I'm studding for my math exam and I found some question I can not answer. I have to find Taylor series of $\ln(xy)$ near $(1,1)$ So I know that $$\ln(1-t) = -\sum_{n=1}^{\infty} \frac{t^n}{n}$$ assume $1-t=xy$ then $t=1-xy$ and so $$\ln(xy)=-\sum_{n=1}^{\infty} \frac{(1-xy)^n}{n}$$
But, I also can write $\ln(xy)=\ln(x)+\ln(y)$ suppose $1-t=x$ (or $y$) then $$\ln(xy)=-\sum_{n=1}^{\infty} \frac{(1-x)^n}{n}-\sum_{n=1}^{\infty} \frac{(1-y)^n}{n}$$
Two of those are not equal.How/why?
Thanks!
The Taylor series near $(1,1)$ should be in powers of $(x-1)$ and $(y-1)$. Your second series is (essentially) of that form. Your first series is not. From the first series, you will need to do $$ 1-xy = -(x-1) - (y-1) - (x-1)(y-1) $$ and substitute that in. A complicated calclulation. The mixed terms $(x-1)^k(y-1)^l$ all cancel, of course.