Problem 22 - Thomas Calculus

Question

Use power series operations to find the Taylor series at \(x=0\) for the functions in Exercises \(13-30\) . $$x \ln (1+2 x)$$

Step-by-Step Solution

Answer

The Taylor series for \(x \ln(1+2x)\) is \(2x^2 - 2x^3 + \frac{8x^4}{3} - 2x^5 + \cdots\).

Step 1: Recall the Taylor series for ln(1 + u)

The Taylor series expansion of \(\ln(1+u)\) at \(u=0\) is \(u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots\). This will be useful to find the series for \(\ln(1+2x)\).

Step 2: Substitute to find ln(1 + 2x)

Set \(u = 2x\) in the series for \(\ln(1+u)\). Then, the series for \(\ln(1+2x)\) is \(2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \frac{(2x)^4}{4} + \cdots\).

Step 3: Simplify the expression

Simplify each term: \(2x - 2x^2 + \frac{8x^3}{3} - 2x^4 + \cdots\). Notice that the coefficients are derived by calculating \(2^n/n\) for each respective term of power \(x^n\).

Step 4: Multiply by x

Multiply each term of the series \(\ln(1+2x)\) by \(x\) to find \(x\ln(1+2x)\). This yields \(2x^2 - 2x^3 + \frac{8x^4}{3} - 2x^5 + \cdots\).

Step 5: Write the Taylor series

The Taylor series for \(x\ln(1+2x)\) at \(x=0\) is the sum of all multiplied terms: \(x\ln(1+2x) = \sum_{n=1}^\infty \frac{(-1)^{n+1} \cdot 2^n}{n} x^{n+1} \).

Key Concepts

Power Series Operations

The concept of power series operations is fundamental in calculus and involves manipulating infinite sums to approximate functions. A power series is a series of the form:

• \( a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots \)

where the coefficients \(a_n\) depend on the function you're attempting to expand.
Using power series operations, you can perform addition, subtraction, differentiation, and integration on power series term by term. This characteristic makes them versatile for many mathematical applications, including estimating functions, solving differential equations, and evaluating integrals.
In this exercise, power series operations are used to expand \( x \ln(1 + 2x) \) into a form that is easier to analyze. We start by recognizing patterns in simpler series like \( \ln(1+u) \) and then substitute to find the desired expansion. By doing so, we transform the function into a series to solve or approximate its value across a domain.

Logarithm Functions

Logarithm functions are essential in mathematics for processing exponential relationships and transforming them into linear ones. The natural logarithm, represented by \( \ln \), is a logarithm with base \(e\) (approximately 2.71828), a number that frequently appears in growth-related contexts.
For example, the Taylor series for \( \ln(1+u) \) is:

• \( u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots \)

This formula comes from integrating the geometric series and is valid for \(-1 < u \le 1\).
When dealing with \( \ln(1+2x) \), substitute \( u = 2x \) into the known series for \( \ln(1+u) \). This substitution is illustrated in the solution steps provided and allows us to express a non-linear logarithmic function as an infinite polynomial, making it easier to compute and understand.

Series Expansion

Series expansion is a powerful mathematical tool used to express complex functions as infinite sums. This approach is particularly useful when direct calculation is difficult.
The goal of a series expansion is to approximate a function to a certain degree of accuracy by using polynomial terms. In the Taylor series, specifically, a function is represented as a sum of its derivatives at a particular point.

• The general form is given by: \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \)

In our exercise, the series expansion for \( x\ln(1+2x) \) is developed by first expanding \( \ln(1+2x) \) through substitution and then multiplying each term by \( x \). This results in a series that converges to the desired function around \( x=0 \).
This series expansion technique allows us to approximate \( x\ln(1+2x) \) in a simpler polynomial form, facilitating computation without directly using the logarithmic form.