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\begin{align} f(\theta)& = \log \frac{1}{1-\theta}-\frac{\theta}{2-\theta}+\frac{\theta^{2}}{(2-\theta)^{2}} \\ & = \log (1-\theta)+\frac{\theta}{2-\theta} \end{align} How are the two equations above equal to each other, i.e. reduce the first into the second?

How can one term be outside of the logarithm on the second line, whereas all 3 terms fall inside the logarithm on the first line? Actually, I'm not even sure if all the terms in the first line are supposed to all fall inside the logarithm, no brackets were given, so it could be either or

Source of first equation:

Source of second equation:

If they don't equal each other like how the second author claims, which author made a mistake?

Edit

I found yet another equality presented by the same second author. This time,

$$ f(\theta) = \log (1-\theta)+\frac{2 \theta(1-\theta)}{(2-\theta)^{2}} $$

develarist
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  • This reads as $\log \left( \frac{1}{1 - \theta} \right)$ where the brackets enclose $\frac{1}{1 - \theta}$. On the first line there are three terms, and on the second line, there are two terms. – Toby Mak Nov 22 '20 at 06:05
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    yes, i said that without giving the counts of 2 and 3. we are seeing the same thing. whichever one is right for the first line, as I discuss in the second paragraph, I still don't see how they reduce for either – develarist Nov 22 '20 at 06:06

1 Answers1

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The first line is not equal to the second line:

$$f(\theta) = \log \left( \frac{1}{1-\theta} \right)-\frac{\theta}{2-\theta}+\frac{\theta^{2}}{(2-\theta)^{2}}$$ $$= \log( (1 - \theta)^{-1}) - \frac{\theta (2 - \theta)}{(2 - \theta)^2} + \frac{\theta^{2}}{(2-\theta)^{2}}$$ $$= -\log(1 - \theta) + \frac{-(2\theta - \theta^2) + \theta^2}{(2 - \theta)^2}$$ $$= -\log(1 - \theta) + \frac{2 \theta^2 - 2 \theta}{(2 - \theta)^2}$$ $$= -\log(1 - \theta) + \frac{(2\theta)(\theta - 1)}{(2 - \theta)^2}$$

Toby Mak
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