Taylor polynomial 3rd degree in 3 variables? I have to calculate taylor polynomial 3rd degree in 3 variables for this function in point (0,0,0):
$f\left(x,y,z\right)=\left(x^{2}+z \right)\cdot e^{xz+y^{2} }    $
I dont know how to expand formula for this. I found some general formulas but I just got lost when I started. Could someone write that formula?
 A: You don't need to use the general Taylor's formula:
$$f(x,y,z)=f(0,0,0)+\sum_{k=1}^n\Bigl(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}\Bigr)^{\rlap k}\cdot f \;{(0,0,0)}+o\bigl(\lVert(x,y,z)\rVert^n\bigr).$$
First, as $x^2+z$ is its own Taylor's polynomial, it is enough to expand $\mathrm e^{xz+y2}$ at order $2$.
Second, as the argument of the exponential is homogeneous of degree $2$, we only need to expand $\mathrm e^u$ at order $1$. So 
\begin{align*}(x^{2}+z)\,\mathrm e^{xz+y^{2}}&=(x^2+z)(1+xz+y^2+o\bigl(\lVert(x,y,z)\rVert^2\bigr)\\&=
z+x^2+xz^2+y^2z+ x^3z+x^2z^2+(x^2+z)o\bigl(\lVert(x,y,z)\rVert^2\bigr)\\
&=z+x^2+xz^2+y^2z+o\bigl(\lVert(x,y,z)\rVert^3\bigr).
\end{align*}
A: I will not calculate it. I will just write a formula.
Let $T_f^{(3)} (x)$ denote the Taylor polynomial of order 3 in $(0,0,0)$. For convenience, let us denote $x=x_1 , y=x_2, z=x_3$. Then 
\begin{align*}
T^3 f(x) & = f(0,0,0)+\sum_{k=1}^{3} \frac{\partial f}{\partial x_k} (0,0,0)(x_i -0)\\
&\phantom{=..}+\frac{1}{2!}\sum_{i,j=1}^3 \frac{\partial^2 f}{\partial x_i\partial x_j}(0,0,0)(x_i-0)(x_j-0)\\
&\phantom{=..}+\frac{1}{3!}\sum_{i,j,k=1}^3 \frac{\partial^3 f}{\partial x_i\partial x_j\partial x_k}(0,0,0)(x_i-0)(x_j-0)(x_k-0)
\end{align*}
To think this formula, recall the one-dimensional one. This multi-dimensional one is proved by using one-dimensional Tayler expansion and chain rule. 
Proof gives a way to calculate something in this case. 
A: Assuming you never heard of the multivariable Taylor formula, you can anyway solve as follows.
First consider $y,z$ as parameters and find the Taylor development around $x=0$:
$$f(x)=(x^2+z)e^{xy+y^2},\\
f'(x)=(2x+y(x^2+z))e^{xy+y^2},\\
f''(x)=(y^2(x^2+z)+4xy+2)e^{xy+y^2},\\
f'''(x)=((x^2+z)y^3+6xy^2+6y)e^{xy+y^2}.$$
Then
$$f_{yz}(0)=ze^{y^2},\\
f'_{yz}(0)=yze^{y^2},\\
f''_{yz}(0)=(y^2z+2)e^{y^2},\\
f'''_{yz}(0)=(zy^3+6y)e^{y^2}.$$
This allows you to write
$$f_{yz}(x)=ze^{y^2}+yze^{y^2}x+(y^2z+2)e^{y^2}\frac{x^2}2+(zy^3+6y)e^{y^2}\frac{x^3}{3!}+\cdots$$
Now you can repeat the process with $y$ as a variable and $x,z$ as two parameters, giving
$$ze^{y^2}=z+zy^2+z\frac{y^4}2+\cdots,\\
yze^{y^2}x=zyx+zy^3x+\cdots\\\cdots$$
As it turns out, after this step all transcendental functions of $z$ have disappeared and you end-up with the requested third degree polynomial.
