1. ## Taylor's Theorem

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.

2. Originally Posted by xcluded

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.
[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

3. yeah. it is kind of weird.

anyone with different views on it ?

4. 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.