(PS3) Power Series Representation#

In this lesson we are going to see how to represent certain functions as a power series. Specifically, functions that can be generated from \(\tfrac{1}{1-x}\) through:

  • multiplication by \(x^k\)

  • composition of functions

Review Videos#

Geometric Series#

Recall that we have previously investigated the geometric series:

\[ \sum_{n=0}^{\infty} ar^n = a+ar+ar^2+ar^3+\cdots = \dfrac{a}{1-r} \qquad \text{for } |r|<1 \]

By setting \(a=1\) and \(r=x\) we have the power series and sum:

\[ \sum_{n=0}^{\infty} x^n = 1+x+x^2+x^3+\cdots = \dfrac{1}{1-x} \qquad \text{for } |x|<1 \]

Using the geometric series, we know this power series:

  • converges when \(|x|<1\)

  • diverges when \(|x|\geq 1\)

Which means that the Interval of Convergence is \((-1,1)\) and the Radius of Convergence is \(R=1\).

Representation#

There are two ways to view this equation:

Series Sum

The term \(\tfrac{1}{1-x}\) can be viewed as the sum of the geometric series with \(a=1\) and \(r=x\).

\[ \sum_{n=0}^{\infty} x^n = \dfrac{1}{1-x} \]
Representation

The term \(\tfrac{1}{1-x}\) can be viewed as a known function that can be written as a power series.

\[ \dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n \]

Our Prototype#

Our Prototype

\[ \dfrac{1}{1-x} = 1+x+x^2+x^3+\cdots = \sum_{n=0}^{\infty} x^n \qquad \text{for } |x|<1 \]

  • Interval of Convergence: \((-1,1)\)

  • Radius of Convergence: \(R=1\)

When we consider this prototype there are two important pieces in the fraction \(\tfrac{1}{1-x}\). We need the \(1\) and the negative in the denominator.

Example 1#

Find a power series representation for the following function. Also find the radius of convergence and interval of convergence.

\[ \dfrac{1}{1+x^3} \]

Try for yourself and then click to see the steps.

We start by writing out the prototype series:

\[ \dfrac{1}{1-x} = 1+x+x^2+x^3+\cdots = \sum_{n=0}^{\infty} x^n \]

In order to use this, we need to first rewrite our given function to match the format:

\[ \dfrac{1}{1-\text{variable}} \]

Right now we have addition: \(\tfrac{1}{1+x^3}=\tfrac{1}{1+\text{variable}}\)

We see that we need a negative in the denominator to match the correct format. We can get this by rewriting the addition sign as a double negative:

\[ \dfrac{1}{1+x^3} = \dfrac{1}{1- (-x^3)} = \dfrac{1}{1-\text{variable}} \]

This is now the correct format, so we can plug the variable part \((-x^3)\) into our prototype equation.

Starting with our prototype equation:

\[ \dfrac{1}{1-x} = 1+x+x^2+x^3+\cdots = \sum_{n=0}^{\infty} x^n \]

We can plug \((-x^3)\) into this (wherever we see an \(x\)) to give us:

\[ \dfrac{1}{1- (-x^3)} = 1+(-x^3)+(-x^3)^2+(-x^3)^3+\cdots = \sum_{n=0}^{\infty} (-x^3)^n \]

And then simplifying, to get:

\[ \dfrac{1}{1+x^3} = 1-x^3+x^6-x^9+\cdots = \sum_{n=0}^{\infty} (-1)^n x^{3n} \]

where the power series is expressed in two formats: (1) with the first few terms written out and (2) in sigma notation.

Next we want to find the interval of convergence for this power series. We know a geometric series converges when \(|r|<1\) and here \(r=-x^3\). So this power series converges when:

\[ |r|< 1 \iff |-x^3|<1 \iff |x^3|<1 \iff |x|^3<1 \iff |x|<1 \]

Since we know a geometric series diverges at both endpoints, we say:

\[ \text{Interval of Convergence} \quad \left( -1,1 \right) \]

Now we can determine the radius of convergence by calculating the width of the interval \(\left( -1, 1 \right)\) to be \(2\). And then half of this is \(1\). (Think of circles, the radius is half the diameter.)

\[ \text{Radius of Convergence} \quad R=1 \]

Example 2#

Find a power series representation for the following function. Also find the radius of convergence and interval of convergence.

\[ \dfrac{1}{x+4} \]

Try for yourself and then click to see the steps.

We start by writing out the prototype series:

\[ \dfrac{1}{1-x} = 1+x+x^2+x^3+\cdots = \sum_{n=0}^{\infty} x^n \]

In order to use this, we need to first rewrite our given function to match the format:

\[ \dfrac{1}{1-\text{variable}} \]

Right now we have addition and a \(c\neq 1\) constant: \(\tfrac{1}{x+4}=\tfrac{1}{\text{variable}+c}\)

We see that we need a few things in the denominator, primarily a negative and a constant of \(1\), to match the correct format.

To get the constant correct, we start by factoring out a \(4\):

\[ \dfrac{1}{x+4}= \dfrac{1}{4+x}= \dfrac{1}{4+\tfrac{4x}{4}} = \dfrac{1}{4\left(1+\tfrac{x}{4}\right)}= \dfrac{1}{4}\cdot \dfrac{1}{1+\tfrac{x}{4}} \]

Then we can get the subtraction by rewriting the addition sign as a double negative:

\[ \dfrac{1}{4}\cdot \dfrac{1}{1-\left(-\tfrac{x}{4}\right)} \]

This is now the correct format, so we can plug the variable part \(\left(-\tfrac{x}{4}\right)\) into our prototype equation.

Starting with our prototype equation:

\[ \dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n \]

We can plug \(\left(-\tfrac{x}{4}\right)\) into this (wherever we see an \(x\)) to give us:

\[ \dfrac{1}{4}\cdot \dfrac{1}{1- \left(-\tfrac{x}{4}\right)} = \dfrac{1}{4}\cdot \sum_{n=0}^{\infty} \left(-\tfrac{x}{4}\right)^n \]

And then simplifying, to get:

\[\begin{split} \dfrac{1}{x+4} & = \dfrac{1}{4}\cdot \sum_{n=0}^{\infty} \dfrac{(-1)^n x^n}{4^n}\\[10pt] & = \sum_{n=0}^{\infty} \dfrac{1}{4}\cdot \dfrac{(-1)^n x^n}{4^n}\\[10pt] & = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^n}{4^{n+1}}\\[10pt] \end{split}\]

where the power series is expressed in sigma notation.

Next we want to find the interval of convergence for this power series. We know a geometric series converges when \(|r|<1\) and here \(r=-\tfrac{x}{4}\). So this power series converges when:

\[ |r|< 1 \iff \left|-\tfrac{x}{4}\right|<1 \iff \tfrac{|x|}{4}<1 \iff |x|<4 \]

Since we know a geometric series diverges at both endpoints, we say:

\[ \text{Interval of Convergence} \quad \left( -4,4 \right) \]

Now we can determine the radius of convergence by calculating the width of the interval \(\left( -4, 4 \right)\) to be \(8\). And then half of this is \(4\). (Think of circles, the radius is half the diameter.)

\[ \text{Radius of Convergence} \quad R=4 \]

Example 3#

Find a power series representation for the following function. Also find the radius of convergence and interval of convergence.

\[ \dfrac{x^2}{x+4} \]

Try for yourself and then click to see the steps.

We can start by splitting up this expression:

\[ \dfrac{x^2}{x+4} = x^2\cdot \dfrac{1}{x+4} \]

And then use the power series representation for \(\tfrac{1}{x+4}\) that we found in example 2.

\[ \dfrac{x^2}{x+4} = x^2\cdot \sum_{n=0}^{\infty} \dfrac{(-1)^n x^n}{4^{n+1}} \]

We can then bring the \(x^2\) inside the summation to get:

\[ \sum_{n=0}^{\infty} x^2 \cdot \dfrac{(-1)^n x^n}{4^{n+1}} \]

And then combine the powers of \(x\) by adding exponents:

\[ \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{n+2}}{4^{n+1}} \]

Multiplying the power series by \(x^2\) does not change the interval of convergence. Therefore the interval of convergence for this power series is the same as in example 2:

\[ \text{Interval of Convergence} \quad \left( -4,4 \right) \]

Multiplying by \(x^2\) also doesn’t change the radius of convergence, so this is the same as well:

\[ \text{Radius of Convergence} \quad R=4 \]