
for every x, in the interval minus pi to the
plus pi, the Fourier series, a naught plus
summation n is running from one to infinity
a n cos n x plus b n sin n x, with coefficients
a naught as one upon 2 pi minus pi to plus
pi f x d x, an as 1 upon pi minus pi to plus
pi f x cos n x d x and b n as 1 upon pi minus
pi to plus pi f x sin n x d x, for all n 1
2 3 and so on, is convergent.
Moreover, it is sum is f x except at a point
x naught at which f x is discontinuous, because
we had assumed that f x is piecewise continuous
and the sum of the series is average of left
hand and right hand limit of f x at x naught.
That is, the series a naught plus summation
n is running from 1 to infinity, an cosine
n x plus b n sin n x is equal to, would be
seen as f x, when at all those points x where
the function is continuous.
And at x naught where it is discontinuous,
at that point it would be equal to half of
f x minus f x naught minus plus f x naught
plus that is, the left hand limit at x naught
and right hand limit at x naught, you take
the average and it would be equal to that.
Proof of course, as I said we will not be
able to do here, but we can see this result
from our examples which we have done for the
Fourier series. So, let us verify these results
with our examples.
Now, let us see, we have done one example
where our function was minus a in the interval
minus pi to 0 and plus a in the interval 0
to pi and the function was periodic with the
period 2 pi that is f x plus 2 pi is same
as f x for all x, do you remember that, in
the last lectures we had find out, the Fourier
series of this as 4 a upon pi, sin x plus
1 by 3 sin 3 x plus 1 by 5 sine 5 x and so
on.
We had seen that, at x is equal to 0 if I
just see here in this particular example,
the function is discontinuous at x is equal
to 0, from the left hand, if I try to reach
to the a 0, I would get the value as minus
a, while from the right hand side, if I try
to reach to the point 0, I would get the value
as plus a, which is not matching with the
value of the, and the function is not defined
at 0. So, this function is discontinuous at
0. Now, what is the average at this point,
that is f 0 minus, plus f 0 plus, a plus minus
a, that would be 0. Now, see the graph which
we have drawn that of the partial sums.
This was the first partial sum, that is where
we have taken only 4 a upon pi sin x, you
see here, that this is the function of, in
the minus pi to plus pi, minus pi to 0 it
is negative, of minus a, plus 0 to plus pi
it is, plus a. And the Fourier series, if
you see is, that is, this is how we are going
on now, at the point 0. Let us see here, of
course, the same thing because it is 2 pi
periodic, so you will be getting is, minus
pi plus pi and all those things as such, see
here at 0, average is 0.
Now, this Fourier series we are saying is,
that is at the point 0 we are getting it 0,
while as, at a points here we are seeing is
at some points we are matching, now you see
this is my first partial sum, it is not the
whole Fourier series. Let us move to the second
partial sum, what we are seeing is that, as
n is increasing, now this is that n is equal
to 1 we are having so much deviation from
the actual values, let us move to the second
partial sum.
Now you see, we have reached here, now the
sum is, let us concentrate with minus pi to
plus pi, we are here, we are seeing is that
in between these points have been less, from
the previous one, you see in the previous
slide ((Refer Time: 07:21)) we are having
so many points in between. Now, we are having
less points in between and at 0, it is 0,
the function is not acting over there because
here we are reaching to 0, that minus a , when
we are reaching to, from this side to 0 it
is plus a, plus 1, we are matching it certain
more points over there. Now, if I go with
the third partial sum, what will happen ((Refer
Time: 07:49)).
This is your third partial sum, now we are,
at more points where we are, making that we
are equal to the function, at 0 we are all
the time matching. Now, you see in the third
partial sum, the points in between all the
points they have been disappeared because
the function is not anywhere in between this,
at 0 we have not defining the function. Then,
if I move to the.
Some, five or six terms here I have taken
because I cannot try it for very large number
of this thing, some six or seven terms has
been taken over here, now you see, we are
matching at many points. So, what we are saying
is, the function is continuous in the interval
0 to pi, the function is continuous in the
interval minus pi to 0, the sum of the Fourier
series, that is the partial sums which I am
talking over here they are matching with the
points.
At 0, where it is discontinuous, the value
that the Fourier series, we having this is
the red points are telling as the point on
the Fourier series and this whites are the
function values, this is 0. So, we are seeing
is that, as n is increasing at a continuous
points, the sum of the Fourier series is equal
to the I ((Refer Time: 09:21)). Here, we could
not say as actually equal to, but it is reaching
towards the equal to the value of the function,
while is, when it is continuous it is coming
at the average of these two limit points.
Another example, you remember that we have
done this example also, minus pi to plus pi
the function is x, again periodic with period
2 pi, f x plus 2 pi is equal to f 2 x, if
you do remember the Fourier series of this
function was two times sin x minus sin 2 x
by 2 plus sine 3 x by 3 and so on, nth term
is minus 1 to the power n plus 1 sin n x upon
n and so on. Now, let us see it is partial
sum and the graph of the function, so first
term 2 sin x, second term 2 sin x minus sin
2 x by 2 and so on.
Now, with first partial sum, this is what
is our function, from minus pi to plus pi
it is x, then it is periodic. So, when we
are reaching from minus pi to plus pi this
is the function, so when I take it again this
one, then it should start from here, this
point which we are having is this line, this
line I had purposefully put it, so that you
could see is, that is here is this function
is not in between here, this is not the function
value, there is no point on this line. You
see all the white points we are having over
here, this just I am this trying to show you
that is how you see now in the Fourier series,
this first partial sum.
We are moving towards now, at one point only
I think is we have, matching with the function
and then we are not matching with the function,
we are matching with the function at the two
points say maximum and at 0, the function
is discontinuous at the limit points that
is, at minus pi to plus pi. So, when this
Fourier series is changing, we are finding
it out from here to here, what is the average,
average is, plus pi plus minus pi that is
0.
You say is, that is at this end points the
value is 0, now here is one point, we are
getting some points in between also, this
is the Fourier series is not going to up to
here, it is moving from here because after
that it is changing so, because when we draw
the graph it would just take a curve like
this one, this is first partial sum. Let us
take the second partial sum that is, first
two terms.
Now you see, we have matched here more than
those two three points and the first one we
were matching only one point here, one point
here, one point here. Now you see, we have
matched a little bit more points over there,
you see is, that is more overlapping with
the function, again the points of discontinuity
or this end points. So, here also, now you
see is, that is at last one we are having
more points over here, now we are having less
points over here and similarly over here,
again. Because, when we are drawing this function,
so its curving also we are not again reaching
towards this ((Refer Time: 12:51)). Now, come
to the third partial sum.
Now you see, we are matching little bit more,
the function is overlapping a quite nicely
over here, the only thing is that is we are
having is, that you just come again discontinuity
points, here the behaviour is different. We
are not still moving till the end because
of this discontinuity and at this point it
is making this one and because of that only,
because we are just talking about first three
terms only. So, we are drifting from here
itself.
Now, let us move to the graph with some eight
or nine terms, it is not more than 10, but
still you can see, how we are overlapping
with the function, this white beads and red
beads. You just see is, that is how at the
points where the function is continuous, our
Fourier series is matching with this, some
of the Fourier series that is the partial
sum, that is because here it is not complete
sum, it is partial sum.
We are matching with the sum, now you see,
we are reaching towards end points as well
and now in this between I do have only one
point of the Fourier series which says is
this average of the right hand limit and the
left hand limit that point is 0, we are not
having any other one, that is what we say
is that is the at the point of continuous,
all those points the Fourier series sum would
be equal to the value of the function but
at the point of discontinuity, it would be
equal to the average of the limit points.
Let us see one more example because those
two examples were of the sin series.
Now, here I do have this example 0 from minus
2 to 0 and x from 0 to 2, now you see, this
is not periodic with the 2 pi, this is periodic
with minus l to plus l and my l is here, 2
l is 4 periodic with f x plus 4 is equal to
f x, this function, this examples also we
have done in our last lectures. And we do
know that from there, we have drawn that the
Fourier series is half plus summation n is
running from 1 to infinity 2 by n, minus 1
to the power n minus 1 upon n square pi cos
n pi x by 2 Plus minus 1 to the power n plus
1 sin, sin n pi x by 2 or if I write it more
elaborately.
I would get half minus 2 upon pi whole square
outside and cos pi x upon 2 square, cos 2
pi x upon 4 square so on, cos n pi x upon
2 n square and then the sin terms 2 upon pi
times sin pi x by 2 upon 1 minus sin pi x
by 2 plus sine 3 pi x by 2 and so on. The
nth term over in this b n series, would be
minus 1 to the power n plus 1 sin n x pi by
2 over n. Now, here what we were the partial
sum, when we were talking about only sin series
or cosine series, we are taking only first
term, second term and so on.
Now, here when I would be defining the partial
sum or be talking about the n is equal to
1, n is equal to 1 means is, that is I would
be taking both the terms, this cosine term
as well as the sin term. So, my first partial
sum would be, half minus 2 upon pi whole square
cos pi x upon 2 square plus 2 upon pi sin
pi x by 2 upon 1 and the second partial sum,
taking up to this point and this point, it
is not the else we have to take first all
these points and then go all these points.
We are just moving with n, n means that is
a n cos n pi x and b n sin n pi x by n. So,
and this is, a naught, so half this is there
Let us see the graph, so the first partial
sum, now the function is 0 from minus 2 to
0 and from 0 to 2, this is x. So, these white
beads are showing you the ((Refer Time: 17:23))
points, at certain discrete points, this is
the first partial sum, we do have that the
function is continuous, but still it takes
a step over here we are, this is first partial
sum. We are at drifting away at certain points
only over, here we are matching on the down
side and then at certain points here we are
matching. And because at end points, again
at the limit points the function is discontinuous,
so we are coming, drifting away from there
and we are not reaching towards the end of
this one.
Let us move to the second partial sum little
bit more better presentation, but you see,
here we have drifted away and then we have
drifted away over here also and then we are
matching towards these points. Now you see
here the points we are finding out that is
continuous all right, we are coming over here,
while at here we are coming this is the point
of discontinuity. Now, you see it is many
points in between, at a point of discontinuity
it should have been the point should have
been somewhere in the between that is, from
2 to 0, I should have got some where at a
1 some, somewhere the point should have been.
But this is only the second partial sum, we
are drifting it one, let us see some more.
Third partial sum, we are again coming up,
but because of your finding it out the function
is, as we are having here is a discontinuous
at a end points and here in the between also
that is, we are finding it out that it is
0 and from here it is actually looking as
continuous, but at this end point it becomes
discontinuous. You find it out that, even
when we were talking about only sin series
by the time of the third partial sum we were
very near to the convergence, we have able
to come that is a, the my function is matching
with the Fourier series. But here you are
seeing is, that is even at the third partial
sum, we are not having that nice kind of behaviour
of the Fourier series with respect to the
function. Let us move to the fourth one.
Again, we are drifting this side also to my,
as we are making is that, here it is coming
up very nicely. Here it is coming closer but
we are drifting it away, but it is simply
saying is that, here with the smaller number
of n s, we will not be able to get that much
nice behaviour of the Fourier series with
respect to the function. Or here for, showing
that the Fourier series is matching we require,
more n or this convergence requires a large
number of n. I think is, we do have here till
fourth one only.
So, you go till some twenty terms or something,
only then we would be able to see this behaviour,
but theoretically it says is, that for large
n, as n approaches to infinity. You would
get that the sum of the Fourier series that
all continuous points, would be equal to the
function and at the point of discontinuity,
it should be with the average of the two limit
points. So, with certain simple functions
we could see it very nicely, but here it is
example I have taken that you could see is
not happening very orderly also, that is still
fourth partial sum we are not near to the,
our result.
Now, one more terminology for this thing,
that for the Fourier series of any periodic
function f x, which we are writing as a naught
summation n is running from 1 to infinity
an cosine n pi x over l plus b n sin n pi
x over l. This term, capital A n which is
nothing but the square root of the sum of
a squares of the coefficients a n and b n.
This is called the amplitude of the nth harmonic,
that what it says is, a in all those things
what we have got is, that is the nth term,
nth term means not a naught, nth term means
is that is, if I take n is equal to 3, a3
cos3 pi x by l plus b 3 sin 3 pi by l x.
This third term, this would be called third
harmonic. Because this is a periodic series
we do know that, cosine function and sine
function both are periodic. So, they would
have some loop kind of things and that is
called the harmonic and when this is called
the amplitude of the harmonic.
Now, let us see an application to the force
oscillation, we have done such a Fourier series
that is, where we are applying it in the differential
equation. we have remembering that is, what
we mean by the force oscillation; that means,
a differential equation is governing my double
dash plus c y dash plus k y is equal to r
t, where m is the mass, c is the damping coefficient,
k is the coefficient of the spring and r t
is that your input function.
You remember, that we have already solve this
kind of equations and there we have done this
equation where this r t was of some special
form. Let us, just try one example which we
have done, that spring mass system with governing
force as y double dash plus 0.02 y dash plus
25 y is equal to r t, where your damping coefficient
is point 0.02, a spring constant is 25 and
the mass is a 1 unit. This r t, if you do
remember, we have, now in this example, now
my r t is this, input function is t plus pi
by 2 in a interval minus pi to 0 and minus
t plus pi by 2 in the interval 0 to pi.
Moreover, this function is periodic, that
is r t plus 2 pi is r t for all t, if you
do remember we have done, that when r t is
of the form that it is cos t or sin t or it
is a some function, e to the power t x kind
of things. Now, how do we find out the solution
of this equation, let us just try, see what
is my function r t?
This is, t plus minus pi by 2 plus t 2 this
one. So, we are getting is at 0 we are getting
is plus pi by 2 at minus pi by 2 would be
getting it as minus pi by 2 again at plus
pi we would be getting as a minus pi by 2.
If you do remember, we have done one function
t.
In the interval minus pi to 0 and minus t
in the interval 0 to pi and my r t is nothing
but, pi by 2 plus f t. Now, you remember that
we have done one result in the Fourier series
which says is that, if we do have the two
functions, periodic functions for which the
Fourier series can be written, then the Fourier
series of the sum function is nothing but
the sum of those Fourier series. So, now here
we see is, this is the constant one, this
kind of examples we have done, so we just
want the Fourier series for this function,
f t.
Let us find it out, see this function f t,
we are getting is f of minus t would be same
as minus of, would be same as f of t, that
says is that, it is an even function. So,
we would get only the cosine series; that
means, all the coefficients b n would be 0,
what will be now a naught, 1 upon 2 pi integral
minus pi to plus pi f x d x, f x here is or
this, if you replace this t by x is a x in
the minus pi to 0 and minus x in 0 to pi.
So, just break in to the two parts, get this
integral solved, here it would be x square
by 2, this integral and you just put the values,
minus pi to 0 and 0 to pi, you get the answer
as pi by 2. So, this is your a naught.
a n, 1 upon pi, minus pi to plus pi f x cos
n x d x, f x again we are breaking into two
parts minus pi to 0 it is x and 0 to pi it
is minus x, this is x cos n x. So, we will
just go with this integration by parts, the
first integral would be minus 2 upon pi because
the function is now x is odd and this cosine
x is even, we are getting is that, minus 2
upon pi 0 to pi x cos n x d x, which is minus
2 upon pi x by n that is again integration
by part sin n x evaluated 0 to pi minus 1
upon n, 0 to pi sin n x d x, which is minus
2 upon n square pi cos n x, 0 to pi.
Evaluation and we get finally that, minus
2 upon n square pi, 1 plus, minus 1 to the
power n which we would be, that is for n 1
3 and 5, it would be 4 upon n square pi for
n is equal to 2 4 6 it would be 0, now that
is what the Fourier series.
We have got, r t as 4 upon pi, summation n
is running from 1 to infinity, 1 upon n square
cos n t, which is 4 upon pi cos t plus 1 upon
3 square cos 3 t and so on. Now, solving the
differential equation, you come to this, modify
our differential equation, y double dash plus
0.02 y dash plus 25 y is equal to 4 upon n
square pi cos n t, that is nth term we have
written from here, our original differential
equation was with r t, rather than solving
it with r t.
Now, we are solving it with this nth term,
that says is since what we had got just now
that is from the convergence of the Fourier
series, that the value of the function at
continuous points is same as the sum of this
series. So, now we are using that idea, where
just solving at nth point, that is nth term,
now if I solve it, now this is of a special
form, we do know how to get the particular
solution or that if you do remember we have
called it a steady state solution.
The particular solution will be of the form,
A n cos n t plus B n sin n t, how to get this
coefficients A n B n, we just substitute y
n dash y n double dash in this given equation
and equate it to the right hand side and find
it out, so usual method, y n dash would be
minus n times a n sin n t minus b n cos n
t and y n double dash would be minus n square,
a n cos n t plus b n sin n t.
Substitute it in given differential equation,
we would get minus n square A n cos n t plus
B n sin n t minus 0.02 c times, that so c
is actually 0.02 A n sin n t minus B n cos
n t plus 25 A n cos n t plus b n sin n t,
this is should be equal to 4 upon n square
pi cos n t. Now, simplify the things, we do
get, 25 minus n square times a n plus 0.02
n b n cos n t plus 25 minus n square B n minus
0.02 n A n sin n, is same as, it should be
equal to 4 upon n square pi cos n t. Now,
equate the coefficients of cosine n t and
sin n t on both the sides, we would get the
two equations, one is 25 minus n square A
n plus 0.02 n B n is equal to 4 upon n square
pi, the coefficient of cosine n t have been
equated. Second term is that, coefficient
of sin n t, 25 minus n square b n minus 0.02
n A n, right hand side it is 0. So, it should
be 0, now these are 2 linear equations solve
them.
We get the solution as, A n as 4 upon n square
pi, 25 minus n square upon 25 minus n square
whole square plus 0.02 n whole square and
B n as 0.08 upon n pi into 25 minus n square
whole square plus 0.02 n square. Now, we see
the denominator in the both the A n and B
n same, lets substitute it as D, then what
we could write A n as, 4 into 25 minus n square
upon n square pi D and B n as 0.08 upon n
pi D.
So, we get the steady state solution for this
modified differential equation as, or actual
equation that is for this modified one we
would get is that the y p is, we are getting
where this coefficient is A n and B n are
been given by these values and the solution
would be, A n cos n t and B n sin n t. Now,
since we are talking about the linear equations
and the linear equations again we do know
that the solution of linear equations if their
two solutions for linear equations then, y
1 plus y 2 were also be a solution of the
linear equation.
Now, what we are having is that, we said is
that rather than that function f r t, we had
substituted the nth term of the Fourier series,
corresponding Fourier series into the term.
So, our what we are getting is that, Fourier
series we are getting the terms with 1 3 5
and so on, all the even terms were 0. So,
now, for our original differential equation,
the particular solution or what is called
the steady state solution, I could add up
as y 1 plus y 3 plus y 5 plus y n, where my
y n is nothing but, A n cosine n t plus B
n sin n t, with A n as this and B n as this.
So, we will get the amplitude as, for the
y n that is the solution part for this differential,
modified differential equation, the y n is
A n cos n t plus B n sin n t, amplitude for
this would be, A n square plus B n square,
if I do calculate you have seen that is in
the last one, that is A n was 4 upon n pi,
n square pi D and then some terms for 25 minus
n square, whole square and so on, and then
I had added up the second B n square, what
we do get if you see, this term, this is again
your D.
So, what we have got, 4 square D upon n square
pi square D square, that could be just simplified
as 4 square upon n square pi square, whole
square D. So, what will be the amplitude,
square root of A n square plus B n square
that is C n and now we are calling this amplitude
as C n, for this nth differential equation
you could call, as a square root of A n square
plus B n square that would be 4 upon n square
pi, square root D. Now, put the values for
different values of n, so that, we could see
that what would be the amplitude of the different
solutions.
A n is 4 into 25 minus n square upon n square
pi D, B n is 0.08 upon n pi D and C n we have
got as 4 upon n square pi root D. Let us evaluate
it for certain values of n, so here, I have
done some calculations, I am showing you here,
you can do it this calculations by yourself
as well. Let us see here, If n is 1 that is
if I am having only the first term, then D
n is this much, A n is this much, B n is this
much and accordingly C n is this value, just
go and see in this table.
We are getting our D n, that is D n means
is the D corresponding to this nth partial
one, that is we are getting 576, 256 then
0.01 then again we are coming from large values
then again we are going to the large value,
what we are finding it out, at the n is equal
to 5, this D n is very small, making it very
small here because of this approximation what
I am getting is, that A n is approximated
as 0, B n is some value, so C n is contribution
of the, in the C n is only of B n, that you
see is that C n is same as B n.
But of course, they would be little bit contribution,
it depends upon that is how much calculation
you could do, but whereas, what a final conclusions
you can do. Now, what it says is, that in
the solution of the differential equation
or the steady state solution, which are we
were having y p as y 1 plus y 2 plus y 3 and
so on, this term y 5 that is ruling because
of this one. Let us see that is what, I am
saying is and how it is working.
Here in the function and y 5, if you do remember
this was my function r t and if I just make
this graph y 5, you are getting this graph.
Now, let us come to this function and y t,
now my y t is actually y 1 plus y 3 plus y
5 plus y 7 and I think is in the table, I
have taken till y 11, so here are also I think
this is the term till y 11, I have added it
up. Now, you see, all other terms are not
giving or they are not dominating the nature
of the solution, as this y 5 is dominating,
you see, this is the term till 11th term I
had seen and you see here, only y 5 with fifth
term or we could say is y 1, y 3 so y 5, so
the third term.
You see is this, whatever is this nature oscillation
which we are having here that is dominated
by this y 5, rather than with any other one.
So, this is very much dominating, this is
what we are seeing is that, amplitude is deciding
that which one is much better approximation
for that function. So, rather than using that
solution of all those once and solving it
and then seeing is again, you see here, the
convergence cannot be achieved in a small
number of ns, for achieving the convergence
you require a large once, this is your input
function. This is actually, this y as not
actually the Fourier series for your r t,
this is actually input r t and this is your
output y t, but this output is dominated by
one function only that was, y 5.
So, you see here, we are having 3 1 I think
is, not very much missing now, I could see
because you had seen this separately now,
here I am superimposing all the 3, this is
my input function that is very clear. The
output you are finding it out that, both the
lines are matching actually, the only point
of the differences is, here little bit which
if you could see it out, that is your this
pink colour line is coming as, your till here
and this another colour this one is, it is
more or less white one, this is coming till
here.
So, you see is that y 5 is actually dominating
because of the amplitude only, now we have
learnt this Fourier series, it was very easy
for any periodic function or even if it is
non periodic to get half a range extensions,
to get Fourier coefficients approximate it
as a Fourier series. Again we had learn, that
is with some partial sums or the Fourier series
sum is being equal to the value of the function
mathematically, practically also we had seen,
that for certain functions, very orderly with
this third or fourth term only we are able
to reach to the required place.
For some other function, this is takes little
bit longer, that is in the value has to be
some 20 or 25, but you do find it out that
is we are reaching over that function. Then
we had seen, in solution in the force oscillation
where we, where you are going to solve only
for functions which are of that form of cosine
or sin kind of functions, but where we said
is, if input function is any periodic function
rather than the sin or cosine, is still we
could give the output function and it is again
oscillating.
There we had also in our particular example
we had seen that it is, was not necessary
that whole, all the terms of the Fourier series
would be dominating, but only we would be
getting is that is one or two terms depending
upon that, this harmonics and this amplitude
we could find it out, that which term is dominating
and only with that we can go ahead with the
analysis. There is one more topic which I
want to cover over here that is called complex
Fourier series, is anything different over
here, no.
Complex Fourier series when we are saying,
it is simply I am using this, as it simplifies
our many calculations. So, you could say is
that, this complex Fourier series we are using
many times, where the calculations are much
simplified than the real Fourier series and
you will find it out it further classes that,
this is how the things we are doing it and
when we want the real Fourier series, we just
go with the real components. The logic or
basis behind this is, the complex exponential
function, do you remember it.
It is, if you do remember or not, e to the
power i n x, this is cos n x plus i sin n
x, this is again as called as Euler’s formula
and e to the power minus i n x as cos n x
minus i sin n x. Now, from here what we could
get, cos n x I can write as half of e to the
power i n x plus e to the power, minus i n
x and sine n x as 1 upon 2 i, e to the power
i n x minus e to the power, minus i n x, what
is this i, remember that we are talking about
the complex. So, this i is nothing but the
square root of minus 1 that, imaginary value.
Now, let us see the Fourier series for any
function, we are writing as, a naught plus
summation n is running from 1 to infinity
a n cos n x plus b n sin n x, now change this
cos n x and the sin n x with these, just now
obtain the results in the terms of complex
exponential functions, what I would get, substitute
it. As, a naught plus summation n is running
from 1 to infinity, a n cos n x have been
substituting e to the power i n x plus e to
the power minus i n x by 2 and with b n, again
here, by 2 is missing. So, it should be b
n by 2, e to the power i n x minus e to the
power minus i n x.
So, we are getting this series, now this is
the correct one, a naught summation n is running
from 1 to infinity, a n upon 2 e to the power
i n x plus e to the power minus i n x plus
b n upon 2 i e to the power i n x minus e
to the power minus i n x. Do little bit simplification,
I can write it as 1 upon 2, a n minus i b
n, e to the power i n x plus half times a
n plus i b n times e to the power minus i
n x. That is, I have collected the coefficients
of e to the power i n x and e to the power
minus i n x.
Rewrite it, as c naught plus summation n is
running from 1 to infinity, c n e to the power
i n x plus k n, e to the power minus i n x.
Obviously, c naught is nothing but a naught,
c n is nothing but 1 by 2 a n minus i b n
and k n is nothing but 1 by 2 a n plus i b
n. So, now we have got a new Fourier series,
new series where my terms are not cosine x
and sin x rather than, I am having e to the
power i n x and e to the power minus i n x.
Now, the coefficients this c n, k n and c
naught, we have derived it from this original
a n, b n and a naught. So, we can use the
Euler’s formula over here.
Where, c naught is same as a naught, so it
is same as 1 upon 2 pi, minus pi to plus pi
f x d x, c n would be nothing but as, is 1
upon 2 a n minus i b n. So, if I write the
Euler’s formula for a n and b n, I would
get it as 1 upon 2 pi, minus pi to plus pi
f x cos n x, that is for an and f x sin n
x, that is for b n. So, now here minus i b
n, so that is, how we have got f x into cos
n x minus i sin n x whole integral with minus
pi to plus pi into 1 upon 2 pi.
Similarly, we can now, if you do remember
e to the, this is nothing but, e to the power
minus i n x, then go for the k n, half a n
plus i b n. Again using the same kind of formula
I would get it as 1 upon 2 pi, minus pi to
plus pi, f x cosine n x plus i sin n x d x.
Now, I can write this as, e to the power plus
i n x, so I am getting is 1 upon 2 pi minus
pi to plus pi f x i n x d x. Now you see,
c naught I am getting f x d x, c n I am getting
f x e to the power minus i n x, this constant
1 upon 2 pi is same, the range of the integral
is same minus pi to plus pi, the difference
is here.
Then, when I am coming at k n, I am getting
the difference as e to the power i n x, now,
if I write k n as nothing but c of minus n,
that is, if I replace here n with minus n,
you will get that, my k of n is nothing but,
c of minus n and when I replace n is equal
to 0, I would get it as 1 and that would be
c naught. So, now you see what I have got.
I have got the Fourier series, minus infinity
to plus infinity, c n times e to the power
i n x, because k n is nothing but c of minus
n, c 0 is nothing but, c n ,where n is replaced
by 0. So, now you find it out that, this is
very compact form I can write and that is
what I said, is it, is easy in the calculation.
Where c n can be given as the formula, 1 upon
2 pi minus pi to plus pi f x, e to the power
i n x d x, for n, 0, plus minus 1, plus minus
2 and so on or you could write in another
manner that is, minus 3 minus 2 and so on,
0 1 2 and so on.
This is called the complex Fourier series
for function of, this is we have obtained
for function f period 2 pi. Now, on the similar
trend, we can write for any function of the
period of any 2 L, what I have to change is
to you, do you remember that is there it,
you use to be cosine n pi by x by L and so,
kind of things. So, I would get it, minus
infinity to plus infinity c n, i n pi x by
L where c n is nothing but 1upon 2 L integral
minus L to plus L, f x e to the power minus
i n x pi L, n pi x by L d x for n is taking
value 0, plus minus 1, plus minus 2 and so
on.
So, this is, now you see till now I was writing
the Fourier series which has containing such
a space and then you have to the three formulae’s
1 for a naught, 1 for a n and another for
b n. Now, we are writing only 1 formula this
is easy and even the calculation is easy,
I have to integrate only 1 integral and most
probably 2. Let us see, that is with the help
of an example.
Find the complex Fourier series for f x is
equal to x, from minus pi to plus pi and its
periodic with f x plus 2 pi is equal to f
x. Remember, we have already find out the
Fourier series for this function, let us see,
how we are doing it in the complex one and
is it really making our calculations easy.
So, let us see, we want a Fourier series of
the form, summation n is running from minus
infinity to plus infinity c n, i n x where
c n should be of the form 1 upon 2 pi, minus
infinity to plus infinity f x e to the power
minus i n x d x.
Now, calculate this one, if I do have here,
this n is equal to 0, I will not be having
the function and the function is x and if
I am having n naught 0, then I would be having
x times e to the power minus n x. We do know
this exponential function, that we can integrate
and all those things, but if there is some
multiple of the x, then we have to go with
the partial one. So, we will break this into
two parts, one is when n is 0, that is c naught,
it would 1 upon 2 pi minus pi to plus pi x
d x or this an odd function it should be 0.
Now, c n for all n not 0, 1 upon 2 pi minus
pi to plus pi, x times e to the power minus
i n x d x. Integrate by part, taking x as
the first function that is the differentiating
function and e to the power minus i n x as
the integrating function, I will get, 1 upon
2 pi, x upon minus i n, e to the power minus
i n x evaluated from minus pi to plus pi plus
1 upon i n minus pi to plus pi integral e
to the power minus i n x d x. The first evaluation,
when we keep x is equal to pi and x is equal
to minus pi, we would get it as 1 upon minus
i is the i.
So, i upon 2 n pi, pi e to the power i n minus
i n pi plus pi, e to the power plus i n pi
minus this integral 1 upon i square, n square
2 pi, e to the power minus i n x, evaluated
from minus pi to plus pi. Again, evaluate
it, we get it as i upon 2 n, this pi I am
taking inside e to the power i n x plus e
to the power minus i n x plus 1 upon 2 n square
pi, because 1 upon i square is minus 1, e
to the power minus i n x minus e to the power
minus i n x.
Again simplify, we get i upon n cos n pi,
minus i upon pi n square sin n pi, what the
simplification we have done, that is cos i
n pi plus, sorry e to the power in pi x plus
e to the power minus i n pi by 2, that is
cos n pi. Similarly, here this divided by
2 i would be, if I am dividing it by 2 i,
then I would have here minus i it would be,
sine nx, so now, you see we have got a complex
function or a complex series. Since, it is
cos n pi is, minus 1 to the power n and sin
n pi is 0 for all n, we are getting i upon
n minus 1 to the power n the coefficient c
n for n not 0.
So, you see, what is the complex Fourier series,
minus infinity to plus infinity i upon n,
minus 1 to the power n, e to the power i n
x, where n should not be or n should be not
0. So, if I expand it, it would be minus infinity
to minus 1, i upon n minus 1 to the power
n, i cos n x minus sin n x and e to the power
and from summation n is running from 1 to
infinity because n is equal to 0 is 0, minus
1 to n, power n upon n i cos n x minus sin
n x.
Finally I would be getting it as, n is running
from 1 to infinity, minus 1 upon 1 to the
power n upon n minus cosine n x minus sin
n x, which we could write like this one. Hence,
finally we would get, summation n is running
from 1 to infinity, two times minus 1 to the
power n plus 1 upon n, sin n x. If you do
remember, that, this was our Fourier, real
Fourier series which we had obtained in our
earlier example. So, here if a thing is only
that, calculations I do not have to go through
that, two kind of different integrals.
Here this, you could not have guess that is
very simple one because it was giving us only
sin series. But if it is giving both sin and
cosine terms then, the terms becomes real
easy and the calculations are easy, that is
why we are using this complex Fourier series
and from there, we get what is the real Fourier
series. That is the all in, this today’s
lecture about the Fourier series.
So, we had learnt the Fourier series, we had
learn that is what is the convergence of the
Fourier series and the terms, that the sum
of the Fourier series is equal to the value
of the function at a point of continuity and
at a point of discontinuity, the sum of the
Fourier series is equal to the average of
the limits. That is, left limit and right
limit, then we had learn one more thing, the
complex Fourier series, the complex Fourier
series is not anything else, only thing is,
that is using this complex Fourier series
many times the formulation becomes easy, the
calculations becomes easy and compact form
and from there we can come back to our real
Fourier series. So, that is all in this Fourier
series and today’s lecture.
Thank you.
