Problem 6 - Thomas Calculus

Question

Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a .\) \(f(x)=1 /(x+2), \quad a=0\)

Step-by-Step Solution

Answer

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 1: 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 \(a=0\).

Step 2: 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.

Step 3: Evaluate Derivatives at \(a=0\)

For each derivative, substitute \(x = 0\):- \(f(0) = 0.5\)- \(f'(0) = -0.25\)- \(f''(0) = 0.125\)- \(f'''(0) = -0.375\).

Step 4: 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: \(P_3(x) = 0.5 - 0.25x + 0.0625x^2 + \frac{-0.375}{6}x^3 = 0.5 - 0.25x + 0.0625x^2 - 0.0625x^3\).

Key Concepts

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 \(a = 0\):

• \(f(0) = 0.5\)
• \(f'(0) = -0.25\)
• \(f''(0) = 0.125\)
• \(f'''(0) = -0.375\)

Function evaluation ensures that each term of the Taylor polynomial captures how the function behaves around the point \(a\), helping to create an accurate polynomial representation. This step is often critical for building reliable approximations using Taylor polynomials.