# Continuum between addition, multiplication and exponentiation?

I noticed this old post which attempts to find the shades of grey between a linear and log scale where results are between zero and one.

However, I was looking for the more general case where we find the continuum between the operators themselves - add, multiply, and exponentiation or even higher.

The function could be a hyperoperation$(n,x,y)$, where $x$ and $y$ are the numbers to operate on, and $n$ is allowed to be a non-integer number for the type of operator. So $n=1$ would be 'addition', $n=2$ would be multiplication, and $n=3$ would be exponentiation. But one would also have $n=2.5$ to be the shade of grey BETWEEN multiplication and exponentiation, or $n=3.5$ or $4$ for beyond exponentiation and tetration).

I want the formula to work with real numbers, not just integers. Any thoughts?

These related posts may also be of interest:

Why are addition and multiplication commutative, but not exponentiation?

Algorithm for tetration to work with floating point numbers

• See tetration.org/Ackermann/index.html for my look a continuous Ackermann function. For example, A(2,2,x)=4 is likely an identity. – Daniel Geisler May 7 '15 at 2:39
• @DanielGeisler: I can follow around half of what's on that page. Interesting the function converges to 2 or less under various scenarios. – Dan W May 7 '15 at 8:58
• I guess that this is directly linked with the "non-integer ranks" problem, we can also call it, as @DanielGeisler says, continuous Ackermann Function. I wrote a detailed answer about this few weeks ago: math.stackexchange.com/questions/1227761/… – MphLee May 9 '15 at 13:22

There is (at least) one option - however depending on whether a solution for the problem of "fractional iteration of logarithm" is available (aka: "Tetration". Note that in the tetration-forum there is an exchange on various approaches to this problem).

Assume for the moment that a version of that fractional iteration is available, and let us denote $$\log_b^{\circ h}(x)$$ for the $$h$$-times iterated logarithm to some given base $$b$$, including the opportunity to have $$h$$ also fractional.
Assume also that we have this well-defined for the interval $$h=-1 \cdots +1$$ (so for $$h=-1$$ this means actually exponentiation) then $$f_b(x,y,0) = \exp_b^{\circ 0} ( \log_b^{\circ 0} (x)+\log_b^{\circ 0} (y)) = x+y \\ f_b(x,y,1) = \exp_b^{\circ 1} ( \log_b^{\circ 1} (x)+\log_b^{\circ 1} (y)) = x \cdot y$$ and for $$0 we have then some operation in between well defined: $$f_b(x,y,h) = \exp_b^{\circ h} ( \log_b^{\circ h} (x)+\log_b^{\circ h} (y)) = x \{\circ _h \} y$$ where the operation-symbol of the circle and the index $$h$$ means that interpolated operation between "+" (for $$h=0$$) and "*" (for $$h=1$$).

Now let's see, whether our assumption of the existence of such fractional iterations is justified/can be realized:

• If the base $$b$$ is $$\exp(1)$$ then one can resort to Hellmuth Kneser's solution for the fractional iteration of $$\exp(x)$$, which provides an analytic expression for this, real for real arguments, as far as I understood things correctly.
• But one can also use the base $$b$$ from the interval $$1 and might use the easier and better understandable Schröder-mechanism for a solution for the iterative loagrithm/exponentiation. Especially the procedere for the base $$b=\sqrt{2}$$ is discussed and described in broadth & width in the tetration-forum.
• In the wikipedia-entry "tetration" there is some introductory information for that basic problem.

(Note: this is just one ad-hoc-approach; it does not, for instance, provide a solution comparable with one which might be expected, when we aproach the problem via the interpolation of the Ackermann-function as indicated by Daniel Geisler)

Addition, multiplication and "half multiplication" tables for $$x,y=1 \ldots 10$$ . I used the base $$b=t^{1/t} =\sqrt2 \approx 1.414$$ with $$t_l=2$$ resp $$t_u=4$$ for the logarithm and the exponentiation. For the half-exponentials/half-logarithms the formulae relate on lower fixpoint $$t_l=2$$ for $$x \le 3$$ and on upper fixpoint $$t_u=4$$ for $$x>3$$ and use the well known Schröder-mechanism for the half-iterate.

[update]: In the comment MphLee provided a link into wikipedia, where this "hyperoperation"-scheme is mentioned as due to A. Bennet (but not then generalized to fractional orders / fractional iteration heights of the $$\log()$$ and the $$\exp()$$ )

addition $$h=0$$ ($$\text{"+"}=\circ _0$$)

  o_0="+"|        1        2        3        4        5        6        7        8        9       10  |
-  +        -        -        -        -        -        -        -        -        -        -  +
1  |  2.00000  3.00000  4.00000  5.00000  6.00000  7.00000  8.00000  9.00000  10.0000  11.0000  |
2  |  3.00000  4.00000  5.00000  6.00000  7.00000  8.00000  9.00000  10.0000  11.0000  12.0000  |
3  |  4.00000  5.00000  6.00000  7.00000  8.00000  9.00000  10.0000  11.0000  12.0000  13.0000  |
4  |  5.00000  6.00000  7.00000  8.00000  9.00000  10.0000  11.0000  12.0000  13.0000  14.0000  |
5  |  6.00000  7.00000  8.00000  9.00000  10.0000  11.0000  12.0000  13.0000  14.0000  15.0000  |
6  |  7.00000  8.00000  9.00000  10.0000  11.0000  12.0000  13.0000  14.0000  15.0000  16.0000  |
7  |  8.00000  9.00000  10.0000  11.0000  12.0000  13.0000  14.0000  15.0000  16.0000  17.0000  |
8  |  9.00000  10.0000  11.0000  12.0000  13.0000  14.0000  15.0000  16.0000  17.0000  18.0000  |
9  |  10.0000  11.0000  12.0000  13.0000  14.0000  15.0000  16.0000  17.0000  18.0000  19.0000  |
10  |  11.0000  12.0000  13.0000  14.0000  15.0000  16.0000  17.0000  18.0000  19.0000  20.0000  |
-  +        -        -        -        -        -        -        -        -        -        -  +


"half multiplication" $$h=0.5$$ ($$\circ _{0.5}$$) with base $$b=\sqrt2$$ using Schröder-mechanism

 o_0.5     |        1        2        3        4        5        6        7        8        9       10  |
-----  +        -        -        -        -        -        -        -        -        -        -  +
1  |  1.42469  2.55554  3.67186  4.77842  5.87792  6.97204  8.06187  9.14818  10.2315  11.3123  |
2  |  2.55554        4  5.39214  6.75005  8.08354  9.39847  10.6987  11.9868  13.2648  14.5340  |
3  |  3.67186  5.39214  7.02490  8.60114  10.1372  11.6429  13.1245  14.5863  16.0315  17.4625  |
4  |  4.77842  6.75005  8.60114  10.3749  12.0939  13.7715  15.4163  17.0342  18.6296  20.2057  |
5  |  5.87792  8.08354  10.1372  12.0939  13.9820  15.8183  17.6136  19.3754  21.1090  22.8185  |
6  |  6.97204  9.39847  11.6429  13.7715  15.8183  17.8034  19.7397  21.6361  23.4990  25.3332  |
7  |  8.06187  10.6987  13.1245  15.4163  17.6136  19.7397  21.8094  23.8332  25.8183  27.7704  |
8  |  9.14818  11.9868  14.5863  17.0342  19.3754  21.6361  23.8332  25.9783  28.0799  30.1443  |
9  |  10.2315  13.2648  16.0315  18.6296  21.1090  23.4990  25.8183  28.0799  30.2932  32.4652  |
10  |  11.3123  14.5340  17.4625  20.2057  22.8185  25.3332  27.7704  30.1443  32.4652  34.7408  |
-  +        -        -        -        -        -        -        -        -        -        -  +


multiplication $$h=1$$ ($$\text{"*"}=\circ _{1}$$)

  o_1="*"|        1        2        3        4        5        6        7        8        9       10  |
-  +        -        -        -        -        -        -        -        -        -        -  +
1  |  1.00000  2.00000  3.00000  4.00000  5.00000  6.00000  7.00000  8.00000  9.00000  10.0000  |
2  |  2.00000  4.00000  6.00000  8.00000  10.0000  12.0000  14.0000  16.0000  18.0000  20.0000  |
3  |  3.00000  6.00000  9.00000  12.0000  15.0000  18.0000  21.0000  24.0000  27.0000  30.0000  |
4  |  4.00000  8.00000  12.0000  16.0000  20.0000  24.0000  28.0000  32.0000  36.0000  40.0000  |
5  |  5.00000  10.0000  15.0000  20.0000  25.0000  30.0000  35.0000  40.0000  45.0000  50.0000  |
6  |  6.00000  12.0000  18.0000  24.0000  30.0000  36.0000  42.0000  48.0000  54.0000  60.0000  |
7  |  7.00000  14.0000  21.0000  28.0000  35.0000  42.0000  49.0000  56.0000  63.0000  70.0000  |
8  |  8.00000  16.0000  24.0000  32.0000  40.0000  48.0000  56.0000  64.0000  72.0000  80.0000  |
9  |  9.00000  18.0000  27.0000  36.0000  45.0000  54.0000  63.0000  72.0000  81.0000  90.0000  |
10  |  10.0000  20.0000  30.0000  40.0000  50.0000  60.0000  70.0000  80.0000  90.0000  100.000  |
-  +        -        -        -        -        -        -        -        -        -        -  +


halfway between multiplication and commutative(!) exponent $$h=1.5$$($$\text{"??"}=\circ _{1.5}$$)

  3/2  |         1        2        3        4        5        6        7        8        9       10  |
-  +         -        -        -        -        -        -        -        -        -        -  +
1  |  0.856374  1.42469  1.95136  2.45125  2.93192  3.39780  3.85179  4.29587  4.73152  5.15984  |
2  |   1.42469  4.00000  7.02490  10.3749  13.9820  17.8034  21.8094  25.9783  30.2932  34.7408  |
3  |   1.95136  7.02490  13.8017  21.9113  31.1374  41.3357  52.4022  64.2572  76.8377  90.0920  |
4  |   2.45125  10.3749  21.9113  36.4401  53.5757  73.0498  94.6618  118.254  143.699  170.889  |
5  |   2.93192  13.9820  31.1374  53.5757  80.7547  112.281  147.853  187.226  230.200  276.603  |
6  |   3.39780  17.8034  41.3357  73.0498  112.281  158.529  211.396  270.556  335.735  406.696  |
7  |   3.85179  21.8094  52.4022  94.6618  147.853  211.396  284.818  367.723  459.769  560.660  |
8  |   4.29587  25.9783  64.2572  118.254  187.226  270.556  367.723  478.278  601.830  738.030  |
9  |   4.73152  30.2932  76.8377  143.699  230.200  335.735  459.769  601.830  761.492  938.375  |
10  |   5.15984  34.7408  90.0920  170.889  276.603  406.696  560.660  738.030  938.375  1161.30  |
-  +         -        -        -        -        -        -        -        -        -        -  +


Commutative(!) expon $$h=2$$($$\text{"??"}=\circ _{2}$$)

   2  |        1        2        3        4        5        6        7          8          9         10  |
-  +        -        -        -        -        -        -        -          -          -          -  +
1  |  1.00000  1.00000  1.00000  1.00000  1.00000  1.00000  1.00000    1.00000    1.00000    1.00000  |
2  |  1.00000  4.00000  9.00000  16.0000  25.0000  36.0000  49.0000    64.0000    81.0000    100.000  |
3  |  1.00000  9.00000  32.5416  81.0000  164.317  292.874  477.416    729.000    1058.95    1478.85  |
4  |  1.00000  16.0000  81.0000  256.000  625.000  1296.00  2401.00    4096.00    6561.00    10000.0  |
5  |  1.00000  25.0000  164.317  625.000  1761.64  4107.92  8404.54    15625.0    27000.1    44040.9  |
6  |  1.00000  36.0000  292.874  1296.00  4107.92  10543.5  23393.4    46656.0    85775.3    147885.  |
7  |  1.00000  49.0000  477.416  2401.00  8404.54  23393.4  55587.9    117649.    227926.    411822.  |
8  |  1.00000  64.0000  729.000  4096.00  15625.0  46656.0  117649.    262144.    531441.  1.00000E6  |
9  |  1.00000  81.0000  1058.95  6561.00  27000.1  85775.3  227926.    531441.  1.12138E6  2.18701E6  |
10  |  1.00000  100.000  1478.85  10000.0  44040.9  147885.  411822.  1.00000E6  2.18701E6  4.40409E6  |
-  +        -        -        -        -        -        -        -          -          -          -  +


(Remark: because double-log of number $$1$$ would be negative infinity, I replaced this by approximation and insertion of $$1+1e-20$$ instead of $$1$$)

We can even define negative-half index of operation:

  -1/2  |        1        2        3        4        5        6        7        8        9       10  |
-  +        -        -        -        -        -        -        -        -        -        -  +
1  |  2.55554  3.32163  4.13229  4.97809  5.85143  6.74644  7.65862  8.58454  9.52153  10.4676  |
2  |  3.32163  4.00000  4.73255  5.50980  6.32348  7.16675  8.03408  8.92107  9.82421  10.7407  |
3  |  4.13229  4.73255  5.39214  6.10270  6.85632  7.64601  8.46581  9.31075  10.1767  11.0604  |
4  |  4.97809  5.50980  6.10270  6.75005  7.44491  8.18072  8.95160  9.75241  10.5787  11.4268  |
5  |  5.85143  6.32348  6.85632  7.44491  8.08354  8.76647  9.48824  10.2439  11.0289  11.8394  |
6  |  6.74644  7.16675  7.64601  8.18072  8.76647  9.39847  10.0719  10.7823  11.5252  12.2968  |
7  |  7.65862  8.03408  8.46581  8.95160  9.48824  10.0719  10.6987  11.3644  12.0652  12.7973  |
8  |  8.58454  8.92107  9.31075  9.75241  10.2439  10.7823  11.3644  11.9868  12.6460  13.3386  |
9  |  9.52153  9.82421  10.1767  10.5787  11.0289  11.5252  12.0652  12.6460  13.2648  13.9184  |
10  |  10.4676  10.7407  11.0604  11.4268  11.8394  12.2968  12.7973  13.3386  13.9184  14.5340  |
-  +        -        -        -        -        -        -        -        -        -        -  +


and the negative $$-1$$- operation

  -1  |        1        2        3        4        5        6        7        8        9       10  |
-  +        -        -        -        -        -        -        -        -        -        -  +
1  |  3.00000  3.54311  4.16993  4.87350  5.64386  6.46968  7.33985  8.24439  9.17493  10.1248  |
2  |  3.54311  4.00000  4.54311  5.16993  5.87350  6.64386  7.46968  8.33985  9.24439  10.1749  |
3  |  4.16993  4.54311  5.00000  5.54311  6.16993  6.87350  7.64386  8.46968  9.33985  10.2444  |
4  |  4.87350  5.16993  5.54311  6.00000  6.54311  7.16993  7.87350  8.64386  9.46968  10.3399  |
5  |  5.64386  5.87350  6.16993  6.54311  7.00000  7.54311  8.16993  8.87350  9.64386  10.4697  |
6  |  6.46968  6.64386  6.87350  7.16993  7.54311  8.00000  8.54311  9.16993  9.87350  10.6439  |
7  |  7.33985  7.46968  7.64386  7.87350  8.16993  8.54311  9.00000  9.54311  10.1699  10.8735  |
8  |  8.24439  8.33985  8.46968  8.64386  8.87350  9.16993  9.54311  10.0000  10.5431  11.1699  |
9  |  9.17493  9.24439  9.33985  9.46968  9.64386  9.87350  10.1699  10.5431  11.0000  11.5431  |
10  |  10.1248  10.1749  10.2444  10.3399  10.4697  10.6439  10.8735  11.1699  11.5431  12.0000  |
-  +        -        -        -        -        -        -        -        -        -        -  +


And to see that different ways to define the fractional iterate of exponentiation lead to different multiplication-tables I show here the "half-multiplication" taken by base $$b=2$$ and the implementation via something which I called "polynomial tetration" (which is based on eigendecomposition of the truncated carlemanmatrices, and seems to approximate the Kneser-solution when the truncation allows larger matrix-sizes)

  "half-multiplication" (h=0.5) by "polynomial tetration", base b=2, matrixsize=16x16
1.55799  2.75402  3.92565  5.08186  6.22771  7.36568  8.49732  9.62384  10.7460  11.8646
2.75402  4.31898  5.81047  7.25673  8.67206  10.0641  11.4377  12.7963  14.1424  15.4777
3.92565  5.81047  7.57736  9.27212  10.9176  12.5260  14.1053  15.6608  17.1964  18.7150
5.08186  7.25673  9.27212  11.1905  13.0426  14.8449  16.6080  18.3393  20.0440  21.7259
6.22771  8.67206  10.9176  13.0426  15.0851  17.0659  18.9982  20.8910  22.7508  24.5824
7.36568  10.0641  12.5260  14.8449  17.0659  19.2138  21.3042  23.3478  25.3524  27.3236
8.49732  11.4377  14.1053  16.6080  18.9982  21.3042  23.5440  25.7302  27.8714  29.9745
9.62384  12.7963  15.6608  18.3393  20.8910  23.3478  25.7302  28.0521  30.3235  32.5519
10.7460  14.1424  17.1964  20.0440  22.7508  25.3524  27.8714  30.3235  32.7197  35.0681
11.8646  15.4777  18.7150  21.7259  24.5824  27.3236  29.9745  32.5519  35.0681  37.5322


[update]: Appendix.

In the (in WP) cited article of A Bennett (1915) for the interpolation to fractional orders (there $$n$$ here $$h$$) A.Bennet misses the possibility to take another base $$b$$ for iterated logarithm and exponentiation, where $$b \ne e$$ and actually $$1 where we can iterate infinitely and also have a "regular" method for interpolation to fractional orders/iterates using E. Schröder's method which I've used here. See the screenshot of some paragraph in A. Bennett's short paper:

• Note that both $t$ and $b$ are close to 1. In dynamics it is helpful to locate the fixed points of a system. Let $A(x,y,1)=x+y$ and $A(x,y,2)=x*y$. $A(1,1,n)=1$ except when $n=1$. So for small values of $z$ we might expect the dynamics of $A(1+z,1+z,n)$ to be relatively simple. – Daniel Geisler May 8 '15 at 18:34
• @DanW Define $\log^{\circ n}(x)$ in this way: $\log^{\circ 0}(x)=x$, $\log^{\circ 1}(x)=\log(x)$ and $\log^{\circ n+1}(x)=\log(\log^{\circ n}(x))$. – MphLee May 10 '15 at 13:44
• @MphLee: Perfect thanks! (so the latter of what I was thinking was correct). I just stumbled as the usual definition of the iterated logarithm is different, but he did use h-times which made me pause for thought. – Dan W May 10 '15 at 13:48
• @GottfriedHelms: Oh btw, I'm the guy who discovered the Mandelbulb formula. You mentioned that here - glad you like the pics! :) – Dan W May 10 '15 at 14:29
• @MPhLee - I've take your link to WP into my answer and also added a link to (and a little screenshot of) an article of A. Bennet (1915) which was linked to in WP. Nice to see historical approaches and even overcome/workaround historical limitations... – Gottfried Helms Jan 9 '18 at 0:09

I was looking for the more general case where we find the continuum between the operators themselves - add, multiply, and exponentiation or even higher.

I think that this is stictly related to the extension of the Ackermann Function to non-integer values: I guess we can call the problem "non-integer ranks problem" because it has to do with Hyperoperations with non integer rank/index $s$

$$H_s(x,y)=x[s]y=x\uparrow^{s-2}y=G(s,x,y)$$

where

$H_1(x,y)=x[1]y=x\uparrow^{-1}y=G(1,x,y)=x+y$

$H_2(x,y)=x[2]y=x\uparrow^{0}y=G(2,x,y)=xy$

$H_3(x,y)=x[3]y=x\uparrow^{1}y=G(3,x,y)=x^y$

$H_4(x,y)=x[4]y=x\uparrow^{2}y=G(4,x,y)={}^{y}x$ (Tetration)

Example $x$, $y$ and $z$ values for $x\uparrow^\alpha y=z$ where $\alpha\in \Bbb R-\Bbb N$

Anyways in short words what I say there is that there is not, as far as I know, a known and accepted method to find non-integer ranks hyperoperations.

• You're right in saying my question is practically the same as the one asked in the link you've supplied. Sad in one way to see that it hasn't been solved yet. Happy in another way that it provides mystery and adventure for mathematicians like you, and also in regards to the kind of progress math might see if it was solved. A tick to any answer so far remains elusive for now, but let me know if a solution arises! (And have an upvote in the meantime!) – Dan W May 10 '15 at 14:11