Chapter 18: Problem 2
Calculate the first-order Taylor polynomial generated by \(y(x)=\sin x\) about (a) \(x=0\) (b) \(x=1\) (c) \(x=-0.5\)
Short Answer
Expert verified
(a) \(T_1(x) = x\); (b) \(T_1(x) = \sin(1) + \cos(1)(x-1)\); (c) \(T_1(x) = \sin(-0.5) + \cos(-0.5)(x+0.5)\).
Step by step solution
01
Understand the Taylor Polynomial Formula
To find the first-order Taylor polynomial of a function \(y(x)\) centered at a point \(c\), you use the formula: \[ T_1(x) = f(c) + f'(c)(x-c) \] where \(f(x)\) is the function, and \(f'(x)\) is its derivative.
02
Compute Values for Part (a)
For \(y(x) = \sin x\) and centered at \(x=0\): - The function value at \(x=0\) is \(f(0) = \sin(0) = 0\).- The derivative of \(y(x)\) is \(y'(x) = \cos x\), so \(f'(0) = \cos(0) = 1\).Substitute into the Taylor formula: \[ T_1(x) = 0 + 1\cdot(x-0) = x \]
03
Compute Values for Part (b)
For \(y(x) = \sin x\) and centered at \(x=1\): - The function value at \(x=1\) is \(f(1) = \sin(1)\).- The derivative of \(y(x)\) is \(y'(x) = \cos x\), so \(f'(1) = \cos(1)\).Substitute into the Taylor formula:\[ T_1(x) = \sin(1) + \cos(1)\cdot(x-1) \]
04
Compute Values for Part (c)
For \(y(x) = \sin x\) and centered at \(x=-0.5\): - The function value at \(x=-0.5\) is \(f(-0.5) = \sin(-0.5)\).- The derivative of \(y(x)\) is \(y'(x) = \cos x\), so \(f'(-0.5) = \cos(-0.5)\).Substitute into the Taylor formula:\[ T_1(x) = \sin(-0.5) + \cos(-0.5)\cdot(x+0.5) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order approximation
The first-order approximation is the simplest form of a Taylor polynomial. It is a way to approximate a function using its value and derivative at a specific point. This method is helpful when you want to understand how a function behaves near that point with minimal effort. You can think of it as drawing a straight line (tangent) that touches the function at that particular point.
For any function, the first-order Taylor polynomial at a point \(c\) is given by:
For any function, the first-order Taylor polynomial at a point \(c\) is given by:
- \( T_1(x) = f(c) + f'(c)(x-c) \)
- \( f(c) \) represents the function's value at the center \(c\).
- \( f'(c) \) denotes the slope or the derivative of the function at \(c\).
- \( (x-c) \) shows the distance from the point \(c\).
Derivative calculation
Understanding derivative calculation is fundamental to computing Taylor polynomials. Derivatives provide information about the slope or rate of change of a function at any given point. For the function \( y(x) = \sin x \), knowing its derivative helps in creating the Taylor polynomial.
The derivative of \( \sin x \) is \( \cos x \). This means at any point \( x \):
The derivative of \( \sin x \) is \( \cos x \). This means at any point \( x \):
- The value of \( y'(x) = \cos x \) gives us the slope of the tangent line at that point.
- \( f'(c) = \cos(c) \)
Taylor series expansion
The Taylor series expansion is a powerful tool that extends the concept of the first-order approximation to higher approximations. While a first-order Taylor polynomial uses the function value and its derivative, a full Taylor series accounts for higher derivatives too.
In general, a Taylor series expansion of a function \( f(x) \) around a point \( c \) is given by:
In general, a Taylor series expansion of a function \( f(x) \) around a point \( c \) is given by:
- \[ f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \ldots \]
- \( T_1(x) = f(c) + f'(c)(x-c) \)