- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,756

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,756

- Jan 31, 2012

- 253

$ \log(1+n) = n-\dfrac{n^2}{2}+\dfrac{n^3}{3}-\cdots $ $( -1 < n \le 1)$

$ \displaystyle \frac{1}{m}- \frac{1}{n} = 1 \implies n = \frac{m}{1-m} $

Because of the restriction on $m$, the values of $n$ fall between $0$ and $1$ including $1$.

And we have

$ \displaystyle \log(1+n) = \log \left( 1 + \frac{m}{1-m} \right) = \log \left(\frac{1}{1-m} \right) = - \log(1-m) $

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,756

Thank you for participating, **Random Variable**! Your proof is correct, well done!