Wikipedia:Reference desk/Archives/Mathematics/2007 May 29

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May 29

Klein Bottle

Is Solid Klein bottle correct? I thought the Klein bottle had no meaningful interior. Black Carrot 07:46, 29 May 2007 (UTC)

I'm pretty sure it's fine, and Google seems to agree with me. The Klein bottle has no meaningful interior when immersed in three-space (and possibly in euclidean space in general, I'm not sure), but who cares about euclidean space? Algebraist 09:01, 29 May 2007 (UTC)

This raises some issues, though, at least for me, with the popular description given in the Klein bottle article as the bottle having no distinction between the "inside" and "outside" surfaces. Embedded in R⁴, also an orientable surface like the traditional sphere S² has no distinction between the "inside" and "outside" of the surface. As the solid Klein bottle shows, its embedding in the quotient space of R³ using the equivalence relation induced by gluing two full planes using the same reflection (e.g., so that (x, y, 1−ε) is close to (−x, y, ε)) has a boundary with a distinct inside and outside: as x²+y² approaches 1, a point (x, y, z) is either in the solid bottle or outside, and this is independent of the value of z. The claimed indistinctness appears to hold specifically for the immersion in R³, and appears not to be an intrinsic property of the manifold. Am I right, and if so, should this be corrected?  --Lambiam^Talk 11:29, 29 May 2007 (UTC)

By "no distinction between inside and outside", I believe it means that any point in the space can be reached from any other without crossing the boundary. In the case of a sphere embedded in Euclidean space of one higher dimension, that's not so - it divides the space in two. Black Carrot 12:57, 29 May 2007 (UTC)

But S² embedded in R⁴ does not divide the space in two.  --Lambiam^Talk 15:02, 29 May 2007 (UTC)

Correction: "no distinction between inside and outside" means, according to the orientability article, that an "ant" crawling on one side at a particular spot can get to the other side of that same spot without leaving the surface or flipping over an edge. Black Carrot 13:21, 29 May 2007 (UTC)

That article contains no mention of "inside" or "outside" and the presence or absence of a "distinction" between the two, so how can this be the meaning of "no distinction between inside and outside" according to that article?  --Lambiam^Talk 15:02, 29 May 2007 (UTC)

"In general, the property of being orientable is not equivalent to being two-sided; however, this holds when the ambient space (such as R3 above) is orientable. For example, a torus embedded in can be one-sided, and a Klein bottle in the same space can be two-sided; here K2 refers to the Klein bottle." It doesn't go to the trouble of labeling one side "in" and another "out", but it is talking about the distinction between them. Black Carrot 15:20, 29 May 2007 (UTC)

It may be talking about shapes like the Mobius strip, which are one-sided and nonorientable. There is a reason why it doesn't discuss issues of inside and outside. nadav (talk) 17:50, 29 May 2007 (UTC)

What would you call a solid Klein bottle (as defined in the article) packed into Euclidean 3space in the same way as the first picture of Klein bottle? It wouldn't be an embedding, since it'd have to overlap itself, and I don't think it'd be an immersion either. That flat bit where the thing suddenly turns around seems like it would be a problem. Black Carrot 15:25, 29 May 2007 (UTC)

Yes, but so what? We were discussing "conventional" non-solid Klein bottles. Notions like "inside" and "outside" are meaningless even for (for example) an orientable surface like S². It is only when it is embedded or immersed or encompassed in an orientable three-dimensional space that a surface may give rise to "sides", separated by the surface. The formulation in the article has no such reservation, and may easily give the impression that the two notions are equivalent in general.  --Lambiam^Talk 19:29, 29 May 2007 (UTC)

I see now. Yeah, you're right, that might need some rephrasing. Do you have any suggestions on my second question, about immersion? Black Carrot 09:47, 30 May 2007 (UTC)

I used "encompassment" above, but that was in semi-jest. The map is continuous and nowhere constant (meaning not constant in any environment of any point). These are the only properties of interest I see. It could be made smooth without change to any topological properties, but I am inclined to conjecture it can't be made to have an everywhere-injective derivative/differential. I don't know a specific name for nowhere constant continuous manifold-to-manifold maps.  --Lambiam^Talk 10:25, 30 May 2007 (UTC)

Another point to add to this is that the Solid Klein bottle is nonorientable. A nonorientable manifold may have nonorientable (or orientable) boundary, whereras the boundary of an orientable manifold must always be orientable. (Nonorientable is the mathematical concept you're probably thinking of when you say "no meaningful interior.") Kfgauss 09:04, 1 June 2007 (UTC)

geometric progressions

find the result of

1 + 1/3 - 1/5 + 1/7 - ....... upto infinity 122.163.81.209 17:30, 29 May 2007 (UTC)

Okay, I've found it. -- Meni Rosenfeld (talk) 17:40, 29 May 2007 (UTC)

I am afraid that a hint could give too much away here, and that this could very well be a HW problem. I'll just say that the person should carefully review the article Taylor series, and see if this particular series is similar to anything there. nadav (talk) 17:42, 29 May 2007 (UTC)

This is not a geometric progression. One hint that, hopefully, does not give too much away, is to look first at 1 − 1/3 + 1/5 − 1/7 + ... For checking that you found the right result: it should have a numeric value of about 1.2146.  --Lambiam^Talk 19:07, 29 May 2007 (UTC)

1 + 1/3 - 1/5 + 1/7 - ...

BECOMES

1

PLUS

+ 1/3 + 1/7 + 1/11 + ...

MINUS

+1/5 +1/9 +1/13 + ...

SO

1 + A - B

Find A, A=Sum[n=0,inf,1/(3+4n)]

Find B, B=Sum[n=0,inf,1/(5+4n)]

Now bind "A - B" together as C

1 + C where C = Sum[n=0,inf,1/(3+4n) - 1/(5+4n)] = Sum[n=0,inf,2/(15+32n+16n^2)]

Ate Negative Pie over Floor is the final answer.

202.168.50.40 00:45, 30 May 2007 (UTC)

How will puking pie on the floor solve a math problem!? =P --Wirbelwindヴィルヴェルヴィント (talk) 01:57, 30 May 2007 (UTC)

${\frac {8-\pi }{4}}=1.214601837\,$

202, in general we try not to "give away" the solution to homework problems, which this might well be.  --Lambiam^Talk 09:55, 30 May 2007 (UTC)

It should be pointed out that 202's observation that $1+{\tfrac {1}{3}}-{\tfrac {1}{5}}+{\tfrac {1}{7}}-\cdots =1+({\tfrac {1}{3}}+{\tfrac {1}{7}}+{\tfrac {1}{11}}+\cdots )-({\tfrac {1}{5}}+{\tfrac {1}{9}}+{\tfrac {1}{13}}+\cdots )$ is not correct under any notion of summation I am familiar with, as those series diverge. -- Meni Rosenfeld (talk) 13:54, 30 May 2007 (UTC)

It's (more or less) a correct statement in terms of the disordered sum (aka Lebesgue integral) of a series, but all it does is shows that this sum is not defined (i.e. this series is not absolutely convergent/integrable), so it's not of any use. Algebraist 14:42, 30 May 2007 (UTC)

I think he was employing telepathy to somehow visualize the answer. It appeared to him in a chaotic jumble of disparate images. nadav (talk) 15:01, 30 May 2007 (UTC)

$\arctan x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}\quad {\mbox{ for }}\left|x\right|\leq 1$

And God Said "Let x be 1"

$\arctan 1=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}$

$\arctan 1={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-...$

$-1\times \arctan 1={\frac {-1}{1}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{7}}-{\frac {1}{9}}+...$

$2+(-1\times \arctan 1)=2+{\frac {-1}{1}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{7}}-{\frac {1}{9}}+...$

$2+(-1\times \arctan 1)={\frac {1}{1}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{7}}-{\frac {1}{9}}+...$

$2+(-1\times {\frac {\pi }{4}})={\frac {1}{1}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{7}}-{\frac {1}{9}}+...$

${\frac {8-\pi }{4}}={\frac {1}{1}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{7}}-{\frac {1}{9}}+...$

Ate Negative Pie over Floor is still the final answer.

202.168.50.40 22:50, 30 May 2007 (UTC)

Yes it is, but please don't give away detailed solutions to problems that may have been assigned for homework. nadav (talk) 22:57, 30 May 2007 (UTC)