Show $G^2=2\sum o \log \frac{o}{e}$ is approximately $X^2=\sum \frac {(o-e)^2}{e}$
$o_i$ = observed $e_i$=expected (I removed $i$'s for ease)
The solution is:
$$G^2=2\sum o \log \frac{o}{e}$$ $$=2\sum e (1+d/e)\log(1+d/e)$$ where $d=o-e$ $$=2\sum (d+d^2/2e)=X^2$$
I don't understand these equlaities.
I tried myself to write a $\log$ expansion:
$\log \frac{o}{e}=\log o - \log e$, hence
$$\log o= \log e +(o-e)/e-\frac{(o-e)^2}{2e^2}$$
Then I tried to do some manipulations:
$$\log \frac{o}{e}=\frac{oe-e^2-0.5o^2+oe+0.5e^2}{e^2}$$
$$\log \frac{o}{e}=\frac{2oe-0.5e^2-0.5o^2}{e^2}=-\frac{o^2-4oe+e^2}{2e^2}$$
This is as close as I got Can you help me?