Question

Find the first five non-zero terms of Maclaurin series (Taylor series centered at x=0) for the function below as well as radius of convergence.

f(x)=xe^x

Answer 1

To find the Maclaurin series for $f(x) = xe^x$, we can use the formula for the Maclaurin series of $e^x$ and differentiate the product term by term:

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$

$$xe^x = x \sum_{n=0}^\infty \frac{x^n}{n!} = \sum_{n=0}^\infty \frac{x^{n+1}}{n!}$$

So the first few terms of the Maclaurin series for $f(x)$ are:

$$f(x) = xe^x = x + x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \frac{x^5}{24} + \cdots$$

To find the radius of convergence, we can use the ratio test:

$$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{x^{n+2}/(n+1)!}{x^{n+1}/n!} \right| = \lim_{n\to\infty} \fracx{n+1} = 0$$

Since the limit is less than 1 for all values of $x$, the Maclaurin series converges for all values of $x$, and the radius of convergence is infinite.