Question

Use the methods of this section to find the power series centered at 0 for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{-3 x}$$

Short answer

Question: Find the power series centered at 0 for the function $$f(x) = e^{-3x}$$ and determine its interval of convergence. Answer: The power series representation of the function $$f(x) = e^{-3x}$$ centered at 0 is given by $$f(x) = \sum_{n=0}^{\infty} \frac{(-3)^n}{n!} x^n$$. The interval of convergence is $$(-\infty, \infty)$$, which means the series converges for all real values of $$x$$.

Step-by-step solution

Find the Taylor series of the function

Find the Taylor series for $$f(x) = e^{-3x}$$ centered at 0: To do this, we need to find the first few derivatives of the function and evaluate them at 0. Then, we will substitute these values into the Taylor series formula.

Calculate derivatives and evaluate at 0

Derivative, $$f'(x) = -3e^{-3x}$$, $$f'(0) = -3$$ Second derivative, $$f''(x)= 9e^{-3x}$$, $$f''(0)=9$$ Third derivative, $$f^{(3)}(x) = -27e^{-3x}$$, $$f^{(3)}(0) = -27$$ The pattern continues, alternating the signs: $$f^{(n)}(x) = (-3)^n e^{-3x},\quad f^{(n)}(0) = (-3)^n$$

Plug values into Taylor series formula

Now, we can plug the pattern we found into the Taylor series formula: $$f(x) = \sum_{n=0}^{\infty} \frac{(-3)^n}{n!} x^n$$

Use the ratio test to find the interval of convergence

For the given series, let $$a_n = \frac{(-3)^n x^n}{n!}$$. To find the interval of convergence, apply the ratio test: $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\left|\frac{\frac{(-3)^{n+1} x^{n+1}}{(n+1)!}}{\frac{(-3)^n x^n}{n!}}\right|$$ Simplify the expression inside the limit: $$\lim_{n\to\infty}\left|\frac{(-1) x(n!)}{(n+1)!}\right| = \lim_{n\to\infty}\left|\frac{(-1) x}{n+1}\right|$$ To converge, this limit must be less than 1: $$\left|\frac{(-1) x}{n+1}\right| < 1$$ Since the limit does not depend on $$n$$, the series converges for all $$x$$.

Write down the power series and interval of convergence

The power series representation of the function $$f(x) = e^{-3x}$$ centered at 0 is given by: $$f(x) = \sum_{n=0}^{\infty} \frac{(-3)^n}{n!} x^n$$ The interval of convergence is $$(-\infty, \infty)$$, which means the series converges for all real values of $$x$$.