# Thread: help me estimate a number with TAYLOR method

1. ## help me estimate a number with TAYLOR method

Assume that f is a function with |f^(n)(x)|<or= 1 for all n and all real x. (The sine and cosine functions have this property.

Find the least integer n for which you can be sure that Pn(2) approximates f(2) within 0.001.

Thanks for the help!

2. Originally Posted by seth55
Assume that f is a function with |f^(n)(x)|<or= 1 for all n and all real x. (The sine and cosine functions have this property.

Find the least integer n for which you can be sure that Pn(2) approximates f(2) within 0.001.

Thanks for the help!
$\displaystyle |P_n(2) - f(2)| \leq \frac{2^{n+1}}{(n+1)!}$

3. Thanks for the quick response, however, I don't quite understand your answer. How do I find the least possible integer?

4. Originally Posted by seth55
Thanks for the quick response, however, I don't quite understand your answer. How do I find the least possible integer?
Because if we can make $\displaystyle \frac{2^{n+1}}{(n+1)!}$ smaller than $\displaystyle .001$ then it would mean $\displaystyle |T_n(2) - f(2)| \leq .001$.

5. Originally Posted by ThePerfectHacker
$\displaystyle |P_n(2) - f(2)| \leq \frac{2^n}{n!}$
Which form or the remainder are you using? Which point are you expanding about?

Expanding about $\displaystyle 0$, and using the Lagrange form of the remainder I get:

$\displaystyle |P_n(2) - f(2)| \leq \frac{2^{n+1}}{(n+1)!}$,

so the $\displaystyle n$ required is the smallest positive integral solution of:

$\displaystyle \frac{2^{n+1}}{(n+1)!}\le 0.001$

Now expanding about $\displaystyle 2$ you will requre only one term, that is $\displaystyle P_0(2)$ is exact when expansion is about $\displaystyle 2$.

RonL

6. Originally Posted by CaptainBlack
Which form or the remainder are you using? Which point are you expanding about?
I edited before you got a chance to reply.