Timeline for Function that is 1 for all positive integers but 0 at 0
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2020 at 4:22 | comment | added | Albert Renshaw | In my opinion 0^0 should be 1, it makes logical since, but even further it would be very useful if it were 1 as the delta function would become closed form, being simply $0^x$ | |
| Jun 4, 2016 at 13:58 | comment | added | Paul Sinclair | @ΈρικΚωνσταντόπουλος - so you are making a completely pointless complaint with no relevance to the question, based on an arbitrary and nonsensical division of the calculation that you've come up with?? | |
| Jun 3, 2016 at 23:37 | comment | added | Paul Sinclair | @PeterLeFanuLumsdaine - In analytic situations, $1$ is almost always the natural choice for $0^0$ as well. Only in rare occasions do you have use for anything else. For a smooth path $(x(t), y(t))$ with $x > 0$ everywhere and converging to $(0,0)$ to have $x^y$ converge to anything other than $1$, you have to have $x' = 0$ at the limit. That is, you only get divergence or values other than $1$ by coming in tangent to the $y$-axis. | |
| Jun 3, 2016 at 23:27 | comment | added | Paul Sinclair | @ΈρικΚωνσταντόπουλος - What does "has 2 items" have to do with anything? Indeed, what does that even mean? I only discovered Business Cat's post after I added the comment. So I upvoted his post as well. But in any case, it is only a minor variant of this answer, and this one was first of the two. | |
| Jun 2, 2016 at 19:47 | comment | added | Peter LeFanu Lumsdaine | In combinatorial and algebraic settings, it’s very natural to consider $0^0 = 1$ — essentially, anywhere that one’s considering the function $x^n$, with $n$ taking integer values. But in analytic settings, where one’s considering $x^y$ with real or even complex values of $y$, then it is much more natural to take $0^0$ to be either 0 or undefined. This is I think the general reason why different authors make the choices they do. | |
| Jun 1, 2016 at 18:37 | comment | added | Joshua | There are paths to 0^0 for which the answer is zero, and 0^n where 0 is constant and n is variable is the most important of them. | |
| Jun 1, 2016 at 17:55 | comment | added | Nick Matteo | @Joshua: what on earth are you talking about? Anyway, it works fine: as you say, 0^0^n is 0^0, which is 1, for all positive n; and 0^0^0 is 0^1 which is 0, for n = 0. There are no "rules in algebra" about statements meaning one thing only if you "got here from cumulative roundoff", whatever that means. | |
| Jun 1, 2016 at 17:37 | comment | added | Joshua | Does not work. 0^0 != 1 with this generator. The reduction rule in algebra says you may do this if you got here from cumulative roundoff. This isn't cumulative roundoff as we can simplify 0^0^n to 0^0 for all n not zero. | |
| Jun 1, 2016 at 17:06 | comment | added | Paul Sinclair | @LuisMasuelli - Only if one is so criminally sloppy in one's mathematics that similar mistakes would be made no matter how $0^0$ is handled (as your example includes with $(a-b)^{-1}$). But this is not the place to argue which is better. The disagreement was acknowleged in the post itself. Cases one way or another should be given in a more appropriate forum. | |
| Jun 1, 2016 at 16:55 | comment | added | Luis Masuelli | Say: a=b=1 -> aa = ab -> aa-bb = ab-bb -> (a+b)(a-b) = b(a-b) -> (a+b)(a-b)(a-b)^(-1) = b(a-b)(a-b)^(-1) -> (a+b)(a-b)^(1 + -1) = b(a-b)^(1 + -1) -> (a+b)(0)^(0) = b(0)^(0). If we equate 0^0=1, then we break the real numbers by saying a+b=b, which makes 2=1, 3=2, 4=3, 1=3 by transitivity... and every possible linear transform over the numbers making numbers totally meaningless with such definition. | |
| Jun 1, 2016 at 16:52 | comment | added | Luis Masuelli | I prefer to disagree 0^0=1. Such result is isolated from divide-by-zero jokes which lead to think stuff like 2=1 | |
| Jun 1, 2016 at 16:50 | comment | added | Paul Sinclair | @Eridan - the term "indeterminate form" refers to how the function behaves under limits. It places no restriction on the value of the function at the point. A function does not have to be continuous to be defined. Thus there is no contradiction between defining $0^0 = 1$ and $0^0$ being an indeterminate form. | |
| Jun 1, 2016 at 16:43 | comment | added | Paul Sinclair | $f(n) = 1 - 0^n$ could be considered simpler than $0^{0^n}$ and accomplishes the same purpose. However, it's the same key idea as this answer. | |
| Jun 1, 2016 at 11:05 | comment | added | Adnan | @Olba12 Oh I didn't know, I'm glad you liked the answer :) | |
| Jun 1, 2016 at 11:00 | comment | added | Olba12 | Oh i am sorry, I really liked your answer. My comment did not have anything with your answer, instead a "contribution" to the discussion with $0^0$. :) | |
| Jun 1, 2016 at 10:57 | comment | added | Adnan | @Olba12 Yes, that is correct. Luckily, OP has stated that $n\in \mathbb{N}$, which only contains positive integers. | |
| Jun 1, 2016 at 10:52 | comment | added | Olba12 | $0^0^{+}=1$, For 0^{-} we have a problem. | |
| Jun 1, 2016 at 7:59 | comment | added | shA.t |
For non-integer, be aware of n = -1 ;).
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| Jun 1, 2016 at 7:48 | comment | added | hvd | Link about $0^0$: "The discussion of 0^0 is very old. [...] Consensus has recently been built around setting the value of 0^0 = 1." | |
| Jun 1, 2016 at 2:34 | comment | added | Sophie Swett | I don't think there's really a "correct answer" to the question of whether $0^0$ is defined or not. Some authors define it, and for those authors, it is defined; other authors do not define it, and for those authors, it is undefined. The question of whether or not it should be defined is something else. | |
| Jun 1, 2016 at 2:05 | comment | added | Arcturus | $0^0$ is not equal to $1$. It is an indeterminate form. | |
| May 31, 2016 at 23:26 | comment | added | Milo Brandt | I'm not entirely convinced that $0^x$ is actually an elementary function; the linked Wikipedia page is ambiguous, but if you take exponentiation as "including the function $e^x$" or "satisfying $\frac{d}{dx}a^x = c\cdot a^x$, then $0^x$ is not elementary if you take $0^0=1$." | |
| May 31, 2016 at 21:43 | history | edited | Adnan | CC BY-SA 3.0 |
added 10 characters in body
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| May 31, 2016 at 21:37 | comment | added | user3294068 | Your problem is not a problem. Google is correct in this case. | |
| May 31, 2016 at 21:28 | history | answered | Adnan | CC BY-SA 3.0 |