Chapter 10: Problem 6
Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a .\) \(f(x)=1 /(x+2), \quad a=0\)
Short Answer
Expert verified
The Taylor polynomials are: Order 0: 0.5, Order 1: 0.5 - 0.25x, Order 2: 0.5 - 0.25x + 0.0625x^2, Order 3: 0.5 - 0.25x + 0.0625x^2 - 0.0625x^3.
Step by step solution
01
Understanding Taylor Polynomials
The Taylor polynomial of degree "n" for a function \(f(x)\) about the point \(a\) is given by \(P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots + \frac{f^n(a)}{n!}(x-a)^n\). To find the polynomials, we need the function and its derivatives evaluated at \(a=0\).
02
Compute Derivatives
Start by finding the derivatives of \(f(x) = \frac{1}{x+2}\). Calculate the first few derivatives:- \(f(x) = (x+2)^{-1}\)- \(f'(x) = -(x+2)^{-2}\)- \(f''(x) = 2(x+2)^{-3}\)- \(f'''(x) = -6(x+2)^{-4}\)- Continue if needed.
03
Evaluate Derivatives at \(a=0\)
For each derivative, substitute \(x = 0\):- \(f(0) = 0.5\)- \(f'(0) = -0.25\)- \(f''(0) = 0.125\)- \(f'''(0) = -0.375\).
04
Construct Taylor Polynomials
Use the evaluated derivatives to construct the Taylor polynomials:- Order 0: \(P_0(x) = 0.5\)- Order 1: \(P_1(x) = 0.5 - 0.25x\)- Order 2: \(P_2(x) = 0.5 - 0.25x + \frac{0.125}{2}x^2 = 0.5 - 0.25x + 0.0625x^2\)- Order 3: \(P_3(x) = 0.5 - 0.25x + 0.0625x^2 + \frac{-0.375}{6}x^3 = 0.5 - 0.25x + 0.0625x^2 - 0.0625x^3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
A Taylor series is a mathematical tool that lets you approximate more complex functions using polynomials. Imagine it like breaking down a function into smaller, easier parts that you can work with. The Taylor series attempts to represent a function as an infinite sum of terms. Each term in the series involves derivatives of the function at a single point, known as the expansion point. For a function \(f(x)\) expanded around a point \(a\), the Taylor series is expressed as:- \(f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots + \frac{f^n(a)}{n!}(x-a)^n\)The series can continue indefinitely, but in practical situations, we use finite Taylor polynomials for approximation.
Derivatives
Derivatives represent the rate at which a function is changing at any point. They form the backbone of calculus. When finding a Taylor polynomial, derivatives of various orders are needed.Consider the function \(f(x) = \frac{1}{x+2}\) in our exercise. Here's a simple breakdown of its derivatives:- The first derivative, \(f'(x)\), shows how \(f(x)\) changes as \(x\) changes.- The second derivative, \(f''(x)\), looks at how the rate of change itself changes, and so on.For our function:- \(f'(x) = -(x+2)^{-2}\)- \(f''(x) = 2(x+2)^{-3}\)- \(f'''(x) = -6(x+2)^{-4}\)These derivatives are plugged into the Taylor polynomial formula to approximate the function around a specific point.
Polynomial Approximation
Polynomial approximation is the process of estimating a more complicated function using a polynomial. In simpler terms, you convert a complex function into an easier "polynomial form". This method is often used in calculus to approximate functions that are otherwise tough to work with.Taylor polynomials are a form of polynomial approximation. In the given exercise, Taylor polynomials of order \(0, 1, 2,\) and \(3\) were constructed using both the value of the function and its derivatives at a particular point. Here’s what each order polynomial looks like for \(f(x) = \frac{1}{x+2}\):- **Order 0:** A constant function (horizontal line)- **Order 1:** A linear function (straight line)- **Order 2:** A quadratic function (parabola)- **Order 3:** A cubic function (cubic curve)Higher-order polynomials provide a better approximation over a broader range of values near the point \(a\).
Function Evaluation
Function evaluation in the context of Taylor series involves determining the value of a function and its derivatives at a specific point, known as the expansion point. This is crucial because these values build the foundation of the Taylor polynomial.In the example of \(f(x) = \frac{1}{x+2}\), all the derivatives were evaluated at the point \(a = 0\):
- \(f(0) = 0.5\)
- \(f'(0) = -0.25\)
- \(f''(0) = 0.125\)
- \(f'''(0) = -0.375\)