Problem 23 - Calculus Early Transcendentals

Question

a. Find the first four nonzero terms of the Taylor series for the given function centered at \(a\) b. Write the power series using summation notation. $$f(x)=1 / x, a=1$$

Step-by-Step Solution

Answer

The first four nonzero terms of the Taylor series are: $$f(x) \approx 1 - (x-1) + 2(x-1)^2 - 6(x-1)^3$$ The power series in summation notation is: $$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n (n!)}{x^{n+1}}$$

Step 1: Find the derivatives of the function

Find the first few derivatives of \(f(x)=\frac{1}{x}\) at \(x=1\) to get the Taylor series coefficients. $$f(x)=\frac{1}{x} \Rightarrow f'(x)=-\frac{1}{x^2} \Rightarrow f''(x)=\frac{2}{x^3} \Rightarrow f^{(3)}(x)=-\frac{6}{x^4} \Rightarrow f^{(4)}(x)=\frac{24}{x^5}$$

Step 2: Evaluate derivatives at \(a=1\)

Compute the first, second, third, and fourth derivatives of the function at the point \(a=1\) to get the coefficients of the Taylor series. $$f(1)=1, \quad f'(1)=-1, \quad f''(1)=2, \quad f^{(3)}(1)=-6, \quad f^{(4)}(1)=24$$

Step 3: Write the first four nonzero terms of the Taylor series

Write down the first four Taylor series terms using the coefficients found in the previous steps. $$f(x) \approx 1 - (x-1) + 2(x-1)^2 - 6(x-1)^3$$

Step 4: Write the power series using summation notation

Represent the series obtained in step 3 using summation notation. $$\sum_{n=0}^{\infty} \frac{(-1)^n (n!)}{x^{n+1}} = 1 - (x-1) + 2(x-1)^2 - 6(x-1)^3 + \cdots$$ Therefore, the first four nonzero terms of the Taylor series are: $$f(x) \approx 1 - (x-1) + 2(x-1)^2 - 6(x-1)^3$$ and the power series in summation notation is: $$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n (n!)}{x^{n+1}}$$