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Find the Taylor polynomial \(p_{2}\) centered at \(a=8\) for \(f(x)=\sqrt[3]{x}\).

Short Answer

Expert verified
Answer: The Taylor polynomial of order 2 for the function f(x) = ∛x centered at a=8 is p₂(x) = -\frac{1}{72}x² + \frac{1}{4}x + \frac{1}{3}.

Step by step solution

01

Find the first and second derivatives of \(f(x)\)

The given function is \(f(x) = \sqrt[3]{x} = x^{\frac{1}{3}}\). Let's find its first and second derivatives. The first derivative, \(f'(x)\), can be found using the power rule: \(f'(x) = \frac{1}{3}x^{-\frac{2}{3}}\). Now let's find the second derivative, \(f''(x)\): \(f''(x) = -\frac{2}{9}x^{-\frac{5}{3}}\).
02

Evaluate the derivatives at \(a=8\)

Now we need to evaluate \(f(8)\), \(f'(8)\), and \(f''(8)\): \(f(8) = \sqrt[3]{8} = 2\) \(f'(8) = \frac{1}{3}8^{-\frac{2}{3}} = \frac{1}{3}(2^{-2}) = \frac{1}{12}\) \(f''(8) = -\frac{2}{9}8^{-\frac{5}{3}} = -\frac{2}{9}(2^{-5}) = -\frac{1}{36}\)
03

Use the Taylor polynomial formula to find \(p_{2}(x)\)

The formula for the second-degree Taylor polynomial \(p_{2}(x)\) is given by: $$ p_{2}(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 $$ Substitute the values of \(a\), \(f(8)\), \(f'(8)\), and \(f''(8)\) into the formula: $$ p_{2}(x) = 2 + \frac{1}{12}(x-8) - \frac{1}{72}(x-8)^2 $$
04

Simplify \(p_{2}(x)\)

Now, let's simplify \(p_{2}(x)\): $$ p_{2}(x) = 2 + \frac{1}{12}x - \frac{2}{3} - \frac{1}{72}x^2 + \frac{1}{3}x - 1 $$ By combining like terms, we get: $$ p_{2}(x) = -\frac{1}{72}x^2 + \frac{1}{4}x + \frac{1}{3} $$ So, the Taylor polynomial \(p_{2}(x)\) centered at \(a=8\) for \(f(x) = \sqrt[3]{x}\) is: $$ p_{2}(x) = -\frac{1}{72}x^2 + \frac{1}{4}x + \frac{1}{3} $$

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Key Concepts

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

Taylor Series
Taylor Series offer an invaluable tool for approximating complex functions with simpler polynomial expressions. The main idea is to take a function, like our cubic root function, and expand it around a center point, called 'a', using its derivatives.
This series provides a means to express the function as a sum of polynomials of increasing degree, becoming a closer approximation of the original function at and near the center point.

**What makes it so powerful?**
  • Convergence: Often, adding more terms can make the polynomial converge to the actual function.
  • Ease of Calculation: It translates complex functions into simpler polynomial form.
  • Wider Applications: Widely used in physics, engineering, and computer science to solve differential equations and perform numerical analysis.
In our case, to find a second-degree Taylor polynomial, or \( p_{2}(x) \), we consider up to the second derivative of the original function around the center \( a = 8 \). This offers a smooth polynomial approximation of \( \sqrt[3]{x} \).
Derivatives
Derivatives enable us to understand how a function changes at any given point. They are the building blocks of Taylor Series. In the exercise, we use both the first derivative \( f'(x) \) and the second derivative \( f''(x) \) to form the Taylor polynomial.
Derivatives capture the rate of change. Here's a simple breakdown:
  • First Derivative \( f'(x) \): This is the slope of the tangent line to \( f(x) \). For our function \( \sqrt[3]{x} \), the derivative reflects how steeply the function rises or falls.
  • Second Derivative \( f''(x) \): This measures the curvature. It tells us how the slope changes, which impacts the concavity of the function.
In practical terms, derivatives form the coefficients of the Taylor Series because they give specific values that describe how the function behaves around the center point \( a \). These values are plugged into the Taylor formula to establish the efficient polynomial's terms.
Polynomial Approximation
Polynomial Approximation is a method for estimating function values using simpler, polynomial expressions. The Taylor polynomial is a specific type of polynomial approximation.
Why use this approach?
  • Simplification: It's easier to work with polynomials than complex functions in calculations.
  • Efficiency: When evaluating functions repeatedly, polynomials are much faster to compute.
  • Accuracy: Especially near the center point \( a \), the approximation is quite close to the actual function.
Through the exercise, we've transformed \( \sqrt[3]{x} \) around \( a = 8 \) into \( p_{2}(x) = -\frac{1}{72}x^2 + \frac{1}{4}x + \frac{1}{3} \). This polynomial provides a good estimate for \( \sqrt[3]{x} \) close to 8 and illustrates how polynomial approximations can effectively model real-world functions.

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Most popular questions from this chapter

Representing functions by power series Identify the functions represented by the following power series. $$\sum_{k=2}^{\infty} \frac{x^{k}}{k(k-1)}$$

Errors in approximations Suppose you approximate \(f(x)=\sec x\) at the points \(x=-0.2,-0.1,0.0,0.1,0.2\) using the Taylor polynomials \(p_{2}(x)=1+x^{2} / 2\) and \(p_{4}(x)=1+x^{2} / 2+5 x^{4} / 24 .\) Assume the exact value of sec \(x\) is given by a calculator. a. Complete the table showing the absolute errors in the approximations at each point. Show three significant digits. b. In each error column, how do the errors vary with \(x ?\) For what values of \(x\) are the errors largest and smallest in magnitude? $$\begin{array}{|c|c|c|} \hline x & \left|\sec x-p_{2}(x)\right| & \left|\sec x-p_{4}(x)\right| \\ \hline-0.2 & & \\ \hline-0.1 & & \\ \hline 0.0 & & \\ \hline 0.1 & & \\ \hline 0.2 & & \\ \hline \end{array}$$

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

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

Dependence of errors on \(x\) Consider \(f(x)=\ln (1-x)\) and its Taylor polynomials given in Example \(8 .\) a. Graph \(y=\left|f(x)-p_{2}(x)\right|\) and \(y=\left|f(x)-p_{3}(x)\right|\) on the interval \([-1 / 2,1 / 2]\) (two curves). b. At what points of \([-1 / 2,1 / 2]\) is the error largest? Smallest? c. Are these results consistent with the theoretical error bounds obtained in Example \(8 ?\)

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