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Let me first state an example: Let $X$ be the multiplication operator on the polynomials in $x$ defined by $Xf(x)=xf(x)$ and let $D$ be the differentiation operator defined by $Df(x) = f'(x).$ Recently I noticed that apparently there are two different "traditions" to express $(X + D)^n $. In one tradition, concerned with special polynomials, a theorem of Burchnall relates it to a variant of the Hermite polynomials in the form $$(X + D)^n = \sum_{j = 0}^{n }{n\choose j} H_{n-j}(X)D^j. $$ Another tradition is concerned with "normal ordering" of operators. Here a theorem of Mikhailov states that

$$(X + D)^n = \sum_{m = 0}^{n }\sum_{j = 0}^{Min(m,n-m) }{n\brack m}_{j} X^{m-j}D^{n-m-j}, $$

where ${n\brack m}_{j}$ are called "Weyl binomial coefficients". It turns out that both theorems say precisely the same thing. But seemingly the different traditions are unaware of this fact.

I would be interested in other examples of this sort.


I think the real reason for this situation is the use of different "languages" describing the same things. Originally the theorem about normal ordering has been derived for "abstract" operators satisfying $ DX=XD+1 $ instead of using the equivalent "concrete" differentiation and multiplication operators for which the situation seems rather evident.

So my question could also be formulated to look for examples of different "languages" describing the same facts.

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If you're looking for a particular answer, I think you should refine the question to say so. If you are looking for a sorted list of answers, then you should use the "edit" link and change the question to "Community Wiki" mode. – S. Carnahan Apr 29 '11 at 14:28
Most if not all theorems in mathematics can be formulated as "the following statements are equivalent: ...". So you are looking for equivalent statements that some people do not recognize as being equivalent? – Franz Lemmermeyer Apr 29 '11 at 16:24
Yes in some sense. I am interested in theorems proved in different research areas which have been found before someone saw their equivalence. – Johann Cigler Apr 30 '11 at 7:43

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