I'm trying to to make the below expression simpler, and it would be great if it could be expressed as something like $(x+y)^k$.

$$ \sum_{i=0}^k\binom{n+1}i\binom{m+1}{k-i}x^iy^{k-i} $$

The number $\binom{n+1}i$ is the coefficient of $x^i$ in $(1+x)^{n+1}$, and similarly $\binom{m+1}{k-i}$ is the coefficient of $y^{k-i}$ in $(1+y)^{m+1}$. The expression above is in fact the $k$th Chern class $c_k$ of the direct sum of two projective spaces ($CP^n$ and $CP^m$), but the problem may be viewed as completely separate from differential geometry.

I've tried to see if the coefficients match up with coefficients in expansions like $(1+x+y)^{n+m}$, but I can't see any immediate pattern. So, my question is, is it possible to simply the above expression?

  • 2
    $\begingroup$ Something I notice right off is that the two-variable nature of this is pretty slim, since you can factor out an overall $y^k$ and define $z=x/y$ to get a generating function in $z$ alone. (Obviously the final expression will probably look more symmetric with $x$ and $y$, but for purposes of analysis I think $z$ is preferable.) $\endgroup$ – Semiclassical Jul 29 '14 at 17:41
  • $\begingroup$ Got something from the answer below? $\endgroup$ – Did Aug 13 '14 at 12:01
  • $\begingroup$ @Did not really, I mean that's the position that I started from, since the Chern class of a direct sum is the product of the Chern classes of the summands. I abandoned this question since I didn't get any other answers, but I guess yours is as close as I can get. $\endgroup$ – Jānis Lazovskis Aug 13 '14 at 12:14

This is the coefficient of $t^k$ in the polynomial $(1+tx)^{n+1}\cdot(1+ty)^{m+1}$.

Not sure one can go further in full generality.

  • $\begingroup$ One can play a bit with something like rewriting this into a product of powers of $\dfrac{1+tx}{1+t y}$ and $(1+t x)(1+ty)$ which might be useful if $n$ and $m$ are far apart v. nearly the same. But beyond that... $\endgroup$ – Semiclassical Jul 29 '14 at 17:51

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