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Magazine 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
\[ \ln 3 + \frac{1}{3}(x-3) - \frac{1}{18}(x-3)^2 \]
Step by step solution
01
Understand the Definition
The Taylor series expansion of a function \( f(x) \) about a point \( a \) is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \] We need to find this series for \( \ln x \) centered at \( a = 3 \).
02
Find the First Derivative
The first derivative of \( \ln x \) is \( f'(x) = \frac{1}{x} \). Evaluate the first derivative at \( a = 3 \): \( f'(3) = \frac{1}{3} \).
03
Find the Second Derivative
The second derivative is \( f''(x) = -\frac{1}{x^2} \). Evaluate this at \( a = 3 \): \( f''(3) = -\frac{1}{9} \).
04
Find the Third Derivative
The third derivative is \( f'''(x) = \frac{2}{x^3} \). Evaluate this at \( a = 3 \): \( f'''(3) = \frac{2}{27} \).
05
Construct the First Three Terms
Using the formula for the Taylor series, plug in the derivatives: - The first term is \( \ln 3 \). - The second term, using the first derivative, is \( \frac{1}{3}(x-3) \). - The third term, using the second derivative, is \( -\frac{1}{18}(x-3)^2 \) (since \( \frac{-1}{9} \cdot \frac{1}{2!} = -\frac{1}{18} \)).
06
Combine the Terms
Combine these to get the first three nonzero terms of the series: \[ \ln 3 + \frac{1}{3}(x-3) - \frac{1}{18}(x-3)^2 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative Calculation
Calculating derivatives is a core part of creating a Taylor series expansion. A derivative measures how a function changes as its input changes. When we find a Taylor series, we start with the derivatives of the function at a specific point. In our case, this is the natural logarithm function, ln(x), centered at 3.
To find the first few terms of a Taylor series, you need the first few derivatives:
To find the first few terms of a Taylor series, you need the first few derivatives:
- The first derivative of ln(x) is \( f'(x) = \frac{1}{x} \). Evaluating at \( a = 3\), we get \( f'(3) = \frac{1}{3} \).
- The second derivative is \( f''(x) = -\frac{1}{x^2} \). At \( x = 3 \), this becomes \( f''(3) = -\frac{1}{9} \).
- The third derivative is \( f'''(x) = \frac{2}{x^3} \). Evaluating at the point gives \( f'''(3) = \frac{2}{27} \).
Natural Logarithm Function
The natural logarithm function, denoted as \( \ln(x) \), is a fundamental concept in calculus and logarithmic calculations. The natural logarithm is the logarithm to the base \( e \), where \( e \) is approximately 2.71828. It arises quite naturally in many areas of mathematics and physics.
A special property of the natural logarithm is that it is the inverse operation to exponentiation with \( e \). This means that if we have \( x = e^y\), then \( \ln(x) = y \).
When constructing a Taylor series expansion involving \( \ln(x) \), the derivatives play a crucial role:
A special property of the natural logarithm is that it is the inverse operation to exponentiation with \( e \). This means that if we have \( x = e^y\), then \( \ln(x) = y \).
When constructing a Taylor series expansion involving \( \ln(x) \), the derivatives play a crucial role:
- We begin by understanding that \( \ln(x) \) is undefined for \( x \leq 0 \), limiting our series to positive values of \( x \).
- The Taylor expansion allows us to express \( \ln(x) \) around \( a=3 \), giving a polynomial that approximates \( \ln(x) \) near \( x=3 \).
Mathematical Series
A mathematical series is a sum of terms that follow a specific rule or pattern, often derived from a function. The Taylor series is a particularly important type of series in calculus, used to approximate functions. It represents a function as the sum of its derivatives at a single point, each multiplied by powers of the variable.
For the Taylor series of \( \ln(x) \) around \( a = 3 \):
For the Taylor series of \( \ln(x) \) around \( a = 3 \):
- The general formula for a Taylor series is \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \)
- Each term incorporates higher derivatives and higher powers of \((x-a)\), converging to the original function.
