Problem 72 - 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

Question: Approximate the value of $$\frac{1}{\sqrt{3}}$$ using the first four terms of the Taylor series for the function $$f(x) = \frac{1}{\sqrt{x}}$$ centered at $$a = 4$$. Answer: 0.577

Step 1: Find the first four derivatives of the given function

To obtain the Taylor series, we need to calculate four successive derivatives of the function: $$f(x) = \frac{1}{\sqrt{x}}$$ $$f'(x) = -\frac{1}{2}x^{-\frac{3}{2}}$$ $$f''(x) = \frac{3}{4}x^{-\frac{5}{2}}$$ $$f'''(x) = -\frac{15}{8}x^{-\frac{7}{2}}$$ $$f^{(4)}(x) = \frac{105}{16}x^{-\frac{9}{2}}$$

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

Now evaluate the derivatives at the point $$a = 4$$: $$f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2}$$ $$f'(4) = -\frac{1}{2}(4)^{-\frac{3}{2}} = -\frac{1}{16}$$ $$f''(4) = \frac{3}{4}(4)^{-\frac{5}{2}} = \frac{3}{128}$$ $$f'''(4) = -\frac{15}{8}(4)^{-\frac{7}{2}} = -\frac{15}{2048}$$ $$f^{(4)}(4) = \frac{105}{16}(4)^{-\frac{9}{2}} = \frac{105}{65536}$$

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

We can now write the Taylor series using the first four terms: $$T_4(x) = \frac{1}{2} - \frac{1}{16}(x-4) + \frac{3}{128}(x-4)^2 - \frac{15}{2048}(x-4)^3$$

Step 4: Approximate the given value using the Taylor series

Now use the Taylor series to approximate the given value $$\frac{1}{\sqrt{3}}$$ by plugging in $$x = 3$$: $$T_4(3) = \frac{1}{2} - \frac{1}{16}(3-4) + \frac{3}{128}(3-4)^2 - \frac{15}{2048}(3-4)^3$$ $$T_4(3) = \frac{1}{2} - \frac{1}{16}(-1) + \frac{3}{128}(1) - \frac{15}{2048}(-1)$$ $$T_4(3) = \frac{1}{2} + \frac{1}{16} + \frac{3}{128} + \frac{15}{2048}$$ $$T_4(3) \approx 0.577$$ Therefore, an approximation of $$\frac{1}{\sqrt{3}}$$ using the first four terms of the Taylor series is $$0.577$$.