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May 20

biased circle

Given three points $z_{1},z_{2},z_{3}$ and three constants $h_{1},h_{2},h_{3}$, find $z_{0}$ such that

${\frac {|z_{1}-z_{0}|}{h_{1}}}={\frac {|z_{2}-z_{0}|}{h_{2}}}={\frac {|z_{3}-z_{0}|}{h_{3}}}$

Each equation defines a circle (add: or line), and I could find the intersection of the three circles; but there may be a more direct way (add: perhaps involving a Möbius transformation?). —Tamfang (talk) 03:18, 20 May 2021 (UTC)

While Möbius transformations preserve circlehood, they do not preserve centres.  --Lambiam 22:50, 20 May 2021 (UTC)

Let $(x_{i},y_{i})=z_{i}.$ Each circle equation is a quadratic equation in the $(x,y)$ coordinates, but by taking their pairwise differences, you get three linear equations. These can be solved like any system of linear equations. After the smoke cleared up, I ended up with this:

Define

$d_{i}=h_{i}^{2}-(x_{i}^{2}+y_{i}^{2}),$

$a=\textstyle {\sum (x_{i}-x_{j})^{2}},$

$b=\textstyle {\sum (x_{i}-x_{j})^{2}(y_{i}-y_{j})^{2}},$

$c=\textstyle {\sum (y_{i}-y_{j})^{2}},$

$s=\textstyle {\sum (x_{i}-x_{j})^{2}(d_{i}-d_{j})^{2}},$

$t=\textstyle {\sum (y_{i}-y_{j})^{2}(d_{i}-d_{j})^{2}},$

where the summations are over pairs of indices $1\leq i<j\leq 3.$ Then

$z_{0}=\left({\frac {bt-cs}{2(ac-b^{2})}},{\frac {bs-at}{2(ac-b^{2})}}\right).$

 --Lambiam 07:10, 20 May 2021 (UTC)

I haven't gone through this in detail, but I think there's something off about it. For one thing, you're finding the intersection of two circles so there should be 2 points in common. As a concrete example, take z₁=(0,0), z₂=(1,0), z₃=(0,1), h₁=1, h₂=2, h₃=√2. Then one solution is z₀=(-1,0) with |z₁-z₀|/h₁ = |z₂-z₀|/h₂ = |z₃-z₀|/h₃ = 1. Another solution is z₀=(.2,.4) with |z₁-z₀|/h₁ = |z₂-z₀|/h₂ = |z₃-z₀|/h₃ = √.2 --RDBury (talk) 12:20, 20 May 2021 (UTC)

You're right; I have solved a different and more specific problem:

${\frac {|z_{1}-z_{0}|}{h_{1}}}={\frac {|z_{2}-z_{0}|}{h_{2}}}={\frac {|z_{3}-z_{0}|}{h_{3}}}=1$

under the assumption that this system has a solution. The first of your two solutions for the counterexample satisfies the extra condition and is the one found by the formulas I gave. The second will be found after dividing each $h_{i}$ by ${\sqrt {5}}.$  --Lambiam 12:49, 20 May 2021 (UTC)

That would explain it. I tried rewriting the system as |z_i-z₀| = k h_i where k is some fixed but unknown value. With your method you can find z₀ if you know k, but I don't see an easy way of finding k in the general case. I suppose you could use the Cayley–Menger determinant to get quadratic equation in k², but I don't know if that's an improvement over the OP's method. --RDBury (talk) 14:19, 20 May 2021 (UTC)

Another, more geometric approach: By this system, the points $z_{1},z_{2},z_{3}$ for a triangle with side lengths $\lambda h_{1},\lambda h_{2},\lambda h_{3}$ for an unknown $\lambda$. Hence (essentially by the side-side-side congruence theorem), the solution space is four-dimensional and is given by rotating, translating, and scaling an ur-triangle (which can be found in many ways, e.g. assuming $z_{1}$ to be at the origin and $z_{2}$ to be on the positive x-axis, then finding $z_{3}$ using stuff like Heron's formula). Duckmather (talk) 04:34, 21 May 2021 (UTC)

not quite that

Going over my algebra again, I think that what I want is instead

$|z_{n}-z_{0}|=e^{ks_{n}}$

where $k$ is also unknown; which looks not nearly so easy. Oh well, I'll probably have to do it iteratively. —Tamfang (talk) 16:48, 20 May 2021 (UTC)

Based on those equations alone, I think I know how to parametrize the solution space: it's $z_{n}=z_{0}+e^{ks_{n}+i\theta _{n}}$, where the free parameters are $z_{0},k,\theta _{n}$. Duckmather (talk) 04:43, 21 May 2021 (UTC)

It's easy to solve with s all equal; from there, maybe Newton's method will get somewhere. —Tamfang (talk) 22:14, 21 May 2021 (UTC)

Er, make that $|z_{n}-z_{0}|={\frac {e^{ks_{n}}}{k}}$. —Tamfang (talk) 21:58, 21 May 2021 (UTC)

Too bad there's no obvious way to extract k as a function of z₀. —Tamfang (talk) 22:40, 21 May 2021 (UTC)

But see Lambert W function – it depends on what one considers obvious. The quite specific form of the rhs suggests that you have an equally quite specific application in mind.  --Lambiam 22:59, 21 May 2021 (UTC)

relative distance between {n/2} gram inner and outer points?

For a {n/2} gram, is there an easy formula for the ratio of the distances between a) the distance from the center to the inner points and b) the distance from the center to the outer points. It is clear that the ratio goes to 1 as n goes to infinity, but I don't know if there is a (relatively) clean formula.Naraht (talk) 12:29, 20 May 2021 (UTC)

By the law of sines, $a/b=\sin(\pi /2-2\pi /n)/\sin(\pi /2+\pi /n)=\cos(2\pi /n)/\cos(\pi /n)=2\cos(\pi /n)-1/\cos(\pi /n)$. --116.86.4.41 (talk) 14:36, 20 May 2021 (UTC)

The triangle

What is a {n/2} gram? Which centre is being discussed - things like triangles have more than one. -- SGBailey (talk) 10:10, 22 May 2021 (UTC)

See Regular polygon § Regular star polygons. --116.86.4.41 (talk) 11:21, 22 May 2021 (UTC)

Thanx. I have a shower with the jets in the shape of an N / 2 gram and was wondering how many points to the gram were needed for the inner points to be at least 90% of the way out to the outer points. This gives me way to calculate!Naraht (talk) 12:07, 24 May 2021 (UTC)

May 23

Acceptability of Cantor's Transfinite numbers

Hi all, I saw the article on controversy surrounding Cantor's theory (Controversy over Cantor's theory) but am looking for a more practical answer. That is, are things like Transfinite numbers now widely accepted? If you were in college, studying in a subject area that included a topic like that, would it almost always be introduced to you as a widely accepted thing (like Darwin's theory, say) or does it depend on the professor? Best - Aza24 (talk) 00:04, 23 May 2021 (UTC)

They are fully accepted in "mainstream" mathematics, which includes everything that can be built on ZFC. There remain, however, mavericks who think the arguments for accepting ZFC as a foundation are not particularly strong. There is no common basis that could be used for demonstrating that they are wrong (or right). For much of mathematical practice, and definitely for applied mathematics, the issues are irrelevant.  --Lambiam 00:26, 23 May 2021 (UTC)

Finitist mathematics (math that does not accept the axiom of infinity) does have its proponents. One prominent one on YouTube is N J Wildberger, and he explains his reasons in his video Why infinite sets don't exist. As you say, for most practical purposes these issues are irrelevant, but there is still room for disagreement among the people who specialize in foundations and philosophy, and there are many schools of thought as to which axioms should or should not be accepted. --RDBury (talk) 01:44, 23 May 2021 (UTC)

Constructivist mathematicians generally have no issue with the notion that there are infinitely many natural numbers – or, rather, that if one starts counting 0, 1, 2, ..., one will never run into some "finish" from which one cannot go further, a dead end – but the definition of cardinality used by Cantor (or classical mathematicians in general) is not meaningful to them, which undermines the whole transfinite edifice beyond $\aleph _{0}$. See Constructivism (philosophy of mathematics) § Cardinality for a (not totally enlightening) discussion.  --Lambiam 10:56, 23 May 2021 (UTC)

I wouldn't say "controversies" on this topic really exist any more, and there aren't real mathematical disagreements either. There are disagreements over philosophy but that's a little bit different. E.g. asking whether transfinite numbers "exist" presupposes a level of mathematical Platonism which is a philosophical viewpoint, in which which there are various differing sub-viewpoints. The current situation is that most mathematics can be translated into a formal system called ZFC, that in turn proves theorems about sets that can be interpreted as transfinite ordinals, and nobody afaict seriously doubt that it "works" if you do it that way. Of course just because something can be done doesn't automatically make it a good idea. 2601:648:8200:970:0:0:0:752 (talk) 07:14, 24 May 2021 (UTC)

• Per the above, there are really two questions here: 1) How accepted are Cantor's main notes on the difference between the infinity of counting numbers and the infinity of real numbers? and 2) How accepted are the explanations, theories, and solutions to any problems created by question #1? The answer to the first is pretty much universal acceptance. There really is NOT a one-to-one correspondence between the natural counting numbers (i.e. 1, 2, 3, 4, etc.) and the real numbers, i.e. all of the possible numbers between 0 and 1 (i.e. 0.1, 0.11, 0.111... etc). Cantor's diagonal argument really is almost brilliantly bulletproof in its application at least proving that idea. So #1 is well established. The issue comes with #2: How do you incorporate this idea of non-correspondence into existing mathematics in a coherent way? How do you generalize and formalize this? What kinds of new mathematics does this introduce? What are useful ways to think about this? We're still mostly on board with the answers to these follow on questions, which is to say there is a clear "mainstream" viewpoint within mathematics, but this viewpoint is not universal, and there are still good-faith disagreements over this part. This is where this becomes a "philosophical" rather that a "mathematical" argument. --Jayron32 14:02, 25 May 2021 (UTC)

  I do have to say, in addition, that these philosophical arguments are also not just trivial or academic exercises. There are all sorts of real applications that depend on these mathematical concepts, including computer science (i.e. the halting problem), cryptography (i.e. P versus NP problem), quantum mechanics issues, etc. that all come down to the ways we resolve these set theory problems. The Russell paradox problem keeps showing up in logic systems where self-reference is allowed, and avoiding self-reference is itself fraught with complexities; Russell's best attempt a creating a system of logic robust enough to avoid self-reference issues introduced by naive set theory is Principia Mathematica. ZFC solves the problem with the axiom of choice. --Jayron32 14:11, 25 May 2021 (UTC)

  I don't really see how the axiom of choice "solves the problem". I would put it rather that the problem is solved by conceiving of sets as objects contained in the cumulative hierarchy edit: I followed the link and that's actually not quite the article I want — see von Neumann universe instead, in which the axiom of choice is "obviously" true (scare quotes because it's obvious to me, but not to everyone). It's actually the axiom of foundation (which our article unfortunately calls "regularity", a silly name for it) that encapsulates the claim that all sets are contained in the cumulative hierarchy.

  But I wouldn't say it's the axiom of foundation that "solves the problem" either; you can't get rid of a contradiction by adding more assumptions. It's rather a matter of getting the informal description correct; then you can see that the axioms that lead to the contradiction just aren't true. --Trovatore (talk) 17:22, 25 May 2021 (UTC)

May 24

Rational approximations of real ratios

Using continued fractions, it is easy to find rational approximations of the ratio between two real numbers, such as

$e:\pi \approx 777:898.$

What about the same for more than two numbers, such as

${\sqrt {2}}:e:\pi \approx 167:321:371.$

Is there theory that gives a better approach than brute force?  --Lambiam 11:44, 24 May 2021 (UTC)

The introduction of [1] says that "The method of Padé approximation has been widely used to obtain effective lower bounds for simultaneous rational approximations to algebraic numbers" and then goes on to elaborate. There is also apparently a simultaneous version of Dirichlet's approximation theorem. --JBL (talk) 19:06, 24 May 2021 (UTC)

Thanks. I'll have a look at the first, but apparently it cannot be used for transcendental numbers, or to numbers obtained by measurement, such as the atomic masses of the elements. I do not immediately see how the simultaneous version of Dirichlet's theorem can be used to improve on brute force. So the search goes on.  --Lambiam 20:22, 24 May 2021 (UTC)

In the simple ratio case, let the ratio sought be (x:y). One way to look at continued fractions, is that they generate pairs of ratios, say (a₁:b₁), (a₂:b₂) so that a₁ b₂ - a₂ b₁ = 1 and (a₁:b₁)<(x:y)<(a₂:b₂). Given one such pair, you can find an improved pair by examining (a₁+a₂:b₁+b₂). (Please let + take precedence over : in this notation so I don't have to write a lot of parentheses.) Either (a₁+a₂:b₁+b₂)<(x:y)<(a₂:b₂) or (a₁:b₁)<(x:y)<(a₁+a₂:b₁+b₂), so choose (a₁+a₂:b₁+b₂), (a₂:b₂) or (a₁:b₁), (a₁+a₂:b₁+b₂) depending on which, as the next pair in a sequence. Given a suitable starting pair, every optimal pair will be in this sequence and every optimal approximate ratio will be in an optimal pair. This can be shown, for example, using well known properties of Farey sequences. Continued fractions just accelerate this process.

One can mimic this idea for complex ratios, but unfortunately the results are less satisfactory. Let the ratio sought be (x:y:z) and let (a₁:b₁:c₁), (a₂:b₂:c₂), (a₃:b₃:c₃) be a triple of ratios with |(a₁:b₁:c₁) (a₂:b₂:c₂) (a₃:b₃:c₃)| = 1 and (x:y:z) in the interior of the triangle formed by the three ratios considered as point in the projective plane. Here | * | denotes the determinate. The plan is similar; pick a point on one of the edges of the triangle, say (a₁+a₃:b₁+b₃:c₁+c₃), and use it to split the triangle into two subtriangles. The desired ratio (x:y:z) must be in exactly one of them, so replace one of the original vertices with this new point according to which, thus replacing the original triangle with one of the subtriangles. The problem is that there's no clear method of choosing which side the new point should be on. Some methods are better than others, for example you should always try a side which is "distant" from the desired ratio, where the exact definition of "distant" is flexible because we're dealing with projective space rather than Euclidean space. The result is that instead of getting a uniquely defined sequence, you get one somewhat arbitrary choice in a collection of such sequences.

One issue is that an exact idea of "closeness" hasn't been defined for complex ratios. The generalization to Dirichlet's theorem uses the uniform norm, max(|y/x - b/a|, |z/x - c/a|), but there's no reason to consider this better than the Euclidean norm √((y/x - b/a)² + (z/x - c/a)²). It doesn't seem to matter much though since it seems unlikely that an optimal approximation in either sense will appear as a vertex of any triangle in the sequence. For example, I tried the ratio (1:√2:√3). By brute force, the best approximation by either Euclidean or uniform norm for ratios with denominator (the first coordinate) less than 1000 is (828:1171:1434). According to one side selection scheme I got the triple

(280:396:485) (903:1277:1564) (944:1335:1635),

all of which are good approximations but not the optimal one. Using another selection scheme I got

(944:1335:1635) (519:734:899) (758:1072:1313)

again good approximations but not optimal and only one point in common between the two.

I think the problem is that there is no analog to the theory of Farey sequences in higher dimensions, at least not a simple one that I've heard of. This seems to be related to the fact that the submonoid of SL(2,Z) generated by

${\begin{pmatrix}1&1\\0&1\end{pmatrix}}$ and ${\begin{pmatrix}1&0\\1&1\end{pmatrix}}$

is free, but afaik there is no analogous fact for SL(3,Z). Another issue that seems to be related is that while a set of points in a line segment divide it into subsegments in a unique way, a set of points in a triangle does not divide the triangle into subtriangles in a unique way. So as I see it there are several options: 1) Use a reasonably good side selection scheme with the method above and be content with the fact that you get a good approximation instead of the best. 2) Try to find a selection scheme that's guaranteed to find an optimal solution. 3) Find a way to "sharpen" a good approximation from (1) to get an optimal approximation. 4) Just abandon this approach altogether. I don't think it's entirely impossible that this problem is NP-hard, at least in arbitrarily large dimension, even though it doesn't really have an NP-hard "feel" to it. But IMO you can say the the same thing about the Lattice reduction problem and that's known to be NP-hard. --RDBury (talk) 13:33, 25 May 2021 (UTC)

Did you try using the more symmetric $a_{1}+b_{1}+c_{1}:a_{2}+b_{2}+c_{2}:a_{3}+b_{3}+c_{3}$ instead? --JBL (talk) 21:30, 25 May 2021 (UTC)

That's actually equivalent to two steps with choosing an edge, with the second step constrained to pick the edge most recently added. I did try that though and several schemes that seemed attractive. I think the main problem with this kind of multistep approach is that not every possible point can be reached. For example with this a+b+c idea, if you start with a triangle where the all coordinate sums are odd, then all coordindate sum remain odd in the next step, so you might miss an optimal solution where the coordinate sum is even. Not that this might not be a problem with the original scheme; that the optimal point is a vertex of some possible triangle generated by the above process is questionable and the fact that it's true for simple ratios relies again on Farey sequences. It does seem to be true in the examples I've tried, for example the triangle (239:338:414), (198:280:343), (828:1171:1434) does contain the point (1:√2:√3) and has an optimal approximation (828:1171:1434) as a vertex, and this triangle is reachable in some sequence starting with (1:0:0), (0:1:0), (0;0:1), but finding this sequence seems to require that you already know the final triangle.

I realize the above scheme is sketchy and incomplete (not to mention over-long), but the short version is: yes, there are generalizations of continued fractions to higher dimensions, but they are more complicated than one might hope, not deterministic, and apparently not guaranteed to produce an optimal answer. Maybe the idea can be fixed up to create an actual working algorithm which finds an optimal approximation, but I didn't see how. --RDBury (talk) 01:04, 26 May 2021 (UTC)

As a measure of "goodness of fit" I used the inner product of the two normed rays (the cosine of the angle they form), which has the nice property of being insensitive to permutations of the n-tuples.  --Lambiam 23:20, 25 May 2021 (UTC)

Yes, that does seem like a better measure. I suppose you could also use the sum of the coordinates instead of the the first one to bound the values. As a PS to the above, I re-did the brute force computation and found that (903:1277:1564) is actually the closest point to (1:√2:√3). I'm not sure what happened; do most of these calculations in a spreadsheet, so perhaps I sorted on the wrong column. --RDBury (talk) 02:14, 26 May 2021 (UTC)

May 25

sufficient condition for two set to be sepreate-able (topology)

Hi, I look for a sufficient condition that for the next claim to hold;

Let A be a compact subset $\mathbb {R} ^{n}$
and let B1 and B2 two different connected component. So there are 2 Open subset $O_{1},O_{2}$

1. $O_{1}\cup O_{2}\supseteq A$
2. $cl(O_{1})\cap cl(O_{2})=\emptyset$
3. $B_{1}\subseteq O_{1}$
4. $B_{2}\subseteq O_{2}$

Thanks!--Exx8 (talk) 19:01, 25 May 2021 (UTC)

Is there supposed to be some relation between $A$ and the pair $\{B_{1},B_{2}\}$, or are all three just given?  --Lambiam 23:44, 25 May 2021 (UTC)

B1 and B2 are connected component of A.--Exx8 (talk) 05:14, 26 May 2021 (UTC)

No additional conditions are necessary. The statement is true as given.--2406:E003:855:9A01:74B0:C329:6D75:B8CD (talk) 06:49, 26 May 2021 (UTC)