Lagrange error bound to estimate sin4 to five decimal places( maclaurin series)


by hangainlover
Tags: bound, decimal, error, estimate, lagrange, maclaurin, places, series, sin4
hangainlover
hangainlover is offline
#1
Mar21-10, 01:28 PM
P: 83
1. The problem statement, all variables and given/known data

Estimate sin4 accurate to five decimal places (using maclaurin series of sin)

2. Relevant equations



3. The attempt at a solution
Lagrange error bound to estimate sin4 to five decimal places( maclaurin series)

4=pi/45 radians

|Rn(pi/45)<1*(pi/45)^n+1/(n+1)! < 5*10^-6

and the answer key says n should be greater than or equal to 3.



It doesn't make sense .



Because, if you write out derivatives, the ones with sines will disappear in the polynomial. So, don't we have to ignore sin (since it is maclaurin series)

So if it is 7rd order polynomial it should be x-(1/3!)x^3 + (x^5)/5!) -(x^7)/7!.

and therefore we need to look at 9th derivative.



It seems the answer key just applied the remainder theorem.
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hangainlover
hangainlover is offline
#2
Mar21-10, 01:30 PM
P: 83
Plus why do they say that the remainder should be less than or equal to .000005
I think it is due to the fact that the questions says it should be accurate to five decimal places.
but what about .000009 or something?
they are still less than .00001
hangainlover
hangainlover is offline
#3
Mar21-10, 01:34 PM
P: 83
http://www.docstoc.com/docs/29577783...r-Series-Error
look at the example number 2
they looked at the very first term that is not included in the maclurin series

hangainlover
hangainlover is offline
#4
Mar21-10, 01:47 PM
P: 83

Lagrange error bound to estimate sin4 to five decimal places( maclaurin series)


i think in this case our error bound inequality should be
1*(pi/45)^(2n+3)/(2n+3)! greater than or equal to .000005
and i get n=1 which means i only need the first two terms in the maclurin series to have a value accurate to the fifth decimal place.


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