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January 22

Fibre Bundles and Actions

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Consider a fibre bundle π : E → B with fibre F. In a problem that I've been working on, F is an abelian group (actually a commutative ring). As an abelian group, F has a natural action on E. I'm assuming that this has been studied before. Does this kind of set up have a name? — Fly by Night (talk) 02:04, 22 January 2011 (UTC)[reply]

I'm not sure, but if I understand you correctly, the condition "F has a natural action on E" implies that what you have is actually just the trivial bundle B×F. The natural action fixes a preferred isomorphism between F and π^-1(b) for every b in B.

This is not true each time you have a fiber bundle with an abelian group as fiber. For example, the tangent bundle of a two-dimensional manifold has R², an abelian group under vector addition, as fiber, but that doesn't mean that R² itself acts on the entire bundle -- it does not make sense to add the vector (1,4) to a vector field everywhere. –Henning Makholm (talk) 03:33, 23 January 2011 (UTC)[reply]

When I said natural, I didn't mean canonical. I just meant natural as in natural, i.e. neither contrived nor convoluted. (When I say convoluted I mean convoluted in the regular sense, i.e. nothing to do with integration.) The definition of a fibre bundle is that π⁻¹(b) ≃ F for all b ∈ B, and that for any open neighbourhood U ⊊ B one has π⁻¹(U) ≃ U × F. — Fly by Night (talk) 05:09, 23 January 2011 (UTC)[reply]

Hm, I assumed that when one talks about "fiber bundle with fiber X" where X is a space with more structure than just a topology, then each π⁻¹(b) must carry the same kind of structure as X and be isomorphic to X (though possibly not uniquely) with respect to that structure. Now, reading the article closely I can see that it does not actually contain such a requirement; it merely wants π⁻¹(b) and X to be homeomorphic. Is it equivalent to my naive assumption to require that the structure group is a subgroup of the automorphism group for X? –Henning Makholm (talk) 06:29, 23 January 2011 (UTC)[reply]

I found the article I was looking for. It was Principal bundle. Thanks anyway. — Fly by Night (talk) 04:17, 23 January 2011 (UTC)[reply]

Limit of a sequence of complex polynomials discontinuous at a single point

Hello everyone,

I am looking into the existence of a sequence of polynomials p_n in C for which p_n(w)=1 for some fixed w, all n (wlog I suppose w=0), and for all z not equal to w, p_n(z) -> 0 as n -> infinity. (A sort of finite 'delta function' in C)

Does there exist such a sequence? I am unaware of any theorems which would help me prove the nonexistence, but at the same time, I'm not having any luck deriving a contradiction by assuming such a sequence exists - could anyone suggest anything to me? Of course perhaps there is one, but in my mind I find it very hard to conceptually construct such a sequence. Thankyou very much for the help, Simba31415 (talk) 04:23, 22 January 2011 (UTC)[reply]

Just let me rewrite the question in LaTeX. You want a sequence of polynomials

p_{n}\in \mathbb {C} [z]\ {\text{such that}}\ p_{n}(z_{0})=1\ {\text{for all}}\ n\in \mathbb {N}

{\text{and, for all}}\ z\neq z_{0},\,p_{n}(z)\to 0\ {\text{as}}\ n\to \infty \,.

I'd be really surprised if such a functions did exist. All polynomials are holomorphic functions which is a very strong condition regarding differentiability. Your limit function is not holomorphic (no non-constant function from C to R can be). But on top of that, you want pointwise convergence to a discontinuous function. And it's not like the discontinuity is an accumulation point, it's isolated in the range. — Fly by Night (talk) 04:46, 22 January 2011 (UTC)[reply]

Be really surprised. A sequence satisfying the stated conditions can be constructed using Runge's theorem. For each

k\geq 1

, apply Runge's theorem with empty A and

K_{k}=\{re^{i\theta }\mid r\in [2/k,2k],\theta \in [1/k,2\pi ]\}\cup \{0\}

f_{k}(z)={\begin{cases}1&|z|<1/k\\\mathrm {undefined} &|z|=1/k\\0&|z|>1/k\end{cases}}

This gives a series of polynomials

(p_{k,n})_{n}

that approximate

f_{k}

uniformly on

K_{k}

. One of these, say

p_{k,n_{k}}

, will differ by at most

1/k

from

f_{k}

on

K_{k}

. Call this polynomial

q_{k}

.

Now it is easy to see that the sequence

(q_{k})_{k\geq 1}

converges pointwise towards 1 at 0 and towards 0 everywhere else.

However, pointwise convergence is not really enough to meet Simba31415's intentions, so the resulting sequence will not behave very delta-like. For example, for any holomorphic f, the average value of

f

on any annulus centered on the origin must be

f(0)

, as a consequence of Cauchy's integral formula. –Henning Makholm (talk) 00:20, 23 January 2011 (UTC)[reply]

You seem to be using the following from our article: "If K is a compact subset of C such that C\K is a connected set, and ƒ is a holomorphic function on K, then there exists a sequence of polynomials (p_n) that approaches ƒ uniformly on K." This has K as a fixed set, it doesn't depend in n. Also, think about ƒ₁. We have ƒ₁(z) = 1 for all | z | < 1 and ƒ₁(z) = 0 for all | z | > 1. So it's discontinuous alone the unit circle | z | = 1. That means that ƒ₁ fails to be holomorphic on K₁ and that ƒ₁ fails to be a polynomial. — Fly by Night (talk) 04:14, 23 January 2011 (UTC)[reply]

I stand by my construction.

I'm using Runge's theorem once for each k. It gives me an entire (anonymous) sequence of polynomials; I select one of those (not necessarily the kth one, but one that approximates f_k to within the tolerance I specify) and throw out the rest. Then for the next k I make a new K_k+1 and a new f_k+1 and apply Runge anew, pick out some polynomial from the resulting sequence and call it q_k+1, and so forth.

The article on Runge's theorem calls the sequence produced by the theorem "(p_n)", but I don't have to use that name for them (and I can't, because p_n already means something else in the OP's description of the problem). In an attempt to minimize confusion I don't call them anything but just use the fact that one of them will approximate f_n well enough to become one of my approximators for the OP's function. Know what, I'll just edit everything such that they are p_n, at the cost of calling the final result of the construction q_k instead of p_n.

It is correct that f₁ is not defined on the unit circle, and so not continuous there. But is is holomorphic in its entire domain (which is C minus the unit circle), and that domain is open and a superset of K₁. Note that K₁ does not contain the unit circle either; it only contains 0 and part of the circle with radius 2 (because 2/k=2k=2 when k=1. In the general case K_k contains 0 and part of an annulus with inner radius twice the radius of the circle that is missing from f_k). –Henning Makholm (talk) 05:44, 23 January 2011 (UTC)[reply]

Henning, I see you made quite a few changes ([1] and [2]) to your orginal post after it had been commented on. You shouldn't make such changes once people have started to comment; it can be very misleading. As your post stood, before the alterations, ƒ₁ was not holomorphic on its domain. You had some pairwise disjoint neighbourhoods that you deleted. Meaning C\K₁ would have included open intervals on the unit circle; where ƒ₁ is not continuous. — Fly by Night (talk) 17:25, 23 January 2011 (UTC)[reply]

The definition of f₁ has not changed, except for renaming the subscript from n to k. Every definition of f₁ in the history is identical except for this name change, and f₁ has always been holomorphic on its domain.

The T_k had not been commented on by the time I simplified them away; they were never used in the definition of f₁.

C\K₁ has always contained the entire unit circle and still does. That is why it doesn't matter that f₁ is not holomorphic there. It only has to be holomorphic on some open superset of K₁.

I stand by each of the historical versions of the construction:

this original version,
this with n renamed to k everywhere but no substantial changes, or
this with a fixed choice of T_k hardcoded.

–Henning Makholm (talk) 18:00, 23 January 2011 (UTC)[reply]

Regarding: "The T_k had not been commented on by the time I simplified them away...", it is bad form to edit a post when someone has commented on it. Full stop! You can't say "Oh well he didn't mention the third sentence in his post so I'll change that as I choose. You post was read, and commented upon, as a whole. I'm please you cleared up the problems. — Fly by Night (talk) 20:30, 23 January 2011 (UTC)[reply]

For reference here are the original and the current versions side by side
original	current
Be really surprised. A sequence satisfying the stated conditions can be constructed using Runge's theorem. First select countably many pairwise disjoint open subintervals $T_{n}$ of $[0,2\pi ]$ . Then for each $n$ , apply Runge's theorem with empty A and $K_{n}=\{re^{i\theta }\mid r\in [2/n,2n],\theta \in [0,2\pi ]\setminus T_{n}\}\cup \{0\}$ $f_{n}(z)={\begin{cases}1&\|z\|<1/n\\\mathrm {undefined} &\|z\|=1/n\\0&\|z\|>1/n\end{cases}}$ This gives a series of polynomials that approximate $f_{n}$ uniformly on $K_{n}$ ; select $p_{n}$ to differ by at most $1/n$ from $f_{n}$ on $K_{n}$ . Then it is easy to see that the $p_{n}$ 's converge pointwise as required.	Be really surprised. A sequence satisfying the stated conditions can be constructed using Runge's theorem. For each $k\geq 1$ , apply Runge's theorem with empty A and $K_{k}=\{re^{i\theta }\mid r\in [2/k,2k],\theta \in [1/k,2\pi ]\}\cup \{0\}$ $f_{k}(z)={\begin{cases}1&\|z\|<1/k\\\mathrm {undefined} &\|z\|=1/k\\0&\|z\|>1/k\end{cases}}$ This gives a series of polynomials $(p_{k,n})_{n}$ that approximate $f_{k}$ uniformly on $K_{k}$ . One of these, say $p_{k,n_{k}}$ , will differ by at most $1/k$ from $f_{k}$ on $K_{k}$ . Call this polynomial $q_{k}$ . Now it is easy to see that the sequence $(q_{k})_{k\geq 1}$ converges pointwise towards 1 at 0 and towards 0 everywhere else.

–Henning Makholm (talk) 18:31, 23 January 2011 (UTC)[reply]

Thanks for adding those tables. They look very impressive. — Fly by Night (talk) 20:27, 23 January 2011 (UTC)[reply]

sine function

we know that sin 90 =1or sin pi/2=1.......and we also know taylors exapansion of sine funtion,in that expansion if we put x=pi/2 do we get exact value 1??what is the equivalence of both the definitions of sine function Meghna goswami (talk) 04:51, 22 January 2011 (UTC)[reply]

Short answer: Yes. The sine function is equal to its Taylor expansion everywhere on the real line. Remember that the Taylor expansion has infinitely many terms, thus to substitute pi/2 and get an exactly value of 1 one would have to carry out infinitely many operations (to an infinite number of decimal figures). Zunaid 06:19, 22 January 2011 (UTC)[reply]

The general equivalence may be shown as follows: Start with this definition of sine: On the unit circle centered on the origin for any given arc-length t (positive if taken in the counterclockwise direction, negative otherwise), starting from (1,0) and ending at a point (x,y) the value of sine(t) is the y-coordinate of that point, i.e. sine(t)=y. This also implies via similar triangles the definition done usually in school of sine to be the ratio of the perpendicular to the hypotenuse in a right triangle. Next we define cosine similarly to be the x-coordinate. We now proceed to develop the trigonometric identities and show that the derivative of sine is cosine and that of cosine is -sine. Thus sine satisfies the ODE y'' + y = 0 ; y(0) = 0, y'(0) = 1.

Now define a new function, call it Sine via the Taylor series expansion. Again its properties may be developed so that it is clear that it satisfies the above ODE. The existence and uniqueness theorem of differential equations now implies that sine=Sine and we are done.-Shahab (talk) 06:39, 22 January 2011 (UTC)[reply]

Filters in Topology

HI! I am having a problem in understanding the difference between the notion of an Ultrafilter to that of a proper filter. I mean, an Ultrafilter is a maximal proper filter so, doesn't it have to be principal? Can someone give me an example (a simple one if possible) for an Ultrafilter which is not principal? Thanks! Topologia clalit (talk) 06:46, 22 January 2011 (UTC)[reply]

~~Consider the collection of cofinite subsets of an infinite set. —Bkell (talk) 08:14, 22 January 2011 (UTC)~~[reply]

(ec)That's a good example of a filter that's not maximal, but I think the original poster wanted an ultrafilter that's not principal.

The short (but not clearly correct) answer is that you can't have an "example" of such a thing, because it requires the axiom of choice to prove that it exists at all. As I say, it's not clear that this really follows — there could be an explicitly definable ultrafilter on an infinite set, depending on certain set-theoretic propositions that are independent of ZFC. However, there can't be one with a simple definition (e.g. an ultrafiliter on the natural numbers cannot be a Borel subset of P(N), where we take the usual topology that makes P(N) into a Polish space). --Trovatore (talk) 08:20, 22 January 2011 (UTC)[reply]

(edit conflict) Hmm, the collection of cofinite subsets of an infinite set is not itself an ultrafilter, but it is a filter. Extend that filter to an ultrafilter in any way you like (you might need the axiom of choice to do this). Then that ultrafilter is not principal. In fact, an ultrafilter on an infinite set X is nonprincipal if and only if it contains all cofinite subsets of X. —Bkell (talk) 08:21, 22 January 2011 (UTC)[reply]

Yes, you definitely need AC. For example, the axiom of determinacy implies that there are no non-principal ultrafilters on the natural numbers. Interestingly, it implies that there is a countably complete nonprincipal ultrafilter on

\aleph _{1}

, which is false in the real world where AC holds. It even gives an explicit example of one (a subset of

\aleph _{1}

is in the ultrafilter just in case it has a subset that is closed and unbounded in

\aleph _{1}

). In the real world that's not an ultrafilter, because using AC you can show there's a subset of

\aleph _{1}

such that neither it nor its complement has a club subset. --Trovatore (talk) 08:41, 22 January 2011 (UTC)[reply]

You don't actually need the whole of AC to get the existence of a non-principal ultrafilter, though it is sufficient. The ultrafilter lemma, that every filter can be completed to an ultrafilter, implies the existence of a non-principal ultrafilter on the natural numbers. But the UF lemma is equivalent to the boolean prime ideal theorem. This is strictly weaker than AC, though it is enough to produce a non-measurable set of reals. It (the BPI) is equivalent to the compactness theorem, and the completeness theorem, for 1st order logic (if I remember rightly). Tinfoilcat (talk) 14:53, 22 January 2011 (UTC)[reply]

Latus rectum of a parabola

Hello, If a parabola is defined by parametric coordinates $x=at^{2}+bt+c$ and $y=a't^{2}+b't+c'$ , how would I find its latus rectum (length of focal chord perpendicular to its axis). Thanks - DSachan (talk) 08:52, 22 January 2011 (UTC)[reply]

I get

{\frac {(ab'-a'b)^{2}}{(a^{2}+a'^{2})^{3/2}}}

Ignore the c′s by translation. Then write u=ax+a′x, v=a′x-ax, so everything in the uv-plane is rotated a bit and scaled by a factor of √(a²+a′²). It's now easy to eliminate t to get, after another translation,

v^{2}={\frac {(ab'-a'b)^{2}}{(a^{2}+a'^{2})}}u

From which you can read off the LR in the uv-plane. Applying the scaling factor gives the formula for the LR in the xy-plane. I left out a lot of detail and I haven't had breakfast yet so someone might want to check the result.--RDBury (talk) 15:22, 22 January 2011 (UTC)[reply]

Thanks a lot RDBury, that was a great help. I checked it out.

{\frac {(ab'-a'b)^{2}}{(a^{2}+a'^{2})^{3/2}}}

is indeed the right answer. - DSachan (talk) 12:46, 25 January 2011 (UTC)[reply]

Polynomial approximation of a non-analytic complex function

Hello everyone, i'm trying to construct a sequence of polynomials converging uniformly to 1/z on the upper-half-plane semicircle (|z|=1, re(z)>=0), but I'm having a lot of problems. I'm aware the fact that the complement of the semicircle is connected (whereas the full circle is not) is probably somehow intrinsic to the problem - however, I don't want to just prove it can be done, i want to actually exhibit such a construction, so existence theorems aren't much use here.

All my constructions so far have failed (I also tried treating 1/z as the complex conjugation function on the semicircle but that didn't work), and I'd really appreciate some help - I can see that we want all the 'messy parts' of the approximation to be in the lower half plane semicircle (since if we could do it without any convergence problems anywhere we could do it on the full circle |z|=1, which we certainly can't), but I don't know what we want our approximation to be. Could anyone lend a hand? Thank you, 131.111.16.20 (talk) 10:58, 22 January 2011 (UTC)[reply]

I'll work with the right half-plane instead of the upper one, for notational convenience. What about setting

w=g(z)={\sqrt {4z+3}}

1/z=h(w)=4/(w^{2}-3)

The semicircle of course lies entirely to the right of g's singularity at z=-3/4. In w space, the semicircle is represented by an arc from w=2-i to w=2+i, which lies entirely to the right of h's singularity at w=sqrt(3). So you can approximate either of g and h uniformly for the relevant values by developing them as power series around a point sufficiently far out on the real axis. Compose successive approximations for g and h. –Henning Makholm (talk) 02:39, 23 January 2011 (UTC)[reply]

Latex: texorpdfstring with explicit superscript sign ^ in it

Hello,

I have a question about Latex instead of mathematics itself... Until recently I did notice that one must take with the bookmarks in a pdf file. I often have math symbol in Section titles, but they look bad in those (clickable) titles in the pdf file...

the solution is

\section{\texorpdfstring{the size of $a^b$}{the size of a^b}}

But in this case, one just sees ab in the title. Is there a way to really see "a^b" (so I want to see the superscript sign)

Many thanks!

Does \^{} give you the right symbol? I got that from [3], which is an incredibly helpful list of how to do just about any symbol in latex. 81.98.38.48 (talk) 20:59, 22 January 2011 (UTC)[reply]

Hello, thanks for the list but that does not help, it just shows q2 in the bookmarks.

\^{} doesn't seem to work in the MediaWiki LaTeX emulation either. Surely there must be some way in LaTeX to type a literal superscript symbol, after all, LaTeX seems to include ridiculously many mathematical symbols. JIP | Talk 19:59, 23 January 2011 (UTC)[reply]

The \^{} command is for non-standard letters, e.g. â ê î ô û. The string $a \, \hat{} \, b$ seems to work. It puts a hat over the space, but that looks crowded, so the \, are added to give half spaces. If you compile it, it looks like this:

a\,{\hat {}}\,b

Is that what you were looking for? — Fly by Night (talk) 16:04, 24 January 2011 (UTC)[reply]

Try \string^. This also works for other special characters (#, &, $, ~, {, }, _; note that the last three will need a \tt font if you want to actually typeset them in the text, but this is not an issue with pdf bookmarks).—Emil J. 15:34, 25 January 2011 (UTC)[reply]

Actually, the complex multi-pass processing of the bookmarks seems to introduce some funny issues. Don't ask me why, but plain \string# won't work. You can get a # inside the bookmark with \#, and a backslash with \string\\.—Emil J. 16:26, 25 January 2011 (UTC)[reply]

Got stuck on a limit problem

Dear Wikipedians:

I got stuck while trying to show the following limit using first principles:

$\lim _{x\to \infty }{\sqrt {x}}-x=-\infty$

I first rewrote the above statement as follows:

For every negative number $N$ , there exists a corresponding positive number $M$ such that ${\sqrt {x}}-x<N$ whenever $x>M$

I then proceeded to prove the above statement as follows:

Let $M=N^{4}$ , then

${\sqrt {x}}-x$

$={\sqrt {x}}(1-{\sqrt {x}})$

$<{\sqrt {N^{4}}}(1-{\sqrt {N^{4}}})$

$<1-{\sqrt {N^{4}}}$

$=1-N^{2}$

It is apparent here that the statement $1-N^{2}\leq N$ must hold in order for the above statements to be true. We have

$1-N^{2}\leq N$

$N^{2}+N-1\geq 0$

Solving the corresponding equation, we have:

$N^{2}+N-1=0$

$N={\frac {-1\pm {\sqrt {1+4}}}{2}}={\frac {-1\pm {\sqrt {5}}}{2}}$

Since $N$ is a negative number, we discard the positive root and end up with

$N\leq {\frac {-1-{\sqrt {5}}}{2}}\approx -1.618$

Therefore, for $N\leq {\frac {-1-{\sqrt {5}}}{2}}\approx -1.618$ , let $M=N^{4}$ , then ${\sqrt {x}}-x<N$ whenever $x>M$

Now the question is, what is the expression for M, in terms of N, when ${\frac {-1-{\sqrt {5}}}{2}}<N<0$ ?

Thanks,

174.88.240.5 (talk) 17:39, 22 January 2011 (UTC)[reply]

Notice that if

M

works for some

N\leq {\frac {-1-{\sqrt {5}}}{2}}

(for example, if

M

works for

N=-2

), then

M

will work for every

N>{\frac {-1-{\sqrt {5}}}{2}}

. So take

M=16

.

You also have an implicit assumption that

N\leq -1

when you replace

x

with

N

, by the way. You can also make a much simpler argument if you take

M=N^{2}

and assume

N<-2

.--203.97.79.114 (talk) 22:11, 22 January 2011 (UTC)[reply]

Thanks, M = 16 works out well. And yes, i did make the implicit assumption that

N\leq -1

when I replaced

x

with

N

, so I should have stated that assumption explicitly at that very step. But in retrospect everything turned out to be fine since

N

not only has to be less than or equal to -1, it had to actually be less than or equal to -1.618. But still, like you said, I should have explicitly stated my assumption. 174.88.242.47 (talk) 01:44, 25 January 2011 (UTC)[reply]

Resolved