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 between addition and multiplication lloyd Junior Fellow  Posts: 10 Threads: 1 Joined: Mar 2011 03/10/2011, 09:10 PM This has probably been thought of before, but here goes anyway. I was thinking about the "sesqui" operation intermediate between adding and multiplying; I'll write "@" here. Obviously a @ b should lie between a+b and ab. Maybe we should take the mean. But which one, arithmetic or geometric? Since one applies to addition and the other to multiplication, why not take both? Then we'll take the mean of these two. But which mean? Again, take both; the proposed value for the sesqui-operation is then the limit of this process when iterated many times. In fact the two values converge quite quickly and for 10-digit precision we usually have convergence within 3 or 4 iterations. Here are some values for a @ a: 1 @ 1 = 1.456791031 2 @ 2 = 4.000000000 3 @ 3 = 7.424041309 4 @ 4 = 11.654328248 5 @ 5 = 16.644985716 6 @ 6 = 22.363401399 7 @ 7 = 28.784583111 8 @ 8 = 35.888457285 9 @ 9 = 43.658368718 10 @ 10 = 52.080163811 11 @ 11 = 61.141591230 12 @ 12 = 70.831889817 13 @ 13 = 81.141493853 14 @ 14 = 92.061815491 15 @ 15 = 103.585079914 16 @ 16 = 115.704197683 17 @ 17 = 128.412664031 18 @ 18 = 141.704478131 19 @ 19 = 155.574077463 20 @ 20 = 170.016283797 21 @ 21 = 185.026258257 22 @ 22 = 200.599463552 23 @ 23 = 216.731631979 24 @ 24 = 233.418738077 25 @ 25 = 250.656975101 26 @ 26 = 268.442734648 27 @ 27 = 286.772588895 28 @ 28 = 305.643275047 29 @ 29 = 325.051681631 30 @ 30 = 344.994836377 31 @ 31 = 365.469895439 32 @ 32 = 386.474133787 I discovered this forum after asking a question recently on sci.math. It looks like people here have been thinking about the same thing: I asked if the next operation after exponentiation should require new numbers, the way that addition/subtraction, multiplication/division, exponentiation/root-taking/logarithms lead from the counting numbers to negative, real and complex numbers respectively. lloyd Junior Fellow  Posts: 10 Threads: 1 Joined: Mar 2011 03/10/2011, 10:51 PM D'oh, I looked at the FAQ and the mean I proposed is covered in detail as a possible basis for the sesqui operation--it is the agm, ArithmeticalGeometricMean in mathematica, also called the Gauss mean. Oh well if I had to invent something that already exists, at least it's something that Lagrange and Gauss also thought of. Back to lurking. Lloyd. tommy1729 Ultimate Fellow     Posts: 1,354 Threads: 328 Joined: Feb 2009 03/10/2011, 10:56 PM i think you just posted the ancient Arithmetic-Geometric Mean ? tommy1729 tommy1729 Ultimate Fellow     Posts: 1,354 Threads: 328 Joined: Feb 2009 03/10/2011, 10:58 PM lol , seems i was typing my reply while you were posting ... what a waste of time ... JmsNxn Long Time Fellow    Posts: 291 Threads: 67 Joined: Dec 2010 03/10/2011, 11:42 PM (This post was last modified: 03/10/2011, 11:44 PM by JmsNxn.) What would happen if we created this: x {0} y = x + y x {0.5} y = x @ y x {1} y = x * y x {2} y = x ^ y And then {0.25} will be the same arithmetic-geometric algorithm of {0} and {0.5}; {0.75} will be the arith-geo-algo of {1} and {0.5}, so on and so forth. We could then solve for x {1.5} n, n E N, since: x {1.5} 2 = x {0.5} x Perhaps Taylor series will be derivable giving us complex arguments. It'd also be very interesting to see what happens with logs, i.e: log(x {1.5} 2) = ? since normal operators undergo a transformation I wonder if something happens for these. lloyd Junior Fellow  Posts: 10 Threads: 1 Joined: Mar 2011 03/11/2011, 12:35 AM Surely, though, {0.25} should be weighted 3/4s towards the arithmetic mean, and 1/4 towards the geometric mean. Ah but is the weighting carried out arithmetically or geometrically? Apply a 3/4 arithmetic : 1/4 geometric weighting there too! And take the limiting case again. In other words, for a {0.25} b, with a a [t] b is smooth (or better analytic) for fixed a and b. I mean you define it on the interval [1,2], i.e. between addition and multiplication, and then you would continue it to the higher operations t>2 by a [t+1] (b+1) = a [t] ( a [t+1] b ) And then it would be interesting whether the curve is (infinitely) differentiable at t=2, t=3, etc. Of course it would between the endpoints, i.e. on (2,3) and (3,4), etc. JmsNxn Long Time Fellow    Posts: 291 Threads: 67 Joined: Dec 2010 03/11/2011, 06:12 PM (03/11/2011, 12:35 AM)lloyd Wrote: Surely, though, {0.25} should be weighted 3/4s towards the arithmetic mean, and 1/4 towards the geometric mean. Ah but is the weighting carried out arithmetically or geometrically? Apply a 3/4 arithmetic : 1/4 geometric weighting there too! And take the limiting case again. In other words, for a {0.25} b, with a a [t] b is smooth (or better analytic) for fixed a and b. I mean you define it on the interval [1,2], i.e. between addition and multiplication, and then you would continue it to the higher operations t>2 by a [t+1] (b+1) = a [t] ( a [t+1] b ) And then it would be interesting whether the curve is (infinitely) differentiable at t=2, t=3, etc. Of course it would between the endpoints, i.e. on (2,3) and (3,4), etc. IIRC, this is exactly what I tried to do, some three years ago or so... It never really worked out, though some sample graphs for 2  x and 3  x looked quite good. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/11/2011, 10:59 PM (03/11/2011, 10:07 PM)martin Wrote: (03/11/2011, 12:42 PM)bo198214 Wrote: For me it seems important that the curve t |-> a [t] b is smooth (or better analytic) for fixed a and b. I mean you define it on the interval [1,2], i.e. between addition and multiplication, and then you would continue it to the higher operations t>2 by a [t+1] (b+1) = a [t] ( a [t+1] b ) And then it would be interesting whether the curve is (infinitely) differentiable at t=2, t=3, etc. Of course it would between the endpoints, i.e. on (2,3) and (3,4), etc. IIRC, this is exactly what I tried to do, some three years ago or so... It never really worked out, though some sample graphs for 2  x and 3  x looked quite good. Hey Martin, we are not talking about the function f(x) = a  x but about the function g(x) = a [x] b. But you are right that we expect the same smoothness also for f(x), which is how Andrew Robbins came to his tetration extension. « Next Oldest | Next Newest »

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