alancalvitti
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 Apr22 comment How to define a compatible relation using a cover Binary reflexive, symmetric relations are also called tolerances: math.stackexchange.com/questions/268726/… Apr16 awarded Yearling Mar20 awarded Curious Mar8 comment What exactly *is* the Riemann zeta function? You wrote "makes sense for complex numbers as long as $\text{Re } s > 1$". Doesn't it converge for all complex numbers except 1? Feb27 comment What are some theorems that currently only have computer-assisted proofs? @TonyK, I reviewed some papers by Schaffer et al, searching for "forced" and only found 1 mention eg in (cs.ucr.edu/~eamonn/205/checkers.pdf) "Independent research has discovered a 10-piece database position requiring a 279-ply move sequence to demonstrate a forced win (a ply is one move by one player) (15). This is a conservative bound; the win length has not been computed for the more difficult (and more interesting) database positions." // most matches for "forced" refer to forced moves - so again, I ask, what makes forced wins a crucial factor in computer assisted proofs? Feb26 comment What are some theorems that currently only have computer-assisted proofs? @TonyK, of course, perfect play means non-losing, but that's a basic definition, why is that the important part? Jan31 comment What is the square root of 1 cm @TheChaz2.0, what unit equals 1 when squared? Jan31 comment What is the square root of 1 cm "don't have a ready physical interpretation, but can sometimes be useful as intermediates" - sounds the same as complex numbers (where btw, physical interpretation lagged the algebra). Jan29 comment Intuition behind functional dependence I didn't assume anything that you didn't assume, since $C^1$ is a subcategory of $C^0$ which is in turn a subcat of $Set$, the category of sets and functions. Jan28 comment Intuition behind functional dependence Yes, a continuous, differentiable function is still a function. For example, Jittorntrum's Implicit Function Theorem (J Opt Th App 1978) does not assume differentiability: if a continuous function $F(x,y)$ is locally injective in some neighborhood of a point, then $F(x,y)=0$ has a unique solution $x=G(y)$ with $G$ also continuous. I don't think there's a combinatorial proof w/o continuity assumption though. Jan27 comment Intuition behind functional dependence Check the table here: en.wikipedia.org/wiki/Relation_algebra - functional is the first item. Functions need not be differentiable or even continuous, but need to be defined everywhere ("left total", or simply "total", or "preserving elements") and the image of each element of a function is also a single element of the codomain ("single valued" or "reflecting distinctions"). Jan25 comment Intuition behind functional dependence Functional dependence means left-total and single-valued relation. Differential structure is not part of the definition. Jan22 comment Intuition behind functional dependence What if the gradients don't exist? It's implicit in your assumption, why not mention it from the outset, eg $C^1$ or similar Jan21 comment Relating categorical properties of arrows @magma: a well known categorist told me n-lab is the opposite of category theory, and I agree: whereas the implication diagrams in this question are posets/DAGs, n-lab is a sprawling graph w/o beginning or end. Jan21 comment Convert universal quantification to existential quantification What if there are no students? Jan21 comment Relating categorical properties of arrows @magma, edited title and added clarification in body, is that ok? Jan21 revised Relating categorical properties of arrows Rephased title, clarified body Jan21 comment Relating categorical properties of arrows Marco, thanks, what about split-? In categories w/ choice, "epics split" but otherwise? Jan19 asked Relating categorical properties of arrows Jan18 awarded Popular Question