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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
11장, 문제 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

검증된 단계별 안내
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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
08:42
Taylor Series

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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가이드 코스
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Intro to Series: Partial Sums

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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가이드 코스
05:36
Piecewise Functions