# Upper Bound For Maclaurin Approximation

• September 22nd 2009, 12:44 PM
soma
Upper Bound For Maclaurin Approximation
There is a homework problem that I don't even understand what they're asking. Can any of you help clarify what they mean?

Quote:

f(x) = sin(x)

Find an upper bound for |[f(x)-[p9(x)]]|
-7 <= x <= 7

p9(x) is the nth order Maclaurin approximaxion to f
• September 22nd 2009, 01:57 PM
HallsofIvy
Quote:

Originally Posted by soma
There is a homework problem that I don't even understand what they're asking. Can any of you help clarify what they mean?

Where should we start? Obviously this problem expects you to know what a "MacLaurin series" is. Do you know that? It is simply the Taylor's series at x= 0. Do you know what that is? If not look up "MacLaurin series" or "Taylor's series" in the index of your book.

And, I strongly suspect that in the section where this problem is given, there is a formula for the error you get in using cutting a McLaurin or Taylor's polynomial of degree n (that is, cutting off the McLaurin or Taylor's series at the nth power). It should look something like this:
$E\le \frac{M_{n+1}}{(n+1)!}x^{n+1}$ where " $M_{n+1}$" is an upper bound on the absolute value of the n+1 derivative of the function.

Since this problem asked about the "9th order MacLaurin series for sin(x)" for x between -7 and 7. Okay you need $M_{10}$. What is the 10th derivative of sin(x)? What is its largest value between -7 and 7 (radians: $2\pi$ is about 6.28 so this is from less than $2\pi$ to larger than $2\pi$).
• September 22nd 2009, 02:39 PM
soma
Ok, so the tenth derivative of sin(x) is -sin(x), and the maximum bound of -sin(x) is when x = -pi/2

-sin(-pi/2) = 1

so

E <= [1/(9+1)!]*[(-pi/2)^(9+1)]
E <= .0000252020

Am I understanding this problem correctly?