Welcome back, in this video we will see sufficient
conditions on the function F(x) so that it's
Fourier series converges point-wise to the
function F(x), okay. So there are many sufficient
conditions, so we'll start with one sufficient
condition on the signal, that is a piecewise
differentiable function, that means it is
a differentiable function, F(x) is differentiable
that means F-(x) exists, and that F-(x) is
a piecewise continuous function.
So let's take, so if we have such a conditions
on the signal F(x) so we will just show that
the Fourier series composed of Fourier coefficients
and fundamental signals in terms of cosines
and sines so that Fourier series composed
of all this, use a Fourier series, we will
show that this series, this function series
converges to the function F(x) point-wise,
that means you fix X then the function series
becomes a number series, that number series
converges to the value of the function at
X, so to do this if you have such a function
if F(x) is piecewise differentiable function
then it has certain properties so we'll do
that properties first, so let's consider let
F(x) be 
piecewise continuous period.
So let's start with the piecewise continuous
function, so one of the sufficient conditions
we have is a piecewise differentiable function,
so if it is piecewise differentiable function
implies it's a continuous function, so for
such a function piecewise continuous, actually
periodic function, so we only consider periodic
functions, periodic function with period,
period let us say L then, then what you have
is this result, so this is called Bessel inequality,
N is from minus infinity to infinity, this
is your CN complex, Fourier coefficients,
this is less than or equal to 1/L - L/2 to
L/2 absolute value, square of the absolute
value of this signal with DX, this is what
you have. So this is called Bessel's inequality.
So we'll first prove this as a theorem, let's
say let's call this theorem and if you proof
us, will first prove this and use this and
the sufficient conditions to show the convergence
of Fourier series of F.
So we'll start with, so you have a, once you
have this CN's as a complex Fourier coefficients,
so you have the Fourier series so the Fourier
series is, Fourier series, this N is from
minus infinity to infinity, and you have CN,
E power IN omega naught X, omega naught is
2 pi/L because it's a, signal is of period
L, so this is your Fourier series so let's
consider the partial sum of this, partial
sums let's represent SN, SN(x) you take the
partial sum of the series there is you take
it from - N to N, it depends on N so if I
change this index as K so it's running from
minus N to N, CK E power IK W naught X, so
this is your partial sum.
So once you assume this is your partial sum
CN is 1/L integral -L/2 to L/2, F(x) E power
- IN omega naught X DX, okay so if you integrate,
integral - L/2 to L/2, F(x) - SN(x) this you
multiply with E power - IN omega naught X
DX, what is this value? This value is actually,
this is integral - L/2 to L/2, you expand
it, so this is F(x) E power - IN omega naught
X DX - integral -L/2 to L/2, SN(x) this you
write it as K is from - N to N, CK E power
IK omega
naught into E power - IN omega naught X, this
with DX, so this is nothing but, this is your
CN and here this is your, so you can, this
is a finite sum you can take it out, K is
from -N to N CK this is the integral - L/2
to L/2, and what you are left with is E power
I K-N omega naught X DX, so what happens when
K and K is from minus N to N, and K is N only
when K equal to N or K - N is nonzero, and
KN - N is nonzero you can see that this quantity
is the same, so it is going to be 0.
So when it is done it's only when it is equal
to 0 so that is actually CN -, so let us choose
this is equal to 1, for example this is always
number so let us take as 1, what happens to
- L/2 to L/2? E power I omega naught X DX,
what is omega naught? Omega naught is 2 pi
by period is L, so this is equal to I omega
naught E power I omega naught X for which
you take this L/2 to L/2 limits you apply
you get E power I times, omega naught is 2
pi/L and X is at L/2, so 2 2 goes L by so
E power I pi - E power, again - I pi, so what
is this one? This is actually E power I pi
is, cos pi is -1 and here also minus off,
cos pi is - 1, so it's going to be + 1 which
is 0, if it's 2, if I take any N value here,
so if K - N is some number, let us say some
L which is nonzero, so what you have is IL
omega naught, so you have IL omega naught,
so you still have IL omega naught IL pi, so
when L is integer, so this is still - 1 power,
- N power L, okay, into minus of minus 1 power
L, so it's any case it is 0, so you can see
that when K - N is a nonzero integer, this
is always 0, so this quantity is 0 so everything
else will go except when K = N, so that is
corresponds to CN, and in that case this will
be E power 0, that is 1 so you have L/2 - L/2,
that's going to be, of course so this is not
exactly CN, CN is L into CN, so you have L
into CN this is actually, by definition this
is the Fourier coefficient L into CN, 1/L
of this is CN, so you have L into CN, and
here CN, and K = N it contributes integral
- L/2 to L/2 and K = 1, this is 1 DX that
is L, so finally it is 0, okay.
So we see that, we observe that this is the
case F(x) - SN multiplied with E power - IN
omega naught X, that quantity is 0, so because
of that we note that - L/2 to L/2 F(x) - SN(x),
so
because you multiply any of this quantity,
okay, so what is your SN? SN is this, you
take any of this quantity that is 0, okay,
so instead of, so you take a linear combination
of them, so that is SN, I have chosen minus
of this, so SN is with plus so you can just
take SN bar that will have exponential with
- I sine, so SN(x) conjugate of this DX this
is nothing but sigma, K is from - N to N,
so SN's you can replace with CK bar, so integral
- L/2 to L/2, F(x) - SN(x) that is as it is
into E power - IK omega naught X DX and this
is exactly what we have seen that it is 0,
okay.
So this simply in the place of N you can put
anything K so it doesn't matter, so you have,
because this quantity is 0, so what you have
is every time it is 0, whatever may be K value
it is
0, this quantity is 0 so that means is completely
0. So we first note this one, then we observe
that if you simply integrate from - L/2 to
L/2, F(x) - SN(x) and this is a complex quantity
and you have F(x) bar - SN(x), the conjugate
of this whole thing that is actually absolute
value square, DX this is always positive,
okay.
So and what is this one? This is actually
equal to - L/2 to L/2 and because this with
SN bar that is 0 so you are left with only
F(x) - SN(x) into F(x) bar DX which is, that
is always, this is same as this because the
other part will be is shown to be 0 earlier
here, okay, so this is equal to integral - L/2
to L/2 this is F(x) into F bar that is mode
F(x) whole square DX -, - L/2 to L/2 SN(x)
F(x) bar DX, okay.
And you can also observe here this gives,
we have noted that this quantity is 0, okay,
this quantity is 0 implies - L/2 to L/2 by
2 F(X) SN(x) bar DX is nothing but - L/2 to
L/2 SN(x) bar modulus square DX, okay, so
that is exactly we will write here, so this
is equal to - L/2 to L/2 F(x) bar modulus
square DX - this is L/2 to L/2 SN(x) square
DX, and clearly this quantity is positive
quantity, okay.
And what is this one? So integral - L/2 to
L/2 mode SN(x) or rather SN(x) into SN(x)
bar DX, if you write this you actually see
that this is quantity, K is from - N to N
CK E power I, so what is this SN? SN would
define as E power IK omega naught X 
into its bar, so that is CK bar, so rather
this into, now I change this is different
that index you can write in a different way
or you can use the same index - N to N CK
bar E power - IK omega naught X DX this is
what it is, this is SN(x), this is a SN bar,
this is equal to so only contribution is when
they are equal, when they are same, that is
and also integral from - L/2 to L/2 any exponential
function, and this one, and this is different,
okay. So rather let's use that different K,
different, so I use this instead of this index
I use a different index that is let us say
M, so this M, M is from - N to N see this
is a product of this finite sums as well as
so, as long as this exponentially multiply
this with this you get E power I K-M, and
only when K = M that is becoming E power 0,
so that is I times 0, that is the only contribution
you will have, otherwise that integral is
0 that is what you have seen earlier.
So what you get is sigma K is -N to N CK mode
square only when M = K that is what is the
result, and it will become 0 so that is exactly
you'll get so that integral will be simply
L, so each one so you have this is - L/2 to
L/2 DX this is nothing but L times sigma K
is from - N to N modulus of CK square, this
is what you get, so this implies sigma K is
from - N to N modulus of CK square is you
can replace this quantity here minus of that,
so if you use this inequality 0 less than
or equal to integral - L/2 to L/2 mode F(x)
whole square DX, in the place of this now
I'll replace this that is actually L times
- L times sigma K is from - N to N CK square,
this is true for every N okay, so this implies,
you can bring this this side so you get sigma
CK whole square, K is from -N to N is always
less than or equal to 1/L integral - L/2 to
L/2 mode of F(x) square DX, so this is true
for every N okay, so this implies, because
this is true for every N, you can allow this
to be N goes to infinity, so you can see that
this is actually true for every N so that
means you can allow N goes to be infinity
so you have, so K is from minus infinity to
infinity mode CK square is less than or equal
to 1/L - L/2 to L/2 mode F(x) square DX, so
this is exactly what you want to prove that
is Bessel's inequality.
So that you can see here, this is your Bessel's
inequality, so this is what we are going to
use to show that if the signal is piecewise
a differentiable function, and we'll show
the Fourier series actually converges, so
the process and the proof of that we will
use this inequality, so one important corollary
of this, one implication of this result is
that, so if this quantity is, the right hand
side quantity is finite that means if F(x)
is, or if - L/2 to L/2 if it is square integrable
function DX, if this is finite, okay, then
this quantity N is from minus infinity to
infinity, I am just changing the index here,
so CN square is always finite that means,
okay, so what does this mean? So this means
that implies, it's clear right this is a straight
forward, if this is finite, this is finite,
that is what I have written, so this implies
a limit of CN, as N goes to infinity, this
is
going to be 0, so this is straight forward
from the number series if a limit, if sigma
AN, N is from 0 to infinity or minus infinity
infinity, if this quantity is finite then
this limit of AN has to go to 0, if it's not
see the proof is why is this so? So we'll
use contrapositive argument that if limit
of this is nonzero, if suppose this limit
AN is not equal to 0, okay, then what happens?
So this converges to some nonzero quantity
some L okay. Then what is the result? AN - L
you can put them in some arbitrary small quantity
whenever N is bigger than some big N okay,
so that means AN is always lying between L
- epsilon to L + epsilon, for every N bigger
than or equal to N, so you can choose epsilon
as 1, so you can choose any epsilon, given
epsilon there is a big N that depends on it,
so given epsilon there is an N beyond which
this AN's will strictly lying here, so if
we choose epsilon as 1, your AN's will be
lying so for some N, N is this capital N is
actually depending on what you give as epsilon,
here we have given as 1.
If L is nonzero let's say you can think of
L as, L is a positive quantity so let's choose
positive quantity so assume that L is positive,
so I can choose epsilon as L/2, okay, so if
I choose L/2 there exist corresponding some
N beyond which you have L + L/2, so this is,
because L - L/2 this is always positive, so
this implies sigma AN, because these are all
positive quantities you are adding up every
time, okay, so for N is running from 0, so
A is 0, A1 every time you are adding a positive
quantity all the time so this has to go to
infinity, but that is not the case, but
we know that this is actually finite, so that
means this has to go to 0, this is a small
result in the on sequences on series of real
numbers, so using that you can see that this
is our limit quantities has to go to 0, so
this is a Riemann Lebesgue Lemma, this is
Riemann Lebesgue, okay.
So now we give, we state this sufficient conditions
as a theorem main result that if F(x) is a
piecewise differentiable function, periodic
function 
with period L, then this what you have is
this Fourier series, complex Fourier series
which I'm writing CN E power IN omega naught
X, omega naught is the fundamental frequency
which is 2 pi/L, okay, this is equal to, it
is actually converges to the value, once we
fix this value X this converges to 1/2 of
F(x+) that means if you have, this is your
X on the real line between - L/2 to L/2, X+
is a limit of F(x+) is a limit of F(t), now
T is going to X from the positive side that
is the meaning, so the limit of this function
as T, if T is here, T is approaching X right
hand side. Similarly F(x-) is a limit of F(t),
T is going to X negative side so this is the
meaning of this two function values, F(x+)
F(x-), so whose average value is what this
series, converges to that, if you fix your
X, X I fix may be inside, okay, because it's
a differentiable function which is a continuous
function, implies it is a continuous function
so F(x) so you can always choose like that,
okay. As of now I am choosing X in between
here, so not at the endpoints, so if it is
endpoints is actually, we will see what it
is okay, so X belongs to - L/2 to L/2.
So let me give you the proof now, this is
the sufficient condition for this Fourier
series converges to function, because it's
a differential function is actually you can
say that this is exactly equal to F(x), because
this is nothing but F(x) because it is of
its continuity both are same, so average value
is itself, so X belongs to -L/2 to L/2, so
proof is again we will consider the partial
sums of this left-hand side, so partial sums
we represent SN(x), okay, as in the earlier
case, so you have K is from - N to N CK E
power IK omega naught X, so this is equal
to sigma K is from -N to N CK, so CK I'm going
to define so what it is? CK is a Fourier coefficient
that is 1/L -L/2 to L/2 F(t) E power - IK
omega naught T that is my CK into, this I
write it as it is so this is with DT and E
power minus this power this is as it is, that
is E power IK omega naught X, so this will
give me, I take this sum inside so you have
1/L - L/2 to L/2, F(t), and now I write the
sum here, finite sum, because it's a finite
sum I can take it inside, so what you have
is E power - IK omega naught T - X DT, so
what is this one? This is equal to 1/L, so
you can see this one, you can just replace,
you can break this integral into two parts,
first from 0 to L/2 F(t) this finite sum,
K is from - N to N E power minus IK omega
naught T - X DT + 1/L - L/2 to 0, F(t) sigma,
K is from -N to N E power -IK omega naught,
T - X.
So what I do here is, so let's call this some
quantity which is messy, so let's define let
DN be, I'll call it some DN(x) this is a function
of X okay, so let me define what is this DN?
DN(x) if I define it as K is from - N to N,
E power -IK omega naught X, if I define it
like this, then what is this quantity? This
quantity is actually equal to so E power IN
omega naught X, when I put K equal to - N
+ E power IN - 1 omega naught X and so on,
we'll go on 0 that is 1, 1 also come here
somewhere comes in between and then it's becoming
negative, when I put K = N is finally end
up with IK, K is N, IN omega naught X, so
this is equal to if you take E power IN omega
naught X out, so you have 1 + E power -I omega
naught X, and the E power -I to omega naught
X and so on, finally E power -I 2N omega naught
X, that is what is that thing, so this sum,
this is in GP so you can just have it, so
E power you can write that sum with A is 1,
into 1- R is E power -I omega naught X, power
2N+1, so you have 2N+1, so N is 2N here, so
N+1 is 2N+1 divided by 1 - E power -I Omega
naught X, that is what it is.
So if I now take it inside, so this is E power,
this is a product for everything, so if you
take this inside E power IN omega naught X
- E power - IN + 1, E power IN+1 omega naught
X divided by 1 - E power - I omega naught
X, what I do is I take, the denominator also
I rewrite like this, I multiply E power IN
omega naught by 2, E power I omega naught
by 2X, if I multiply and divide, IN+1 omega
naught X divided by, so you also do it here,
so if I do both sides, so you take it inside
the denominator one you take it inside here,
so you have E power I omega naught 2X - E
power - I omega naught/2X this is what it
happens, it becomes.
Now take this one inside, so what you see
is E power I (N+1/2) omega naught X - E power
is actually minus, - I (N+1/2) this one so
- 1, -I, +I become that's - 1/2, so that's
what happens, omega naught X divided by, this
is actually nothing but this is as it is,
so I omega naught/2X - E power - I omega naught
/2X, so what is this one? This is exactly
equal to sine 2I sine of (N+1/2) omega naught
X divided by sine omega naught/2X, this is
what is my DN(x), if I define like this as
DN, DN is this so clearly DN(-x) is actually
equal to DN(x), this is an even function,
even function, so if you use this one what
happens to your SN? SN is this one, so what
happens to your SN? SN(x) is 1/L integral
0 to L/2, so what happens to F(t), DN of,
what is your DN(t-x), DN(t-x) DT + 1/L - L/2
to 0, F(t) DN(t-x) DT, that is also same,
right, so you can see that both sides are
same, so you just split that integral, so
this is what it is.
Now put T-X as some X dash in this, in this
integral if you do this you have DT is DX
dash and that becomes, and here T-X as, first
of all put T = -T, so in both the integrals
you can put it as T-X as X dash, then what
happens to SN(x)? SN(x) is 1/L integral 0
to, so if I put T = 0, and this is going to
be minus, so T-X is, so you need not split
it, before splitting itself you can do, you
can split it later so minus L/2 to L/2 this
is what is this SN, you put this one, so you
can see that it's going to be -L/2 -X, because
X is fixed to +L/2 -X, because it's a periodic,
the translation will be same as again -L/2
to L/2, so both will be same, so this will
not change because it's a periodic function,
so - L/2 to L/2, F(t) is X+X dash or DN(x
dash) DX dash, this is what it is.
Now you can split this into two parts so 1/L
-L/2 to 0, F(x+ x dash) DN(x dash) DX dash
+ 1/L times 0 to L/2, F(x+ x dash) DN(x dash)
DX dash, so what happens here, so here you
change X dash by, put some X dash by -X, -T
if you say if you do so you get 1/L, this
becomes L/2 to 0, F(x-t) DN(-t) which is,
DN is an even function, so DN(-t) is DN(t)
and you have - DT, so that is minus, this
minus so you can change the limits, so L/2,
so T you replace as it is, so X dash itself,
actually this is what, okay let's write DT
only, so DT and this one is +1/L integral
0 to L/2 F(x+t) DN(t) DT, this is what you
have, okay, so this is exactly it has become
my SN(x), so partial sum of the Fourier series.
So the next step is I'll just write what is
your DN(t), that is 1/L, 0 to L/2 you can
combine these two integrals there is going
to be F(x+t) + F(x-t) this one I wrote first
and into, this plus this or DN(t) is common,
so DN(t) we have seen already that this is
sine N+1/2 omega naught X divided by sine
omega naught X/2, okay, this is what we have,
this is a function of T so you write T, DT.
So what I do is, I do small trick here, I
just add and subtract because so I am going
to assume that this quantity is one function,
okay, this is one function if I take this
as one function so you will see that, so what
we are going to show is eventually maybe I'll
see in the next video that I will show that
if I choose, if QN, so let's call this some
QN of, let's say a quantity QN(t) is F(x+t)
+ F(x-t), X is fixed, so that's why I'm calling
it T, function of T and divided by sine omega
naught T/2, so if I call this as my function
this is, this has to be well-defined, okay,
you see that T is actually is, T belongs to
0 to L/2, so that is what is the integral
is between, so as T goes to 0 I have a trouble
here, so this is 0 and this quantity, so something
divided by 0, so that is why to avoid this
something some quantity F(x), 2 times F(x)
divided by 0 that is going to be infinity,
so to avoid that so I will just rewrite this
Q as F(x+t) - and I have X a positive side
function, okay, and then + F(x-t) I will also
subtract a negative side function, divided
by, I divide with X so that I have to multiply
X divided by this sine omega naught T/2.
Okay, this is what I will do, okay, so this
integrand value this is what I saw, because
I added and subtracted, so I will have to
add, I will have to take, I'll have some addition
that is F(x+) + F(x-) because I subtracted
both times, XX I don't use divided by sine
omega naught, I have
subtracted only this one, right, so okay so
we'll see maybe in the next video we don't
have time, so from here we will go on next,
we'll see in the next video, we're trying
to prove that the Fourier series converges
to the function F when it is piecewise differentiable
function, so we are halfway through so we
have taken the partial sum of this Fourier
series, and we manipulated up to some point,
so the remaining part to show that this actually
converges to the function F(x) in the next
video. Thank you very much. [
