# Nick Kidman

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Actress with an interest in the philosophy of science.

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 2d comment how do you read: “$\lim (n-1)/(n-2) = 1$” @adam: I'm not sure if the limit can be approaching something. 2d asked Can we capture all domains of discouse in the predicate logic within categorical logic? Dec16 accepted How or why does intutionistic logic proof negations from within the theory, constructively? Dec16 comment How or why does intutionistic logic proof negations from within the theory, constructively? @MJD: Yeah, is the consistent with "Another way to understand $⊥$ that you might find more intuitive is that it is the type of a program that doesn't return a value, perhaps because it threw a fatal exception that terminated the computation, or because it went into an infinite loop and never completed." Dec15 comment How or why does intutionistic logic proof negations from within the theory, constructively? @Hurkyl: Haha, it gets involved. Well the void function is also interesting, but difference from the $P$'s above is that they can't even run. The "absurdness" comes from it being formally designed for computation once given argument, in a clean and syntactically correct manner, while it can easily be inferred that it can never actually run any arguments. I see the "constructiveness" obscured by the Curry-Howard isomorphism here, we obscure the function idea. But I don't want to beat a dead horse - I understand the situation much better now and will accept an answer tomorrow evening. Dec15 comment How or why does intutionistic logic proof negations from within the theory, constructively? @Hurkyl: Yo, that I know, I used $⊥$ in the question above. It's just surprising he wants it to be a function type in the first line. Not specifying a specific domain, it would be all non-terminating programs over Type. Some lines later he seems to say loop, with domain $a$, should have type $a\to ⊥$. Dec15 comment How or why does intutionistic logic proof negations from within the theory, constructively? Thanks! Is the "$⊥$" in the first line rather supposed to be a negation of a type? Otherwise, if $⊥$ itself is the type of programs which turns forever, then it shouldn't be empty. Regarding the last paragrgraphs, I want to emphasise that I make the case that the proof of $((A→0)×(B→0))→(A+B)→0$, considered as function, doesn't just never output a value like a loop, it can't even be given the totality of its arguments. Related to this, you also say in Haskell $\neg A$ is inhabited for all $A$. So can general Haskell have a reasonable proof interpretation if all negations are already inhabited? Dec15 comment How or why does intutionistic logic proof negations from within the theory, constructively? @ZhenLin: Absurd? I feel it's like a "painting" which can't be looked at. But if you say so. It's probably a semantical question. Dec15 revised How or why does intutionistic logic proof negations from within the theory, constructively? edited body Dec14 revised How or why does intutionistic logic proof negations from within the theory, constructively? added 1441 characters in body Dec14 asked How or why does intutionistic logic proof negations from within the theory, constructively? Dec13 answered What is the derivative of $\ln\left(x^2 + (3/4)x\right)$? Dec13 comment How prove this $\displaystyle\lim_{\tau\to t}f(t,\tau)=\frac{1}{2\pi}\frac{x'_{1}(t)x''_{2}(t)-x'_{2}(t)x''_{1}(t)}{[x'_{1}(t)]^2+[x'_{2}(t)]^2}$ Just a remark: At first glance, the last expression looks like $\left[\frac{1}{4\pi}(x_2'/x_1'-x_1'/x_2')\right]'$, and that outer derivative is a limit $g'=\lim_{h\to 0}\tfrac{1}{h}(g(h,t)-g(0,t))=\lim_{h\to t}[\tfrac{1}{t-h}(g(t-h,t)-g(0,t))]$. Remains to bring $f$ to the implied form in brackets. Dec13 comment Is there a connection between contravariant functors and the axiom of choice? @MartinBrandenburg: It's not even close in the sense that it doesn't provide specific functions? Dec13 comment Is there a connection between contravariant functors and the axiom of choice? @MartinBrandenburg: I agree and it's mostly because I don't want to use names for things I don't properly understand. My idea is that if I postulate that the axiom of choice is true, this implies the existence of kind of an endofunctor which turns arrows between objects $A,B$ in a category $\mathcal C$ in the reversed direction $\mathcal C(A,B)\to\mathcal C(B,A)$. Dec13 asked Is there a connection between contravariant functors and the axiom of choice? Dec12 comment Simple Functional Equation $\frac{f(a)-f(b)}{a-b}\cdot(-a)+f(a)=-ab$ @BeniBogosel: I'm not sure if you understood what I mean. I ment that saying it's important, followed by the sentence about it's relevance in math competitions, makes it sounds like the verification is important because of it's value in math competitions. (I should add that I might just have less premises than you here: If or if it's not importantant in math competition is, for me, a priori independend of anything - they are rules of a game and could just as well be made up). Dec11 comment Simple Functional Equation $\frac{f(a)-f(b)}{a-b}\cdot(-a)+f(a)=-ab$ +1, I like the semantic ambiguity of the last sentence: It sounds like the reason it's important is the fact that you lose points at the Olympiad, not the other way around. Dec10 comment Poems related to mathematics Roses are red, Violets are too, If moving away, At near lightspeed from you. Dec9 comment How can I find this integral? It looks simpler if you push down the $\cos^2(x)$ in the denominator. You get $1$ over $\tan^2(x)+4\tan(x)$.