Wikipedia:Reference desk/Archives/Mathematics/2011 April 28

= April 28 =

Help with a question of Limit
Let $$f\,\colon\,R\to R$$ be a positive increasing function with $$\lim_{x\to \infty}\frac{f(3x)}{f(x)}=1$$, then what will be the $$\lim_{x\to \infty}\frac{f(2x)}{f(x)}$$. I can straightaway think $$f(x)=\mathrm{ln}(x)$$, which satisfies some of the given conditions and satisfies the given limit condition. But how do we go about it without knowing the function? Thanks - DSachan (talk) 13:59, 28 April 2011 (UTC)


 * The condition that f is an increasing function is important. You can conclude that f(2x) must lie between f(x) and f(3x). How can you use this ? Gandalf61 (talk) 14:31, 28 April 2011 (UTC)
 * $$f(x) \le f(2x) \le f(3x) \Rightarrow \lim_{x\to \infty}\frac{f(x)}{f(x)} \le \lim_{x\to \infty}\frac{f(2x)}{f(x)} \le \lim_{x\to \infty}\frac{f(3x)}{f(x)} \Rightarrow 1 \le \lim_{x\to \infty}\frac{f(2x)}{f(x)} \le 1 \Rightarrow \lim_{x\to \infty}\frac{f(2x)}{f(x)}=1$$ Widener (talk) 01:47, 30 April 2011 (UTC)
 * The second implication is valid because f is a positive function. Widener (talk) 01:53, 30 April 2011 (UTC)
 * That is certainly the right idea, but you can express it more carefully as a consequence of the squeeze theorem. The problem with your argument as it stands is that you write the symbol $$\lim_{x\to \infty}\frac{f(2x)}{f(x)}$$ before you actually know that this quantity exists.   Sławomir Biały  (talk) 12:46, 30 April 2011 (UTC)

Arzelà-Ascoli theorem and second order differential equations
How would one use the Arzelà-Ascoli theorem to show the set $$S_n \subset C([0,1])$$ of thrice-differentiable solutions to the D.E. $$f'' = f + (f')^2$$ with $$|f(0)|\le N,\,|f(1)|\le N,\,|f'(0)|\le N,\,|f'(1)|\le N$$ totally bounded as a subset of $$C([0,1])$$? I assume I'm meant to be using Arz-Asc, since this is a 2-part problem the first part of which is state and prove the Arzelà-Ascoli theorem. However, it is possible i'm not meant to - they certainly seem related though. However, the only thing I could think of using Arz-Asc for would be to show that the set was uniformly bounded and equicontinuous (therefore totally bounded): but it isn't clear to me even why the set should be uniformly bounded, let alone equicontinuous, and in my experience UB is the easier of the two conditions to prove. I am told "it may be useful to consider interior maxima", but thus far that has failed to get me anywhere. Could anyone suggest anything? Spalton232 (talk) 14:49, 28 April 2011 (UTC)


 * At an interior maximum, $$f=f\le 0$$. At an interior minimum $$f=f\ge 0$$, so $$|f|\le N$$ throughout the domain.  You also need to get a uniform bound of the derivative of $$f$$ in order to conclude equicontinuity.  For this, you can get a lower bound from $$f(x)\ge f(x)$$ by integrating over [0,x].  Get an upper bound from the same estimate by integrating over [x'',1] instead.   Sławomir Biały  (talk) 15:30, 28 April 2011 (UTC)
 * Wow - it would have taken me absolutely hours to come up with that, very simple and elegant, thank you ever so much! Spalton232 (talk) 16:04, 28 April 2011 (UTC)

Maths of the PID Controller algorithm
In the pseudocode description PID_controller, if I am measuring things at intervals of a second (or a year) could I just use a value of "1" for dt? Or is there more to it than that? Thanks 92.29.127.30 (talk) 16:36, 28 April 2011 (UTC)


 * Yes. Of course for any particular system, changing the interval (dt) will change the values needed for Kp, Ki and Kd. -- SGBailey (talk) 19:53, 29 April 2011 (UTC)

Fibre Bundle Question
Let M be a smooth, real, n-dimensional manifold. Let h be a bilinear form on M. We can interpret h as a cross-section of the vector bundle &pi; : Hom(TM&thinsp;⊗&thinsp;TM,&thinsp;R) → M. Can anyone think of a bundle-theoretic statement that is equivalent to h being "non-degenerate" at a particular point of M? Can anyone think of any geometrical properties? If h fails to be non-degenerate then is there a geometric interpretation with respect to the section h : M → Hom(TM&thinsp;⊗&thinsp;TM,&thinsp;R) at that point? — Fly by Night  ( talk )  16:49, 28 April 2011 (UTC)


 * Perhaps not very satisfactory, but to any such section h is associated a canonical density (or volume element) which is a section $$|\Omega|$$ of $$|\wedge^n T^*M|$$ defined on a collection of vectors $$X_1,\dots,X_n$$ by
 * $$|\Omega(X_1,\dots,X_n)|^2 = |\det h(X_i,X_j)_{1\le i,j\le n}|.$$
 * (In the presence of an orientation, $$|\Omega|$$ is just the absolute value of the canonical volume form.) Then h is degenerate at precisely those points where the volume element vanishes.  The "geometrical" interpretation is that any reference parallelotope in the tangent space will have zero volume if measured using the bilinear form h.


 * Depending on the application, a more suitable answer might be that positive definite symmetric forms can be identified with sections of the quotient bundle GL(M)/O(n) of the principal bundle GL(M) of linear frames on M by the action of the orthogonal group. A similar quotient space description is available for indefinite signatures (just replace O(n) by O(p,q)).  I don't know of any immediate way to accommodate all non-degenerate forms at once.  One way would be to complexify, i.e., look at GL(M,C)/O(n,C), but then you would need to allow for complex-valued forms as well.   Sławomir Biały  (talk) 17:29, 28 April 2011 (UTC)


 * It's a pretty good effort Sławomir. The first part about volume elements relates to some work I've already done, and that I'm trying to generalise. But I'm working with projective transformations now, and these volume elements aren't preserved. The condition that h be non-degenerate seems to be a projective invariant. I was trying to prove the non-degeneracy of h does not depend on the normal bundle NM. I can prove it by doing using Koszul connections, but I wanted to understand it geometrically. The second part is a bit too general form my purposes. I know it's very powerful stuff, but I think I'll lose some of the subtle structure if I use such general methods. — Fly by Night  ( talk )  18:01, 28 April 2011 (UTC)


 * So you don't have a particular h, is that what you're saying?  Sławomir Biały  (talk) 18:13, 28 April 2011 (UTC)


 * With regards to the original post, no. Just a section. With regards to the specific application, I have a family of them. They're all non-zero, smooth, functional multiples of the second fundamental form. But this is only evident after using connections. In fact, proving they're all multiple of the SFF proves the fact that their non-degeneracy only depends on M up to projective transformation. But I'd really like to understand my first question. — Fly by Night  ( talk )  18:18, 28 April 2011 (UTC)


 * I would settle for a section (that's actually what I meant). But it seems from what you're saying that you don't even have that: you just have a section defined up to rescaling by some nonzero function.  Do I have this right?   Sławomir Biały  (talk) 18:24, 28 April 2011 (UTC)


 * Very good. That's actually what I do do next. To put it concreetly. I embed M into Rn+1. So Imagine M as living in Rn+1. Consider the tangent bundle TRn+1 restricted to M and split it as a Whitney sum:
 * $$ \left.T\R^n\right|_M \cong TM \oplus NM \, . $$
 * A choice of normal bundle NM is like choosing a transverse vector field. (E.g. If that transverse vector is the unit normal, then the normal bundle is the orthogonal complement to TM.) We use the covariant derivative on Rn+1 to give a bilinear form, via the Gauss–Codazzi equations:
 * $$ D_XY = \nabla_XY + h(X,Y) \, \xi \,, $$
 * where the fibre of NM over p is identified with the span of &xi; at p. Here h is the bilinear form mentioned above and X and Y are two vector fields on M. It turns out, that no matter which &xi; we choose, i.e. which Whitney sum splitting we choose, the bilinear form is always a non-zero, functional multiple of the second fundamental form (which itself corresponds to &xi; being the unit normal). So the splitting doesn't matter. The non-degeneracy of h is a property of M and not the splitting. In fact it's invariant under projective transformations of Rn+1. It seems to me that h is only interested in what the tangent bundle does. For me, the fibre bundle &pi; : Hom(TM&thinsp;⊗&thinsp;TM,&thinsp;R) → M is the key, and is independent of the embedding of M in Rn+1. To understand what I need to understand, I boiled all that down to understanding what a section of &pi; : Hom(TM&thinsp;⊗&thinsp;TM,&thinsp;R) → M being degenerate on a fibre means geometrically, and not abstractly, as I have proven. (Sorry for the long post, I was trying my best to keep it short). — Fly by Night  ( talk )  18:58, 28 April 2011 (UTC)
 * Thanks for the explanation. So this actually seems to confirm my suspicion that the correct object may not be a section h, but rather a section defined up to some sort of ambiguity.  It's a little unclear to me exactly what that ambiguity is.  For instance, presumably the scale of h will depend on the scale of &xi;.  A more worrisome concern is that it isn't clear from the definition whether this is the only ambiguity in h.  Assuming I understand the problem correctly, there is a lot more freedom than just scaling in how &xi; is selected.  Also, doesn't the connection also change under projective transformations in ways that are going to mess this up?   Sławomir Biały  (talk) 19:13, 28 April 2011 (UTC)

Sławomir, about the original question about non-degenerate bilinear forms, and sections of the bundle &pi; : Hom(TM&thinsp;⊗&thinsp;TM,&thinsp;R) → M. Consider a fixed point p in M and the fibre over p. That will be a linear map hp : TpM&thinsp;⊗&thinsp;TpM → R. I think the bilinear form h is non-degenerate at p if and only if the restriction hp : {v}&thinsp;⊗&thinsp;TpM → R is a submersion for all non-zero v in TpM. We can extend this by letting p vary. Let &sigma; : M → TM be a section, and consider h : {&sigma;}&thinsp;⊗&thinsp;TM → R. If &sigma; is a non-zero section then all of those maps will be submersions if and only if h is non-degenerate on M. For non-zero &sigma;, this translates into a restricted bundle &pi; : Hom({&sigma;}&thinsp;⊗&thinsp;TM,&thinsp;R) → M, which is isomorphic, as a vector bundle, to &pi; : Hom(TM,&thinsp;R) → M, which is the cotangent bundle. So I think I need all non-zero, single factor restrictions to give submersion bundles. But those choices of &sigma; look like they're giving an isomorphism with the cotangent bundle. Am I making any sense? Can you add to, and maybe clarify, what I've tried to say? — Fly by Night  ( talk )  00:16, 1 May 2011 (UTC)