Wikipedia:Reference desk/Archives/Mathematics/2008 January 7

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January 7

the empty set

Contemplate it! —Tamfang (talk) 07:17, 7 January 2008 (UTC)

This is not a question. Your efforts to save this day from eternal questionlessness were in vain. -- Meni Rosenfeld (talk) 11:33, 7 January 2008 (UTC)

${\bar {U}}$, with U being the universal set. Contemplate that!

Detroit / Homeslessness

After searching several websites, as well as contacting the police and records department, my search has been fruitless.

I am looking for statistics about homeless people in Detroit. Something along the lines of -how many are hit by car accidents in a given year -how many are attack by vicious animals (and which ones)

and any other statistic pertinent to homelessness in Detroit.

My team and I are designing a video game to help increase awareness of the issue, and need the statistics to help educate the players.

Thanks in advance.

SallyZ (talk) 01:52, 7 January 2008 (UTC)Sally

A noble cause indeed, though I don't know where to go to get that information, the humanities reference desk would probably be the best place to ask this question, despite it's being a statistic. A math-wiki (talk) 08:53, 7 January 2008 (UTC)

limit 2 dimensions w/ polar coordinates

What am I doing wrong here? I have this limit: lim (x,y)->(0,0) (x^2+y^2)/(x+y), which, according to the teachers and the solutions doesn't have a limit. However, if I use polar coordinates, don't I get r*(1/(cos(f)+sin(f))) which is 0 for every direction? Thanks. —Preceding unsigned comment added by 62.57.239.234 (talk) 14:59, 7 January 2008 (UTC)

What happens when $f={\frac {3\pi }{4}}=135^{\circ }$? -- Meni Rosenfeld (talk) 15:03, 7 January 2008 (UTC)

For extra credit: You probably know that if the limit exists, you will obtain it no matter how you approach the point, even through a curve which is not a straight line. What, then, will be the limit when approached through the curve $x=t,\ y=t+\alpha t^{2}\;\!$ for some number $\alpha \neq 0$? -- Meni Rosenfeld (talk) 15:20, 7 January 2008 (UTC)

I dont see your point: isn't the limit simply 0 along that curve? Perhaps you meant $y=\alpha t^{2}-t$ instead, so that the limit is ${\frac {2}{\alpha }}$ and depends on your choice of parameter? --Tardis (talk) 17:11, 7 January 2008 (UTC)

You are of course correct, it should be $y=-t+\alpha t^{2}$. My point was that this is not just some sort of removable discontinuity, but rather the behavior of the function is truly erratic at the origin. -- Meni Rosenfeld (talk) 17:33, 7 January 2008 (UTC)

Basic problem is that (x²+y²)/(x+y) isn't even defined along the line x+y=0. Gandalf61 (talk) 16:27, 7 January 2008 (UTC)

I don't see that as much of a problem. The similar ${\frac {x^{2}-y^{2}}{x+y}}$ isn't defined along that line, but it is everywhere else equal to $x-y$, of course, so I would say it has a limit of 0 at the origin. Does it really have to exist for all approaches, or is it sufficient that, say, for any ray from the point, an approach whose angle from the ray approaches 0 yields the common value? --Tardis (talk) 17:11, 7 January 2008 (UTC)

Problem is even if you exclude the line x+y=0, or maybe set f(x,y)=0 along this line, you still have to deal with paths which asymptotically approach this line as they approach the origin. Meni has given one example (yes, I assume he meant $x=t,\ y=-t+\alpha t^{2}\;\!$); an even more pathological example is $x=t,\ y=-t+t^{3}\;\!$. See Limit of a function of more than one variable. Gandalf61 (talk) 17:30, 7 January 2008 (UTC)

[ec]The traditional definition, usually phrased in the language of neighborhoods, insists that the limit exist with any approach (and in particular, that the function will be defined at a neighborhood of the point). However, there is nothing stopping you from discussing more general definitions which are resistant to these kinds of removable discontinuities. -- Meni Rosenfeld (talk) 17:33, 7 January 2008 (UTC)

Quite an interesting issue. But the direction x+y=0 is out of domain, so the targeted limit is zero, IMHO.

"...you still have to deal with paths which asymptotically approach this line as they approach the origin."

This is no big deal, since the limit there (asymptotically near the line x+y=0) is undoubtbly zero. Pallida  Mors 17:40, 7 January 2008 (UTC)

No it is not. Have you read Meni and Gandalf's examples above? Algebraist 17:54, 7 January 2008 (UTC)

Alright, I was speaking of linear directions. Let's look at the above non-linear examples:

• $x=t,\ y=-t+\alpha t^{2}\;\!$.

The limit over this line is $t(2-t\alpha )$

• $x=t,\ y=-t+t^{3}\;\!$

Here the limit is $-t\left(t^{2}-2\right)$

Please correct me if I'm wrong. But for x->0, t->0 and both expressions tend to zero. Pallida  Mors 18:06, 7 January 2008 (UTC)

You are talking about the function Tardis gave, while the rest of us are talking about the function the OP gave.

BTW, if a direction is linear then it is not asymptotically near the line x+y=0. -- Meni Rosenfeld (talk) 18:10, 7 January 2008 (UTC)

PS. Did you notice that in your last post, you have dismissed the examples of Gandalf and me as uninteresting, rejected Tardis' calculation of the limit, and ignored Algebraist's defense of the above? I would think we all deserve enough credit for you to think a bit longer (and re-examining the OP's question) before doing so. -- Meni Rosenfeld (talk) 18:19, 7 January 2008 (UTC)

I apologize Meni, Gandalf, Tardis and Algebraist for having given such impressions. I should have read the section more carefully. I feel really sorry for my performance with this answer. And yes, linear directions may be as near as possible to the line given by x+y=0, while asymptotically near makes no real sense for what I was trying to say. Furthermore, my English is not perfect. I will try to be up to the task next time. Just let me again say sorry if I upset anyone with my answers.

190.1.10.220 (talk) 18:58, 7 January 2008 (UTC)

That was me, of course. Pallida  Mors 19:01, 7 January 2008 (UTC)

It's okay, I don't think anyone is really that upset. I just felt this needs to be said. -- Meni Rosenfeld (talk) 19:12, 7 January 2008 (UTC)

Let's clarify this a little. Whether a limit

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

exists or not, of course, also depends on the domain D of f, which, if needed, should be made clear writing something like:

$\lim _{(x,y)\in D,\ (x,y)\to (0,0)}f(x,y)$

If we agree that here the domain is the set D:= {(x,y) in R² : x+y ≠0 }, then the limit does not exist, simply because in *any* neighborhood of (0,0) that function takes *all* real values. This excludes the possibility of having a limit (either finite or not). Notice also that in some smaller domain D' you would have in fact a limit =0. PMajer (talk) 18:21, 7 January 2008 (UTC)

To the original poster: As Meni Rosenfeld says, if the angle were 135º you would still have an indetermination of the type 0*infinity. Try the directions he gave above and you will be able to refute its existence. --Taraborn (talk) 19:06, 7 January 2008 (UTC)

It's not really necessary to look at it this way - as discussed above, with the definition of limit the OP is expected to work with, it is enough that the function isn't defined for $x=-y$ (or equivalently, in polar coordinates, angle of 135°). -- Meni Rosenfeld (talk) 19:20, 7 January 2008 (UTC)

Well, that's true too :) --Taraborn (talk) 20:48, 7 January 2008 (UTC)

Hopefully someone else will find this interesting: the idea of taking limits along straight lines can be expressed in terms of the topology of 'radially open' sets, that is subsets of the plane which, when they contain a point x, also contain open line segments in all directions through that point. This is strictly finer than the usual topology on the plane. 172.202.11.50 (talk) 01:14, 8 January 2008 (UTC)

^that's me. Algebraist 01:15, 8 January 2008 (UTC)