For the given function and interval, find the smallest value of n such that the remainder estimate Rn < |x"+¹, where M is the maximum value of |ƒ (¹+¹) (x) on the interval, yields M (n + 1)! 1 |R₁| < 1000 6x f(x) = e on [1,1] n =
For the given function and interval, find the smallest value of n such that the remainder estimate Rn < |x"+¹, where M is the maximum value of |ƒ (¹+¹) (x) on the interval, yields M (n + 1)! 1 |R₁| < 1000 6x f(x) = e on [1,1] n =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![For the given function and interval, find the smallest value of \( n \) such that the remainder estimate
\[ |R_n| \leq \frac{M}{(n+1)!} |x|^{n+1}, \]
where \( M \) is the maximum value of \( |f^{(n+1)}(x)| \) on the interval, yields
\[ |R_n| \leq \frac{1}{1000}. \]
Given:
\[ f(x) = e^{-6x} \]
on the interval \([-1, 1]\).
Determine:
\[ n = \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1dbff557-c987-45fc-8dd6-6084416c9157%2Fa1ba46b1-e422-44f0-8ebb-aaf6b901cea3%2F7ejgch_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For the given function and interval, find the smallest value of \( n \) such that the remainder estimate
\[ |R_n| \leq \frac{M}{(n+1)!} |x|^{n+1}, \]
where \( M \) is the maximum value of \( |f^{(n+1)}(x)| \) on the interval, yields
\[ |R_n| \leq \frac{1}{1000}. \]
Given:
\[ f(x) = e^{-6x} \]
on the interval \([-1, 1]\).
Determine:
\[ n = \]
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