Thread: i have no idea what this question is asking (find logarithmic approximations)

i remember doing something like this last year, but it was just glanced over, and hardly mentioned this year at all. it seems like the kind of thing thats easy once you figure out what to do, but figuringthat out is the whole battle. so, not nly do i not know what to do, i don't even know how to begin to show my work

anyways we have:
Use the following theorem with x=1/3 and m=5 to calculate approximations to log 2. retain 9 decimal places and optain the inequalities .6931460<x<.6931476


and ill just put the whole theorem because its a doozy and i cant make heads or tails of it:
If 0<x<1 and m>_1 we have
log((1+x)/(1-x))=2(x+x^3/3 + .... + (x^ (2m-1))/(2m-1)+ R(subscript m) (x)
where the error term, R(subscript m)(x) satisfies the inequalities
(x^(2m+1))/(2m+1)< R (subscript m)(x)< [(2-x)/(1-x)]*[(x^(2m-1))/(2m+1)]


so, as you can see thats quite a mouthfull...i am pretty sure i copied it down perfectly, but im sure it would be a plus if you are familiar with that theorem, as i had to use so many parenthesis, it must be very hard to understand just by looking at it

thanks in advance for any help with solving this

You have a Taylor series to log( (1+x)/(1-x) ). If you put in x=1/3 you'll see that (1+x)/(1-x) = 2. So you're actually approximating log(2). The m=5 says to evaluate 5 terms. When you add up the terms, you'll get a good guess for log(2). The remainder term tells you how bad your guess could possibly be - that's where you get the bounds on log(2).

i still dont get what 5 terms to use?/
that clarifies a little, but i still really don't know how to get those numbers, especially when we literally havent used a calculator all year

Look at your series for .
The first term is .
The 2nd term is ...
The + signs separate terms.