Chapter 30: Problem 13
Let \(f(x)=\ln (1+x)\). Find the \(n\) th degree Taylor polynomial generated by \(f\) about \(x=0 .\)
Short Answer
Expert verified
The \(n^{th}\) degree Taylor polynomial for the function \(f(x) = \ln(1+x)\) generated about \(x=0\) is \(0\).
Step by step solution
01
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 \(x=0\).
02
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.
03
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.