Problem 61 - Calculus Early Transcendentals

Question

Identify the functions represented by the following power series. $$\sum_{k=1}^{\infty}(-1)^{k} \frac{k x^{k+1}}{3^{k}}$$

Step-by-Step Solution

Answer

Answer: The given power series represents the function $$f(x)=-\frac{3 - 2x}{(3 - x)^2}, \text{ when } |-x/3| < 1.$$

Step 1: Initial Observation

The given power series is: $$\sum_{k=1}^{\infty}(-1)^{k} \frac{k x^{k+1}}{3^{k}}$$ Now let's observe the series and identify the patterns: 1. It has alternating signs given by \((-1)^k\) 2. It has a \(k\) in the numerator 3. The term \(x^{k+1}\) is the power of \(x\) 4. The term \(3^{k}\) in the denominator

Step 2: Manipulate the Series

To make it easier to compare with known functions, let's rewrite the series. Notice that we can write: $$\frac{k x^{k+1}}{3^{k}} = k \cdot \frac{x^{k+1}}{3^{k}}$$ Now let's rewrite the series: $$\sum_{k=1}^{\infty}(-1)^{k} k \cdot \frac{x^{k+1}}{3^{k}}$$

Step 3: Identify Known Functions

Notice that the series looks like a geometrical series whose common ratio is \(\frac{x}{3}\). If we recall a regular geometric series, we have: $$\sum_{k=1}^{\infty} \left(\frac{x}{3}\right)^{k} = \frac{\frac{x}{3}}{1 - \frac{x}{3}} = \frac{x}{3 - x} \text{, when } |-x/3| < 1$$

Step 4: Derive and Compare

The given series has a \(k\) multiplying the geometric series. We will find the derivative of the known function \(\frac{x}{3 - x}\) to see if we can match the given series. Compute the derivative: $$\frac{d}{dx}\left(\frac{x}{3 - x}\right) = \frac{(1)(3-x) - (1) x}{(3 - x)^2} = \frac{3-x-x}{(3-x)^2}$$ which simplifies to: $$\frac{d}{dx}\left(\frac{x}{3 - x}\right) = \frac{3 - 2x}{(3 - x)^2} \text{, when } |-x/3| < 1$$

Step 5: Adjust Signs

The given series has alternating signs. To match that, we can multiply the derivative of the function by \(-1\): $$-\frac{3 - 2x}{(3 - x)^2} \text{, when } |-x/3| < 1$$ Now, let's compare the two functions: $$\sum_{k=1}^{\infty}(-1)^{k} k \cdot \frac{x^{k+1}}{3^{k}} = -\frac{3 - 2x}{(3 - x)^2} \text{, when } |-x/3| < 1$$ The power series is equal to: $$ f(x)=-\frac{3 - 2x}{(3 - x)^2} \text{, when } |-x/3| < 1 $$