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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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Convergence of Taylor & Maclaurin Series

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

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)

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Textbook Question

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


lim ₙ → 0 (x²/2 - 1 + cos x)/x⁴

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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. There is more than one way to choose the center of the series.


sinh (-1)

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Textbook Question

Approximating ln 2 Consider the following three ways to approximate

ln 2.

d. At what value of x should the series in part (c) be evaluated to approximate ln 2? Write the resulting infinite series for ln 2.

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Textbook Question

Approximating ln 2 Consider the following three ways to approximate

ln 2.

c. Use the property ln a/b = ln a - ln b and the series of parts (a) and (b) to find the Taylor series for ƒ(x) = ln (1 + x)/(1 - x) b centered at 0.

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Textbook Question

A differential equation Find a power series solution of the differential equation y'(x) - 4y + 12 = 0, subject to the condition y(0) = 4. Identify the solution in terms of known functions.

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