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).
Notice that in the (comparatively easy) 4th root case, your conjecture catches 1/4 3/4 1/4 sqrt(161 - 12 5 ) = 2 5 - 3 sqrt(5) + 4 5 + 6 but not 1/4 1/4 1/4 3/4 3/4 1/4 sqrt(2) sqrt(31 - 4 3 5 ) = 3 5 - 2 sqrt(5) + 3 5 + 2 sqrt(3).
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.
Indeed, but the only one I've found so far is perversely pentanomial: 1/7 Sqrt(7 (2 - 2 )) 1/7 3/7 5/7 6/7 ------------------ = -1 + 2 2 + 2 + 2 - 2 . 1/14 2
Just my 2 cents.
Warut
You didn't propose 6th roots. Did you have reason to suspect that they are as infertile as they seem to be? --rwg PRECHRISTMAS PETRARCHISMS ALGORISMIC MICROGLIAS