For the two examples I got from some math-fun posts sometime ago: sqrt(161 - 12*5^(1/4)) = 2*5^(3/4) - 3*sqrt(5) + 4*5^(1/4) + 6 and sqrt(31 - 4*15^(1/4)) = (3^(1/4)*5^(3/4) - 2*sqrt(5) + 3^(3/4)*5^(1/4) + 2*sqrt(3))/sqrt(2), they corresponds to k = 9 and k = 4, respectively. Let's try a rational k. For k = 3/2, we have sqrt(7 - 2*45^(1/4)) = (sqrt(3) - sqrt(15) + 5^(1/4) + 5^(3/4))/2. In short, sqrt((2k^2-1) - 2*(k^4-k^2)^(1/4)) = (sqrt(k^2-k) - sqrt(k^2+k) + (k+1)^(1/4)*(k-1)^(3/4) + (k-1)^(1/4)*(k+1)^(3/4))/sqrt(2). Beautiful, isn't it? On Mon, Jan 5, 2015 at 9:43 AM, Warut Roonguthai <warut822@gmail.com> wrote:
Should be further simplified to:
sqrt(7 - 2*12^(1/4)) = 1 - sqrt(3) + (3/4)^(1/4) + (27/4)^(1/4).
On Mon, Jan 5, 2015 at 7:21 AM, Warut Roonguthai <warut822@gmail.com> wrote:
Another example for R is 192/2401 which leads to this beauty:
sqrt(7 - 2*12^(1/4)) = 1 - sqrt(3) + sqrt(3 + 2*sqrt(3)).
There are plenty of such R's. For example, just take
R = 16*k^2*(k^2 - 1)/(2*k^2 - 1)^4.
My R is from k = 2, and Bill Gosper's is from k = 8.
However, I don't know if this covers all R's that factor P_4(R,x).
On Sun, Jan 4, 2015 at 12:38 PM, Bill Gosper <billgosper@gmail.com> wrote:
√(1 - ⁿ√R) denests when a certain polynomial P_n(R,x) factors. E.g. P₄(R,x) =1 - 256 x^2 + 128 R x^2 + 1024 R x^4 + 3200 R^2 x^4 - 8192 R^3 x^6 - 28672 R^4 x^8 + 34816 R^5 x^8 - 131072 R^6 x^10 + 131072 R^7 x^12 + 1048576 R^10 x^16 . These grow *fast* with n. I'm really interested in n=6 and 7, for each of which I have one "splitting" R. So I decided to collect a bunch of "toy" R for n=4 from old Macsyma notebooks. I can't find them!! I did find a small one converted to an html. It gave
R= 64512/260144641
In[748]:= Factor[%728[[1]] /. R -> %] (using a[3] for x) Out[748]= ((-33038369407 + 458752 a[3]^2) (-33038369407 + 589824 a[3]^2) (1091533853073393531649 + 33569097773154048 a[3]^2 + 262193283072 a[ 3]^4) (1191446152405248657777607437681912764659201 - 304972394845902975940055612302553191173586944 a[3]^2 + 302486329754988497319546160627366515179520 a[3]^4 + 18453680701462988948598008315904 a[ 3]^8))/14195439340812710119888169814159419160980592549659303797\ 19569954760828848317673958401
(How the he<< did I find that?) Ok, let's just try a million R and look for a pattern: Do[If[Mod[i, 10000] == 0, Print[i]]; If[{} =!= #, Print[pos2newman[i], #]] &@ Cases[Factor[%728[[1]] /. R -> pos2newman[i]], p_Plus /; Length[p] == 2], {i, 1, 999999}] // tim
10000
20000 ... 890000
900000
Nada! OK, how long do we need to wait for Mr. Newman to reach 64512/260144641?
In[742]:= newman2pos[64512/260144641]//Short Out[742]//Short= 3490308671510526617822104163<<6014>>9495630609150025673335635968 In[743]:= N[%] Out[743]= 3.490308671510527*10^6069
A mere googol^60. Maybe I'm just impatient, but I'd really like a newmanoid bijection with more highly composite rationals. E.g. 69 ↔︎ {1,0,-2,1}, where the list is exponents of 2,3,5,7,... . Or better yet, somebody just tell me how to find the magic Rs. --Bill _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun