Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| \u003c 1, to determine the Maclaurin series and the interval of convergence for the following functions.ƒ(x) = ln (1 - 4x)

Ch. 11 - Power Series

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 11 - Power Series

문제 11.R.34

11장, 문제 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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주요 개념

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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.

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.

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.

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)