Chapter 9: Problem 50
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 \(45-50,\) find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0.\) $$f(x)=\tan x$$
Short Answer
Step by step solution
Understand the Problem
Find Function Values
Compute First Derivative
Compute Second Derivative
Linearization
Quadratic Approximation
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Approximation
- \( P_2(x) = f(a) + f'(a) \cdot (x-a) + \frac{f''(a)}{2} \cdot (x-a)^2 \)
In the exercise given, the quadratic approximation simplifies to \( P_2(x) = x \) because the second derivative evaluated at \( x = 0 \) is zero. This shows that higher-order behavior beyond linear isn’t contributing at the point of interest, making the approximation effectively linear at this point.
Linearization
- \( P_1(x) = f(a) + f'(a) \cdot (x - a) \)
This means the linearized version simplifies to \( x \), demonstrating that near \( x = 0 \), the curve of tangent is practically a line. Linearization is useful because it simplifies the error and complexity in calculations for applications where precise exactness over large intervals isn't necessary.
Derivatives
- The first derivative, \( f'(x) = \sec^2 x \), measures how \( \tan x \) changes with a small change in \( x \).
- The second derivative is derived from \( f'(x) = \sec^2 x \), giving us \( f''(x) = 2 \sec^2 x \cdot \tan x \).
- \( f'(0) = 1 \)
- \( f''(0) = 0 \)