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Chapter 9: Problem 41

a. Approximate the given quantities using Taylor polynomials with \(n=3\) b. Compute the absolute error in the approximation assuming the exact value is given by a calculator.$$\tan (-0.1)$$ c.\(T_{3}(z)=0.33333333 x^{3}+x\)

Short Answer

Expert verified
Question: Approximate the value of \(\tan(-0.1)\) using a Taylor polynomial of degree 3. Calculate the absolute error between the approximation and the exact value. Answer: The approximation of \(\tan(-0.1)\) using the Taylor polynomial of degree 3 is -0.1 - 1/500. The absolute error between the approximation and the exact value is approximately 0.00016666664683.

Step by step solution

01

Recall Taylor polynomial formula

Taylor polynomial of degree n for a function \(f(x)\) is given by: $$ T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k $$ where \(n\) is the degree of the polynomial, \(f^{(k)}(x)\) is the k-th derivative of the function, and \((x-a)^k\) is the difference between the point \(x\) and the point \(a\) at which the Taylor polynomial is centered. In our case, \(f(x)=\tan(x)\), n=3 and we focus on the point \(x=-0.1\).
02

Compute the first few derivatives of \(f(x)\)

We need to find the first few derivatives of \(\tan(x)\): 1st derivative: \( f'(x) = \frac{d}{dx} (\tan(x)) = \sec^2(x) \) 2nd derivative: \( f''(x) = \frac{d}{dx} (\sec^2(x)) = 2\sec(x)\tan(x)\sec^2(x) \) 3rd derivative: \( f'''(x) = \frac{d}{dx} (2\sec(x)\tan(x)\sec^2(x)) = 4 \sec(x)(\tan^2(x)+1)\sec^2(x) \)
03

Evaluate the derivatives at x=0

Since we want to evaluate the tangent function around the point x=0, we must evaluate these derivatives at x=0: \(f(0)=\tan(0)=0\) \(f'(0)=\sec^2(0)=1\) \(f''(0)=2\sec(0)\tan(0)\sec^2(0)=0\) \(f'''(0)=4 \sec(0)(\tan^2(0)+1)\sec^2(0)= 4\)
04

Calculate the Taylor polynomial

Now, calculate the Taylor polynomial of degree 3 using the derived values: $$ T_3(x) = f(0) +\frac{f'(0)}{1!}(x-0) +\frac{f''(0)}{2!}(x-0)^2 +\frac{f'''(0)}{3!}(x-0)^3 $$ $$ T_3(x) = 0 +\frac{1}{1!}x +\frac{0}{2!}x^2 +\frac{4}{3!}x^3 $$ $$ T_3(x) = x + \frac{2}{3}x^3 $$
05

Compute the approximation

Approximate the given quantity using the computed Taylor polynomial for \(x = -0.1\) : $$ \tan(-0.1) \approx T_3(-0.1) = (-0.1) + \frac{2}{3}(-0.1)^3 $$ $$ \tan(-0.1) \approx -0.1 - \frac{1}{500} $$
06

Compute the absolute error

The absolute error is defined as the absolute difference between the exact value and the approximation: $$ \text{Absolute error} = \left| \text{Exact value} - \text{Approximation} \right|$$ Use a calculator to find the exact value of \(\tan(-0.1)\), then compute the absolute error: $$ \text{Absolute error} = \left| -0.099833416646...\ -\ (-0.1 + \frac{1}{500}) \right|$$ $$ \text{Absolute error} = 0.00016666664683... $$

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Key Concepts

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

Understanding Approximation Error
When using Taylor polynomials to estimate values, it's important to grasp what approximation error is. Simply put, approximation error shows how close or far an estimated value is from the true value. In our exercise, the Taylor polynomial of degree 3 provides an approximation for \( \tan(-0.1) \). The formula used is:
  • \( T_3(x) = x + \frac{2}{3}x^3 \)
To determine the accuracy of this approximation, we calculate the absolute error, which is the difference between the actual value and the approximated value. This is done using:
  • \( \text{Absolute error} = \left| \text{Exact value} - \text{Approximation} \right| \)
In this context, a smaller error indicates a more accurate approximation.
This accuracy is essential when approximations are used in critical applications like engineering and sciences.
The Role of Derivatives in Taylor Polynomials
Derivatives are the backbone of Taylor polynomials. They inform us how a function behaves around a certain point. In a Taylor polynomial, each derivative plays a crucial part in shaping the polynomial's curve, thereby influencing the accuracy of the approximation. For the tangent function \( f(x) = \tan(x) \), the first few derivatives are:
  • 1st derivative: \( f'(x) = \sec^2(x) \)
  • 2nd derivative: \( f''(x) = 2\sec(x)\tan(x)\sec^2(x) \)
  • 3rd derivative: \( f'''(x) = 4 \sec(x)(\tan^2(x)+1)\sec^2(x) \)
These expressions become simpler when evaluated at \( x = 0 \), a common technique to simplify calculations. Understanding derivatives helps us create more precise Taylor polynomials by correctly incorporating changes in the function's slope.
Understanding the Tangent Function and Its Taylor Polynomial
The tangent function, well known for its periodic and asymptotic behavior, offers fascinating challenges and insights in calculus. Approximating \( \tan(x) \) using a polynomial involves calculating its Taylor polynomial, which acts like a shortcut to complex function evaluation around a point.Given \( T_3(x) = x + \frac{2}{3}x^3 \), this polynomial is centered around \( x=0 \) for our approximation task. The choice of degree \( n=3 \) reveals more about the function's behavior than a simple linear approximation. The Taylor polynomial provides a curve mimicking \( \tan(x) \), allowing us to estimate values for \( x \) close to zero, such as \(-0.1\). This approximation becomes feasible because the polynomial incorporates enough derivative information to simulate the function's shape effectively. Understanding these fundamental concepts of the tangent function and its Taylor polynomial equips students to tackle challenging real-world problems.

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Most popular questions from this chapter

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

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