Ch. 11 - Power Series

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 11 - Power Series

Problem 11.4.54

Chapter 11, Problem 11.4.54

Evaluating an infinite series Write the Maclaurin series for f(x) = ln (1+x) and find the interval of convergence. Evaluate f(−1/2) to find the value of ∑ₖ₌₁∞ 1/(k 2ᵏ)

Verified step by step guidance

Recall that the Maclaurin series for a function \( f(x) \) is its Taylor series expansion at \( x = 0 \), given by \( f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n \). For \( f(x) = \ln(1+x) \), start by finding the derivatives and evaluating them at 0 to identify the pattern of coefficients.

Use the known Maclaurin series expansion for \( \ln(1+x) \), which is \( \ln(1+x) = \sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{k} \). This series converges for \( -1 < x \leq 1 \).

Determine the interval of convergence by applying the ratio test or by recalling the domain of the logarithm series. The series converges when \( |x| < 1 \) and conditionally at \( x = 1 \), but diverges at \( x = -1 \).

To evaluate \( f(-\frac{1}{2}) = \ln(1 - \frac{1}{2}) = \ln(\frac{1}{2}) \), substitute \( x = -\frac{1}{2} \) into the series: \( \ln(1 - \frac{1}{2}) = \sum_{k=1}^\infty (-1)^{k+1} \frac{(-\frac{1}{2})^k}{k} \). Simplify the powers and signs to express the sum in terms of \( \sum_{k=1}^\infty \frac{1}{k 2^k} \).

Recognize that the series \( \sum_{k=1}^\infty \frac{1}{k 2^k} \) corresponds to the negative of the evaluated series (due to the alternating signs), so rearrange the expression to isolate \( \sum_{k=1}^\infty \frac{1}{k 2^k} \) and express it in terms of \( \ln(\frac{1}{2}) \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Maclaurin Series Expansion

The Maclaurin series is a Taylor series expansion of a function about x = 0. It expresses a function as an infinite sum of its derivatives at zero, multiplied by powers of x and divided by factorial terms. For ln(1+x), the series is derived by differentiating and evaluating at zero, providing a polynomial approximation valid near x = 0.

Interval of Convergence

The interval of convergence defines the range of x-values for which the infinite series converges to the function. It is found using tests like the ratio or root test. For ln(1+x), the series converges when -1 < x ≤ 1, ensuring the series sum accurately represents the function within this domain.

Evaluating Series at Specific Points

Substituting specific values of x into the Maclaurin series allows evaluation of infinite sums related to the function. For example, plugging x = -1/2 into the ln(1+x) series helps find the sum of ∑ₖ₌₁∞ 1/(k 2ᵏ). This technique connects function values to series sums, enabling exact or approximate calculations of infinite series.

Recommended video:

06:00

Geometric Series

Related Practice

Textbook Question

Approximating real numbers Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers.

cos 2

views

Textbook Question

Use of Tech Linear and quadratic approximation

a. Find the linear approximating polynomial for the following functions centered at the given point a.

b. Find the quadratic approximating polynomial for the following functions centered at a.

c Use the polynomials obtained in parts (a) and (b) to approximate the given quantity.

Find the Taylor polynomials p₁, p₂, and p₃ centered at a=1 for f(x)=x³.

views

Textbook Question

Combining power series Use the geometric series

f(x) = 1/(1-x) = ∑ₖ₌₀∞ xᵏ, for |x| < 1,

to find the power series representation for the following functions (centered at 0). Give the interval of convergence of the new series.

g(x) = x³/(1 − x)

views

Textbook Question

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.

g(x) = 2/(1 − 2x)² using f(x) = 1/(1 − 2x)

views

Textbook Question

Radius of convergence Find the radius of convergence for the following power series.

∑ₖ₌₁∞ (k!xᵏ)/(kᵏ)

views

Textbook Question

Combining power series Use the power series representation

f(x ) =ln (1 − x) = −∑ₖ₌₁∞ xᵏ/k, for −1 ≤ x < 1,

to find the power series for the following functions (centered at 0). Give the interval of convergence of the new series.

p(x) = 2x⁶ ln(1 − x)

views