Hello, welcome back, the last video we have
seen how to see the convergence of Fourier
series, when the signal is piecewise smooth
function that means if the function is at
finitely many points is discontinuous but
solving of jumped discontinuity and other
places its continuous function, it's actually
differentiable function, so in that case you
can have the convergence of the Fourier series
to the function at the continuous points,
if it is discontinuous points you take the
average value of jump, that is half of the
jump, average of the value of the jump of
the function, that means function at both
the ends you consider and take half of it,
you sum of it and take the half of the sum,
so that is what you have seen in the sufficient
conditions for the convergence of the Fourier
series.
In this video what we will have is, when the
signal is piecewise continuous function and
up here you know that if the Fourier series
is converging uniformly, okay, we're asking
more than piecewise, more than convergence,
point wise convergence that means if you assume
that it's a uniform convergence of the series,
the Fourier series that we have is uniformly
converging to a function, two certain function
then that function must be, the function F
at continuous points, at other points we are
at discontinuous, you have half of the jump,
okay, that means average value of the jump
that's what we will see here, so it's not,
result is that not that significant because
we are not looking at the convergence of the
Fourier series rather if it converges, it
has to converges to the function value or
the average value of the jump, both sides
of the function at the jump, okay.
So we are actually assuming that it's a Fourier
series itself is converging so why we're still
doing is in the process we can make use of
a delta function that is useful when you are
doing the Fourier transforms and Laplace transform
later, so you will understand what is a delta
function and it's use in the convergence of
this series, Fourier series when apriori you
know that it's actually converging uniformly,
okay, so we will write it as sufficient conditions
before I do this what is the convergence of
uniform convergence, on sufficient condition
is so if you consider this as a Fourier series
CN E power IN omega naught X, so this is your
Fourier series,
okay, N is from minus infinity infinity, so
this is your Fourier series, so if this converges
uniformly means this sequence of partial sums,
that is N is from -K to K CN E power IN omega
naught X this has your SN, this sequence converges
to the function which is finite, so it's E
power IN omega naught X so uniformly.
Uniformly means irrespective of the value
of X so you have the convergence is uniform,
so a pace of convergence is you have a minimum
pace of convergence at all values that converges,
so you have a minimum pace of convergence
that works for all values of X, that's crudely
saying uniform convergence, point-wise convergence
means for different X values you have a different
pace, and so as when you consider some X values,
so if you consider the limit of these X values
it may go to the minimum pace may not be positive,
so that may go to 0, okay, so that means it
may not be convergence going with the same
pace, okay, there's no uniform pace at which
this converges for all values of X, so what
are the sufficient conditions for this uniform
convergence is if I assume that these coefficients,
Fourier coefficients, N is from minus infinity
infinity, if this is finite, if this is finite
I can say that this is uniformly convergent,
then this SN converges uniformly.
So this is easy to see because if you look
at the SN, modulus of SN this is less than
or equal to sigma CN, N is from -K to K, okay,
mode CN so you can have this mode of CN because
of this so SN is a function of X, so I can
write it as a function of X, this is significant
to this, so this converges by M test, so each
of this is bounded with this, okay, so and
this is bounded with this because this series
is a finite as a number series and this is
from minus infinity infinity, so this is finite,
so once you have this so you can easily see
that this is by M-test of convergence SN(x)
converges uniformly, irrespective of X value
this is still less, always converging, so
that's what you can easily see, okay. So FN(x)
converges to F(x) uniformly, so this M-test
is, if you have FN(x) is bounded with some
MN for each N, and MN converges to M, then
you can see that
FN(x) converges to or rather sigma, sigma
MN is finite, if sigma MN is finite then FN(x)
sigma FN converges uniformly, so that is M-test,
so we using from calculus, so if you use this
so you can see that if this is the sufficient
conditions to ensure that this Fourier series
is converging uniformly, so I make use of
this and to state this result, so let me write
as small theorem, so if the signal F(x) is
piecewise smooth, not smooth, so piecewise
only continuous function 
and sigma CN, N is from minus infinity infinity,
if this is finite then the series the Fourier
series converges, N is from minus infinity
to infinity CN E power IN omega naught X = F(x),
if F is continuous at X, otherwise you will
have to take the jump value, okay, those average
of the jump, so that is sum of the limiting
values, take the half of it, if F is having
jumped this discontinuous, if at all is discontinuous
it should be jump because it's a piecewise
continuous function, okay.
So the if at all this, if you assume that
this is true then you have this Fourier series
converging uniformly, so if it is a piecewise
continuous function or rather just continuous
function you cannot say that Fourier series
need not converge, okay, you cannot say that
Fourier is converge, so the Fourier series
need not converge for a continuous function
that is in the literature, we
will not look into those details, but why
when you choose certain piecewise continuous
functions which we normally assume, if you
just write it as elementary functions and
certain piece how you represent as an elementary
function, another certain piece, another certain
piece you have a elementary function that
is represented as a function, then these are,
because these are elementary functions they
are actually not just continuous, they are
also differentiable, that is why it is actually
piecewise a differentiable function, or piecewise
smooth function that is why the Fourier series
converging, that is what you are seeing all
these examples so far, okay, and you will
also see many examples because by considering,
when you consider function with elementary
functions it's actually piecewise a smooth
function so that your Fourier series converging
actually to the function F or its average
value of the jump.
So if it is a piecewise continuous function
or a continuous function, the Fourier series
need not converge, okay, we have examples
in the literature but here if the Fourier
series converging and it has to converge to
the value, it cannot converge to some other
thing except but it has to converge to the
function F(x), if F is continuous at discontinuous
point that's the jump value, average jump
value of the function at the discontinuous
point, okay, so to show this result is, as
I said earlier result is not that significant,
the proof we make use of delta function so
that will be useful later, okay, so how do
I see this, this is actually this series is
converging to F(x) so let S(x), let me call
this full series as S(x), N is from minus
infinity to infinity CN E power IN omega naught
X, okay, so this you can represent as, now
we know because it's a Fourier series and
we know what is the CN, CN's are where CN's
are 1/L -L/2 to L/2 F(x), these are the Fourier
coefficients - IN omega naught X DX, so if
you substitute here and you see that is going
to be minus infinity to infinity, CN I substitute
here these Fourier coefficients F(x) E power
-IN omega naught X DX times, so let me put
it as, because I already have X let me use
this dummy variable T, so this is T, E power
IN omega naught X, so if you clubbed together
so
what you get is N is from minus infinity to
infinity, 1/L - L/2 to L/2 F(t) E power -N
omega naught or + omega naught X-T DT, so
here I split this I take N = 0, I write separately,
N = 0 if you take this is 1, so you have 1/L
- L/2 to L/2, average value of F(t) plus,
now if you write this as N is from minus infinity
to infinity, but N is not equal to 0 what
is left here, so that is - L/2 to L/2 F(t)
E power IN Omega naught X-T DT, so this is
equal to, if you actually combine negative
and positive, okay, so negative part if so
this is where I'm using the uniform convergence,
okay.
So uniform convergence so this is what I'm
using, so if this is true this is a kind of
series is uniform convergent, if it is uniform
convergent it's actually uniformly and absolutely
convergent, so that means I can do the, I
can shuffle all the terms and I can add N
term to, for example here minus infinity term,
plus infinity term, - K term, + K term and
so are all kinds of sum, you reshuffle it,
rearrange the terms and we can add it you
still converge to the same value, so if I
do that I have 1/L - L/2 to L/2 F(t) DT +
here if - infinity to + infinity if we add,
so it's going to be from 1 and -1 if I add,
so what I have is -L/2 to L/2, so you have
F(t) E power, N is from, N is 1 and N is -1,
so we have a 2 times, you have a cos instead
of this thing you have cos N omega naught
X-T DT, so this is what you have, because
of rearrangement and then addition of those
2 each, addition of -K and +K terms if you
do, this is what you get, okay, so that we
have this nice form, this is possible first
of all this step is possible, because of uniform
convergence.
We're not proving the convergence of the Fourier
series here, if it converges it has to go
to only a function not anything else, so if
I pull this together, so again I am trying
to pull this integral out of this series,
that is also possible because of uniform convergence,
if I do this you see there it's going to be
-L/2 to L/2 F(t), if I write it 1/L +2/L sigma,
N is from 1 to infinity cos N omega naught
X-T DT, so this is the series to DT, again
this step is also because of uniform convergence
you have this.
So let me call this as some new function so
let D capital D(x-t) as 1/L + 2/L sigma, N
is from 1 to infinity cos N omega naught X-T,
let me represent like this, now in this here
I'd take the digression, digress little bit
to define what is a delta function here, a
delta function is not a function, so delta
function so first of all unusual functions
which you have here continuous functions,
differentiable function which you, all these
usual functions which you see there are functions
that are defined, it doesn't take any infinite
value, okay, so any the generalized functions
are a limit of usual functions, okay.
So first of all if I define what is generalized
function? Generalized 
function if I define as a limit of usual functions,
limit of usual functions 
in 
an average sense is called a generalized function,
limit of usual functions in an average sense
is called a generalized function. For example
I'll just give you, so I choose a delta function
itself, let me choose this as my first function
0 here, suddenly there is a jump, there's
a 0 again here from here to here, okay, so
this is 1, again next this is your F1, F2
is 0 up to here, go up and you go up and again
0 and this is your F2, like this you go on
getting it like this, okay, 0 if it's 1 here,
2 here, 3 here and so on, okay, and if you
go on as N goes to infinity as like that you
go on, you have FN, FN if you consider for
any each N this is a usual function because
0 at some place it's a finite number and other
places it is 0, okay, this is a usual function
but you consider the limit as N goes to infinity
what you get is 0 for every value X is not
equal to 0, okay, because eventually this
is 0, around 0 both your eventual is 0, at
0 you are getting as you see there is a 3,
3, N is value is N at 0, that's some small
neighborhood of 0 it's N.
So at X = 0 that is infinity, so such a function,
so this is not a function what you get the
result of this usual function, but this is
called the generalized function and it's a
limit to, what do you mean by this limit,
this converges to this function? This is a
some average sense you have to look at it
that is F(x) you consider some G(x) take the
average wherever this integral makes sense,
now you look at this N goes to infinity this
average minus infinity to infinity, so that
this integral makes sense, you consider all
G(x) such that this integral makes sense,
this is equal to, if I call this as some new
function let us say generalized function capital
G(x), so what you have is G(x) into your any
test function if I call this G, any G(x) that
makes sense, okay, so this converges to the
finite quantity like this, so this limit converges
on an average FN converges to G(x) that is
the meaning of this, so that is a weak convergence,
okay, on an average means a weak convergence,
so that means FN(x) converges weakly on an
average to G(x), that is the meaning, okay,
so that is what happens for certain limit
of functions such as this.
So delta function is one such thing, so delta
function is generalized function, so how do
I define this delta function as a delta(x),
which is 0, at X = 0 at X equal to, X naught
= 0, infinity at X = 0, so this I can represent
this as a limit of some usual functions FN(x)
as N goes to infinity, where FN(x) I define
it as, I can define this as, how do I define?
Yourself construct, if you take this -1/N
to 1/N for example, okay, so if X is between
-1/N to 1/N this is value is, its value is
N for example, so what is the difference?
This integral value is, and so I define like
this N outside otherwise 0, for example let
me put it, it's not, so what happens? For
each N I maintain, and integral value for
minus infinity to infinity FN(x) DX if you
take the integral value that I maintain as
a constant, okay, so such a limit of functions
is called delta functions.
So delta function not only, this is a limit
of usual functions but you also maintain this
area for example this integral value has to
be 1, so if I choose this here FN(x) N times
-1/N to 1/N DX, so this is N times, what is
the value of this? One, so that is 1/N, 1/N
- +1/N, so you have 2/N, so you have 2/N,
so in order to get as 1 so this should be
N/2, so I define this as N/2, if I define
N/2, so this will be simply 1, value is maintained
for every N, okay, so because of this you
have this properties true, okay.
So you have many representations for such
delta function, so you can have a different
FN values, different sequences converges to
this property, converges to this type of function
and also keeping the average value is 1, integral
value is equal to 1, such a function is delta
function, that place an important role in
the physical and engineering sciences, for
example when you consider instant instantaneous
loads in a civil engineering so you consider,
you use these delta functions, okay, so many
other places you can use this delta function,
so we also use, make use of this, we also
use this delta function, you can also apply
Fourier transform, you can also do Fourier
transforms on these delta functions in our
course so these are also useful, you can also
apply Laplace transform on this delta function.
So once you have this delta function so I
will try to prove here that one of the property
of the delta function is, once you have such
a delta function what you have is average
value of this FN(x) if you multiply any F(x)
or rather a delta(x) you multiply with the
F(x), this value equal to DX = F(0), so how
do I prove this? This is the property, property
of delta function, so this proof you can easily
see as minus infinity to infinity, delta you
have seen that is usual limiting value of,
this is actually you can write it as limiting
as N goes to infinity, FN(x) F(x) D(x) okay,
so LHS is this, so this you can write it as
a limit N goes to infinity, so this is from
- 1/N to 1/N and you have FN(x) is N/2 and
F(x) D(x) okay, so this is N/2 which of your
constant comes out, limit N goes to infinity,
this is from -1/N to 1/N F(x) D(x), so by
calculus you can have this average value you
can consider N/2 as 2/N times F at some value,
so you have if F is the usual function so
F is a, this here multiplication function
that we have for any usual function F(x) okay,
for any continuous function, let us say, continuous
function or a differentiable function, okay,
F(x), so if I choose such a F(x) you have
this theorem in your calculus F(x) D(x) this
value actually B - A times F(c), as C is,
F is continuous, C is the value between A
and B, okay for some C like that, so you have
you can write like here, so you can have some
C where C is between -1/N to 1/N.
So if you have this now as you, now you have
the limit, so you still have the limit, so
you can try at outside so this is the limit
okay, so this limit once you have this limit
N goes to infinity. so this N, N goes so what
you have is this limit, N goes to infinity
F(c), C as N goes to infinity
this becomes -1/N to 1/N you end up getting,
as N goes to infinity this is nothing but
F(0), so this is the property that we have,
this is your right-hand side, so with this
preamble this delta function we can have,
we can make use of this delta function here.
so what we have is this Fourier series that
we have S(x) is actually F(t) times this if
I show that this D(X of T) is delta function
behaves like, if this is actually kind of
a delta function then I can show that this
is value is equal to F(x), okay, so that's
what we will see.
So one immediate thing which we have is, if
I have instead of 0 if my instantaneous load
is, my concentration is only at 0, okay this
is how you are putting your concentration,
limit of sequences, so what is the delta function
of X-T is a concentration, as a function says
the limit of FN(x-t), the concentration it's
some value T, okay, so you can have such a
thing.
Then you have some immediate corollary of
these properties, integral delta(x-t) into
F(x) DX as minus infinity to infinity is actually
equal to F(t) okay for continuous F(x) continuous
function for any continuous function F(x)
so this is what exactly we use there your
Fourier series, what we have is one by, what
we get is -L/2 to L/2, F(t) so what we have
is a D(x-t), if I show that this is actually
kind of delta function, so we have DT this
becomes F(x) if F is continuous, okay, so
that is exactly what we are going to show.
You can also use make use of the delta functions,
if it is a discontinuous points, if it's discontinuous,
function is discontinuous you can also write
this as, you can have one more property that
if you use 0 to infinity, delta F(x) DX this
will give you because of your concentration
only, you're only looking at the positive
side so you have only this part if you look
at it, it contribution will be half so you
have half of F(0), so if I do like this, if
F is discontinuous point you can use this,
if F is continuous this one, otherwise F(x+)
+ F(x-) divided by 2, half of F(x+) for F(x+2)
infinity, other one is, other part is minus
infinity to F(x-) part up to here will have
half contribution here, half contribution
here, because of this property, so only thing
I have to show is D(x-t) is actually delta(x-t),
so how do I show this? So to do this let D(x-t)
as, I put it as a limit, X goes to, sorry,
so I introduce some new parameter here as
DR(x-t), so where DR(x-t) is, you have this
series L/R 2/L times, 1/2 + this is actually
1/L I am writing like 2/L times 1/L so that
I can take this series, 2/L out this series,
N is from 1 to infinity, so instead of simply
cosines that we have for D(XT) you write introduce
R power N times cos N omega naught X-T DT,
X-T, so this is what you have.
So by introducing this as R is less than 1
what you have is this modulus of this inner
product in this sum is always less than 1,
if mode or less than 1 then what you have
is mode R power N
cos N omega naught X-T is also less than 1,
so because of this I can sum because this
becomes a geometric sum so I can sum it up
easily, so for that reason only for R less
than 1, more or less than 1 this makes sense,
is finite, so as R goes to 1- because less
than one side you approach the limit that
you define it as D(X of T), so if you allow
R goes to, limit of R goes to 1- this is actually
D(x-t) that you can easily see, put R = 1
that is exactly what you have, that is your
D(xt).
So by defining like this your D and then you
can write, you can sum it up this you can
add it and you can write this as limit R goes
to 1- and you have a real part of 2/L times
1/2 + sigma N is from 1 to infinity, R power
N so I write simply instead of cosine I write
exponential function so that I write real
part of it, E power IN omega naught X-T okay,
so remember omega naught is 2 pi/L here, so
now if you can add this 1/2, so this becomes,
so if you add this series, this series I can
add and you can easily see that 2/L that is
our DR okay, DR(x-t) if you add this series
you'll see that 2/L, so if you add this what
you see is real part of 2/L times, this becomes
1 - R square + I 2R sine X-T omega naught
divided by 2 times 1+R square -2R cos omega
naught X-T, you can write here as sine omega
naught X-T as one term for sine.
Similarly for cos omega naught as put it as
one term, so this is what you can easily see
by adding this as geometric series that is
like sigma R power N, N is from 1 to infinity
is actually R power 1 is R, so R divided by
1 - this geometric thing is that is R, okay,
something like this or you have N is from
0 to infinity, R power N is simply the first
term is 1, 1-R, okay, this is if I use the
sum you can easily see that this is the result,
okay.
So this you have to take the real part, if
you nicely see so 2 2 goes so you end up getting
1/L, 1-R square/1+R square -2R cos omega naught
X-T, okay, now what you do is you know that
D(x-t) is a limit of our DR(x-t), as R goes
to 1 -, so you take this limit R goes to 1
-, so if R=1 you can easily say if X equal
to, so you can easily see if X is not equal
to T this cosine function never be 1, if X
is not equal to T cos omega naught, cos 2
pi/L times X-T as long as this is not equal
to 0 this never be 1, if X is not equal to
T, so for that reason this denominator never
be 0 and what you get is numerator is, but
numerator is going to be 0 so you have 0 here,
otherwise what you have is, if it is 1, if
X = T you have this becomes 1 and you have
this becomes 1-R whole square, numerator is
1-R square, so you have this is nothing but
1+R divided by 1-R, so as R goes to 1 - is
actually given to be infinity, so this is
exactly like your delta function.
You cannot still say unless I show that integral
other property of the delta function, integral
value of delta, and integral value of D(x-t)
DX minus infinity infinity, so you have to
show that this value is equal to 1, if I just
multiply with any function 1, and if any continuous
function if you multiply it has to become
F that function value at T, so here I choose
this as 1, so if I choose this this is sufficient,
so if I choose this, if equal to 1 then then
D(x-t) is actually delta function, actual
delta function, okay, here are still here
I cannot say though it's taken 0 at X =T X
is not equal to T, at X = T it's taking infinity,
it looks like delta function, but if I provide
only this then we clearly say that D is actually
delta function.
So how do I say this? Again because we know
that this is a D(x-t) DX is actually a limit
of R goes to 1 -, okay, so this is minus infinity
infinity, and this is your DR(x-t), because
of the uniform convergence of this DR(x-t)
which is a series, and I am just putting R,
I can bring this integral inside and you end
up getting this limit, limit you can take
it as inside okay, so I can exchange these
limits and you have this DX, so if I show
that this value is equal to fixed, if it is
equal to 1 then I am done, so to show this,
this is equal to let me look at only this
one, okay, to show this, this is equal to
2, to show that this is equal to 1 let me
choose this DR(x-t) DX, this if you calculate
minus infinity infinity, as you see DR(x-t)
outside because it's coming from the Fourier
series, so outside you can assume that it
is 0, okay, DR(x-t) or D(x-t), between X-T
is, if X-T is between -L/2 to L/2 only this
is true, this is outside you can safely assume
that is 0, okay, just like you have a periodic
signal which is only periodic between -L/2
to L/2, outside you can take it as 0, so if
I do this -L/2 to L/2 this will become -L/2
to L/2 and the DR(x-2) is 1-R square/L times
1 over 1+R square -2R cos omega naught, that
is 2 pi/L X-T DX, okay, so this is equal to,
so you can easily show that this quantity
is equal to 1.
How do I show this? This is equal to, so you
can take this out 1-R square/L -L/2 to L/2
and here before I do this, so I'll write X-T
as X dash so that you have DX is DX dash divided
by 1+R square -2R cos 2 pi/L times, X-T as
X dash and this becomes when you put X = -L/2
to -T, this -T because of this periodicity
and I can simply, this is again same as -L/2
to L/2, okay, because DR(x-t) is 0 outside,
so it's anyway periodic so it doesn't matter,
so you can, this is cosine function is periodic
so you can have this, this is same as this,
okay.
So this is actually -T and -T because of this
cosine function which is periodic so that
is actually same as this one, you have 2 pi/L
so you have, that's why, that is the reason,
okay, so X is a dummy variable I can remove
these dashes, so this is what you have.
And the next step is let's choose 2 pi/L X
as T, so if I do this you have 1 - R square/L
this becomes when I put this, this becomes
-pi to pi, DX is L/2 pi times DT 1+R square
-2R cos T, so L L goes and you have this 1
- R square/2 pi, so again this is even function,
integrand is even function so you have 2 times
0 2 pi DT/1+R square -2R cos T, so 2 2 goes
here so you end up simply pi.
So again this one I can split this into 1/pi
part pi/2, and pi/2 to pi DT/1+R square -2R
cos T, so again if I choose in this integral,
if I choose T as X-pi/2 and we get X = T - pi/2,
at T = pi/2 so it becomes 0, and T = pi, it
is pi/2, and cos T becomes - cos X, okay,
so if I use this, this change of variable
in the second integral, so you end up getting
1-R square/pi times, so you see that this
is going to be 0 to pi/2, so if you add this
you get 1+R square whole square -4R square
cos square T, DT what you have is 2(1+R square)
times DT, so you have 1-R square, 1+R square
into 2 /pi times integral 0 to pi/2, DT by
and here one more step I do is I write cos
square T as 1/secant squared T, so if I put
this as 1+R square whole square, 1 secant
square that I write it as 1 + tan square X,
and tan square T -4R square so that you can
take this secant square T DT okay, so this
is what you have.
Now put tan T as X in this, so you end up
getting 1-R square, 1+R square 2/pi times
here, 0 to pi/2, so if I do this T = 0, tan
0 is 0, T = pi/2, tan T there is X is, X becomes
infinity so you have DX 1+R square whole square
-4R square + 1+R square whole square times
tan square T that is X square, this is what
you have so this if you pull it (1-R square)
(1+R square) 2/pi, if I pull this part outside
what you get is this is actually equal to
1-R square whole square, okay, so this if
I pull it apart and you have 0 to infinity,
DX divided by 1 + 1+R square/1-R square whole
square
X square, so this gets cancelled here, so
you end up getting 2/pi times 1+R square/1-R
square integral 0 to infinity DX /1 + 1+R
square/1 +1-R square X whole square, so this
is actually equal to 2/pi times tan inverse
1+R square/1-R square X, this you apply 0
to infinity limits, this is at infinity this
is pi/2 at 0 this is 0, so you have 2/pi times
pi/2 - 0, so this is actually equal to 1.
So this is what we have shown, so this implies
the DR, so this means when you come back here
and write it here so this integral value have
shown that for any irrespective of value of
R this limit R goes to 1- is actually constant
,so it is actually equal to 1, so delta function
is this and delta function integral value
is 1, so this implies D(x-t) is, there's actually
a delta function, so that means what we have
is this Fourier series if you come back and
use here this is actually if you use the property
of the delta function is -L/2 to L/2 F(t),
now you have this, in the place of this you
have a delta function delta(x-t) DT, so this
is nothing but F(x), if it's a continuous,
okay, otherwise because again using the delta
function property if it is a discontinuous
you can
have, if F is discontinuous at X, X+ and X-
both the places you have a jump, such a jump
you have if it is a discontinuous jump here
to here, okay, so it's minus infinity to here
1, and here to infinity 1 if you consider
so each contribution by the delta is 1/2 F(x+)
+ 1/2 F(x-), so this is exactly what we have
if at discontinuous point, discontinuous point,
continuous point, at continuous point this
is what is the case, okay.
So that's what we have seen that if it is
continuous function this Fourier series need
not converge, and if it converges actually
we strongly assume that it is uniformly convergent,
a Fourier series actually converges to F(x)
in the process we have defined what is a generalized
function and such an example is the delta
function, we made use of this delta function
to show that the series actually converges
to F(x) and average jump value, okay.
So with this we'll do more examples in the
next video for the convergence of the Fourier
series when you consider such examples usually
they are in terms of elementary functions,
they are piecewise smooth so that a Fourier
series convergence is guaranteed, okay. Thank
you very much.
