October 2018 - mathfun@mailman.xmission.com

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 05 Oct '18

05 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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