Determining a Function Representing a Power Series Examples 1

We will now look at some more examples of determining a function that represents a specific power series. Be sure to check the Determining a Function Representing a Power Series page beforehand though and recall the following power 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 $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n+3}$.

(1)

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

Notice that if $x = 0$, then $C = 0$, and so:

(2)

\begin{align} x^3 f(x) = -x - \frac{x^2}{2} - \ln (1 - x) \\ f(x) = -\frac{1}{x^2} - \frac{1}{2x} - \frac{\ln(1 - x)}{x^3} \end{align}

Example 2

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

To make this question less complicated, let's perform an index swap. Let $m = n + 1$ so that $n = m - 1$ and $2n = 2m - 2$, and so:

(3)

\begin{align} f(x) = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{n+1} \\ f(x) = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}x^{2m-2}}{m} \\ x^2f(x) = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}x^{2m}}{m} \\ \frac{d}{dx} \left ( x^2 f(x) \right ) = \sum_{m=1}^{\infty} \frac{d}{dx} \frac{(-1)^{m-1}x^{2m}}{m} \\ \frac{d}{dx} \left ( x^2 f(x) \right ) = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}2mx^{2m-1}}{m} \\ \frac{d}{dx} \left ( x^2 f(x) \right ) = \sum_{m=1}^{\infty} (-1)^{m-1}2x^{2m-1} \\ \frac{\frac{d}{dx} \left ( x^2 f(x) \right )}{x} = \sum_{m=1}^{\infty} (-1)^{m-1}2x^{2m-2} \\ \frac{\frac{d}{dx} \left ( x^2 f(x) \right )}{x} = 2\sum_{m=1}^{\infty} (-1)^{m-1}(x^2)^{m-1} \\ \frac{\frac{d}{dx} \left ( x^2 f(x) \right )}{x} = 2\sum_{m=1}^{\infty} (-x^2)^{m-1} \\ \end{align}

Now we will change our indices back. $n = 0$ and $m - 1 = n$:

(4)

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

Notice that if $x = 0$, then $C = 0$, and so:

(5)

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

We have this $f(x)$ represents the series $\sum_{n=0}^{\infty} \frac{x^n}{n+3}$ for $\mid -x^2 \mid < 1$, or in other words for $-1 < x < 1$. Notice that if we test $x = -1$ and $x = 1$ (the end points of the interval of convergence), we get that the series converges to the alternating harmonic series both ways, and so in fact we have that $-1 ≤ x ≤ 1$. Notice however that $x = 0$ is undefined with the current formulation of our function $f$. Using L'Hospital's Rule we have that:

(6)

\begin{align} \quad \lim_{x \to 0} \frac{\ln (1 + x^2)}{x^2} \overset{H} = \lim_{x \to 0} \frac{\frac{2x}{1 + x^2}}{2x} = \lim_{x \to 0} \frac{1}{1 + x^2} = 1 \end{align}

Therefore we will define our function as follows:

(7)

\begin{align} f(x) = \left\{\begin{matrix} \frac{\ln(1 + x^2)}{x^2} & -1 ≤ x ≤ 1 \: , x \:\neq 0\\ 1 & x = 0 \end{matrix}\right. \end{align}