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You might be thinking oh great another stupid question, and you're probably right. Coming from minimal education to math I find it hard to figure out notation, recently I have been trying to learn about bell polynomials and have come across the following definition:

$$ B_{n,k}(x_1,x_2......,x_{n-k+1})= \sum \frac{n!}{j_1!j_2!.....j_{n-k+1}!}(\frac{x_1}{1!})^{j_1}(\frac{x_2}{2!})^{j_2}.....(\frac{x_{n-k+1}}{(n-k+1)!})^{j_{n-k+1}} $$

Where:

$j_1+j_2.....+j_{n-k+1} = k$ and $j_1+2j_2+.....+(n-k+1)j_{n-k+1} = n$

So what are the bounds? Why would you make a definition of something this intriguing so vague? I looked for what this means , It says that $\sum = \sum_{I=1}^{n}$ according to wiki, so how would this apply to the summation above?

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  • $\begingroup$ I guess that the sum is taken over $k$ from $1$ to $n$. $\endgroup$ – Sigur Oct 30 '14 at 0:14
  • $\begingroup$ @Sigur Then to determine the values of $j$ I would do what? Also where would I plug in a $k$ value? $\endgroup$ – Eric L Oct 30 '14 at 0:15
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    $\begingroup$ @EricLawson, hum... so I guess that for fixed $k$ you have to consider the sum over all the possible partitions of $k$, that is, all the possible ways to decompose $k$ as sums. $\endgroup$ – Sigur Oct 30 '14 at 0:19
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    $\begingroup$ The first line of the wikipedia article answers your question. en.wikipedia.org/wiki/Bell_polynomials $\endgroup$ – PVAL-inactive Oct 30 '14 at 0:23
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The sigma notation you are used to, $\sum\limits_{k=1}^n$, is just one convention.

It can also be expressed using a set constructor convention as: $\sum\limits_{k\,: k\geq 1, k\leq n}$

This simply means: "sum using integer values of the iterator $k$ which satisfy these equations."

Any equations can be used, as long as they construct a set of values for the iterator(s) of the sum.

In the posted question, the sum is iterated using all the values of the vector $[j_1, j_2, \ldots, j_{n-k+1}]$ that are integer solutions for the two simultaneous equations.

The condition for the sum is added in the "where" clause, because it's just too big to fit neatly under the sigma symbol.

That is all.

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  • $\begingroup$ No wonder it didn't make sense haha $\endgroup$ – Eric L Oct 31 '14 at 20:55

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