[math-fun] While playing with A103314
In attacking A103314, I decided to try to plot the sums of subsets of the roots of 1. For square roots, I got 1 2 1 where numbers count multiplicities of points. The 2 at the origin corresponds to A103314(2) = 2. For cube roots, I got 1 1 1 2 1 1 1 so A103314(3) = 2. For fourth roots, I got 1 2 1 2 4 2 1 2 1 so that A103314(4) = 4. Now, notice that the square root plot includes the vertices you would see sighting a square along its diagonal. The cube root plot includes the vertices you would see sighting a cube along its great diagonal. Is the fourth root plot what you would (might) see sighting a tesseract along its great diagonal? Does the analogy extend to higher roots? - David W. Wilson "Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"
Yes, naturally, this is a pretty picture. If you take the unit cube in n dimensions to be product of the unit intervals [0,1] on each coordinate, then there is a linear map sending the unit vectors anywhere you please. You've mapped them to the nth roots of unity, so that the vertices map to the various values of partial sums of roots of unity. The linear map is not literally a view "as you would see it", in perspective, but it is the limit of perspective viewsas you move far away and rescale to keep the image the same size. The smoothed out shape will converge to a 2-dimensional Gaussian distribution. Bill Thurston On May 8, 2005, at 4:36 AM, David Wilson wrote:
In attacking A103314, I decided to try to plot the sums of subsets of the roots of 1.
For square roots, I got
1 2 1
where numbers count multiplicities of points. The 2 at the origin corresponds to A103314(2) = 2.
For cube roots, I got
1 1
1 2 1
1 1
so A103314(3) = 2.
For fourth roots, I got
1 2 1
2 4 2
1 2 1
so that A103314(4) = 4.
Now, notice that the square root plot includes the vertices you would see sighting a square along its diagonal. The cube root plot includes the vertices you would see sighting a cube along its great diagonal. Is the fourth root plot what you would (might) see sighting a tesseract along its great diagonal? Does the analogy extend to higher roots?
- David W. Wilson
"Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings" _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
And I realized something else from the little figures I drew. The sum of the numbers in the nth root plot is 2^n. The center number is a(n) = A103314(n). If you remove the center number from the figure, the remaining numbers are arranged with n-way symmetry, so their sum, 2^n - a(n), must be divisible by n. This implies. [1] a(n) == 2^n (mod n). which, along with a(p) = 2 (prime p), I will note on A103314. [1] gives a(30) == 4 (mod 30), which is consistent with Scott's value a(30) = 146854 (though it is weak confirmation). ----- Original Message ----- From: "Bill Thurston" <wpt4@cornell.edu> To: <ham>; "math-fun" <math-fun@mailman.xmission.com> Sent: Sunday, May 08, 2005 7:29 AM Subject: Re: [math-fun] While playing with A103314
Yes, naturally, this is a pretty picture. If you take the unit cube in n dimensions to be product of the unit intervals [0,1] on each coordinate, then there is a linear map sending the unit vectors anywhere you please. You've mapped them to the nth roots of unity, so that the vertices map to the various values of partial sums of roots of unity.
The linear map is not literally a view "as you would see it", in perspective, but it is the limit of perspective viewsas you move far away and rescale to keep the image the same size.
The smoothed out shape will converge to a 2-dimensional Gaussian distribution.
Bill Thurston
On May 8, 2005, at 4:36 AM, David Wilson wrote:
In attacking A103314, I decided to try to plot the sums of subsets of the roots of 1.
For square roots, I got
1 2 1
where numbers count multiplicities of points. The 2 at the origin corresponds to A103314(2) = 2.
For cube roots, I got
1 1
1 2 1
1 1
so A103314(3) = 2.
For fourth roots, I got
1 2 1
2 4 2
1 2 1
so that A103314(4) = 4.
Now, notice that the square root plot includes the vertices you would see sighting a square along its diagonal. The cube root plot includes the vertices you would see sighting a cube along its great diagonal. Is the fourth root plot what you would (might) see sighting a tesseract along its great diagonal? Does the analogy extend to higher roots?
- David W. Wilson
"Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings" _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (2)
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Bill Thurston -
David Wilson