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 Operations with fractional index between + and * ? Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 10/09/2009, 08:38 PM This is only a sketch, yet; I didn't compute values. Also there is one conceptional problem. Consider the binary operations add(x,b) = x+b , mult(x,b)=x*b , pow(x,b) = x^b (the version with left associativity) and their iterable notations add(x,b,h)=x+b*h, mult(x,b,h) = x*b^h , pow(x,b,h) = x^b^h It may look overartificial to find the schröder-functions for fractional iterates of add(), mult() and pow(), since it is trivial to insert fractional heights h into the formulae. But for more intense study let's try to express this with the use of schröder-functions. Always we have the general form of composition g ° b^h ° f or g( b^h * f(x)) where f is the appropriate schröder-function and g its inverse. add(x,b,h) can be expressed as logb ( b^(b*h) * (b^x)) where the schröder-function is b^x and its inverse the logb (=log to base b) Explicitely we have by this logb( b^(b*h) * (b^x)) = logb( b^(x+b*h)) = x + b*h Now we look at the mult-function. The schröder-function is simply the id-function id(b^h*id(x)) = x*b^h The pow-function has the logb-function as schröder-function: b^( b^h * logb(x)) = b^(logb( x^b^h)) = x^(b^h) Now we see, that the schröder-functions for the three basic operations are the iterates of the exp-function: add() --> exp_b°1(x) = b^x mul() --> exp_b°0(x) = id(x) pow() --> exp_b°-1(x) = logb(x) where the iteration-heights of the exp-function may serve as integer index with an unlucky offset. But let's ignore that unusual offset. What we immediately see is, that we could study which sense it makes to define operations fractionally indexed between add() and mul() based on fractional iterates of the exp-/log-function. However, for one inconsistency I didn't find a smooth workaround yet: for add() the cofactor to the schröder-function is b^(b*h), while for mult() and pow() it is b^h only. I also have not yet numerical examples, and think I'll need some more time for this (I'm short, courses begin next week), but thought, perhaps someone other could already look at this. What do you think? Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 10/10/2009, 12:11 AM (This post was last modified: 10/10/2009, 12:17 AM by Gottfried.) Hmm, couldn't resist.... I just tried this for some numerical data without further checking, whether this all can eventually give a meaningful concept. Anyway. I've considered the ten fractionally indexed oprations between mul() and pow(). Base is sqrt(2), initial value x0 is 2.0 because with x0=1 the power-operation would give trivial result. The iteration-height h is 1,2 and 3. The (fractional) operation-index is k; k=0 means mul(x0,sqrt(2),h), k=1 means pow(x0,sqrt(2),h) Here are the results: Code:´          h=1                h=2             h=3 ´    |  2.0* b^1.0      2.0* b^2.0      2.0* b^3.0                             k  | 2.82842712475   4.00000000000   5.65685424949  <--- mul(2,sqrt(2),h)  , direct computation     -----------------------------------------------------                           0.0  2.82842712475   4.00000000000   5.65685424949  = mul()                   0.1  2.81156857833   4.00000000000   5.75322682321                         0.2  2.79480389679   4.00000000000   5.85853861394                         0.3  2.77814247936   4.00000000000   5.97272443872                         0.4  2.76159351333   4.00000000000   6.09569942066                         0.5  2.74516595391   4.00000000000   6.22787081195                         0.6  2.72886850525   4.00000000000   6.37054764165                         0.7  2.71270960247   4.00000000000   6.52607412279                         0.8  2.69669739492   4.00000000000   6.69762735731                         0.9  2.68083973075   4.00000000000   6.88877250162                         1.0  2.66514414269   4.00000000000   7.10299330132  = pow()             -----------------------------------------------------                          | 2.66514414269   4.00000000000   7.10299330132    <--- pow(2,sqrt(2),h)  , direct computation      |  2.0^ b^1.0       2.0^ b^2.0      2.0^ b^3.0 Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 10/15/2009, 08:44 PM (This post was last modified: 10/15/2009, 08:45 PM by bo198214.) Gottfried, this is really an interesting approach. So your suggestion is for the ladder of left-bracketed hyperoperations. I need to ponder about in some quiet hour, if I get some time. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/17/2009, 09:20 PM Let me rewrite this so I understand it better. (1) $({+}b)^h(x) = x + bh = \mathcal{S}[{+}b]^{-1}(b^h\mathcal{S}[{+}b](x))$ (2) $({\times}b)^h(x) = xb^h = \mathcal{S}[{\times}b]^{-1}(b^h\mathcal{S}[{\times}b](x))$ (3) $(\!{\uparrow}b)^h(x) = x^{b^h} = \mathcal{S}[\!{\uparrow}b]^{-1}(b^h\mathcal{S}[\!{\uparrow}b](x))$ and (4) $\mathcal{S}[{+}b](x) = a^{x/b}$ for almost any a. (5) $\mathcal{S}[{\times}b](x) = x$ (6) $\mathcal{S}[\!{\uparrow}b](x) = \log_a(x)$ for almost any a. so what I think you mean is that you can set $a=b^b$ in (4), and $a=b$ in (6) so that there is a common base? That is quite interesting. Honestly, I had never seen that before. This effectively reduces the problem of "nice" noninteger lower hyperoperations to the problem of "nice" noninteger tetration. I am well aware of another similar situation with the commutative hyperoperations, where the general form can be written: $H_{n+2}(a, b) = \exp^{n}\left(\log^{n}(a)\log^{n}(b)\right)$ this can also be considered a way of reducing noninteger hyperoperations to noninteger tetration. Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 10/17/2009, 09:41 PM (This post was last modified: 10/17/2009, 11:29 PM by Gottfried.) @henryk - well, take your time @andrew - yes, i think you got this right. @all... well, if the multiplicative coefficient to the schröder-function, which is of the form b^h, would be consistent, I would have made "a thing of it". But since their forms for the three operations are not constant and also not "linear" (in some sense) I doubt this all is eventually much helpful. Just try to continue that sequence to higher integer indexes.. it looks to me as some well known try to extend the sequence of operations with some symmetric approach, as Andy has referred to. So my first impression was, that this all might be -besides its nice appeal from the fractional indexes between mul() and pow() - somehow trivial. Actually, if I don't get an intriguing and practicable idea how to "linearize" the structure of the cofactor of the schröder-function, then I feel this all as somehow "suboptimal" (though extremely interesting & animating...). But may be, my notorious trend to perfectionisnm keeps me away from recognizing some present useful aspect... You see: I'm not having the feeling of a proper standing here... [... including some updating...] Gottfried Helms, Kassel andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/20/2009, 08:42 PM Just had another thought on this matter. In order for this interpolation to work, you should also check to see if it applies for ranks above 3, for example (to use a combination Knuth-Bromer-Mueller arrow system), we have the first 5 lower hyperoperations are: $\{a + b, ab, a{\uparrow}b, a{\downarrow}{\uparrow}b, a{\downarrow}{\downarrow}{\uparrow}b\}$, and so we should check that: $\mathcal{S}[{\downarrow}{\uparrow}b](x) = \log_a(\log_a(x))$, and $\mathcal{S}[{\downarrow}{\downarrow}{\uparrow}b](x) = \log_a(\log_a(\log_a(x)))$ and I have a feeling that this is not the case. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/21/2009, 01:30 AM (This post was last modified: 10/21/2009, 01:31 AM by andydude.) So I checked up on it, and no, it is not the case. Also, I have some bad news. If we assume the Schroder functions (accordingly, the Abel functions) are iterated exponentials, then that means that the associated hyperoperation hierarchy would have the recursion rule: $HE_n(x, y) := \exp(HE_{n-1}(\log(x), y))$ which (if n=2 is multiplication) requires that (n=1) is $HE_1(x, y) = x + \log(y)$, which is not addition, so this scheme does not form a hyperoperation hierarchy. However, if I might draw your attention to the Tetration Ref. section 2.3.4, the recursion rule: $HR_n(x, y) := \exp(HR_{n-1}(\log(x), \exp^{n-3}(y)))$ has many of the same characteristics, and satisfies addition (n=1), mul (n=2), and powers (n=3). The only difference is in the heighest exponent. $HR_n(x, y) = HE_n(x, \exp^{\circ\left({n-2 \atop 2}\right)}(y))$ « Next Oldest | Next Newest »

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