Problem 63 - Essential Calculus

Question

Explain how to use the series \(g(x)=e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}\) to find the series for each function. Do not find the series. (a) \(f(x)=e^{-x}\) (b) \(f(x)=e^{3 x}\) (c) \(f(x)=x e^{x}\) (d) \(f(x)=e^{2 x}+e^{-2 x}\)

Step-by-Step Solution

Answer

The series for the four functions are: (a) \(\sum_{n=0}^{\infty} \frac{(-x)^{n}}{n !}\), (b) \(\sum_{n=0}^{\infty} \frac{(3x)^{n}}{n !}\), (c) \(x \sum_{n=0}^{\infty} \frac{x^{n}}{n !}\), and (d) \(\sum_{n=0}^{\infty} \frac{(2x)^{n}}{n !} + \sum_{n=0}^{\infty} \frac{(-2x)^{n}}{n !}\)

Step 1: The Series for \(f(x) = e^{-x}\)

For this function, replace \(x\) with \(-x\) in the given series. The series for \(f(x) = e^{-x}\) therefore becomes \(\sum_{n=0}^{\infty} \frac{(-x)^{n}}{n !}\)

Step 2: The Series for \(f(x) = e^{3x}\)

This time, replace \(x\) with \(3x\). The series for \(f(x) = e^{3x}\) is \(\sum_{n=0}^{\infty} \frac{(3x)^{n}}{n !}\)

Step 3: The Series for \(f(x) = xe^{x}\)

In this case, differentiate both sides of the series for \(g(x)\) to get the series for \(f(x) = xe^{x}\). This yields \(\sum_{n=0}^{\infty} \frac{nx^{n-1}}{(n-1) !} = x \sum_{n=0}^{\infty} \frac{x^{n}}{n !}\)

Step 4: The Series for \(f(x) = e^{2x} + e^{-2x}\)

This function is the sum of two series obtained by replacing \(x\) with \(2x\) and \(-2x\) respectively in the series for \(g(x)\). Thus, the series for \(f(x) = e^{2x} + e^{-2x}\) is \(\sum_{n=0}^{\infty} \frac{(2x)^{n}}{n !} + \sum_{n=0}^{\infty} \frac{(-2x)^{n}}{n !}\)