Question

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

Short 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-by-step solution

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 \(\frac{1}{(1-x^2)}\).

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.

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: $$\frac{2x}{(1-x^2)^2}= \sum_{k=1}^{\infty}2kx^{2 k - 1}$$

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: $$x\left(\frac{2x}{(1-x^2)^2}\right)=\sum_{k=1}^{\infty}2kx^{2 k}$$

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 \( \frac{1}{2}\times \frac{x^2}{(1-x^2)^2} \).