<s>[INST] Is the taylor polynomial of degree $2$ near $(0,0)$ of $(, ) = \frac{1}{ 2 - ( + ^2)}$ the following: $ P(, ) = \frac{1}{2} + \frac{}{4} + \frac{^2}{4} + \frac{^2}{2}$
Is this right?
I can't tell, as I can't seem to see the remainder going to $0$ when divided by $x^2 + y^2$ as $(x, y) → (0,0)$
 [/INST] $$\begin{align}\frac{1}{2-(x+y^2)}&=\frac{1}{2}\frac{1}{\left(1-\frac{x+y^2}{2}\right)}\\&=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(x+y^2)^k}{2^k}\\&=\frac{1}{2}+\frac{x+y^2}{4}+\frac{(x+y^2)^2}{8}+\text{ (terms that give powers }>2)\\&=\frac{1}{2}+\frac{x}{4}+\frac{y^2}{4}+\frac{x^2}{\color{red}{8}}+\text{ (terms with powers }>2)\end{align}$$
I got an $8$ there.
And $$\frac{1}{2-(x+y^2)}-\frac{1}{2}-\frac{x}{4}-\frac{y^2}{4}-\frac{x^2}{8}=\frac{-x^3-x^2 y^2-4 x y^2-2 y^4}{8 (x+y^2-2)}$$
</s>