Use the Alternating Series Estimation Theorem or Taylor's Formula to estimate the range of values of a for which the given approximation is accurate to within the stated error. Check your answer graphically. (Round your answers to three decimal places.) (error] < 0.001) We know that sin xx I sin x = x -164 120 x³ 6 2²³ + 3! By the Alternating Series Estimation Theorem, the error in the approximation sin x = x= less than Chapter 6 Prot notx 1. x <ょく 1 5! Since we want [error] < 0.00000001, then we mus solve -25 - 25 <0.00012 X Taking the fifth root of a we have that x < 0.065439. When rounded to three decimal places, this gives a range of values for a as 164 X 12 Chapter 6 Prot < 0.00000001, which gives us notx

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Subject: calculus 

 

Use the Alternating Series Estimation Theorem or Taylor's Formula to estimate the range of values
of a for which the given approximation is accurate to within the stated error. Check your answer
graphically. (Round your answers to three decimal places.)
(error] < 0.001)
We know that
sin xx
1
sin x = x-
-.164
25
120
1
3!
By the Alternating Series Estimation Theorem, the error in the approximation sin x = x-
less than
2³ +
x <<
Chanter & Drnt. nntv
1.
Since we want [error] < 0.00000001, then we mus solve
1
5!
-
25 <0.00012 X
Taking the fifth root of a we have that x < 0.065439. When rounded to three decimal places,
this gives a range of values for a as
.164
12 Chanter 6. Drot nntv
1
A
< 0.00000001, which gives us
Transcribed Image Text:Use the Alternating Series Estimation Theorem or Taylor's Formula to estimate the range of values of a for which the given approximation is accurate to within the stated error. Check your answer graphically. (Round your answers to three decimal places.) (error] < 0.001) We know that sin xx 1 sin x = x- -.164 25 120 1 3! By the Alternating Series Estimation Theorem, the error in the approximation sin x = x- less than 2³ + x << Chanter & Drnt. nntv 1. Since we want [error] < 0.00000001, then we mus solve 1 5! - 25 <0.00012 X Taking the fifth root of a we have that x < 0.065439. When rounded to three decimal places, this gives a range of values for a as .164 12 Chanter 6. Drot nntv 1 A < 0.00000001, which gives us
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