https://www.youtube.com/watch?v=N-KXStupwsc offers an attractive expression for (one) solution of the general cubic, but I can't make it work. Have I entered this incorrectly? In[488]:= Power[#1 - Sqrt[#1^2 + #2^3], (3)^-1] + Power[#1 + Sqrt[#1^2 + #2^3], (3)^-1] - b/3/a &[-b^3/27/a^3 + b c/6/a/a - d/2/a, c/3/a - b^2/9/a/a] Out[488]= -(b/(3 a)) + (-(b^3/(27 a^3)) + (b c)/(6 a^2) - d/(2 a) - Sqrt[(-(b^2/(9 a^2)) + c/(3 a))^3 + (-(b^3/(27 a^3)) + (b c)/( 6 a^2) - d/(2 a))^2])^( 1/3) + (-(b^3/(27 a^3)) + (b c)/(6 a^2) - d/(2 a) + Sqrt[(-(b^2/(9 a^2)) + c/(3 a))^3 + (-(b^3/(27 a^3)) + (b c)/( 6 a^2) - d/(2 a))^2])^(1/3) —rwg
Hello Math-Fun, The digits of a(n) and a(n+1), interleaved in a proper manner, produce a prime. This is the lexicographically earliest infinite sequence of distinct positive terms with this property (I hope): S = 1,3,2,9,5,11,8,21,4,7,6,13,10,19,12,23,18,... We accept two ways of interleaving the digits of A and B (with A ad B sharing the same number of digits – or differing by one). (...) 15 more ligns (and color marks) here on my weblog: https://bit.ly/3aE4wgi Best, É.
Hello Math-Fun The first line of this table is well known; what about extending the table with line #b, line #c, etc. --------------------------------------------- line #a: 10 71 32 23 14 15 16 27 18 19, so far line #b: 20 81 72 53 44 35 26 47 38 29, so far line #c: 40 101 82 73 64 65 56 77 58 39, so far --------------------------------------------- Example: the integer 101 must be read "There are 10 digits 1 so far in the whole table". Which is true (I hope). Best, É.
Hello Math-Fun If you've seen this: https://www.futilitycloset.com/2020/03/13/a-self-describing-table/ ... I'm happy to show here the same task done in base 2, with a 37-line array computed by Gilles Esposito-Farèse: 11 0, 100 1. 1001 0, 1000 1. 10000 0, 1100 1. 10101 0, 10011 1. 11001 0, 11011 1. 11110 0, 100011 1. 100101 0, 101010 1. 101011 0, 110010 1. 110001 0, 111010 1. 111010 0, 1000000 1. 111111 0, 1001010 1. 1001001 0, 1010000 1. 1001111 0, 1011010 1. 1010110 0, 1100011 1. 1011011 0, 1101110 1. 1011111 0, 1111010 1. 1101001 0, 10000001 1. 1110010 0, 10001001 1. 1111001 0, 10010011 1. 10000100 0, 10011010 1. 10001110 0, 10100010 1. 10011000 0, 10101010 1. 10100010 0, 10110010 1. 10101001 0, 10111101 1. 10110001 0, 11000111 1. 10111010 0, 11010000 1. 11000100 0, 11011000 1. 11001011 0, 11100011 1. 11010001 0, 11101111 1. 11011000 0, 11111010 1. 11100100 0, 100000001 1. 11110000 0, 100001000 1. 11111001 0, 100010010 1. 100000101 0, 100011010 1. 100010010 0, 100100001 1. 100011100 0, 100101011 1. 100100110 0, 100110101 1. Gilles backtracked the array on the last 10 lines, at every step, when he was blocked. He tells me that 37 might be easy to beat... and why not with an infinite array! Best, É. (my page is here, with more results (to beat): http://cinquantesignes.blogspot.com/2020/03/peche-la-ligne.html)
Presumably In[488] was intended to relate to the expression at 1:53 in the video; but I have to confess failure to establish any actual correspondence! Incidentally, I recall that the formula is practicably applicable only when there is a single real root. WFL On 3/10/20, Bill Gosper <billgosper@gmail.com> wrote:
https://www.youtube.com/watch?v=N-KXStupwsc offers an attractive expression for (one) solution of the general cubic, but I can't make it work. Have I entered this incorrectly? In[488]:= Power[#1 - Sqrt[#1^2 + #2^3], (3)^-1] + Power[#1 + Sqrt[#1^2 + #2^3], (3)^-1] - b/3/a &[-b^3/27/a^3 + b c/6/a/a - d/2/a, c/3/a - b^2/9/a/a]
Out[488]= -(b/(3 a)) + (-(b^3/(27 a^3)) + (b c)/(6 a^2) - d/(2 a) - Sqrt[(-(b^2/(9 a^2)) + c/(3 a))^3 + (-(b^3/(27 a^3)) + (b c)/( 6 a^2) - d/(2 a))^2])^( 1/3) + (-(b^3/(27 a^3)) + (b c)/(6 a^2) - d/(2 a) + Sqrt[(-(b^2/(9 a^2)) + c/(3 a))^3 + (-(b^3/(27 a^3)) + (b c)/( 6 a^2) - d/(2 a))^2])^(1/3) —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
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Bill Gosper -
Fred Lunnon -
Éric Angelini