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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 11 - Power SeriesProblem 11.R.31
Chapter 11, Problem 11.R.31

Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| < 1, to determine the Maclaurin series and the interval of convergence for the following functions.


ƒ(x) = 1/(1 + 5x)

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Recall the geometric series formula: \(\sum_{k=0}^{\infty} x^k = \frac{1}{1 - x}\) for \(|x| < 1\).
Rewrite the given function \(f(x) = \frac{1}{1 + 5x}\) in a form similar to the geometric series by expressing the denominator as \(1 - (-5x)\).
Identify the term inside the geometric series as \(-5x\), so the series becomes \(\sum_{k=0}^{\infty} (-5x)^k\).
Write the Maclaurin series explicitly as \(\sum_{k=0}^{\infty} (-1)^k 5^k x^k\).
Determine the interval of convergence by applying the condition \(| -5x | < 1\), which simplifies to \(|x| < \frac{1}{5}\).

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Key Concepts

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

Geometric Series and Its Sum Formula

A geometric series is an infinite sum of the form Σ x^k, where each term is a constant ratio times the previous term. Its sum converges to 1/(1 - x) for |x| < 1. This formula is fundamental for expressing functions as power series by rewriting them in a geometric series form.
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Geometric Series

Maclaurin Series

The Maclaurin series is a Taylor series expansion of a function about x = 0. It represents the function as an infinite sum of powers of x with coefficients derived from the function's derivatives at zero. This series helps approximate functions near zero and is used to find power series representations.
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Interval of Convergence

The interval of convergence is the set of x-values for which a power series converges to the function. For geometric series, it is determined by the condition |x| < 1 after rewriting the function in the form 1/(1 - r). Identifying this interval ensures the power series accurately represents the function within that range.
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Related Practice
Textbook Question

Taylor polynomials Find the nth-order Taylor polynomial for the following functions centered at the given point a.

ƒ(x) = e^(sin x), n = 2, a = 0

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Textbook Question

Approximating ln 2 Consider the following three ways to approximate

ln 2.

b. Use the Taylor series for ln (1 - x) centered at 0 and the identity ln 2 = -ln 1/2. Write the resulting infinite series.

Textbook Question

Limits by power series Use Taylor series to evaluate the following limits.


lim ₙ → ₄ ln (x - 3)/(x² - 16)

Textbook Question

Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| < 1, to determine the Maclaurin series and the interval of convergence for the following functions.


ƒ(x) = 1/(1 - x²)

Textbook Question

Taylor polynomials Find the nth-order Taylor polynomial for the following functions centered at the given point a.

ƒ(x) = sinh (-3x), n = 3, a = 0

Textbook Question

Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| < 1, to determine the Maclaurin series and the interval of convergence for the following functions.


ƒ(x) = ln (1 - 4x)