Wikipedia:Reference desk/Archives/Mathematics/2008 November 7

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November 7

Distance between currves

Say, I have two curves in $R^{d}$ parameterized as $x(t),y(t),t\in [0,1]$. I know Hausdorff distance, Frechet distance, but is there some metric which takes into account smoothness of the curves, or other conditions, like boundary conditions? Something similar to metric in Sobolev spaces $W_{p}^{k}$? (Igny (talk) 04:13, 7 November 2008 (UTC))

If you want to take into account smoothness (say, $C^{1}$) you could consider the corresponding liftings in the tangent bundle, that is just $\textstyle (\ x(t),x'(t)\ )$ and $\textstyle (\ y(t),y'(t)\ )$ as continuous curves $R^{2d}$, e.g. with Hausdorff distance. Also: you can consider the infimum of the functional distance ($C^{k}$, or Sobolev) over all reparametrizations of x and of y, (.e. a quotient distance) --PMajer (talk) 09:30, 7 November 2008 (UTC)

So does the following metric make sense

$\inf _{u:[0,1]\to [0,1]}\inf _{v:[0,1]\to [0,1]}\|x(u(t))-y(v(t))\|$

infimum of some suitable norm in some space over some class of re-parametrizations. Although the re-parametrizations are in some way equivalent to substitutions in the integrals, and this somehow has to be taken into account.(Igny (talk) 17:46, 7 November 2008 (UTC))

Yes; to be precise that does not give a distance in general, because the triangular inequality may fail to hold. It's just the minimal distance to the orbit of x and the orbit of y through the action of the (say, group of) re-parametrizations. It may happen that the minimal distance from the orbit of x and y is small, and also the minimal distance from the orbit of y and the orbit of z, but the minimal distance from the orbit of x and z is large. To get a distance you need that this action by re-paramentrizations be an isometry; in which case in the formula you wrote, you can take equivalently the infimum just over the u's, taking fixed any v. Good luck! --PMajer (talk) 20:28, 7 November 2008 (UTC)

I rewrote the article on Frechet distance as it didn't seem to give a correct definition of distance (namely, it was sup(d(x,y)) for x and y on their respective curves). The definition now given on the article seems quite close to the one given above by Igny. --Xedi^Talk 22:02, 7 November 2008 (UTC)

Notice that the formula written by Igny, that I would rewrite in a bit more clear form:

$\inf _{u:[0,1]\to [0,1]}\inf _{v:[0,1]\to [0,1]}\|x\circ u-y\circ v\|$

gives a true distance (i.e. the Frechet) in the case $\|x-y\|$ is the uniform distance. The reason is, as I mentioned, that in that case there is an isometric action: $\|x-y\|=\|x\circ u-y\circ v\|$. This is not true for a generic functional distance, and the triangular inequality may drop. Also note that in the definition of the Frechet distance you can take the infimum just over all the reparametrizations of the first curve, keeping fixed the other --PMajer (talk) 10:52, 8 November 2008 (UTC)

Technically, it is not true that $\|x\circ u-y\circ v\|_{\infty }=\|x-y\|_{\infty }$ for all u, v. But it is true that

$\sup _{u}\sup _{v}\|x\circ u-y\circ v\|_{\infty }=\sup _{v}\|x-y\circ v\|_{\infty }$

is the Frechet distance. (Igny (talk) 03:04, 9 November 2008 (UTC))

sorry I meant of course $\|x-y\|=\|x\circ u-y\circ u\|$, which is exactly the isometricity of the action I mentioned, and that implies what you wrote.--PMajer (talk) 09:59, 10 November 2008 (UTC)

You actually can use Sobolev norms in this situation. For example, for $W_{p}^{0}=L^{p}$ (i.e. the Lp_space), the distance between x(t) and y(t) would be $\left(\int _{0}^{1}\|x(t)-y(t)\|^{p}dt\right)^{1/p}$. For $W_{p}^{1}$, the distance is $\left(\int _{0}^{1}\|x(t)-y(t)\|^{p}+\|x'(t)-y'(t)\|^{p}dt\right)^{1/p}$, and so on for $W_{p}^{k}$. All we've done is replaced the absolute value in the standard one-dimensional definition with the norm in $R^{n}$.kfgauss (talk) 13:13, 13 November 2008 (UTC)

What is the integral of the Dirichlet function?

What is the integral of the Dirichlet function (i.e. the characteristic function of rational numbers)?

Many years ago I had a math class in which the instructor used the Dirichlet function as an example of a function that's discontinuous everywhere and not Riemann-integrable. The instructor also remarked that the function was Lebesgue-integrable (but that's something beyond the syllabus). I never asked what the result would be if a definite integral of the function was evaluated, but based on intuition (that $\mathbb {R}$ is of a higher cardinality than $\mathbb {Q}$), I think the only reasonable value of, say, $\int _{0}^{1}D(x)dx$ is 0.

Is my intuition correct? —Preceding unsigned comment added by 98.114.98.183 (talk) 07:46, 7 November 2008 (UTC)

Yes, that's correct. See Lebesgue integration, begin with a picture in the Intuitive interpretation section.
However the cardinality difference is not the only reason you should consider here. Much more important is the fact that the set of irrational numbers is dense in $\mathbb {R}$ – this implies that $\mathbb {Q}$ does not contain any interval, so its measure is zero and consequently L-integral of $D(x)$ is zero. --CiaPan (talk) 08:59, 7 November 2008 (UTC)

This isn't correct at all. The irrationals have full measure (i.e. the same measure as the reals) but this same argument would work for them, because their complement, the rationals, is dense as well. Countability is the point: any countable set has zero Lebesgue measure, because measures satisfy countable additivity and points have zero measure in the Lebesgue measure. kfgauss (talk) 13:25, 13 November 2008 (UTC)

To help your intuition you could observe that your D(x) is the carachteristic function of the set $S:=\mathbb {Q} \cap [0,1]$, so your integral has the meaning of the "total lenght" of this set, and you can change your question into: what is the only reasonable value of the "total lenght" of $\mathbb {Q} \cap [0,1]$. The answer is zero, provided you agree that the lenght has the properties of a measure, and that the lenght of a point is 0. Notice that the complement of S in [0,1] has consequently measure 1 (the last sentence of CiaPan is not correct, or unclear :) in that for a subset of $\mathbb {R}$, having empty interior does not imply having zero Lebesgue measure)--PMajer (talk) 09:16, 7 November 2008 (UTC)

Right, if that were the case, then by the same argument the set of irrationals would have measure 0.

But it's true that it's not clear that cardinality is necessarily enough to decide the issue. At least it's consistent with ZFC that there's a set of reals having cardinality less than that of R, but that is not of measure zero. (It can't actually have positive measure -- it would have to be non-measurable.) My intuition is that there's no such set, but it's a fairly weak intuition. (You can prove from certain forcing axioms, for example Martin's axiom, that there's no such set.) --Trovatore (talk) 09:28, 7 November 2008 (UTC)

Beyond the observation that countable sets have zero Lebesgue measure, one can't say much relating measure and cardinality. For example, the Cantor_set has the same cardinality as the reals, but has zero Lebesgue measure. kfgauss (talk) 13:25, 13 November 2008 (UTC)

An other argument to help your intuition for "$\mathbb {Q} \cap [0,1]$ has lenght zero" is: what is the probability that a real number picked at random in $[0,1]$ be rational? The point of course is, to define in a suitable sense "picked at random" (I must confess that when I submit this kind of mental experiments to my intuition, she stares at me like I were talking nonsense) --PMajer (talk) 09:46, 7 November 2008 (UTC)

See:

Simple function Indicator function

Useful (easy) facts:

• Any simple function can be expressed as the linear combination of indicator functions
• The above fact is crucial in the definition of the Lebesgue integral of a function

(If you know everything above already, you can ignore my post)

Topology Expert (talk) 12:22, 8 November 2008 (UTC)

---

Lower cardinality is not enough, but countability is. The integral is just the sum of the the integrals of the indicator functions of the individual rationals. They're all equal to 0. Michael Hardy (talk) 03:10, 9 November 2008 (UTC)

halogen derivatives of alkanes

what is meant by +ve inductive effect —Preceding unsigned comment added by Kunal pdj (talk • contribs) 11:25, 7 November 2008 (UTC)

See Wikipedia:Reference desk/Science. I think you should probably ask you question there.

Topology Expert (talk) 01:38, 8 November 2008 (UTC)

Measuring buildings from photos

Hello, is there a method, formula or tool that allows us to make measurements (mostly the height) of a building with only a (or several) photo taken of it available, while taking distortion due to perspective into consideration? 85.112.95.14 (talk) 15:47, 7 November 2008 (UTC)

If you can distinguish floors, then you can estimate the size after guestimating the floor height. 195.35.160.133 (talk) 16:03, 7 November 2008 (UTC) Martin.

I don't know if this is what you mean, but I suppose you are talking of a photo of a tower taken from below like this:

..........__ t

......../___\

....../_____\

..../_______\

../_________\ a......(ignore the dots :)

now suppose that in the picture the top level t appears 1/10 shorter that the level a : this means that the height of the top level from the ground (or better from the camera) is 10 times the height of the level a . Say that you know the height of level a, then you are done --PMajer (talk) 19:49, 7 November 2008 (UTC)

The cross-ratio in projective geometry can be used to flatten out a projective drawing and calculate lengths photos like that tower above. You can extend a rectangle to other similar rectangles for instance by finding where the diagonals cross, and where two sides meet at infinity and joining the two. This gives the halfway point on a side and connecting to an opposite corner and extending gives the corner of an adjacent rectangle. Much easier if I'd drawn it - there's probably a diagram somewhere on wiki. However I'd just look for bricks or blocks and calculate the height by counting them. Dmcq (talk) 22:03, 7 November 2008 (UTC)

If the building's exterior contains a planar surface that extends from the bottom to the top and you can identify some right angles in that plane (say, the sides of a window), you can use a panorama stitching program like hugin to calculate the picture you would have gotten if your film/sensor had been parallel to that plane and large enough to capture the whole building. Distances in pixels in that image are proportional to physical distances within the plane, so you can work out the height of the building if there's any feature of known size in the plane (like a window again, or a brick). To do this in hugin you need to select the rectilinear projection and define some horizontal and vertical control points (as well as ordinary feature-matching control points if you're stitching more than one photo). Depending on how much accuracy you need and how expensive your lens is, you might also want to have hugin estimate the barrel/pincushion distortion so that it can eliminate it from the output (select "optimize everything" in the Optimizer tab). -- BenRG (talk) 18:29, 8 November 2008 (UTC)

You might consider PhotoModeler Lite, an old but free (though limited) photogrammetry program. It's not supported or distributed by PhotoModeler.com anymore but it's still "available". Saintrain (talk) 20:19, 8 November 2008 (UTC)

Thank you for these answers, I was going to ask for a mathematical formula to solve this problem, but from your description, hugin seems to me like the solution. I'll try it and I may come back for questions. 85.112.95.14 (talk) 04:49, 9 November 2008 (UTC)

Equivalence relations on square matrices

Hello,

I was looking at different equivalence relations we can define on matrices : first of all, say A ~ B iff there exists an invertible P such that A = P^-1BP. If we're working over an algebraically closed field, we get as a simple representative of each equivalence class a matrix in Jordan Normal Form. What do we get in general though, how much can we do if we see that the characteristic polynomial doesn't split ? Can we just say that each matrix is of the form $J\oplus M_{1}\oplus \cdots \oplus M_{k}$ where J is a matrix in Jordan Normal Form and the M_i are some sort of fundamental irreducible matrices (in the same way we can split any polynomial over $\mathbb {R}$ as a product of linear factors and irreducible polynomials of degree 2).

Secondly, considering A ~ B iff there exists an invertible P such that A = P^TBP, and here the representative for each class is a diagonal matrix, and depending on the field we can get it to the form $I_{p}\oplus -I_{q}\oplus 0_{s}$ or $I_{k}\oplus 0_{s}$ for example.

One obvious way of bringing those two together is considering A ~ B iff there exists an orthogonal such that A = P^TBP, as P^T=P^-1. I have no idea what we get here though ?

Another thing I find interesting, is that, the first case corresponds to a transformation law of linear maps V->V, so sort of constitutes a transformation law for tensors of type (1,1), whereas the second example is a transformation law for bilinear forms, that is, tensors of type (0,2). How does this generalise to tensors of any type?

Thanks. --Xedi^Talk 21:28, 7 November 2008 (UTC)

I'm too drunk to process all your questions, but the answer to the first is rational canonical form. Algebraist 02:46, 8 November 2008 (UTC)

I think the answer to your last question is simple enough. In the first case you're taking A ~ B if Aⁱ_j Λ^i'_i Λ^j_j' = B^i'_j' for some Λ, where the two Λs are matrix inverses of each other because the primed and unprimed indices are swapped. In the second case you're taking A ~ B if A_ij Λⁱ_i' Λ^j_j' = B_i'j' for some Λ. The corresponding equivalence on (1,3) tensors would be R ~ R' if Rⁱ_jkl Λ^i'_i Λ^j_j' Λ^k_k' Λ^l_l' = R'^i'_j'k'l' for some Λ.

Canonical form#Linear algebra lists some canonical forms. You probably want the adjoint (conjugate transpose) rather than the transpose, since these results extend nicely to $\mathbb {C}$ with the adjoint, but I don't think they do with the transpose.

You can't get an arbitrary matrix into diagonal form with a congruence transformation, since it preserves symmetry. You can get an arbitrary self-adjoint matrix into the form you mentioned. The spectral theorem says that any self-adjoint matrix is orthogonally (real-)diagonalizable. I don't know if there's a nice canonical form for arbitrary matrices modulo congruence or orthogonal transformations—I'm guessing probably not. -- BenRG (talk) 14:03, 8 November 2008 (UTC)

Ah yes, thanks. I suppose non-Hermitian matrices are not that useful when considering bilinear forms anyway, seeing as any real quadratic form corresponds to a Hermitian matrix (except in characteristic 2), so that was mainly what I was interested in. Thanks again. --Xedi^Talk 16:52, 8 November 2008 (UTC)