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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.29
Chapter 11, Problem 11.R.29

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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1
Recall the geometric series formula: \(\sum_{k=0}^{\infty} x^k = \frac{1}{1 - x}\) for \(|x| < 1\).
Notice that the given function is \(f(x) = \frac{1}{1 - x^2}\), which can be seen as \(\frac{1}{1 - (x^2)}\).
Apply the geometric series formula by substituting \(x\) with \(x^2\), so the series becomes \(\sum_{k=0}^{\infty} (x^2)^k = \sum_{k=0}^{\infty} x^{2k}\).
Write the Maclaurin series for \(f(x)\) as \(\sum_{k=0}^{\infty} x^{2k}\), which expands to \(1 + x^2 + x^4 + x^6 + \cdots\).
Determine the interval of convergence by applying the condition \(|x^2| < 1\), which simplifies to \(|x| < 1\).

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Key Concepts

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

Geometric Series

A geometric series is an infinite sum of the form Σ x^k from k=0 to ∞, which converges to 1/(1 - x) when |x| < 1. This formula provides a foundation for expressing functions as power series by recognizing patterns similar to geometric series.
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Geometric Series

Maclaurin Series

The Maclaurin series is a special case of the Taylor series centered at zero, representing a function as an infinite sum of its derivatives at zero multiplied by powers of x. It allows us to express functions as power series expansions around x = 0.
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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 it represents. For geometric series, this interval is determined by the condition |x| < 1, and it must be adjusted when the series involves expressions like x².
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Related Practice
Textbook Question

Approximating ln 2 Consider the following three ways to approximate

ln 2.

e. Using four terms of the series, which of the three series derived in parts (a)–(d) gives the best approximation to ln 2? Can you explain why?

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

1
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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

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.

1
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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 + 5x)

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)