Problem 63 - Calculus Early Transcendentals

Question

Identify the functions represented by the following power series. $$\sum_{k=2}^{\infty} \frac{k(k-1) x^{k}}{3^{k}}$$

Step-by-Step Solution

Answer

Question: Identify the function represented by the given power series: $$\sum_{k=2}^{\infty} \frac{k(k-1) x^{k}}{3^{k}}$$ Answer: The function represented by the given power series is the second derivative of the function \(f(x) = e^{x/3}\), which is $$f''(x) = e^{x/3}$$

Step 1: Identify a familiar function/power series representation

We need to find a function which, when represented as a power series, can be related to our given power series. A familiar function that might help is the exponential function \(e^x\). Its power series representation is given by: $$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$

Step 2: Modify the Exponential Function

Now, we will try to modify the exponential function in order to get a function that has a power series representation that matches our given power series. Let's consider the function $$f(x) = e^{x/3}$$ Its power series representation is: $$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k! 3^k}$$

Step 3: Differentiate the Function Twice

We can see that our given power series has a \(k(k-1)\) term. We can get this term by differentiating the function \(f(x)\) twice. First derivative: $$f'(x) = \sum_{k=1}^{\infty} \frac{kx^{k-1}}{k! 3^k} = \sum_{k=1}^{\infty} \frac{x^{k-1}}{(k-1)! 3^k}$$ Second derivative: $$f''(x) = \sum_{k=2}^{\infty} \frac{k(k-1)x^{k-2}}{(k-1)! 3^k} = \sum_{k=2}^{\infty} \frac{k(k-1)x^k}{k! 3^k}$$ Now, our derived power series matches the given power series.

Step 4: Identify the Function

Since we've derived the given power series by differentiating the function \(f(x)\) twice, the function represented by the given power series is none other than the second derivative of \(f(x)\). Thus, $$\sum_{k=2}^{\infty} \frac{k(k-1) x^{k}}{3^{k}} = f''(x)$$ Where the function \(f(x) = e^{x/3}\).