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"circular" operators, "circular" derivatives, and "circular" tetration. - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Hyperoperations and Related Studies (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=11) +--- Thread: "circular" operators, "circular" derivatives, and "circular" tetration. (/showthread.php?tid=665) Pages:
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"circular" operators, "circular" derivatives, and "circular" tetration. - JmsNxn - 06/20/2011 Well, before I can begin our trip into abstract algebra, I must first explain what a "circular" operator is, but to do that, I must first explain what a "hyperbolic" operator is. hyperbolic exponentiation, is what we normally refer to as exponentiation. I've coined it "hyperbolic" because of the following: and then therefore, hyperbolic multiplication is normal multiplication; it being that hyperbolic exponentiation is the super function of hyperbolic multiplication. And also therefore hyperbolic multiplication is the superfunction of hyperbolic addition. Very, simple. Circular exponentiation, is a different form of exponentiation, defined by the following: Interestingly, for imaginary arguments, it behaves the same as hyperbolic exponentiation, only reversed. Now that was all fairly simple to understand, but now we come to the part where we must define circular multiplication. It's still fairly simple, but may seem a bit awkward. but seeing as therefore and now, to make things simple, since Also, we know that circular multiplication, and circular exponentiation behave like normal multiplication and exponentiation. and so therefore we get the beautiful, circular logarithm laws, (cln is taken to mean natural circular logarithm, which is just the circular logarithm base where: And now, with these laws, defining circular multiplication is as simple as: circular division, the inverse of circular multiplication is: And circular exponentiation is as simple as: And now, with all three of these defined, we come to the issue of defining circular addition. Since circular multiplication is it's superfunction we can write the equation as such: and or generally: This gives the very strange equation that "converts" addition into circular addition: which also implies: Circular addition has identity These equations become necessary to observe when defining the circular derivative. Which we shall do now. Therefore: and: and also: We'll probably find: Therefore, with this, we can now create an infinite termed "circular" polynomial that will be it's own circular derivative. if and If: And this is where I'm stuck and I need help. If J(x) turns out to equal I have the equations sort of worked out, but they rely on confirmation that The only way I can properly confirm it is if I had a way of calculating If anybody is curious, the test I'd be doing is seeing if: if it does, then we're in business and there's a whole lot more I can post. RE: "circular" operators, "circular" derivatives, and "circular" tetration. - mike3 - 06/23/2011 This one can be solved via trigonometric identities. We have If we can find a value Then, the inverse, the "circular logarithm", is given by Of course, since RE: "circular" operators, "circular" derivatives, and "circular" tetration. - JmsNxn - 06/24/2011 (06/23/2011, 12:58 AM)mike3 Wrote: This one can be solved via trigonometric identities. We have Wow that's incredible how you did that. That's awesome that cxp can be a closed form expression of only cos, that means for imaginary arguments it should be purely positive. I thought it would be way harder to solve for it... I guess sometimes the answers just so simple and right infront of you that you can't think of it. Thanks a lot for your help. I'll try to see if this pans out to anything ![]() RE: "circular" operators, "circular" derivatives, and "circular" tetration. - Catullus - 06/17/2022 (06/23/2011, 12:58 AM)mike3 Wrote: Then, the inverse, the "circular logarithm", is given byYou did not rationalize your denominator. Can you imagine trying to compute x/√(2) on slide ruler, without the denominator being rationalized? RE: "circular" operators, "circular" derivatives, and "circular" tetration. - tommy1729 - 06/18/2022 this is a curious thread. But cxp(x) satisfies the nice equation f ' (x) = f( - x) now I know you like fractional derivatives so D^z f(x) = f( (-1)^z x) or D^z f(x) = f( cos(pi z) x) ??? regards tommy1729 RE: "circular" operators, "circular" derivatives, and "circular" tetration. - JmsNxn - 06/18/2022 (06/18/2022, 12:43 AM)tommy1729 Wrote: this is a curious thread. Lol, I was like 18 years old when I made this thread tommy. It's hot garbage! RE: "circular" operators, "circular" derivatives, and "circular" tetration. - tommy1729 - 06/18/2022 (06/18/2022, 01:22 AM)JmsNxn Wrote:(06/18/2022, 12:43 AM)tommy1729 Wrote: this is a curious thread. ah yes but how about the fractional derivatives ? ![]() is that garbage too ? why prefer one garbage over another ? ![]() Im not convinced by fractional derivatives nor their applications like interpolation and such. I find it funny that you can define the fractional derivatives in infinitely many ways but only makes sense for integers imo ... RE: "circular" operators, "circular" derivatives, and "circular" tetration. - JmsNxn - 06/19/2022 (06/18/2022, 11:48 PM)tommy1729 Wrote: ah yes but how about the fractional derivatives ? Tommy, for you to so explicitly call fractional derivatives is garbage shows me how little you understand about it. Yes, you can also define exponentiation INFINITELY many ways, but there is one which satisfies a uniqueness condition. There's only one mellin transform. Don't be a dunce. RE: "circular" operators, "circular" derivatives, and "circular" tetration. - tommy1729 - 06/25/2022 (06/19/2022, 08:19 PM)JmsNxn Wrote:(06/18/2022, 11:48 PM)tommy1729 Wrote: ah yes but how about the fractional derivatives ? what are you talking about ? Maybe I know little about fractional derivatives. But how is it not an arbitrary thing ? What uniqueness criterion ? And why that one ? I know Mellin transform. so ? You sure you dont mean laplace transforms ? enlighten me. regards tommy1729 RE: "circular" operators, "circular" derivatives, and "circular" tetration. - JmsNxn - 06/26/2022 There is ONE fractional derivative to be used for interpolation, as I've said it. It's nothing more than the Mellin transform. $$ \frac{d^{-z}}{dw^{-z}}\vartheta(w) = \frac{1}{\Gamma(z)} \int_0^\infty \vartheta(w-x)x^{z-1}\,dx\\ $$ Which converges in a vertical strip \(0 \le \Re(z) \le b\). We assume that \(\vartheta\) is entire, and additionally has some kind of \(O(x^{-b})\) growth as \(x \to \infty\). This fractional derivative can be extended to \(-b \le \Re(z)\): $$ \Gamma(-z)\frac{d^{z}}{dw^{z}}\vartheta(w) = \sum_{n=0}^\infty \vartheta^{(n)}(w) \frac{(-1)^n}{n!(z-n)} + \int_1^\infty \vartheta(w-x)x^{-z-1}\,dx\\ $$ This is absolutely a unique operator, so long as you assume that \(F(z) = \frac{d^{z}}{dw^{z}}\Big{|}_{w=0}\vartheta(w) \) is in the exponential space: $$ |F(z)| \le Ce^{\rho|\Re(z)| + \tau|\Im(z)|}\,\,\text{for}\,\,\rho,\tau \in \mathbb{R}^+\,\,\tau < \pi/2\\ $$ There then exists a correspondence of functions \(\vartheta\) which produce \(F\)'s. What I have shown, is that certain functions produce \(\vartheta\), which then produce \(F\) again through this formula. It's nothing really more than Ramanujan's Master Theorem, described using Fractional calculus. THIS IS UNIQUE. NOW, what I think you are getting at, is that there are uncountably many infinite fractional derivatives. This is true. We can iterate the derivative uncountably many ways. BUT! There is only one way to do this such that: $$ \frac{d^z}{dw^z} e^w = e^w\\ $$ THERE'S ONLY ONE FRACTIONAL DERIVATIVE THAT DOES THIS! It's known as the Riemann-Liouville, or the Exponential, fractional derivative. It's also known as the Weyl fractional derivative. IT IS UNIQUE. So it is not arbitrary to interpolate sequences using fractional calculus, if you only use this operator. It's absolutely the exact opposite of arbitrary. It's unique. It either works or it doesn't. And if it works, it's unique. I don't know what to say, tommy. You've never liked the fractional calculus approach. But you clearly also don't understand it. And I don't know what else to say. |