Re: [math-fun] Fast factoring ??
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
[ ... ]
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?
verification doesn't require computer algebra. it took me less than half a page to calculate and simplify (3^(1/5) + 4^(1/5) - 54^(1/5) + 72^(1/5))^3 with pen and paper. contrast this to FINDING the identity! mike
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
[ ... ]
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?
verification doesn't require computer algebra. it took me less than half a page to calculate and simplify (3^(1/5) + 4^(1/5) - 54^(1/5) + 72^(1/5))^3 with pen and paper. contrast this to FINDING the identity!
mike
Amen. I've now tried ~ 10^8 cases of (a^(1/5)-b^(1/5))^(1/3) without further success, and for sqrt, only 3/5 4/5 3/5 2/5 2/5 4/5 1/5 1/5 1/5 1/5 - 2 3 + 3 + 2 2 3 - 2 3 + 2 sqrt(4 - 3 ) = ---------------------------------------------------. 5 And, of course, these can only give pentanomials. --rwg ALGORISMIC MICROGLIAS
On Fri, Dec 19, 2008 at 12:20 PM, <rwg@sdf.lonestar.org> wrote:
Amen. I've now tried ~ 10^8 cases of (a^(1/5)-b^(1/5))^(1/3) without further success, and for sqrt, only
3/5 4/5 3/5 2/5 2/5 4/5 1/5 1/5 1/5 1/5 - 2 3 + 3 + 2 2 3 - 2 3 + 2 sqrt(4 - 3 ) = ---------------------------------------------------. 5
And, of course, these can only give pentanomials. --rwg ALGORISMIC MICROGLIAS
I have no technical knowledge on denesting, but here is how I look at the problem: Since sqrt( a^(1/5) + b^(1/5) ) = a^(-2/5) * sqrt( a + (a^4*b)^(1/5) ), I'll consider only the form: sqrt( a + b^(1/5) ). If sqrt( a + b^(1/5) ) could be denested, I'd expect it to be denested into the form: x0 + x1*b^(1/5) + x2*b^(2/5) + x3*b^(3/5) + x4*b^(4/5). That's why I don't think you could find anything longer than pentanomial. OTOH, sqrt( a^(1/7) + b^(1/7) ) is a better bet for finding hexanomials or longer. Just my 2 cents. Warut
participants (3)
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Michael Reid -
rwg＠sdf.lonestar.org -
Warut Roonguthai