Problem 57 - Essential Calculus

Question

Comparing Maclaurin Polynomials (a) Compare the Maclaurin polynomials of degree 4 and degree \(5,\) respectively, for the functions \(f(x)=e^{x}\) and \(g(x)=x e^{x} .\) What is the relationship between them? (b) Use the result in part (a) and the Maclaurin polynomial of degree 5 for \(f(x)=\sin x\) to find a Maclaurin polynomial of degree 6 for the function \(g(x)=x \sin x\). (c) Use the result in part (a) and the Maclaurin polynomial of degree 5 for \(f(x)=\sin x\) to find a Maclaurin polynomial of degree 4 for the function \(g(x)=(\sin x) / x\).

Step-by-Step Solution

Answer

The relationship between the Maclaurin polynomials of \(e^x\) and \(xe^x\) is that each term in the series for \(g(x),\) is the derivative of the corresponding term in \(f(x),\) multiplied by \(x.\) Based on this, the Maclaurin series for \(x \sin x\) to the 6th degree is \(x^2 - x^4/4! + x^6/6!\), and the series for \((\sin x) / x\) to the 4th degree is \(1 - x^2/3! + x^4/5!\).

Step 1: Find the Maclaurin polynomials

Start with expressing the given functions as their Maclaurin polynomials. Recall the series expansions for \(e^x\) and \(x e^x\). For \(f(x)=e^x\), the Maclaurin series up to the 5th degree is given by: `\(f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\)`For \(g(x) = x e^x\), the Maclaurin series expansion can be found by taking the derivative of the expansion for \(e^x\) and then multiplying by \(x\). So we get: `\(g(x) = x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + \frac{x^5}{4!} + \frac{x^6}{5!}\)`

Step 2: Identify the relationship

We can see from the formulas in Step 1 that the relation between the two series is that the Maclaurin series for \(g(x)\) is simply the derivative of the series for \(f(x)\), with each term multiplied by \(x\).

Step 3: Apply the relationship to find new Maclaurin series

Given the expansion for \(f(x) = \sin(x)\), which is `\(f(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\)` apply the derived relationship to find the series for `\(g(x) = x\sin(x)\), which gives us `\(g(x) = x^2 - \frac{x^4}{4!} + \frac{x^6}{6!}\)`.

Step 4: Find the Maclaurin series for \((\sin x) / x\)

Likewise for \(g(x) = (\sin x) / x\), note that \(xg(x)\) is simply \(f(x) = \sin(x)\). So, \(g(x)\) can be gotten from \(f(x)\) by dividing every term by \(x\), giving `\(g(x) = 1 - \frac{x^2}{3!} + \frac{x^4}{5!}.\)`

Key Concepts

Taylor Series

The Taylor series is a powerful mathematical tool for approximating functions that are otherwise hard to calculate directly. A Taylor series represents a function as an infinite sum of terms that are calculated from the values of the function’s derivatives at a single point. Typically, a Taylor series is centered around some value, often zero, which turns it into a Maclaurin series. The general concept hinges on approximating a function with polynomials, where each subsequent term offers a more precise estimate.

The key formula for the Taylor series of a function is:

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

Here, \(a\) is the point around which the function is approximated. Understanding how to derive these terms involves computing derivatives, which can be complex but is crucial for accurate approximations.

The ability to represent a function accurately near a specific point and then smoothly extend the approximation over a larger interval is where the Taylor series shines. It's widely used in engineering, physics, and computer science for function estimation and solving differential equations, enabling calculations that would otherwise be intractable.

Polynomial Approximation

Polynomial approximation is a method that uses polynomials to estimate more complicated functions within a certain interval. It's all about finding a polynomial that closes in on the original function's behavior as closely as possible. Such approximations can simplify problems, making them easier to solve both theoretically and computationally.

The essence of polynomial approximation lies in using a finite number of terms from a function's Taylor series. By selecting enough terms, one can achieve a desired level of accuracy for the approximation.

A Maclaurin polynomial is a specific type of polynomial approximation where the function is approximated around the point zero. For example, the exponential function \(e^x\) can be approximated by a polynomial:

• \[ P(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]

This polynomial grows more accurate with each added term, though often only the first few terms are needed for practical purposes. Such approximations are invaluable in numerical methods, where simple calculations can offer insights and solutions into complex problems.

Series Expansions

Series expansions are a fundamental concept for dissecting and approximating functions. In essence, a series expansion expresses a function as a sum of its components. Each component usually represents a derivative or a related operation evaluated at a particular point. The most common type of series expansion is the Taylor series expansion, which provides a polynomial representation of functions.

Series expansions offer a way to break complex functions into manageable parts. For instance, the Maclaurin series, a type of series expansion, allows us to expand functions like \(e^x\), \(\sin(x)\), and \(\cos(x)\) readily around zero. This involves calculating derivatives, which evolve the series sequentially by degree.

The beauty of series expansions is their flexibility and utility in analysis, especially in solving differential equations and performing integrations that are otherwise difficult. It's crucial, however, to determine the convergence of the series to ensure it gives a reliable approximation within a desired range. A well-understood series expansion can transform complex problems into solvable puzzles.