# Taylor's Theorem

• Oct 27th 2009, 09:33 AM
xcluded
Taylor's Theorem
http://img.photobucket.com/albums/v2...5/Untitled.jpg

I tried to write out the taylor's theorem formula , but from there onwards i have no idea how to continue to show $\displaystyle 1\leq f(\frac{1}{2}) \leq \frac{5}{4}$

Can anyone guide me along? Thanks in advance. (Nod)
• Oct 27th 2009, 11:39 AM
tonio
Quote:

Originally Posted by xcluded
http://img.photobucket.com/albums/v2...5/Untitled.jpg

I tried to write out the taylor's theorem formula , but from there onwards i have no idea how to continue to show $\displaystyle 1\leq f(\frac{1}{2}) \leq \frac{5}{4}$

Can anyone guide me along? Thanks in advance. (Nod)

[font=arial][size=4] From the given info it seems like they expect you to write out a MacClaurin series = a Taylor series around x = 0, but then I don't get it:

$\displaystyle f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\ frac{f'''(0)}{3!}x^3 +....= 2+0+\frac{1}{2}x^2+\frac{f'''(0)}{6}x^3+...$

As all the visible terms are non-negative $\displaystyle f(\frac{1}{2}) \geq 2$...

Tonio
• Oct 27th 2009, 09:29 PM
xcluded
yeah. it is kind of weird.

anyone with different views on it ?
• Oct 28th 2009, 08:27 AM
HallsofIvy
Quote:

Originally Posted by xcluded
yeah. it is kind of weird.

anyone with different views on it ?

Looks to me like you are expected to use the LaGrange estimate of the error:
If you truncate a Taylor's series at the $\displaystyle n^{th}$ power term, then the error is less than $\displaystyle \frac{M}{(n+1)!}(x- a)^{n+1}$ where M is an upper bound on the n+1 derivative.