Problem 11 - Calculus Early Transcendentals

Question

a. Find the first four nonzero terms of the Maclaurin series for the given function. b. Write the power series using summation notation. c. Determine the interval of convergence of the series. $$f(x)=\left(1+x^{2}\right)^{-1}$$

Step-by-Step Solution

Answer

Answer: The Maclaurin series for the function $f(x)=(1+x^2)^{-1}$ is $1+x^2$. The interval of convergence for this series is $(-\infty, \infty)$.

Step 1: Find the first four nonzero terms of the Maclaurin series for the given function

To find the first four nonzero terms of the Maclaurin series, we will first find the first four derivatives of the function and evaluate them at x=0: $$f(x) = \left(1+x^{2}\right)^{-1}$$ $$f'(x) = -2x\left(1+x^{2}\right)^{-2}$$ $$f''(x) = 2(1-3x^{2})\left(1+x^{2}\right)^{-3}$$ $$f'''(x) = -24x(1+x^{2})^{-4}+24x(5x^{2}-3)(1+x^{2})^{-5}$$ Now, let's evaluate these derivatives at x=0: $$f(0)=1$$ $$f'(0)=0$$ $$f''(0)=2$$ $$f'''(0)=0$$ The first four nonzero terms of the Maclaurin series are 1, 2, and 0.

Step 2: Write the power series using summation notation

We have the general formula for the Maclaurin series: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$ Using the nonzero terms we found previously: $$f(x) = \frac{f(0)}{0!}x^0 + \frac{f''(0)}{2!}x^2 = 1+\frac{2}{2}x^2 = 1+x^2$$

Step 3: Determine the interval of convergence of the series

With our current power series, we can use the Ratio Test or Root Test to determine the interval of convergence: For the Ratio Test, we need to find the limit: $$L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$$ In our case, since our series has a finite number of terms, the limit L will be 0. This means that our series converges on the entire real number line: $$\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 0 < 1$$ Therefore, the interval of convergence is: $$(-\infty, \infty)$$