Drop everything and watch
1.7M views in one day. Derek wisely avoided the following wrinkle: A cubic can have all three roots real, yet inexpressible with real radicals! E.g.,
In[]:=
Solve[8x^3==6x+1]
Out[]=
x,x,x
Cops out completely and says to do it numerically! But we can force it to use radicals:
In[]:=
ToRadicals[x/.%]
Out[]=
-(1-(1+-(1+,--,(1++
1
4
3
)1/3
1
2
3
)1
2
2/3
1
2
3
)1-
3
2
2/3
2
1/3
(1+
3
)4/3
(1+
3
)4
1/3
2
1
2
1/3
1
2
3
)1
2/3
2
1/3
(1+
3
)Yikes, cube roots of complex cube roots of unity! But these three quantities are actually real—their imaginary parts are zero!
In[]:=
FullSimplify@Im@%
Out[]=
{0,0,0}
So just tell us the real parts already!
In[]:=
FullSimplify@Re@%%
Out[]=
,,Cos
π
9
Two copouts and a trig function. I just want radicals!
In[]:=
ToRadicals@%
Out[]=
-(1-(1+-(1+,--,-1+
1
4
3
)1/3
1
2
3
)1
2
2/3
1
2
3
)1-
3
2
2/3
2
1/3
(1+
3
)4/3
(1+
3
)4
1/3
2
1
2
8/9
(-1)
2/9
(-1)
Besides the , we now have (-1)^(8/9) ! Don’t blame Mathematica—there is simply no way to express these real numbers with real radicals! Maybe this is the real reason someone shot Galois. Actual proof the roots are real:
-1
In[]:=
FullSimplifyx-Cosx+Cosx+Sin==--+
π
9
2π
9
π
18
1
8
(3x)
4
3
x
Out[]=
True
In[]:=
FullSimplify/@x-Cosx+Cosx+Sin==--+
π
9
2π
9
π
18
1
8
(3x)
4
3
x
Out[]=
x-Cosx+Cosx+Sin--+
π
9
2π
9
π
18
1
8
3x
4
3
x
In[]:=
SimplifyCount[p_]:=Which[Head[p]===Symbol,1,IntegerQ[p],If[p==0,1,Floor[N[Log[2,Abs[p]]/Log[2,10]]]+If[p>0,1,2]],Head[p]===Rational,SimplifyCount[Numerator[p]]+SimplifyCount[Denominator[p]]+1,Head[p]===Complex,SimplifyCount[Re[p]]+SimplifyCount[Im[p]]+1,NumberQ[p],2,True,SimplifyCount[Head[p]]+If[Length[p]==0,0,Plus@@(SimplifyCount/@(List@@p))]]
In[]:=
SimplifyCount/@List@@%%
Out[]=
{29,14}