[math-fun] quintic question
I believe the saying goes: A quintic with rational coeffs is solvable in radicals iff it factors or the Lagrange resolvent sextic (discussed here with JHC decades ago) has a rational root. What about if the quintic has radical coeffs? E.g., the center disk in the D_5 packing if 31 disks has radius a satisfying (43875 + 37180*I) + 74360*(-1)^(1/10) - 97460*(-1)^(3/10) - 70760*(-1)^(2/5) + 70760*(-1)^(3/5) + 23100*(-1)^(7/10) + ((47063255 + 40030096*I) + 80060192*(-1)^(1/10) - 104803824*(-1)^(3/10) - 76142960*(-1)^(2/5) + 76142960*(-1)^(3/5) + 24743632*(-1)^(7/10))*a + ((-4765010 - 4048824*I) - 8097648*(-1)^(1/10) + 10603816*(-1)^(3/10) + 7702560*(-1)^(2/5) - 7702560*(-1)^(3/5) - 2506168*(-1)^(7/10))*a^2 + ((1630 + 336*I) + 672*(-1)^(1/10) - 1584*(-1)^(3/10) - 1040*(-1)^(2/5) + 1040*(-1)^(3/5) + 912*(-1)^(7/10))*a^3 + ((-225 + 1116*I) + 2232*(-1)^(1/10) - 2964*(-1)^(3/10) + 200*(-1)^(2/5) - 200*(-1)^(3/5) + 732*(-1)^(7/10))*a^4 - 5*a^5 Can we say that such a quintic is solvable in radicals iff the resolvent has a root expressible in radicals? If not can we at least say that no resolvent solution implies no quintic solution? -rwg I have the resolventin Mma (LeafCount 3029) and Macsyma, if anyone wants. OLEO-GUM-RESIN NUMEROLOGIES
No replies? I expected a blizzard of unintelligible Galwology. Anyway, if the resolvent won't crack, we have the mildly surprising result that the D_5 packing of 31 disks <http://gosper.org/subopt31D5.png> is the first for which the radii are inexpressible in radicals. (Well, this may take some checking. E.g., there are D_12 and D_4 suboptimal "13"s, at least.) --rwg PS, David remarked on the absence of nondihedral rotational symmetry in his solutions. Corey remarks that he probably hasn't looked nearly far enough. For optimality in propeller type symmetry, you'd probably need >= 2 rings of disks with repeating size patterns abc and cba (or longer). On Fri, Jul 22, 2011 at 9:06 PM, Bill Gosper <billgosper@gmail.com> wrote:
I believe the saying goes: A quintic with rational coeffs is solvable in radicals iff it factors or the Lagrange resolvent sextic (discussed here with JHC decades ago) has a rational root.
What [] if the quintic has radical coeffs? E.g., the center disk in the D_5 packing [o]f 31 disks has radius a satisfying
(43875 + 37180*I) + 74360*(-1)^(1/10) - 97460*(-1)^(3/10) - 70760*(-1)^(2/5) + 70760*(-1)^(3/5) + 23100*(-1)^(7/10) + ((47063255 + 40030096*I) + 80060192*(-1)^(1/10) - 104803824*(-1)^(3/10) - 76142960*(-1)^(2/5) + 76142960*(-1)^(3/5) + 24743632*(-1)^(7/10))*a + ((-4765010 - 4048824*I) - 8097648*(-1)^(1/10) + 10603816*(-1)^(3/10) + 7702560*(-1)^(2/5) - 7702560*(-1)^(3/5) - 2506168*(-1)^(7/10))*a^2 + ((1630 + 336*I) + 672*(-1)^(1/10) - 1584*(-1)^(3/10) - 1040*(-1)^(2/5) + 1040*(-1)^(3/5) + 912*(-1)^(7/10))*a^3 + ((-225 + 1116*I) + 2232*(-1)^(1/10) - 2964*(-1)^(3/10) + 200*(-1)^(2/5) - 200*(-1)^(3/5) + 732*(-1)^(7/10))*a^4 - 5*a^5
Can we say that such a quintic is solvable in radicals iff the resolvent has a root expressible in radicals? If not can we at least say that no resolvent solution implies no quintic solution? -rwg I have the resolvent in Mma (LeafCount 3029) and Macsyma, if anyone wants. OLEO-GUM-RESIN NUMEROLOGIES
Corey may well be right. I had merely been surprised by the absence of such symmetry through N = 32. David ----- Original Message ----- From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Sent: Sunday, July 24, 2011 6:13:24 AM Subject: Re: [math-fun] quintic question PS, David remarked on the absence of nondihedral rotational symmetry in his solutions. Corey remarks that he probably hasn't looked nearly far enough. For optimality in propeller type symmetry, you'd probably need >= 2 rings of disks with repeating size patterns abc and cba (or longer).
Bill, Dave Dummit has a paper http://www.emba.uvm.edu/~dummit/quintics/solvable.pdf called "Solving Solvable Quintics". It appeared in Math. Comp. As far as I can see although his theorem 1 is stated for quintics with coefficients over Q, the proof works with Q replaced by any number field. Victor On Sun, Jul 24, 2011 at 2:13 AM, Bill Gosper <billgosper@gmail.com> wrote:
No replies? I expected a blizzard of unintelligible Galwology. Anyway, if the resolvent won't crack, we have the mildly surprising result that the D_5 packing of 31 disks <http://gosper.org/subopt31D5.png> is the first for which the radii are inexpressible in radicals. (Well, this may take some checking. E.g., there are D_12 and D_4 suboptimal "13"s, at least.) --rwg PS, David remarked on the absence of nondihedral rotational symmetry in his solutions. Corey remarks that he probably hasn't looked nearly far enough. For optimality in propeller type symmetry, you'd probably need >= 2 rings of disks with repeating size patterns abc and cba (or longer).
On Fri, Jul 22, 2011 at 9:06 PM, Bill Gosper <billgosper@gmail.com> wrote:
1040*(-1)^(3/5) + 912*(-1)^(7/10))*a^3 + ((-225 + 1116*I) + 2232*(-1)^(1/10) - 2964*(-1)^(3/10) + 200*(-1)^(2/5)
76142960*(-1)^(2/5) + 76142960*(-1)^(3/5) + 24743632*(-1)^(7/10))*a + ((-4765010 - 4048824*I) - 8097648*(-1)^(1/10) + 10603816*(-1)^(3/10) + 7702560*(-1)^(2/5) - 7702560*(-1)^(3/5) - 2506168*(-1)^(7/10))*a^2 + ((1630 + 336*I) + 672*(-1)^(1/10) - 1584*(-1)^(3/10) - 1040*(-1)^(2/5)
I believe the saying goes: A quintic with rational coeffs is solvable in radicals iff it factors or the Lagrange resolvent sextic (discussed here with JHC decades ago) has a rational root.
What [] if the quintic has radical coeffs? E.g., the center disk in the D_5 packing [o]f 31 disks has radius a satisfying
(43875 + 37180*I) + 74360*(-1)^(1/10) - 97460*(-1)^(3/10) - 70760*(-1)^(2/5) + 70760*(-1)^(3/5) + 23100*(-1)^(7/10) + ((47063255 + 40030096*I) + 80060192*(-1)^(1/10) - 104803824*(-1)^(3/10)
200*(-1)^(3/5) + 732*(-1)^(7/10))*a^4 - 5*a^5
Can we say that such a quintic is solvable in radicals iff the resolvent has a root expressible in radicals? If not can we at least say that no resolvent solution implies no quintic solution? -rwg I have the resolvent in Mma (LeafCount 3029) and Macsyma, if anyone wants. OLEO-GUM-RESIN NUMEROLOGIES
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The paper "Commentary on an unpublished lecture by G.N. Watson" http://www.math.carleton.ca/~williams/papers/pdf/244.pdf is also relevant. Victor On Sun, Jul 24, 2011 at 9:46 AM, Victor Miller <victorsmiller@gmail.com>wrote:
Bill, Dave Dummit has a paper http://www.emba.uvm.edu/~dummit/quintics/solvable.pdf called "Solving Solvable Quintics". It appeared in Math. Comp. As far as I can see although his theorem 1 is stated for quintics with coefficients over Q, the proof works with Q replaced by any number field.
Victor
On Sun, Jul 24, 2011 at 2:13 AM, Bill Gosper <billgosper@gmail.com> wrote:
No replies? I expected a blizzard of unintelligible Galwology. Anyway, if the resolvent won't crack, we have the mildly surprising result that the D_5 packing of 31 disks <http://gosper.org/subopt31D5.png> is the first for which the radii are inexpressible in radicals. (Well, this may take some checking. E.g., there are D_12 and D_4 suboptimal "13"s, at least.) --rwg PS, David remarked on the absence of nondihedral rotational symmetry in his solutions. Corey remarks that he probably hasn't looked nearly far enough. For optimality in propeller type symmetry, you'd probably need >= 2 rings of disks with repeating size patterns abc and cba (or longer).
On Fri, Jul 22, 2011 at 9:06 PM, Bill Gosper <billgosper@gmail.com> wrote:
1040*(-1)^(3/5) + 912*(-1)^(7/10))*a^3 + ((-225 + 1116*I) + 2232*(-1)^(1/10) - 2964*(-1)^(3/10) + 200*(-1)^(2/5)
76142960*(-1)^(2/5) + 76142960*(-1)^(3/5) + 24743632*(-1)^(7/10))*a + ((-4765010 - 4048824*I) - 8097648*(-1)^(1/10) + 10603816*(-1)^(3/10) + 7702560*(-1)^(2/5) - 7702560*(-1)^(3/5) - 2506168*(-1)^(7/10))*a^2 + ((1630 + 336*I) + 672*(-1)^(1/10) - 1584*(-1)^(3/10) - 1040*(-1)^(2/5)
I believe the saying goes: A quintic with rational coeffs is solvable in radicals iff it factors or the Lagrange resolvent sextic (discussed here with JHC decades ago) has a rational root.
What [] if the quintic has radical coeffs? E.g., the center disk in the D_5 packing [o]f 31 disks has radius a satisfying
(43875 + 37180*I) + 74360*(-1)^(1/10) - 97460*(-1)^(3/10) - 70760*(-1)^(2/5) + 70760*(-1)^(3/5) + 23100*(-1)^(7/10) + ((47063255 + 40030096*I) + 80060192*(-1)^(1/10) - 104803824*(-1)^(3/10)
200*(-1)^(3/5) + 732*(-1)^(7/10))*a^4 - 5*a^5
Can we say that such a quintic is solvable in radicals iff the resolvent has a root expressible in radicals? If not can we at least say that no resolvent solution implies no quintic solution? -rwg I have the resolvent in Mma (LeafCount 3029) and Macsyma, if anyone wants. OLEO-GUM-RESIN NUMEROLOGIES
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It seems unlikely, but AFAIK, no current CAS provides a radicals-where-possible quintic solver. Years ago CNWH remarked on math-fun that it was high time somebody wrote out the complete solution. Victor Miller kindly alerted me to a 2002 Intelligencer article by Berndt, Spearman, and Williams in which G. N. Watson, in a 1948 lecture, claimed to give a complete solution, and also claimed to have solved perhaps five times as many quintics as anyone else. Unfortunately the article, reconstructed from the lecture notes, appears to contain too many misprints to serve as a basis for implementation. But coauthor Kenneth Williams has more than rectified this situation with http://www.math.carleton.ca/~williams/papers/pdf/276.pdf, which is not only nearly errorless, but also more general, handling *three classes* (plus subcases) where Watson divides by zero! (Did CNWH know about Watson's method, and its deficiencies?) I was going to append my Mma implementation, but I just noticed it's a little shaky. Maybe I can push it to take radical and symbolic coefficients, so as to handle those multiparameter solvable quintics in http://en.wikipedia.org/wiki/Quintic_function . Meanwhile, here it goes on a randomly generated case: In[530]:= squintic[-5-2 #-12 #^2-2 #^3-12 #^4-7 #^5] (You can use any variable instead of #.) Out[530]= -(12/35)+1/35 37^(2/5) (1/2 (-481+35 Sqrt[185]))^(1/5) E^((2 I \[Pi] #1)/5)-24/35 (-1)^(2/5) (2/(6189+455 Sqrt[185]))^(1/5) E^((4 I \[Pi] #1)/5)+3/35 (-1)^(2/5) 2^(4/5) (-6189-455 Sqrt[185])^(1/5) E^((6 I \[Pi] #1)/5)-1/35 (-1)^(3/5) 37^(2/5) (1/2 (-481-35 Sqrt[185]))^(1/5) E^((8 I \[Pi] #1)/5)& The answer is a function mapping integers mod 5 onto the five solutions. Testing: In[534]:= Union[MinimalPolynomial/@%530/@Range[5]] Out[534]= {5+2 #1+12 #1^2+2 #1^3+12 #1^4+7 #1^5&} --rwg On Fri, Jul 22, 2011 at 9:06 PM, Bill Gosper <billgosper@gmail.com> wrote:
I believe the saying goes: A quintic with rational coeffs is solvable in radicals iff it factors or the Lagrange resolvent sextic (discussed here with JHC decades ago) has a rational root.
What about if the quintic has radical coeffs? E.g., the center disk in the D_5 packing if 31 disks has radius a satisfying
(43875 + 37180*I) + 74360*(-1)^(1/10) - 97460*(-1)^(3/10) - 70760*(-1)^(2/5) + 70760*(-1)^(3/5) + 23100*(-1)^(7/10) + ((47063255 + 40030096*I) + 80060192*(-1)^(1/10) - 104803824*(-1)^(3/10) - 76142960*(-1)^(2/5) + 76142960*(-1)^(3/5) + 24743632*(-1)^(7/10))*a + ((-4765010 - 4048824*I) - 8097648*(-1)^(1/10) + 10603816*(-1)^(3/10) + 7702560*(-1)^(2/5) - 7702560*(-1)^(3/5) - 2506168*(-1)^(7/10))*a^2 + ((1630 + 336*I) + 672*(-1)^(1/10) - 1584*(-1)^(3/10) - 1040*(-1)^(2/5) + 1040*(-1)^(3/5) + 912*(-1)^(7/10))*a^3 + ((-225 + 1116*I) + 2232*(-1)^(1/10) - 2964*(-1)^(3/10) + 200*(-1)^(2/5) - 200*(-1)^(3/5) + 732*(-1)^(7/10))*a^4 - 5*a^5
Can we say that such a quintic is solvable in radicals iff the resolvent has a root expressible in radicals? If not can we at least say that no resolvent solution implies no quintic solution? -rwg I have the resolventin Mma (LeafCount 3029) and Macsyma, if anyone wants. OLEO-GUM-RESIN NUMEROLOGIES
participants (3)
-
Bill Gosper -
dwcantrell＠comcast.net -
Victor Miller