Problem 60 - Essential Calculus

Question

Let \(f(x)=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}\) (a) Find the interval of convergence of \(f\). (b) Show that \(f^{\prime}(x)=f(x)\). (c) Show that \(f(0)=1\). (d) Identify the function \(f\).

Step-by-Step Solution

Answer

The interval of convergence for this series is \(-\infty < x < \infty\), the derivative of the function equals the function itself, the value of the function at zero is 1, and the function is the exponential function \(e^x\).

Step 1: Find the Interval of Convergence

To find the interval of convergence of the series, the Ratio Test can be used. For the Ratio Test, take the ratio of the (n+1)-th term to the n-th term, and then find the limit as n approaches infinity. If this limit is less than 1, the series converges. The ratio of the (n+1)-th term to the n-th term is \(\frac{x^{n+1}/(n+1)!}{x^n/n!} = \frac{x}{n+1}\). The limit as n approaches infinity of this ratio is 0, which is less than 1, so the series converges for all x. Therefore, the interval of convergence is \(-\infty < x < \infty\).

Step 2: Show that the Derivative of the Function equals the Function

To find the derivative of the function, differentiate the n-th term of the series with respect to x, using the Power Rule. This gives \(f'(x) = \sum_{n=1}^{\infty} n*x^{n-1}/n! = \sum_{n=1}^{\infty} x^{n-1}/(n-1)!\), which can be rewritten as \(f'(x) = \sum_{n=0}^{\infty} x^n/n!\), which is the same as \(f(x)\), so \(f'(x)=f(x)\).

Step 3: Find the Value of the Function at Zero

The value of the function at zero is calculated by substituting x=0 into the function. As \(0^n/n!\) equals 1 when n=0 and 0 for all other n, the infinite sum becomes 1. So \(f(0)=1\).

Step 4: Identify the function

The function is a well-known mathematical function. It is the series expansion of the exponential function \(e^x\), which has the property that its derivative equals itself and that \(e^0 = 1\).