Apr
3
accepted $\Delta^d m^n =d! \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$ Is this a new formula?
Mar
29
comment $\Delta^d m^n =d! \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$ Is this a new formula?
I have posted more details on the MO Question
Mar
29
comment Can this sum be shown to be zero? $\sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k} = 0$
I think I can show it: use the identity $\left[ m+1 \atop k \right] = \left[ m+2 \atop k+1\right] - (m+1)\left[ m+1 \atop k+1\right]$ repeatedly until all terms are zero by the inversion formula. Anyone wants to show the details can get a free accept from me.
Mar
29
revised Can this sum be shown to be zero? $\sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k} = 0$
edited title
Mar
29
comment Can this sum be shown to be zero? $\sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k} = 0$
Oh well, I guess I have not found a simpler proof then. For the other post that is.
Mar
28
revised $\Delta^d m^n =d! \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$ Is this a new formula?
Inversion formula did not apply where I thought it did.
Mar
28
asked Can this sum be shown to be zero? $\sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k} = 0$
Mar
28
answered $\Delta^d m^n =d! \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$ Is this a new formula?
Mar
28
revised $\Delta^d m^n =d! \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$ Is this a new formula?
notation tweak
Mar
28
comment $\Delta^d m^n =d! \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$ Is this a new formula?
Quite cool and interesting (+1). It may take me some time to feel like I fully understand your answer though (after studying more with generating functions of Stirling numbers). I will add later the path which led me to the formula, it was a fun one.
Mar
27
comment $\Delta^d m^n =d! \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$ Is this a new formula?
Notice that this is the inversion formula when $n=0$ and $z=0$, which gives the correct restult of $1$.
Mar
27
comment $\Delta^d m^n =d! \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$ Is this a new formula?
I should have mentioned I am reading the book right now, and have not seen the identity. It has a fairly similar one $\sum_{k} \left[ n \atop k\right] { k \brace m } = {n \choose m}(n-1)^\underline{n-m}$ on page 265. I would be interested in an actual proof as opposed to "probably there is a proof/could be done". Good book though, yes.
Mar
27
asked $\Delta^d m^n =d! \sum_{k} \left[ m \atop k \right] { {k+n} \brace m + d}(-1)^{m+k}$ Is this a new formula?
Mar
14
comment Rank-one modification of graph Laplacian
Eigenvectors should not be considered to have magnitude since they represent a class of vectors, a direction not a point. Are you then asking only about the eigenvalues? If so, if it is a fact then it is just a fact, are you asking how it relates to the perturbations?
Mar
14
comment Concrete Mathematics, Newton Series and Inversion
Though I am not too sure still since I do not see where $x$ went in the second formula here for $\Delta^k g(x)$. The authors may have done better to just not even mention formula 5.45 there since it does not seem to matter anyway, they just prove things without using it.
Mar
14
answered Concrete Mathematics, Newton Series and Inversion
Mar
14
comment Concrete Mathematics, Newton Series and Inversion
Okay, he shows proof for the statement (your "first one") but I don't see how it is a special case either...
Mar
14
comment Concrete Mathematics, Newton Series and Inversion
He shows how it is a special case on the next page (193). It is not supposed to be readily obvious since it takes five lines to show it.
Mar
12
awarded  Announcer
Mar
3
awarded  Good Answer