Talk:Bell polynomials
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Touchard polynomials
I just noticed that exponential polynomials redirects to Touchard polynomials, but exponential polynomial redirects to Bell polynomials. From what I can tellm the two are closely related, but the relationship is not really spelled out. Can someone who knows more a) make the redirect unambiguous (and as specific as possible) and b) take whatever other measures are useful? Should the articles possibly be merged? Thanks! --Stephan Schulz 09:19, 29 July 2007 (UTC)
: I've just redirected exponential polynomial to Touchard polynomials. Michael Hardy 14:42, 29 July 2007 (UTC)
: I've added an explanation that Touchard polynomials are simply complete Bell polynomials with all arguments being x. Maxal (talk) 19:51, 8 July 2015 (UTC)
Revert
I've reverted [http://en.wikipedia.org/w/index.php?title=Bell_polynomials&curid=572498&diff=291940303&oldid=285853419 this edit] because it left the article worse. The two section headings involved were supposed to be WITHIN the "applications" section, as subsections. Michael Hardy (talk) 11:22, 24 May 2009 (UTC)
Recurrence relation for partial exponential Bell polynomials
Currently, the recurrence formula for the partial exponential Bell polynomials has the form
B_{n,k} = \sum_{i=1}^{n-k+1} \binom{n-1}{i-1} x_i B_{n-i,k-1}
.
I suggest to instead rewrite it in the equivalent, yet slightly more elegant form
B_{n+1,k+1} = \sum_{i=0}^{n-k} \binom{n}{i} x_{i+1} B_{n-i,k}
.
This is certainly a matter of taste, but I believe that the latter form provides better insight on the involved combinatorics. It is also more consistent with with the recurrence relation given above for the complete exponential Bell polynomials, namely
B_{n+1}(x_1, \ldots, x_{n+1}) = \sum_{i=0}^n \binom{n}{i} B_{n-i}(x_1, \ldots, x_{n-i}) x_{i+1}
.
In addition, consistently with the above formula and the rest of the article, it might be preferable to include the arguments of the partial polynomials, eventually giving
B_{n+1,k+1}(x_1, \ldots, x_{n-k+1}) = \sum_{i=0}^{n-k} \binom{n}{i} x_{i+1} B_{n-i,k}(x_1, \ldots, x_{n-k-i+1})
.
I'm doing the edit, please let me know or revert it is you think this is not suitable. Loj19 (talk) 20:54, 17 August 2023 (UTC)