Problem 43 - 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 \(x=0\). \(f(x)=1 / \sqrt{1-x^{2}}\)

Step-by-Step Solution

Answer

(a) Linearization: 1. (b) Quadratic Approximation: \(1 + \frac{1}{2} x^2\).

Step 1: Identify the Function and Its Derivatives

We start with the function \( f(x) = \frac{1}{\sqrt{1-x^2}} \). To find the Taylor polynomials, we need the derivatives of this function. The first derivative \( f'(x) \) is calculated using the chain rule and is \( f'(x) = \frac{x}{(1-x^2)^{3/2}} \). The second derivative \( f''(x) \) gives us \( f''(x) = \frac{1+2x^2}{(1-x^2)^{5/2}} \). We'll use these derivatives in the next steps.

Step 2: Evaluate Function and Derivatives at x=0

Evaluate \( f(x) \), \( f'(x) \), and \( f''(x) \) at \( x=0 \). We find \( f(0) = \frac{1}{\sqrt{1-0^2}} = 1 \), \( f'(0) = \frac{0}{(1-0^2)^{3/2}} = 0 \), and \( f''(0) = \frac{1+2\times0^2}{(1-0^2)^{5/2}} = 1 \). These values represent the coefficients used in linear and quadratic Taylor polynomials.

Step 3: Compute the Linearization (Order 1 Taylor Polynomial)

The Taylor polynomial of order 1 (linearization) at \( x=0 \) is given by the formula: \( L(x) = f(a) + f'(a)(x-a) \). Substituting the evaluated values at \( x=0 \), we get: \( L(x) = 1 + 0 \times x = 1 \). The linearization is simply \( 1 \).

Step 4: Compute the Quadratic Approximation (Order 2 Taylor Polynomial)

The Taylor polynomial of order 2 (quadratic approximation) at \( x=0 \) is given by: \( Q(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 \). Substituting the evaluated derivatives at \( x=0 \), we have: \( Q(x) = 1 + 0 \times x + \frac{1}{2} x^2 = 1 + \frac{1}{2} x^2 \). Thus, the quadratic approximation is \( 1 + \frac{1}{2} x^2 \).

Key Concepts

Taylor polynomial

A Taylor polynomial is an approximation of a function around a specific point, using the function's derivatives. It's like a snapshot of your curve at a given point. Imagine you took pieces of information about the function—such as its slope or how it curves—and used them to recreate the function. The Taylor polynomial of order 2 includes up to the second derivative. This order 2 polynomial is known as the quadratic approximation.
Consider Taylor polynomials as a sequence of increasingly better approximations as you add more terms. The more terms you include, the more accurate the approximation. The Taylor polynomial of order 2 at a point \( x = a \) is:

• \( f(a) \): Function value at \( a \), gives the starting point.
• \( f'(a)(x-a) \): Linear part, accounts for the slope.
• \( \frac{f''(a)}{2}(x-a)^2 \): Quadratic part, accounts for how the curve bends.

For the function \( f(x) = \frac{1}{\sqrt{1-x^2}} \) at \( x = 0 \), we calculated these in the original solution and found the quadratic approximation \( 1 + \frac{1}{2} x^2 \). This gives us a parabola representing the curve near \( x = 0 \).

linearization

Linearization is like the simplest form of Taylor polynomials and is known as the first-order approximation. When we linearize a function at a particular point, we're essentially creating a straight-line approximation—tangential to the curve at that point. This is where derivatives play an essential role.
Think of linearization as taking a magnifying glass to a small segment around the point of interest on your function and drawing the tangent line there. This line (or linear approximation) captures both the value and the slope of the function at that specific point. We usually write it as:

• \( L(x) = f(a) + f'(a)(x-a) \)

In the original exercise, for \( f(x) = \frac{1}{\sqrt{1-x^2}} \) at \( x = 0 \), the linearization ended up as simply \( L(x) = 1 \). This indicates that near \( x = 0 \), a flat line closely approximates the function's behavior. This simplicity makes linearization very powerful in analyzing and solving problems where detailed precision isn't critical.

derivatives

Derivatives are mathematical tools to understand how a function changes. Simply put, they give you the slope of a function at any given point. This concept is immensely powerful and underpins the process of creating Taylor polynomials and linearization.
The first derivative \( f'(x) \) tells us how the function rises or falls—just like you question how steep a hill is. Meanwhile, the second derivative \( f''(x) \) provides information on the function's concavity, or how the slope itself is changing, much like evaluating how the hill's steepness changes.

• \( f'(x) \): First derivative, indicates the slope or rate of change.
• \( f''(x) \): Second derivative, shows how the slope itself changes (concavity).

When constructing the Taylor polynomial or quadratic approximation for \( f(x) = \frac{1}{\sqrt{1-x^2}} \) at \( x = 0 \), we specifically calculated these derivatives. They acted as critical components for both linearizing and deriving the quadratic polynomial, showing their pivotal role in approximating complex functions.