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jack
  • Member for 9 years, 8 months
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asked
Decomposing a planar graph
comment
Covering the surface below a convex function
I think your solution works for any function $f$ continuous and nonincreasing.
accepted
Covering the surface below a convex function
comment
Covering the surface below a convex function
(+1) Maybe the area of the rectangle $[0, a_j] \times [b_{j+2}, b_{j+1}]$ can be considered to improve your method.
comment
Covering the surface below a convex function
Thank you! I have some doubts, doesn't the diagonal of the leftmost rectangle always intersect $F$? Also could you please provide more details why the procedure stops after a finite number of steps?
revised
Covering the surface below a convex function
added 18 characters in body
asked
Covering the surface below a convex function
awarded
 Notable Question
comment
Tiling a Jordan polygon
@Wolfgang No, it's not. The parallelograms are not necessarily equiangular.
revised
Tiling a Jordan polygon
added 12 characters in body
comment
Tiling a Jordan polygon
Thanks for the reference. However in this problem it's assumed that this specific simple polygon $P$ can be tiled with parallelograms. It's an initial condition.
revised
Tiling a Jordan polygon
added 4 characters in body
asked
Tiling a Jordan polygon
awarded
 Yearling
comment
$2$-adic order bound for $P(x)$
Thank you, @Will Sawin.
accepted
$2$-adic order bound for $P(x)$
comment
$2$-adic order bound for $P(x)$
@LSpice: question edited according to Will Sawin comment.
revised
$2$-adic order bound for $P(x)$
added 13 characters in body; edited title
asked
$2$-adic order bound for $P(x)$
accepted
Are there infinitey many $n$ dividing $\epsilon_11^n + \epsilon_22^n + \cdots + \epsilon_kk^n$?
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