Aug
12
revised Determinant of block matrices with non square matrices
edited body
Jul
28
answered Method of orthogonalization that preserves invertibility
Jul
28
comment Method of orthogonalization that preserves invertibility
Also, you would have to work on columns of one and rows of the other. Is that what you mean by acting the same either way?
Jul
28
comment Method of orthogonalization that preserves invertibility
I imagine just about any matrix is a counter example, and any method that chooses one vector from the matrix to keep and orthogonalize the other vectors with will not do what you ask. Because choosing a vector from the inverse in a similar manner will definitely not coincide with any choice in the main matrix (unless of course that vector is already orthogonal from all others). I would say no such method exists unless it calculates and works with the inverse as well.
Jul
2
awarded  Curious
Jul
1
awarded  Necromancer
May
16
comment What are some good iPhone/iPod Touch/iPad Apps for mathematicians?
looks like the for iOS link is gone.
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?