Skip to main content
Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 11, Problem 11.3.25a

Taylor series and interval of convergence


a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.


f(x) = ln (x − 2), a = 3

Verified step by step guidance
1
Recall the definition of the Taylor series of a function \(f(x)\) centered at \(a\): \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n,\] where \(f^{(n)}(a)\) is the \(n\)-th derivative of \(f\) evaluated at \(x = a\).
Identify the function and center: here, \(f(x) = \ln(x - 2)\) and the center is \(a = 3\). We will find derivatives of \(f\) at \(x=3\).
Compute the first derivative: \[f'(x) = \frac{1}{x - 2}.\] Evaluate at \(x=3\): \[f'(3) = \frac{1}{3 - 2} = 1.\]
Find higher order derivatives by differentiating repeatedly: - Second derivative: \[f''(x) = -\frac{1}{(x - 2)^2}\] Evaluate at \(x=3\): \[f''(3) = -1.\] - Third derivative: \[f^{(3)}(x) = \frac{2}{(x - 2)^3}\] Evaluate at \(x=3\): \[f^{(3)}(3) = 2.\] - Fourth derivative: \[f^{(4)}(x) = -\frac{6}{(x - 2)^4}\] Evaluate at \(x=3\): \[f^{(4)}(3) = -6.\]
Write the first four nonzero terms of the Taylor series using the formula: \[f(x) \approx f(3) + f'(3)(x - 3) + \frac{f''(3)}{2!}(x - 3)^2 + \frac{f^{(3)}(3)}{3!}(x - 3)^3 + \frac{f^{(4)}(3)}{4!}(x - 3)^4.\] Substitute the values found for \(f(3)\) and the derivatives to express the series up to the fourth term.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
7m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Taylor and Maclaurin Series

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point a. When a = 0, it is called a Maclaurin series. Each term involves the nth derivative evaluated at a, multiplied by (x - a)^n and divided by n!. This series approximates the function near the point a.
Recommended video:
08:26
Convergence of Taylor & Maclaurin Series

Derivatives of Logarithmic Functions

To find the Taylor series of f(x) = ln(x - 2), you need to compute successive derivatives of the logarithmic function. The first derivative is 1/(x - 2), and higher derivatives involve powers of (x - 2) in the denominator with alternating signs. Understanding these derivatives is essential to form the terms of the series.
Recommended video:
05:18
Derivative of the Natural Logarithmic Function

Interval of Convergence

The interval of convergence is the range of x-values for which the Taylor series converges to the function. For ln(x - 2) centered at a = 3, the series converges where |x - 3| is less than the distance to the nearest singularity (x = 2). Determining this interval ensures the series accurately represents the function.
Recommended video:
08:44
Interval of Convergence