Example

Even a sequence of differentiable functions $\{f_n\}$ converging uniformly to a differentiable function, it may not be true that
MATH

Example If
MATH
then $\{f_n\}$ converges uniformly to the function $f(x)=0$, but
MATH
and MATH does not always exist (for example, it does not exist if $x=0).$

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 $\{f_n^{\prime }\}$ converge uniformly (to some continuous function).

Convergence of $f_n^{\prime }$

Theorem. Suppose that $\{f_n\}$ is a sequence of functions which are differentiable on $[a,b],$ with integrable derivatives $f_n^{\prime }$ and that $\{f_n\}$ converges (pointwise) to $f.$ Suppose, moreover, that $\{f_n^{\prime }\}$ converges uniformly on $[a,b]$ to some continuous function $g$. Then $f$ is differentiable and
MATH

Proof. Applying a previous theorem (MATH) to the interval $[a,x]$, we see that for each $x$ we have
MATH
Since $g$ is continuous, it follows that MATH $f_n^{\prime }(x)$ for all $x$ in the interval $[a,b].$

Now that the basic facts about uniform limits have been established, it is clear how to treat functions defined as infinite sums,
MATH
This equation means that
MATH
our previous theorems apply when the new sequence
MATH
converges uniformly to $f$. Since this is the only case we shall ever be interested in, we single it out with a definition.

Uniform Convergence of MATH

Definition. The series MATH converges uniformly (more formally: the sequence is uniformly summable) to $f$ on $A$, if the sequence
MATH
converges uniformly to $f$ on $A.$

We can now apply the previous three theorems to uniformly convergent series; the results may be stated in one common corollary.

Corollary. Let MATH converge uniformly to $f$ on $[a,b].$

Corollary

(1) If each $\,f_n$ is continuous on $[a,b]$, then $f$ is continuous on $[a,b].$

(2) If $f$ and each $\,f_n$, is integrable on $[a,b],$ then
MATH
Moreover, if MATH converges (pointwise) to $f$ on $[a,b],$ each $\,f_n$ has an integrable derivative $\,f_n^{\prime }$ and MATH converges uniformly on $[a,b]$ to some continuous function, then

(3) MATH for all $x$ in $[a,b].$

PROOF (1) If each $f_n$ is continuous, then so is each MATH, and $f$ is the uniform limit of the sequence MATH, so $f$ is continuous.

(2) Since MATH converges uniformly to $f,$ it follows that
MATH

(3) Each function MATH is differentiable, with derivative MATH and MATHconverges uniformly to a continuous function, by hypothesis. It follows that
MATH

At the moment this corollary is not very useful, since it seems quite difficult to predict when the sequence MATH 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 $\{f_n\}$ be a sequence of functions defined on $A$, and suppose that $\{M_n\}$ is a sequence of numbers such that
MATH
Suppose moreover that MATH converges. Then for each $x$ in $A$ the series MATH converges (in fact, it converges absolutely), and MATH converges uniformly on $A$ to the function
MATH

PROOF. For each $x$ in $A$ the series MATH converges, by the comparison test; consequently MATH converges (absolutely). Moreover, for all $x$ in $A $ we have
MATH

Since MATH converges, the number MATH can be made as small as desired, by choosing $n$ sufficiently large.

Region of Convergence

Theorem. If MATH converges at $x_1\ne 0,$ then the two series
MATH
and
MATH
both converges absolutely for MATH

2. If the series MATH diverges at $x=x_1$ then MATH diverges for MATH

Proof. If the series MATH converges then MATH Thus there exists $N$ so that MATH for $n\ge N.$ Fix $x$ with MATH Choose $s$ such that MATH Then
MATH
for $n\ge N.$ Since the geometric series MATH converges, it follows that MATH converges. The inequality also gives
MATH
for $n\ge N.$ Since the series MATH converges (comparison test), it follows that MATH converges and so MATH converges.

From this theorem we see that there are exactly three possibilities for a power series:

case 1) MATH converges only for $x=0.$ For example MATH

case 2) MATH converges everywhere absolutely. For example MATH

case 3) There is a unique positive real number $r$ such that MATH converges absolutely for $\left| x\right| <r$ and MATH diverges for MATH For example MATH

This unique number is called the radius of convergence of the power series MATH The maximal interval on which a power series converges is called the interval of convergence. The possible intervals of convergence are:

case 1) MATH

case 2) $\{0\}.$

case 3) MATH

Examples:

1) MATH converges for $x=1$ and diverges for $x=-1,$ so it has $(-1,1]$ as the interval of convergence.

2) MATH converges for $x=1$ and $x=-1.$ If $\left| x\right| >1$ then MATH so MATH diverges. Therefore it has $[-1,1]$ as the interval of convergence.

3) Consider MATH From the ratio test we see that
MATH
as MATH If MATH the series converges. If MATH the series diverges. For $x=\pm 6$ the series diverges. Therefore it has $(-6,6)$ as the interval of convergence.

4) Consider MATH From the ratio test we see that
MATH
as MATH so the series converges for any $x.$ Therefore it has $(-\infty ,\infty )$ as the interval of convergence.

5) Consider MATH For MATH we have MATH so the series diverges for any $x.$ Therefore it has $\{0\}$ as the interval of convergence.

The Derivative of a Power Series

Theorem If the power series MATH has radius of convergence $r>0,$ then $f$ is differentiable at $x$ with MATH and
MATH

Proof. Let $\epsilon >0.$ We write
MATH
as
MATH
we show that the absolute value of each of the three parts is less than $\epsilon /3.$

(1) Choose $x_0$ with MATH We will only consider $h$ with MATH From the identity
MATH
we see that
MATH
Since
MATH
we have
MATH
But the series MATH converges, so if $N$ is sufficiently large, then
MATH
This mean that
MATH

(2) Since MATH converges, it follows that if $N$ is sufficiently large, then
MATH

(3) Now choose $N$ so that both (1) and (2) are true. Then
MATH
since the polynomial function $\sum_{n=0}^Na_nx^n$ is differentiable at $x.$ Thus
MATH
for $\left| h\right| $ sufficiently small.

Binomial Series

Theorem If $m$ is any real number, then
MATH

MATH
provided that MATH

Proof. Let $f(x)=(1+x)^m.$ Then
MATH

MATH

MATH

MATH
From Taylor's Theorem, we have
MATH

MATH
where
MATH

MATH

It remains to show that
MATH
Now
MATH
The function MATH attains maximum at $t=0$ and so MATH Therefore
MATH

MATH
From the ratio test
MATH
as MATH we see that the series
MATH
converges; therfore
MATH
and so MATH

Consequences of the Binomial Expansion

For $\left| x\right| <1$ we have
MATH

MATH

MATH

MATH

MATH

MATH

MATH

MATH

MATH

MATH

MATH

MATH

Replacing $x$ by $-x$ we have
MATH

Series Expansion of $\ln (1+x)$

Theorem For $-1<x\le 1$ we have
MATH

Proof. For $\left| t\right| <1$ we have
MATH
Therefore
MATH

MATH

From the identity
MATH
we have
MATH
for $x\ge 0.$ For $x\ge 0$ we have the estimate
MATH
As MATH this last term tends to $0$ for $x=1,$ so
MATH

Analytic Functions

Definition A real function $f$ is called analytic at $x_0$ if there exists a power series MATH with positive radius of convergence $r$ such that
MATH

Uniquess of Coefficients

Theorem If MATH for MATH then $a_n=b_n$ for all $n.$

Proof. Compare the $n$th derivative of both sides.

Series Solution of Differential Equations

Example Find the analytic solution to
MATH
Here we are asking for a solution of the form
MATH
so let us assume that $y$ is such solution. Then
MATH
Therefore
MATH
The left hand side equals
MATH
It follows from the uniqueness of the coefficients that
MATH
for all $n.$ From this we arrive at the recurrence relation
MATH
The initial condition $y(0)=1$ gives
MATH
and the initial condition $y^{\prime }(0)=0$ gives
MATH
Therefore
MATH
and
MATH
From this we have
MATH

Example Let $k\ge 0$ be a constant. Assume that the power series MATH satisfies the Bessel equation
MATH
Find the coefficients $a_n.$

Write
MATH
We then have
MATH

MATH

MATH
Since $y$ is a solution, we have
MATH
For $n=k:$
MATH
we allow $a_k$ to be an arbitrary constant.

For $n=k+1:$
MATH
this shows that $a_{k+1}=0.$

For $n\ge k+2:$ from the relation
MATH
we have
MATH
Thus
MATH

MATH

MATH

MATH

MATH

MATH

MATH
The solution
MATH

MATH
is known as the Bessel function of the first kind of order $k$ $.$

Exercise

Expand $g(x)$ as indicated and specify the values of x for which the expansion is valid.

1. MATH in powers of $x-1.$

2. MATH in powers of $x-2.$

3. $g(x)=2x^5+x^2-3x-5$ in powers of $x+1.$

4. $g(x)=x^{-1}$ in powers of $x-1.$

5. $g(x)=(1+x)^{-1}$ in powers of $x-1.$

6. $g(x)=(b+x)^{-1}$ in powers of $x-a,a\ne -b.$

7. $g(x)=(1-2x)^{-1}$ in powers of $x+1.$

8. $g(x)=e^{-4x}$ in powers of $x+1.$

9. $g(x)=\sin x$ in powers of $x-\pi .$

10. $g(x)=\sin x$ in powers of $x-\frac \pi 2.$

11. $g(x)=\cos x$ in powers of $x-\pi .$

12. $g(x)=\cos x$ in powers of $x-\frac \pi 2.$

13. MATH in powers of $x-1.$

14. $g(x)=\sin \pi x$ in powers of $x-1.$

15. $g(x)=\ln (1+2x)$ in powers of $x-1.$

16. $g(x)=\ln (2+3x)$ in powers of $x-4.$

Expand $g(x)$ as indicated.

17. $g(x)=x\ln x$ in powers of $x-2.$

18. $g(x)=x^2+e^{3x}$ in powers of $x-2.$

19. $g(x)=x\sin x$ in powers of $x.$

20. $g(x)=\ln (x^2)$ in powers of $x-1.$

21. $g(x)=(1-2x)^{-3}$ in powers of $x+2.$

22. $g(x)=\sin ^2x$ in powers of $x-\frac \pi 2.$

23. $g(x)=\cos ^2x$ in powers of $x-\pi .$

24. $g(x)=(1+2x)^{-4}$ in powers of $x-2.$

25. $g(x)=x^n$ in powers of $x-1.$

26. $g(x)=(x-1)^n$ in powers of $x-1.$

27. $g(x)=e^{-x}$ in powers of $x-a.$

Find the interval of convergence.

1. $\sum kx^k$ 2. $\sum \frac 1kx^k$ 3. MATH 4. MATH

5. MATH 6. MATH 7. MATH 8.MATH

9. MATH 10. MATH 11. MATH 12.MATH

13. MATH 14. $\sum ka^kx^k$ 15. MATH 16.MATH

17. MATH 18. MATH 19. MATH 20.MATH

21. MATH 22. MATH

23. MATH 24. MATH

Expand in powers of $x.$

1. $xe^{5x^2}$ 2. $\ln (1+x^2)$ 3. MATH

4. $a^x$ 5. $(x+x^2)(\sin x^2)$ 6. MATH

This document created by Scientific WorkPlace 4.0.