Chapter 11: Problem 35
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 \(x=a .\) In Exercises \(33-38\) , find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0\) . $$ f(x)=1 / \sqrt{1-x^{2}} $$
Short Answer
Step by step solution
Identify the Function and its Derivatives
Compute Higher Order Derivatives
Evaluate the Derivatives at x=0
Find the Linearization (Order 1 Polynomial)
Find the Quadratic Approximation (Order 2 Polynomial)
Verification of the Approximations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linearization
To linearize a function, we need its derivative at the point of approximation. In the example given, for the function \( f(x) = \frac{1}{\sqrt{1-x^2}} \), this process starts by obtaining its derivative:
- Compute the first derivative: \( f'(x) = \frac{x}{(1-x^2)^{3/2}} \).
- Evaluate at \( x = 0 \): \( f'(0) = 0 \).
- Use the linear approximation formula: \( f(x) \approx f(0) + f'(0) \cdot x \).
- \( f(x) \approx 1 \).
Quadratic Approximation
The quadratic approximation, or the Taylor polynomial of order two, is given by:\[ f(x) \approx f(a) + f'(a) \cdot (x - a) + \frac{f''(a)}{2} \cdot (x - a)^2 \]For the function \( f(x) = \frac{1}{\sqrt{1-x^2}} \) at \( x = 0 \):
- \( f(0) = 1 \)
- \( f'(0) = 0 \)
- Compute the second derivative: \( f''(0) = 1 \)
- \( f(x) \approx 1 + 0 \cdot x + \frac{1}{2} \cdot x^2 = 1 + \frac{x^2}{2} \).
Derivatives
To understand derivatives in the context of Taylor polynomials, let's consider this broken down:
- The **first derivative**, \( f'(x) \), gives the slope of the tangent line at a point. This is used for the linear term in the Taylor series.
- The **second derivative**, \( f''(x) \), provides information about the curvature or concavity of the function. It is used in the quadratic term of the Taylor series.
- First derivative: \( f'(x) = \frac{x}{(1-x^2)^{3/2}} \) - At \( x=0 \), \( f'(0) = 0 \).
- Second derivative: \( f''(x) = \frac{1 + 2x^2}{(1-x^2)^{5/2}} \) - At \( x=0 \), \( f''(0) = 1 \).