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visits | member for | 2 years, 10 months |
seen | 1 hour ago | |
stats | profile views | 636 |
Dec 22 |
comment |
Why do we teach calculus students the derivative as a limit?
I've read through the whole dialog in comments about the sine function, and I'm still mystified. Among the mathematicians who got it wrong, what was the wrong answer that they gave? What was the reasoning that they offered? |
Dec 10 |
comment |
Probability that a self-avoiding walk on $\mathbb{Z}^3$ closes to a polygon
@JosephO'Rourke: I would bet you a nickel that the length before reaching a cul de sac has a distribution that drops off at least as fast as an exponential. As time goes on, the probability per unit time of being trapped should increase, because you've built up a trail that could help to trap you. Based on your simulation, I'd guess that the expected length is on the order of $10^4$. |
Dec 8 |
revised |
Probability that a self-avoiding walk on $\mathbb{Z}^3$ closes to a polygon
added 18 characters in body |
Dec 8 |
revised |
Probability that a self-avoiding walk on $\mathbb{Z}^3$ closes to a polygon
deleted 2 characters in body |
Dec 8 |
revised |
Probability that a self-avoiding walk on $\mathbb{Z}^3$ closes to a polygon
added 1 character in body |
Dec 8 |
revised |
Probability that a self-avoiding walk on $\mathbb{Z}^3$ closes to a polygon
added 13 characters in body |
Dec 8 |
answered | Probability that a self-avoiding walk on $\mathbb{Z}^3$ closes to a polygon |
Dec 4 |
comment |
Is non-existence of the hyperreals consistent with ZF?
Abraham Robinson suggested that ZF and ZFC were in some sense constructed exactly so as to allow us to do analysis on the reals. From that point of view, it's not surprising that in ZF(C) the reals exist and are unique, whereas ZF doesn't make the hyperreals exist, and ZFC doesn't make the hyperreals unique. |
Dec 1 |
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Prove existence of different programs printing each other code
@JoelDavidHamkins: OK, maybe it would be helpful if you could edit the question and put in what you think is the correct statement of the hypothesis. Currently it doesn't quantify over all languages, and it doesn't refer to Turing completeness. |
Dec 1 |
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Prove existence of different programs printing each other code
[...] begins with a character that is never a legal first character for source code. |
Dec 1 |
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Prove existence of different programs printing each other code
Shouldn't the hypothesis to be proved be of the form "for any programming language with property P, programs exist that print each other's code?" And should P be Turing-completeness? If you omit the initial quantifying phrase, then a programming language can be any function that takes a string (source code) as input and gives a string as output. One can then come up with both trivial examples and trivial counterexamples, simply by defining the programming language in a special way. Turing-completeness probably isn't sufficient, either, e.g., you could have a language whose output always [...] |
Oct 14 |
awarded | Good Question |
Oct 14 |
accepted | What did Rolle prove when he proved Rolle's theorem? |
Oct 14 |
awarded | Popular Question |
Oct 14 |
awarded | Nice Question |
Oct 14 |
revised |
What did Rolle prove when he proved Rolle's theorem?
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Oct 14 |
asked | What did Rolle prove when he proved Rolle's theorem? |
Oct 6 |
comment |
The distribution of the shortest path through $n$ points
Maybe I'm just dense, but I don't understand what you mean by the shortest path through a set of points. |
Oct 4 |
awarded | Popular Question |
Oct 3 |
comment |
Why do roots of polynomials tend to have absolute value close to 1?
Here's a different way of stating the heuristic. Say the coefficients $a_i$ are IID with mean 1 and variance $\sigma^2$. Then for fixed $z$, $P(z)$ has a (complex) variance given by a geometric sum. For large $n$ and $|z|\gtrsim1$ this variance has real and imaginary parts that grow approximately like $|z|^{2n}$, and therefore the distribution gets too wide, and the probability of $|P|<\epsilon$ falls like $|z|^{-2n}$. This drops off so fast that the prob. of having $|z|$ significantly greater than 1 is small. But the problem is isomorphic under circle inversion, so the same holds for $|z|<1$. |