Problem 27 - Thomas Calculus

Question

Use power series operations to find the Taylor series at \(x=0\) for the functions in Exercises \(11-28 .\) $$\frac{x}{3} \ln \left(1+x^{2}\right)$$

Step-by-Step Solution

Answer

The Taylor series is \( \frac{x^3}{3} - \frac{x^5}{6} + \frac{x^7}{9} - \ldots \).

Step 1: Identify the inner function

First, we identify the function for which we need to find the Taylor series: \( \frac{x}{3} \ln(1 + x^2) \). The task is to express this function as a series around \( x = 0 \). We begin by considering the part of the function inside the logarithm: \( 1 + x^2 \).

Step 2: Recall the series expansion for ln(1+u)

The Taylor series expansion for \(\ln(1+u)\) is given by \( u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \ldots \). Here, \( u = x^2 \). Thus, substituting \( x^2 \) for \( u \), the expansion becomes: \( x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots \).

Step 3: Multiply the series by x/3

Our next step is to multiply the entire series obtained in Step 2 by \(\frac{x}{3}\). This yields: \[ \frac{x}{3}(x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots) = \frac{x^3}{3} - \frac{x^5}{6} + \frac{x^7}{9} - \ldots \].

Step 4: Write the final Taylor series

The final step is to collect the terms and write the full Taylor series for \( \frac{x}{3} \ln(1 + x^2) \): \[ \frac{x^3}{3} - \frac{x^5}{6} + \frac{x^7}{9} - \cdots \] This series captures the behavior of the original function around \( x = 0 \).

Key Concepts

Power Series

A power series is an infinite series of the form \( a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots \), where the coefficients \( a_n \) are constants. This series is a powerful tool for representing functions as an infinite sum of terms, allowing us to explore and analyze functions in a simpler way. In the context of Taylor series, a specific type of power series is used to approximate functions near a given point.

When we work with power series, each term's power increases incrementally, allowing us to express complex functions in a simple polynomial-like form. One significant advantage of power series is that, for values close enough to the point of expansion, these series converge to the function they represent. This makes them incredibly useful for both theoretical and practical purposes, such as solving differential equations or evaluating difficult integrals.

• The power series is centered at a specific point, often written as \( (x - c) \) for a series centered at \( x = c \).
• Taylor series, a form of power series, provides a way to represent functions like logarithmic functions at a point, like \( x = 0 \), by expanding into an infinite sum.
• The convergence of the series depends on the interval around the point, where the series approximates the function accurately.

Logarithmic Functions

Logarithmic functions, such as \( \ln(x) \), play an important role in mathematics, with applications across many disciplines. In our exercise, we examine an expression involving the logarithm of \( 1 + x^2 \). The challenge here lies in expressing this function as a series to approximate its behavior around \( x = 0 \).

The Taylor series expansion for \( \ln(1+u) \) helps us achieve this. By substituting \( u = x^2 \) into this series, we convert the expression within the logarithm into a manageable form. This series, \( x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots \), represents \( \ln(1 + x^2) \) as an infinite sum, making it easier to combine with other terms, such as multiplying by \( \frac{x}{3} \).

• Logarithmic expansions can simplify complex functions by breaking them down into more straightforward terms.
• This behavior becomes particularly useful in calculus and analysis, allowing us to work with otherwise difficult-to-manage expressions.
• The logarithmic series highlights how an infinite series can approximate a function over a specific interval.

Multiplying Series

Multiplying series is a common technique used to manipulate series representations of functions. After finding the series expansion for the logarithmic part, the next step is to multiply it by another expression, such as \( \frac{x}{3} \). This operation essentially multiplies each term of the series by the given expression, altering the powers and coefficients accordingly.

In our example, multiplying \( \frac{x}{3} \) by the series \( x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots \), results in a new series: \( \frac{x^3}{3} - \frac{x^5}{6} + \frac{x^7}{9} - \ldots \). Each term's power starts as \( x^3 \), \( x^5 \), and so on, depending on the original series' powers.

• Multiplication adjusts each coefficient in the series, affecting the series convergence and accuracy.
• By doing this step, we align the series representation closely with the behavior of the original function.
• This technique enables us to transform and manipulate series to achieve the desired form for further analysis or computation.