Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Plot on the same axes the given function along with the Maclaurin polynomials of orders \(1,2,3\), and \(4 .\) $$ \frac{1}{1+x^{2}} $$

Step-by-Step Solution

Answer

Plot \( f(x) = \frac{1}{1+x^2} \) along with polynomials: constant 1, \( 1-x^2 \), and \( 1-x^2+x^4 \).

Step 1: Identify the Function

The function we need to work with is given as \( f(x) = \frac{1}{1+x^2} \). This is a rational function that decreases as \( x \) moves away from zero.

Step 2: Understand Maclaurin Series

The Maclaurin series is a type of Taylor series expansion centered at \( x = 0 \). It expresses a function as a sum of its derivatives evaluated at zero, multiplied by powers of \( x \): \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \).

Step 3: Calculate Derivatives at x=0

To find the Maclaurin polynomials, we need derivatives of \( f(x) = \frac{1}{1+x^2} \) evaluated at \( x = 0 \): - \( f(0) = 1 \) - \( f'(x) = -\frac{2x}{(1+x^2)^2} \rightarrow f'(0) = 0 \)- \( f''(x) = -\frac{2(1-x^2)}{(1+x^2)^3} \rightarrow f''(0) = -2 \)- \( f'''(x) = \frac{8x(x^2-3)}{(1+x^2)^4} \rightarrow f'''(0) = 0 \)- \( f^{(4)}(x) = \frac{24(1-6x^2+x^4)}{(1+x^2)^5} \rightarrow f^{(4)}(0) = 24 \)

Step 4: Obtain Maclaurin Polynomial Terms

Using the derivatives from Step 3, we compute:- Order 1: \( P_1(x) = 1 \)- Order 2: \( P_2(x) = 1 - x^2 \)- Order 3: \( P_3(x) = 1 - x^2 \) (same as Order 2 since \( f'''(0) = 0 \))- Order 4: \( P_4(x) = 1 - x^2 + \frac{24}{24}x^4 = 1 - x^2 + x^4 \)

Step 5: Plot the Function and Polynomials

Using a graphing tool, plot \( f(x) = \frac{1}{1+x^2} \) and the Maclaurin polynomials:- \( P_1(x) = 1 \), a horizontal line.- \( P_2(x) = 1 - x^2 \), a parabola opening downwards.- \( P_3(x) = 1 - x^2 \), same parabola as \( P_2 \).- \( P_4(x) = 1 - x^2 + x^4 \), a slightly altered parabola with an upward component.

Key Concepts

Taylor series

The Taylor series is an infinite series that approximates a function. It uses derivatives of the function at a specific point, often around zero. When it's centered at zero, it’s known as a Maclaurin series. The basic idea is that you can express complex functions as sums of simpler polynomial terms.
For any function that’s smooth and continuous, its Taylor series at point \( a \) is given by:

• \( f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \)

Each term in the series involves a derivative of the function evaluated at \( a \), and is weighted by a corresponding power of \((x-a)\). This series can closely approximate the function over a range.
In the specific exercise mentioned, we're focusing on Maclaurin polynomials, which are Taylor series centered at \( x = 0 \). This means, for the function \( f(x) = \frac{1}{1+x^2} \), we evaluate derivatives at \( x = 0 \) to build the polynomial terms.

polynomial approximation

Polynomial approximation is a fundamental concept in calculus and numerical analysis, where a complex function is represented by a polynomial. Polynomials are easier to handle and compute, and thus they make approximations practical for solving problems.The Maclaurin series, a special case of the Taylor series, provides a polynomial approximation of a function at \( x = 0 \). In the exercise with \( f(x) = \frac{1}{1+x^2} \), each degree of the Maclaurin polynomial increases the accuracy of the approximation.
Here are the polynomial forms:

• Order 1: \( P_1(x) = 1 \), which is a constant approximation.
• Order 2: \( P_2(x) = 1 - x^2 \), introduces curvature to match the function's behavior more closely.
• Order 3: Remains \( 1 - x^2 \) as the next term in the series vanishes due to zero derivative.
• Order 4: \( P_4(x) = 1 - x^2 + x^4 \), which refines the curve, providing a better match.

The higher the order, the better a polynomial can capture the function’s shape, especially near the center of expansion at \( x = 0 \). However, as higher-order terms are added, the calculations become more complex.

rational functions

Rational functions are mathematical expressions of the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. The Maclaurin series expansion often involves simplifying these functions into polynomial form for easier analysis.
In the exercise, the function \( f(x) = \frac{1}{1+x^2} \) is a classic example of a rational function. It behaves smoothly and is symmetric around \( x = 0 \), but its original form is not a polynomial. Hence, using the Maclaurin series to approximate it with polynomial terms allows us to examine how the function changes near the origin without the complexity of fraction expressions.

• Rational functions decrease more rapidly than polynomial functions as \( x \) moves away from the center, making local polynomial approximations very practical.
• Despite simplification, these polynomials have a limited range of accuracy, which improves with higher-order terms.

Overall, transforming a rational function into a polynomial approximation, as seen in this exercise, helps in analytical and numerical computations, especially for integrals or differential equations.