Problem 35 - 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

The first three nonzero terms are \( x^2, -\frac{x^3}{2}, -\frac{x^4}{6} \) and the series converges for \( |x| < 1 \).

Step 1: Recall the Maclaurin series

The Maclaurin series for a function \( f(x) \) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \dots \) The series is a special case of the Taylor series evaluated at \( a=0 \).

Step 2: Find the Maclaurin series for \( \sin(x) \) and \( \ln(1+x) \)

The Maclaurin series for \( \sin x \) is \( x - \frac{x^3}{3!} + \frac{x^5}{5!} + \dots \). The Maclaurin series for \( \ln(1+x) \) is \( x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \).

Step 3: Determine product of series terms up to required order

To find the Maclaurin series of \( f(x) = \sin x \ln(1+x) \), multiply the first few terms of both series:\[\left(x - \frac{x^3}{6} + \frac{x^5}{120}\right) \times \left(x - \frac{x^2}{2} + \frac{x^3}{3}\right)\]

Step 4: Distribute and collect terms

Calculate the product term by term:- First nonzero term: \( x \cdot x = x^2 \)- Next term: \( x \cdot -\frac{x^2}{2} = -\frac{x^3}{2} \)- Next terms: \( -\frac{x^3}{6} \cdot x = -\frac{x^4}{6} \) and combine with other terms of same order to continue with next powers.Focus on collecting terms up to the cubic term.

Step 5: Identify the first three nonzero terms

By distributing and collecting terms, the first three nonzero terms are identified as:- \( x^2 \)- \( -\frac{x^3}{2} \)- \( -\frac{x^4}{6} \)Adding these gives the first three nonzero terms of the series.

Step 6: Determine the interval of absolute convergence

The series \( \sin(x) \) converges for all \( x \) and \( \ln(1+x) \) converges for \( |x| < 1 \). Hence the product \( f(x) \) converges absolutely for \( |x| < 1 \).

Key Concepts

Convergence of Series

A series is a sum of terms that follows a specific rule or pattern. Converging series are the ones where the sum approaches a finite value as you add more terms. This behavior is vital in mathematics because it tells us whether a series gives us a meaningful result when extended infinitely. In the context of series convergence, it is important to focus on absolute convergence. A series is said to converge absolutely if the series formed by taking the absolute value of each term also converges.
For instance, in the given exercise, the function involves the product of two series: the Maclaurin series expansions of \( \sin(x) \) and \( \ln(1+x) \). While the \( \sin(x) \) series converges for all \( x \), the \( \ln(1+x) \) series only converges for \( |x| < 1 \).
By multiplying these series, it was determined that the product also converges absolutely within the interval \( |x| < 1 \). This insight is crucial in understanding which values of \( x \), the series sum is reliable and meaningful.

Taylor Series

The Taylor series is a way to represent functions as infinite sums of terms derived from their derivatives at a single point. The Maclaurin series is a special case of the Taylor series created when the expansion point is zero. This makes Maclaurin series especially useful for approximating functions near the origin.
Mathematically, a function \( f(x) \) can be expressed using its Taylor series as:

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

When \( a=0 \), this becomes the Maclaurin series:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \dots \)

In our exercise, the product of the Maclaurin series for \( \sin(x) \) and \( \ln(1+x) \) helps us find the series expansion for a more complex function, \( f(x) = \sin(x) \ln(1+x) \).
The function is represented around \( x=0 \), allowing for efficient approximation and calculation of the function values nearby zero.

Series Expansion

Series expansion is about breaking down complex functions into simpler components, expressed as a sum of terms. This process makes analysis and computation more manageable. The series expansion transforms functions into infinite polynomials.
For example, the Maclaurin series expansion for \( \sin(x) \) is:

• \( x - \frac{x^3}{6} + \frac{x^5}{120} + \dots \)

And for \( \ln(1+x) \), it is:

• \( x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \)

To find the Maclaurin series of \( f(x) = \sin(x) \ln(1+x) \), we multiply and collect terms between these two expansions. The resulting terms describe \( f(x) \) as a series expansion.
This multiplication and combination, outlined in the steps of the exercise, yield the first three non-zero terms of the resulting Maclaurin series: \( x^2 \), \( -\frac{x^3}{2} \), and \( -\frac{x^4}{6} \). These steps help compactly and efficiently express \( f(x) \), providing valuable approximations around zero.