Cauchy Product of Power Series

# Cauchy Product of Power Series

The following theorem will give us a way to in a sense, "multiply" two power series together.

 Theorem 1 (The Cauchy Product of Power Series): 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 whenever both of these power series convergent we have that $\left ( \sum_{n=0}^{\infty} a_nx^n \right ) \left ( \sum_{n=0}^{\infty} b_nx^n \right ) = \sum_{n=0}^{\infty} \left ( \left ( \sum_{j=0}^{n} (a_jb_{n-j}) \right ) x^n \right ) = \sum_{n=0}^{\infty} \left (\left (a_0b_n + a_1b_{n-1} + ... + a_{n+1}b_{1} + a_nb_0 \right)x^n \right )$. This power series has a radius of convergence $R$ such that $R ≥ \mathrm{min} \{ R_1, R_2 \}$.

It is common to reformulate the theorem for the Cauchy Product above as follows. Let $\sum_{n=0}^{\infty} a_nx^n$ and $\sum_{n=0}^{\infty} b_nx^n$ be two power series with radii of convergence $R_1$ and $R_2$ respectively. Then the Cauchy Product of these series can be defined as $\sum_{n=0}^{\infty} c_nx^n$ where:

(1)
\begin{align} \quad c_n = a_0b_n + a_1b_{n-1} + ... + a_{n-1}b_1 + a_nb_0 = \sum_{j=0}^{n} a_jb_{n-j} \end{align}

Furthermore, the Cauchy product $\sum_{n=0}^{\infty} c_nx^n$ has a radius of convergence $R$ at least larger or equal to the smaller of the two radii $R_1$, $R_2$, that is $R ≥ \mathrm{min} \{ R_1, R_2 \}$.

We will now look at some examples of applying the Cauchy product to some power series.

## Example 1

Using the Cauchy Product determine a power series representation of the function $f(x) = \frac{1}{(1 - x)^2}$ for $-1 < x < 1$.

We know that for $\mid x \mid < 1$, $\frac{1}{1 - x} = 1 + x + x^2 + x^3 + ... = \sum_{n=0}^{\infty} x^n$. So to determine a power series representation of the function $f$, we will simply take the Cauchy product of $\sum_{n=0}^{\infty} x^n$ with itself.

First note that:

(2)
\begin{align} \sum_{j=0}^{n} (a_jb_{n-j} ) = \sum_{j=0}^{n} 1 \cdot 1 = n + 1 \end{align}

Therefore it follows by the Cauchy product that $f(x) = \frac{1}{(1 - x)^2} = \sum_{n=0}^{\infty} (n + 1)x^n$ is a representation for $-1 < x < 1$.

## Example 2

Using the Cauchy Product determine a power series representation of the function $f(x) = \frac{1}{(1 - x)^4}$.

From example 1 we have that for $\mid x \mid < 1$, $\frac{1}{(1 - x)^2} = \sum_{n=0}^{\infty} (n + 1)x^n$. We will take the Cauchy Product of this series with itself once again as follows:

(3)
\begin{align} \quad \quad \sum_{j=0}^{n} (j + 1)(n - j +1) = \sum_{j=0}^{n} jn - j^2 + n + 1 = \sum_{j=0}^{n} n(j + 1) - j^2 + 1 = \sum_{j=0}^n n(j+1) - \sum_{j=0}^n j^2 + \sum_{j=0}^n 1 \end{align}

And so:

(4)
\begin{align} \quad \sum_{j=0}^{n} (j + 1)(n - j +1) = n \left ( \frac{n^2 + n}{2} + (n +1) \right ) - \left ( \frac{n(n+1)(2n+1)}{6} \right ) + (n + 1) \\ \quad \sum_{j=0}^{n} (j + 1)(n - j +1) = \frac{n^3 + n^2}{2} + (n^2 +n) - \left ( \frac{n(n+1)(2n+1)}{6} \right ) + (n + 1) \\ \quad \sum_{j=0}^{n} (j + 1)(n - j +1) = \frac{n^3 + 3n^2 + 2n}{2} - \left ( \frac{n(n+1)(2n+1)}{6} \right ) + (n + 1) \\ \quad \sum_{j=0}^{n} (j + 1)(n - j +1) = \frac{3n^3 + 9n^2 + 6n}{6} - \left ( \frac{n(n+1)(2n+1)}{6} \right ) + (n + 1) \\ \quad \sum_{j=0}^{n} (j + 1)(n - j +1) = \frac{3n^3 + 9n^2 + 6n}{6} - \left ( \frac{2n^3 + 3n^2 + n}{6} \right ) + (n + 1) \\ \quad \sum_{j=0}^{n} (j + 1)(n - j +1) = \frac{n^3 + 6n^2 + 5n}{6} + (n + 1) \\ \quad \sum_{j=0}^{n} (j + 1)(n - j +1) = \frac{n^3 + 6n^2 + 11n + 6}{6} \end{align}

Therefore it follows that $f(x) = \frac{1}{(1 - x)^4} = \sum_{n=0}^{\infty} \left ( \frac{n^3 + 6n^2 + 11n + 6}{6} \right) x^n$ for $-1 < x < 1$.