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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 11 - Power SeriesProblema 11.R.34
Capitolo 11, Problema 11.R.34

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)

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Recall the geometric series formula: \(\sum_{k=0}^{\infty} x^k = \frac{1}{1 - x}\) for \(|x| < 1\). This will be the foundation for finding the Maclaurin series of \(f(x) = \ln(1 - 4x)\).
Recognize that the derivative of \(f(x)\) is \(f'(x) = \frac{d}{dx} \ln(1 - 4x) = \frac{-4}{1 - 4x}\). This derivative can be expressed using the geometric series by rewriting \(\frac{1}{1 - 4x}\) as a power series.
Express \(f'(x)\) as a power series: \(f'(x) = -4 \sum_{k=0}^{\infty} (4x)^k = -4 \sum_{k=0}^{\infty} 4^k x^k = - \sum_{k=0}^{\infty} 4^{k+1} x^k\) for \(|4x| < 1\), which simplifies to \(|x| < \frac{1}{4}\).
Integrate the power series term-by-term to find \(f(x)\): \(f(x) = \int f'(x) \, dx = - \sum_{k=0}^{\infty} 4^{k+1} \int x^k \, dx = - \sum_{k=0}^{\infty} 4^{k+1} \frac{x^{k+1}}{k+1} + C\).
Determine the constant of integration \(C\) by evaluating \(f(0) = \ln(1 - 0) = 0\), which implies \(C = 0\). Thus, the Maclaurin series for \(f(x)\) is \(- \sum_{k=0}^{\infty} \frac{4^{k+1}}{k+1} x^{k+1}\), valid for \(|x| < \frac{1}{4}\).

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Geometric Series and Its Sum Formula

The 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 is fundamental for expressing functions as power series and serves as a starting point for deriving more complex series expansions.
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Geometric Series

Maclaurin Series Expansion

A Maclaurin series is a Taylor series centered at zero, representing a function as an infinite sum of its derivatives at 0 multiplied by powers of x. It allows approximation of functions near zero and is essential for expressing ln(1 - 4x) as a power series.
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Convergence of Taylor & Maclaurin Series

Interval of Convergence

The interval of convergence defines the set of x-values for which a power series converges. Determining this interval involves analyzing the radius of convergence, often using the ratio or root test, and is crucial to ensure the validity of the series representation.
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Interval of Convergence
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Domanda del libro di testo

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?

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Domanda del libro di testo

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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Domanda del libro di testo

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.

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Domanda del libro di testo

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


lim ₙ → ₄ ln (x - 3)/(x² - 16)

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Domanda del libro di testo

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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Domanda del libro di testo

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

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