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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 5 months |
| seen | Jan 11 at 22:35 | |
| stats | profile views | 78 |
beginning
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Jan 11 |
comment |
Substitution tilings with parallelograms if rhombus is ok you could skew a mathworld.wolfram.com/PerfectRectangle.html |
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Jan 11 |
awarded | Commentator |
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Jan 11 |
comment |
Check if line intersects with circles perimeter I dont think your code cares about the end points of line segments. I added code to my answer. |
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Jan 11 |
revised |
Check if line intersects with circles perimeter added 342 characters in body |
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Jan 11 |
answered | Check if line intersects with circles perimeter |
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Jan 10 |
comment |
Cayley table group visualization this is really really nice! sadly there are too few colors for bigger groups. |
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Jan 10 |
comment |
Back to cauchy sequences. do you mean all caucy sequences of reals are convergent? |
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Jan 10 |
accepted | How does Fraenkel's urelement proof show choice is independent of ZF? |
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Jan 10 |
asked | How does Fraenkel's urelement proof show choice is independent of ZF? |
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Jan 9 |
accepted | Cayley table group visualization |
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Jan 9 |
answered | Substitution in lambda calculus |
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Jan 9 |
comment |
Substitution in lambda calculus [N/x]M is standard notation. |
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Jan 8 |
asked | Cayley table group visualization |
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Jan 8 |
accepted | complete partial order by adjoint functor theorem |
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Jan 8 |
comment |
Why does $(2/p)=\prod_{k=1}^{(p-1)/2}2\cos\left(\frac{2\pi k}{p}\right)$? thanks for the proof @DavidSpeyer |
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Jan 8 |
comment |
Why does $(2/p)=\prod_{k=1}^{(p-1)/2}2\cos\left(\frac{2\pi k}{p}\right)$? why I edelted it: How does this answer the question? – hmIII 6 hours ago |
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Jan 8 |
comment |
complete partial order by adjoint functor theorem wait a second you did it backwards (or forwards if you consider me backwards) starting with $\downarrow$ (what I called $G$)? it seems like knowing explicitly what $\downarrow$ is (downward closed sets) the stuff about adjoint functor theorem distracts from the core of the proof, is that right? Thanks for your answer. In fact all this category theory is irrelevant if we somehow knew the idea of sup union downarrow s = join s. |
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Jan 8 |
revised |
complete partial order by adjoint functor theorem edited body |
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Jan 8 |
asked | complete partial order by adjoint functor theorem |
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Jan 8 |
awarded | Critic |
