Problem 24 - 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. $$f(x)=1 / x, a=2$$

Step-by-Step Solution

Answer

Answer: The first four nonzero terms of the Taylor series are \(\frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{4}(x-2)^2 - \frac{3}{8}(x-2)^3\), and the power series using summation notation is: $$f(x) \approx \sum_{n=1}^\infty \frac{(-1)^{n+1}(x-2)^{n-1}}{2^{n+1}n}$$

Step 1: Determine the First Four Derivatives

Starting with the original function \(f(x) =\frac{1}{x}\), we will find the first through the fourth derivatives of the function. First derivative: $$f'(x) = -\frac{1}{x^2}$$ Second derivative: $$f''(x) = \frac{2}{x^3}$$ Third derivative: $$f^{(3)}(x) = -\frac{6}{x^4}$$ Fourth derivative: $$f^{(4)}(x) = \frac{24}{x^5}$$

Step 2: Evaluate Derivatives at a = 2

Next, we need to substitute \(a=2\) into each of the derivatives. $$f(2)=\frac{1}{2}$$ $$f'(2) = -\frac{1}{4}$$ $$f''(2) = \frac{2}{8}$$ $$f^{(3)}(2) = -\frac{6}{16}$$ $$f^{(4)}(2) = \frac{24}{32}$$

Step 3: Calculate the Taylor Series Terms

Now, we will use the Taylor series formula with the derivatives of f evaluated at \(a=2\) to find the first four nonzero terms of the series: $$f(x) \approx f(2) + f'(2)(x-2) + \frac{f''(2)(x-2)^2}{2!} + \frac{f^{(3)}(2)(x-2)^3}{3!} + \frac{f^{(4)}(2)(x-2)^4}{4!}$$ $$f(x) \approx \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{4}(x-2)^2 - \frac{3}{8}(x-2)^3 + \frac{3}{4}(x-2)^4$$

Step 4: Write the Power Series using Summation Notation

The power series can be represented using summation notation as follows: $$f(x) \approx \sum_{n=1}^\infty \frac{(-1)^{n+1}(x-2)^{n-1}}{2^{n+1}n}$$ So, the first four nonzero terms of the Taylor series for this function centered at \(a=2\) are \(\frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{4}(x-2)^2 - \frac{3}{8}(x-2)^3\), and the power series using summation notation is: $$f(x) \approx \sum_{n=1}^\infty \frac{(-1)^{n+1}(x-2)^{n-1}}{2^{n+1}n}$$