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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 11.3.39
Capitolo 11, Problema 11.3.39

Manipulating Taylor series Use the Taylor series in Table 11.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0.


{(eˣ−1)/x if x ≠ 1, 1 if x = 1

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1
Recall the Taylor series expansion of the exponential function centered at 0: \(e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \cdots\).
Substitute this series into the given function \(\frac{e^{x} - 1}{x}\) for \(x \neq 0\). This gives: \(\frac{e^{x} - 1}{x} = \frac{\left(1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots \right) - 1}{x}\).
Simplify the numerator by canceling the 1's: \(\frac{x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots}{x}\).
Divide each term in the numerator by \(x\): \(1 + \frac{x}{2!} + \frac{x^{2}}{3!} + \frac{x^{3}}{4!} + \cdots\).
Write out the first four nonzero terms explicitly: \(1 + \frac{x}{2} + \frac{x^{2}}{6} + \frac{x^{3}}{24}\).

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Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives at a single point, usually centered at zero (Maclaurin series). It approximates functions locally and helps express complicated functions as polynomials, making analysis and computation easier.
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Manipulating Series and Handling Indeterminate Forms

When a function involves expressions like (e^x - 1)/x, direct substitution at x=0 leads to an indeterminate form 0/0. Using the Taylor series for e^x, we expand numerator and denominator terms, then simplify by dividing series to find a valid series representation around zero.
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Piecewise Function and Continuity at a Point

The function is defined differently at x=1 and elsewhere, requiring careful consideration of limits and continuity. Understanding how the series expansion matches the function's value at the point ensures the series correctly represents the function, especially when the function is defined piecewise.
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