Determining a Function Representing a Power Series

We will now look at some examples of determining a function that represents a given power series. We will extensively use algebraic operations, differentiation, and integration of power series. It will also be useful to remember the following power series derived from the geometric series:

• $\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + ... = \frac{1}{1 - x}$ for $-1 < x < 1$.

• $\sum_{n=0}^{\infty} nx^{n-1} = 1 + 2x + 3x^2 + ... = \frac{1}{(1 - x)^2}$ for $-1 < x < 1$.

• $\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = - \ln(1 - x)$ for $-1 ≤ x < 1$.

Example 1

Determine a function $f(x)$ such that $f(x) = \sum_{n=0}^{\infty} nx^{n+2}$.

In solving problems such as this one, our goal is essentially to use any power series operations we have under our belt to modify the series into a more recognizable series that we can substitute in to find $f$.

(1)

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

We note that since we used the derivative of the geometric series in our substitution we have that $\frac{x^3}{(1 - x)^2} = \sum_{n=0}^{\infty} nx^{n+2}$ for $\mid x \mid < 1$.

Example 2

Determine a function $f(x)$ such that $f(x) = \sum_{n=0}^{\infty} (n+2)x^{n+1}$.

This example will be a little more complicated as we will use integration and differentiation as follows:

(2)

\begin{align} f(x) = \sum_{n=0}^{\infty} (n+2)x^{n+1} \\ \int f(x) \: dx = \sum_{n=0}^{\infty} \int (n+2)x^{n+1} \: dx \\ \int f(x) \: dx = C + \sum_{n=0}^{\infty} \frac{(n+2)x^{n+2}}{(n+2)} \\ \int f(x) \: dx = C + \sum_{n=0}^{\infty} x^{n+2} \\ \frac{1}{x^2} \int f(x) \: dx = \frac{1}{x^2} \left ( C + \sum_{n=0}^{\infty} x^{n+2} \right ) \\ \frac{1}{x^2} \int f(x) \: dx = \frac{C}{x^2} + \sum_{n=0}^{\infty} x^n \\ \frac{1}{x^2} \int f(x) \: dx = \frac{C}{x^2} + \frac{1}{1 - x} \\ \int f(x) \: dx = C + \frac{x^2}{1 - x} \\ \frac{d}{dx} \int f(x) \: dx = \frac{d}{dx} \left ( C + \frac{x^2}{1 - x} \right ) \\ f(x) = \left ( \frac{x^2}{1 - x} \right )' \\ f(x) = \frac{2x(1 - x) + x^2}{(1 - x)^2} \\ f(x) = \frac{2x - 2x^2 + x^2}{(1 - x)^2} \\ f(x) = \frac{-x^2 + 2x}{(1 - x)^2} \end{align}

We note that we utilized the geometric series here, and so our function $\frac{-x^2 + 2x}{(1 - x)^2} = \sum_{n=0}^{\infty} (n+2)x^{n+1}$ for $\mid x \mid < 1$.

Example 3

Determine a function $f(x)$ such that $f(x) = \sum_{n=0}^{\infty} (-1)^n(n+1)(n+3)x^n$.

Once again we will try to get rid of the series by moulding it into a recognizable series and making an appropriate replacement.

(3)

\begin{align} f(x) = \sum_{n=0}^{\infty} (-1)^n(n+1)(n+3)x^n \\ \int f(x) \: dx = \sum_{n=0}^{\infty} \int (-1)^n(n+1)(n+3)x^n \: dx \\ \int f(x) \: dx = C + \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)(n+3)x^{n+1}}{(n+1)} \\ \int f(x) \: dx = C + \sum_{n=0}^{\infty} (-1)^n(n+3)x^{n+1} \\ x \int f(x) \: dx = x \left ( C + \sum_{n=0}^{\infty} (-1)^n(n+3)x^{n+1} \right ) \\ x \int f(x) \: dx = Cx + \sum_{n=0}^{\infty} (-1)^n (n+3) x^{n+2} \\ \int x \int f(x) \: dx \: dx = \int \left ( Cx + \sum_{n=0}^{\infty} (-1)^n (n+3) x^{n+2} \right ) \: dx \\ \int x \int f(x) \: dx \: dx = D + \frac{Cx^2}{2} + \sum_{n=0}^{\infty} \int (-1)^n (n+3) x^{n+2} \: dx \\ \int x \int f(x) \: dx \: dx = D + \frac{Cx^2}{2} + \sum_{n=0}^{\infty} \frac{(-1)^n (n+3) x^{n+3}}{(n+3)} \\ \int x \int f(x) \: dx \: dx = D + \frac{Cx^2}{2} + \sum_{n=0}^{\infty} (-1)^n x^{n+3} \\ \int x \int f(x) \: dx \: dx = D + \frac{Cx^2}{2} + (1 - x + x^2) - \sum_{n=0}^{\infty} (-1)^n x^{n} \\ \int x \int f(x) \: dx \: dx = D + \frac{Cx^2}{2} + (1 - x + x^2) - \frac{1}{1 + x} \\ \quad \frac{d}{dx} \int x \int f(X) \: dx \: dx = \frac{d}{dx} \left ( D + \frac{Cx^2}{2} + (1 - x + x^2) - \frac{1}{1 + x} \right ) \\ x \int f(x) \: dx = Cx - 1 + 2x - \left ( \frac{1}{1 + x} \right ) ' \\ x \int f(x) \: dx = Cx - 1 + 2x - \left ( - \frac{1}{(1 + x)^2} \right ) \\ x \int f(x) \: dx = Cx - 1 + 2x + \frac{1}{(1 + x)^2} \\ \int f(x) \: dx = C - \frac{1}{x} + 2 + \frac{1}{x(1+x)^2} \\ \frac{d}{dx} \int f(x) \: dx = \frac{d}{dx} \left ( C - \frac{1}{x} + 2 + \frac{1}{x(1+x)^2} \right ) \\ f(x) = - \left ( -\frac{1}{x^2} \right ) + \left ( \frac{1}{x(1+x)^2} \right ) ' \\ f(x) = \frac{1}{x^2} + \frac{-(1 + x)^2 - 2x(1+x)}{x^2(1+x)^4} \\ f(x) = \frac{1}{x^2} + \frac{-(1 + x) - 2x}{x^2(1 + x)^3} \\ f(x) = \frac{(1 + x)^3 -(1 + x) - 2x}{x^2(1 + x)^3} \\ f(x) = \frac{(1 + x)^3 -1 - 3x}{x^2(1 + x)^3} \\ f(x) = \frac{1 + 3x + 3x^2 + x^3 - 1 - 3x}{x^2(1 + x)^3} \\ f(x) = \frac{3x^2 + x^3}{x^2(1 + x)^3} \\ f(x) = \frac{x^2(3 + x)}{x^2(1 + x)^3} \\ f(x) = \frac{3 + x}{(1 + x)^3} \end{align}

We used the geometric series in deriving this function and so $\frac{3 + x}{(1 + x)^3} = \sum_{n=0}^{\infty} (-1)^n(n+1)(n+3)x^n$ for $\mid x \mid < 1$.