Timeline for Interesting examples of vacuous / void entities
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1 at 11:56 | comment | added | user3840170 | The fact that $\lim_{x\to 0} \lim_{y \to 0} \exp(y \ln(x))$ and $\lim_{y\to 0} \lim_{x \to 0} \exp(y \ln(x))$ are different suggests that there are settings in which either choice is a more natural generalization, and the definition of $\exp(y \ln(x))$ gives us no reason to prefer one over another. p-norms are a great example here: the blog post I linked before characterizes the mean, median and mode as minimizers of $x \mapsto \exp(p \ln(\Vert x \Vert_p))$ for $p$ = 2, 1 and 0. The last of which relies on continuity wrt $p$, and not wrt $x$. | |
| Apr 1 at 11:36 | comment | added | user3840170 | I would be fine with $\sum_{k=0} c_k (z \uparrow k)$; it’s just one more symbol (well, three with the parentheses). Though perhaps a better notation for $\exp(y \ln(x))$ should exist. And I would not be so dismissive of continuity: continuity is often a reflection of logical uniformity, generalization without special cases, parametricity. (I seem to recall there are even settings where this is literal.) | |
| Nov 12, 2021 at 14:50 | comment | added | Pietro Majer | The convention $0^0=1$ in real or complex setting is implicitly assumed when we write a power series $\sum_{k=0}c_kz^k$. Noboby wants to write a generic power series as $c_0+\sum_{k=1}c_kz^k$. | |
| Dec 20, 2020 at 12:48 | comment | added | user3840170 | The conflaton of $x \uparrow y := \prod_{k\in(y, 0]\cap\mathbb{Z}} \frac 1 x \cdot \prod_{k\in[0, y)\cap\mathbb{Z}} x$ with $\exp(y\ln(x))$ by the notation $x^y$ is a mistake, and the question of how to define $0^0$ is just one piece of evidence against it: some situations demand one convention, others demand another. Another is that $(-1) \uparrow k$ is perfectly straightforward, but $\exp(k\ln(-1))$ is meaningless. | |
| Dec 20, 2020 at 3:14 | comment | added | user76284 | @user3840170 And yes, real exponentiation “out to be” an extension of natural exponentiation (in particular, it “ought to be” defined wherever the latter is defined). This is the most natural convention, IMO, and there is no downside to it. The continuity point is moot, since $x^y$ will never be continuous anyway. | |
| Dec 20, 2020 at 3:04 | comment | added | user76284 | @user3840170 Of course it’s not continuous at 0. That doesn’t mean it has to be undefined at 0. The sign function is discontinuous at 0, but there are good reasons to let it be 0 there. | |
| Jul 11, 2020 at 9:13 | comment | added | user3840170 | Real exponentiation is not an extension of natural exponentiation, they just happen to coincide where both are defined. $\prod_{k \in [0, y) \cap \mathbb Z} x$ is the empty product for $y = 0$ and therefore equal to 1 regardless of the value of $x$, but $\exp(y \ln(x))$ is undefined for $x = 0$ and cannot be defined at $x = y = 0$ while maintaining continuity with respect to both variables. | |
| Jul 10, 2020 at 18:23 | comment | added | user76284 | -1 for “It is not true for real zero.” I’d like to see an argument for this. Real exponentiation is an extension of natural exponentiation, i.e. it ought to have the same value wherever the latter is defined. | |
| Jun 30, 2020 at 12:53 | comment | added | user3840170 | I think that's an argument for setting the value at zero — as it is the number of generating elements of $\mathbb Z/0\mathbb Z$. | |
| May 7, 2018 at 17:27 | comment | added | LSpice | @YemonChoi, I think that "the number of generating elements" means "the number of elements that form singleton generating sets", rather than "the minimum cardinality of a generating set". | |
| May 7, 2018 at 15:38 | comment | added | Yemon Choi | I've just seen this. Surely ${\bf Z}$ is singly generated | |
| Feb 10, 2014 at 18:24 | comment | added | Sh.M1972 | Yes you are right, only exponent should be Natural. | |
| Feb 10, 2014 at 16:58 | comment | added | Michael Hardy | If the base is the real or complex zero and the exponent is the natural zero, then $0^0=1$. For example: $\displaystyle e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$. If $z=0$, then the first term in this expansion is $0^0/0!=1$. | |
| S Feb 10, 2014 at 7:10 | history | answered | Sh.M1972 | CC BY-SA 3.0 | |
| S Feb 10, 2014 at 7:10 | history | made wiki | Post Made Community Wiki by Sh.M1972 |