Re: [math-fun] Fast factoring ??
From: "Fred lunnon" <fred.lunnon@gmail.com>
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
Ug, I hope I've stitched that back together and yahoo mail doesn't murder it on egress.
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 ?!
The ^1/3 has already been highlighted. So far, I've only seen the numeric verifications, but surely a symbolic one would be more appropriate? ? f2=Mod(a,a^5-2) Mod(a, a^5 - 2) ? f3=Mod(b,b^5-3) Mod(b, b^5 - 3) I cubed both sides, which is fair as we're entirely within the reals. ? lift(lift((-f2*f3^3 + f2^3*f3^2 + f3 + f2^2)^3)) -25*a + 25*b^3 ? lift(lift((f3^3-f2)*25)) -25*a + 25*b^3 Phil
lost a Mathematica 7.0 that had gone 2.5 days on
Factor[x^3-3^(3/5)+2^(1/5), Extension -> {5^(1/3),2^(1/5),3^(1/5)}] [rwg:]
On 12/18/08, Phil Carmody <thefatphil@yahoo.co.uk> wrote:
.... So far, I've only seen the numeric verifications, but surely a symbolic one would be more appropriate?
? f2=Mod(a,a^5-2) Mod(a, a^5 - 2) ? f3=Mod(b,b^5-3) Mod(b, b^5 - 3)
I cubed both sides, which is fair as we're entirely within the reals.
? lift(lift((-f2*f3^3 + f2^3*f3^2 + f3 + f2^2)^3)) -25*a + 25*b^3 ? lift(lift((f3^3-f2)*25)) -25*a + 25*b^3
Well, I didn't have 2.5 days to spare --- and anyway, I couldn't find my spectacles! WFL
participants (2)
-
Fred lunnon -
Phil Carmody