Problem 62 - Calculus Early Transcendentals

Question

Identify the functions represented by the following power series. $$\sum_{k=1}^{\infty} \frac{x^{2 k}}{k}$$

Step-by-Step Solution

Answer

Answer: The power series \(\sum_{k=1}^{\infty} \frac{x^{2 k}}{k}\) represents the integral of the function \( \frac{1}{2}\times \frac{x^2}{(1-x^2)^2} \).

Step 1: Identify the well-known power series

First, let's analyze the given power series: $$\sum_{k=1}^{\infty} \frac{x^{2 k}}{k}$$ We observe that the power series involves even powers of x. The power series that contain even powers of x are typically those of cosine or sine functions, or any power series containing \(\frac{1}{(1-x^2)}\).

Step 2: Compare with the power series of \(\frac{1}{1-x^2}\)

Consider the geometric series for \(\frac{1}{1-x^2}\), which is: $$\sum_{k=0}^{\infty} x^{2 k}$$ In our given power series, each term has an extra factor of \(\frac{1}{k}\) compared to the geometric series, so we know it is not the function \(\frac{1}{1-x^2}\). Let's take the derivative of the geometric series to see if that resembles the given power series.

Step 3: Take the derivative of the geometric series

Let's take the derivative of both sides of the geometric series with respect to x: $$\frac{d}{dx}\frac{1}{1-x^2} = \sum_{k=0}^{\infty} \frac{d(x^{2 k})}{dx}$$ By applying the power rule of differentiation, we obtain: $$\frac{2x}{(1-x^2)^2}= \sum_{k=1}^{\infty}2kx^{2 k - 1}$$

Step 4: Multiply both sides by x

We are almost close to the given power series. To make it identical, let's multiply both sides of the equation by x: $$x\left(\frac{2x}{(1-x^2)^2}\right)=\sum_{k=1}^{\infty}2kx^{2 k}$$

Step 5: Divide by 2k

The last step is to divide both sides by 2k and account for the sum term: $$\frac{1}{2}\int \frac{x^2}{(1-x^2)^2} dx = \sum_{k=1}^{\infty}\frac{x^{2 k}}{k}$$ The provided power series, therefore, represents the integral of the function \( \frac{1}{2}\times \frac{x^2}{(1-x^2)^2} \).