Ok. So, having proved the previous ah theorem,
so, we will investigate slightly into the
convergence at the at a point of ah continuity
ok.
So, a Fourier ah series of some function f
converges to f 
at a point x, if f of x is written in the
form a naught plus limit N going to infinity
summation k equals 1 to capital N a k cos
k x plus b k sine k x. So, there are various
notions of ah convergence whether it is point
wise convergence, or uniform convergence etcetera
etcetera right. So, I would basically refer
you to analysis lectures for for these various
notions.
Now, what is the saying that I can take a
series infinite series expansion and that
series expansion is very close to the true
function that is it converges to this function
as N goes to infinity right capital N goes
to infinity. So, with this we can state a
theorem. Suppose f is continuous and a 2 pi
periodic function, it 
is continuous and it is 2 pi periodic, then
at each point x, there f dash of x that is
derivative of x is defined the Fourier series
of f at x so, at every point x you have a
an expansion right at x converges to f of
x. So, suppose f is continuous and 2 pi periodic,
then at each point x where the derivative
of the function is defined, the Fourier series
of f at x converges to the true function f
of x. So, we will prove this result there
are many steps in the proof of this result
and it is a long way long theorem.
But I hope you appreciate the details in this
in this result. Now, for a positive integer
capital N let us define the sum S N S N of
x is a naught plus summation k equals 1 to
capital N a k cos k x plus b k k sin k x we
saw this summation. Now what we need, what
we need to show is very important right, we
have to precisely put it it in in mathematical
terms S N of x to f of x as N goes to infinity
right, this is what we need to show. Now for
this 
let us rewrite S N in a slightly different
way ok. So, we do not see this through a sequence
of steps.
So, step 1 so, we substitute the Fourier coefficients
and, when we do that we can write S N of x
a naught is 1 upon 2 pi integral minus pi
2 pi f of t dt and, then a N is 1 upon pi
integral minus pi to pi f of t cos k t dt.
So, basically I write as summation k equals
1 to capital N, this is a series sum f t cos
k t times cos k x. You see deliberately I
have gotten this t and ah x because, the definite
integral right. So, I can I can always write
it in this form.
So, this 1 upon pi could be here I just pulled
it outside, but you can see this is basically
your A N right and, then your B N is you can
put this here itself, this is basically integral
minus pi to pi f of t sin k t 
sin k x and I have to put a dt here, whenever
I write this in integral a dt.
So, now ah we can pull um so, we can we can
we can pull all the terms together right and,
I can I can say I have a cos k t cos k x,
then there is a sin k t sin k x right, I can
pull all the terms together. So, I can say
S N of x. So, and I can use this formula cos
A minus B is cos A cos B, this is the compound
angle formula cos A cos B plus sin A sine
B using this, I call this equation 1, we can
simplify S N of x this is 1 upon pi integral
minus pi to pi f of t there is a half factor
outside. So, 1 1 by 2 pi integral minus pi
to pi f t dt is this term. So, I will put
a dt here ok. So, 1 upon pi f of t is half
factor that dt is this term ok. Now I have
f of t outside I can pull this out right,
I have a cos k t cos k x sin k t sin k x and,
I can write this as summation k equals 1 to
capital N cos k into t minus x right. The
same Fourier series is written like this,
slightly different way ok, this is sort of
straightforward.
Now, in step 2. We have the following lemma,
but writing is a little more unconventional
I could start with sequences of definitions
lemmas and theorems and I say yes invoke that
lemma here, but I do not want to do this this
way because, I want you to get a feel of how
you want to progress in proving the main theorem
and, a step could be potentially a lemma,
or it could be just some portion of a derivation.
Now for any number u belonging to minus pi
to plus pi 1 half plus cos u plus cos 2 u
plus 1 cos N times u, this is basically sin
N plus half u over 2 sin u upon 2 for u naught
equal to 0 and equals N plus half, when u
equals 0.
Then u equals 0 you can just apply L'Hospital
rule and you can; obviously, see that this
is sort of valid to this equation. Let us
um or if you just put u equals 0 directly,
you have N terms here N plus half both the
ways, you can immediately check the veracity
of this result for this condition. Let us
look at this part this is also very easy to
straight forward just a trigonometric sum.
ah
So, whenever you see this series sums you
have to get the Euler formula in your mind
ok. So, e power j u power n is cos n u plus
j sin n u. Now consider 1 half plus summation
k equals 1 to capital N cos k u, now when
k equals ah if I want to replace k equals
1 to N by k equals 0 to N, then what I do
is I put m minus half factor out plus real
part of the summation equals 0 to capital
N e power j u power k this is not really difficult
to see write k equals 0 this is 1.
So, 1 minus half is half you have half here
and the rest is basically take the real part
you are going to get the cos no big deal right,
this is minus half plus real part, now this
is a geometric sum. So, you can write this
in this form over j times N plus 1 u divided
by 1 minus e power j u right. The common ratio
is e power j u and the first term is 1. So,
A into 1 minus R power N over ah m A into
1 minus R power N, or 1 upon 1 minus R R power
N plus 1 over 1 1 minus R ok, you applied
the geometric series sum here is what you
this is what your are going to get.
Now, you simplify this further it is minus
half that is a real part, I will get e power
minus j u upon 2 minus e power j N plus half
you, if I just multiply and divide by e power
minus j u upon 2 for both numerator and denominator
I get it in this form ok. So, what do I get
here? So, I get a minus half plus a real part
now e power minus j theta is cos u upon 2
minus j sin u upon 2 that minus and minus
will cancel it is going to be a plus right
cos theta ok. This is a minus here ah minus
j sin u upon 2 minus, this is a cos N plus
half u 
minus j sin N plus half u ok.
Now this is again cos u upon 2 minus j sin
u upon 2 minus. So, there is some big minus
here and then, I just have to put a brace
here appropriately and there is a plus here
ok. Now, here it is cos u upon 2 plus j sin
u upon 2 it is just routine routine algebra
right. If you are very shrewd you will notice
that just group the j terms appropriately
here, right and then just it will cancel out.
So, this cos u by 2 minus cos u upon 2 will
cancel out. So, the denominator u will be
landing up with minus 2 j sin u upon 2 right.
So, you have it is simplified this carefully,
it is minus 1 half plus 2 sin u upon 2 and
in the numerator, it is sin N plus half u
plus sin u upon 2 and how do I get this, there
is a minus j, there is a minus j, this is
minus 2 j sin u upon 2 right because, it is
cos u by 2 and minus cos u by 2 will cancel
out and, then you will here have cos u upon
2 minus something. And then you grouped j
terms you will have a minus j times sin u
upon 2 plus sin n plus half u right in the
j and j will cancel the negative will cancel
out and, then you will end up with taking
the real part, you will land up with this
equation here.
It is sort of straightforward and, this you
could simplify right because, the sin u upon
2 and this will cancel out this half and half
will cancel out. So, therefore you will land
up with sin n plus half u divided by 2 sin
u upon 2 and, this is when u is not equal
to 0, when u equal to 0 you apply L'Hospital
rule and you will land up with N plus half
because, denominator will cancel out ok.
So, this is indeterminate form this is 0 by
0 form apply L'Hospital rule differentiate.
The numerator and denominator and take the
limits as u goes to 0 and you land up with
N plus half. And this is consistent because,
you just add the terms when u equals 0 you
have N terms N plus half is what we have.
Now, this proves this lemma, but we have just
addressed the second part, second step of
the proof of this theorem.
Then we go to step 3. Let us evaluate the
partial sum of the Fourier series, we will
evaluate the partial sum of the Fourier series.
So, now we have S N of x is 1 upon pi integral
minus pi to plus pi f of t and we have half
plus sum of k equals 1 to capital N cos k
times t minus x dt. Now I write this in this
form 1 upon 2 pi integral minus pi to plus
pi f of t using the previous lemma right,
I can simplify this quantity as sin N plus
half t minus x divided by sin t minus x upon
2 dt. The 2 factor that appeared I just pulled
it here ok, just observe this to here ok.
The factor of 1 1 upon 2 here, write that
do I am pulling it at this step.
Now, if we define define P N of u as 1 upon
2 pi sin N plus half u divided by sin u upon
2, if I define this function P N of you as
this form.
Then I can write S N of x as integral minus
pi to plus pi f of t P N t minus x dt right.
This is this is our sort of simplified form
right, but let us go one step further, let
u equals t minus x just a change of variable.
So, therefore d u is dt, I can write S N of
x is integral minus pi to pi f of u plus x
P N of u d u. So, that it is in this form
this is like the kernel ok, I ponder why the
limits on the definite integral, remained
unchanged though it had to be pi minus x and
minus pi minus x, ah when you have to do a
change of a variable here.
Now, let us go to step 4, there is an interesting
property call this as a lemma minus pi to
pi it is a trivial trivial result P N of u
d u equals 1 that is if you integrate this
kernel it it gives 1. Now, how would you like
prove this result so, basically you write
P N of u in the original form 1 upon pi 1
half plus cos u plus cos 2 u plus so, on till
cos N times u. Now where you integrate all
these functions cos u cos 2 u so, on till
cos n u that is that just vanishes, over the
cycle. And 1 upon 2 pi you integrate from
minus pi to pi it is 1 right, this is basically
straightforward no big deal here.
Now we will go one, one last step which is
step 5. We start where we were in step 3.
So, from the previous lemma right from step
4 lemma I can write f of x is integral minus
pi to plus pi f of x P N u du trivial right,
I just folded f of x inside to integral minus
pi to pi P N of u du is 1. So, it is a very
trivial result just observe that I am integrating
with respect to u not x right I am integrating
with respect to u and not x.
Now, this trick is important. We are home
if we show integral minus pi 2 plus pi f of
u plus x minus f of x times P N of u du heads
to 0 as N goes to infinity, this is the important
part. The second part is no big deal because,
I am integrating with respect to u and f of
x is a constant so, therefore, that is no
big deal. This is important because, then
I take f of u plus x and then I fold it because
now there is no argument to you, which is
appearing in the expansion of the function
and, I have to show that this heads to 0 as
N goes to infinity. Now you appreciate why
this kernel and all these things are playing
your own right.
So, now let us simplify this now 1 upon 2
pi is what we have integral minus pi to pi
f of u plus x minus f of x divided by there
is a negative sign here, divided by sin. So,
what I do is I just replace it by the kernel
sin N plus half times u du. Now invoking 
previous theorem, limit k going to infinity
integral minus pi to plus pi f of x sin k
x heads to 0 as k goes to that is s k goes
to infinity right, except at u equals 0 pi
minus pi, that is some function I am basically
expanding this and when this is basically
modulating as N goes large this basically
vanishes.
So, now let us see what is happening here.
So, let me write this function g of u as f
of u plus x minus f of x over sin u upon 2.
So, this function is continuous 
except at 2 equal to 0 over minus pi and pi
except at u equals 0 right, when u equals
0 it is in the form 0 by 0 form therefore,
it is it is continuous except at this point.
Now the derivative of this function can be
written as limit u going to 0 f of u plus
x minus f of x over u right, it is the f of
x plus u minus f of x over you as you going
to infinity that is the derivative of t of
t function that is the reason we say f dash
exists by hypothesis, where the derivative
exists this is where I mentioned because,
we will make use of this fact.
now consider limit u going to 0 f of u plus
x minus f of x divided by u times u upon 2
divided by sin u upon 2 times 2. So, this
is by definition this is f dash of x times
this is 1 this is 2 this is 2 times f dash
of x. Now let us define g of 0 because, g
of u is what is that function write g of u
is that function which we defined like this
and, we want to look at that limit as u goes
to 0 because that is the point which we are
interested in because, it was discontinuous
point.
So, if you define g of 0 as 2 times f dash
of x at those points x g of u extends across
u equals 0, as a continuous function right
at that point it is discontinuous, if I put
this in this form it extends as a continuous
function right. So, therefore we can we can
conclude ah that this is at at all all the
points that is u equals 0 and the rest of
the points, integral minus pi 2 plus pi f
of u plus x minus f of x times P N u du this
heads to 0 as N goes to infinity right, we
close this theorem here ok.
So, there are 2 subtle points, we had to note
1 is u being equal to 0 and when u is not
equal to 0 and, when u not equal to 0 then
ah we have no problem because, we invoke the
previous theorem and therefore, that that
integral basically vanishes right because
you have some term here which is of this form
times some frequency large frequency here,
and that basically dives to 0 and we invoke
the previous theorem right. So therefore,
this is true but when u is equal to 0 we have
a trouble because we have a 0 by 0 form so,
we have to consider it carefully and using
the derivative definition for the function
f of x, we show that if this is possible and
we extend g of u as a continuous function
with u of 0 point then still we are ok and
therefore, this um this expansion ah the Fourier
series expansion in the limit form it converges
to the function itself ok. So, in the next
lecture we will discuss the convergence at
a point of discontinuity and, then we will
subsequently build ah our notions of various
other forms of convergence of the Fourier
series we will stop here.
