Problem 37 - Thomas Calculus

Question

Find the first three nonzero terms of the Maclaurin series for each function and the values of \(x\) for which the series converges absolutely. \(f(x)=(\sin x) \ln (1+x)\)

Step-by-Step Solution

Answer

First three terms: \(x^2, -\frac{x^3}{2}, -\frac{x^4}{6}\). Converges for \(-1 < x \leq 1\).

Step 1: Recall the Maclaurin Series of Component Functions

The Maclaurin series is the Taylor series expansion of a function about 0. For \( \sin x \), the series is \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \). For \( \ln (1+x) \), the series is \( \ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \).

Step 2: Expand the Maclaurin Series for the Product of Two Functions

To find the Maclaurin series for \( f(x) = (\sin x)(\ln (1+x)) \), we multiply the series for \( \sin x \) and \( \ln (1+x) \) and combine terms up to the necessary degree for the first three nonzero terms. Start by multiplying the first few terms: \((x - \frac{x^3}{3!})(x - \frac{x^2}{2})\).

Step 3: Compute and Combine Terms

Multiply and combine the terms: 1. \(x \cdot x = x^2\)2. \(x \cdot (-\frac{x^2}{2}) = -\frac{x^3}{2} \)3. \(-\frac{x^3}{6} \cdot x = -\frac{x^4}{6} \)4. \(-\frac{x^3}{6} \cdot (-\frac{x^2}{2}) = \frac{x^5}{12} \)So, the combination of the first few terms is \(x^2 - \frac{x^3}{2} - \frac{x^4}{6} + \cdots \).

Step 4: Identify the First Three Nonzero Terms

From the computations, the first three nonzero terms of the series are \(x^2, -\frac{x^3}{2}\), and \(-\frac{x^4}{6}\).

Step 5: Determine the Interval of Convergence

Each component series \(\sin x\) and \(\ln(1+x)\) has specific intervals of convergence. The series \( \ln(1+x) \) converges for \(-1 < x \leq 1\) and \(\sin x\) converges for all \(x\). Thus, \(f(x)\) converges absolutely where both conditions are true: \(-1 < x \leq 1\).

Key Concepts

Taylor series

The Taylor series is a fundamental concept in calculus that allows us to represent functions as infinite sums of terms calculated based on the derivatives of the function at a certain point. When this point is zero, the series is referred to as a Maclaurin series, which is a special case of the Taylor series.
For any infinitely differentiable function, the Taylor series expanded about a point a can be written as:
\[T(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]This provides a powerful tool for approximating functions with polynomials, which are often easier to handle than the functions themselves.
The usefulness of Taylor series comes in its ability to approximate complex functions with simpler polynomial expressions, particularly significant in fields such as physics, engineering, and computer science.

interval of convergence

The interval of convergence for a power series is a range of values for which the series converges to a finite value. Understanding this concept ensures that we use the series safely without venturing into ranges where the series might diverge.
For the Maclaurin series of a function, which is a special type of power series expansion, the interval of convergence is critical in confirming the valid operational range of the series representation.

• The series \(\ln(1+x)\) converges for the interval \(-1 < x \leq 1\).
• The series \(\sin x\) converges for all real numbers \(x\).

For composite functions like \((\sin x)(\ln(1+x))\), the interval of convergence is limited by the most restrictive convergence interval among the component functions. In this case, the convergence of \(\sin x\) is everywhere, but because \(\ln(1+x)\) is limited from \(-1\) to \(1\), \(f(x)\) converges absolutely within the same range, specifically \(-1 < x \leq 1\).

power series expansion

A power series is a series in the form \( \sum_{n=0}^\infty a_n(x - c)^n \), where \(a_n\) are coefficients, \(x\) is the variable, and \(c\) is the center of the series (0 for a Maclaurin series). The power series expansion is a method of expressing a function as an infinite sum of terms that are powers of \(x\).
Power series expansions, like Taylor and Maclaurin series, offer a way to represent complex functions with polynomial terms, allowing for easier computation and analysis. The main advantage of using these expansions is simplification, as polynomials are usually simpler to manipulate than their original function forms.
To construct a power series expansion, we often rely on known expansions like those of basic functions \(\sin x\) and \(\ln(1+x)\), and manipulate them through algebraic operations to obtain expansions for more complex functions.
The beauty of these series lies in their precision and efficiency in approximation, particularly crucial in numerical computation.

absolute convergence

Absolute convergence is a stronger form of convergence for series, particularly power series. A series is said to converge absolutely if the series formed by taking the absolute value of each term also converges.
This concept is critical because when a series converges absolutely, it guarantees convergence irrespective of the order of the terms. This differs from conditional convergence, where rearranging terms can result in different sums or divergence.
In the context of our Maclaurin series for \(f(x) = (\sin x)(\ln(1+x))\), we determine absolute convergence by analyzing the intervals where each component series converges absolutely.

• The function \(\ln(1+x)\) converges absolutely over \(-1 < x \leq 1\).
• Since \(\sin x\) converges for all of \(x\), the absolute interval remains constrained by \(\ln(1+x)\).

Thus, the series \(f(x)\) converges absolutely within the interval \(-1 < x \leq 1\), merging the conditions of its components to function correctly across the identified range.