Find the smallest value of n such that the remainder estimate |R s (x – a)n + 1, where M is the maximum value of |f(n + 1)(z)| on the interval between a and the indicated point, yields (n + 1)! IR,I s 1 on the indicated interval. 1,000 f(x) = e* on [-3, 3], a = 0
Find the smallest value of n such that the remainder estimate |R s (x – a)n + 1, where M is the maximum value of |f(n + 1)(z)| on the interval between a and the indicated point, yields (n + 1)! IR,I s 1 on the indicated interval. 1,000 f(x) = e* on [-3, 3], a = 0
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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how do u do this problems
![Find the smallest value of n such that the remainder estimate |R.l <
M
-(x - a)" + ', where M is the maximum value of |f(" + 1(z)| on the interval between a and the indicated point, yields
(n + 1)!
1
on the indicated interval.
1,000
f(x) %3D е Х on [-3, 3), а %3D 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb849a623-4c59-4625-854d-b538b3e0e746%2Fff22a737-9daa-440c-9ce7-7d841dae44f8%2Fpxflj4c_processed.png&w=3840&q=75)
Transcribed Image Text:Find the smallest value of n such that the remainder estimate |R.l <
M
-(x - a)" + ', where M is the maximum value of |f(" + 1(z)| on the interval between a and the indicated point, yields
(n + 1)!
1
on the indicated interval.
1,000
f(x) %3D е Х on [-3, 3), а %3D 0
![M
Find the smallest value of n such that the remainder estimate |Rl <
a)" + ', where M is the maximum value of |f(n + 1)(z)| on the interval between a and the indicated point, yields
(n + 1)!
1
on the indicated interval.
1,000
f(x) = sin(x) on [-x, n], a
= 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb849a623-4c59-4625-854d-b538b3e0e746%2Fff22a737-9daa-440c-9ce7-7d841dae44f8%2Fmvh768_processed.png&w=3840&q=75)
Transcribed Image Text:M
Find the smallest value of n such that the remainder estimate |Rl <
a)" + ', where M is the maximum value of |f(n + 1)(z)| on the interval between a and the indicated point, yields
(n + 1)!
1
on the indicated interval.
1,000
f(x) = sin(x) on [-x, n], a
= 0
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