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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 11, Problem 11.1.49

{Use of Tech} Estimating errors Use the remainder to find a bound on the error in approximating the following quantities with the nth-order Taylor polynomial centered at 0. Estimates are not unique.


e⁰ᐧ²⁵, n=4

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Identify the function to approximate: here, it is \(f(x) = e^x\), and we want to approximate \(e^{0.25}\) using the 4th-order Taylor polynomial centered at 0.
Recall the Taylor polynomial of order \(n\) for \(f(x)\) centered at 0 is given by: \[T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k,\] where \(f^{(k)}(0)\) is the \(k\)th derivative of \(f\) evaluated at 0.
The remainder (error) term for the Taylor polynomial approximation is given by the Lagrange form: \[R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}\] for some \(c\) between 0 and \(x\).
Since \(f(x) = e^x\), all derivatives are \(e^x\), which are positive and increasing. To find an upper bound on the error, evaluate the maximum of \(|f^{(n+1)}(c)|\) on the interval between 0 and 0.25, which is \(e^{0.25}\) because \(e^x\) is increasing.
Substitute \(n=4\), \(x=0.25\), and the maximum derivative value into the remainder formula to get the error bound: \[|R_4(0.25)| \leq \frac{e^{0.25}}{5!} (0.25)^5.\] This expression gives a bound on the error when approximating \(e^{0.25}\) with the 4th-order Taylor polynomial centered at 0.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Taylor Polynomial Approximation

A Taylor polynomial approximates a function near a point using a finite sum of its derivatives at that point. The nth-order Taylor polynomial centered at 0 (Maclaurin polynomial) uses derivatives up to order n to estimate the function's value close to zero.
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Remainder (Error) Term in Taylor Series

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