
#1
Feb1408, 01:52 AM

PF Gold
P: 4,182

I was curious if there was a noncommutative version of Pascal's triangle for operators (such as those used in brahket notation)
The important note is that (a + b)^2 = a^2 + b^2 + ab +ba where ba != ab 



#2
Feb1408, 02:13 AM

PF Gold
P: 4,182

I'm guess if it's an even power such as (x + y)^4
you can always assume the crossterms are split between their orderings, so from pascal's triangle, which gives the coefficients 1 4 6 4 1, the middle coefficient corresponds to 3x^2y^2 + 3y^2x^2. Not quite sure how to handle odd degrees without doing pages of algebra. I'm posting in this forum because tensors have the noncommutative property and I happen to be applying brahket notation so I thought it fit. Apologies if not. 



#3
Feb1408, 07:07 AM

P: 406

This is interesting.
Going by (a+b)^3 = a^3 + a^2*b+aba + ab^2 + ba^2 + bab + b^2*a + b^3 It would seem that the only possible row in the "triangle" for general noncommuntative numbers would be "1 1 1 1 1 1 1 1". The exact form may depend on just how uncommutative the number are. 



#4
Feb1508, 02:49 AM

PF Gold
P: 4,182

Pascal's Triangle for noncommutative? 



#5
Feb1508, 06:55 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

The point of Pascal's triangle is that the i,j entry simply counts the number of ways you can order i "x"s and ji "y"s. If your multiplication is not commutative, those do not add. All terms are distinct. That's why ObsessiveMathFreak says you just get "1 1 1 ...".



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