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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 11.1.46

Remainders Find the remainder Rₙ for the nth−order Taylor polynomial centered at a for the given functions. Express the result for a general value of n.


f(x) = 1/(1 - x), a=0

Guida verificata passo dopo passo
1
Identify the function and the center of the Taylor polynomial: here, the function is \(f(x) = \frac{1}{1 - x}\) and the Taylor polynomial is centered at \(a = 0\).
Recall the formula for the remainder (Lagrange form) of the nth-order Taylor polynomial centered at \(a\): \[R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - a)^{n+1}\] where \(c\) is some value between \(a\) and \(x\).
Find the \((n+1)\)th derivative of \(f(x)\). Since \(f(x) = (1 - x)^{-1}\), use the general formula for derivatives of this form: \[f^{(k)}(x) = k! \cdot (1 - x)^{-(k+1)}\] with appropriate sign considerations.
Substitute \(a = 0\) into the remainder formula and express \(R_n(x)\) in terms of \(f^{(n+1)}(c)\) and \((x - 0)^{n+1} = x^{n+1}\).
Write the final expression for the remainder \(R_n(x)\) as: \[R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}\] where \(c\) lies between \(0\) and \(x\), and \(f^{(n+1)}(c)\) is the \((n+1)\)th derivative evaluated at \(c\).

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