Chapter 30: Problem 14
Find the first three nonzero terms of the Taylor expansion for the given function and given value of a. $$\ln x \quad(a=3)$$
Short Answer
Expert verified
First three nonzero terms: \( \ln 3 + \frac{x-3}{3} - \frac{(x-3)^2}{18} \).
Step by step solution
01
Understand the Taylor Series Formula
The Taylor series for a function \( f(x) \) centered at \( a \) is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] We need to find the derivatives of \( \ln x \) and evaluate them at \( x = 3 \).
02
Calculate the First Derivative
Determine the first derivative of \( \ln x \), which is \( f'(x) = \frac{1}{x} \). Evaluate at \( x = 3 \): \[ f'(3) = \frac{1}{3} \]
03
Calculate the Second Derivative
The second derivative is \( f''(x) = -\frac{1}{x^2} \). Evaluate at \( x = 3 \): \[ f''(3) = -\frac{1}{9} \]
04
Calculate the Third Derivative
The third derivative is \( f'''(x) = \frac{2}{x^3} \). Evaluate at \( x = 3 \): \[ f'''(3) = \frac{2}{27} \]
05
Evaluate the Function at \( a \)
Evaluate \( \ln x \) at \( x = 3 \): \[ f(3) = \ln 3 \]
06
Construct the Taylor Series
Substitute the derivatives and \( f(3) \) into the Taylor series formula: \[ \ln x = \ln 3 + \frac{x-3}{3} - \frac{(x-3)^2}{18} + \frac{(x-3)^3}{81} + \cdots \] The first three nonzero terms are \( \ln 3 + \frac{x-3}{3} - \frac{(x-3)^2}{18} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
In calculus, derivatives are a fundamental concept that describe how a function changes as its input changes. They provide us with the slope of the function at any given point.
Especially in the context of a Taylor series, derivatives are crucial as they determine the terms in the series expansion.
For the function \( \ln x \), we begin by finding its derivatives.
This step-by-step calculation makes sure we include all necessary details to approximate the function accurately.
Especially in the context of a Taylor series, derivatives are crucial as they determine the terms in the series expansion.
For the function \( \ln x \), we begin by finding its derivatives.
- First Derivative: \( f'(x) = \frac{1}{x} \). This shows how the natural logarithm changes with respect to \( x \).
- Second Derivative: \( f''(x) = -\frac{1}{x^2} \). This illustrates the concavity of the natural logarithm function.
- Third Derivative: \( f'''(x) = \frac{2}{x^3} \). As we move to higher derivatives, we are capturing more precise aspects of the function's behavior.
This step-by-step calculation makes sure we include all necessary details to approximate the function accurately.
Natural Logarithm Function
The natural logarithm function, denoted as \( \ln x \), is a logarithmic function with the base \( e \) (approximately 2.718) and is heavily used in mathematics.
This function is defined only for positive numbers and provides a unique and continuous transformation of numbers.
The slope or rate of change of the natural logarithm function is described by its first derivative, \( \frac{1}{x} \), indicating that as \( x \) increases, the rate of change of \( \ln x \) decreases, approaching zero.
Its calculus properties, like the derivatives we calculated, help in expanding the function into a Taylor series.
This function is defined only for positive numbers and provides a unique and continuous transformation of numbers.
The slope or rate of change of the natural logarithm function is described by its first derivative, \( \frac{1}{x} \), indicating that as \( x \) increases, the rate of change of \( \ln x \) decreases, approaching zero.
Its calculus properties, like the derivatives we calculated, help in expanding the function into a Taylor series.
- This series allows us to approximate the function around a particular point \( a \), making it easier to compute \( \ln x \) for values near \( a \).
- This method is particularly useful because the exact computation of the logarithm function might not be straightforward for all values.
Evaluating Series at a Point
A Taylor series allows us to express a complicated function as an infinite sum of terms calculated from its derivatives at a particular point.
This process is known as evaluating the series at a point, and it makes understanding and computing functions much simpler.
For \( \ln x \) around \( a = 3 \), we start by evaluating the function and its derivatives at this point.
The Taylor series essentially builds a polynomial expression that mimics \( \ln x \) near \( x=3 \), offering a much simpler calculation than direct evaluation.
This process is known as evaluating the series at a point, and it makes understanding and computing functions much simpler.
For \( \ln x \) around \( a = 3 \), we start by evaluating the function and its derivatives at this point.
- Original Function Value: \( f(3) = \ln 3 \). This becomes the starting point or the constant term in the series.
- First Derivative \( f'(3) = \frac{1}{3} \), multiplied by \( (x - 3) \) to estimate how \( \ln x \) changes near \( x = 3 \).
- Higher derivatives \( f''(3) = -\frac{1}{9} \) and \( f'''(3) = \frac{2}{27} \) adjust the series with more terms like \( (x-3)^2 \) and \( (x-3)^3 \).
The Taylor series essentially builds a polynomial expression that mimics \( \ln x \) near \( x=3 \), offering a much simpler calculation than direct evaluation.