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Re: [math-fun] least-squares, again
by Dan Asimov 05 Oct '18

05 Oct '18

The question of the best approximating line to a set of points in Euclidean n-space (a "point cloud") is interesting. This can be thought of as a side-effect of applying the statistical method called analysis of variance to a set of points P = {p_k in R^n | 1 <= k <= r}. Think of r as being large. The general method works like this: 1. Approximate the point cloud by the normal density d : R^n —> R on R^n having the same mean mu in R^n as the sample mean and the same covariance matrix Sigma in Mat(n x n; R) as the sample covariance matrix. Sigma can be any symmetric positive semi-definite matrix. To avoid degenerate cases: Assume Sigma is actually *positive definite*. Then d is defined by d(x) = (1/K) * exp(-(1/2)<x, Sigma(x)>) where K = Integral_{R^n} exp(-(1/2)<x, Sigma(x)>) dvol, so that the integral of d(x) over R^n comes out to unity. 2. Now find the eigenvalues and eigenvectors of Sigma. The eigenvalues must all be positive reals, and the eigenvectors are mutually perpendicular. Assume they are ordered by size of eigenvalue: L_1 >= L_2 >= ... >= L_n > 0 with v_k the eigenvector corresponding to eigenvalue L_k. 3. Finally in a least-squares sense, the k-dimensional affine subspace of R^n A^k(P) = mu + <v_1, v_2, ..., v_k> that is the mean mu plus the span of the first k eigenvectors (call it A^k(P)) is "the closest approximating k-dimensional subspace" to the original set P of points, for any k in the range 1 <= k <= n. —Dan

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[math-fun] least-squares derivation ?
by Henry Baker 04 Oct '18

04 Oct '18

In common discussions of least squares, the parameters (m,b) are estimated for the equation y = m*x+b using as data various datapoints [x1,y1], [x2,y2], [x3,y3], etc. For example, in Wikipedia (where m=beta2 and b=beta1): https://en.wikipedia.org/wiki/Linear_least_squares#Example So far, so good. Now, if I merely exchange x and y, then my equation is x = m'*y+b', where should be m' = 1/m and b' = -b/m. (Let's ignore the case where the best m=0.) However, if I then estimate (m',b') using the same least squares method, I don't get (1/m,-b/m) ! So either I'm doing something wrong, or perhaps there is a more symmetric least squares method that treats x and y symmetrically ??

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Re: [math-fun] least-squares derivation ?
by Henry Baker 04 Oct '18

04 Oct '18

After additional Googling, it would appear that the appropriate search terms are "total least squares" (as opposed to "ordinary least squares") and "orthogonal distance regression" for handling (x,y) data symmetrically. There is also this discussion, where we subtract off the centroid of all the (x,y) points, and then find the principal axis, which this author calls the "perpendicular regression": http://www.mathpages.com/home/kmath110.htm *** I assume that the "principal axis" is parallel to the "principal moment of inertia", which I assume is that axis of *minimum* moment of inertia. *** At 09:28 AM 10/4/2018, you wrote: >Standard least squares measures the *vertical* distance to the line from >each point and minimizes the sum of the squares. Interchanging x and y is >equivalent to switching to measuring the *horizontal* distance. You can >instead measure the perpendicular distance, which is invariant under >interchanging x and y (or any other isometry for that matter) but the >solution is correspondingly hairier. > >On Thu, Oct 4, 2018 at 9:24 AM Henry Baker <hbaker1(a)pipeline.com> wrote: >> In common discussions of least squares, the parameters >> (m,b) are estimated for the equation y = m*x+b using as >> data various datapoints [x1,y1], [x2,y2], [x3,y3], etc. >> >> For example, in Wikipedia (where m=beta2 and b=beta1): >> >> https://en.wikipedia.org/wiki/Linear_least_squares#Example >> >> So far, so good. >> >> Now, if I merely exchange x and y, then my equation is >> x = m'*y+b', where should be m' = 1/m and b' = -b/m. >> (Let's ignore the case where the best m=0.) >> >> However, if I then estimate (m',b') using the same least >> squares method, I don't get (1/m,-b/m) ! >> >> So either I'm doing something wrong, or perhaps there is >> a more symmetric least squares method that treats x and >> y symmetrically ??

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[math-fun] Continuous roots
by françois mendzina essomba2 04 Oct '18

04 Oct '18

Hello I searched for such a formula with integers, but it may be impossible, a small consolation for me. 1/4*Pi = (arctan(1/3)^2+arctan(2)*(arctan(2/9)^2+arctan(13/16)*(arctan(4/33)^2+arctan(49/128)*(arctan(8/129)^2+arctan(193/1024)*sqrt((a[4])^2+v[4]*sqrt((a[5])^2+v[5]*cdots)))^(1/2))^(1/2))^(1/2))^(1/2); a[n]=arctan(1/(2^n)/(2+1/(2^n)^2))^2 ; v[n]=arctan(1/2/(2^n)^3+3/2/(2^n)); For an exact value, we can associate v [n] with c[n] = arctan (1/2 ^ (n + 1)) Pi/4=(arctan(1/3)^2+arctan(2)*(arctan(2/9)^2+arctan(13/16)*(arctan(4/33)^2+arctan(49/128)*(arctan(8/129)^2+arctan(193/1024)*arctan(1/16))^(1/2))^(1/2))^(1/2))^(1/2); FME...

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[math-fun] Atiyah RH claim?
by rcs＠xmission.com 01 Oct '18

01 Oct '18

[from Bill DuBuque -- rcs] Date: Wed, 26 Sep 2018 11:40:57 -0400 Subject: Re: [math-fun] Atiyah RH claim? From: Bill Dubuque <lin.dubuque(a)gmail.com> See also the following MathOverflow thread on Atiyah's RH "proof" (which should help one to understand why few professional mathematicians are willing to discuss its correctness publicly) Is there a way to discuss the correctness of the proof of the RH by Atiyah in MO? https://meta.mathoverflow.net/questions/3894/is-there-a-way-to-discuss-the-… On Thu, Sep 20, 2018 at 12:39 PM Michael Kleber <michael.kleber(a)gmail.com> wrote: Anyone have opinions, rumors, hearsay, etc. about Atiyah's claimed Riemann Hypothesis proof talk on Monday? Circulating rumor: https://mobile.twitter.com/troublewithecon/status/1042611713839751169 Confirmation of what's in the abstracy by Heidelberg Laureate Forum: https://twitter.com/HLForum/status/1042793378629541889 John Baez's counter, that Aityah's other recent big claims have not held up: https://mobile.twitter.com/johncarlosbaez/status/1042609961216335874 "Everyone who knows him well has been too embarrassed to publicly discuss the reasons." --Michael -- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Branko Grünbaum (1929-2018)
by James Buddenhagen 30 Sep '18

30 Sep '18

Perhaps already mentioned on this list, but I don't remember it. https://math.washington.edu/news/2018/09/18/branko-grunbaum-1929-2018

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Re: [math-fun] Curve-fitting methods ?
by Keith F. Lynch 27 Sep '18

27 Sep '18

Brent Meeker <meekerdb(a)verizon.net> wrote: > Keith F. Lynch wrote: >> But any three points that obey the triangle inequality are >> compatible with any radius of curvature whatsoever. (Except for a >> radius less than that which makes them coplanar with the center of >> the circle they're on.) > No they aren't. The spherical law of cosines is > cos a = cos b cos c + sin b sin c cos A > where lower case are edges and upper case are interior angles. I know. I should have said any three *sides* that obey the triangle inequality. For one thing, speaking of three *points* that obey the triangle inequality makes little sense, as no three points can fail to do so. Suppose all you have are the lengths of three sides of a triangle, and the knowledge that the three vertices of the triangle are equidistant from some "center" point in space, possibly a point at infinity, possibly a point in the same plane with the triangle. You don't know how curved the sides are, except that they all have the same radius of curvature, the curvature such that every point on all the sides is also equidistant from that same center point. Wouldn't you agree that with the information you have, you can't rule out any radius above a certain minimum? I'm not a visual thinker, so I won't guess at whether the locus of possible center points is a line or some other kind of curve. If the three sides of the triangle were equal, I assume it would have to be a line, by symmetry, unless it splits into three curves. > So you guess some radius R and solve for A and by permutation also B > and C. Then you see whether the law of sines is satisfied. > sin A/sin a = sin B/ sin b = sin C/sin c > In general it won't be but one side, say (sin a/sin b)? depends on > R while the other side (sin A/sin B) doesn't. So you adjust R and > iterate. > I guess I should try to this see if you really works. :-) > I tried it. Doesn't work in a practical sense as written...just not > enough resolution. Maybe I can fix it. Again, I don't think you have enough information. The law of sines should *always* be satisfied, unless you choose a radius less than that of the circle the three points are on. This is complicated by the fact that you can't assign any definite coordinates for the points or any straight-line distances between them until you know the radius. Any three points in space are on a unique plane and are on a unique circle in that plane. They're also on an infinite number of spheres, with all radii starting with that of the circle. Again, this is complicated by the fact that our points aren't fixed, but move around as the fixed-length sides change curvature, but I think the principle is the same. For extra credit, extend the scenario to negative curvature. See https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines

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Re: [math-fun] Curve-fitting methods ?
by Keith F. Lynch 26 Sep '18

26 Sep '18

Brent Meeker <meekerdb(a)verizon.net> wrote: > What you do is iterate. You can take three points and using > spherical geometry determine a radius and a center. But any three points that obey the triangle inequality are compatible with any radius of curvature whatsoever. (Except for a radius less than that which makes them coplanar with the center of the circle they're on.)

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Re: [math-fun] Curve-fitting methods ?
by Keith F. Lynch 26 Sep '18

26 Sep '18

Henry Baker <hbaker1(a)pipeline.com> wrote: > Keith F. Lynch wrote: >> One of my projects for satisfying extreme skeptics is a way to >> construct a crude picture of the Earth from years of data on the >> brightness of the Earth-lit part of the crescent moon as seen from >> one's backyard. Of course the skeptic would have to trust my >> software, so there's little point in my writing such software >> unless I am the skeptic. Still, it's an interesting project. > I'll bite. Have you constructed this crude picture of Earth, or are > you suggesting that it could be done? The latter, as I thought would be clear from, "there's little point in my writing such software...." In addition to the software, I'd need hardware with which I could measure the brightnesses of the earthlit part of the moon, the sunlit part of the moon, and an equal area of the sky near the moon. I'd subtract out the brightness of the sky near the moon so I'm only paying attention to light from the moon. I'd look at the ratio of brightnesses of the sunlit and earthlit parts (as adjusted for lunar phase and for how the lunar albedo varies due to lunar topography) to compensate for absorption as the light approaches me through Earth's atmosphere. Ideally this would be done in full color, so as to get a full color image of Earth. > I would be interested in such a picture, as it is an analogy to the > "1 pixel camera" discussed here previously. Years ago, I saw images of Pluto based entirely on its brightness measured from Earth. The north-south resolution was almost nil, presumably because all the observations were made at nearly the same Pluto latitude. Since Pluto has a high obliquity, this can be corrected by watching it for a whole orbit, 248 years. I doubt anyone will bother, since today we have high resolution images from New Horizons, a Pluto flyby probe. Indeed, I can't find those old low-res images online, since nearly any request for Pluto images returns images from that space probe. We can do better with the moon, since its terrestrial latitude varies from about -30 to +30 degrees. And we only have to wait 19 years for the moon to be in every possible state relative to the sun and our planet, not 248. That should give enough data that averaging would remove the clouds from the image and let you clearly see the continents, oceans, ice caps, and major islands. I'm a fairly radical skeptic. I believe trusted authorities are often lying to us, are mistaken, or are repeating others' lies or innocent mistakes. (As I myself recently did here with a mistaken claim about smooth functions. Sorry.) So I prefer to observe, measure, calculate, reason out, and prove as many things for myself as possible. If anyone's curious why I believe this, please ask me off-list. Thanks.

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Re: [math-fun] Curve-fitting methods ?
by Keith F. Lynch 26 Sep '18

26 Sep '18

Brent Meeker <meekerdb(a)verizon.net> wrote: > Keith F. Lynch wrote: >> A much easier project is using a table of distances between cities >> to calculate the radius of the Earth. (Has anyone done that?) > I'm sure it's been done. I've seen it illustrated in textbooks. I'm curious what the best way is. Here are my first thoughts: The first city can be given any coordinates. The second city can be placed anywhere on a sphere around the first city with a radius equal to their distance. The third city can be placed anywhere on the circle where the spheres of the correct distances from the first two cities intersect. The fourth city can be placed on either of the two points where the three correct-distance spheres intersect. But what about the fifth city? Unless the data are of infinite precision and accuracy, there probably won't be an exact intersection point of the correct-distance spheres around the first four cities. You could do a least-squares fit, but why should all of the error budget go to the fifth city? You could do simulated annealing, or something like it, where all the cities are free to wiggle around by smaller and smaller amounts so as to minimize the sum of the squares of the errors in their pairwise distances. Anyhow, after you have N cities, you simply find a point equidistant from all of them, or as close to it as possible. That will be the center of the Earth. Again, you'd probably want to do simulated annealing to distribute the error budget. Maybe combine it with the first annealing. What is wrong with this picture? The main problem is that the distances from the table were never intended to be straight line distances, but distances along the curve of the Earth. You can convert from one to the other easily enough, but only if you already know the Earth's radius, which is what we set out to find out. Maybe if you repeated the process several times, using the radius from the previous run to get the curvatures for the current run, it would converge to the correct value? > Of course cities which are far from all the others provide better > estimates, since the proportional distance error is smaller. Another approach is to stop with four cities. Which four? As you said, the ones that are furthest from each other. (Assuming all table entries are equally accurate and precise.) But four cities have six mutual distances. Which one(s) should be maximized? The longest? The shortest? The sum of all six? The product of all six? The volume of the resulting tetrahedron? If I recall correctly, just as any three points (if not all on a line) are on a circle, i.e. are equidistant from a unique point, any four points (if not all on a plane) are on a sphere, i.e. are equidistant from a unique point, the center of that sphere. Of course you'd still need to repeat the process with successive estimates of the curvature. And you'd want to do it for more than one set of four cities. Then there's the simplest approach of all: Find the maximum distance between any two cities, add a little, and call that half the circumference of the Earth.

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