* R. William Gosper <rwg@osots.com> [Apr 14. 2005 22:18]:
Fourier series for Snowflake curve. E.g., www.tweedledum.com/rwg/cog.htm [...]
1/4 1/8 (d6) tanh(x) sqrt(tanh(2 x)) tanh (4 x) tanh (8 x) =
- 2 x 2 (1 - %e )
setting x=arctanh(y) gives prod( f@@n(y)^(1/2^n), n=0..infinity) == (y+1)^2/4 f@@0= y f@@1= 2 y + 1 ------ 2 y f@@2= 4 2 y + 6 y + 1 ------------- 3 4 y + 4 y f@@3= 8 6 4 2 y + 28 y + 70 y + 28 y + 1 ------------------------------ 7 5 3 8 y + 56 y + 56 y + 8 y f@@n = the iteration for sqrt(1) of order 2^n Using f((1+e)/(1-e))=(1+e^2)/(1-e^2) one finds prod( ((1+e^(2^k))/(1-e^(2^k)))^(1/2^k) ) == 1/(1-e)^2 (see page 350 of today's fxtbook draft). The first nontrivial sum/prod expressions with iterations I have seen so far. Nice!
[...]
All the best, jj -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.