Stillwell's paper `Eisenstein's Footnote' (link below) gives a lucid explanation of how Eisenstein (at the tender age of 14) obtained a solution y for y^5+y=x as an infinite series in x of the form
y=x-x^5+(10*x^9)/2!-(14*15*x^13)/3!+.... ... Nonetheless, I don't think this gives the whole picture since the range of convergence for x isn't that big. Namely, the above series is exactly x*4F3(1/5, 2/5, 3/5, 4/5; 5/4, 2/4, 3/4|5^5*x^4/4^4), which converges (absolutely) for |x| less than or equal to (0.8)*(0.2)^(1/4), which is approximately 0.53499.
--incidentally, this is best possible. The discriminant of y^5 + y - x is 256+3125*x^4 which becomes 0 when |x|=4*5^(-5/4)=0.53499..., i.e. a double root y of y^5+y=x springs into existence for five different x with |x|=0.53499... via a bifurcation, whereas for all x with |x|<0.53499.... the root is isolated. Hence, the function itself becomes undefined (or more precisely is no longer uniquely defined) if |x|>0.53499... So, if one is interested in solving y^5+y=1, say, then the series above won't converge. Is there an analytic continuation of the 4F3 that would be defined for large values of x and which would still give a solution of that quintic, or is there another way to get around that?
6. Re: hypergeometric solution of quintic (Warren D Smith)
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Message: 1 Date: Wed, 12 Mar 2014 18:07:59 +0100 From: Joerg Arndt <arndt@jjj.de> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Lewis Carroll's determinant method and some (new?) uses Message-ID: <20140312170759.GB14741@jjj.de> Content-Type: text/plain; charset=us-ascii
* Fred Lunnon <fred.lunnon@gmail.com> [Mar 10. 2014 19:46]:
On 3/10/14, Joerg Arndt <arndt@jjj.de> wrote:
* Warren D Smith <warren.wds@gmail.com> [Mar 10. 2014 07:22]:
[...]
For Toeplitz matrices, there should be FFT methods.
Can you say a bit more about how this would work?
I wish I could. It's many years ago I looked into that and ended up with no notes, that indicates lack of success on my side.
I vaguely remember that one has to use a larger matrix that somehow makes the problem look like the one for circulants. No more than that, sorry.
Best, jj
[...]
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Message: 2 Date: Wed, 12 Mar 2014 16:47:14 -0400 From: Warren D Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Subject: [math-fun] hypergeometric solution of quintic Message-ID: <CAAJP7Y0b2oa6ZfmXjkTW=1k_H_z+4AHMg_4=W5Hdn34Pw4YgZA@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
Of course 4F3(x) and indeed any pFq(x) with p<=q+1, is defined for all complex x except for a set of measure zero, via analytic continuation. And this particular 4F3 is of course an algebraic function which makes that obvious.
The solution of the quintic via this method is however nuts in the sense that you will be far better off solving quintic by typical numerical methods, than by applying the "exact solution," i.e. you will obtain more digits faster. Indeed the best way to evaluate that 4F3 is by solving the quintic, not the other way round.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Message: 3 Date: Wed, 12 Mar 2014 15:14:01 -0600 From: rkg <rkg@cpsc.ucalgary.ca> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] G4G11 Message-ID: <alpine.LRH.2.03.1403121419550.26323@cpsc.ucalgary.ca> Content-Type: text/plain; format=flowed; charset="US-ASCII"
Dear all, I've also (as well as Jim Propp) been assembling something for the Gathering for Gardner next week. This is honoring John Conway, so the following, which may be distributed in any way one wishes, is dedicated to him, with the hope that he will complete The Triangle Book. It would be flattering if he can use any of this.
Guyration is the combination of `quadration' and `twinning' with Conway's `extraversion'. It yields a striking generalization of the triangle, sometimes with as many as 32 items appearing for the price of one.
Quadration grants the same status to the orthocentre of a triangle as it does to the vertices, so that each of the four points is the orthocentre of the triangle formed by the other three. It regards the triangle as an orthocentric quadrangle, which now has 4 vertices, 6 edges and 3 diagonal points.
Twinning draws the perpendicular bisectors of each of the six edges, producing a quadrangle of circumcentres, congruent to the original quadrangle. Alternatively, the two twins are related by reflexion in, or rotation through 180 deg about, a common `centre' .
Our triangle now has eight vertices, which are also both circumcentres, and orthocentres. It has six pairs of parallel edges which form three rectangles whose twelve vertices lie on the `50-point circle' [9-point? Euler-Feuerbach?] The other 38 points are the points of contact with the 32 touch-circles and six points of contact with the double-deltoid [Steiner, with the appearance of a Star-of-David].
There are 12 edge-circles, 24 medial circles, 32 touch-circles, 32 Gergonne points, 32 Nagel points, 144 Morley triangles, 64 Pavillet tetrahedra, 256 radpoints, 384 guylines [haven't checked if these are distinct; maybe they coincide in sets of four], and much more.
In case you'd like to do-it-yourself, here are the makings of a picture of the 32 touch-circles touching the 12 edges at 8 points, and each touching the 50-point circle.
Vertices V1 (36,51), V2 (-204,-77), V4 (132,-77), V8 (36,103) Ve (-36,-51), Vd (204,77), Vb (-132,77), V7 (-36,-103)
[hexadecimal subscripts; the 12 lines are ViVj with i,j in {1,2,4,8} and in {e,d,b,7} = {14,13,11,7}.]
The 32 circles are 16 \put(x,y){\circle{d}} and 16 with (-x,-y) d, and the following centres (x,y) and diameters d.
(20,-21) 112, (-92,-525) 896, (180,19) 192, (-252,115) 384, (48,55) 24, (660,259) 1248, (20,99) 32, (88,-105) 104, (16,63) 40, (-484,363) 1040, (60,91) 48, (-120,-209) 312, (12,-5) 144, (216,63) 280, (-288,175) 504, (-84,-437) 720.
32 touch-circles touch the 12 edges at 8 points, and each touches the 50-point circle.
There are 3 or 4 largish files which can be sent to interested individuals. R.
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Message: 4 Date: Wed, 12 Mar 2014 14:37:04 -0700 From: Henry Baker <hbaker1@pipeline.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] G4G11 Message-ID: <E1WNqps-0002wj-Mv@elasmtp-kukur.atl.sa.earthlink.net> Content-Type: text/plain; charset="us-ascii"
Very cool!
I wonder how this stuff "intersects" with a posting I made almost 2 years ago regarding incircles/excircles. In particular, an elegant quartic is formed by considering the incenter and 3 excenters as roots of a complex polynomial, where the triangle "lives" in the complex plane.
I'd be interested in getting the files. Thx.
At 02:14 PM 3/12/2014, rkg wrote:
Dear all, I've also (as well as Jim Propp) been assembling something for the Gathering for Gardner next week. This is honoring John Conway, so the following, which may be distributed in any way one wishes, is dedicated to him, with the hope that he will complete The Triangle Book. It would be flattering if he can use any of this.
Guyration is the combination of `quadration' and `twinning' with Conway's `extraversion'. It yields a striking generalization of the triangle, sometimes with as many as 32 items appearing for the price of one.
Quadration grants the same status to the orthocentre of a triangle as it does to the vertices, so that each of the four points is the orthocentre of the triangle formed by the other three. It regards the triangle as an orthocentric quadrangle, which now has 4 vertices, 6 edges and 3 diagonal points.
Twinning draws the perpendicular bisectors of each of the six edges, producing a quadrangle of circumcentres, congruent to the original quadrangle. Alternatively, the two twins are related by reflexion in, or rotation through 180 deg about, a common `centre' .
Our triangle now has eight vertices, which are also both circumcentres, and orthocentres. It has six pairs of parallel edges which form three rectangles whose twelve vertices lie on the `50-point circle' [9-point? Euler-Feuerbach?] The other 38 points are the points of contact with the 32 touch-circles and six points of contact with the double-deltoid [Steiner, with the appearance of a Star-of-David].
There are 12 edge-circles, 24 medial circles, 32 touch-circles, 32 Gergonne points, 32 Nagel points, 144 Morley triangles, 64 Pavillet tetrahedra, 256 radpoints, 384 guylines [haven't checked if these are distinct; maybe they coincide in sets of four], and much more.
In case you'd like to do-it-yourself, here are the makings of a picture of the 32 touch-circles touching the 12 edges at 8 points, and each touching the 50-point circle.
Vertices V1 (36,51), V2 (-204,-77), V4 (132,-77), V8 (36,103) Ve (-36,-51), Vd (204,77), Vb (-132,77), V7 (-36,-103)
[hexadecimal subscripts; the 12 lines are ViVj with i,j in {1,2,4,8} and in {e,d,b,7} = {14,13,11,7}.]
The 32 circles are 16 \put(x,y){\circle{d}} and 16 with (-x,-y) d, and the following centres (x,y) and diameters d.
(20,-21) 112, (-92,-525) 896, (180,19) 192, (-252,115) 384, (48,55) 24, (660,259) 1248, (20,99) 32, (88,-105) 104, (16,63) 40, (-484,363) 1040, (60,91) 48, (-120,-209) 312, (12,-5) 144, (216,63) 280, (-288,175) 504, (-84,-437) 720.
32 touch-circles touch the 12 edges at 8 points, and each touches the 50-point circle.
There are 3 or 4 largish files which can be sent to interested individuals. R.
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Message: 5 Date: Wed, 12 Mar 2014 16:53:55 -0700 From: Bill Gosper <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: [math-fun] Name that octadecahedron Message-ID: <CAA-4O0FHTgCB5M=dNUbtwbPrcZCQf2U3F=TcksEEaGxUumw51A@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
http://gosper.org/IMG_0216.JPG http://gosper.org/IMG_0217.JPG --rwg
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Message: 6 Date: Wed, 12 Mar 2014 20:03:48 -0400 From: Warren D Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] hypergeometric solution of quintic Message-ID: <CAAJP7Y0-pkny8zymJVMGg4a-jzaDB9XQZtJTFzq9sHW_109LnQ@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
On 3/12/14, Warren D Smith <warren.wds@gmail.com> wrote:
Of course 4F3(x) and indeed any pFq(x) with p<=q+1, is defined for all complex x except for a set of measure zero, via analytic continuation. And this particular 4F3 is of course an algebraic function which makes that obvious.
--and http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/ gives explicit formulas such as integral representations. Pade and Borel summation should also work.
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