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June 22

0^0 other values...

The standard explanation of the fact that 0^0 is undefined is the fact that lim x->0 of x^0 remains 1 and lim y->0 of 0^y remains 1. Considering this as part of the graph z=x^y, is there an easy way to figure out the line of the form y=kx that approaching (0,0), the limit is 1/2 (to pick a number).Naraht (talk) 13:12, 22 June 2017 (UTC)[reply]

The limit of x^(kx) is always 1. This can be found in the exercises of any calculus textbook in the chapter on limits. A good rule of thumb when dealing with limits of the form f(x)^g(x) is to begin by taking a logarithm. JBL (talk) 13:24, 22 June 2017 (UTC)[reply]

@Naraht: Take

y={\frac {k}{\ln x}}

for any real constant k and you'll get a curve approaching (0, 0) which maintains

x^{y}=e^{k}={\text{const.}}

Example graph at Wolfram Alpha: link. --CiaPan (talk) 13:40, 22 June 2017 (UTC)[reply]

You ment to write

y={\frac {\ln k}{\ln x}}

for any positive constant k. Bo Jacoby (talk) 06:16, 24 June 2017 (UTC).[reply]

@Bo Jacoby: No, I didn't. --CiaPan (talk) 12:02, 26 June 2017 (UTC)[reply]

Exponentiation#Zero_to_the_power_of_zero talks about this. In fact entering 0^0 gets there. Dmcq (talk) 14:21, 22 June 2017 (UTC)[reply]

The function

f(x,y)=x^{y}

is defined for all $y$ when $x$ is positive, and for all $x$ when $y$ is a nonnegative integer, and for nonzero $x$ when $y$ is a negative integer. However $f(x,y)$ is discontinuous when $(x,y)=(0,0)$ . The limit

\lim _{(x,y)\rightarrow (0,0)}f(x,y)

is undefined even if

f(\lim _{x\rightarrow 0}x,\lim _{y\rightarrow 0}y)=0^{0}=1

is defined. Bo Jacoby (talk) 06:07, 24 June 2017 (UTC).[reply]

Thanx to all. This gave me what I was looking for.Naraht (talk) 04:42, 26 June 2017 (UTC)[reply]

@Naraht: You can find some more notes about

00

in the Math Ref.Desk archive from June 30, 2015, the day you asked about the least symmetric triangle :)

Wikipedia:Reference desk/Archives/Mathematics/2015 June 30#Usefulness vs. naturalness

CiaPan (talk) 13:36, 26 June 2017 (UTC)[reply]

Symmetric closure of transitive relation

Is there a transitive relation whose symmetric closure is not transitive? GeoffreyT2000 (talk) 14:50, 22 June 2017 (UTC)[reply]

Ok, SemanticMantis, Meni Rosenfeld, RDBury, John Z, there seems to be consensus that my initial comment can be read in a much sharper way than I intended it. I have struck it out below and replaced it with something that hopefully conveys the point I was trying to make more clearly. --JBL (talk) 15:20, 27 June 2017 (UTC)[reply]

@GeoffreyT2000:, you post a lot of substantive questions on the ref desk. Would you please consider including with your questions some indication of what you have attempted before posting the question, to save time and effort for potential answerers? For example, in this instance it is clear that you did not investigate small relations before posting your question, and that would have been a useful piece of information. Thanks. --JBL (talk) 15:20, 27 June 2017 (UTC)[reply]

Do you put any thought into your questions before you post them? If so, perhaps you could adapt your posting style to demonstrate it. If not, perhaps you should. --JBL (talk) 17:17, 22 June 2017 (UTC)[reply]

Please either help, or do not; being bitey is wasting everyone's time, and reflects poorly on our reference desks. SemanticMantis (talk) 18:10, 22 June 2017 (UTC)[reply]

I think JBL's comment is legitimate. IIRC the OP posts a lot of questions. I haven't followed this closely enough to recognize if there's an offending pattern, but if there is, it's justified to call them out on it. The reference desk does have rules. -- Meni Rosenfeld (talk) 18:23, 22 June 2017 (UTC)[reply]

Thank you, Meni. I do not think it is unreasonable to expect repeat users to include a sentence of the form "Here is what I have tried before posting this problem: ...." (In this case, such a sentence would not include "looking at relations on a 2 or 3 element set.") --JBL (talk) 18:41, 22 June 2017 (UTC)[reply]

It seems to me we don't need such a policy; if you don't like a question then don't answer it. Probably more people will find a question interesting enough to answer if the poster says what they've tried, so maybe it's to the poster's advantage to do that, but it shouldn't be a requirement. --RDBury (talk) 12:25, 23 June 2017 (UTC)[reply]

I am not asking for a policy, I am asking for a cultural norm: people shouldn't post thoughtlessly here. --JBL (talk) 12:49, 23 June 2017 (UTC)[reply]

While I could not oppose such a norm, one question to ask before responding to posts is whether the response(s) transgress this norm more than the post. To my eyes, the response here that transgresses least is Semantic Mantis's, less than Meni Rosenfeld's (whose transgression is admitted) and rather less than yours. The OP has asked many questions over the years, but they are usually quite intelligent questions, especially considering lack of advanced age. I am often the only one to answer (sometimes only in the archives). In one case I recall, he had obviously put much more thought in the question than my first rather obtuse answer or anyone else's - and I feel I owe him a number of answers to his never answered questions. People have different points of view, things obvious from one may not be obvious from another & the phenomenon of "dumb questions" leading to "interesting discussions" is universal, essential in science and illustrated here. IMHO those expectations of repeat users are not reasonable, and are therefore not "in the rules". "Berating" might be too strong - my candidate adjective is "waspish", but it is a better description than a biased positive self-judgment. So again, the questions in the first response are better addressed to the first responder, than the OP.John Z (talk) 04:24, 24 June 2017 (UTC)[reply]

If JBL or anyone else wants to discuss OP's behavior, or suggest any rules on how much thought is required before posting, or rant about "culture" the place to do that is either the OP's talk page or the ref desk talk page. This space is for providing references and helping people get answers, not for berating OP. I remind JBL that his participation here is voluntary, and if he doesn't like OP's questions, he is free to ignore them. The math desk is very slow these days anyway, it's not like we are overloaded with questions. SemanticMantis (talk) 19:03, 23 June 2017 (UTC)[reply]

On the same principle, you are not obligated to read or respond to my comments. Moreover, it is clear from your own comments in this thread that you do not believe in a general principle forbidding personal comments on the ref desk. Finally, my comment is not "berating" -- it is critical but polite and to the point, and I am in fact interested in the question of whether GeoffreyT2000 puts any thought at all into the questions that he (I hope this is not presumptive) posts. --JBL (talk) 20:50, 23 June 2017 (UTC)[reply]

I have no idea what you're talking about, nor do I care. Asking OP "Do you put any thought into your questions before you post them?" Is rude and unprofessional. Simple as that. I ignore a lot of your comments, but I will continue to speak up when I see you acting like a WP:DICK. SemanticMantis (talk) 16:53, 26 June 2017 (UTC)[reply]

I haven't thought about this stuff in a long time but symmetric closure would seem to preserve transitivity. If R is the transitive relation, and S is one symmetric closure, then you know (aRb & bRc) => aRc, and you know that bSa => (aRb or bRa). From here you need only to work out that (aSb & bSc) => aSc. Does that clear it up? SemanticMantis (talk) 18:10, 22 June 2017 (UTC)[reply]

I disagree, I think the answer to the OP's question is positive. I can think of at least three very nice examples. -- Meni Rosenfeld (talk) 18:23, 22 June 2017 (UTC)[reply]

I think the smallest number of for the underlying set is two. Specifically, if S={0, 1} and R={(0, 0), (0, 1)}. R is transitive but R∪R^t is not. An easy way to check transitivity is to think of R as a Boolean matrix, then R is transitive iff RR⊊R. --RDBury (talk) 00:41, 23 June 2017 (UTC)[reply]

Since by now the OP has hopefully had some time to think about it, I'll share what I had in mind (which are less artificial than yours):

Real numbers (or any other totally ordered set) and $<$ (has to be strict inequality).
Positive integers and $|$ .
Sets and $\subseteq$ .

As for your example, it can be trimmed down further if you're ok with R being transitive vacuously -

R=\{(0,1)\}

. -- Meni Rosenfeld (talk) 08:55, 23 June 2017 (UTC)[reply]

Nice one, even more minimal. There are more permutations to think about: Is the transitive closure of symmetric relation still symmetric? Is the symmetric closure of a reflexive relation still reflexive? There are six basic combinations but there are more complex ones too. It looks like the answer is yes for most combinations. --RDBury (talk) 12:25, 23 June 2017 (UTC)[reply]

So corrected, thank you. SemanticMantis (talk) 15:27, 23 June 2017 (UTC)[reply]