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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
Capitolo 11, Problema 11.4.46

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.
√e

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Recognize that \( \sqrt{e} \) can be rewritten as \( e^{1/2} \). This allows us to use the Taylor series expansion for \( e^x \) centered at \( x = 0 \).
Recall the Taylor series expansion for \( e^x \) about 0 is given by: \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
To find the series for \( e^{1/2} \), substitute \( x = \frac{1}{2} \) into the series: \[ e^{1/2} = 1 + \frac{1}{2} + \frac{(1/2)^2}{2!} + \frac{(1/2)^3}{3!} + \cdots \]
Calculate each term up to the first four nonzero terms without simplifying the numerical values completely, so the terms are: \[ 1, \quad \frac{1}{2}, \quad \frac{(1/2)^2}{2!}, \quad \frac{(1/2)^3}{3!} \]
Write the first four nonzero terms of the infinite series as the approximation for \( \sqrt{e} \): \[ \sqrt{e} \approx 1 + \frac{1}{2} + \frac{1}{8} \cdot \frac{1}{2} + \frac{1}{16} \cdot \frac{1}{6} + \cdots \] (You can leave the terms in factorial and power form to keep the expression clear.)

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Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the function's derivatives at a single point. It approximates functions near that point, allowing complex expressions to be expressed as polynomials. Understanding how to construct and use Taylor series is essential for approximating values like √e.
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To find the Taylor series of a composite function like √e, one often rewrites the expression in a form suitable for expansion, such as √(e) = (e^x)^(1/2). Substituting variables and manipulating the function helps apply known series expansions effectively, enabling the extraction of the first few nonzero terms.
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