Problem 13 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=\ln (1+x)\). Find the \(n\) th degree Taylor polynomial generated by \(f\) about \(x=0 .\)

Step-by-Step Solution

Answer

The \(n^{th}\) degree Taylor polynomial for the function \(f(x) = \ln(1+x)\) generated about \(x=0\) is \(0\).

Step 1: Identify the Function

The function \(f(x)\) is given as \(\ln(1+x)\). The task is to find the \(n^{th}\) degree Taylor polynomial for this function around \(x=0\).

Step 2: Compute the Derivatives

In order to obtain the Taylor polynomial, it's necessary to calculate the first \(n\) derivatives of \(f(x)\) at \(x=0\). Using the recursive rule of \( \frac{{d^n}}{{dx^n}} \ln(1+x) = (-1)^{n+1}(n-1)!x^{-(n-1)} \) for \(n ≥ 1\) with \(\frac{{d^0}}{{dx^0}} \ln(1+x) = \ln(1+x)\), successive derivatives of \(f(x)\) at \(x=0\) can be obtained. Since \(f^n(0) = 0\) for all \(n ≥ 1\), this means that all terms in the Taylor sequence will be zero except for the \(n= 0\) term.

Step 3: Insert the Derivatives into the Taylor Polynomial Formula

It's now possible to substitute the calculated derivative values into the Taylor polynomial formula: \(P_n(x) = \sum_{k=0}^{n} \frac{{f^k(0)}}{{k!}} x^k = f(0) = \ln(1+0) = 0\). This implies that the Taylor polynomial of any degree \(n\) for \(f(x) = \ln(1+x)\) about \(x=0\) is the constant zero function.

Key Concepts

Derivatives in Taylor Polynomials

When we're working with Taylor polynomials, derivatives play a crucial role. A derivative tells us how a function behaves at a small scale near a specific point. To create a Taylor polynomial, we calculate the successive derivatives of the function at a point which, in this case, is around \( x=0 \). For the function \( f(x) = \ln(1+x) \), calculating derivatives helps us understand how the function changes as \( x \) changes.

• The first derivative of \( \ln(1+x) \) gives the function's slope at a point.
• The second derivative provides information about the concavity – whether the function curves upwards or downwards.
• Higher-order derivatives continue to provide deeper insights into the "bends" of the function.

For \( f(x) = \ln(1+x) \), interestingly, at \( x=0 \), the higher-order derivatives equal zero, meaning there are no changes beyond the initial value at this point. This results in the Taylor polynomial being extremely simple: a zero function when expanded about this point.

Understanding the Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental concept in mathematics, particularly useful for scaling problems and analyzing growth. Its growth rate itself is determined by its derivative, which is \( \frac{1}{1+x} \).

• \( \ln(1+x) \) describes how a quantity grows relative to another as \( x \) increases from zero.
• Its domain is restricted to \( x > -1 \) because the logarithm is undefined for non-positive numbers.
• When \( x = 0 \), \( \ln(1+x) = \ln(1) = 0 \).
• This simplicity at \( x=0 \) is one reason why Taylor polynomials at this point are straightforward – there's nothing of the function beyond its initial value: zero.

The natural logarithm is tied closely to e, the base of natural logarithms, which makes it an indispensable part of calculus and advanced mathematical functions.

Exploring the Maclaurin Series

The Maclaurin series is a specific type of Taylor series that is expanded around \( x=0 \). It provides a way to represent complex functions using infinite sums of powers of \( x \). Every polynomial term contains derivative information of a function at the expansion point.

• A Maclaurin series allows us to approximate functions using simpler polynomial expressions.
• For \( \ln(1+x) \), the Maclaurin series reveals that all terms with a degree \( n \geq 1 \) become zero at \( x=0 \). This is because all derivatives of the function at that point are zero, as described earlier.
• The simplicity of having \( f^n(0) = 0 \) for higher-order derivatives results in the complete series for \( \ln(1+x) \) boiling down to zero, as every term apart from the initial one adds no value.

Understanding this series helps simplify complex calculations and provides insight into the behavior of functions near specific points, making them an essential concept in calculus and mathematical analysis.