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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.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

Guida verificata passo dopo passo
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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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.
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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.
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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.
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Percorso guidato
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Alternating Series Remainder
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ƒ(x) = eˣ, a = 0; e-0.08


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Radius and interval of convergence Use the Ratio Test or the Root Test to determine the radius of convergence of the following power series. Test the endpoints to determine the interval of convergence, when appropriate.



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