Power Series and Uniform Continuity

Power Series

Convergence

A sequence $x=(x_n)$ converges if:

\begin{align*} \forall \varepsilon > 0, \quad \exists N:& \ \\ &n \geq N \implies |x_n - x | < \varepsilon \end{align*}

and it is hence expressed:

\lim(x_n) = x

A series is generated by a sequence,

If $(a_n)$ is a sequence, the series is $(S_n)$ :

\begin{align*} S_1 &= a_1 \\ s_2&= S_1 + a_2 \\ S_3&= S_2 + a_3\\ S_4&= S_3+ a_4\\ \cdots \end{align*}

The series is convergent if:

\begin{align*} \forall \varepsilon > 0, \exists N:&\\ & n \geq N \ \implies \left| S_n - L \right| < \varepsilon \end{align*}

The series is absolutely convergent if $\left( \left| S_n \right| \right)$ is convergent.

The notation for series used is:

\begin{align*} \sum^{\infty}_{n=1} \left[ \left( a_n \right) \right] = \sum \left( a_n \right) = \lim_{}\left[ S_n \right] \end{align*}

Be mindful that this notation is used ambiguously to represent both:

The infinite Series
The limit value of the series.

In practice however the ambiguity is a non-issue because context will discern the difference.

Sequences of Functions¹

We can have sequences of real numbers, and similarly we can have sequences of functions.

Sequences of functions can converge in two ways:

Pointwise
uniformly

Uniform convergence is important because it preserves term properties to the limit function which will be seen.

Define a sequence of functions:

Let $A \subseteq \mathbb{R} , n \in \mathbb{N}$ and take some function $f$ :

\begin{align*} f_n: A \rightarrow \mathbb{R} \end{align*}

It is said that $\left( f_n \right)$ is a sequence of functions on $A$ to $\mathbb{R}$ .

For every $x \in A$ there will be a sequence of real numbers:

\begin{align*} f_n\left( x \right) \end{align*}

For some values of $x$ , the sequence may converge, for others it may not.

The point of convergence is $\lim_{}\left[ f_n\left( x_n \right) \right]$ which depends on the choice of $x$ .
Thus we could Create a set of all $x \in A$ forwhich $\left( f_n\left( x \right) \right)$ converges.
- This set would be a domain for a function $f\left( x \right)$ that would act as the limit of the sequence $\left( f_n\left( x \right) \right)$ .

Pointwise Convergence

Take some function:

\begin{align*} f: A_0 \rightarrow \mathbb{R} \qquad \left( A_0 \subseteq A \subseteq \mathbb{R} \right) \end{align*}

We say that the sequence is pointwise convergent if,

for every $x \in A_0$
- The sequence $\left( f_n\left( x \right) \right)$ converges to $f\left( x \right)$ .
e.g. $\ \$ consider $g_n\left( x \right) : = x^n$ ;

\begin{align*} \left( g_n\left( x \right) \right) = \left( x, x^2, x^3, x^4 \cdots \right) \end{align*}

If $- 1 < x < 1 \quad$ , then $x$ is a fraction or zero so:

\begin{align*} \left( x, x^2, x^3, x^4 \cdots 0 \right) \qquad \text{ Converges to 0} \end{align*}

if $x = 1$ , then:

\begin{align*} \left( x, x^2, x^3 \cdots \right) = \left( 1, 1^2, 1^3, \cdots 1 \right) \qquad \text{Converges to 1} \end{align*}

if $x= - 1$ , then:

\begin{align*} \left( x, x^2, x^3, \cdots \right) = \left( 1, - 1, 1, - 1, \cdots \pm 1 \right) \qquad \text{Divergent} \end{align*}

if $\left| x \right| > 1$ , then:

\begin{align*} \left( x, x^2, x^3, \cdots \infty \right) \qquad \text{Divergent} \end{align*}

So $\lim_{}\left[ g_n\left( x \right) \right]$ on the set $(- 1, 1]$ where:

\begin{align*} g\left( x \right) = \begin{cases} 0, \qquad \text{for} \enspace \left( - 1<x<1 \right) \\ 1,\qquad \text{for} \enspace \left( x = 1 \right) \end{cases} \end{align*}

Definition

An alternative, but equivalent definition for pointwise convergence is:

\begin{align} \forall \varepsilon>0, \enspace \forall &x \in A_0, \enspace \exists N:\\ \ \\ &n \geq N \implies \left| f_n\left( x \right) - f\left( x \right) \right| < \varepsilon \end{align}

Where $N$ is a function of $x$ and $\varepsilon$ , i.e.:

$N = N\left( \varepsilon, x \right)$ .

###Definition of Power Series A power series is a series of the form:

\begin{align} \sum^{\infty}_{n= 0} \left[ c_nx^n \right] = c_0+ c_1x+ c_2x^2+ c_3x^3 + \cdots \end{align}

More generally a series will be of the form:

\begin{align} \sum^{\infty}_{n= 0} \left[ c_n \left( x- a \right) ^n \right] = c_0 + c_1\left( x- a \right) + c_2\left( x- a \right) ^2 + \cdots \end{align}

Convergence of Power Series

For any given power series of the form $\sum^{\infty}_{n= 0} \left[ c_n \left( x- a \right) ^n \right]$ there are only three possibilities:

The series converges only when $x = a$
The series converges for all x.
There is a positive number $R$ such that the series converges if $\left| x- a \right| < R$ and diverges if $\left| x - a \right| > R$ .

Why power Series

The whole idea of power series is represengint a known function as an infinite series, this is useful for integrating functions that don't have elementary anti8derivatives.

Take for example the geometric series:

\begin{align} \sum^{\infty}_{n= 0} \left[ ax^n \right] &= \frac{a}{1 - x}\\ & \implies f\left( x \right) = \frac{a}{1-x} = \sum^{\infty}_{n= 0} \left[ ax^n \right] \end{align}

Example

Find a power series representation for $f\left( x \right) = \frac{x^3}{x + 2}$

\begin{align} \frac{1}{2 + x} = \frac{1}{2\left( 1+ \frac{x}{2} \right) }\\ &= \frac{1}{2\left( 1 - \left( - \frac{x}{2} \right) \right) }\\ &= \frac{1}{2} \cdot \sum^{\infty}_{n= 0} \left[ - \frac{x}{2} \right] ^n\\ &= \sum^{\infty}_{n= 0} \left[ \frac{\left( - 1 \right) ^n}{2^{n + 1}} \cdot x^n \right] \\ \implies \frac{x^3}{2 + x } &= \sum^{\infty}_{n= 0} \left[ \frac{\left( -1 \right) ^n}{2^{n+ 1}}\cdot x^{n+ 3} \right] \end{align}

because this is a geometric series, it converges when:

\begin{align} \left| - \frac{x}{2} \right| &< 1\\ & \implies x \in \left( - 2, 2 \right)\\ & \tiny{ \text{ This is known as the radius of convergence $\left( R= 2 \right) $} } \end{align}

So from this we could show something like:

\begin{alignat}{5} f\left( x \right) = &\frac{x^3}{2+ x} &=& &\sum^{\infty}_{n= 0} &\left[ \frac{\left( - 1 \right) ^n}{2^{n+ 1}}\cdot x^{n+ 3} \right]&\\ \int^{}_{} &\frac{x^3}{2+ x} \operatorname{d}x \enspace &=& \enspace C + &\sum^{\infty}_{n= 0} &\left[ \frac{\left( - 1 \right) ^n}{2^{n+ 1}} \cdot \frac{x^{n+ 4}}{n+ 4} \right] & \end{alignat}

The important part with all of this is that it works with Taylor Series.²

Convergence of Power Series

if $\sum^{\infty}_{n= 0} \left[ a_n\cdot \left( z- z_0 \right)^n \right]$ is convergent for some $z= \alpha$ then,

It will converge absolutely for all values of $\left| z- z_0 \right| < \left| z- \alpha \right|$

Uniform Convergence of Power Series

if $\sum^{\infty}_{n= 0} \left[ a_n\cdot \left( z- z_0 \right) ^n \right]$ converges when $z = \alpha$ but $\left( \alpha \neq z_0 \right)$ :

Then the series converges uniformly in any open-neighbourhood:

\begin{align*} \left| z- z_0 \right| \leq r&\\ & \text{Where} \enspace r = \left| z_0 - \alpha \right| \end{align*}

The sum of the series represents an analytic function, i.e.:

\begin{align*} f\left( z \right) = \sum^{\infty}_{n= 0} \left[ a_n\cdot \left( z- z_0 \right) ^n \right] \end{align*}

Such that $f\left( z \right)$ is an analytic function that can also be represented by the power series

Circle of Convergence

$\sum^{\infty}_{n= 0} \left[ a_n\cdot \left( z- \alpha \right) ^n \right]$ is convergent only for $\left| z- \alpha \right| <R$ :

if $R= 0$ , the series is only convergent for $z= \alpha$
if $r = \infty$ , the series is convergent $\forall z\in \mathbb{C}$
if $R \in \mathbb{R} ^+$ , the series is convergent on some open disc centred at $\alpha$ of radius R.

Taylor Series³

If a function is analytic at $\alpha$ and in an open disc $\left| z- \alpha \right| <R$ , there will always be a power series representation of $f\left( z \right)$ :

\begin{align*} f\left( z \right) = \sum^{\infty}_{n= 0} \left[ a_n\cdot \left( z- \alpha \right) ^n \right] , \qquad \left| z- z_0 \right| < r&\\ \text{where:}\enspace a_n &= \frac{f^{\left( n \right) }\left( z_0 \right) }{n!} \end{align*}

This can be shown using the Cauchy Integral Formula, however a clearer justification is:

If a function a power series representation for a function exists:

\begin{align*} f\left( z \right) &= \sum^{\infty}_{n= 0} \left[ c_n\cdot \left( z- \alpha \right) ^n \right] \qquad {\tiny \text{$C_n$ and $\alpha$ are complex constants}}\\ &= c_0 + c_1\left( z- \alpha \right) + c_2\left( z- \alpha \right) ^2 \cdots\\ \implies f \left( \alpha \right) &= c_0\times 0^0 + c_1\times 0 + 0 + \cdots \\ &= c_0\\ \end{align*}

Consider the first derivative:

\begin{align*} f'\left( z \right) &= c_1 + 2\cdot c_2\left( z- \alpha \right) + 3\cdot c_2\cdot \left( z\alpha \right) ^2\cdots\\ \implies f'\left( \alpha \right) &= c_1 \end{align*}

Consider the second derivative:

\begin{align*} f''\left( z \right) &= 2\cdot c_2+ 3\times 2\cdot \left( z- \alpha \right) + 4\times 3\times 2\cdot \left( z- \alpha \right) ^2+ \cdots\\ &= 2!c_2 + 3!\left( z- \alpha \right) + 4!\cdot \left( z- \alpha \right) ^2+ \cdots\\ f''\left( \alpha \right) &= 2c_2 \end{align*}

By this logic the nth derivative will be:

\begin{align*} f^{\left( n \right) }&= n!\cdot c_n\\ \implies c_n&= \frac{f^{\left( n \right) }\left( \alpha \right) }{n!} \end{align*}

We will always be able to find $c_n$ where $f\left( z \right)$ is analytic and clearly the power series will be convergent (i.e. to $f\left( z \right)$ ) on a radius of convergence where $f\left( z \right)$ is analytic, so:

\begin{align*} f\left( z \right) &= \sum^{\infty}_{n= 0} \left[ \frac{f^{\left( n \right) }\left( \alpha \right) }{n!} \times \left( z- \alpha \right) \right] , \qquad & \left| z- z_0 \right| <R \\ &&{\tiny \text{$R$ is the radius of the open disc of analyticity}} \end{align*}

The Cauchy Hadamard Theorem

If we have a power series:

\begin{align*} \sum^{\infty}_{n= 0} \left[ a_n\cdot \left( z- z_0 \right) ^n \right] \end{align*}

Then we can find the radius of convergence:

\begin{align*} l&= \limsup{ \left[ \left| a_n \right| ^{\frac{1}{n}} \right] }\\ \ \\ & \implies R = \frac{1}{l} \end{align*}

Further if

$\left| a_n \right| \neq_0$ and $\lim_{}\left[ \left| \frac{a_{n+ 1}}{a_n} \right| \right] = R$ Then:
$\lim_{}\left[ \left| a_n \right| ^{\frac{1}{n}} \right] = \lim_{}\left[ \left| \frac{a_{n+ 1}}{a_n} \right| \right] = R$

Generally the $n^{\text{th}}$ root test is more powerful than the ratio test, however the ratio test is the only test that can deal with factorials⁴, so it is important to have it in our toolbox.

Uniform Continuity

What is uniform continuity

Imagine the function $y= \frac{1}{x}$ and consider the interval $\left( 0,\infty \right)$ ,

Consider the limit value of 2 (e.g. the line $y= 2$ )

drawing

If I choose some $\varepsilon$ value, there is always a corresponding $\delta$ -value, this $\delta$ -value will depend on the $\varepsilon$ -value:

drawing2

so the function $f\left( x \right) = \frac{1}{x}$ is continuous because:

$\varepsilon$ can be chosen anywhere
Any $\varepsilon$ value will have a corresponding $\delta$ value.

The function would be uniformly continuous, if:

$\varepsilon$ can be chosen anywhere
The Corresponding $\delta$ value will exist AND not change size wherever $\varepsilon$ is chosen.

So in this case the function is not uniformly continuous, because if I move $\varepsilon$ down, $\delta$ would have to get larger, so it IS NOT uniformly continuous:

drawing3

So basically:

A function is continuous if a $\delta$ -value always exists and can be described as a function:
- $\delta = \delta\left( \varepsilon, x \right)$
A function is uniformly continuous if a $\delta$ -value always exists and can be described as a function only of $\varepsilon$ :
- $\delta = \delta\left( \varepsilon \right)$

Cantor's Theorem

If a function is continuous on an interval $\left[ a,b \right]$ , it is uniformly continuous on that interval.

The reasoning being that basically you could chose the smallest $\delta$ -value that will work at all points on that interval

Why is uniform continuity important?

Something like,

if $f\left( x \right)$ is uniformly continuous
Then:

\int^{}_{} \lim_{}\left[ f\left( x \right) \right] \operatorname{d}x \iff \lim_{}\left[ \int^{}_{} f\left( x \right) \operatorname{d}x \right]

but I'm not sure about this so don't quote me, we don't do it here so don't worry about it too much, we just need to show that we understand it the idea of uniformly continuous functions.

Problem Example

Prove that the function $f\left( x \right) = \frac{1}{1+ x^3}$ is uniformly continuous on the interval $[1, \infty)$

So the first thing to notice is that Cantor's Theorem cannot be applied because it is an open interval.

State the Definition $f\left( x \right)$ is uniformly convergent if:

\begin{align*} \forall x,y \in [1, \infty), \enspace \forall \varepsilon>0,& \enspace \exists \delta\left( \varepsilon \right) :\\ &0< \left| x- y \right| <\delta \implies \left| f\left( x \right) - f\left( y \right) \right| < \varepsilon \end{align*}

Work Backwards from the $\varepsilon$ Definition

\begin{align*} \left| f\left( x \right) - f\left( y \right) \right| &= \left| \frac{1}{1+ x^3} - \frac{1}{1+ y^3}\right| \\ &= \frac{ \left| y^3- x^3 \right| }{ \left| 1+ x^3 \right| \times \left| 1+ y^3 \right| }\\ &\leq \frac{ \left| y^3- x^3 \right| }{ \left| 1+ x^3 \right| }\\ &= \frac{ \left| y- x \right| \cdot \left| y^2+ xy+ x^2 \right| }{ \left| 1+ x^3 \right| }\\ \text{Without loss of generality assume that $y<x$}\\ &\leq \left| y- x \right| \times 3\cdot \frac{ \left| x^2 \right| }{ \left| 1+ x^3 \right| }\\ \text{Recall that $x\geq 1 \implies \frac{1}{ \left| 1+ x^3 \right| }< \frac{1}{ \left| x \right| ^3}$}\\ &\leq \left| y- x \right| \cdot \frac{1}{ \left| x \right| } \cdot 3\\ &\leq \left| y- x \right| \cdot 3\\ &\leq 3\cdot \delta\\ \end{align*}

So choose \delta :

\begin{align*} 3\delta &\leq \varepsilon\\ \delta &\leq \frac{\varepsilon}{3} \end{align*}

$\therefore$ it is sufficient to choose $\delta\leq \frac{\varepsilon}{3}$ .

Proof

\begin{align*} \forall x,y \in [1,\infty), \enspace \forall \varepsilon>0,& \enspace \exists \delta \leq \frac{\varepsilon}{3} \end{align*}

\begin{align*} \left| x- y \right| < \delta \implies \left| f\left( x \right) - f\left( y \right) \right| &\leq \left| \frac{1}{1+ x^3} - \frac{1}{1+ y^3} \right| \\ &\leq \left| y- x \right| \cdot \left( \frac{ \left| y^2 \right| + \left| xy \right| + \left| x^2 \right| }{ \left| 1+ x^3 \right| \cdot \left| 1+ y^3 \right| } \right) \\ &\leq \left| y- x \right| \cdot \left( \frac{y^2}{ \left| 1+ y^3 \right| } + \frac{ \left| x \right| \cdot \left| y \right| }{ \left| 1+ x^3 \right| \cdot 1+ y^3} + \frac{ \left| x^2 \right| }{ \left| 1+ x^3 \right| } \right) \\ &\leq \left| y- x \right| \cdot \left( \frac{ \left| y \right| ^2}{ \left| y^3 \right| } + \frac{ \left| x \right| \cdot \left| y \right| }{ \left| x \right| ^3\cdot \left| y \right| ^3} + \frac{ \left| x \right| ^2}{ \left| x \right| ^3} \right) \\ &\leq \left| y- x \right| \cdot \left( \frac{1}{ \left| y \right| } + \frac{1}{ \left| x \right| \cdot \left| y \right| }+ \frac{1}{ \left| x \right| } \right) \\ &\leq \left| y- x \right| \cdot \left(\frac{1}{1}+ \frac{1}{1\times 1}+ \frac{1}{1} \right) \\ &\leq \left| y- x \right| \cdot 3\\ &< 3\cdot \delta\\ &<3\times \frac{\varepsilon}{3}\\ &<\varepsilon\\ \square \end{align*}

1 This is in the Bartle and Sherbert Textbook at Chapter [8.1.7] p. 246↩

2 Refer to Ch. 58, p. 190 of Churchill’s Complex Variables 9thEd.↩

3 3Blue1Brown does a nice video on this ↩

4 How to choose a test by pauls Online Notes ↩

Power Series and Uniform Continuity

Power Series

Convergence

Sequences of Functions1

Define a sequence of functions:

Pointwise Convergence

Definition

Convergence of Power Series

Why power Series

Example

Convergence of Power Series

Uniform Convergence of Power Series

Circle of Convergence

Taylor Series3

The Cauchy Hadamard Theorem

Uniform Continuity

What is uniform continuity

Cantor's Theorem

Why is uniform continuity important?

Problem Example

Sequences of Functions¹

Taylor Series³