Question

Determine the Taylor series about the point \(x_{0}\) for the given function. Also determine the radius of convergence of the series. \(\ln x, \quad x_{0}=1\)

Short answer

Answer: The Taylor series for the function \(\ln x\) centered at \(x_{0}=1\) is given by \(\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n\). The radius of convergence for this series is 1, and the interval of convergence is \((0, 2)\).

Step-by-step solution

Find the nth Derivative of the Function

We will start by finding the nth derivative of the given function, \(\ln x\): 1st derivative: \(f'(x) = \frac{1}{x}\) 2nd derivative: \(f''(x) = -\frac{1}{x^2}\) 3rd derivative: \(f'''(x) = \frac{2}{x^3}\) By observing the pattern, we can find the nth derivative: nth derivative: \(f^{(n)}(x) = \frac{(-1)^{n+1}(n-1)!}{x^n}\)

Evaluate the Derivatives at \(x_{0}=1\)

Now, we'll evaluate these derivatives at the center point \(x_0=1\). \(f(1) = \ln(1) = 0\) \(f'(1) = \frac{1}{1} = 1\) \(f''(1) = -\frac{1}{1^2} = -1\) \(f'''(1) = \frac{2}{1^3} = 2\) Generally, \(f^{(n)}(1) = (-1)^{n+1}(n-1)!\)

Write Out the Taylor Series

The Taylor series centered around \(x_0=1\) for the function \(\ln x\) is given by: \[\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n\] Which simplifies to: \[\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n\]

Determine the Radius of Convergence

To find the radius of convergence, we'll use the ratio test, which involves finding the limit: \[\lim_{n \to \infty} \frac{|a_{n+1}(x-1)^{n+1}|}{|a_n(x-1)^n|} \] Considering only the coefficients of the series: \[\lim_{n \to \infty} \frac{\frac{1}{n+1}}{\frac{1}{n}} = 1\] For the series to converge, this limit must be less than 1: \[|x-1| < 1\] Thus, the radius of convergence is 1, and the interval of convergence is \((0, 2)\), i.e., the series converges for \(x\) in the interval \((0, 2)\).

Key concepts

Understanding Taylor Series

A Taylor series is a powerful tool in calculus that allows us to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is named after the mathematician Brook Taylor, who introduced the concept in the early 18th century. The series provides a way to approximate complex functions using polynomials, which are easier to compute and analyze.

The general form of a Taylor series for a function \( f(x) \) about a point \( x_0 \) is given by:
\[ f(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n + \cdots \]
In this representation, each term involves the \( n \)th derivative of the function evaluated at the point \( x_0 \), denoted as \( f^{(n)}(x_0) \). The factorial term \( n! \) serves as a normalization factor.

When calculating the Taylor series for \( \ln x \) about the point \( x_0=1 \), we observe that each successive term depends on the nth derivative of the function at this point and factorials that grow with the order of the derivative. The series for \( \ln x \) presented in the exercise demonstrates this concept beautifully.

nth Derivative and its Role in Taylor Series

The \( n \)th derivative of a function plays a crucial role in the formation of a Taylor series. As we compute the first, second, third, and so on derivatives of \( f(x) \), we obtain the coefficients for the corresponding terms in the series. Understanding the pattern of these derivatives can greatly simplify the calculation of the series.

For the function \( \ln x \), an important observation is made: as the order of the derivative increases, the derivative's formula involves a recognizable alternating sign pattern and factorial terms. Specifically, the nth derivative at \( x_0=1 \) is \( f^{(n)}(1) = (-1)^{n+1}(n-1)! \), showing how each term in the Taylor series will look. This explicit pattern allows us to write the entire Taylor series without needing to compute each derivative individually past a certain point, saving time and effort.

Ratio Test for Radius of Convergence

The radius of convergence is a measure of the extent to which a Taylor series represents a function accurately within a certain interval around the center of expansion, \( x_0 \). It is crucial in determining where the series converges to the function it represents. The ratio test is a commonly used method to find the radius of convergence for a given Taylor series.

The ratio test involves taking the limit of the absolute values of the ratio of successive terms of the series. Formally, the test states that for the series \( \sum a_n(x) \), if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \), then the series converges if \( L < 1 \). Applying the ratio test to the Taylor series of \( \ln x \), we find that the radius of convergence is 1. This means the Taylor series for \( \ln x \) converges when \( x \) is within the interval \( (0, 2) \), centered around \( x_0=1 \). Outside of this interval, the series does not reliably represent the function. Understanding how to apply the ratio test is crucial for students as it aids in comprehending the behavior of the series and ensuring its proper use in problem-solving.