# Thread: Error bound for Maclaurin series sin(x)?

1. ## 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?

2. ## Re: Error bound for Maclaurin series sin(x)?

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 $\displaystyle |x| < 1$.....

3. ## Re: Error bound for Maclaurin series sin(x)?

HiILikeSerena,
Thank you for your reply. But I think the last part is missing?

4. ## Re: Error bound for Maclaurin series sin(x)?

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?

5. ## 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?

6. ## Re: Error bound for Maclaurin series sin(x)?

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!

7. ## Re: Error bound for Maclaurin series sin(x)?

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 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!

8. ## Re: Error bound for Maclaurin series sin(x)?

Originally Posted by LLLLLL
I see now 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!

9. ## Re: Error bound for Maclaurin series sin(x)?

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!