Editing Bell polynomials (section)

===Convolution identity===

For sequences ''x''<sub>''n''</sub>, ''y''<sub>''n''</sub>, ''n'' = 1, 2, ..., define a [[convolution]] by:

:<math>(x \mathbin{\diamondsuit} y)_n = \sum_{j=1}^{n-1} {n \choose j} x_j y_{n-j}.</math>

The bounds of summation are 1 and ''n''&nbsp;−&nbsp;1, not 0 and ''n'' .

Let <math>x_n^{k\diamondsuit}\,</math> be the ''n''th term of the sequence

:<math>\displaystyle\underbrace{x\mathbin{\diamondsuit}\cdots\mathbin{\diamondsuit} x}_{k \text{ factors}}.\,</math>

Then{{Sfn|Cvijović|2011}}

:<math>B_{n,k}(x_1,\dots,x_{n-k+1}) = {x_n^{k\diamondsuit} \over k!}.\,</math>

For example, let us compute <math> B_{4,3}(x_1,x_2) </math>.  We have

:<math> x = ( x_1 \ , \ x_2 \ , \ x_3 \ , \ x_4 \ , \dots ) </math>

:<math> x \mathbin{\diamondsuit} x = ( 0,\  2 x_1^2 \ ,\  6 x_1 x_2 \ , \  8 x_1 x_3 + 6 x_2^2 \ , \dots ) </math>

:<math> x \mathbin{\diamondsuit} x \mathbin{\diamondsuit} x = (  0 \ ,\ 0 \  , \ 6 x_1^3 \ , \ 36 x_1^2 x_2 \ , \dots ) </math>

and thus,

:<math> B_{4,3}(x_1,x_2) = \frac{ ( x \mathbin{\diamondsuit} x \mathbin{\diamondsuit} x)_4 }{3!} = 6 x_1^2 x_2. </math>