Algebraic Operations of Power Series

Algebraic Operations of Power Series

We will now begin to look at some algebraic properties of power series.

 Theorem 1: Consider the power series $\sum_{n=0}^{\infty} a_nx^n$ with a radius of convergence $R_1$, and let $k$ be some constant. Then the power series $\sum_{n=0}^{\infty} (ka_n)x^n$ has the radius of convergence of $R_1$ and whenever $k \sum_{n=0}^{\infty} a_nx^n$ converges, $\sum_{n=0}^{\infty} (ka_n)x^n = k \sum_{n=0}^{\infty} a_nx^n$.
 Theorem 2: Consider the power series $\sum_{n=0}^{\infty} a_nx^n$ with a radius of convergence $R_1$, and the power series $\sum_{n=0}^{\infty} b_nx^n$ with a radius of convergence $R_2$. Then $\sum_{n=0}^{\infty} (a_n + b_n)x^n$ has a radius of convergence $R$ such that $R ≥ \mathrm{min} \{ R_1, R_2 \}$ and whenever $\sum_{n=0}^{\infty} a_nx^n$ and $\sum_{n=0}^{\infty} b_nx^n$ converge, $\sum_{n=0}^{\infty} (a_n + b_n)x^n = \sum_{n=0}^{\infty} a_nx^n + \sum_{n=0}^{\infty} b_nx^n$.

Another important algebraic operation of power series is known as the Cauchy Product of Power Series which we we look at subsequently.

Let's now look at some examples of applying Theorems 1 and 2.

Example 1

Determine a power series representation of the function $f(x) = \frac{2}{(1 - x)}$.

We know that for $-1 < x < 1$, $\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n$, so then is we multiply both sides by $2$ and utilize Theorem 1, we get that for $-1 < x < 1$ that $f(x) = \frac{2}{1 - x} = \sum_{n=0}^{\infty} 2x^n$. Example 2

Determine a power series representation of the function $f(x) = \frac{2}{(1 - x^2)}$.

Let's use partial fractions to reduce this function down such that:

(1)
\begin{align} \frac{2}{1 - x^2} = \frac{1}{1 - x} + \frac{1}{1 + x} \end{align}

We thus have that $\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n$ and $\frac{1}{1 + x} = \sum_{n=0}^{\infty} (-1)^nx^n$ for $\mid x \mid < 1$. Applying Theorem 2 we have that:

(2)
\begin{align} \frac{2}{1 - x^2} = \sum_{n=0}^{\infty} (1 + (-1)^n)x^n \quad \mathrm{for \: \mid r \mid < 1 } \end{align}

Alternatively we could evaluate the following series as follows:

(3)
\begin{align} \quad \frac{2}{1 - x^2} = 2 \frac{1}{1 - x^2} = 2 \sum_{n=0}^{\infty} (x^2)^n = \sum_{n=0}^{\infty} 2x^{2n} \end{align}