Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
Come to think of it, how can we prove that the mapping (0,oo)^3 —> (0,oo)^3 of positive octants via (a,b,c) |—> (girth_ab, girth_bc, girth_ca) is one-to-one? Or is it? —Dan
I think some girth combinations are impossible. For example, the girths 1, eps, eps where eps is some very small epsilon. If a girth is very small, then both of the semiaxes in its plane must be very small, so the remaining semiaxis will be close to 1/4, in which case both of the remaining girths will be close to 1. Dan Asimov writes:
Come to think of it, how can we prove that the mapping (0,oo)^3 —> (0,oo)^3 of positive octants via
(a,b,c) |—> (girth_ab, girth_bc, girth_ca)
is one-to-one? Or is it?
—Dan
If we lengthen axis a, we must shorten axis b and c to preserve the ab and ac girths. Consequently the bc girth decreases, a contradiction. The inverse holds for shortening axis a. A root-solving algorithm should converge to arbitrary precision given enough time and a good enough initial guess. —Brad
On Jun 23, 2019, at 5:24 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Come to think of it, how can we prove that the mapping (0,oo)^3 —> (0,oo)^3 of positive octants via
(a,b,c) |—> (girth_ab, girth_bc, girth_ca)
is one-to-one? Or is it?
—Dan
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Excellent proof! By the way, for the 'obvious' lemma that the girth is an increasing function of the two semi-axes, it follows from: Theorem: If L is a compact subset of R^n and K is a convex compact subset of L, then the surface area of K is bounded above by the surface area of L, with equality if and only if K = L. Proof: The map f : boundary(K) --> boundary(L) given by outward normal projection cannot decrease area. QED. Incidentally, does anyone have a reference for this beautiful theorem? It's well known, but I don't know what it's called, and I'd like to cite it in my thesis. Thank you! Best wishes, Adam P. Goucher
Sent: Monday, June 24, 2019 at 12:19 AM From: "Brad Klee" <bradklee@gmail.com> To: "Dan Asimov" <dasimov@earthlink.net>, math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
If we lengthen axis a, we must shorten axis b and c to preserve the ab and ac girths. Consequently the bc girth decreases, a contradiction. The inverse holds for shortening axis a.
A root-solving algorithm should converge to arbitrary precision given enough time and a good enough initial guess.
—Brad
On Jun 23, 2019, at 5:24 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Come to think of it, how can we prove that the mapping (0,oo)^3 —> (0,oo)^3 of positive octants via
(a,b,c) |—> (girth_ab, girth_bc, girth_ca)
is one-to-one? Or is it?
—Dan
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Hi Adam, I don't know about your beautiful theorems, but do consider this lemma obvious. The Law of Sines can be applied to obtuse triangles in the infinitesimal limit, while the tangent vectors of the thinner ellipse flatten out more readily. Cheers, Brad On Sun, Jun 23, 2019 at 7:09 PM Adam P. Goucher <apgoucher@gmx.com> wrote:
Excellent proof!
By the way, for the 'obvious' lemma that the girth is an increasing function of the two semi-axes, it follows from:
I believe the two-dimensional version of this assertion was made by Archimedes, and was intended to serve as a way of characterizing arc-length. Jim Propp On Sun, Jun 23, 2019 at 8:09 PM Adam P. Goucher <apgoucher@gmx.com> wrote:
Excellent proof!
By the way, for the 'obvious' lemma that the girth is an increasing function of the two semi-axes, it follows from:
Theorem: If L is a compact subset of R^n and K is a convex compact subset of L, then the surface area of K is bounded above by the surface area of L, with equality if and only if K = L.
Proof: The map f : boundary(K) --> boundary(L) given by outward normal projection cannot decrease area. QED.
Incidentally, does anyone have a reference for this beautiful theorem? It's well known, but I don't know what it's called, and I'd like to cite it in my thesis.
Thank you!
Best wishes,
Adam P. Goucher
Sent: Monday, June 24, 2019 at 12:19 AM From: "Brad Klee" <bradklee@gmail.com> To: "Dan Asimov" <dasimov@earthlink.net>, math-fun < math-fun@mailman.xmission.com> Subject: Re: [math-fun] Is it me, or is math.stackexchange.com controlled by morons?
If we lengthen axis a, we must shorten axis b and c to preserve the ab and ac girths. Consequently the bc girth decreases, a contradiction. The inverse holds for shortening axis a.
A root-solving algorithm should converge to arbitrary precision given enough time and a good enough initial guess.
—Brad
On Jun 23, 2019, at 5:24 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Come to think of it, how can we prove that the mapping (0,oo)^3 —> (0,oo)^3 of positive octants via
(a,b,c) |—> (girth_ab, girth_bc, girth_ca)
is one-to-one? Or is it?
—Dan
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In case that wasn’t clear, I was replying to Adam Goucher’s question, saying that I believe that Archimedes said something that in modern language would be that if L is a compact subset of R^2 and K is a convex compact subset of L, then the perimeter of K is bounded above by the perimeter of L, with equality if and only if K = L. That is: Rather than giving a definition of arc length (as we moderns would feel obliged to do), Archimedes gave a characterization of arc length that is just as useful. Jim Propp
Hi Jim, Is your reference "Archimede's Axioms for Arc-Length and Area" by Scott Brodie [1]? If yes, you need another condition on L. This example with the two ellipses shows none of the pathology of Brodie's Figure 2, so (unless Adam wants to tell us more about his thesis) we really don't have context to need to worry about Archimede's Axiom 2. However, Brodie's Figure 2 is a nice challenge to the law of sines technique. Yet if we approximate FBG as a triangle, and ignore small deviations, then FB is shorter than FG + GB by the triangle inequality. As the angles FBG and BFG are significantly greater than the small angle at A, we can be fairly confident that the contained curve is the shorter. --Brad [1] https://www.jstor.org/stable/2690029?seq=1#page_scan_tab_contents On Sun, Jun 23, 2019 at 9:49 PM James Propp <jamespropp@gmail.com> wrote:
In case that wasn’t clear, I was replying to Adam Goucher’s question, saying that I believe that Archimedes said something that in modern language would be that if L is a compact subset of R^2 and K is a convex compact subset of L, then the perimeter of K is bounded above by the perimeter of L, with equality if and only if K = L.
That is: Rather than giving a definition of arc length (as we moderns would feel obliged to do), Archimedes gave a characterization of arc length that is just as useful.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I was rashly extrapolating from the original statement by Archimedes, which I learned from Andy Gleason when I was an undergraduate. I’m working from memory here, but it went something like this: if you have a line segment from point P to point Q, a convex curve from P to Q, and another convex curve from P to Q, such that the first curve lies between the line segment and the second curve, then the length of the first curve is intermediate between the length of the line segment and the length of the second curve. Jim Propp On Mon, Jun 24, 2019 at 2:44 AM Brad Klee <bradklee@gmail.com> wrote:
Hi Jim,
Is your reference "Archimede's Axioms for Arc-Length and Area" by Scott Brodie [1]? If yes, you need another condition on L.
This example with the two ellipses shows none of the pathology of Brodie's Figure 2, so (unless Adam wants to tell us more about his thesis) we really don't have context to need to worry about Archimede's Axiom 2.
However, Brodie's Figure 2 is a nice challenge to the law of sines technique. Yet if we approximate FBG as a triangle, and ignore small deviations, then FB is shorter than FG + GB by the triangle inequality. As the angles FBG and BFG are significantly greater than the small angle at A, we can be fairly confident that the contained curve is the shorter.
--Brad
[1] https://www.jstor.org/stable/2690029?seq=1#page_scan_tab_contents
On Sun, Jun 23, 2019 at 9:49 PM James Propp <jamespropp@gmail.com> wrote:
In case that wasn’t clear, I was replying to Adam Goucher’s question, saying that I believe that Archimedes said something that in modern language would be that if L is a compact subset of R^2 and K is a convex compact subset of L, then the perimeter of K is bounded above by the perimeter of L, with equality if and only if K = L.
That is: Rather than giving a definition of arc length (as we moderns
would
feel obliged to do), Archimedes gave a characterization of arc length that is just as useful.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Hi Jim, No worries. Your latest statement sounds equivalent to the one from "Mathematics Magazine". So we've got it all straightened out except that the spelling "Archimede's" is really not an acceptable approximation to "Archimedes' ". Ouch! I guess I need to watch those apostrophes more closely. --Brad On Mon, Jun 24, 2019 at 7:26 AM James Propp <jamespropp@gmail.com> wrote:
I was rashly extrapolating from the original statement by Archimedes, which I learned from Andy Gleason when I was an undergraduate. I’m working from memory here, but it went something like this: if you have a line segment from point P to point Q, a convex curve from P to Q, and another convex curve from P to Q, such that the first curve lies between the line segment and the second curve, then the length of the first curve is intermediate between the length of the line segment and the length of the second curve.
Jim Propp
On Mon, Jun 24, 2019 at 2:44 AM Brad Klee <bradklee@gmail.com> wrote:
Hi Jim,
Is your reference "Archimede's Axioms for Arc-Length and Area" by Scott Brodie [1]? If yes, you need another condition on L.
This example with the two ellipses shows none of the pathology of Brodie's Figure 2, so (unless Adam wants to tell us more about his thesis) we really don't have context to need to worry about Archimede's Axiom 2.
However, Brodie's Figure 2 is a nice challenge to the law of sines technique. Yet if we approximate FBG as a triangle, and ignore small deviations, then FB is shorter than FG + GB by the triangle inequality. As the angles FBG and BFG are significantly greater than the small angle at A, we can be fairly confident that the contained curve is the shorter.
--Brad
[1] https://www.jstor.org/stable/2690029?seq=1#page_scan_tab_contents
On Sun, Jun 23, 2019 at 9:49 PM James Propp <jamespropp@gmail.com> wrote:
In case that wasn’t clear, I was replying to Adam Goucher’s question, saying that I believe that Archimedes said something that in modern language would be that if L is a compact subset of R^2 and K is a convex compact subset of L, then the perimeter of K is bounded above by the perimeter of L, with equality if and only if K = L.
That is: Rather than giving a definition of arc length (as we moderns
would
feel obliged to do), Archimedes gave a characterization of arc length that is just as useful.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (5)
-
Adam P. Goucher -
Brad Klee -
Dan Asimov -
James Propp -
Tom Karzes