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From: leox on 20 Dec 2009 16:21
1/((1-x)(1-y) (1-xy))=\sum_{i,j=0} min(i+1,j+1) x^i y^j
From: TCL on 20 Dec 2009 16:27
On Dec 20, 4:21 pm, leox <leonid...(a)gmail.com> wrote:
> 1/((1-x)(1-y) (1-xy))=\sum_{i,j=0} min(i+1,j+1) x^i y^j

What are the constrains on x,y? Like |xy|<1,....
From: leox on 20 Dec 2009 16:53
On 20 çÒÄ, 23:27, TCL <tl...(a)cox.net> wrote:
> On Dec 20, 4:21špm, leox <leonid...(a)gmail.com> wrote:
>
> > 1/((1-x)(1-y) (1-xy))=\sum_{i,j=0} min(i+1,j+1) x^i y^j
>
> What are the constrains on x,y? Like |xy|<1,....

|x|<1, |y|<1
From: Henry on 20 Dec 2009 17:30
On 20 Dec, 21:53, leox <leonid...(a)gmail.com> wrote:
> On 20 çÒÄ, 23:27, TCL <tl...(a)cox.net> wrote:
>
> > On Dec 20, 4:21Å¡pm, leox <leonid...(a)gmail.com> wrote:
>
> > > 1/((1-x)(1-y) (1-xy))=\sum_{i,j=0} min(i+1,j+1) x^i y^j
>
> > What are the constrains on x,y? Like |xy|<1,....
>
> |x|<1, |y|<1

Does (\sum_{i,j>=0} x^i y^j) * (\sum_{k>=0} x^k y^k) help?
From: achille on 20 Dec 2009 22:29
On Dec 21, 5:21 am, leox <leonid...(a)gmail.com> wrote:
> 1/((1-x)(1-y) (1-xy))=\sum_{i,j=0} min(i+1,j+1) x^i y^j

Same trick, split the sum into two cases
i <= j (introduce index l = 0..\infty : j = i+l )
and j < i (introduce index m = 1..\infty : i = j+m ) to get

\sum_{i=0} (i+1) (xy)^i \sum{l=0} y^l
+ \sum_{j=0} (j+1) (xy)^j \sum{m=1} x^m

and simplify....
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