Error bound for Maclaurin series sin(x)?

• Mar 18th 2013, 12:10 PM
LLLLLL
Error bound for Maclaurin series sin(x)?
The question goes like this:
If the approximate formula sin(x)=x-(x^3)/3! is used and |x|＜1 (radian), then the error is numerically less than ____

I got confused because even terms are omitted in Maclaurin series for sin(x) (such as a2(x^2) and a4(x^4)
If Lagrange error bound's applied, the term after -(x^3)/3! would actually be a4*x^4= 0
If the series were treated as an alternating series, there's no specific number given for x since the question only says |x|＜0

How should I approach this question?
• Mar 18th 2013, 02:13 PM
ILikeSerena
Re: Error bound for Maclaurin series sin(x)?
Quote:

Originally Posted by LLLLLL
The question goes like this:
If the approximate formula sin(x)=x-(x^3)/3! is used and |x|＜1 (radian), then the error is numerically less than ____

I got confused because even terms are omitted in Maclaurin series for sin(x) (such as a2(x^2) and a4(x^4)
If Lagrange error bound's applied, the term after -(x^3)/3! would actually be a4*x^4= 0
If the series were treated as an alternating series, there's no specific number given for x since the question only says |x|＜0

How should I approach this question?

Hi LLLLLL!:)

Since it is an alternating series with decreasing magnitude, the remainder is less than the next term in the series.
The next term is $\displaystyle \frac {x^5}{5!}$.
So the error is less than $\displaystyle |\frac {x^5}{5!}|$, and since $\displaystyle |x| < 1$.....
• Mar 18th 2013, 08:33 PM
LLLLLL
Re: Error bound for Maclaurin series sin(x)?
HiILikeSerena,
Thank you for your reply. But I think the last part is missing?
• Mar 18th 2013, 08:34 PM
LLLLLL
Re: Error bound for Maclaurin series sin(x)?
Quote:

Originally Posted by ILikeSerena
Hi LLLLLL!:)

Since it is an alternating series with decreasing magnitude, the remainder is less than the next term in the series.
The next term is $\displaystyle \frac {x^5}{5!}$.
So the error is less than $\displaystyle |\frac {x^5}{5!}|$, and since $\displaystyle |x| < 1$.....

HiILikeSerena,
Thank you for your reply. But I think the last part is missing?
• Mar 18th 2013, 10:28 PM
LLLLLL
Re: Error bound for Maclaurin series sin(x)?
I just realized that the answer where be different by applying Lagrange error bound and alternating series but I can't figure out why.
When applying Lagrange error bound, I get:
R3(x)＜max|f^4(c)*(x-a)^4/4!| a=0 and 0＜c＜1
c=1 and f^4(c)=sin(c)
R3(x)=sin(1)/4!

When viewing the function as alternating series,
R=1/5!

how come?
• Mar 18th 2013, 11:32 PM
ILikeSerena
Re: Error bound for Maclaurin series sin(x)?
Quote:

Originally Posted by LLLLLL
I just realized that the answer where be different by applying Lagrange error bound and alternating series but I can't figure out why.
When applying Lagrange error bound, I get:
R3(x)＜max|f^4(c)*(x-a)^4/4!| a=0 and 0＜c＜1
c=1 and f^4(c)=sin(c)
R3(x)=sin(1)/4!

When viewing the function as alternating series,
R=1/5!

how come?

The 4th order term of the series is zero.
So you can also take the 5th order term for your Lagrange error bound:
R4(x)＜max|f^5(c)*(x-a)^5/5!| a=0 and 0＜c＜1
c=0 and f^5(0)=cos(0)
R4(x)<1/5!

This result is (in this case) the same as an upper bound for the next term in the series,
R<|x^5/5!| and |x|<1
R<1^5/5!
• Mar 19th 2013, 05:03 PM
LLLLLL
Re: Error bound for Maclaurin series sin(x)?
Quote:

Originally Posted by ILikeSerena
The 4th order term of the series is zero.
So you can also take the 5th order term for your Lagrange error bound:
R4(x)＜max|f^5(c)*(x-a)^5/5!| a=0 and 0＜c＜1
c=0 and f^5(0)=cos(0)
R4(x)<1/5!

This result is (in this case) the same as an upper bound for the next term in the series,
R<|x^5/5!| and |x|<1
R<1^5/5!

I see now :D So it's because that the 4th term is zero and doesn't affect the sum.
Sorry for bothering but I have one last question. When viewing the series as an alternating series, how do you know that you should set x=1 instead of other numbers such 0.5, or 0.3?
Thank you so much!
• Mar 19th 2013, 11:24 PM
ILikeSerena
Re: Error bound for Maclaurin series sin(x)?
Quote:

Originally Posted by LLLLLL
I see now :D So it's because that the 4th term is zero and doesn't affect the sum.
Sorry for bothering but I have one last question. When viewing the series as an alternating series, how do you know that you should set x=1 instead of other numbers such 0.5, or 0.3?
Thank you so much!

It's an upper bound.

R <_ |x^5/5!| = |x|^5/5! < 1^5/5!
• Mar 20th 2013, 09:52 PM
LLLLLL
Re: Error bound for Maclaurin series sin(x)?
Quote:

Originally Posted by ILikeSerena
It's an upper bound.

R <_ |x^5/5!| = |x|^5/5! < 1^5/5!

Oh so it's like taking the maximum?
Thank you so much for helping me! I get it now!