Chapter 9: Problem 71
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. $$f(x)=\sqrt[3]{x} \text { with } a=64 ; \text { approximate } \sqrt[3]{60}$$
Short Answer
Expert verified
Question: Using a Taylor series, approximate the value of $$\sqrt[3]{60}$$ centered at $$a=64$$ with the first four terms of the series.
Answer: Approximately $$3.852$$.
Step by step solution
01
Compute the first four derivatives of the function
We will need to compute the first four derivatives of the function $$f(x) = \sqrt[3]{x}$$, which can be also written as $$f(x) = x^{\frac{1}{3}}$$ for simplification.
Here are the derivatives:
$$f'(x) = \frac{1}{3}x^{-\frac{2}{3}}$$
$$f''(x) = -\frac{2}{9}x^{-\frac{5}{3}}$$
$$f^{(3)}(x) = \frac{10}{27}x^{-\frac{8}{3}}$$
$$f^{(4)}(x) = -\frac{80}{81}x^{-\frac{11}{3}}$$
02
Evaluate the derivatives at $$a=64$$
Now we need to evaluate each derivative at the point $$x=a=64$$:
$$f(64) = 64^{\frac{1}{3}} = 4$$
$$f'(64) = \frac{1}{3}(64)^{-\frac{2}{3}} = \frac{1}{3}(4) = \frac{4}{3}$$
$$f''(64) = -\frac{2}{9}(64)^{-\frac{5}{3}} = -\frac{2}{9}(2) = -\frac{4}{9}$$
$$f^{(3)}(64) = \frac{10}{27}(64)^{-\frac{8}{3}} = \frac{10}{27}(1/4) = \frac{10}{108} = \frac{5}{54}$$
$$f^{(4)}(64) = -\frac{80}{81}(64)^{-\frac{11}{3}} = -\frac{80}{81}(1/8) = -\frac{80}{648} = -\frac{10}{81}$$
03
Construct the Taylor series
Now we can construct the Taylor series by plugging in the computed coefficients:
$$T(x) = f(a) + f'(a)(x - a) + \frac{f''(a)(x - a)^2}{2!} + \frac{f^{(3)}(a)(x - a)^3}{3!} + \frac{f^{(4)}(a)(x - a)^4}{4!} + \cdots$$
Plugging in the computed values at $$a=64$$, we get:
$$T(x) = 4 + \frac{4}{3}(x - 64) - \frac{4}{9}(x - 64)^2 + \frac{5}{54}(x - 64)^3 - \frac{10}{81}(x - 64)^4 + \cdots$$
04
Approximate $$\sqrt[3]{60}$$ using the first four terms of the Taylor series
Now we can plug in $$x=60$$ into the Taylor series and compute the approximation using the first four terms:
$$T(60) \approx 4 + \frac{4}{3}(60 - 64) - \frac{4}{9}(60 - 64)^2 + \frac{5}{54}(60 - 64)^3$$
$$T(60) \approx 4 - \frac{4}{3}(4) + \frac{4}{9}(16) - \frac{5}{54}(64)$$
$$T(60) \approx 4 - \frac{16}{3} + \frac{64}{9} - \frac{320}{54}$$
$$T(60) \approx 4 - 5.333 + 7.111 - 5.926$$
Rounding the result to three decimal places, we get:
$$T(60) \approx 3.852$$
Therefore, the approximation of $$\sqrt[3]{60}$$ using the first four terms of the Taylor series for the function $$f(x)=\sqrt[3]{x}$$ centered at $$a=64$$ is approximately $$3.852$$.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Polynomial
The Taylor polynomial is a powerful method used in calculus to approximate functions near a specific point, known as the expansion point. This method involves using the derivatives of a function to build a series that allows us to estimate the function's value at points near or at the expansion point. In this exercise, the function \(f(x) = \sqrt[3]{x}\) is expanded about the point \(a = 64\), and the first four terms are used to approximate the cube root of 60, which lies close to 64.
Taylor series is formed by a polynomial whose coefficients are derived from the function’s derivatives at a specific point:
Taylor series is formed by a polynomial whose coefficients are derived from the function’s derivatives at a specific point:
- The zeroth term is simply the function value at the expansion point, \(T_0(x) = f(a)\).
- The first derivative gives us the linear approximation, \(T_1(x) = f(a) + f'(a)(x-a)\).
- Higher order derivatives allow us to refine this approximation by adding more terms: quadratics, cubics, and so on.
Derivatives
Derivatives play a crucial role in constructing the Taylor polynomial. Derivatives are the foundation upon which each term of the series is built. They measure the rate at which a function changes and are vital in determining the polynomial's coefficients. For the function in this exercise, \(f(x) = \sqrt[3]{x}\), derivatives of various orders were calculated.
Here's how the derivatives work in crafting the Taylor Series:
Here's how the derivatives work in crafting the Taylor Series:
- The first derivative, \(f'(x) = \frac{1}{3}x^{-\frac{2}{3}}\), provides the slope of the tangent line which serves as the best linear approximation around \(a\).
- Subsequent derivatives \(f''(x) = -\frac{2}{9}x^{-\frac{5}{3}}\), \(f^{(3)}(x) = \frac{10}{27}x^{-\frac{8}{3}}\), and \(f^{(4)}(x) = -\frac{80}{81}x^{-\frac{11}{3}}\) contribute to quadratic, cubic, and quartic terms respectively.
- Each derivative gives a new term, recalculating the function's curvature and refining approximation accuracy.
Function Approximation
Function approximation with a Taylor polynomial provides a simplistic expression to estimate a function's value near an expansion point. This method is particularly useful for functions that are difficult to compute directly, such as taking cube roots of non-perfect cubes. By using the derivatives to form a polynomial, the Taylor series estimator allows us to predict values quickly with reasonable accuracy.
The power of this approximation is visible when we approximate \(\sqrt[3]{60}\) by substituting \(x = 60\) into the previously constructed Taylor polynomial:
The power of this approximation is visible when we approximate \(\sqrt[3]{60}\) by substituting \(x = 60\) into the previously constructed Taylor polynomial:
- First, recognize that the value 64 was chosen because its cube root is a simple integer, allowing for easier manual calculation of derivatives.
- Substituting in this series, as shown in the solution, results in a sequence of additions and subtractions that mathematically converge to an approximate value.
- While only the first four terms were used here, adding more terms further improves accuracy.
Cubic Root
Finding cube roots, like \(\sqrt[3]{60}\), can be challenging without technological aids. Taylor series offer a method to approximate these roots by evaluating a polynomial derived from a function's derivatives.
For functions like \(\sqrt[3]{x}\), applying Taylor series near a known root simplifies calculations:
For functions like \(\sqrt[3]{x}\), applying Taylor series near a known root simplifies calculations:
- The point \(a = 64\) was chosen because of the ease of calculating \(\sqrt[3]{64} = 4\), a calculation easily done by hand.
- From \(x = 64\), we can modify the function's value slightly to gain a close approximation of cube roots of nearby values.
- This method handles the root-finding process by deconstructing the function about a simpler equivalent.