Taylor polynomial approximation (HELP ME)

1. Nov 27, 2007

frasifrasi

Ok, we are asked to determined the degree of the the taylor polynomial about c =1 that should be used to approximate ln (1.2) so the error is less than .001

the book goes throught the steps and arrives at:

|Rn(1.2)| = (.02)^(n+1)/(z^(n+1)*(n+1)

but then, it states that

(.02)^(n+1)/(n+1) < .001

and there is an error pointing this expression to

1000 < (n+1)(5^(n+1))

--> I have no idea how the book arrived at this second expression, could anyone please explain?

- Also, if anyone knows, why was the z term left out in the second step?

Thank you.

2. Nov 27, 2007

frasifrasi

why was the z term dropped?

does anyone know?

3. Nov 28, 2007

HallsofIvy

Staff Emeritus
I just don't believe your book has what you wrote. For one thing, |Rn(1.2)| = (.02)^(n+1)/(z^(n+1)*(n+1) is missing a ")". For another, the remainder formula has z in the numerator not the denominator- You can't make the error smaller by choosing z larger!

4. Nov 28, 2007

frasifrasi

OK, in general how does this process work? I am having a little trouble following the steps...

*Yeah, I forgot the last parenthesis