[math-fun] solving, denesting, and reading comprehension
reg>http://www.math.harvard.edu/~elkies/trinomial.html <http://www.math.harvard.edu/%7Eelkies/trinomial.html> gives a (rational) continuum of irreducible, supposedly solvable sextics: (125 - u)*x^6 + 12*u*(u + 3)^2*x + u*(u - 5)*(u + 3)^2 but my sextic solver fails for u≠0. (≠-3,≠5) Can anybody propose a correction? CG>I have no correction, but I concur with your result. For example, with u = 4 I get 121x^6 + 2352x - 196 with Galois group PSL(2,5): polgalois(121*x^6 + 2352*x - 196) %1 = [60, 1, 1, "L(6) = PSL(2,5) = A_5(6)"] which is not solvable. Charles Greathouse Analyst/Programmer Case Western Reserve University Mea gufa. With more patience than I deserve, Elkies privately explained that his paper claims only that the Galois groups are atypically small, not necessarily small enough to solve. He also had no news re the possibility of a solvable, irreducible septic trinomial. Come to think of it, I don't even recall seeing a solvable, irreducible x^6+ax+b. --rwg
OK, so now I'm curious: Consider the space of all monic polynomials of degree n >= 5 over the reals or complexes. They are in bijective correspondence with R^n (resp. C^n), so let's give them that topology. What does the set of solvable monic polynomials look like within the space of all of them? Even more basic: What is the *dimension* of the space of solvable monic polynomials in the space of all of them (over R or C) ? --Dan On 2013-08-30, at 11:59 AM, Bill Gosper wrote: . . . . . .
a (rational) continuum of irreducible, supposedly solvable sextics: (125 - u)*x^6 + 12*u*(u + 3)^2*x + u*(u - 5)*(u + 3)^2 . . . . . .
Bill Gosper <billgosper@gmail.com> writes:
Mea gufa. [...]
He also had no news re the possibility of a solvable, irreducible septic trinomial. Come to think of it, I don't even recall seeing a solvable, irreducible x^6+ax+b. --rwg
Here's the full list up to 100 according to gp: ? for(a=1,100,for(b=-100,100,p=x^6+a*x+b;if(polisirreducible(p),if(polgalois(p)[1]%5,print(p," ",polgalois(p)))))) x^6 + 3*x + 3 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 3*x + 5 [48, -1, 1, "2S_4(6) = [2^3]S(3) = 2 wr S(3)"] x^6 + 8*x + 20 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 8*x + 89 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 10*x + 5 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 14*x + 35 [48, -1, 1, "2S_4(6) = [2^3]S(3) = 2 wr S(3)"] x^6 + 30*x + 93 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 40*x + 82 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 44*x + 55 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 45*x + 55 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 49*x - 49 [24, -1, 2, "2A_4(6) = [2^3]3 = 2 wr 3"] x^6 + 56*x + 62 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 65*x + 13 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] (I didn't use a<0 because a,b and -a,b are equivalent under x <--> -x.) Malle's paper [M1] should give the general formulas. [M1] G. Malle: Polynomials for primitive nonsolvable permutation groups of degree d<=15. J. Symbolic Computation 4 (1987) #1, 83-92. NDE
Wow, 78 roots, two real. Here's one of those: Out[101]= 67/246 Sqrt[7] Sin[1/3 ArcCsc[2 Sqrt[7]]] + 245/738 Sin[1/3 ArcSin[71/98]] + 469/738 Sin[1/3 (4 π - ArcSin[13/14])] + 35/246 Sin[1/3 (4 π + ArcSin[13/14])] + 70/123 Sqrt[7/3] Sin[1/3 (2 π + ArcSin[(3 Sqrt[3/7])/2])] + 67/246 Sqrt[7/3] Sin[1/3 (4 π + ArcSin[(3 Sqrt[3/7])/2])] + ( 469 Sin[1/3 (4 π - ArcSin[(3 Sqrt[3])/14])])/(246 Sqrt[3]) - ( 35 Sin[1/3 ArcSin[(3 Sqrt[3])/14]])/(82 Sqrt[3]) + ( 245 Sin[1/3 (2 π - ArcSin[(39 Sqrt[3])/98])])/( 246 Sqrt[3]) - Sqrt[ 7/6 (-1 + Sqrt[13] Sin[1/3 (2 π - ArcSin[293/(182 Sqrt[13])])])] In[104]:= N[%%%^6 + 49*%%%, 49] Out[104]= 49.00000000000000000000000000000000000000000000000 (RootReduce wimped out.) --rwg On Fri, Aug 30, 2013 at 4:12 PM, Noam Elkies <elkies@math.harvard.edu>wrote:
Bill Gosper <billgosper@gmail.com> writes:
Mea gufa. [...]
He also had no news re the possibility of a solvable, irreducible septic trinomial. Come to think of it, I don't even recall seeing a solvable, irreducible x^6+ax+b. --rwg
Here's the full list up to 100 according to gp:
? for(a=1,100,for(b=-100,100,p=x^6+a*x+b;if(polisirreducible(p),if(polgalois(p)[1]%5,print(p," ",polgalois(p))))))
x^6 + 3*x + 3 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 3*x + 5 [48, -1, 1, "2S_4(6) = [2^3]S(3) = 2 wr S(3)"] x^6 + 8*x + 20 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 8*x + 89 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 10*x + 5 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 14*x + 35 [48, -1, 1, "2S_4(6) = [2^3]S(3) = 2 wr S(3)"] x^6 + 30*x + 93 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 40*x + 82 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 44*x + 55 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 45*x + 55 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 49*x - 49 [24, -1, 2, "2A_4(6) = [2^3]3 = 2 wr 3"] x^6 + 56*x + 62 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"] x^6 + 65*x + 13 [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"]
(I didn't use a<0 because a,b and -a,b are equivalent under x <--> -x.)
Malle's paper [M1] should give the general formulas.
[M1] G. Malle: Polynomials for primitive nonsolvable permutation groups of degree d<=15. J. Symbolic Computation 4 (1987) #1, 83-92.
NDE
Bill Gosper <billgosper@gmail.com> writes:
Wow, 78 roots, two real. Here's one of those: [...]
A trinomial can never have more than two real roots (Descartes' rule). For the 72-element group, a general formula is 27*x^6 + 9*t^3*(t+8)*x + t^3*(t+8)*(t^2+7*t+1) = 3*A^2 + t*(t+8)*B^2 where A = 3*x^3 - t*(t+8)*x/2 - t^2*(t+8)/2 B = (6*x^2 + 3*t*x - (t^2+2*t)) / 2 it seems that this has a real pair of roots *iff* -8 < t < 0. NDE
On Sat, Aug 31, 2013 at 10:33 AM, Noam Elkies <elkies@math.harvard.edu>wrote:
Bill Gosper <billgosper@gmail.com> writes:
Wow, 78 roots, two real. Here's one of those: [...]
A trinomial can never have more than two real roots (Descartes' rule).
With what disclaimer? 4 x - 5 x^3 + x^5 has five. rwg>In[104]:= N[%%%^6 + 49*%%%, 49] Out[104]= 49.00000000000000000000000000000000000000000000000 (RootReduce wimped out.) Likewise MinimalPolynomial. But at least In[161]:= Timing[ MinimalPolynomial[ 67/246 Sqrt[7] Sin[1/3 ArcCsc[2 Sqrt[7]]] + 245/738 Sin[1/3 ArcSin[71/98]] + 469/738 Sin[1/3 (4 \[Pi] - ArcSin[13/14])] + 35/246 Sin[1/3 (4 \[Pi] + ArcSin[13/14])] + 70/123 Sqrt[7/3] Sin[1/3 (2 \[Pi] + ArcSin[(3 Sqrt[3/7])/2])] + 67/246 Sqrt[7/3] Sin[ 1/3 (4 \[Pi] + ArcSin[(3 Sqrt[3/7])/2])] + (469 Sin[ 1/3 (4 \[Pi] - ArcSin[(3 Sqrt[3])/14])])/(246 Sqrt[ 3]) - (35 Sin[1/3 ArcSin[(3 Sqrt[3])/14]])/(82 Sqrt[ 3]) + (245 Sin[ 1/3 (2 \[Pi] - ArcSin[(39 Sqrt[3])/98])])/(246 Sqrt[3]) - Sqrt[7/6 (-1 + Sqrt[13] Sin[1/3 (2 \[Pi] - ArcSin[293/(182 Sqrt[13])])])]]] Out[161]= {1365.427547, -49 + 49 #1 + #1^6 &}
NDE>For the 72-element group, a general formula is
27*x^6 + 9*t^3*(t+8)*x + t^3*(t+8)*(t^2+7*t+1) = 3*A^2 + t*(t+8)*B^2
where
A = 3*x^3 - t*(t+8)*x/2 - t^2*(t+8)/2 B = (6*x^2 + 3*t*x - (t^2+2*t)) / 2
it seems that this has a real pair of roots *iff* -8 < t < 0.
NDE
Lotsa nice ones in there: In[178]:= Factor[ 27*x^6 + 9*t^3*(t + 8)*x + t^3*(t + 8)*(t^2 + 7*t + 1) /. t -> -7] Out[178]= -343 - 3087 x + 27 x^6 In[179]:= Ssexy[%][[1]] Out[179]= {x -> Sqrt[7/3]/3 - (7^(2/3)*(1 + 3*Sqrt[21]))/ (3*(6*(378 - 140*Sqrt[21] + Sqrt[14*(41310 - 5832*Sqrt[21])]))^ (1/3)) + (7*(378 - 140*Sqrt[21] + Sqrt[14*(41310 - 5832*Sqrt[21])]))^(1/3)/ (3*6^(2/3))} In[180]:= ContinuedFraction[-%[[1, 2]], 105] Out[180]= {0, 9, 750141, 1, 1, 4, 50009, 4, 3, 4, 6, 1, 389, 1, 5, 7, 7, 2, 5, 1, 1, 1, 4, 64, 2, 1, 1, 1, 1, 1, 1, 2, 7, 1, 1, 4, 1, 1, 3, 1, 2, 1, 1, 10, 1, 5, 1, 1, 1, 1, 2, 12, 1, 32, 1, 13, 1, 47, 1, 2, 5, 1, 1, 1, 1, 1, 4, 6, 4, 2, 2, 2, 5, 1, 130, 1, 48, 1, 2, 4, 2, 3, 1, 1, 1, 2, 1, 16, 1, 1, 1, 34, 1, 1, 1, 4, 9, 1, 3, 33, 10, 1, 2, 4, 1} NDE>... and there are infinitely many solvable trinomials ax^8+bx+c, such as 4x^8+8x+13 (this must be well-known too)... My recipe for solving irreducible octics: Run and hide. Based on Tito Piezas's x^8-x^7+29x^2+29, whose solution, some of you may recall, involved solving the auxiliary septic 8903+47647v+39672v^2+7192v^3-522v^4-174v^5+v^7 for which he gave "v_i = 2(w^11+w^13+w^16+w^18)-2(w+w^12+w^17+w^28)-(w^2+w^5+w^24+w^27)+ (w^3+w^7+w^22+w^26)+(w^4+w^10+w^19+w^25)-(w^8+w^9+w^20+w^21) and one can set w_i = {t, t^7, t^23, t^25, t^16, t^20, t^24}, and t = exp(2Pi*I/29)." My septic solver instead returns something so unspeakably cumbersome that MinimalPolynomial can't even recover the septic. --rwg
Bill Gosper <billgosper@gmail.com> writes:
On Sat, Aug 31, 2013 at 10:33 AM, Noam Elkies <elkies@math.harvard.edu>wrote:
A trinomial can never have more than two real roots (Descartes' rule).
With what disclaimer? 4 x - 5 x^3 + x^5 has five.
Sorry, one needs the right parities. P(x) = x^n + a x^m + b with n>m, n even, m odd. (Allowing a factor of x is cheating...) P(x) has at most one more root than its derivative, which has at most 2 because m is odd, but since n is even so is the total number of roots, etc. Or, for x^6+ax+b, just observe that the graph of x^6+b is convex and thus meats a straight line y=ax in at most two points.
In[178]:= Factor[ 27*x^6 + 9*t^3*(t + 8)*x + t^3*(t + 8)*(t^2 + 7*t + 1) /. t -> -7]
Out[178]= -343 - 3087 x + 27 x^6
In[179]:= Ssexy[%][[1]]
Out[179]= {x -> Sqrt[7/3]/3 - (7^(2/3)*(1 + 3*Sqrt[21]))/ (3*(6*(378 - 140*Sqrt[21] + Sqrt[14*(41310 - 5832*Sqrt[21])]))^ (1/3)) + (7*(378 - 140*Sqrt[21] + Sqrt[14*(41310 - 5832*Sqrt[21])]))^(1/3)/ (3*6^(2/3))}
In[180]:= ContinuedFraction[-%[[1, 2]], 105]
Out[180]= {0, 9, 750141, 1, 1, 4, 50009, 4, 3, 4, 6, 1, 389, 1, 5, 7, [...]
Are you suggesting there's some significance to the large convergents?
NDE>... and there are infinitely many solvable trinomials ax^8+bx+c, such as 4x^8+8x+13 (this must be well-known too)...
My recipe for solving irreducible octics: Run and hide. Based on Tito Piezas's x^8-x^7+29x^2+29, whose solution, some of you may recall, involved
solving the auxiliary septic [...]
No septics here. This family of octics factors over quadratic extensions, so it's "just" solving a quartic over Q(sqrt(d)).
"v_i = 2(w^11+w^13+w^16+w^18)-2(w+w^12+w^17+w^28)-(w^2+w^5+w^24+w^27)+ (w^3+w^7+w^22+w^26)+(w^4+w^10+w^19+w^25)-(w^8+w^9+w^20+w^21) and one can set w_i = {t, t^7, t^23, t^25, t^16, t^20, t^24}, and t = exp(2Pi*I/29)."
That's no fair (and not really feasible once the cyclotomic extension is at all large...).
My septic solver instead returns something so unspeakably cumbersome that MinimalPolynomial can't even recover the septic.
I don't speak Mathematica, but you can always guess the polynomial by computing a root to high precision and then applying one of the lattice-reduction algorithms to find an integer relation among the first n powers of the root. NDE
participants (3)
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Bill Gosper -
Dan Asimov -
elkies＠math.harvard.edu