322 reputation
39
bio website lit-et-raire.blogspot.com.es
location Bilbao, Spain
age 60
visits member for 1 year, 10 months
seen Apr 28 '13 at 13:41

Just an amateur with very limited general knowledge; not only limited math knowledge.


Jul
2
awarded  Curious
Jan
7
awarded  Yearling
Sep
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awarded  Necromancer
Feb
26
comment Cumulative probabilities
I did not; thanks.
Feb
26
revised Cumulative probabilities
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Feb
26
revised Cumulative probabilities
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Feb
26
asked Cumulative probabilities
Feb
19
comment Least power. Squares again
Ok; i see know. But it would have been clearer to say Product () -Product () = 2f(0) = constant. I thought, first, c was an arbitray external constant. This is why i did not iunderstand.
Feb
19
comment Least power. Squares again
Though there is the short posibility of g(0)=2(2k+1)=4k+2 with no solution.
Feb
19
comment Least power. Squares again
I do not understand quite well the step where it is said that Product(x+ai)-Product(x+bi) is a constant
Feb
19
comment Least power. Squares again
Andreas, your proof seems correct to me. Just you should not say g(0) is odd "avant la lettre". In my opinion g(0) must be odd because u and v have distint parity and we must have uv = 2^(k/2-1)g(0)^2 so g(0) must be odd. Also i agree k must be a multiple of 4, because half the multiplied terms are negative and a square is always positive. The only small doubt i have is the validity/generality of the choosen x= (k-1)/2, a non integer value for x; to give the full proof.
Feb
18
revised Least power. Squares again
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Feb
18
comment Least power. Squares again
Thank you for the answers. I thought i had a proof by induction for the case k=4 but it was not a correct proof. Sorry for it.
Feb
17
revised Least power. Squares again
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Feb
17
asked Least power. Squares again
Feb
17
comment Examples of apparent patterns that eventually fail
See:lit-et-raire.blogspot.com.es/2013/02/…
Feb
17
revised Examples of apparent patterns that eventually fail
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Feb
17
revised Examples of apparent patterns that eventually fail
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Feb
5
awarded  Enthusiast
Feb
3
comment Many kinds of Infinitely many
Your math argument begins well, the inductive part, but then there are errors. But it could also be i am unable to understand your math style.