Re: Taylor expansion

Rushlan schrieb:
On Oct 13, 8:03 pm, The World Wide Wade <aderamey.a...@xxxxxxxxxxx>
wrote:
In article <1192317284.444180.269...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Michael Ejercito <mejer...@xxxxxxxxxxx> wrote:
No, log [(1+x/(1-x)]= log (1+x)- log (1-x).
Michael
Of course. Thanks.

Thanks for your replies guys. But is that all I have to do ? How do i
expand that ? I use the x = f(x)/f'(x) and find the roots ?
I am a bit confused how to approach the problem. Thank you .

You computed the T.-expansion of log(1+x), OK? Then, what is the T.-expansion of log(1-x) (you get it very easily from the T.-expansion of log(1-x))? So now, how do you compute the T.-expansion for log(1+x) - log(1-x), given the t.-expansions for log(1+x) and log(1-x)?

--
Thomas Nordhaus
.

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