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Chapter 11: Problem 4

In Exercises \(1-8,\) find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a .\) $$ f(x)=1 /(x+2), \quad a=0 $$

Short Answer

Expert verified
The Taylor polynomials are \( \frac{1}{2} \), \( \frac{1}{2} - \frac{1}{4}x \), \( \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2 \), and \( \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2 - \frac{1}{16}x^3 \).

Step by step solution

01

Understand the Taylor Polynomial Formula

The Taylor polynomial of order \( n \) for a function \( f(x) \) about \( a \) is given by:\[ T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n \]where \( f^{(k)}(a) \) is the \( k \)-th derivative of \( f(x) \) evaluated at \( x = a \).
02

Calculate the Function and Its Derivatives at a=0

Calculate the function and its derivatives at \( x = 0 \):- \( f(x) = \frac{1}{x+2}, \) so \( f(0) = \frac{1}{2} \).- Derivative: \( f'(x) = -\frac{1}{(x+2)^2}, \) so \( f'(0) = -\frac{1}{4} \).- Second derivative: \( f''(x) = \frac{2}{(x+2)^3}, \) so \( f''(0) = \frac{1}{4} \).- Third derivative: \( f'''(x) = -\frac{6}{(x+2)^4}, \) so \( f'''(0) = -\frac{3}{4} \).
03

Construct the Taylor Polynomial of Order 0

The Taylor polynomial of order 0 is simply the function evaluated at \( a \):\[ T_0(x) = f(0) = \frac{1}{2} \]
04

Construct the Taylor Polynomial of Order 1

Add the first derivative term to the constant term:\[ T_1(x) = \frac{1}{2} - \frac{1}{4}(x-0) = \frac{1}{2} - \frac{1}{4}x \]
05

Construct the Taylor Polynomial of Order 2

Include up to the second derivative:\[ T_2(x) = \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2 \]The \( \frac{1}{8} \) term comes from \( \frac{1}{2!} \cdot \frac{1}{4} \).
06

Construct the Taylor Polynomial of Order 3

Include up to the third derivative:\[ T_3(x) = \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2 - \frac{1}{16}x^3 \]The \( \frac{1}{16} \) term comes from \( \frac{1}{3!} \cdot \left(-\frac{3}{4}\right) \).

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Key Concepts

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

Derivatives
In calculus, a derivative is a measure of how a function changes as its input changes. Derivatives tell us the rate at which a function is changing. This can be particularly useful in understanding dynamic systems in physics or economics.When you calculate the derivative at a given point, you are essentially finding the slope of the tangent line to the function's graph at that point. For a function like \( f(x) = \frac{1}{x+2} \), we can compute derivatives step-by-step:
  • First Derivative \( f'(x) \): Computes the slope of the function's graph, which gives us \( f'(x) = -\frac{1}{(x+2)^2} \). It indicates how quickly \( f(x) \) decreases as \( x \) increases.

  • Second Derivative \( f''(x) \): Measures the rate of change of the first derivative, calculated as \( f''(x) = \frac{2}{(x+2)^3} \). It's related to the concavity of the function鈥攁 positive second derivative suggests the graph is concave up.

  • Third Derivative \( f'''(x) \): Measures the rate of change of the second derivative, given by \( f'''(x) = -\frac{6}{(x+2)^4} \).
Polynomial Approximation
Polynomial approximation helps us understand functions by representing them as simpler polynomial equations. This is especially useful when the exact form of a function is too complex to work with easily.A Taylor polynomial approximates a function using information about its derivatives at a specific point. Here鈥檚 the breakdown for the Taylor series used in the exercise:
  • Order 0 Polynomial \( T_0(x) \): This is the simplest approximation, which is just the function's value at \(a = 0\), calculated as \( T_0(x) = \frac{1}{2} \).

  • Order 1 Polynomial \( T_1(x) \): Incorporates the function's first derivative, giving a linear approximation: \( T_1(x) = \frac{1}{2} - \frac{1}{4}x \).

  • Order 2 Polynomial \( T_2(x) \): Includes the second derivative, leading to a quadratic approximation: \( T_2(x) = \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2 \).

  • Order 3 Polynomial \( T_3(x) \): Adds the third derivative, resulting in a cubic approximation: \( T_3(x) = \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2 - \frac{1}{16}x^3 \).

These approximations provide different levels of accuracy for estimating the original function \( f(x) \), especially near the point \( a = 0 \).
Function Evaluation
In mathematical analysis, function evaluation involves calculating the result of a function at specific values of its variables. This process allows us to understand the behavior and characteristics of functions better.While solving the exercise, function evaluation plays a crucial role:
  • Evaluating \( f(x) \) at \( a = 0 \): First, we compute \( f(0) = \frac{1}{2} \), which is the value of the function at the center of our Taylor expansion.

  • Evaluating Derivatives at \( a = 0 \): We also calculate the first, second, and third derivatives at this point: \( f'(0) = -\frac{1}{4} \), \( f''(0) = \frac{1}{4} \), \( f'''(0) = -\frac{3}{4} \). These values are crucial for constructing the Taylor polynomials.

Easily evaluating \( f(x) \) and its derivatives helps create the polynomial approximations, leading to a deeper understanding of the function's dynamics in a specific neighborhood.

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Most popular questions from this chapter

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{y \rightarrow 0} \frac{y-\tan ^{-1} y}{y^{3}} $$

Is it true that a sequence \(\left\\{a_{n}\right\\}\) of positive numbers must converge if it is bounded from above? Give reasons for your answer.

According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for tan \(^{-1} 1\) would you have to add to be sure of finding \(\pi / 4\) with an error of magnitude less than \(10^{-3} ?\) Give reasons for your answer.

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