Determining a Function Representing a Power Series Examples 2

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 $-1 ≤ x < 1$.

We will now look at some more examples of determining a function representing a power series.

Examples 1

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

(1)

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

If $x = 0$ then $C = 0$ from above, and so:

(2)

\begin{align} \quad xf(x) = -\ln \mid 1 - x \mid - x \\ \quad f(x) = -\frac{\ln \mid 1 - x \mid}{x} - 1 \end{align}

Example 2

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

(3)

\begin{align} \quad f(x) = \sum_{n=0}^{\infty} (n+2)(n+1)2^{n+1}x^n \\ \quad \int f(x) \: dx = C + \sum_{n=0}^{\infty} (n+2)2^{n+2}x^{n+1} \\ \quad \int \left ( \int f(x) \: \right ) \: dx = Cx + D + \sum_{n=0}^{\infty} 2^{n+2} x^{n+2} \\ \quad \int \left ( \int f(x) \: \right ) \: dx = Cx + D + \sum_{n=0}^{\infty} (2x)^{n+2} \\ \quad \int \left ( \int f(x) \: \right ) \: dx = Cx + D + 4x^2 \sum_{n=0}^{\infty} (2x)^{n} \\ \quad \int \left ( \int f(x) \: \right ) \: dx = Cx + D + \frac{4x^2}{1 - 2x} \\ \quad \int f(x) \: dx = C + \frac{(1 - 2x)(8x) - (4x^2)(-2)}{(1 - 2x)^2} \\ \quad f(x) = \frac{[(1 - 2x)^2] [8 - 16x] - [1 - 2x)(8x) - (4x^2)(-2)][2(1 - 2x) (-2)]}{(1 - 2x)^4} \end{align}