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 \(f\).

Short 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-by-step solution

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 \(-\infty < x < \infty\).

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 \(f'(x)=f(x)\).

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 \(f(0)=1\).

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 \(e^0 = 1\).