Re: [math-fun] Fast factoring ??
Fred wrote: << On 12/17/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
... Thereby reenacting one of my favorite denestings:
1/5 3/5 3/5 2/5 1/5 2/5 3/5 1/5 1/3 - 2 3 + 2 3 + 3 + 2 (3 - 2 ) = ------------------------------------- , 2/3 5
a major simplification that looks anything but.
Indeed! u := 3^(1/5); v := 2^(1/5); w := 5^(1/3); l := (u^3 - v)^1/3; r := (u + v^2 - v*u^3 + v^3*u^2)/w^2; evalf(l - r); # -.6607822845 ?!
Mathematica gives a different result: <<n[1]:= u = 2^(1/5) 1/5 Out[1]= 2 In[2]:= v = 3^(1/5) 1/5 Out[2]= 3 In[3]:= w = 5^(1/3) 1/3 Out[3]= 5 In[4]:= x = (v^3 - u)^(1/3) 1/5 3/5 1/3 Out[4]= (-2 + 3 ) In[5]:= y = (-u v^3 + u^3 v^2 + v + u^2)/w^2 2/5 1/5 3/5 2/5 1/5 3/5 2 + 3 + 2 3 - 2 3 Out[5]= ----------------------------------- 2/3 5 In[6]:= N[x-y] -16 Out[6]= -2.22045 10
--Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
The discrepancy here reminds me of an interesting paper called "Entangled Radicals" (it was the colloquium lecture at the math meetings in 2006) by Hendrik Lenstra which is about the problem of computing in extensions given by radicals. Along the way he shows that by taking different n-th roots in a nested radical expression you can even get non-isomorphic fields! Victor On Wed, Dec 17, 2008 at 2:10 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Fred wrote:
<< On 12/17/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
... Thereby reenacting one of my favorite denestings:
1/5 3/5 3/5 2/5 1/5 2/5 3/5 1/5 1/3 - 2 3 + 2 3 + 3 + 2 (3 - 2 ) = ------------------------------------- , 2/3 5
a major simplification that looks anything but.
Indeed!
u := 3^(1/5); v := 2^(1/5); w := 5^(1/3); l := (u^3 - v)^1/3; r := (u + v^2 - v*u^3 + v^3*u^2)/w^2; evalf(l - r); # -.6607822845 ?!
Mathematica gives a different result:
<<n[1]:= u = 2^(1/5)
1/5 Out[1]= 2
In[2]:= v = 3^(1/5)
1/5 Out[2]= 3
In[3]:= w = 5^(1/3)
1/3 Out[3]= 5
In[4]:= x = (v^3 - u)^(1/3)
1/5 3/5 1/3 Out[4]= (-2 + 3 )
In[5]:= y = (-u v^3 + u^3 v^2 + v + u^2)/w^2
2/5 1/5 3/5 2/5 1/5 3/5 2 + 3 + 2 3 - 2 3 Out[5]= ----------------------------------- 2/3 5
In[6]:= N[x-y]
-16 Out[6]= -2.22045 10
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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On 12/17/08, victor miller <victorsmiller@gmail.com> wrote:
The discrepancy here reminds me of an interesting paper called "Entangled Radicals" (it was the colloquium lecture at the math meetings in 2006) by Hendrik Lenstra which is about the problem of computing in extensions given by radicals. Along the way he shows that by taking different n-th roots in a nested radical expression you can even get non-isomorphic fields!
Victor
Maple's roots are (unique) real --- I floated them to check. So presumably Dan's Mathematica computation is actually utilising complex roots at some point; likewise Bill's original claim --- although I had inferred from his final sentence that all quantities were real ?! WFL
On Wed, Dec 17, 2008 at 2:10 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Fred wrote:
<< On 12/17/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
... Thereby reenacting one of my favorite denestings:
1/5 3/5 3/5 2/5 1/5 2/5 3/5 1/5 1/3 - 2 3 + 2 3 + 3 + 2 (3 - 2 ) = ------------------------------------- , 2/3 5
a major simplification that looks anything but.
Indeed!
u := 3^(1/5); v := 2^(1/5); w := 5^(1/3); l := (u^3 - v)^1/3; r := (u + v^2 - v*u^3 + v^3*u^2)/w^2; evalf(l - r); # -.6607822845 ?!
Mathematica gives a different result:
<<n[1]:= u = 2^(1/5)
1/5 Out[1]= 2
In[2]:= v = 3^(1/5)
1/5 Out[2]= 3
In[3]:= w = 5^(1/3)
1/3 Out[3]= 5
In[4]:= x = (v^3 - u)^(1/3)
1/5 3/5 1/3 Out[4]= (-2 + 3 )
In[5]:= y = (-u v^3 + u^3 v^2 + v + u^2)/w^2
2/5 1/5 3/5 2/5 1/5 3/5 2 + 3 + 2 3 - 2 3 Out[5]= ----------------------------------- 2/3 5
In[6]:= N[x-y]
-16 Out[6]= -2.22045 10
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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Maple 12:
FifthRoot2 := evalf(2^(1/5)); 1.148698355
FifthRoot3 := evalf(3^(1/5)); 1.245730940
CubeRoot5 := evalf(5^(1/3)); 1.709975947
Left := (FifthRoot3^3-FifthRoot2)^(1/3); 0.9222768479
Right := (-FifthRoot2*FifthRoot3^3+FifthRoot2^3*FifthRoot3^2+FifthRoot3+FifthRoot2^2)/CubeRoot5^2; 0.9222768470
Maple 7 gives both sides the value 0.92227684711757969, agreeing with your answer, and minimal polynomial: x^75 - 125*x^60 + 40000*x^45 + 2375000*x^30 + 32159375*x^15 - 9765625. On Wed, Dec 17, 2008 at 4:58 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Maple 12:
FifthRoot2 := evalf(2^(1/5)); 1.148698355
FifthRoot3 := evalf(3^(1/5)); 1.245730940
CubeRoot5 := evalf(5^(1/3)); 1.709975947
Left := (FifthRoot3^3-FifthRoot2)^(1/3); 0.9222768479
Right := (-FifthRoot2*FifthRoot3^3+FifthRoot2^3*FifthRoot3^2+FifthRoot3+FifthRoot2^2)/CubeRoot5^2; 0.9222768470
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On Wed, Dec 17, 2008 at 6:29 PM, James Buddenhagen <jbuddenh@gmail.com> wrote:
Maple 7 gives both sides the value 0.92227684711757969, agreeing with your answer, and minimal polynomial:
x^75 - 125*x^60 + 40000*x^45 + 2375000*x^30 + 32159375*x^15 - 9765625.
Which means, I might add, that both sides are also equal to: [25 - 30*24^(1/5) - 45/2*24^(2/5) + 5/2*24^(3/5) + 15/2*24^(4/5) ]^(1/15)
participants (5)
-
Dan Asimov -
Eugene Salamin -
Fred lunnon -
James Buddenhagen -
victor miller