Problem 73 - Calculus Early Transcendentals

Question

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. $$f(x)=1 / \sqrt{x} \text { with } a=4 ; \text { approximate } 1 / \sqrt{3}$$

Step-by-Step Solution

Answer

Based on the provided step-by-step solution, the first four terms of the Taylor series for the function \(f(x) = \frac{1}{\sqrt{x}}\) at the point \(a=4\) have been used to approximate the value of \(f(3) = \frac{1}{\sqrt{3}}\). After following the steps to obtain the Taylor series approximation and plug in \(x=3\), we found that \(f(3) \approx \frac{565}{660}\).

Step 1: Find the derivatives of the function f(x)

Let's find the derivatives of the function \(f(x) = \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}\) up to the 3rd order. \(1\)st derivative: $$f'(x) = -\frac{1}{2} x^{-\frac{3}{2}}$$ \(2\)nd derivative: $$f''(x) = \frac{3}{4} x^{-\frac{5}{2}}$$ \(3\)rd derivative: $$f'''(x) = -\frac{15}{8} x^{-\frac{7}{2}}$$

Step 2: Evaluate the derivatives at the point a=4

Now, let's find the values of the derivatives at the point \(a=4\). $$f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2}$$ $$f'(4) = -\frac{1}{2} (4)^{-\frac{3}{2}} = -\frac{1}{8}$$ $$f''(4) = \frac{3}{4} (4)^{-\frac{5}{2}} = \frac{3}{64}$$ $$f'''(4) = -\frac{15}{8} (4)^{-\frac{7}{2}} = -\frac{15}{512}$$

Step 3: Use the Taylor series formula to obtain the 4-term approximation

The Taylor series formula is given by, for a function \(f(x)\) with derivatives at point \(a\): $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$$ Let's use the first four terms of the Taylor series to approximate the function \(f(x) = \frac{1}{\sqrt{x}}\) at \(a=4\): $$f(x) \approx f(4) + \frac{f'(4)}{1!}(x - 4) + \frac{f''(4)}{2!}(x - 4)^2 + \frac{f'''(4)}{3!}(x - 4)^3$$ Substituting the values obtained in step 2, we get: $$f(x) \approx \frac{1}{2} -\frac{1}{8}(x - 4) + \frac{3}{64 \cdot 2}(x - 4)^2 - \frac{15}{512 \cdot 6}(x - 4)^3$$

Step 4: Approximate f(3) using the 4-term Taylor series

To approximate \(f(3) = \frac{1}{\sqrt{3}}\), we plug in \(x=3\) into the expression obtained in step 3: $$f(3) \approx \frac{1}{2} -\frac{1}{8}(3 - 4) + \frac{3}{64 \cdot 2}(3 - 4)^2 - \frac{15}{512 \cdot 6}(3 - 4)^3$$ Simplifying the expression, we get: $$f(3) \approx \frac{1}{2} + \frac{1}{8} + \frac{3}{64} - \frac{15}{512} \approx \frac{565}{660}$$ So, using the first four terms of the Taylor series, we have approximated \(\frac{1}{\sqrt{3}}\) as \(\approx \frac{565}{660}\).