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Chapter 9: Problem 46

The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a .\) Find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0.\) $$f(x)=\tan x$$

Short Answer

Expert verified
Both the linearization and quadratic approximation at \( x=0 \) are \( P(x) = x \).

Step by step solution

01

Understanding the Taylor Polynomial

The Taylor polynomial of a function \( f(x) \) around \( x = a \) is given by \( P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n \). For our problem, we need the Taylor polynomials of order 1 (linearization) and order 2 (quadratic approximation) of \( f(x) = \tan x \) around \( x = 0 \).
02

Calculating Derivatives

First, compute \( f'(x) \) and \( f''(x) \) for \( f(x) = \tan x \). We know \( f'(x) = \sec^2 x \). For \( f''(x) \), we can differentiate \( f'(x) = \sec^2 x \) again: \( f''(x) = 2 \sec^2 x \cdot \tan x = 2 \sec^2(x)\tan x \).
03

Evaluating Derivatives at \( x = 0 \)

Evaluate \( f(x), f'(x), \) and \( f''(x) \) at \( x = 0 \). We have: \( f(0) = \tan(0) = 0 \), \( f'(0) = \sec^2(0) = 1 \), and \( f''(0) = 2 \sec^2(0)\tan(0) = 0 \cdot 2 = 0 \).
04

Constructing the Linearization

The linearization (order 1 Taylor polynomial) is given by \( P_1(x) = f(a) + f'(a)(x-a) \). Thus, for \( x = 0 \), the linearization \( P_1(x) = 0 + 1 \cdot x = x \).
05

Constructing the Quadratic Approximation

The quadratic approximation (order 2 Taylor polynomial) is given by \( P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 \). Thus, for \( x = 0 \), it simplifies to \( P_2(x) = 0 + x + \frac{0}{2}(x)^2 = x \).
06

Conclusion

We found the Taylor polynomial of order 1 (linearization) and order 2 (quadratic approximation) for \( f(x) = \tan x \) at \( x = 0 \). Both results give \( P(x) = x \) since the second derivative at \( x = 0 \) contributed nothing to the quadratic term.

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Key Concepts

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

Linearization
Linearization is a method used to approximate a function by using a linear polynomial, usually around a specific point. When you linearize a function, you essentially create the best possible straight-line (linear) approximation for that function near the given point.
  • It involves taking the first derivative of the function.
  • This linear approximation is known as the Taylor polynomial of order 1 (also called the first-order Taylor expansion).
  • Mathematically, it is represented as: \( P_1(x) = f(a) + f'(a)(x-a) \).
In our specific problem with the function \( f(x) = \tan(x) \), the linearization at \( x = 0 \) gives \( P_1(x) = x \). This is because \( f'(0) = \sec^2(0) = 1 \), making it a straightforward linear approximation. This means that near the point \( x = 0 \), \( \tan(x) \approx x \).
Quadratic Approximation
When you want a more accurate approximation than a straight line, quadratic approximation can be useful. It is the next step in Taylor series and uses a polynomial of degree two.
  • This involves the second derivative of the function.
  • The quadratic approximation functions as the Taylor polynomial of order 2.
  • It鈥檚 mathematically expressed as: \( P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 \).
For our example with \( f(x) = \tan(x) \), at \( x = 0 \), the quadratic approximation turned out to be the same as the linearization \( P_2(x) = x \). This happened because the second derivative \( f''(0) = 0 \) contributed no additional terms, making it less distinct from the linearization in this specific case.
Differentiation
Differentiation is a powerful tool in calculus used to determine the rate at which a function is changing at any given point. By computing derivatives, we gain insights into the behavior of the function.
  • First derivative \( f'(x) \) provides the slope or rate of change of the function.
  • Second derivative \( f''(x) \) indicates the concavity or the acceleration of the function.
  • Higher-order derivatives can give further insights.
In our exercise, differentiating \( \tan(x) \) gives us:- \( f'(x) = \sec^2(x) \), which tells us how steep \( \tan(x) \) is at different points.- \( f''(x) = 2 \sec^2(x) \tan(x) \), showing how the curvature of \( \tan(x) \) behaves, although it simplifies to 0 at \( x=0 \).
Function Evaluation
Function evaluation means to substitute a specific value into a function and compute the output. This is essential for constructing approximations like Taylor polynomials.
  • This helps in finding the exact value of the function at a specific point \( a \).
  • It involves calculating the function and its derivatives at that point.
In the task with \( f(x) = \tan(x) \), evaluating at \( x = 0 \) yields:- \( f(0) = \tan(0) = 0 \), indicating the function value at that point.- \( f'(0) = \sec^2(0) = 1 \), offering the slope of the tangent line.- \( f''(0) = 0 \), suggesting no contribution to curvature, simplifying the approximation process as it involves fewer terms. This evaluation is foundational in ensuring accuracy when approximating complex functions.

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