Chapter 9: Problem 17
Use sigma notation to write the Taylor series about \(x=x_{0}\) for the function. $$\ln x ; x_{0}=1$$
Short Answer
Expert verified
The Taylor series is \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n\).
Step by step solution
01
Understand the Taylor Series Formula
The general Taylor series expansion for a function \(f(x)\) around \(x_0\) is given by: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \] Here, \(f^{(n)}(x_0)\) is the \(n\)-th derivative of the function \(f(x)\) evaluated at \(x_0\). In our exercise, we have \(f(x) = \ln x\) and the point \(x_0 = 1\).
02
Calculate Derivatives of the Function
Find the derivatives \(f^{(n)}(x)\) for \(f(x) = \ln x\). - \(f(x) = \ln x, \quad f'(x) = \frac{1}{x}, \quad f''(x) = -\frac{1}{x^2}, \quad f'''(x) = \frac{2}{x^3}, \) and so on.These derivatives alternate in sign and factorial denominators emerge for \(n\geq1\).
03
Evaluate Derivatives at \(x_0 = 1\)
Substitute \(x = 1\) into each derivative:- \(f(1) = \ln 1 = 0\)- \(f'(1) = \frac{1}{1} = 1\)- \(f''(1) = -\frac{1}{1^2} = -1\)- \(f'''(1) = \frac{2}{1^3} = 2\)- In general, \(f^{(n)}(1) = (-1)^{n+1}(n-1)!\) for \(n \geq 1\).
04
Write the Taylor Series in Sigma Notation
The Taylor series in sigma notation is given by substituting the fixed derivatives evaluated at \(x = 1\) into the Taylor series expansion formula:\[ \ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n \]The term \((-1)^{n+1}(n-1)!/n!\) reflects the alternating sign, factorial sequence.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sigma Notation
Sigma notation is a powerful tool for representing the sum of a sequence of terms in a compact form. When you see a summation symbol, \(\sum_{n=a}^{b} expression \), it indicates that you should evaluate the **expression** for each integer value of \(n\) from \(a\) to \(b\) and then find the sum of these values.
In the context of a Taylor series, sigma notation allows us to succinctly express the infinite sum of terms that arise when expanding a function about a particular point. For instance, for the natural logarithm expansion around \(x_0 = 1\), the sigma notation form begins from \(n=1\) because the first derivative of \(\ln x\) at that point is zero, simplifying our sum. This transforms the infinite series of terms into a formula that is easier to analyze and compute.
- \(n\) is the index of summation.
- \(a\) is the lower limit where summation starts.
- \(b\) is the upper limit of the summation.
In the context of a Taylor series, sigma notation allows us to succinctly express the infinite sum of terms that arise when expanding a function about a particular point. For instance, for the natural logarithm expansion around \(x_0 = 1\), the sigma notation form begins from \(n=1\) because the first derivative of \(\ln x\) at that point is zero, simplifying our sum. This transforms the infinite series of terms into a formula that is easier to analyze and compute.
Derivative Evaluation
Derivative evaluation involves calculating the derivative of a function at a certain point, which is crucial in constructing Taylor series. For a given function like \(f(x) = \ln x\), derivatives help derive the closer approximation of the function as a power series. Let's look at the process step-by-step:
Once the derivatives are found, the next step is to evaluate them at the point \(x_0 = 1\). This provides the coefficients needed for the Taylor expansion:
- First Derivative: The first derivative of \(\ln x\) is \(f'(x) = \frac{1}{x}\).
- Second Derivative: Following the first, the second derivative is \(f''(x) = -\frac{1}{x^2}\).
- Higher-order Derivatives: This sequence continues as \(f'''(x) = \frac{2}{x^3}\), showing a pattern of alternating signs and increasing powers of \(x\).
Once the derivatives are found, the next step is to evaluate them at the point \(x_0 = 1\). This provides the coefficients needed for the Taylor expansion:
- \(f(1) = 0\), because \(\ln 1 = 0\).
- \(f'(1) = 1\), as substituting \(x = 1\) in \(\frac{1}{x}\) gives 1.
- \(f''(1) = -1\), and so on.
Natural Logarithm Expansion
The natural logarithm expansion is a classic example of an application of Taylor series. By expanding the natural logarithm function \(\ln x\) around a point, we can approximate the function in the form of an infinite series. Generally, this is how Taylor expansions work:
Given that \(f^{(n)}(x_0)\) denotes the \(n\)-th derivative evaluated at \(x_0\), our task is to place these into the sum. Since the logarithm lacks a straightforward pattern at \(x_0=1\), only terms from \(n=1\) onwards contribute:
- The series becomes \(\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n \), reflecting the characteristic alternating sign and factorial denominator.
This expansion is significant for approximating \(\ln x\) near \(x_0 = 1\), capturing its behavior efficiently and elegantly through calculus.
- The base function \(\ln x\) is to be expanded around \(x_0 = 1\).
- Using Taylor's formula, this function is expressed as \( \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \).
Given that \(f^{(n)}(x_0)\) denotes the \(n\)-th derivative evaluated at \(x_0\), our task is to place these into the sum. Since the logarithm lacks a straightforward pattern at \(x_0=1\), only terms from \(n=1\) onwards contribute:
- The series becomes \(\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n \), reflecting the characteristic alternating sign and factorial denominator.
This expansion is significant for approximating \(\ln x\) near \(x_0 = 1\), capturing its behavior efficiently and elegantly through calculus.