meta user
network profile
| bio | website | |
|---|---|---|
| location | California | |
| age | ||
| visits | member for | 2 years, 3 months |
| seen | 5 hours ago | |
| stats | profile views | 694 |
A mediocre professor at a top-notch university making mediocre contributions to this top-notch community
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May 18 |
awarded | Nice Answer |
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Apr 28 |
answered | Pythagorean theorem and its cause |
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Apr 24 |
comment |
Expected length of a sequence that contains all words of a given length. I know I'm getting old when a related question was asked by me and I have no memory of asking it. |
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Apr 24 |
asked | Expected length of a sequence that contains all words of a given length. |
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Apr 21 |
comment |
Prove or disprove isomorphic graphs It should be classified as "group theory" instead of "graph theory." Even if this problem came from graph theory, your presentation of it leaves no trace of that evolution and you've given it as a pure group theory problem. |
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Apr 21 |
answered | Prove or disprove isomorphic graphs |
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Apr 10 |
answered | Classifying Algebraic Structures as Fields |
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Mar 25 |
comment |
Explain to me what these symbols mean. The confusion often arises from the fact that many writers call $dy/dx$ a "symbol" as if it were atomic, but then later start doing algebra with it. This leads to the question, "well, then what is $dy$ really?" |
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Mar 10 |
comment |
Summation of element of a subset and divition I think your two added assumptions are reasonable enough to make the original question interesting. But 8 feels quite arbitrary now. |
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Mar 8 |
comment |
Summation of element of a subset and divition Technically, it is true: the empty set is always a subset of $A$ and 8 divides 0. |
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Mar 6 |
revised |
How to prove $\det(e^A) = e^{\operatorname{tr}(A)}$? edited title |
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Mar 1 |
comment |
Continued Fraction [1,1,1,…] You're welcome. This is the cutest (and most elementary) solution I could think of. Good luck. |
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Feb 26 |
answered | Continued Fraction [1,1,1,…] |
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Feb 26 |
comment |
Continued Fraction [1,1,1,…] I think this is the argument the OP already has but is worried about. But thanks for the lovely typesetting nonetheless. :) |
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Feb 16 |
awarded | Yearling |
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Feb 14 |
answered | Prove that if $n$ is a composite and $p \gt \sqrt[3]n$, then $n/p$ is a prime. |
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Jan 19 |
comment |
Given $N$, find $ab = N$ with $a$ and $b$ as close as possible Correct, which is why I was careful to not claim that it was. If you know an efficient (meaning poly-time) factoring algorithm, please send me a private note and we'll write a paper. Or start a company. Or be captured/assassinated before we can do either... |
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Jan 19 |
comment |
Given $N$, find $ab = N$ with $a$ and $b$ as close as possible Factoring is thought to be hard. Knapsack is NP-Hard, and therefore thought to be hard. If $P = NP$ then all of these are poly-time solvable, but that's unlikely. The best known general factoring algorithms are super-polynomial (on a conventional computer; quantum algorithms are polynomial-time). |
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Jan 19 |
answered | Given $N$, find $ab = N$ with $a$ and $b$ as close as possible |
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Dec 30 |
accepted | Derivation of $e$ |
