Even a sequence of differentiable functions
converging uniformly to a differentiable function, it may not be true that
Example If
then
converges uniformly to the function
,
but
and
does not always exist (for example, it does not exist if
Despite such examples, the Fundamental Theorem of Calculus practically guarantees that some sort of theorem about derivatives will be a consequence of uniform convergence; the crucial hypothesis is that converge uniformly (to some continuous function).
Theorem. Suppose that
is a sequence of functions which are differentiable on
with integrable derivatives
and that
converges (pointwise) to
Suppose, moreover, that
converges uniformly on
to some continuous function
.
Then
is differentiable and
Proof. Applying a previous theorem
()
to the interval
,
we see that for each
we have
Since
is continuous, it follows that
for all
in the interval
Now that the basic facts about uniform limits have been established, it is
clear how to treat functions defined as infinite sums,
This equation means that
our previous theorems apply when the new sequence
converges uniformly to
.
Since this is the only case we shall ever be interested in, we single it out
with a definition.
Definition. The series
converges uniformly (more formally: the sequence is
uniformly summable) to
on
,
if the sequence
converges uniformly to
on
We can now apply the previous three theorems to uniformly convergent series; the results may be stated in one common corollary.
Corollary. Let converge uniformly to on
(1) If each is continuous on , then is continuous on
(2) If
and each
,
is integrable on
then
Moreover, if
converges (pointwise) to
on
each
has an integrable derivative
and
converges uniformly on
to some continuous function, then
(3) for all in
PROOF (1) If each is continuous, then so is each , and is the uniform limit of the sequence , so is continuous.
(2) Since
converges uniformly to
it follows that
(3) Each function
is differentiable, with derivative
and
converges
uniformly to a continuous function, by hypothesis. It follows that
At the moment this corollary is not very useful, since it seems quite difficult to predict when the sequence will converge uniformly. The most important condition which ensures such uniform convergence is provided by the following theorem; the proof is almost a triviality because of the cleverness with which the very simple hypotheses have been chosen.
Theorem. Let
be a sequence of functions defined on
,
and suppose that
is a sequence of numbers such that
Suppose moreover that
converges. Then for each
in
the series
converges (in fact, it converges absolutely), and
converges uniformly on
to the function
PROOF. For each
in
the series
converges, by the comparison test; consequently
converges (absolutely). Moreover, for all
in
we have
Since converges, the number can be made as small as desired, by choosing sufficiently large.
Theorem. If
converges at
then the two series
and
both converges absolutely for
2. If the series diverges at then diverges for
Proof. If the series
converges then
Thus there exists
so that
for
Fix
with
Choose
such that
Then
for
Since the geometric series
converges, it follows that
converges. The inequality also gives
for
Since the series
converges (comparison test), it follows that
converges and so
converges.
From this theorem we see that there are exactly three possibilities for a power series:
case 1) converges only for For example
case 2) converges everywhere absolutely. For example
case 3) There is a unique positive real number such that converges absolutely for and diverges for For example
This unique number is called the radius of convergence of the power series The maximal interval on which a power series converges is called the interval of convergence. The possible intervals of convergence are:
case 1)
case 2)
case 3)
Examples:
1) converges for and diverges for so it has as the interval of convergence.
2) converges for and If then so diverges. Therefore it has as the interval of convergence.
3) Consider
From the ratio test we see that
as
If
the series converges. If
the series diverges. For
the series diverges. Therefore it has
as the interval of convergence.
4) Consider
From the ratio test we see that
as
so the series converges for any
Therefore it has
as the interval of convergence.
5) Consider For we have so the series diverges for any Therefore it has as the interval of convergence.
Theorem If the power series
has radius of convergence
then
is differentiable at
with
and
Proof. Let
We write
as
we show that the absolute value of each of the three parts is less than
(1) Choose
with
We will only consider
with
From the identity
we see that
Since
we have
But the series
converges, so if
is sufficiently large, then
This mean that
(2) Since
converges, it follows that if
is sufficiently large, then
(3) Now choose
so that both (1) and (2) are true. Then
since the polynomial function
is differentiable at
Thus
for
sufficiently small.
Theorem If
is any real number, then
provided that
Proof. Let
Then
From Taylor's Theorem, we have
where
It remains to show that
Now
The function
attains maximum at
and so
Therefore
From the ratio test
as
we see that the series
converges; therfore
and so
For
we have
Replacing
by
we have
Theorem For
we have
Proof. For
we have
Therefore
From the identity
we have
for
For
we have the estimate
As
this last term tends to
for
so
Exercise: Show that
Exercise: Show that for
Exercise: Show that
Exercise: Show that
Definition A real function
is called analytic at
if there exists a power series
with positive radius of convergence
such that
Theorem If for then for all
Proof. Compare the th derivative of both sides.
Example Find the analytic solution to
Here we are asking for a solution of the form
so let us assume that
is such solution. Then
Therefore
The left hand side equals
It follows from the uniqueness of the coefficients that
for all
From this we arrive at the recurrence relation
The initial condition
gives
and the initial condition
gives
Therefore
and
From this we have
Example Let
be a constant. Assume that the power series
satisfies the Bessel equation
Find the coefficients
Write
We then have
Since
is a solution, we have
For
we allow
to be an arbitrary constant.
For
this shows that
For
from the relation
we have
Thus
The solution
is known as the Bessel function of the first kind of order
Expand as indicated and specify the values of x for which the expansion is valid.
1. in powers of
2. in powers of
3. in powers of
4. in powers of
5. in powers of
6. in powers of
7. in powers of
8. in powers of
9. in powers of
10. in powers of
11. in powers of
12. in powers of
13. in powers of
14. in powers of
15. in powers of
16. in powers of
Expand as indicated.
17. in powers of
18. in powers of
19. in powers of
20. in powers of
21. in powers of
22. in powers of
23. in powers of
24. in powers of
25. in powers of
26. in powers of
27. in powers of
Find the interval of convergence.
1. 2. 3. 4.
5. 6. 7. 8.
9. 10. 11. 12.
13. 14. 15. 16.
17. 18. 19. 20.
21. 22.
23. 24.
Expand in powers of
1. 2. 3.
4. 5. 6.
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