Chapter 5: Problem 3
In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point. $$ f(x)=\cos (2 x) \text { at } a=\pi $$
Short Answer
Expert verified
The Taylor polynomial of degree two is \( T_2(x) = 1 - 2(x-\pi)^2 \).
Step by step solution
01
Understand the Taylor Polynomial Formula
A Taylor polynomial of degree two for a function \( f(x) \) centered at \( a \) is given by: \[ T_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 \] where \( f'(x) \) and \( f''(x) \) are the first and second derivatives of \( f(x) \), respectively.
02
Evaluate the Function at the Center
To evaluate \( f(x) = \cos(2x) \) at \( x = a = \pi \), calculate \( f(\pi) = \cos(2\pi) = \cos(0) = 1 \).
03
Find the First Derivative and Evaluate at the Center
Find the first derivative: \( f'(x) = -2\sin(2x) \). Evaluate this at \( x = a = \pi \): \( f'(\pi) = -2\sin(2\pi) = -2(0) = 0 \).
04
Find the Second Derivative and Evaluate at the Center
Find the second derivative: \( f''(x) = -4\cos(2x) \). Evaluate this at \( x = a = \pi \): \( f''(\pi) = -4\cos(2\pi) = -4(1) = -4 \).
05
Construct the Degree Two Taylor Polynomial
Using the Taylor polynomial formula, we have: \[ T_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 \] Substitute the values: \[ T_2(x) = 1 + 0 \cdot (x-\pi) + \frac{-4}{2} (x-\pi)^2 = 1 - 2(x-\pi)^2 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives play a crucial role when working with Taylor polynomials. The derivative of a function gives us the rate at which the function is changing at any point. For Taylor polynomials, you need to find both the first and second derivatives to construct a polynomial of degree two.
Let's break it down:
Let's break it down:
- **First Derivative**: The first derivative, noted as \( f'(x) \), captures the slope of the tangent line to the function at any given point on the curve. For the function \( f(x) = \cos(2x) \), the first derivative is \( f'(x) = -2\sin(2x) \).
- **Second Derivative**: This derivative, \( f''(x) \), informs us about the curvature or concavity of the function. It shows us how the rate of change is itself changing. Here, \( f''(x) = -4\cos(2x) \).
- The derivatives are then evaluated at a specific point \( a \) to be inserted in the Taylor polynomial formula.
Trigonometric Functions
Trigonometric functions like cosine and sine appear widely in applications spanning from physics to engineering. Here, you work with the cosine function. This function is periodic, meaning it repeats its values in regular intervals. This periodicity is crucial because:
- The function \( \cos(2x) \) is crucial for problems involving waves or oscillations.
- The cosine function always produces values between -1 and 1, affecting how its derivatives behave.
- Understanding its properties helps when you're seeking patterns or symmetry in functions either for computation purposes or proof simplifications.
Approximation
The concept of approximation using Taylor polynomials involves estimating the value of a function near a given point using polynomials. This is especially helpful when the exact function is too complex to handle analytically.
In the Taylor polynomial:
In the Taylor polynomial:
- The zero-degree part \( f(a) \) is the value of the function at the center point \( a \). In our case, at \( x = \pi \), this value is \( 1 \).
- The first-degree term \( f'(a)(x-a) \) reflects the initial slope of the function, but here it is zero, indicating no initial linear change near \( a \).
- Higher-order terms, like \( \frac{f''(a)}{2}(x-a)^2 \) indicate adjustments to this approximation, allowing the polynomial to "bend" to match the function closer over more points.