Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Define a transform on polynomials which is linear and acts on each monomial as $$\widehat{z^k} = \frac{(1+z)(2+z)\ldots(k+z)}{k!}.$$ Does anyone know whether this has a name (and therefore has been studied)?

(Also, not sure what to tag this with... please suggest/edit.)

share|improve this question
    
The right hand side is just $ \binom {k+z} k $ or its suitable generalization to noninteger $z$, for what it's worth. –  Jiahao Chen May 13 '12 at 1:46
    
I've seen $(z+1)\cdots(z+k)$ called "$z$ to the $k$ rising", although this also sometimes means $z\cdots (z+k-1)$, and denoted "$z^{\overline k}$". From this perspective, binomial coefficients like $\binom{z+k}{k}$ are divided rising powers. (More common are "falling" powers, so that $\binom{x}{k}$ is a divided falling power.) The reason to invent rising and falling powers is to have good basis for differences. Differentiation of polynomials is locally nilpotent, and so you can write it as a Jordan block with zeros on the diagonal. That picks out the usual (divided) monomial basis. Rising ... –  Theo Johnson-Freyd May 13 '12 at 2:31
    
... (divided) powers have the same structure for $f(z) \mapsto f(z) - f(z-1)$. So the only thing that seems strange to me about your transformation is the division by $k!$. Or rather, to me divided powers are most natural, so I would have expected $z^k/k! \mapsto \binom{z+k}{k}$, and not $z^k \mapsto \binom{z+k}{k}$. The operator that is like differentiation whose basis consists of non-divided powers is the map $f(z) \mapsto \frac1z\bigl(f(z)-f(0)\bigr)$. This map is not translation invariant, but does turn up occasionally. –  Theo Johnson-Freyd May 13 '12 at 2:38
    
I don't know if this terminology is standard, but I would call it an inverse Beta transform because, although off by a factor of 1, your expression above can be rewritten as $$\frac{\Gamma(z+k+1)}{\Gamma(k+1)\Gamma(z+1)}$$ –  Suvrit May 13 '12 at 23:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.