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 Isomorphism of newtonian calculus rules for Non-Newtonian (anti)derivatives of hypers Micah Junior Fellow Posts: 9 Threads: 2 Joined: Feb 2019 02/24/2019, 10:50 PM Dear Forum users sorry for the multiple posts, but I would like to discuss a property that I have suspected for a while.  Putting it into words has been difficult for me, but I think that It can roughly be expressed via the title of this post.  Some time ago, I encountered something called the "product integral" (while reading the original paper on the Cox Proportional Hazards Model).  The notion of the product integral blew my mind, having been given a traditional calculus exposition I quickly hoped to read/learn all I could about the so called multiplicative calculus.   The marriage of these so-called "Non-Newtonian" calculi (Here you can see the wife of one of the originators and proponents of Non-Newtonian calculi, explaining the jist of the product derivative Jane Grossman Discusses Product Calculus - she also makes good piano tutorials for those budding musicians among the tetrologists on the forum ) with hyper-operations could yield some interesting theoretical properties.  For instance, the general definition of an additive derivative is constructed in the following manner:  $\frac{\partial f}{\partial x} = \lim\limits_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}\qquad\qquad(1)$ You may recall the painful (or pleasurable for a weird few like us) activity of evaluating (1) for each function we desired the analytical form of the derivative for.  Then a little later, to much mystery the teacher illuminated the "power rule" which gives the algorithmic procedure that we are more used to when evaluating the derivatives of polynomial expressions.  But when we consider applying the traditional *additive derivative to a function of form (2) the familiar rules are no longer around to help us and we must reduce to applying the limiting definition in (1) in order to construct appropriate analytical representations of the derivative, which is hard because the domain is hard to think about.  $f(x) = x \uparrow \uparrow t = {}^tx\qquad\qquad(2)$  If however we consider the product derivative (a non-newtonian derivative given by (3) below) it may be possibly to use a "product-power rule"  $\partial f^{(\partial x)^{-1}}=\lim\limits_{\Delta x \to 0}\left(\frac{f(x+\Delta x)}{f(x)}\right)^{\frac{1}{\Delta x}}\qquad\qquad(3)$  Digression: If we wish to continue with this particular trend of taking a derivative based on higher and higher orders of hyper operations we could simply continue with this trend of operating to (tetrating to, pentating to, etc...) the multiplicative inverse of the differential in the expression, for instance, we could define a tetrational derivative as follows in (4 - dont you love it when your equation numbers work out) ${}^{\partial x^{-1}}\partial f=\partial f \uparrow \uparrow (\partial x^{-1})= \lim\limits_{\Delta x \to 0} {}^{\frac{1}{\Delta x}}\left(f(x+\Delta x)^{\frac{1}{f(x)}}\right)= \lim\limits_{\Delta x \to 0}(f(x+\Delta x) \uparrow (f(x)^{-1})) \uparrow \uparrow \Delta x^{-1}\qquad\qquad(4)$ Of course, we now can see that the general hyper-k derivative of a function (following this trend) can be expressed using arrow notation as (5).  $\partial f \uparrow^{(k)} (\partial x^{-1}) = \lim\limits_{\Delta x \to 0} (f(x+\Delta x) \uparrow^{(k-1)} f(x)^{-1}) \uparrow^{(k)} \Delta x^{-1}\qquad\qquad(5)$ Back on Track: Back to the original reason for this post (sorry for the momentary digression, I think it may be safely ignored).  If we consider the power rule for the additive derivative of a function it can be expressed succinctly as (6).  $\frac{\partial x^n}{\partial x} = n\cdot x^{n-1}\qquad\qquad(6)$ The primary question I have is whether we can easily derive an expression for this simplifying algorithm in general for all hyper-k derivatives (as discussed above). I conjecture that we can write down the general expression for the analogous generalization to an arbitrary hyper-k derivative power rule as in (7)  $\partial (x \uparrow^{(k+1)} n) \uparrow^{k} (\partial x^{-1}) = (x \uparrow^{(k-1)} n) \uparrow^{(k)} n - 1\qquad\qquad(7)$  Conclusion:  Any proof of the above conjecture, or any thoughts on this topic would be greatly appreciated, I am hoping to make a blog post about this topic soon and would greatly appreciate any insights/corrections/direction.  The above form (7) was derived by inspection and analogy, it is possible that it is errant, any corrections would be greatly appreciated (feel free to either message me, or reply directly to the thread).  Thanks,     Micah « Next Oldest | Next Newest »

 Messages In This Thread Isomorphism of newtonian calculus rules for Non-Newtonian (anti)derivatives of hypers - by Micah - 02/24/2019, 10:50 PM RE: Isomorphism of newtonian calculus rules for Non-Newtonian (anti)derivatives of hypers - by jaydfox - 03/01/2019, 06:13 AM RE: Isomorphism of newtonian calculus rules for Non-Newtonian (anti)derivatives of hypers - by jaydfox - 03/02/2019, 02:50 AM RE: Isomorphism of newtonian calculus rules for Non-Newtonian (anti)derivatives of hypers - by Micah - 03/02/2019, 08:13 PM RE: Isomorphism of newtonian calculus rules for Non-Newtonian (anti)derivatives of hypers - by Micah - 03/02/2019, 08:23 PM

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