A question on Taylor development of multivariate function How can I express the Taylor development of $$f(t+\Delta t, x+h)$$
When one of the variables is fixed, I can find it but with both of them varying, I have no idea what to do.
$$f(t+\Delta t, x)=f(t,x) + \Delta t \cfrac{\partial f}{\partial t}(t,x)+(\Delta t)^2 \cfrac{\partial^2 f}{\partial t^2}(t,x) + O((\Delta t)^3)$$
and 
$$f(t, x+h)=f(t,x) + h \cfrac{\partial f}{\partial x}(t,x)+h^2 \cfrac{\partial^2 f}{\partial x^2}(t,x) + O(h^3)$$
 A: One approach to work out the multi-variable Taylor expansion of $f(\vec{v})$ about a point $\vec{v}_0$ is to reduce it to the single variable case by writing the function
$$ g(s) = f(\vec{v}_0 + s (\vec{v} - \vec{v}_0)) $$
then
$$ g(s) = \sum_{n=0}^{+\infty} \frac{s^n}{n!} g^{(n)}(0) $$
The repeated derivatives of $g$ are repeated directional derivatives of $f$: (if this isn't clear, try working out $g'(s)$ yourself by whatever means you like)
$$ g^{(n)}(s) = (\nabla_{\vec{v} - \vec{v}_0})^n f(\vec{v}_0 + s (\vec{v} - \vec{v}_0)) $$
If you've only seen directional derivatives defined for unit vectors, the generalization to all vectors is the obvious one: $\nabla_{\vec{w}} h = \vec{w} \cdot \nabla h$
So the Taylor series is
$$ f(\vec{v}_0 + s (\vec{v} - \vec{v}_0)) = \sum_{n=0}^{+\infty} \frac{s^n}{n!} \left( \nabla_{\vec{v} - \vec{v}_0}\right)^n f (\vec{v}_0) $$
which simplifies to, if it converges,
$$ \begin{align} f(\vec{v}) &= \sum_{n=0}^{+\infty} \frac{1}{n!} \left( \nabla_{\vec{v} - \vec{v}_0}\right)^n f (\vec{v}_0) 
\\ &= \sum_{n=0}^{+\infty} \frac{1}{n!} \left( (\vec{v} - \vec{v}_0) \cdot \nabla \right)^n f (\vec{v}_0) 
\end{align}
$$
A: $$f(t+\Delta{t},\, x+h)=\sum\limits_{n=0}^{\infty}{{\frac{1}{n!}}\;{\sum\limits_{k=0}^{n}{\binom{n}{k}}\frac{\partial^n{f}}{\partial{x}^k {}\partial{y}^{n-k}}(\Delta{t})^k h^{n-k}}} \tag{*}$$
Denote $u=(t,\,x),\;\; \Delta{u}=(\Delta{t},\,h), \;F(\lambda)=f(u+\lambda\Delta{u}) \;\;(\lambda\in \mathbb{R}).$ 
 Expansion $(*)$ can be obtained from Taylor expansion
$$F(\lambda)=\sum\limits_{n=0}^{\infty}{\frac{1}{n!}}F^{(n)}(0)\,\lambda^n$$
by differentiating $F(\lambda)$ by $\lambda$ and then putting $\lambda=1.$
