I found a tool to compute taylor polynomials for 2-variable functions. In order to test the programm, I tried to find the first order Taylor polynomial of $f(x,y)=x^2+y^2$ in $(0,2)$.
The Taylor polynomial is correct, but I don't understand the formula:
Does $\frac{\partial }{\partial x}$ mean $\frac{\partial f}{\partial x}\big |_{x=0}$?
I am confused!
On
That answer is correct, but if the point is $(2,0)$ rather than $(0,2)$. In fact, note that\begin{align}x^2+y^2&=\bigl((x-2)+2\bigr)^2+y^2\\&=(x-2)^2+4(x-2)+4+y^2\end{align}And the Taylor polynomial that you're after is what you get after deleting those terms with degree greater than $1$; you will get $4(x-2)+4=4x-4$ indeed.
And here $\frac\partial{\partial x}$ is $\left.\frac{\partial f}{\partial x}\right|_{x=x_0}$.
It only means the differential operator of partial differentiation w.r.t. $x$. One major advantage of this condensed notation is that for Taylor's formula at higher orders, it behaves like the binomial formula.
Example: The general Taylor's formula at order $2$ in two variables is written as \begin{align}f(x,y)&=f(x_0,y_0) +\frac1{1!}\Bigl((x-x_0)\frac{\partial}{\partial x}+(y-y_0)\frac{\partial}{\partial y}\Bigr)\cdot f(x_0,y_0)\\r &+\frac1{2!}\Bigl((x-x_0)\frac{\partial}{\partial x}+(y-y_0)\frac{\partial}{\partial y}\Bigr)^{\!2}\!\cdot f(x_0,y_0)+ o\bigl(\|(x-x_0,y-y_0)\|^2\bigr), \end{align} which means precisely that
\begin{alignat}{2}f(x,y)&=&&f(x_0,y_0)+\frac1{1!}\Bigl((x-x_0)\frac{\partial f}{\partial x}(x_0,y_0)+(y-y_0)\frac{\partial f}{\partial y}(x_0,y_0)\Bigr)\cr &&&+\frac1{2!}\biggl((x-x_0)^2\frac{\partial^2f}{\partial x^2}(x_0,y_0)+2(x-x_0)(y-y_0)\frac{\partial^2f}{\partial x\,\partial y}(x_0,y_0)+(y-y_0)^2\frac{\partial^2 f}{\partial y^2}(x_0,y_0)\biggr)+ o\bigl(\|(x-x_0,y-y_0)\|^2\bigr). \end{alignat}