Question:
find the Taylor series of $f(x,y) = 1 / (2 + xy^2) $ around the point $(0,0)$
Answer:
$$ \sum_{n=0}^\infty (-1)^n \frac{x^n y^{2n}}{2^{n + 1}} $$
When I derive the function at $f(0,0)$ i only get that all derivatives are equal to $0,$ so I am definitely doing something wrong here. Help is appreciated :)
On
We have:
$$\left. \frac{\partial f(x,y)}{\partial x^k} \right|_{(x,y)=(0,0)} = \left.\frac{(-1)^k k! y^k}{(xy+2)^{k+1}}\right|_{(x,y)=(0,0)} = 0$$
$$\left. \frac{\partial f(x,y)}{\partial y^k} \right|_{(x,y)=(0,0)} = \left. \frac{(-1)^k k! x^k}{(xy+2)^{k+1}} \right|_{(x,y)=(0,0)} = 0$$
$$\left. \frac{\partial }{\partial y^i} \frac{\partial }{\partial y^i} f(x,y)\right|_{(x,y)=(0,0)} = \left. \frac{ (i!)^2}{(xy+2)^{2i+1}} \left((-1)^i2^i + f(i) \sum_{j=1}^i (xy)^j \right) \right|_{(x,y)=(0,0)} = \frac{(-1)^i(i!)^2}{2^{i+1}}$$
So you'll need to test the other possibilities. But, observe that not all of the combinations are 0.
\begin{align} & \frac 1 {2 + xy^2} = \frac{1/2}{1 + \frac{xy^2} 2} = \frac 1 2 \cdot \frac 1 {1-r} \\[6pt] & \big(\text{where } r = - xy^2/2\big) \\[8pt] = {} & \frac 1 2 \left( 1 + r + r^2 + r^3 + \cdots \right). \end{align}