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Chapter 10: Problem 73

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

Short Answer

Expert verified
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 by step solution

01

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: $$f'''(x) = -\frac{15}{8} x^{-\frac{7}{2}}$$
02

Evaluate the derivatives at the point a=4

Now, let's find the values of the derivatives at the point \(a=4\). $$f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2}$$ $$f'(4) = -\frac{1}{2} (4)^{-\frac{3}{2}} = -\frac{1}{8}$$ $$f''(4) = \frac{3}{4} (4)^{-\frac{5}{2}} = \frac{3}{64}$$ $$f'''(4) = -\frac{15}{8} (4)^{-\frac{7}{2}} = -\frac{15}{512}$$
03

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: $$f(x) \approx \frac{1}{2} -\frac{1}{8}(x - 4) + \frac{3}{64 \cdot 2}(x - 4)^2 - \frac{15}{512 \cdot 6}(x - 4)^3$$
04

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 \(\approx \frac{565}{660}\).

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Most popular questions from this chapter

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. In most cases you do not need to use the definition of the Taylor series coefficients. b. If possible, determine the radius of convergence of the series. $$f(x)=\tan x$$

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

Given the power series $$\frac{1}{\sqrt{1-x^{2}}}=1+\frac{1}{2} x^{2}+\frac{1 \cdot 3}{2 \cdot 4} x^{4}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{6}+\cdots$$ for \(-1< x <1,\) find the power series for \(f(x)=\sin ^{-1} x\) centered at \(0 .\)

Determine whether the following statements are true and give an explanation or counterexample. a. To evaluate \(\int_{0}^{2} \frac{d x}{1-x},\) one could expand the integrand in a Taylor series and integrate term by term. b. To approximate \(\pi / 3,\) one could substitute \(x=\sqrt{3}\) into the Taylor series for \(\tan ^{-1} x\) c. \(\sum_{k=0}^{\infty} \frac{(\ln 2)^{k}}{k !}=2\)

Exponential function In Section 3, we show that the power series for the exponential function centered at 0 is $$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}, \quad \text { for }-\infty < x < \infty$$ Use the methods of this section to find the power series for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{2 x}$$
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