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Chapter 11: Problem 73

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

Expert verified
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

01

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.
02

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: $$f^{(n)}(x) = (-3)^n e^{-3x},\quad f^{(n)}(0) = (-3)^n$$
03

Plug values into Taylor series formula

Now, we can plug the pattern we found into the Taylor series formula: $$f(x) = \sum_{n=0}^{\infty} \frac{(-3)^n}{n!} x^n$$
04

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 $$x$$.
05

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 $$x$$.

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Most popular questions from this chapter

Find the remainder in the Taylor series centered at the point a for the following functions. Then show that \(\lim R_{n}(x)=0\) for all \(x\) in the interval of convergence. $$f(x)=\cos 2 x, a=0$$

Derivative trick Here is an alternative way to evaluate higher derivatives of a function \(f\) that may save time. Suppose you can find the Taylor series for \(f\) centered at the point a without evaluating derivatives (for example, from a known series). Then \(f^{(k)}(a)=k !\) multiplied by the coefficient of \((x-a)^{k}\). Use this idea to evaluate \(f^{(3)}(0)\) and \(f^{(4)}(0)\) for the following functions. Use known series and do not evaluate derivatives. $$f(x)=e^{\cos x}$$

Compute the coefficients for the Taylor series for the following functions about the given point \(a\), and then use the first four terms of the series to approximate the given number. $$f(x)=\sqrt[3]{x} \text { with } a=64 ; \text { approximate } \sqrt[3]{60}$$.

Use the Maclaurin series $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1

Find power series representations centered at 0 for the following functions using known power series. $$f(x)=\frac{1}{1-x^{4}}$$
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