• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • #1

Homework Statement



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

Homework Equations





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.
 

Answers and Replies

  • #2
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
 
  • #3
Last edited by a moderator:
  • #4
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.
 

Related Threads on Lagrange error bound to estimate sin4° to five decimal places( maclaurin series)

Replies
0
Views
1K
Replies
6
Views
2K
Replies
11
Views
7K
Replies
5
Views
2K
  • Last Post
Replies
0
Views
4K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
426
  • Last Post
Replies
7
Views
6K
Replies
1
Views
7K
  • Last Post
Replies
1
Views
14K
Top