This looks like two parameters, but it's homogeneous in the quantity a/b: Out[162]= -(a - b)^5 + 5 (a^4 + 121 a^3 b + 381 a^2 b^2 + 121 a b^3 + b^4) x - 5 (a - b) (2 a^2 - 379 a b + 2 b^2) x^2 + 5 (2 a^2 + 121 a b + 2 b^2) x^3 - 5 (a - b) x^4 + x^5 with the root x In[164]:= FullSimplify[%162 /. x -> b (-1 + (a/b)^(1/5))^5] Out[164]= 0 —rwg On Sat, Oct 31, 2020 at 3:57 PM Bill Gosper <billgosper@gmail.com> wrote:
Another: In[122]:= Factor@CoefficientList[%%%, x].x^Range[0, 5]
Out[122]= -(-1 + a)^5 + 5 (1 + 121 a + 381 a^2 + 121 a^3 + a^4) x - 5 (-1 + a) (2 - 379 a + 2 a^2) x^2 + 5 (2 + 121 a + 2 a^2) x^3 - 5 (-1 + a) x^4 + x^5
Is solved by In[123]:= FullSimplify[% /. x -> (Power[a,1/5] - 1)^5]
Out[123]= 0 —rwg
On Sat, Feb 13, 2016 at 5:51 AM Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Feb 12, 2016 at 6:52 PM, Bill Gosper <billgosper@gmail.com> wrote:
This is probably ho-hum, but while trying to simplify a Chowla-surdberg,
where, with the help of the Chowla-Selberg formula, valuations of Dedekind's η at complex rationals and quadratic irrationals are typically the product of two large expressions, the "Chowla surdberg" and the "Chowla Gammaberg". E.g., NeilB recently found η((1+I √35)/6)==(Power[(1+√5)/(35 π), 1/4] Power[((17+I √35) Γ(1/35) Γ(3/35) Γ(4/35) Γ(9/35) Γ(11/35) Γ(12/35) Γ(13/35) Γ(16/35) Γ(17/35) Γ(27/35) Γ(29/35) Γ(33/35))/(Γ(2/35) Γ(6/35) Γ(8/35) Γ(18/35) Γ(19/35) Γ(22/35) Γ(23/35) Γ(24/35) Γ(26/35) Γ(31/35) Γ(32/35) Γ(34/35)), 1/8])/2^(5/8)
although, in this case, the surdberg is but a surd Maulwurfshügel.
Caution: my bottom-quoted quintic solution appears to fail in the lower half-plane. For some reason, negating a and cleaning up some seems to fix this:
The solutions of x^5 + 20 x^3 + 80 x == a are
{((a - Sqrt[4096 + a^2])^(1/5) i^(4/5))/2^( 1/5) + ((a + Sqrt[4096 + a^2])^(1/5) i^(4/5))/2^( 1/5), (a - Sqrt[4096 + a^2])^(1/5)/( 2^(1/5) i^(8/5)) + (a + Sqrt[4096 + a^2])^(1/5)/( 2^(1/5) i^(4/5)), (a + Sqrt[4096 + a^2])^(1/5)/( 2^(1/5) i^(8/5)) + (a - Sqrt[4096 + a^2])^(1/5)/( 2^(1/5) i^(4/5)), (a + Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((a - Sqrt[4096 + a^2])^(1/5) i^(8/5))/2^( 1/5), (a - Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((a + Sqrt[4096 + a^2])^(1/5) i^(8/5))/2^(1/5)}
Noam Elkies instantly pegged this, with a = 32 (t^5 - t^-5), as a Chebyshev quintic, one of whose solutions is x = 2 (t - 1/t):
In[696]:= Expand[x^5 + 20 x^3 + 80 x /. x -> 2 t - 2/t]
Out[696]= -32/t^5 + 32 t^5
So indeed ho-hum, and probably just one-parameter. But you can use that Chebyshev hack to construct solvable septics, etc: In[706]:= 448 x + 224 x^3 + 28 x^5 + x^7 /. x -> 2 t - 2/t // Expand
Out[706]= 128 t^7 - 128 t^-7 --rwg
I found that
0 = a + 80 x + 20 x^3 + x^5 has solutions {((-a - Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5), ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), (-a + Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^( 1/5), (-a - Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^(1/5)} Is it from a two-parameter family? C.f. https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics --rwg