Problem 26 - Calculus: Early Transcendentals

Question

How many terms of the Maclaurin series for \( \ln(1 + x) \) do you need to use to estimate \( \ln \) 1.4 to within 0.001?

Step-by-Step Solution

Answer

Use 5 terms to achieve the desired precision.

Step 1: Identify the Maclaurin series for \( \ln(1+x) \)

The Maclaurin series for the function \( \ln(1+x) \) is given by the infinite series:\[\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]To approximate \( \ln(1.4) \), we substitute \( x = 0.4 \) into this series.

Step 2: Determine the remainder term from alternating series

The remainder of an alternating series can be estimated by the absolute value of the first omitted term. In this case, we need to find the smallest \( n \) such that the remainder \(|(-1)^n \cdot \frac{0.4^{n+1}}{n+1}| < 0.001\).

Key Concepts

ln(1 + x)

In calculus, one important expansion is the Maclaurin series for the natural logarithm function. This particular series represents the function \( \ln(1+x) \) and is given by the expression:\[\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]This series is an infinite series, meaning it continues forever. However, for practical purposes, we cannot write an infinite number of terms. Instead, we utilize a finite number of terms to approximate the function. These terms include alternating signs and fractions, with each fraction's numerator being a power of \(x\) and the denominator being a natural number. When plugging in a value for \(x\), like \(x = 0.4\) in our example, we can estimate \(\ln(1.4)\). This series only converges for \(-1 < x \leq 1\). This means if \(x\) is within this range, the series will approach a specific number, helping us estimate the log value.

alternating series

An alternating series is a series where the terms change sign alternately. A typical form is given by:\[ a_1 - a_2 + a_3 - a_4 + \cdots\]The Maclaurin series for \( \ln(1+x) \) falls into this category. In each term, the sign alternates between positive and negative. This feature can significantly affect how quickly a series converges. Alternating series are essential because they can be more efficiently summed than non-alternating ones. Typically, an alternating series converges if the absolute values of the terms decrease monotonically—meaning each term is smaller than the term before it. This property is useful in approximating functions, like \( \ln(1.4) \), because it allows us to assess how many terms we need for an acceptable approximation.

remainder estimation

When using a finite number of terms from an infinite series, we must estimate the error introduced by not using the whole series. This error is referred to as the remainder term. For an alternating series, the remainder is estimated by the absolute value of the first omitted term. This is particularly handy with the Maclaurin series for \( \ln(1+x) \). If our task is to use this series to estimate \( \ln(1.4) \) within an error of 0.001, we must find the smallest number of terms where the remainder—\(| (-1)^n \cdot \frac{0.4^{n+1}}{n+1} |\)—is less than 0.001. Calculating this tells us the minimum number of terms necessary to keep our error within the desired limit.

convergence of series

The convergence of a series refers to whether the series approaches a specific value as more terms are added. For the Maclaurin series, we want to know if the series for \( \ln(1+x) \) converges to the true value of the function when substituting \(x = 0.4\). To establish convergence:

• The series must be geometrically shrinking, with terms getting progressively smaller.
• The series must fit within the interval of convergence, \(-1 < x \leq 1\).
• In the case of an alternating series, convergence is ensured by decreasing term size and alternating signs.

Convergence assures us that as we add more terms, our approximation of \(\ln(1.4)\) will get closer and closer to its true value. Understanding convergence is crucial when deciding how many terms are required to reach an adequate estimation within an error margin like 0.001.