Problem 50 - Thomas Calculus

Question

Quadratic Approximations 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 \(x=0\). \(f(x)=\tan x\)

Step-by-Step Solution

Answer

The linearization and quadratic approximation of \(\tan x\) at \(x=0\) both equal \(x\).

Step 1: Understand the Problem

We are asked to find both the linearization (Taylor polynomial of order 1) and the quadratic approximation (Taylor polynomial of order 2) of the function \(f(x) = \tan x\) at \(x=0\). This involves calculating derivatives and evaluating them at \(x=0\).

Step 2: Find the Function and Its Derivatives at x=0

The function given is \(f(x) = \tan x\). The first derivative is \(f'(x) = \sec^2 x\), and the second derivative is \(f''(x) = 2\sec^2 x \tan x\). Now, evaluate these at \(x=0\): \(f(0) = \tan(0) = 0\), \(f'(0) = \sec^2(0) = 1\), and \(f''(0) = 2(\sec^2(0))(\tan(0)) = 0\).

Step 3: Calculate the Linearization

The linearization, or Taylor polynomial of order 1, at \(x = 0\) is given by the formula \(P_1(x) = f(0) + f'(0)x\). Substituting the evaluated values gives us \(P_1(x) = 0 + 1 \cdot x = x\). Thus, the linearization is \(x\).

Step 4: Calculate the Quadratic Approximation

The quadratic approximation, or Taylor polynomial of order 2, at \(x = 0\) is given by \(P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2\). Substituting the values, we get \(P_2(x) = 0 + x + 0 \cdot x^2 = x\). Thus, the quadratic approximation is also \(x\).

Key Concepts

Quadratic Approximation

The concept of quadratic approximation centers around the idea of using a polynomial to estimate the behavior of a function near a specific point. The Taylor polynomial of order 2, which provides the quadratic approximation, represents this approximation. In simpler terms, the quadratic approximation is like a parabola that hugs the curve of the function closely at the point of interest.
To arrive at the quadratic approximation of a function, you need to know the function's value and its first two derivatives at the specified point.

• For the quadratic approximation of the given function, \(f(x) = \tan x\), at \(x = 0\), we assess the function and its derivatives at 0, as explained in the original step-by-step solution.
• This provides information about the function's slope and concavity, enabling us to sketch a good approximation.

Although the quadratic approximation can be more accurate than a linear approximation, it is crucial to recognize that it is still an approximation and best suited for points near the approximation point.

Linearization

Linearization is the process of approximating a function using a linear function or a straight line. It is essentially the Taylor polynomial of order 1.
The linearization of a function involves finding the tangent line to the function at a point, which provides the best linear estimate of the function's behavior near that point.
For a function \(f(x)\), the linearization at \(x = a\) is given by: \( L(x) = f(a) + f'(a)(x-a) \).
In the given exercise, the function is \(f(x) = \tan x\), and we find the linearization at \(x = 0\).

• The first derivative of the function, which is \(f'(x) = \sec^2 x\), evaluated at \(x = 0\) gives 1.
• Thus the linearization of \(f(x)\) around \(x = 0\) is exactly \(x\).

Linearization is particularly useful in providing a simple and quick estimate of non-linear functions for small deviations around the point of interest.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. They are crucial for understanding and calculating both the linearization and quadratic approximation of functions.
For any function, finding its derivative involves understanding how the function's output changes as its input changes.
For instance, if \(f(x) = \tan x\), its first derivative, \(f'(x) = \sec^2 x\), indicates how fast \(\tan x\) changes with respect to \(x\).

• The second derivative, \(f''(x)\), which in this problem is \(f''(x) = 2\sec^2 x \tan x\), tells us about the curvature of the function, or how its rate of change itself changes.
  
• These derivatives are key components when constructing the Taylor polynomial of any order.

By evaluating these derivatives at \(x = 0\), we determine essential coefficients that help form the best polynomial approximation of the original function near that point.

Trigonometric Functions

Trigonometric functions like \(\tan x\), \(\sin x\), and \(\cos x\) are periodic functions that relate angles to side ratios in right-angled triangles. They play a vital role in mathematics, especially in the study of oscillatory behaviors and waves.
In this exercise, the given function \(\tan x\) illustrates the complexities of handling trigonometric functions within calculus.

• Understanding the behavior of \(\tan x\) means recognizing its periodicity and points at which it becomes undefined, such as \(\pi/2\) and \(-\pi/2\).
  
• Approximating \(\tan x\) with polynomials, like the Taylor series, allows us to handle its behavior around points like \(x=0\) where it remains defined.
• Using derivatives of trigonometric functions, such as \(\sec^2 x\) which is \(f'(x)\) for \(\tan x\), helps in developing these polynomial approximations.
  

By utilizing these techniques, mathematicians can simplify complex trigonometric functions into more manageable forms using calculus methods like Taylor series expansions.