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