Timeline for How to define transfinite derivatives of a function?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 27, 2018 at 7:25 | comment | added | Dávid Natingga | @AlecRhea To elaborate a little more, Hessenberg natural sum and product on ordinals are commutative, en.wikipedia.org/wiki/Ordinal_arithmetic#Natural_operations. | |
| Jun 26, 2018 at 19:54 | comment | added | Alec Rhea | @Burak Just to be clear, the relation $\omega=1+\omega$ does not hold if we interpret $+$ as the natural sum of ordinals instead of the recursive sum, which is what I am suggesting we should do. I do think that derivative operators indexed in the Grothendieck ring of the ordinals are more likely to satisfy nice duality properties with integration, since we always have successors in both directions. | |
| Jun 26, 2018 at 19:34 | comment | added | Qfwfq | @James: oh I see, I was not aware of parity of ordinals | |
| Jun 26, 2018 at 19:25 | comment | added | James | @Qfwfq I am using (a generalization of) this notion: Even and Odd Ordinals Or, if you think this is cheating/verbal trickery, you can check each of the four "obvious" possibilities sin(t), cos(t), -sin(t), and -cos(t), and find an inconsistency. The only somewhat reasonable answer is 0 as Logan M suggested (which happens to be the average of these four possibilities), but this kind of reasoning would also lead us to think $D^{\omega}e^t = 0$ as well, which most people would reject. | |
| Jun 26, 2018 at 18:47 | comment | added | nombre | @Burak: It is certainly a matter of definition what type of algebraic relation should the transfinite iterations of the differential operator satisfy. For instance, the iterated integration should satisfy properties dual to that of the iterated differentiation, so at least one of those will be at odds with your common sense argument. (And I think Alec Rhea was refering to Hessenberg arithmetic, i.e. natural arithmetic) | |
| Jun 26, 2018 at 18:44 | comment | added | Qfwfq | "(...) since the ordinal $\omega$ is $0(\mathrm{mod}4)$, if anything" - Really? Why? | |
| Jun 26, 2018 at 18:01 | comment | added | Burak | @AlecRhea: If you have $D^{\beta} \circ D^{\alpha} = D^{\beta+\alpha}$, then $D^{\omega} f = D^{1+\omega}f=D(D^{\omega}f)$ and so the $\omega$-th derivatives have to be of the form $Ce^x$. I think James' choice is the correct one since $\alpha+\beta$ in ordinal arithmetic means "apply $\beta$-many successors after $\alpha$-many times", so $D^{\alpha+\beta}$ should mean "differentiate $\beta$-many times after $\alpha$-many times". | |
| Jun 26, 2018 at 17:08 | comment | added | Logan M | Speaking as a physicist, you can "regulate" $D^\omega \sin t$ by instead considering $D^\omega \sin \lambda t$. While not defined for $\lambda = 1$, for all $|\lambda| <1$ there is no difficulty in defining $D^\omega \sin \lambda t = \lim_{n\rightarrow \infty} D^n \sin\lambda t = 0$. For $|\lambda| > 1$ it is less well-behaved but can be considered as (complex) $\infty$ (there is some difficulty at integral multiples of $\pi/2$ but if more general perturbations are allowed this can be dealt with). I would not call this "robust" but it shows that your example is perhaps a degenerate case. | |
| Jun 26, 2018 at 17:03 | comment | added | Alec Rhea | This argument fails if we use natural addition in the definition of $D^\beta\circ D^\alpha=D^{\beta+\alpha}$ since $1+\omega=\omega+1\neq\omega$, which seems the natural choice to me. I think the main problem you are illuminating here is that noncommutative binary operations don't generalize well to operators indexed over that set if we want the induced operator addition to be commutative, which is fixed by choosing a commutative binary operation to begin with. | |
| Jun 26, 2018 at 16:41 | history | answered | James | CC BY-SA 4.0 |