On 2015-11-24 19:42, Dan Asimov wrote:
I learned (or was reminded) only recently that the real and imaginary parts of any nth root of unity can each be expressed in terms of (an iterated rational expression in) radicals of integers (using only roots lower than the nth).
1) Is there an easy proof of this?
2) How do I get Mathematica to output these expressions? (Everything I've tried so far gets only the dumbest error messages.)
For instance, Wikipedia lists the real part of (exp(2pi*i/7)) as follows:
cos(2pi/7) = (-1 + ((7 + 21*sqrt(-3)/2)^(1/3) + ((7 - 21*sqrt(-3)/2)^(1/3))/6
In[43]:= ToRadicals[Cos[\[Pi]/7]] Out[43]= -(1/2) (-1)^(6/7) (1 + (-1)^(2/7)) In[44]:= Developer`TrigToRadicals[Cos[\[Pi]/7]] During evaluation of In[44]:= Developer`TrigToRadicals::obs: Developer`TrigToRadicals has been superseded by ToRadicals, [I don't see how. --rwg] and is now obsolete. It will not be included in future versions of the Wolfram Language. >> Out[44]= 1/2 (1/ 3 (1/2 (1 - I Sqrt[7]) - ((-1 + I Sqrt[3]) (1/2 (-1 + I Sqrt[7]) + 1/2 (-1 - I Sqrt[7]) (1/2 (-1 + I Sqrt[3]) + 1/4 (-1 + I Sqrt[3])^2)))/(2 (6 + 3/4 (-1 + I Sqrt[3]) (-1 + I Sqrt[7]) + 1/2 (-1 - I Sqrt[7]) (1 + 3/4 (-1 + I Sqrt[3])^2))^( 1/3)) - 1/ 4 (-1 + I Sqrt[3])^2 (6 + 3/4 (-1 + I Sqrt[3]) (-1 + I Sqrt[7]) + 1/2 (-1 - I Sqrt[7]) (1 + 3/4 (-1 + I Sqrt[3])^2))^(1/3)) + 1/3 (1/2 (1 + I Sqrt[7]) - ((-1 + I Sqrt[3])^2 (1/2 (-1 - I Sqrt[7]) + 1/2 (-1 + I Sqrt[7]) (1/2 (-1 + I Sqrt[3]) + 1/4 (-1 + I Sqrt[3])^2)))/(4 (6 + 3/4 (-1 + I Sqrt[3]) (-1 - I Sqrt[7]) + 1/2 (-1 + I Sqrt[7]) (1 + 3/4 (-1 + I Sqrt[3])^2))^( 1/3)) - 1/ 2 (-1 + I Sqrt[3]) (6 + 3/4 (-1 + I Sqrt[3]) (-1 - I Sqrt[7]) + 1/2 (-1 + I Sqrt[7]) (1 + 3/4 (-1 + I Sqrt[3])^2))^(1/3))) In[47]:= MinimalPolynomial[Cos[2 \[Pi]/7]]@x Out[47]= -1 - 4 x + 4 x^2 + 8 x^3 In[50]:= Solve[0 == %] Out[50]= {{x -> 1/6 (-1 + 7^(2/3)/(1/2 (1 + 3 I Sqrt[3]))^( 1/3) + (7/2 (1 + 3 I Sqrt[3]))^(1/3))}, {x -> -(1/6) - ( 7^(2/3) (1 + I Sqrt[3]))/(6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3)) - 1/12 (1 - I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3)}, {x -> -(1/ 6) - (7^(2/3) (1 - I Sqrt[3]))/( 6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3)) - 1/12 (1 + I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3)}}
.
I'm confident that several of the 2^2 3^2 = 36 choices of roots give the correct answer, but: which ones? This may be easy to guess, but how about cos(6pi/23) ?
In[51]:= Developer`TrigToRadicals[Cos[6 \[Pi]/23]] During evaluation of In[51]:= Developer`TrigToRadicals::obs: Developer`TrigToRadicals has been superseded by ToRadicals, and is now obsolete. It will not be included in future versions of the Wolfram Language. >> Out[51]= <Enormous expression> In[52]:= Short[%] Out[52]//Short= 1/2 (1/11 (1/2 (-1+I Sqrt[23])+<<9>>+(<<1>> <<1>>)/(9<<5>>5<<1>><<1>>))+<<1>>) --rwg
3) There ought to be a notation that doesn't leave these choices unspecified.
"The ordering used by Root[f,k] takes real roots to come before complex ones, and takes complex conjugate pairs of roots to be adjacent. "
—Dan