# Tetration Forum

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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.
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.
i think you just posted the ancient Arithmetic-Geometric Mean ?

tommy1729
lol , seems i was typing my reply while you were posting ...

what a waste of time ...
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.
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<b,

m1 of a and b = a + (b-a)*0.25 (0.25 of the way between a and b, judged arithmetically)
m2 of a and b = a * (b/a)^0.25 (0.25 of the way between a and b, judged geometrically)

Now plug m1 and m2 into a and b, and iterate until you get something stable (i.e. m1 = m2 to whatever degree of precision you need)

This seems like a good generalisation to real values between 0 and 1!

(03/10/2011, 11:42 PM)JmsNxn Wrote: [ -> ]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.

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.
(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<b,

m1 of a and b = a + (b-a)*0.25 (0.25 of the way between a and b, judged arithmetically)
m2 of a and b = a * (b/a)^0.25 (0.25 of the way between a and b, judged geometrically)

Now plug m1 and m2 into a and b, and iterate until you get something stable (i.e. m1 = m2 to whatever degree of precision you need)

This seems like a good generalisation to real values between 0 and 1!

Yes! I like that a lot.
(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 [4] x and 3 [4] x looked quite good.
(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 [4] x and 3 [4] x looked quite good.

Hey Martin, we are not talking about the function f(x) = a [4] 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.
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