Problem 16 - Essential Calculus

Question

In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=\frac{x}{x+1}, \quad n=4 $$

Step-by-Step Solution

Answer

The Maclaurin polynomial of degree 4 for the function \( f(x) = \frac{x}{x+1} \) is \(f(x) = x - x^2 + x^3 - x^4 \).

Step 1: Understand the Maclaurin Series

The Maclaurin series of a function can be written as: \(f(x) = f(0) + f'(0)x + f''(0)\left( \frac{x^2}{2!} \right) + f'''(0)\left(\frac{x^3}{3!}\right) + f'''(0)\left(\frac{x^4}{4!}\right) + ...\). Given that we want to find the Maclaurin series to the 4th degree, the rest of the terms will be zero.

Step 2: Finding function's derivatives

Finding the first four derivatives of \(f(x) = \frac{x}{x + 1}\) at \(x = 0\): First derivative: \(f'(x) = \frac{1}{(x+1)^2}\), so \( f'(0) = 1 \). Second derivative: \(f''(x) = \frac{-2}{(x+1)^3}\), so \( f''(0) = -2 \). Third derivative: \(f'''(x) = \frac{6}{(x+1)^4}\), so \( f'''(0) = 6 \). Fourth derivative: \(f^{(4)}(x) = \frac{-24}{(x+1)^5}\), so \( f^{(4)}(0) = -24 \).

Step 3: Substituting into Maclaurin Series Formula

Now, plug these values back into the Maclaurin Series formula. So, currently \(f(x) = f(0) + f'(0)x + f''(0)\left(\frac{x^2}{2}\right) + f'''(0) \left(\frac{x^3}{6}\right) + f^{(4)}(0) \left(\frac{x^4}{24}\right)\).Substituting our values yields: \(f(x) = 0 + 1 \cdot x + (-2) \cdot \left( \frac{x^2}{2} \right) + 6 \cdot \left(\frac{x^3}{6}\right) - 24 \cdot \left(\frac{x^4}{24}\right)\)

Key Concepts

Maclaurin Polynomial

Let's explore the Maclaurin polynomial, a specific type of Taylor polynomial for approximating functions. The Maclaurin polynomial is centered at zero, which means it's a Taylor polynomial where the point of expansion is at the origin, or simply put, when the function is expanded around the point x=0. It is a powerful tool in calculus that helps to approximate more complex functions using a polynomial.

To understand the Maclaurin polynomial, one must look at the expansion formula:\[\[\begin{align*}P(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots + \frac{f^{(n)}(0)x^n}{n!}\end{align*}\]\]Here, the superscript denotes the order of the derivative, and n! represents the factorial of n. The formula includes terms of up to the n-th derivative of the function evaluated at 0. Each term in the series is a product of a derivative evaluated at zero and a power of x, divided by the factorial of the exponent.

In our exercise, the Maclaurin polynomial of degree 4 for the function \(f(x)=\frac{x}{x+1}\) was calculated using derivatives. The polynomial provides an estimate for f(x) near 0 and is expressed by the first five terms of the series since the rest of the terms after the fourth derivative will equate to zero for our degree of n=4.

Calculus Derivatives

Derivatives are a cornerstone of calculus. They measure how a function changes as its input changes, representing the rate of change or slope of the function at any given point. A derivative can be thought of as an instantaneous rate of change, providing a snapshot of movement at a precise moment in time.

In the context of the Maclaurin series, derivatives evaluated at zero are crucial. To find the polynomial, we differentiate the function multiple times and evaluate these derivatives at zero. This exercise required calculating the first four derivatives of \(f(x) = \frac{x}{x + 1}\). The process involved applying rules of differentiation such as the quotient rule and chain rule.

Importance of Derivatives in Maclaurin Series

For a Maclaurin polynomial, each coefficient is determined by derivatives. These are the values that give the polynomial its shape and allow it to closely mimic the behavior of the original function around x=0. Getting these values correct is essential because each derivative will influence the degree of accuracy with which the polynomial approximates the function.

Power Series Expansion

A power series is an infinite series of the form \(\sum_{n=0}^\infty a_nx^n\), where each term increases in power as n increases. A power series expansion is a way to express a function as an infinite sum of terms, each a constant multiplied by a power of the variable. This expansion is invaluable for working with functions that are otherwise difficult to manipulate algebraically.

The Maclaurin series is a special case of the power series expansion when the function is expanded around the point x=0. When working with a power series, it's essential to understand its interval of convergence, meaning the set of x values for which the series converges to a finite number.

Connecting Maclaurin Series to Power Series

In our exercise, the power series expansion is truncated to get the Maclaurin polynomial of degree 4. This polynomial is a finite sum representing the first part of the function's power series and can be used to approximate \(f(x)\) for values of x near zero. Knowing that the power series is actually an infinite expansion can also provide insights into the behavior of functions over a wider range of x values and is fundamental when studying advanced calculus.