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Chapter 9: Problem 76

Suppose you want to approximate \(\sqrt{72}\) using four terms of a Taylor series. Compare the accuracy of the approximations obtained using Taylor series for \(\sqrt{x}\) centered at 64 and 81

Short Answer

Expert verified
A) Taylor series centered at 64 B) Taylor series centered at 81 Answer: A) Taylor series centered at 64

Step by step solution

01

Find the derivatives of the function

To approximate \(\sqrt{x}\), we'll first find its first, second, and third derivatives: $$f(x) = \sqrt{x}$$ $$f'(x) = \frac{1}{2\sqrt{x}}$$ $$f''(x) = -\frac{1}{4x\sqrt{x}}$$ $$f'''(x) = \frac{3}{8x^2\sqrt{x}}$$
02

Evaluate the derivatives at 64 and 81

Now we'll evaluate the derivatives at 64 and 81 to find the Taylor series centered at these points. For \(a = 64\): $$f(64) = \sqrt{64} = 8$$ $$f'(64) = \frac{1}{2\sqrt{64}} = \frac{1}{16}$$ $$f''(64) = -\frac{1}{4(64)\sqrt{64}} = -\frac{1}{1024}$$ $$f'''(64) = \frac{3}{8(64)^2\sqrt{64}} = \frac{3}{262144}$$ For \(a = 81\): $$f(81) = \sqrt{81} = 9$$ $$f'(81) = \frac{1}{2\sqrt{81}} = \frac{1}{18}$$ $$f''(81) = -\frac{1}{4(81)\sqrt{81}} = -\frac{1}{1458}$$ $$f'''(81) = \frac{3}{8(81)^2\sqrt{81}} = \frac{3}{39366}$$
03

Find the first four terms of the Taylor series

Using the formula for the Taylor series, we'll find the first four terms for each approximation: For the Taylor series centered at 64: $$\sqrt{x} \approx 8 + \frac{1}{16}(x-64) - \frac{1}{1024}(x-64)^2 + \frac{3}{262144}(x-64)^3$$ For the Taylor series centered at 81: $$\sqrt{x} \approx 9 - \frac{1}{18}(x-81) - \frac{1}{1458}(x-81)^2 + \frac{3}{39366}(x-81)^3$$
04

Approximate \(\sqrt{72}\) using the Taylor series

Now we'll use the Taylor series to approximate \(\sqrt{72}\): For the series centered at 64: $$\sqrt{72} \approx 8 + \frac{1}{16}(72-64) - \frac{1}{1024}(72-64)^2 + \frac{3}{262144}(72-64)^3 \approx 8.48901367$$ For the series centered at 81: $$\sqrt{72} \approx 9 - \frac{1}{18}(72-81) - \frac{1}{1458}(72-81)^2 + \frac{3}{39366}(72-81)^3 \approx 8.48060512$$
05

Compare the accuracy of the approximations

Let's now compare the accuracy of the approximations. The actual value of \(\sqrt{72}\) is approximately 8.48528137. For the series centered at 64, the error is: $$|8.48901367 - 8.48528137| = 0.0037323$$ For the series centered at 81, the error is: $$|8.48060512 - 8.48528137| = 0.00467625$$ Since the error for the approximation using the Taylor series centered at 64 is smaller, this approximation is more accurate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
To approximate functions using a Taylor series, we need to calculate derivatives. Derivatives are expressions derived from a function that describe its rate of change. In our exercise, the function is \( f(x) = \sqrt{x} \). Each derivative gives us insights into how the function behaves near specific points.

  • The first derivative, \( f'(x) = \frac{1}{2\sqrt{x}} \), describes the slope or inclination of the function.
  • The second derivative, \( f''(x) = -\frac{1}{4x\sqrt{x}} \), gives information on the concavity, telling us if the curve is curving upwards or downwards.
  • The third derivative, \( f'''(x) = \frac{3}{8x^2\sqrt{x}} \), can help estimate the change in concavity.
To form a Taylor polynomial, these derivatives will be evaluated at specific points where they provide the coefficients of different terms of the polynomial.
Taylor Polynomial
A Taylor polynomial is a powerful tool for approximating functions around a specific point, often denoted as \( a \). It uses derivatives to build a polynomial, which locally mimics the function.

The general formula for a Taylor series of a function \( f(x) \) centered at \( a \) is:\[ f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]In the exercise, Taylor polynomials centered at 64 and 81 are used.
  • When centered at 64, the polynomial has terms up to \( (x - 64)^3 \); these use the values of the function and its derivatives at \( x = 64 \).
  • Similarly, when centered at 81, the polynomial relies on the values at \( x = 81 \).
These polynomials provide the four-term approximations for \( \sqrt{72} \). Each term incorporates the information from a derivative to improve the approximation.
Approximation Accuracy
The accuracy of a Taylor polynomial approximation depends on various factors, including the degree of the polynomial and the chosen point \( a \).

  • In our exercise, the actual value of \( \sqrt{72} \) is 8.48528137. Two Taylor series, centered at 64 and 81 respectively, are used to approximate this value.
  • When centered at 64, the polynomial yields an approximation of 8.48901367, with a small error of 0.0037323.
  • The polynomial centered at 81 gives an approximation of 8.48060512, showing a larger error of 0.00467625.
The closer \( x \) is to the center \( a \), the more accurate the approximation is likely to be. Here, centering at 64 was more accurate due to closer proximity to 72. Generally, increasing the number of terms improves accuracy but involves more computational effort.

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Most popular questions from this chapter

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. $$f(x)=\sqrt{x} \text { with } a=36 ; \text { approximate } \sqrt{39}$$

Use composition of series to find the first three terms of the Maclaurin series for the following functions. a. \(e^{\sin x}\) b. \(e^{\tan x}\) c. \(\sqrt{1+\sin ^{2} x}\)

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. b. Determine the radius of convergence of the series. $$f(x)=\left\\{\begin{array}{ll} \frac{\sin x}{x} & \text { if } x \neq 0 \\ 1 & \text { if } x=0 \end{array}\right.$$

Suppose you wish to approximate \(e^{0.35}\) using Taylor polynomials. Is the approximation more accurate if you use Taylor polynomials centered at 0 or \(\ln 2 ?\) Use a calculator for numerical experiments and check for consistency with Theorem 9.2. Does the answer depend on the order of the polynomial?

If the power series \(f(x)=\sum c_{k} x^{k}\) has an interval of convergence of \(|x|
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