Hello friends, welcome to my lecture on uniform
convergence of power series. We have already
discussed that when the power series converges
uniformly some function of the power series
is a continuous function. Now as regards the
term by term and also we discussed the integration
term by term of a power series.
So as regards the term by term differentiation
of a power series; we have the following results.
If f(x) = sigma n = 0 to infinity, cn(x ñ
a) to power n convergence for mod of x - a
< R, then the series sigma n = 1 to infinity
nCn(x ñ a) to the power n - 1 has precisely
the same radius of convergence.
So in this series you can see sigma n = 1
to infinity nCn(x ñ a) to the power n - 1
had been obtained by term by term differentiation
of the series sigma n = 0 to infinity Cn(x
ñ a) to the power of n. So, precisely what
we are saying is that, if the given series
has radius of convergence R, then the differentiated
series also has the radius of convergence
R. So let us prove this result.
Let us take an arbitrary point x0 and let
us fix it. So x0 be an arbitrary and but fixed
point such that 0 < mod of x0 - a < R, and
then choose a point x such that mod of x - a
< mod of x0 - a < R. Now the convergence of
the series, sigma n = 0 to infinity, Cn(x
ñ a) raise to the power n, for mod of x - a
< R because we have assumed that series sigma
n = 0 to infinity, Cn (x ñ a) to the power
n has the radius of convergence R.
So the reason of convergence for the series
is given by the inequality mod of x - a < R,
so the convergence of this series for the
region mod of x - a < R implies that series
sigma n = 0 to infinity Cn(x0 ñ a) to the
power n is convergent. Now let us see how
we get this. You have x0 is a point which
belongs to this interval okay, which belongs
to the region mod of x - a < R. So what we
have?
The convergence of this implies that sigma
n = 0 to infinity Cn(x0 ñ a) raise to the
power n is a convergent series. The convergence
sigma n = 0 to infinity Cn(x ñ a) to the
power n for mod of x - a < R implies that
series sigma n = 0 to the infinity Cn(x0 ñ
a) to the power n is convergent because x0
belongs to this interval a ñ R to a + R by
our hypothesis, Okay. Now since the series
is convergent.
Okay we know that for an infinite series the
convergence of an infinite series implies
that the nth term of the series goes to 0
so this will imply that limit n tends to infinity
Cn(x0 ñ a) raise to the power n = 0. The
nth term of the series goes to 0 that is the
necessary condition for the convergence of
an infinity series. Now, let us look at the
sequence Cn(x0 ñ a) to the power n the limit
of the sequence as n goes to infinity 0 implies
that the sequence is bounded.
So we can find the constant A > 0 such that
Cn(x0 ñ a) to the power of n mod of that
is < = A okay. So this implies that the sequence
(x0 ñ a) raise to the power n is a bounded
sequence. So by the definition of the bounded
sequence, we can find the constant a > 0 such
that mod of Cn (x0 ñ a) to the power of n
is < = power a for all n belonging to n. Now
let us choose real number x such that mod
of x ñ a/mod of x0 ñ a.
We assume rho this ratio to be rho then from
this inequality mod of x - a < mod of x0 ñ
a, we follow that rho is < 1. Now when rho
< 1, we get this inequality mod of nCn(x ñ
a) to the power n - 1 is < = to n * a rho
to the power n ñ 1 * x0 - a mod. Okay. So
let us see how we get this. So nCn (x ña)
to the power of n we can write as nCn(x ñ
a) to power of n * (x0 ñ a) to power n/x0
- a to the power n and this I can write nCn
(x0 ñ a) to the power n.
And then we have x0 - a here and we get here
(x ñ a) to the power we have nCn (x ñ a)
to power of n nCn (x ñ a) and we get nCn/
(x0 ñ a) to the power of n - 1. So we have
this nCn (x ñ a) to the power n - 1 we have
okay. So nCn (x ñ a) to power n - 1 we have.
So we take n - 1 here. Okay. Yeah. So we nCn
(x ñ a) to power n - 1 = nCn (x ñ a) to
power n - 1. Now we multiply and divide by
the x0 - a to the power n and then collect
nCn * (x0 ñ a) to the power n because for
this we have the inequality okay.
We have the inequality that Cn * x0 - a to
the power n this is < = a. So what do we get
now let us see this will be so mod of this
okay mod of this = mod of this okay and then
we get mod of this mod of this so let us take
what I mean is that you take the mod of nCn(x
ñ a) to the power n - 1 nCn * x - a to the
power of n - 1 and then you get mod of nCn(x
ñ a) to the power of (n ñ 1) (x0 ñ a) power
to n/x0 -a power to n.
And you get mod of nCn x0 - a to the power
n upper mod of x0 -a and this is x ñ a okay.
Now this is what this is = mod of nCn (x0
ñ a) raise to the power n/mod of x0 - a,
then this is = and this is a rho okay. So
rho to the power n - 1 okay and rho < 1 and
the mod of nCn (x0 ñ a) so mod of nCn (x0
ñ a) to the power n is < = A.
So this is n * a/mod of x0 -a * rho to the
power n - 1 okay. So what we get mod of nCn
x0 mod of nCn (x ñ a) to the power of n - 1
is < n * a < = okay < = mod of n * a rho to
the power n ñ 1/mod of x0 ñ a okay. This
is the rho to the power n - 1okay and mod
of Cn (x0 ñ a) raise to the power n < = so
this is < = n * a mod of x0 - a * rho to the
power n - 1.
Now what we do, now let us look at the convergence
of this series sigma n = 1 to the infinity
n * a rho to the power n - 1 mod of x0 ñ
a okay. The convergence of this implies that
now why it is convergent now you can apply
ratio test so a limit n tends to infinity
let us apply the ratio test. So limit n tends
to the infinity mod of cn + 1, so that means
mod of n + 1 * a * rho to the power of n/mod
of x0 -a * mod of x0 -a/n * a rho to the power
n ñ 1 okay.
So mod of x0 -a we will cancel the mod of
x0 - a; a will cancel okay; rho to the power
n - 1 when divides rho to the power n you
get rho and n + 1/n okay so what we get is
limit n tends to the infinity n + 1/n * rho
which = rho and rho is < 1. Okay. So the limit
is exists. So the limit exist and is < 1 okay.
So this mean that this series by ratio test
it is convergent. Now then this series is
convergent.
This series sigma n = 1 to the infinity nCn(x
ñ a) to power of n - 1 converges informally
okay by test for mod of x0 - a < mod of x0
ña okay. So since x0 is an arbitrary okay
with 0 < mod of x0 - a < R this series sigma
n = 1 to the infinity nCn x - a to the power
n - 1 converges informally for mod of x - a
< R. Now let us take okay so when this series
differentiated series converges informally
for a mod of x0 x - a < R.
We can say that if R dash is its radius of
convergence then R dash must be at least R.
okay. So R dash must be > = R. Now if the
radius of the convergence R of the given series
is infinity since R dash > R = R. R dash is
also infinity and so R = R dash and therefore
the theorem is proved okay. So in the other
case let us take R dash to be > R okay. Our
aim is to show that R dash = R.
So when R dash > R, let us choose a point
x such that a R < mod of x - a < R dash. Then
for this x okay since the radius of the convergence
of the differentiated series is R dash okay
sigma n = 1 to the infinity nCn x - a to the
power n - 1 converges absolutely while the
series sigma n = 0 to the infinity nCn(x ñ
a) to the power n diverges because it is the
radius of the convergence is R okay and to
the point x lies outside.
I mean the circle of the convergence or the
region of the convergence because we are assuming
the mod of x - a > R so then mod of Cn(x ñ
a) to the power n. So let us look at this
we want to arrive at the contradiction okay.
So mod of Cn(x ñ a) to the power of n we
can write as nCn(x ñ a) power to the n - 1
* mod of x ñ a/n okay. Now this is < = nCn
x - a to the power of n - 1 as soon as mod
of x ñ a/n < 1.
So we have already chosen a point x such that
R is < mod of x - a < R. okay. Now let us
take n to be so large that mod of x ñ a/n
becomes < 1. Then for such n okay what do
we notice? mod of Cn(x - a) to the power n
is < = nCn x - a to the power n - 1. So by
comparison test okay since the series sigma
nCn(x - a) to the power n - 1 is convergent
the series sigma Cn(x - a) to the power n
is also convergent, but that is clearly a
contradiction because the series sigma n = 0
to infinity Cn(x - a) to the power n diverges
for x which is mod of x - a. We satisfy mod
of x - a > R okay.
The sigma n = 0 to infinity Cn(x0 - a) to
the power n converges for this x sigma n = 0
to infinity cn x - a to the power n converges
for this x which is false and hence R dash
should be = R. So that is how we prove this
theorem.
Now under the same hypothesis okay. So what
we have now the series sigma n = 0 to infinity
Cn(x - a) to the power n we have seen if it
has a radius of convergence R then the differential
series also has radius of convergence. The
sum function of that will then by F dash x
okay. So when we assume that if we have an
infinite series sigma n = 0 to infinity Cn
(x - a) to the power n with radius of convergence
are then the differentiated series also has
same radius of convergence and the sum function
which was f(x) now it will be f dash x okay.
Let f(x) be sigma n = 0 to infinity cn(x - a)
to the power n it converges in the region
mod of x - a < R then by induction we can
say because we have seen that f dash x = sigma
n = 0 to infinity cn(x - a) to the power n
okay. F has derivatives of all orders and
the kth derivative of f is obtained by differentiating
the infinite series sigma n = 0 to infinity
cn(x - a) to the power n term by term k times.
All the derived series have the same radius
of convergence. So using mathematical induction
on n we get this corollary.
Okay, so now let us give a remark if the radius
of convergence of the power series is infinite.
Suppose the series converges for all values
of x okay then the convergence is not necessarily
uniform on the entire line. We have said that
the series had radius of convergence R and
then the region of convergence given by mod
of x - a < R then the differentiated series
also has the same radius of convergence that
it says that it also converges in the same
region mod of x - a < R.
So this convergence is not uniform on the
entire line okay, but it will be uniform on
a any closed interval okay. So this remark
which we have to notice now a function f defined
in some interval by a convergent power series
is called an analytic function okay. So the
definition of an analytic function here is
that it should be defined in some interval
by a convergent power series. Now if f is
analytic, then the power series representing
f is known as the Taylor series.
Let us say if we have 2 functions f(x) sigma
n = 0 to infinity an x to the power n and
g(x) sigma n = 0 to infinity bn x to the power
n are analytic okay, then the sigma n = 0
to infinity an x to the power n is the Taylor
series of f okay then sigma n = 0 to infinity
an x to the power n is the Taylor series of
f 
so an must be = f(n) 0/n factorial okay and
sigma n = 0 to infinity bn x to the power
n is the Taylor series 
of the function g so bn must be = g(n)0/n
factorial.
So if 2 functions f(x) and g(x) are analytic
and f(n)0 = g(n)0. So when f(n)0 = g(n)0 okay
then an will be = bn for all n okay = 0, 1,
2, 3, and so on okay and so the 2 functions
f(x) and g(x) will be same.
Now vanishing of all coefficients. If a power
series has a positive radius of convergence
and a sum that is identically 0 throughout
the interval of convergence. So let us consider
the power series sigma n = 0 to infinity cn(x
- a) raise to the power n which has a positive
radius of convergence say R and its region
of convergence let us say is given by mod
of x - a < R. The sum of the series is suppose
f(x) then we have f(x) = sigma n = 0 to infinity
cn(x - a) raise to the power n.
Now it is given that the sum is identically
0 throughout its interval of convergence,
so f(x) = 0 we are given f(x) = 0 for all
x satisfying mod of x - a < R okay. Now cn
are given by f(n)(A)/n factorial for all n
= 0, 1, 2, and so on to infinity okay. So
since f(x) = 0 okay for all x we have f(n)(a)
= 0 for all n because f(x) is identically
0 so all other derivatives of f(x) that x
= a are also 0 and therefore cn = 0, cn is
f(n)/n factorial. So cn = 0 for all n and
so on.
So each coefficient of the series the coefficients
cn each coefficient of the series is also
0.
Let us consider the second order differential
equation x double dot + x = 0 or we can say
d square x/dt square + x = 0 and so how we
can find the power series solution of this
differential equation so let us assume the
solution of this differential equation to
be x(t) = sigma n = 0 to infinity cn t raise
to the power n okay. So let us assume that
this power series is a summation of this given
differential equation which has got the radius
of convergence R.
So the series converges for mod of t < R and
R is positive. So then what we will have?
X dot t will be = sigma. Now the first term
here is c0 so when we differentiate that it
will vanish and term series start with n = 1
onwards. So n = 1 to infinity n * cn * t to
the power n - 1. Now we have seen that the
series can be differentiated term by term
in the region mod of t < R so n has the same
radius of convergence x double dot will be
= sigma.
Now n * n - 1 * cn * t raise to the power
n - 2 okay and now n I am going to start with
n = 2 to infinity. So now let us substitute
because x(t) = sigma n = 0 to infinity cn
* t to the power n we have assumed to the
solution of the given equation so let us substitute
these values in the given equation then what
we will have. Sigma n = 2 to infinity n(n
- 1) Cn t raise to the power n - 2 + sigma
n = 0 to infinity cn * t raise to the power
n = 0.
Here now what we will do let us replace n
by say j + 2 okay. n = j + 2 then this will
be = sigma j = 0 to infinity and then we will
have j + 2, j + 1. So I can write j + 1, j
+ 2, cj + 2 t to the power j + sigma n = 0
to infinity cn t raise to the power n = 0.
I can now change the summation/j2 summation/m
okay. So sigma n = 0 to infinity and I can
collect the terms so (n + 1) (n + 2) cn +
2 okay and then I can write like this okay.
So what we will have.
Now we know that when the power series I mean
some function of the power series is identically
0 over an interval so here this is identically
0. This is valid for all t satisfying mod
of t < R. So every term of the series must
be 0 okay. So when we take okay this will
give you okay and t to the power n and the
coefficient of t to the power n = 0 gives
(n + 1) (n + 2) cn + 2 = - cn or I can say
cn + 2 = - cn/(n +1) ( n + 2).
So we can start with n = 0 when n = 0 what
we get? C2 = - c0 and then n = 0 gives 1 * 2
then I can find c3 this relation is known
as recurrence relation okay. It can be recursively
used to determine the values of the coefficient
cn so for different values of n so when n
= 0 c2 we can find from here. This is - c0/1
* 2 then c3 will be - c1 n = 1 so we have
2 * 3 and then we can find c4. C4 will be
= - c2 we are putting n = 2 so 3 * 4 and is
further = c2 = - c0/1 * 2.
So we have - 1 whole square * c0 and then
1, 2, 3, 4. So we have 4 factorial okay. Then
C5 we can find. C5 will be = when you put
n = 3 - c3/3 + 1 that is 4 and then 3 + 2
= 5. So we have and then we can get 5 in terms
of c1 from using this relation so we get - 1
to the power 2 and then c3 is - c1/2 * 3.
So c1/2, 3, 4, 5 okay. So 
what 
we get? This will be = 5 factorial right.
- 1 to the power 2 c1/5 factorial.
Now so what we get is if we go on like this
we can see that c2k okay c2k will be = - 1
to the power k * c0/2k factorial and k will
take values and k can take values 1, 2, 3,
and so on and c2k + 1 will come out to be
- 1 to the power k * c1/2k + 1 factorial.
So let us verify. We have written the values
of c2k and c2k + 1 okay. So you can put k
= 1 where c2 should be = - c0/2 factorial
okay. So this is - c0/2 factorial then you
put k = 2 then we have c4.
C4 will be - 1 to the power 2 c0/4 factorial.
So we get this is c4 okay and here when we
put k = 1 you get c3 as - c1/3 factorial.
So this is - c1/3 factorial and when you put
k = 2 then you get c5 as - 1 to the power
2 c1/5 factorial so you get this. So in general
we can write the value of c2k and c2k + 1
from here and once we have that okay what
we have? xt will be = so then xt sigma n = 0
to infinity cn t to the power n okay.
So we will have c0. So this series will be
= sigma k = 1 to infinity - 1 to the power
k. If I take k = 0 to infinity - 1 to the
power k. c0/2k factorial + sigma k = when
you take c0 and c1 are arbitrary okay. So
k = 0 to infinity - 1 to the power k is t
to the power 2k here okay because we are writing
the value of c2k okay so - 1 to the power
k * c1/2k + 1 factorial * t raise to the power
2k + 1. Luckily what we are doing is that
xt you can split into 2 parts in 2 series
sigma k = 0 to infinity c2k t to the power
2k + sigma k = 0 to infinity c2k + 1 t to
the power 2k + 1 okay.
One series will contain even powers of t,
the other series will contain odd powers of
t. So we can write the series representing
xt function in this manner and then put the
values of c2k and c2k + 1. So we get this
okay. now what we have, this is c0 * sigma
k = 0 to infinity - 1 to the power k t to
the power 2k/2k factorial and then we have
c1 * sigma k = 0 to infinity - 1 to the power
k/2k + 1 factorial t to the power 2k + 1 okay
and we can see that.
This series sigma k = 0 to infinity - 1 to
the power k * t to the 2k/2 factorial represents
the cosine t function and sigma k = 0 to infinity
-1 to the power k * t to the power 2k + 1/2k
+ 1 factorial represents the sine function
okay so xt = c0 cos t + c1 sine t. Now you
note that in this example in this particular
example we are able to write the solution
of the differential equation in a closed form.
We are getting xt = c0 cos t + c1 sin t since
it is a second order differential equation
then will have 2 arbitrary constants and we
have 2 arbitrary constants here c0 and c1
so it is the general solution. Now in this
particular example of course we have got the
solution in the closed form, but in a general
situation we will obtain power series that
do not necessarily represent known functions.
So that is the remark I want to make. In our
next lecture, we shall see more examples of
differential equations where we will see that
the solution cannot be represented in a closed
form that is the solution of the differential
equation general solution comes in the form
of infinite series which do not represent
known function, known elementary functions.
Thank you very much for your attention.
