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?
Thanks in advance!
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!
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!