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July 15

Bounded variation

So, I'm working on a qualifier problem, as usual. Here, I just want to know if there is a typo. It asks about a function being of bounded variation on (0, 1):

If f is uniformly continuous on the open interval (0, 1), then f is of bounded variation on (0, 1).

Prove or disprove. See, the thing is, Royden, de Barra, Wikipedia all define bounded variation only on closed intervals. So, is it even defined on an open interval? And, if so, what does it mean? There seems to be some sort of minor/major error on every qualifying exam so I would not be surprised if it's wrong. But, I could also see the definition just being f is BV(a, b) when f is BV[c, d] for every a < c < d < b. StatisticsMan (talk) 01:40, 15 July 2009 (UTC)[reply]

A uniformly continuous function on (0, 1) extends uniquely to a uniformly continuous function on [0, 1], so it might just mean that the latter function is of BV. Algebraist 01:46, 15 July 2009 (UTC)[reply]

Exact. The variation of f on an open interval I is defined the same way as in the case of closed bounded intervals. If you(SM) are ok with the definitions on closed intervals, you can just say it is the sup of the variation of f on all bounded closed subintervals. For the easy counterexample on (0,1), the hint is: such an f should oscillate a lot, so as to have unbounded variation, still not too much, because it needs to have limits at 0 and 1. So, look for the right α in f(x):=x^αsin(1/x).....Note: the definition you (SM) suggest is a weaker thing; it means: f is BV_loc(I), locally BV.--pma (talk) 07:59, 15 July 2009 (UTC)[reply]

Also notice that if f is both continuous and of bounded variation on (0,1), then it is uniformly continuous; in any case, if f is BV on (0,1) then it admits limits at 0 and at 1; the corresponding extended function F on [0,1] so that it is continuous on 0 and 1 has of course the same variation as f on (0,1) (the variation of F on [0,x] is continuous at x if F is). So, this gives you another equivalent definition of a BV function f on (0,1) : it is the restriction to (0,1) of a BV function on [0,1], that you can assume continuous on the end-points. --pma (talk) 09:11, 15 July 2009 (UTC)[reply]

BV functions are differentiable almost everywhere, so a continuous nowhere-differentiable function will fail spectacularly. Or, without using known properties or standard weird functions, you could build a counterexample by hand by just making a function with a spike of height 1/n centred at 1/n. Algebraist 10:03, 15 July 2009 (UTC)[reply]

nice example... but be careful not to sit on it ;-) --pma (talk) 10:13, 15 July 2009 (UTC)[reply]

I already know that x sin(1/x) is the counterexample, though I do not have a proof. But, I just looked at the page for uniformy continuity and I found the Heine-Cantor theorem which says a continuous function on a compact space is uniformly continuous. If I define f to be x sin(1/x) except at 0 and 0 at 0, then it is continuous on [0, 1] and thus uniformly continuous. It seems pretty clear from the definition that if I now delete the endpoints from the domain that it's still uniformly continuous as the definition on [0, 1] says for any epsilon, there exists a delta such that for any x, y with d(x, y) < delta, then d(f(x), f(y)) < epsilon. In particular, it is true for any x, y in (0, 1). Then, bounded variation I have seen before and it's just to pick a sequence of points that give the max value of each oscillation. You get a sequence of partial sums that is divergent so it's not of bounded variation.

Speaking of that (since I was thinking it was BV at first but it's not), Royden does absolute continuity only with finite sums. But, the Wiki article says "(finite or infinite)". So, my question is, are the three definitions equivalent if you do it with 1) finite sums only, 2) infinite sums only, 3) both? I think they're all equivalent. StatisticsMan (talk) 14:31, 15 July 2009 (UTC)[reply]

Yes, they're all pretty obviously equivalent. Algebraist 14:44, 15 July 2009 (UTC)[reply]

elliptical tube cuts a plan

The cutting line of an elliptical tube and plan, is it an ellipsis? 88.72.242.144 (talk) 02:17, 15 July 2009 (UTC)[reply]

Yes, unless the axis of the tube is parallel to the plane, in which case you get either nothing, one line, or two parallel lines. --Spoon! (talk) 04:23, 15 July 2009 (UTC)[reply]

Yes, though it's an ellipse because an ellipsis is somehting else.. —Preceding unsigned comment added by 83.100.250.79 (talk) 16:13, 15 July 2009 (UTC)[reply]

Which is strange enough, given that both come from the same Greek word (ἔλλειψις). — Emil J. 16:56, 15 July 2009 (UTC)[reply]

Thank you Spoon!, Emil and 83.100.250.79 for your the quick answers. Do you know where to find a proof or reference to this result? 88.72.242.144 (talk) 21:24, 15 July 2009 (UTC)[reply]

Well, here's a crude sketch: If the tube were perpendicular to the plane, then the cross section is obviously an ellipse. As you tilt the plane, it just "stretches" the cross section (the cross section as you look down the tube is still the same, but now it lies on a tilted plane, so distances are farther in one direction). A stretched ellipse is still an ellipse, because stretching preserves the sign of the discriminant of a conic section. --Spoon! (talk) 05:01, 16 July 2009 (UTC)[reply]

You might find a perusal of Conic sections useful (remembering that a tube is merely a special case of cone). 87.81.230.195 (talk) 00:04, 21 July 2009 (UTC)[reply]

functions

please can anyone give me hints on these questions?

2f(x-1) - f(1/x - 1) = x

what is the value of f(x)?

next....

f(1) = 2005 f(1) + f(2) + f(3) + ...... + f(n) = n² * f(n), n>1

what is the value of f(2004)? —Preceding unsigned comment added by 122.50.137.12 (talk) 12:47, 15 July 2009 (UTC)[reply]

What is the domain of f supposed to be for the first question? Algebraist 12:59, 15 July 2009 (UTC)[reply]

As for the second one, rearranging the equation to (n² − 1)f(n) = f(1) + f(2) + f(3) + … + f(n − 1) gives you a recursive definition of f. In order to solve the recurrence, compute the first few values of the function, and see whether a pattern emerges. (Note that all values of the function are linear in the value of f(1); the picture will be much more clear if you start with f(1) = 1 at first, and only obfuscate it by multiplying it with 2005 after you solve it.) Then prove that your pattern is correct by induction on n, and plug in n = 2004 to get the result. — Emil J. 13:15, 15 July 2009 (UTC)[reply]

Better yet, subtracting the equation for n and n − 1 gives

f(n) = n²f(n) − (n − 1)²f(n − 1).

This recurrence is much easier to solve than the one above. — Emil J. 14:58, 15 July 2009 (UTC)[reply]

As for the first one, substitute 1/x for x to get

2f(1/x − 1) − f(x − 1) = 1/x.

Together with the original equation, you now have a system of two linear equations in two unknowns f(x − 1) and f(1/x − 1). Solve it. — Emil J. 13:34, 15 July 2009 (UTC)[reply]

Equation solving

I have just added the Wikiproject Mathematics template to the talk page of Equation solving. The article seems to have been pretty much ignored until now and it needs a lot of work. I have filled in the bits on ratings etc.. If someone wants to do a more official assessment then please do. Yaris678 (talk) 16:44, 15 July 2009 (UTC)[reply]

Wikipedia talk:WikiProject Mathematics is a more appropriate place for such messages than the Reference desk. — Emil J. 16:49, 15 July 2009 (UTC)[reply]

OK. Thanks. I will post it there. Yaris678 (talk) 18:01, 15 July 2009 (UTC)[reply]

On the wider point, am I right in thinking that Wikipedia talk:WikiProject Mathematics is for bringing up issues with mathematics articles, whereas the helpdesk is for questions about mathematics itself? Perhaps that should be made clear at the top of the article. Yaris678 (talk) 18:21, 15 July 2009 (UTC)[reply]

Yes, that is correct. At the top of which article? If you mean the Mathematics Reference desk (which is not an article), then since the header is shared with the other desks, it should probably be discussed in Wikipedia talk:Reference desk (the irony is not lost on me). -- Meni Rosenfeld (talk) 20:20, 15 July 2009 (UTC)[reply]

Yes, by article I mean desk. I have made a post on Wikipedia talk:Reference desk. Thanks, Yaris678 (talk) 12:40, 17 July 2009 (UTC)[reply]

test edit

\int _{0}^{\infty }{\frac {dx}{x}}>0.

TeX is dead while the image server is down, I think. Algebraist 20:46, 15 July 2009 (UTC)[reply]

Is this alleged fact that is stated by saying "the image server is down" supposed to be somehow known to the Wikipedia public? Where does one find such information? Michael Hardy (talk) 20:48, 15 July 2009 (UTC)[reply]

There should be a note at the top of the page related to the downage, though it doesn't mention maths rendering specifically: 'Uploads and Thumbnails generation have been temporarily halted while we upgrade our image store.' The best way to have up-to-date technical information is via IRC. Algebraist 20:51, 15 July 2009 (UTC)[reply]

The top of the page is where I'd never think of looking. One edits talk pages at the bottom of the page. Michael Hardy (talk) 20:54, 15 July 2009 (UTC)[reply]

Well, it's the established place for sitewide notices. More details here, btw. Algebraist 20:58, 15 July 2009 (UTC)[reply]

Thank you. I'd never have guess that "techblog" exists. I do have some vague suspicion that the hardware and software that makes Wikipedia work were not actually brought down from Heaven by an archangel at the beginning of Time, but it's only a vague suspicion. Michael Hardy (talk) 21:06, 15 July 2009 (UTC)[reply]

And now the initial line I posted above is getting properly rendered. Maybe we're back to normal. Michael Hardy (talk) 21:12, 15 July 2009 (UTC)[reply]