Hello, in continuation of the discussion that
we had on the last video, on the convergence
of Fourier series to the function F, when
F is a periodic differentiable function that
is piecewise, so basically that means function,
when the function F is piecewise differentiable
periodic function that means it's derivative
function that is F dash(x) is a piecewise
periodic function.
So with period L, if such a function F, what
is its Fourier series and it's convergence,
it converges to the function F itself whenever
X belongs to the period interval -L/2 to L/2,
so as of now we'll just, we're only talking
about the open interval -L/2 to L/2 and eventually
we discuss at the end, once we prove this
we will discuss the endpoints also, or the
points of discontinuity, so these things we
will see, so anyway so there's no point of
discontinuity because we are considering differentiable
function that means it's continuous function,
okay.
So we were discussing about the proof of this,
we started with this, we consider the series,
a partial sums of that series and we just
simplified and in between we define what is
DN, and
then when you substitute and you observe that
DN is this, DN of this function is actually
this,
simple sine function involving simple sine
functions and also we observe that it is an
even function and then it becomes this, partial
sum becomes this once you change of variable
in this integrand T-X = X dash will give you,
so this type of function, so this partial
sum will, only so
far we consider only partial sum of that series,
Fourier series and we end up having this expression.
So what we do now, we will just rewrite this,
so we'll rewrite and see that as N goes to
infinity this converges to F(x), so that's
what we are going to show, so we will show
as N goes to infinity this partial sums of
the series, Fourier series it converges to
F(x), let's see how we will do this, so I'm
just going to do some manipulation here, so
we'll write 1/L so it's from 0 to L/2, what
I do is I just subtract something from this
quantity so that I have to add, once you subtract
and you also have to add, so we are fixing
X, X we are fixing, and T is the integrating
variable, so we write F(x+t) - F(x+), X+ and
X- we have defined that means limit of F(t),
as T goes to X+, as T goes to X- that means
coming from right side or left side of X,
that limit is this F at X+ and X-, okay.
And this is what I subtracted plus, from this
also you can do the same, so X-T - F(x-) okay,
and then this divided by, and you have this
sine N+2, N+1/2 W naught T DT divided by sine
W naught/2T DT that's what it is, so see as
T, so T is between 0 to L/2, so as T goes
to 0 and I have the quantity and this is 0
and you have the top quantity that is the
numerator, this part is 0, so as T goes to
0, FX- FX+ +FX- so what happens as T goes
to 0, if you look at it as T goes to 0 this
quantity becomes F(x), basically two times
F(x) - F(x+) + F(x-), because this is continuous
function, because anyway is actually is 0,
if X is in between, if X belongs to - L/2
to L/2, if it is in the open interval, if
it is the endpoints we don't know really,
so if it is like this your function is like
this, if it's repeated from - L/2 to L/2,
so you see this endpoint here is that this
is the different values, okay, so two times
of, this is really not you don't know exactly
what this means, okay, so if X is this, if
X equal to, if you take -L/2 and this this
difference average value is here, so you cannot
say that, so in that case we don't know, so
it's not 0, okay, so you actually show that
is actually 0, it has to be 0 otherwise this
integral doesn't make sense, because denominator
we have 0, so numerator has to be 0.
So anyway so that observation, so just to
avoid that and we also know that, just remove
this here, we also know that these limit F(x+t)
- F(x+) divided by X+T -X + that is T, so
this limit T goes to 0 is nothing but, this
is anyway X side, X positive side minus that
limiting value, so there's something like
this here, so you take some value here this
is X+T, this is X+ so that will come from
a value of the function from this limit, so
this is actually F derivative at X+ at this
value so this is a derivative at X+, if it's
piecewise this may be different from and the
other side X-, that is need not be same because
what we construct is a piecewise a differentiable
function, okay, so this need not be same as
this, so this is what it is, so you take this,
so because we assume that is piecewise differentiable
function so this is finite.
Similarly we have T goes to 0, F(x-t) - F(x-)
divided by T again, X-T -X okay, so this is
also F dash at X-, this is also finite because
that's what we assume, so to get this quantity
we only divide, we need to divide with T,
so if we divide T we have to multiply T, so
you rewrite, there's a most elementary way
of doing, showing the convergence of this
Fourier series, so I write this F(x+t) - F(x+)
+ F(x-t) - F(x-) this I divide with T for
both the parts, and also I write T divided
by sine omega naught by 2 T, okay, into that
whatever remaining part N+1/2 omega naught
DT, so I just multiplied, anyway it's T is
from 0 to L/2, so T is nonzero so you can
always divide.
As T goes to 0 it doesn't mean that T is 0,
so T goes to 0 means T is nonzero and it's
approaching 0, so I can always divide L, divide
with T both sides so this quantity I call
something, the quantity here call it something
else so, so once I add and subtract I have
to, so here I subtracted so I have to add
so that quantity is whatever is left with
is -L 0 to L/2 I subtracted so I have to add
F(x+) + F(x-) into this quantity sine N+1/2
omega naught T/sine omega naught, omega naught/2T
DT this integral is addition so I have to
add here also, 1/L 0 to L/2 this quantity
says anywhere nothing to do with that integrant
variable so I write this outside F(x+) + F(x-)
times integral 0 to L/2 sine N+1/2 omega naught
T divided by sine omega naught T/2 DT, so
this quantity, this integral you can actually
show that is actually 1, okay.
So we look at this integral plus this integral,
so before we allow N goes to infinity, this
is actually our SN, I first showed that I
eventually I allow N goes to infinity, so
before to do that, before doing that we will
just take these 2 integrals, we will handle
these 2 integrals separately, so we will start
with the first this one, so one observe that,
observe that this integral 0 to L/2 sine N+1/2
omega naught T DT divided by sine omega naught
T/2, this value if you directly evaluate in
terms of sines it may be little tough, so
instead you write as sum, so this is simply,
so you just, because this is an even function,
because this is a sine function, this is a
sine function, so it's an even function, so
this is actually, because the integrand is
an even function you can write from -L/2 to
L/2 and this is going to be a sine same function,
okay, omega naught T DT/sine omega naught
T/2, and this is equal to 1/2 I integrate
from L/2 to L/2, so this quantity integrand
will just, this is actually DN(x) which I
use a primitive expression that is finite
sum, so that is partial sums -N to N, E power
-IK omega naught X, okay, omega naught T DT.
Now you can see that, so once you have this
exponential function which is running from
-L/2 to L/2 whenever K is nonzero, whenever
K is nonzero this integral value is actually
equal to 0, okay, so when this is the nonzero
only when K = 0, though which you have -N,
-N, -1 and so on you keep on doing up to 0,
0 is the only contribution and K = 0 that
is only contribution is left, and when K=
0 this is simply 1, other values so you should
write like this, you can take this sum outside
integral -L/2 to L/2 E power - IK omega naught
T DT.
So now you look at the each of this when K
is -N to N, so whenever this exponential function,
when K is nonzero this is going to be 0, this
integral is 0, so for that reason only contribution
will be when K = 0, so when K = 0 so that
is nothing but -L/2 to L/2, K = 0 that is
1 DT, so this is nothing but L/2, this is
what it is, so you have this, this will be
L/2 so you have finally this whole thing is
1/2 of the average value of, that is actually
1/2 of F(x+) + (x-) the average value.
And we also define what is Q, let me define
what is QN(x), it depends on N, so if you
write this, now let's say not QN so it's just
Q, Q(x) if I write it as that first integrand
whatever in these brackets, this brackets
F(x+t) - F(x+) + F(x-t) - F(x-)/T into T/sine
omega naught T/2 that's what if you define
like this then what happens to this integral,
so you can see that it's a well-defined function,
so Q(t) as rather, so let's write Q(t), so
X is fixed, when we fix X should use the function
of T it's well-defined for every, so as T
goes to 0 this quantity is 1, sine X/X type
and this quantity is anyway these derivatives
are exist as T goes to infinity, no other
T value as T between 0 to pi L/2 there is
no issue so it's well-defined function, so
QT is defined for every T is in 0 to L/2,
okay.
If you do that what happens to your SN? SN
is a QN, this is Q(t) that sine function,
so let's write what happens to your SN, SN(x)
is what do you have is 1/L 0 to L/2 Q(t) into
sine N+1/2 omega naught T DT plus the other
integral by the earlier calculation so this
becomes 1/2 of F(x+) + F(x-) okay, this is
what it is.
Now what happens to this? Now we look at this
integral as N goes to infinity, so we'll take
this limit, N goes to infinity so you have
this limit N goes to infinity and this is
anyway constant so that doesn't change with
the limit, so we only have to worry about
this N goes to infinity what happens to this
1/L 0 to L/2 Q(t) sine function sine N+1/2
omega naught T DT, okay, so this is equal
to, so I'll write, so again so Q(t), Q(t)
if you observe this T T goes, this is what
is this one, this becomes, what happens this
is odd function, if you see, replace T by
-T numerator is same, here this this cancel
so you have minus because of sine function,
sine of -T so minus comes out so this is an
odd function, so Q(t) is an odd function,
sine T is an odd function, so together this
is an even function, okay, so this integrand
is an even function so again like earlier
we can write a limit of 1/L rather so 2 /L
times -L/2 to L/2, Q(t) sine N+1/2 omega naught
T DT, so because it is an even function, so
integrant is even so you can write 2 times
that integral so you guys replace, you can
extend it to negative side, so this is equal
to limit N goes to infinity, so you can write
now 1 by, rather half of it right so this
is 1/2 times, so you have 1/4 times 2/L -L/2
to L/2, 2 2 cancels the same, okay, and you
have Q(t), now you can expand this sine function
so that is sine N omega naught T and cos of,
cos omega naught T/2 times sine N+1/2 omega
naught T +, and now Q(t) sine omega naught
T/2 of this function you have cos N+1/2 omega
naught T, so this DT, so this is what you
have so this implies.
So what happens so as limit N goes to infinity,
so what is this one? This is limit, N goes
to infinity 1/4 so this is exactly Fourier
coefficient of this function with sine and
cosine, okay, sorry I think it should be,
this is N, sine N, sine N omega naught, this
is cos I expand it right so you have cos N
omega naught T, so this is a Fourier coefficient
of, sine coefficient of, Fourier coefficient,
that is the coefficient of sine and omega
naught T for this function Q(t) cos omega
naught T/2, and we have seen by Riemann Lebesgue
Lemma if Q(t) is well-defined function, so
we'll start with whatever you have, F is piecewise
periodic function, so F is differentiable
function, and F is, that means F is a continuous
function, piecewise continuous, so it is integrable
function, okay, and what you have seen is
earlier if F is square integrable function
or -L/2 to L/2 then its Fourier coefficient
is, whose Fourier coefficient is actually
is going to 0, okay, so I'll just show you
what it is what I'm going to use, so if you
just go back, so look at
this, if you look at this Bessel inequality,
so you look at the Bessel inequality and you
see this one, so if F is a periodic function,
periodic function which is piecewise differentiable
that means the derivative is, the derivative
exists that means this integral is finite,
so integral mode FX DX which is finite, and
once this is mode X is finite it is also square
integrable, F square, F into F is also piecewise
continuous function, F is piecewise continuous
function, F square is also piecewise continuous
function, that implies that is also integrable
function, okay, so that way what you have
is this quantity the right hand side this
integral is finite, once this integral is
finite this number series sigma CK square
that means that is finite.
So once you have this by this corollary which
is the Riemann Lebesgue Lemma the nth term
has to go to 0, okay, so in that sense if
you use this one, here for this function if
F is such a thing F square is also integrable
function, so if F is integrable, QN is also
integrable you can see
this one, so this will not really, see you
can see that this quantity, this first term
involving this and second term is involving
this, these are a derivative kind of things,
they are derivatives, these are F dash(x)
- X+ and F dash(x-) that is the difference
by T into, T divided by sine omega naught
T/2, so after when you take the limit T goes
to 0, this is what it is, so these are the
derivatives, these are the things so this
is a fixed quantity and you have T divided
by this, as T goes to 0 so that is where only
we have problem only when T = 0 this quantity
is going to 1 so it is a constant, so and
you have Q(t) that is well-defined for all
0 to L/2, even at 0 I don't have problems,
it's not bounded, okay, what it shows is that
the limit Q(t) only worried about T goes to
0, this is a finite number, because you look
at this limit, as a limit this goes to, as
T goes to 0, this is actually first term,
these two terms will become this, and the
other term is eventually becomes that itself
it's 1, so it's not 1, 2/omega naught, okay,
so F dash(x+) - F dash(x-) divided by 2/omega
naught, okay, so 2, omega naught is 2 pi/L,
so you have finally 2 2 goes L/pi, okay, so
this is going to be L/pi, so it's well-defined.
So as T goes to 0 this function Q(t) at T
= 0 it's a finite quantity and it is defined
for all values, so it's a piecewise, so it's
a piecewise continuous function implies Q(t)
is integrable function that means this integral
Q(t) is finite - L/2 to L/2, okay, that is
one.
Basically that you composite with, you multiply
with cosine or sine that is actually is a
piecewise continuous function is finite, and
once this is piecewise, Q is piecewise continuous
function it's multiplication, square is also
like that so this is nothing but this one,
so it is once
this is finite by Bessel's inequality and
its subsequent corollary that is Riemann Lebesgue
Lemma you can see that the Fourier coefficient
that CN CK there is for FK here this whole
quantity, this is your CK for this function
Q(t) into cos omega naught T/2, so this goes
to 0 so this into this goes to 0, this goes
to 0 + again 1/4 times, now 2/L into this
quantity for this function this is your Fourier
coefficient, this is -L/2 to L/2, so this
is DT + 2/L, so this if you make it two brackets,
so this integral is again a Fourier coefficient
say BN, this is AN and this is BN, so all
once CN goes to 0, AN BN also goes to 0 because
AN's are simply nothing but
CN and C-N divided by 2 and 2Y, okay, so because
of that each of this quantity goes to 0, so
by taking the limit so that is actually becoming
this limit SN, SN(x) as N goes to infinity
that is nothing but this Fourier series.
What happens? What is your Fourier series?
That's going to be minus infinity infinity
into F(x) that is CN, so you have CN E power
IN omega naught X which is actually this converges,
that is what happens, right, this limit is
a partial sum and N goes to infinity this
is what it is, and we have seen that this
goes to limit SN is what we have seen is,
if this quantity is going to 0 and finally
this becomes is only this quantity, it is
left is as N goes to infinity this integral
is going to be 0, this integral is 0 though
what I have shown just now, so what is left
is only this one, so this is actually equal
to 1/2 of F(x+) + F(x-) okay, so this is what
you have shown. So now I can remove this one
that is exactly what you want to show, so
where is this one? If X belongs to -L/2 to
L/2, if you choose only interior point of
that open interval -L/2 to L/2, okay.
And now see since F is continuous function
because you have chosen piecewise continuous
function, piecewise a differentiable function
so that means whose derivative exists at all
points, interior points, so F(x) is same as
F(x+) is same as F(x-) so that means 2 times
F at X value, F(x) is defined, so this is
exactly they were 2X, two times, 2 times FX
and half of it so there you get F(x) okay,
so it converges, this converges to F(x) whenever
X is in here.
Now the question is what happens at the endpoints,
if so you remember that you fixed X value
so X you fixed and they worked with this,
if X = -L/2 let us say, what happens you look
at again this expression the limit SN, if
now X = -L/2, okay, if X = -L/2, X is -L/2+T,
so where are you? So that's easy to work out,
similarly you can work out for L/2, let's
choose X = L/2, L/2 you have SN as this one,
so because I wrote SN in terms of 0 to L/2,
I have chosen X = L/2, if you want at X = -L/2
you rewrite, so when you split this into two
parts 0 to L/2, 0 to L/2, what you make is
you make this 0 to L/2 as from - L/2 to 0,
so that eventually you will end up getting
this SN as integral -L/2 to 0, and then you
can fix X equal to, the same expression will
get similar expression, but X = -L/2 you can
work similarly, here for this type of expression
for X equal to, argument is easier if X equal
to, if you choose X = L/2, if you choose X
= L/2, F(x+), F(L/2) + T that is you're going
out of this integral, once you go out of this
integral because it's a periodic, you're going
to, so where is this function value? Once
this function, so let me write this one, this
function F(x+t) L/2, L/2 is the endpoint,
L/2 + T because it is a periodic function
repetitive, this is actually a repetitive
so that means it goes back to, so you have
- L/2 to L/2, so whatever the value, so here
is actually same as a value here, because
it's
repeating, its repetitive, if it is like this
again it's like this, so what is this value?
This value is here which is same as this here,
okay. So this is actually kind of, you will
see that F(L/2 + 2) is nothing but F(-L/2
+ T), that's what you have, okay.
So you have this is -L/2 to L/2, L/2 + T is
nothing but -L/2 + T, that's what you have
here, okay, so this is same as this, so this
one - F of, now L/2 so L/2 you have chosen,
so you have this, this L/2 value, let us say
if it is like this L/2 value that you keep
it as it is, L/2+ right, where is that? Yeah
here, so this is the one, so if L/2+1 F(x+)
X is L/2+, so L/2 +1 means it's coming from,
so it's again the same argument if it is from
-L/2 to L/2 so you want this integral, so
this limit of the function that is same as
the limit of this function, because wherever
the values if you want to look at it they
are nothing but here, so as you approach here
you are approaching right side means you are
approaching here, because it is repetitive,
so this is same as F(-L/2), okay.
And then what happens to the other one? That
is going to be F(L/2 - T) that is same, anyway
inside -F(L/2) that is also inside, that limit
from this side so which is well-defined divided
by T, so what is this one? So this one, when
you write like this eventually so F(x+) is
nothing but F (-L/2) so other part, other
part is what you have is, if you do this one,
so when you add and
subtract it you have to add with F(-L/2) +
F(+L/2) so this anyway again the same argument,
this divided by T into T divided by sine,
when sine omega naught T/2 that whole as a
Q(t) and then you have here, this part will
be 1/L times so this is going to be half of
this, so you will have that one Q(t) + half
of, where Q now, wherever X is you replace
with - L/2, and the second part you have +L/2,
so with this argument if you do the same thing
SN(t), SN(x) that is also going to 0, SN(x)
is going to 0, first integral is going to
be 0, second integral is now converged to
this function, this value.
So when X = L/2, what you have is a limit
SN, as N goes to infinity is nothing but when
X is L/2 is actually equal to F(-L/2) + F(+L/2)/2,
half of the value, okay. If X = -L/2 you want
to look at it, you write this SN, not from
0 to L/2, you take it from minus, you replace
over T/-T and you rewrite from -L/2 to 0,
then you view it similar way, you can see
that SN(x) as N goes to infinity, in both
cases will be same as this, that you can take
it as an exercise, okay. It's the same way
so argument is same fix X and you look at
its periodic thing, so if this is your periodic
thing if you want to look at the values here
they are here because it's repetitive, if
you want to look at this side they are repeating
here, so if you want here they are repeating
here, so there's where is that, if you want
to look at here so it's somewhere here, exactly
half, okay, so
just add if you want wherever you want to
look at some L + some A, A+L, you add up to
L that is A+L, that is the value of L, okay,
so that's how if you look at it you'll see
that this is going to be this.
So the Fourier series that converges to CN
E power IN omega naught X, this is actually
converges to, if F is a piecewise differentiable
function F(x) if X belongs to -L/2 to L/2
open interval, and if it's an endpoints this
is going to be addition of, average value
of the endpoints, half of this if X is equal
to + or - L/2, so when this is our F(x), if
it is -L/2 to L/2, if your function is like
this if the value here, here both are same
then it is function at the endpoints, okay,
so this is what you have, so if F is same,
if F(-L/2) is same as F(L/2) then average
value becomes when at X = + or - L/2 this
Fourier series CN E power IN omega naught
X, N is from minus infinity infinity is F(x)
for every X belongs to minus, including the
points, okay, because if we include, so if
such that is the case, if this is the case
then we have this, this is the more general
one, otherwise we have only this one, so only
interior point you can always say converges
to the function F(x) because F is continuous,
here you're not sure, if your function is
like this value here, here is both are different,
so it converges at this point it converges
to the average value and here also it converges
to the average value, that's what it means,
okay.
So we'll see, we have used Bessel's inequality
and Riemann Lebesgue Lemma the corollary of
it, and we have shown this convergence of
the Fourier series to the function F itself
and it is piecewise differentiable function,
periodic piecewise differentiable function,
we will see the corollary that is this such,
if your Fourier coefficients are same then
you have unique Fourier series, okay, if you
can always define if it is a piecewise periodic
function, piecewise continuous a periodic
function if we choose you can construct your
Fourier coefficients in place you have a Fourier
series, so far we have not seen if it is a
piecewise continuous function we have not
seen the convergence of this Fourier series
to the function F, but if it is a piecewise
differentiable continuous, differentiable
function, piecewise differentiable periodic
function then you have just now seen that
the Fourier series, Fourier coefficients again
you can easily define, and using the Fourier
coefficients you can make a Fourier series
that actually converges a point wise, that
means you fix X in the Fourier series that
converges to as a number series, it converges
to F(x), if it is continuous for, because
anyway continuous function within the open
interval it converges to F(x) at the endpoints
it converges to the average value of the endpoints,
function value at endpoints, that is this
okay.
We will see it's a Fourier series is unique,
if your Fourier coefficients are uniqueness
theorem so we can write the uniqueness of
the Fourier series, so if I choose two functions
F(x) and G(x) be two piecewise differentiable
function, differentiable periodic functions
and with, you have two such piecewise periodic
differential functions, then you can define
Fourier coefficients for each of them, let's
call this FN and so it's earlier we have defined
CN, CN is F(x) into some integral, right integral
with exponential function that is your CN,
so I am calling it for corresponding to F,
I am calling it a FN, so FN with FN being
Fourier coefficients, with Fourier coefficients
let's write with Fourier complex, Fourier
coefficients FN is same as GN, this is true
for every N, okay.
So let this is the case, okay, Fourier coefficients
FN such that FN = GN for every N this is what
it have, then what happens? Then F(x) = G(x)
that means F(x) is nothing but we have seen
just now seen from this theorem that F(x)
is of limits of Fourier series of F, and G(x)
is a Fourier series of G, and we are seeing
that Fourier series are also same if the Fourier
coefficients are same, that is what is the
meaning of this, so a proof is easy, if FN
= GN for every N then CN which is FN - GN
which is equal to 0, so some coefficients
are given which is 0 and we know that CN,
so what is the CN? CN is a Fourier coefficient
of, CN is a Fourier coefficient of F-G, okay,
if FN is a Fourier coefficient of F, so FN
is 1/L, what is the Fourier coefficient? 1/L
the complex Fourier coefficient which is 1/L
-L/2 to L/2 F(x) E power -IN omega naught
X DX.
Similarly GN, F,G is nothing but G,F so you
replace G F with this, these are the Fourier
coefficients GN FN, so wherever you have G
you put G instead of F okay, so FG so FGN
like that, if you do these are the coefficients,
so what happens to FN - GN? This is a Fourier
coefficient of -L/2 to L/2 for F-G(x) E power
-IN omega naught X DX, so a Fourier coefficient
of F-G, okay, so because F and G are continuous
function, piecewise differentiable functions
F-G is also piecewise differentiable function
so you have a, whose Fourier coefficients
are CN E
power IN omega naught X, N is from minus infinity
infinity should converge to by the theorem,
it has to converge to F-G which is equal to,
what happens to this? Which is right F(x)
- G(x), and we know that CN is 0, so that
means the whole thing is 0, so this implies
F(x) = G(x) for every X in -L/2 to L/2, that
is where F is continuous, okay, so we have
to write for every X wherever, if wherever
F and G both are continuous, okay, so that
means we have seen so the
piecewise differential function means let
us say wherever period functions in -L/2 to
L/2, okay, L/2 with Fourier coefficients for
every X, for all X at which F and G are continuous.
If they are piecewise differentiable functions,
and this means obviously it should be continuous
for all X in this, okay, in the open interval
that shows that this is a unique Fourier,
so you get a unique Fourier series once you
have unique Fourier coefficients this is one
corollary, one result
which immediately can say after this convergence
after this theorem and you can also say one
more result that if F is, if F dash(t), if
F dash(x) is piecewise continuous, then limit
N goes to infinity, N times Fourier coefficient
FF you call this FN, it's going to be 0, okay,
so what does it mean? That is FN of, FN behaves
like order of 1/N power 1 + alpha, alpha is
positive, right, otherwise this is not going
to be 0, this limit, right, N power alpha
then it N into FN will be order of 1/N power
alpha, alpha is positive so as N goes to infinity
that goes to 0, so this is true.
So how do we prove this one? This is from
Riemann lebesgue lemma, so if F(x) is piecewise
continuous that Fourier coefficient is, Fourier
coefficient of its derivative DF/DX F dash
N Fourier coefficient is defined as 1/L -L/2
to L/2 F dash(x) E power I omega IN omega
naught X DX, you do the integration by parts,
what you gain is, E power so what you get
F(x) integration by parts if we do IN omega
naught X, okay, from -L/2 to L/2 1/L - integral
1/L -L/2 to L/2 F(x) you do the integration
here, so you'll get IN omega naught into E
power itself IN omega naught X DX, so 1/L
of this is a Fourier coefficient so you'll
get and this one will be 0, so both will be
0 you can see, okay, this will not be contributing
so what you get is -IN omega, omega naught
is 2 pi/L, 2 pi/L square it's a constant,
anyway this is your FN, so you can easily
see that, this quantity NFN by Riemann Lebesgue
lemma we know that if it is a Fourier coefficient
of a function F dash is, which is piecewise
continuous implies is Fourier coefficients
define, and implies it is integrable and it
is square integrable implies it is, whose
Fourier coefficient, Bessel inequality is
true so from that from the Bessel inequality
you have a Riemann Lebesgue Lemma that shows
that this is going to be 0.
What is this FN dash? This is nothing but
N times FN into minus some constant, as N
goes to infinity equal to 0 so that constant
is if you write -IN 2/L square = -IN 2/L square
this constant times this, because of this
is 0 and this has to be 0, NFN as N goes to
infinity so this means this has to go to 0,
this limit NFN, N goes to infinity this is
0, that shows that so alpha is actually 1,
so is equal to 1, okay, strictly speaking
this is like order of 1/N square that is exactly
what you get in the corollary, okay.
So we can go on doing like this if F dash
itself is a piecewise differentiable function
that means F double dash is piecewise continuous
implies N square FN is 0 and so on, okay,
you can go on doing like this recursively.
So what is that we have done so far? So if
F is a piecewise, a differentiable periodic
function the Fourier series converges to the
function itself because F is continuous in
the interval - L/2 to L/2, open interval this
converges to F(x) if the endpoints are same
the Fourier series actually converges to the
function F itself F(x) when X is the endpoints,
if endpoints are different average values
of the function value at the endpoints is
the limiting value of the Fourier series at
the endpoints, okay, this is what we have
seen, this so far we have used only differentiability
of the function F, piecewise differentiability,
piecewise differentiable function F, okay,
so these are actually sufficient conditions
for the Fourier series convergence to the
function here, so we'll define what is a delta
function and we will prove the sufficient
conditions in the next video. Thank you very
much.
