Example

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).

Convergence of

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.

Uniform Convergence of

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

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.

Weierstrass M-test

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.

Region of Convergence

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.

The Derivative of a Power Series

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.

Binomial Series

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

Consequences of the Binomial Expansion

For we have

Replacing by we have

Series Expansion of

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
  
  

Analytic Functions

Definition A real function is called analytic at if there exists a power series with positive radius of convergence such that

Uniquess of Coefficients

Theorem If for then for all

Proof. Compare the th derivative of both sides.

Series Solution of Differential Equations

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

Exercise

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.