Chapter 18: Problem 3
Calculate the tirst-order Taylor polynomial generated by \(y(x)=\cos x\) about (a) \(x=0\) (b) \(x=1\) (c) \(x=-05\)
Short Answer
Expert verified
(a) At x=0: \( P_1(x) = 1 \); (b) At x=1: \( P_1(x) = \, \cos(1) - \sin(1)(x - 1) \); (c) At x=-0.5: \( P_1(x) = \, \cos(-0.5) + \sin(0.5)(x + 0.5) \).
Step by step solution
01
Understanding the Form of Taylor Series
A Taylor series is used to approximate functions near a specified point. For a function \( f(x) \) expanded about the point \( x_0 \), the first-order Taylor polynomial is given by \( P_1(x) = f(x_0) + f'(x_0)(x - x_0) \). Here, the function \( f(x) = \, \cos x \).
02
Calculating the Taylor Polynomial at x=0
At \( x = 0 \), we need the values of \( f(0) \) and \( f'(0) \).Calculate \( f(0) \):\[f(0) = \, \cos(0) = 1\]Calculate \( f'(0) \):The derivative of \( f(x) = \, \cos x \) is \( f'(x) = -\sin x \).\[f'(0) = -\sin(0) = 0\]So, the first-order Taylor polynomial about \( x = 0 \) is:\[P_1(x) = 1 + 0(x - 0) = 1\]
03
Calculating the Taylor Polynomial at x=1
At \( x = 1 \), we need the values of \( f(1) \) and \( f'(1) \).Calculate \( f(1) \):\[f(1) = \, \cos(1)\]Calculate \( f'(1) \):\[f'(1) = -\sin(1)\]So, the first-order Taylor polynomial about \( x = 1 \) is:\[P_1(x) = \, \cos(1) - \sin(1)(x - 1)\]
04
Calculating the Taylor Polynomial at x=-0.5
At \( x = -0.5 \), we need the values of \( f(-0.5) \) and \( f'(-0.5) \).Calculate \( f(-0.5) \):\[f(-0.5) = \, \cos(-0.5)\]Calculate \( f'(-0.5) \):\[f'(-0.5) = -\sin(-0.5) = \sin(0.5)\]So, the first-order Taylor polynomial about \( x = -0.5 \) is:\[P_1(x) = \, \cos(-0.5) + \sin(0.5)(x + 0.5)\]
05
Summary of the Polynomials
We have calculated the first-order Taylor polynomials for \( y(x) = \, \cos x \) at three different points:- At \( x = 0 \): \( P_1(x) = 1 \)- At \( x = 1 \): \( P_1(x) = \, \cos(1) - \sin(1)(x - 1) \)- At \( x = -0.5 \): \( P_1(x) = \, \cos(-0.5) + \sin(0.5)(x + 0.5) \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order Taylor Polynomial
The first-order Taylor polynomial is a simple way to approximate a function around a specific point by considering only the function's value and its first derivative at that point. Imagine you have a point on a curve, and you draw a straight tangent line touching that point. This tangent line is a linear approximation of the curve near that point.The mathematical expression for the first-order Taylor polynomial of a function \( f(x) \) about the point \( x_0 \) is given by:\[P_1(x) = f(x_0) + f'(x_0)(x - x_0)\]Where:
- \( f(x_0) \) is the value of the function at \( x_0 \).
- \( f'(x_0) \) is the first derivative of the function evaluated at \( x_0 \).
Function Approximation
Function approximation is a technique used to find a simple way to estimate the value of a complicated function. There are various methods available, but the Taylor series is one of the most popular ones.The idea is to create a polynomial that mimics the function's behavior near a chosen point. The first-order Taylor polynomial forms the basis of this method by using only the function's value and slope at a given point, which provides a linear approximation of the function.Why is this useful? Function approximation is quite helpful in scenarios where calculating the exact function value is computationally intense or when only a rough estimate is sufficient. It also helps:- Simplify complex calculations- Predict function behavior- Solve equations numericallyIn our specific problem, we’re using the cosine function, \( y(x) = \cos x \). This function, like many trigonometric functions, is periodic and complex across its range. However, with the Taylor series approach, we can approximate its value near specific points like \( x=0 \), \( x=1 \), and \( x=-0.5 \) by simple linear equations.
Derivative Calculation
Derivatives are central to understanding and constructing Taylor polynomials. In simple terms, a derivative measures how a function changes as its input changes. For our cosine function, \( y(x) = \cos x \), the derivative is \( f'(x) = -\sin x \). Derivatives allow us to understand the slope or rate of change of the function at any given point.In the calculation of the first-order Taylor polynomial:
- At \( x = 0 \): The derivative is \( 0 \) because \( -\sin(0) = 0 \).
- At \( x = 1 \): The derivative is \( -\sin(1) \), a specific value needed for the approximation.
- At \( x = -0.5 \): We have \( \sin(0.5) \) as the derivative’s negative negates the negative in \( \sin(-0.5) \).