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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.RE.16

Find the remainder term Rₙ(x) for the Taylor series centered at 0 for the following functions. Find an upper bound for the magnitude of the remainder on the given interval for the given value of n. (The bound is not unique.)


ƒ(x) = ln (1 - x); bound R₃(x), for |x| < 1/2

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1
Identify the function and the point about which the Taylor series is centered. Here, the function is \(f(x) = \ln(1 - x)\) and the series is centered at \(0\) (Maclaurin series).
Recall the general form of the remainder term (Lagrange form) for the Taylor series of order \(n\) centered at \(0\): \[R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}\] where \(c\) is some value between \(0\) and \(x\).
Compute the derivatives of \(f(x)\) up to order \(n+1 = 4\). For \(f(x) = \ln(1 - x)\), the derivatives follow a pattern: - \(f'(x) = -\frac{1}{1-x}\) - \(f''(x) = -\frac{1}{(1-x)^2}\) - \(f^{(3)}(x) = -\frac{2}{(1-x)^3}\) - \(f^{(4)}(x) = -\frac{6}{(1-x)^4}\) Use this to write \(f^{(4)}(c)\) explicitly.
Substitute \(f^{(4)}(c)\) into the remainder formula: \[R_3(x) = \frac{f^{(4)}(c)}{4!} x^4 = \frac{-6}{4! (1 - c)^4} x^4\] Simplify the factorial and constants.
To find an upper bound for \(|R_3(x)|\) on the interval \(|x| < \frac{1}{2}\), note that \(c\) lies between \(0\) and \(x\), so \(|c| < \frac{1}{2}\). Use this to find the maximum value of \(\frac{1}{|1 - c|^4}\) on this interval, then multiply by \(\frac{|x|^4}{4!}\) and the absolute value of the constant to get the bound.

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주요 개념

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

Taylor Series and Remainder Term

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives at a single point. The remainder term Rₙ(x) measures the error between the function and its nth-degree Taylor polynomial, quantifying how well the polynomial approximates the function near the center.
추천 영상:
08:42
Taylor Series

Lagrange Form of the Remainder

The Lagrange remainder provides an explicit formula for the error term Rₙ(x), involving the (n+1)th derivative evaluated at some point between the center and x. It helps estimate the maximum possible error by bounding the derivative on the interval of interest.
추천 영상:
가이드 코스
06:32
Alternating Series Remainder

Bounding the Remainder on an Interval

To find an upper bound for |Rₙ(x)|, identify the maximum absolute value of the (n+1)th derivative on the given interval and use it in the remainder formula. This approach ensures the error estimate holds for all x within the specified range, providing a practical measure of approximation accuracy.
추천 영상:
가이드 코스
06:32
Alternating Series Remainder