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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 2 years, 4 months |
| seen | 7 hours ago | |
| stats | profile views | 463 |
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Apr 27 |
comment |
Challenge in combinatorics Exact duplicate of math.stackexchange.com/questions/370800/combinatorics-challenge |
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Apr 26 |
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Integrate linear function over spherical triangle Surely the answer to your edited question would be that it's just the original average, by induction over the level of subdivision? Or am I missing something? |
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Apr 23 |
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special polynomials Comparing the desired forms of the derivatives with the derivatives of the previous level we can establish the requirements that $a=b=c=d \pmod 4$ or (wlog) $a=b=c+2=d+2 \pmod 4$; $e=f=g \pmod 3$; and $h=i \pmod 2$. So a reasonable starting point would be a brute force check for non-trivial solutions in $\mathbb{Z}/12\mathbb{Z}[x]$. |
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Apr 23 |
revised |
special polynomials added 82 characters in body |
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Apr 20 |
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combinations of Rubik cube And to round this off, the centre pieces are the only ones which are affected by the extra information, because the corners must be correctly rotated if the colours match. |
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Apr 20 |
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The half-life of a radio active substance is 7.94 days. Originally there is 230g of this substance. How long before only 15g remain? Are you given a working definition of half-life? |
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Apr 19 |
answered | Finding the value of a polynomial at zero |
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Apr 18 |
awarded | Custodian |
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Apr 18 |
reviewed | Close Turing machine question |
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Apr 12 |
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Galois group of a quintic The discriminant tells you whether the Galois group is a subgroup of the alternating group, which suffices to distinguish $D_{10}$ (yes) and $F_{20}$ (no). |
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Apr 12 |
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Galois group of a quintic There is a mistake: $x^5 -4x + 12 = (x^2+6x+7)(x^3+7x^2+3x+11) \pmod {13}$ |
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Apr 11 |
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Spherical right triangle question What precisely is your definition of "right spherical triangle" and "hypotenuse"? When you have two right angles, which edge is the hypotenuse? |
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Apr 9 |
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Proving that language is not regular by pumping lemma The complement of a regular language is regular (easily demonstrated by inverting which states are accepting), so if the complement is not regular then neither is the original language. And no infinite arithmetic sequence of integers contains only primes. |
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Apr 9 |
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“Infinito”, a combinatorial game with infinite width game-tree Can you argue that on a board of a given size every game is equivalent to a game in which the stones are bounded by some small polynomial of the board size with an additional constraint on which stones may be placed on a given turn? |
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Apr 5 |
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The last few digits of $0^0$ are $\ldots0000000001$ (according to WolframAlpha). In combinatorics it needs to be 1 to avoid adding special cases to things like the binomial theorem. Concrete Mathematics even has an index entry for $0^0$ pointing to the page which discusses this. |
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Apr 3 |
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Hard proof concerning the periodicity of trigonometrical functions. Is that a challenge or just trivial Yes, once you first prove (e.g. by induction on $k$) that $\sin(k\pi) = 0$, but that takes you further out of your way. Surely it would be more straightforward to prove that $sin(2\pi) = 0$, then prove that $2\pi$ is the period (still a missing step - you need to show that $\cos(2\pi) = 1$!), and finally use that to show that it is minimal. |
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Apr 3 |
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Hard proof concerning the periodicity of trigonometrical functions. Is that a challenge or just trivial @DominicMichaelis, that proves that there can be no zero of $\sin$ between $\pi$ and $2\pi$, which is considerably weaker than the statement that $n$ must be a multiple of $\pi$. (It's also sufficient, though, once you show that $\pi$ isn't a period and $2\pi$ is). |
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Apr 1 |
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Solutions to $\binom{n}{5} = 2 \binom{m}{5}$ Maybe I'm being dense, but why must there be no prime $m \lt p \le n$? I can see why there must be no prime $(m \lt p) \wedge (n-4 \le p \le n)$, but that's a much weaker restriction. |
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Mar 31 |
answered | Conjecture to start a proof |
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Mar 31 |
comment |
Solutions to $\binom{n}{5} = 2 \binom{m}{5}$ Considering possible values of $m$ modulo primes from 7 to 31, it's possible to quick-reject all but 8223271875 values $\pmod {247357937827}$ (i.e. to quick-reject about 29 in 30 values). It might be possible to prove that for prime $p$ at least $f(p)$ values of $m \pmod p$ have no possible value of $n \pmod p$ with a function $f$ which grows fast enough to make a heuristic argument for the number of solutions being vanishingly small. |
