• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 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  jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 03/01/2019, 06:13 AM (This post was last modified: 03/01/2019, 06:19 AM by jaydfox.) Hi Micah, This is something that I hadn't considered formally before, but I've played with similar ideas over the years, at least for the product-based derivative.  Since you posted this a few days ago, I've been playing with the product based derivative, when I can find time.  I want to get a better understanding, before I attempt to tackle the next higher hyperoperation based derivatives. My initial reaction is that the product based derivative bears a strong resemblance to the so-called logarithmic derivative, and indeed, to a first order approximation, they are related.  I've worked out some of the details with pencil and paper, but haven't yet taken the time to write up some formal TeX expressions. My curiosity then is whether the next higher order derivative would resemble some sort of superlogarithmic derivative, at least to a first order approximation?  I haven't even tried to work it out yet, so I could be way off track with my guess.  Anyway, thanks for bringing this to my attention.  More to ponder and try to learn from! Edit: the "logarithmic derivative" isn't as exciting as it sounds.  It's just a regular derivative, like we're all used to, but you take the logarithm first.  I didn't want to convey the impression that it's a different type of derivative, even if it does have some cool properties.  The product based derivative that Micah posted does at first glance appear to be a different type of derivative. Edit 2: link to wikipedia article on logarithmic differentiation: https://en.m.wikipedia.org/wiki/Logarith...rentiation ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 03/02/2019, 02:50 AM Hi Micah, As promised, I took a look at the product-based derivative in your equation (3), and it's very interesting.  As I was working out the derivation, I found that, to a first order approximation, it appears to be resemble the logarithmic derivative.  Of course, those higher order differences really add it.  Interestingly enough, I think this ends up being the exponentiation of the logarithmic integral.  At first, that seems like the exponentiation and the logarithm would cancel, being functional inverses of each other.  But alas, the order of operations prevents them from directly cancelling. Forgive my lack of TeX expertise here.  I couldn't remember how to do some of the formatting, so it's a bit rough. Repeating your equation (3): $\partial f^{(\partial x)^{-1}}=\lim\limits_{\Delta x \to 0}\left(\frac{f(x+\Delta x)}{f(x)}\right)^{\frac{1}{\Delta x}}\qquad\qquad(3)$  Let's compare the logarithmic derivative: $\frac{\partial \ln(f)}{\partial x} = \frac{1}{f}\frac{\partial f}{\partial x}$ ${}={\frac{1}{f(x)}}\lim\limits_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ ${}=\lim\limits_{\Delta x \to 0} \frac{1}{\Delta x}\frac{f(x + \Delta x) - f(x)}{f(x)}$ ${}=\lim\limits_{\Delta x \to 0}{\frac{1}{\Delta{x}}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)}$ Already, there should be a sense of similarity, as both expressions have the term $\frac{f(x+\Delta x)}{f(x)}$. We can take a binomial expansion of (3): $\lim\limits_{\Delta x \to 0}\left(1+\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)\right)^{\frac{1}{\Delta x}$ ${}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left({{\Delta x}^{-1} \choose {k}}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)^{k}\right)}$ The "choose" notation here is shorthand for a product: ${{\Delta x}^{-1} \choose {k}}=\prod_{n=1}^{k}\left(\frac{{\Delta x}^{-1}-(n-1)}{n}\right)$ Only the highest power of (1/dx)^k matters, alternatively the lowest power of dx^{-k}, so eliminate the neglible terms: ${{\Delta x}^{-1} \choose {k}} \approx \frac{{\Delta x}^{-k}}{k!}$ ${}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left(\frac{{\Delta x}^{-k}}{k!}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)^{k}\right)}$ ${}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left(\frac{1}{k!}\left[\frac{1}{\Delta{x}}\left({\frac{f\left(x+{\Delta{x}}\right)}{f(x)}-1}\right)\right]^{k}\right)}$ And now we see that we have the power series of the exponential function, taking the logarithmic derivative as its argument: ${}=\lim\limits_{\Delta x \to 0}\sum_{k=0}^{\infty}{\left(\frac{1}{k!}\left[\frac{\partial \ln(f)}{\partial x}\right]^{k}\right)}$ Therefore: $\partial f^{(\partial x)^{-1}}=\exp\left(\frac{\partial \ln(f)}{\partial x}\right)$ I hope my derivation was correct.  I get so caught up in editing the TeX code, that I sometimes miss mistake in my maths.  However, I did some crude testing in Excel, and my results do seem to bear out this result. If this is true, could we conjecture that the derivative based on the next-higher hyperoperation would be sexp(derivative(slog(f(x))))? ~ Jay Daniel Fox Micah Junior Fellow Posts: 9 Threads: 2 Joined: Feb 2019 03/02/2019, 08:13 PM Dear Jay Fox,       Thankyou so much for your replies and interest!   I also need to spend some time working with the pathway that you provided to see if I can recreate your derivation, it is probably correct, I would just like to also be sure I can follow your derivation.   Since I originally posted this, it occurred to me that the $\partial f$ defined in the geometric derivative and the further derivatives is definitely not isomorphic to the original differential $\partial f$.  Consider the following.  $\lim\limits_{\Delta x \to 0} ( f(x + \Delta x) - f(x) )=\partial f\qquad\qquad (1)$ $\lim\limits_{\Delta x \to 0} \left(\frac{f(x+\Delta x)}{f(x)}\right) =\partial f^{\prime}\qquad\qquad (2)$ Then we must draw a direct relationship between the two $\partial f$'s in order to re-specify the original postings formula properly.  We can see that clearly,  $\lim\limits_{\Delta x \to 0}\ln\left(\frac{f(x+\Delta x)}{f(x)}\right)=\lim\limits_{\Delta x \to 0}(\ln{f(x + \Delta x)} - \ln{f(x)}) \ne \lim\limits_{\Delta x \to 0}\ln{f(x+\Delta x) - f(x)}\qquad\qquad (3)$  Further note that because the definition of the real logarithm function is not continuous at 0 (it is asymptotic), the limit and logarithm are not interchangeable, hence to generalize the expression of $\partial f^{\prime}$, however, the argument of the log in this case is neither zero, nor a function of zero (unless at f(x) = 0 where everything breaks down) hence we must be able to specify an invertable functional form g(.) such that:  $g(\ln f(x + \Delta x) - \ln f(x)) = \ln(f(x + \Delta x) - f(x))\qquad\qquad (4)$ of equivalently,  $\ln f(x + \Delta x) - \ln f(x) = g^{-1}(\ln(f(x + \Delta x) - f(x)))\qquad\qquad (5)$ Micah Junior Fellow Posts: 9 Threads: 2 Joined: Feb 2019 03/02/2019, 08:23 PM (This post was last modified: 03/02/2019, 08:56 PM by Micah. Edit Reason: cite -> site ) Hi Again Jay,      sorry for the quick double post,  after reading through your derivation I believe that I agree with you, the problem is that the log and the exponential cannot cancel because there a limit in the middle of them that we cannot work out.  I would suspect that your conjecture about the relationship between the super-log derivative and the tetration derivative that is expressed in (4) is true, but would need to really consider what the super-log derivative would look like (I haven't spent as much time with these logarithmic derivatives which are also very cool). Thanks again,     -Micah  PS - I also made a post about some work with a theoretical probability distribution defined by the tetration operator on my that you may find interesting to skim if you have the time, there is a mistake in the generalized formula for the generic probability distribution function in my post though I bet you could spot it - I noticed it a few days ago and will update it soon https://people.smu.edu/34915677/2019/02/...ion-on-01/)  PPS - I am not aware of the etiquette for posting links to outside sites here on Tetration forum, and hope that this is appropriate (I will gladly edit this post if it is not). PPPS - Does anyone know what the deal is with the neomaths ankh forum advertisement in the Banner? I tried visiting that forum a week or so back, and noticed that there are very few users or posts, far fewer than there are on this site. « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post On my old fractional calculus approach to hyper-operations JmsNxn 14 3,121 07/07/2021, 07:35 AM Last Post: JmsNxn 1st iterated derivatives and the tetration of 0 Xorter 0 2,849 05/12/2018, 12:34 PM Last Post: Xorter A calculus proposition about sum and derivative tommy1729 1 3,971 08/19/2016, 12:24 PM Last Post: tommy1729 Fractional calculus and tetration JmsNxn 5 12,639 11/20/2014, 11:16 PM Last Post: JmsNxn On the binary partition and taking derivatives tommy1729 0 2,941 09/29/2014, 11:34 PM Last Post: tommy1729 Theorem in fractional calculus needed for hyperoperators JmsNxn 5 11,727 07/07/2014, 06:47 PM Last Post: MphLee Newtonian Tetration fivexthethird 1 3,711 01/24/2014, 10:04 PM Last Post: Gottfried Integer tetration and convergence speed rules marcokrt 5 11,742 12/21/2011, 06:21 PM Last Post: marcokrt "circular" operators, "circular" derivatives, and "circular" tetration. JmsNxn 2 9,482 06/24/2011, 07:21 PM Last Post: JmsNxn very general calculus tommy1729 0 3,320 12/01/2010, 06:16 PM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)