How works multiindex? Can anyone explain me how works multiindex?
For example here:
\begin{equation}\begin{aligned}
&|\overline{r}|= r_0 + r_1 + \cdots + r_n\\
&\sum_{j=1}^{m}\sum_{|\overline{r}|=\mu_j}a_{ij}\left(\frac{\partial \tau}{\partial{x_0}} \right)^{r_0} \left(\frac{\partial \tau}{\partial{x_1}} \right)^{r_1} \cdots \left(\frac{\partial \tau}{\partial{x_n}} \right)^{r_n}
\end{aligned}\end{equation}
And sometimes I see this note: $|\overline{r}| \leq \mu_j$ 
Thank you in advance for explaining! 
 A: The multi-index is an $n$-tuple of numbers $$r = (r_1,r_2,...r_n)\;\;\;\;r_i \in \mathbb{N}\;\;\forall i$$
(so is different from a vector for the fact that all of it's components have to be positive integers) where you can define some operations such as sum and difference, absolute value, power:


*

*$r+s = (r_1+s_1,\dots, r_n+s_n)$

*$|r| = r_1+r_2+\dots+r_n$

*$x^r = x_1^{r_1}x_2^{r_2}\dots x_n^{r_n}$


With this facts you can define, in a compact way, lots of mathematical concept like partial derivatives. I don't want to give you a list of all the things, for that there's a very good page on Wikipedia.
Let us now look at your formula. Without any context is difficult to give it a precise explanation but let me make some hypothesis: $$\vec{\mu} = (\mu_0, \mu_1, \mu_2) = (6,10,6)\;\;\;\text{vector}\\ r = (r_0,r_1,r_2) = (2,1,3)\;\;\;\text{multi-index}\\ \mathbf{x} = (x_0,x_1,x_2)\;\;\text{where}\;\; x_n\;\;\text{is our coordinate system}\\a_{ij}\;\;\text{is a matrix}$$  
From your formula, the index $i$ is not summed over so it remains at the end of the calculation. Bare in mind that $|r| =  2+1+3 = 6$ 
\begin{align}
&B_i=\sum_{j=1}^{3}\sum_{|\overline{r}|=\mu_j}a_{ij}\left(\frac{\partial \tau}{\partial{x_0}} \right)^{r_0} \left(\frac{\partial \tau}{\partial{x_1}} \right)^{r_1}\left(\frac{\partial \tau}{\partial{x_2}} \right)^{r_2} = \\&a_{i1} \left(\frac{\partial \tau}{\partial{x_0}} \right)^{2} \left(\frac{\partial \tau}{\partial{x_1}} \right)^{1}\left(\frac{\partial \tau}{\partial{x_2}} \right)^{3} + \;\;\;\;\;\text{for }j=1: \mu_1 = 6 = |r| 
\\&a_{i2} \left(\frac{\partial \tau}{\partial{x_0}} \right)^{2} \left(\frac{\partial \tau}{\partial{x_1}} \right)^{1}\left(\frac{\partial \tau}{\partial{x_2}} \right)^{3} + \;\;\;\;\;\text{for }j=2: \mu_2 = 10 \neq |r| \text{ so we don't count it} 
\\&a_{i3} \left(\frac{\partial \tau}{\partial{x_0}} \right)^{2} \left(\frac{\partial \tau}{\partial{x_1}} \right)^{1}\left(\frac{\partial \tau}{\partial{x_2}} \right)^{3} \;\;\;\;\;\;\;\;\;\text{for }j=2: \mu_3 = 6 =  |r| 
\end{align}
So in the end we have 
$$
B_i = a_{i1} \left(\frac{\partial \tau}{\partial{x_0}} \right)^{2} \left(\frac{\partial \tau}{\partial{x_1}} \right)^{1}\left(\frac{\partial \tau}{\partial{x_2}} \right)^{3} + a_{i3} \left(\frac{\partial \tau}{\partial{x_0}} \right)^{2} \left(\frac{\partial \tau}{\partial{x_1}} \right)^{1}\left(\frac{\partial \tau}{\partial{x_2}} \right)^{3}
$$
so if, for example $a_{ij} = \delta_{ij}$, the identity matrix, we get that 
$$ B_i = \left\{\begin{matrix} 2\left(\frac{\partial \tau}{\partial{x_0}} \right)^{2} \left(\frac{\partial \tau}{\partial{x_1}} \right)\left(\frac{\partial \tau}{\partial{x_2}} \right)^{3}&\text{if }i=1,3 \\ 0 &\text{if }i=2\end{matrix}\right.$$
I hope this will help you!
Edit
As @Steffen Plunder made me notice, the formula that we were trying to study is quite more complex that how I made it look like. That's because the second summation $\sum_{|r|=\mu_j}$ requires ones to choose beforehand only the value of $|r|$ (that probably is given) and then sum ove all the multi-indices whose absolute value is $|r|$ and it's equal to $\mu_j$ (which from my example is $|r|=6$ with $r=(r_0,r_1,r_2)$) so like $$(0,0,6), (1,0,5), (1,1,4), (6,0,0), (0,5,1), (2,2,2), \dots$$ and so on. But I think that the concept of how does the multi-index works is still given by my previous answer, the problem now is on the formula itself. In the comment to my answer is a more precise formula for the sum over all the multi-indices whose absolute value is $|r|$ and is less or equal to some value $k$. Thanks for the correction!   
