Representation of Functions with Power Series

Section 5.2 Representation of Functions with Power Series

Power series can be used to represent certain types of functions.

Subsection 5.2.1 Motivational Example: Geometric Series

The first and simplest example we already know is the geometric series,

\begin{gather*} \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \end{gather*}

This equation represents the function \(f(x) = \frac{1}{1-x}\) as the power series \(\sum_{n=0}^{\infty} x^n\) (as long as \(-1 \lt x \lt 1\)), and so we say that is the power series representation of \(\frac{1}{1-x}\text{.}\) Consider this Desmos applet: Geometric Series Power Series. Recall that the sum of a series is the limit of its sequence of partial sums. Then, if we write \(s_N(x)\) to be the partial sum,

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\begin{gather*} s_N(x) = \sum_{n=0}^{N} x^n = 1 + x + x^2 + x^3 + \dots + x^N \end{gather*}

Then, as the number of terms increases (as \(N\) increases), the sum becomes a better and better approximation for the function \(\frac{1}{1-x}\) (where \(-1\lt x\lt 1\)).

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Subsection 5.2.2 Power Series as an Approximation

In general, for a function with a power series, a partial sum of the power series can be used to approximate the function (within its interval of convergence). That is, a function \(f\) with a power series can be approximated by its (partial) power series, a polynomial of degree \(n\text{,}\)

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\begin{gather*} f(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n \end{gather*}

In general, the approximation improves if we add more terms, i.e. as \(n \to \infty\text{.}\)

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Subsection 5.2.3 Power Series Representation of Functions

The geometric series can be used to find power series representations of other related functions. The idea is to write the function in the form \(\frac{1}{1-u}\text{,}\) where \(u\) is any expression, and then apply the geometric series formula.

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Example 5.2.1.

For the function \(f(x) = \frac{1}{1+x}\text{,}\) it just has a positive \(x\) instead of a negative \(x\text{.}\) We can write \(x\) as \(-(-x)\text{,}\) and apply the geometric series,

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\begin{align*} \frac{1}{1 - (-x)} \amp = \sum_{n=0}^{\infty} (-x)^n \\\\ \amp = \sum_{n=0}^{\infty} (-1)^n x^n \\\\ \amp = 1 - x + x^2 - x^3 + x^4 - \dots \end{align*}

Note that typically, we split up the \(-1\) from the power of \(x\text{,}\) to more clearly separate the coefficient in front from the power.

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\begin{gather*} \boxed{\frac{1}{1 + x} = 1 - x + x^2 - x^3 + \dots = \sum_{n=0}^{\infty} (-1)^n x^n} \end{gather*}

For the interval of convergence, we could use the ratio test, but it is unnecessary, because we know that a geometric series converges precisely when \(\abs{r} \lt 1\text{.}\) In this case, it is \(-x\text{,}\) so if \(\abs{-x} \lt 1\text{,}\) or \(\abs{x} \lt 1\text{,}\) or \(-1 \lt x \lt 1\text{.}\)

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Example 5.2.2.

Consider \(f(x) = \frac{1}{1+x^2}\text{.}\) Here, there is \(x^2\) instead of \(x\text{,}\) and it is also positive instead of negative. So, we can write,

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\begin{align*} \frac{1}{1+x^2} \amp = \frac{1}{1-(-x^2)} \\\\ \amp = \sum_{n=0}^{\infty} (-x^2)^n \\\\ \frac{1}{1+x^2} \amp = \sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + \dots \end{align*}

This converges if \(\abs{-x^2} \lt 1\text{,}\) or \(x^2 \lt 1\text{,}\) or \(-1 \lt x \lt 1\text{.}\)

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Example 5.2.3.

Consider \(f(x) = \frac{1}{x + 4}\text{.}\) To make the denominator into the form \(1-u\text{,}\) we need to have a 1, so we need to factor out a 4,

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\begin{align*} \frac{1}{4+x} \amp = \frac{1}{4\brac{1 + \frac{x}{4}}} \\\\ \amp = \frac{1}{4} \cdot \frac{1}{1 - \brac{-\frac{x}{4}}} \\\\ \amp = \frac{1}{4} \cdot \sum_{n=0}^{\infty} \brac{-\frac{x}{4}}^n \\\\ \amp = \frac{1}{4} \cdot \sum_{n=0}^{\infty} \frac{(-1)^n}{4^n} x^n \\\\ \frac{1}{4+x} \amp = \sum_{n=0}^{\infty} \frac{(-1)^n}{4^{n+1}} x^n = \frac{1}{4} - \frac{1}{16} x + \frac{1}{64}x^2 - \dots \end{align*}

This converges if \(\abs{-\frac{x}{4}} \lt 1\text{,}\) or \(\abs{x} \lt 4\text{,}\) or \(-4 \lt x \lt 4\text{.}\)

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Example 5.2.4.

Consider \(f(x) = \frac{2x^2}{1 + x^3}\text{.}\) First, separate the \(2x^2\) to the side, and then write the positive \(x^3\) as \(-(-x^3)\text{,}\)

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\begin{align*} f(x) \amp = 2x^2 \cdot \frac{1}{1 + x^3} \\\\ \amp = 2x^2 \cdot \frac{1}{1 - (-x^3)} \\\\ \amp = 2x^2 \sum_{n=0}^{\infty} (-x^3)^n \\\\ \amp = 2x^2 \sum_{n=0}^{\infty} (-1)^n x^{3n} \\\\ \amp = \sum_{n=0}^{\infty} 2(-1)^n x^{3n+2} = 2x^2 - 2x^5 + 2x^8 - \dots \end{align*}

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