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August 22

Nothing?

'Cmon, something? --hydnjo talk 23:17, 22 August 2008 (UTC)[reply]

There was something, but it was immediately reverted by its poster. Algebraist 23:19, 22 August 2008 (UTC)[reply]

But if you really want something, here goes: Is there any place in the world where today, the anniversary of one of the founding fathers of modern mathematics, is celebrated as it deserves? Algebraist 23:24, 22 August 2008 (UTC)[reply]

Lots of people have big celebrations on Newton's birthday, does that count? ;) --Tango (talk) 17:58, 23 August 2008 (UTC)[reply]

Newton's birthday is on the 4th of January. I was speaking of the 21st of August. ~~Where are these celebrations, anyway? I never noticed them, and I spent the last four years at his college.~~ Algebraist 22:07, 23 August 2008 (UTC)[reply]

Walked into that one. I take it you're using old style? Algebraist 22:09, 23 August 2008 (UTC)[reply]

I use whatever calendar is necessary for my jokes to work! --Tango (talk) 02:27, 24 August 2008 (UTC)[reply]

Ah, I misread you - I switched the words "where" and "today" and thought you were using "today" to means modern times and were just asking a general question. I doubt anywhere celebrates Cauchy's birth - he doesn't get all that much recognition within the mathematical community that I've seen, certainly not in the wider population. I suppose his home town might raise a toast to him, but that's probably it. --Tango (talk) 02:34, 24 August 2008 (UTC)[reply]

If not, I shall have to visit Paris specially for the purpose, next year perhaps. On the recognition-in-mathematics issue, he does pretty well on the eponym front, which is about the most you can expect of such a (generalizing wildly) historically ignorant profession. Algebraist 19:34, 26 August 2008 (UTC)[reply]

Wow, that's pretty tough criticism, even for a wild generalization. I take it you mean that mathematicians have been generally ignored by history, not that they are generally ignorant of history. -- JackofOz (talk) 00:21, 27 August 2008 (UTC)[reply]

What I meant (I'm not sure if it's true or not) is that mathematicians tend to be less aware of the history of their own field than people in other branches of academia. Algebraist 18:46, 27 August 2008 (UTC)[reply]

Semplifying my problem above

Since nobody seems to be going to answer to my question above I'll try with a semplified version of the problem which only requires to know standard calculus to be understood.

Let $T$ be a map (from $\mathbb {R} ^{2}$ to itself) and $x_{0}$ be the unique fixed point of $T$ .

Let $x_{1}$ be a point such that the sequence $x_{n+1}=T(x_{n})$ converges to $x_{0}$ .

Let $T_{\epsilon }$ be a map which also has $x_{0}$ as its unique fixed point and such that

|T(x)-T_{\epsilon }(x)|<\epsilon

for any

x

.

Question: if we fix a neighbourhood $B$ of $x_{1}$ is it true that for $\epsilon$ small enough there exists $y_{1}$ inside $B$ such that $y_{n+1}=T_{\epsilon }(y_{n})$ converges to $x_{0}$ ?

--Pokipsy76 (talk) 09:03, 22 August 2008 (UTC)[reply]

At the very least, you're going to need some continuity assumptions on T. What are you assuming? Algebraist 11:42, 22 August 2008 (UTC)[reply]

You are obviously right. Let's assume

T

and

T_{\epsilon }

are continuous on

\mathbb {R} ^{2}

.--Pokipsy76 (talk) 18:32, 22 August 2008 (UTC)[reply]

Smooth injective counterexample (I think):

T

and

T_{\epsilon }

both map the plane to

B_{\epsilon /2}(x_{0})\!

; T is contractive while

T_{\epsilon }

is a rotation around

x_{0}

composed with a monotonic radial map which expands some neighborhood of

x_{0}

. -- BenRG (talk) 13:28, 22 August 2008 (UTC)[reply]

But T is fixed: it cannot have a behavoiur which depends on epsilon.--Pokipsy76 (talk) 18:32, 22 August 2008 (UTC)[reply]

Oops. Then what if

T(x)=x_{0}

? No longer injective but still continuous. -- BenRG (talk) 21:28, 22 August 2008 (UTC)[reply]

For T bijective, how about

T(r,\theta )=({\text{sinh}}^{-1}\,r,\theta +\Delta \theta )

and

T_{\epsilon }(r,\theta )=({\text{sinh}}^{-1}\,r+\epsilon \,{\text{tanh}}\,r,\theta +\Delta \theta )

, for any nonzero angle

\Delta \theta

? -- BenRG (talk) 21:47, 22 August 2008 (UTC)[reply]

I think your counterexample is ok. It works because you consider a degenerate fixed point (T is tangent to the identity in the fixed point). I suppose that I should have considered the C¹ norm and assumed some non-degeneracy condition for the fixed point of T.--Pokipsy76 (talk) 08:24, 25 August 2008 (UTC)[reply]

Yes I'd been wondering what he meant, perhaps he meant maps satisfying the Banach fixed point theorem? Dmcq (talk) 16:43, 22 August 2008 (UTC)[reply]

No, the maps need not be contractions.--Pokipsy76 (talk) 18:32, 22 August 2008 (UTC)[reply]

The page you refer to in the previous post deals only with hyperbolic fixed points for flows, not for maps. Should this be fixed? Oded (talk) 19:19, 22 August 2008 (UTC)[reply]

Not sure what the previous question was but if the map isn't a contraction then for instance

T(x,y)->(x-(|x|+|y|)/(1+|x|),y)

when iterated moves points on the line

(x,0)

for positive x towards

(0,0)

but has no other fixed points and no other points go to

(0,0)

. It is also continuous. Points with non-zero y will never converge to

(0,0)

no matter how close to a point on the line. So

T_{\epsilon }

can be the same as

T

. Does this do the job as a counterexample? Dmcq (talk) 19:54, 22 August 2008 (UTC)[reply]

Sorry I see you want there to be no point within a short distance of

x_{0}

which goes to the fixed point. Yes I see that previous counterexample now. Just have

x_{0}

be the fixed point at the centre of a rotation and that is a counterexample. It always maps to itself and nothing else does no matter how close. Dmcq (talk) —Preceding undated comment was added at 20:12, 22 August 2008 (UTC)[reply]

The the distance between the points $T_{\epsilon }(y_{n})$ and $T(x_{n})$ would have to decrease with increasing n. In fact, if z_n is the distance between $T_{\epsilon }(y_{n})$ and $T(x_{n})$ , then z_n would have to converge to 0. —Preceding unsigned comment added by Topology Expert (talk • contribs) 13:37, 24 August 2008 (UTC)[reply]

Category of monoids

Is the category of monoids (in Set) Cartesian closed? 83.23.209.193 (talk) 15:25, 22 August 2008 (UTC)[reply]

No, while it is cartesian, it isn't even a closed category. Or look at the exponential law applied to the one element monoid.John Z (talk) 20:06, 30 August 2008 (UTC)[reply]

Thanks. 83.23.247.146 (talk) 02:06, 2 September 2008 (UTC)[reply]

About the Möbius strip

I'm reading about topology, and I can't figure this one out.

Why do a paper strip and a Möbius strip have different intrinsic topologies? Or as they ask it, How does a Two-Dimensional creature (Flatlander) living on a Möbius strip logically implies that it's not a paper strip?... —Preceding unsigned comment added by DarkLaguna (talk • contribs) 15:43, 22 August 2008 (UTC)[reply]

I'm not sure what you mean by a paper strip; I shall assume you mean a cylinder, i.e. a Möbius strip without the half twist. In this case, a difference is that the paper strip is orientable while the Möbius strip is not. Concretely, if a flatlander set off on a trip round the Möbius strip, he would end up back where he started, but would find that he had become the mirror-reflection of his former self. Algebraist 15:48, 22 August 2008 (UTC)[reply]

Why would he be a mirror image? If the "circumference" of the mobius strip is the length of the original piece of paper, then after 2 laps he is back where he started as he was. After one lap, he is still "himself" but is just on the other side of the piece of paper - which stops him knowing he is the other way round etc. Are you assuming that the mobius flatland allows interaction between the two "sides" of the strip? If so why is that assumption justified / necessary? -- SGBailey (talk) 22:13, 23 August 2008 (UTC)[reply]

A truly two-dimensional strip doesn't have 'sides', just as our three-dimensional space doesn't, and the flatlanders live in it rather than on it. Of course, the OP used 'on' in the question, so my interpretation may be wrong. If they are living on the surface of the strip, then there's no topological difference between their home and a cylinder. Algebraist 22:16, 23 August 2008 (UTC)[reply]

Unless you consider the sides to be joined by the boundary, allowing you to cross over. Then you would know you were on a Moebius band, not a cylinder, if you could go half way round, cross over the boundary and get back to where you started. I'm not sure how to describe that topologically, but it works in the intuitive sense of an ant walking on the surface. (Actually, if you consider the two sides to be distinct, and joined by the boundary, then I think you have the Klein bottle.) --Tango (talk) 22:44, 23 August 2008 (UTC)[reply]

No, that's not the Klein bottle. Tango, I suggest you read Algebraist's entry again. Oded (talk) 22:51, 23 August 2008 (UTC)[reply]

If you consider a paper Möbius strip as a three-dimensional object, and then consider its boundary (smoothing out that pesky edge), then you just get a torus, not a Klein bottle. The same applies if you start with a non-Möbius strip. Tango's method doesn't work since going 'half way round' doesn't make sense topologically. Algebraist 23:01, 23 August 2008 (UTC)[reply]

Oh yeah, it is a torus - the ends are identified with a half-rotation, not a reflection, that doesn't help then. My method depends on the boundary being recognisable in some way. If it is, then the Moebius band is recognisable because there exists a closed curve which crosses the boundary precisely once. If the boundary isn't recognisable, then it's just a torus and there's no difference. --Tango (talk) 00:20, 24 August 2008 (UTC)[reply]

Ah, I see. Just for fun, I wonder if there's some sensible way to topologize the 'surface' of a Möbius strip so that the boundary is topologically recognizable. Any ideas? Algebraist 00:26, 24 August 2008 (UTC)[reply]

I am not 100% if I understand your question the way you meant it, but the natural thing is to think of it as a laminated 2-manifold. In this case this just means that on the surface there is a curve which you require the transition maps of the manifold atlas to respect. Perhaps it can also be thought of as an orbifold; I'm not sure. Oded (talk) 00:57, 24 August 2008 (UTC)[reply]

I assumed you could do something like that. I was just wondering if there's a sensible answer to the question 'what topological space is the surface of a Möbius strip' (which is, as it stands, not quite well-posed enough to necessarily have a unique answer) which allowed you to retain the boundary. Algebraist 01:18, 24 August 2008 (UTC)[reply]

I guess it comes down to definition. What is required for it to count as the surface of a Moebius strip? I suppose everywhere other than the boundary should be locally Euclidean, otherwise imagining flatlanders living there becomes rather difficult, and that's really all you need since you only need to consider the surface near a section of the boundary, so basically what you need is a way to identify the boundaries of two half-planes in such a way as to make it recognisable. I'll need to think about what that actually means... Obviously, if we consider it as a differentiable manifold, then it's easy - that's probably the best way to do it! By the way, aren't you British? What's with all the z's? --Tango (talk) 01:02, 24 August 2008 (UTC)[reply]

I am a citizen of the Kingdom of Kent, owing no allegiance to anything north of the Thames. Remember Otford! At present, however, I'm preparing to commence my studies at Oxford. Algebraist 01:18, 24 August 2008 (UTC)[reply]

Ok, how about this: Take two closed half-planes, identify the boundary and then, instead of giving it the usual quotient topology, give it the topology formed by taking the union of the topologies of each half-plane as a subbasis. Away from the boundary, everything is as it should be and the boundary is clearly identifiable since it's the only place you can have line segments that are open. Exactly how you do the same thing with a Moebius band, I don't know, but the principle should be exactly the same locally, it's only globally that things are different. --Tango (talk) 02:25, 24 August 2008 (UTC)[reply]

I agree with Algebraist (as usual). Here are a few other ways to think about it. First, the boundary of the cylinder consists of two disjoint circles, while the boundary of the Mobius strip is just one circle. You could object to this by saying that you are interested in the open Mobius trip (without the boundary) and the open cylinder. Another observation is that if you remove from the cylinder a simple closed curve, you disconnect it. However, if you remove the "equator" from the Mobius strip, what you get is a cylindar. Oded (talk) 16:28, 22 August 2008 (UTC)[reply]

On beyond `expm1`

The expm1 function effectively removes the first term of the Taylor series of $e^{x}$ without the cancellation caused by doing it manually. Does anyone know of an extant function (a CS question) or an expression (a numerical math question) that would remove the first two terms and accurately calculate $e^{x}-1-x$ ? (Example: an expression of expm1 in terms of exp and sinh.) --Tardis (talk) 16:07, 22 August 2008 (UTC)[reply]

Ummm, why not just compute the Taylor series minus whichever terms you want to exclude? Dragons flight (talk) 16:56, 22 August 2008 (UTC)[reply]

I would like to avoid switching between multiple algorithms (as the series would converge slowly for large x) and to avoid rounding error in summing the series explicitly. I would trust a simple combination of standard functions (with known, small errors) more than a long sequence of elementary operations. --Tardis (talk) 17:19, 22 August 2008 (UTC)[reply]

The version on the python list expm1(x) = 2*exp(x/2)*sinh(x/2) looks incredibly inefficient and does not improve accuracy. I would not take this as a base for any numerical algorithm. --Salix alba (talk) 09:02, 23 August 2008 (UTC)[reply]

You must have misread the formula. The only area where expm1 is problematic is around 0, and the sinh formula neatly avoids the cancellation there. It should be at worst about 2x or 3x more expensive than exp. It's obviously more efficient to use a custom Taylor series or interpolating polynomial with a fancy argument reduction scheme, but it's up to you whether you want to spend an evening implementing and debugging that or if you're happier typing in a simple formula that just works^TM. Fredrik Johansson 10:38, 23 August 2008 (UTC)[reply]

It might be worth having a look at the source code of some real implementations, for example[1]. It seems that to calculate exp(x) use x=k ln(2) + r, with |r|< ln(2)/2, k integer. So exp(x) = exp(r) * exp(k ln(2)) = exp(r) * 2^k. The use a Reme algorithm to find a 5 degree polynomial in r² which can efficiently and accurately calculate exp(r)-1. --Salix alba (talk) 09:54, 23 August 2008 (UTC)[reply]

Sine function

It seems that $Im(x^{x})\times |x|^{|x|}=sin(x\times \pi )$ for negative numbers. Why is that? Is there a name for this equality? Thanks --helohe (talk) 22:37, 22 August 2008 (UTC)[reply]