On Mon, 04 Nov 2013 14:34:23 -0800, jonas.thornvall wrote: > Den måndagen den 4:e november 2013 kl. 15:27:19 UTC+1 skrev Dave Angel: >> On Mon, 4 Nov 2013 05:53:28 -0800 (PST), jonas.thornv...@gmail.com >> wrote: [...] >> > This is not the solution but this is why it is working. >> >> > 65536=256^2=16^4=***4^8***=2^16
"this" being Jonas' alleged lossless compression method capable of compressing random data. >> > Yes i am aware that 256 is a single byte 8 bits, but the approach >> is valid anyway. I must say, I cannot see the connection between the fact that 256**2 == 2**16 and compression of random data. I might as well state that I have squared the circle, and offer as proof that 3+4 == 5+2. >> And e ^ (I * pi) == -1 >> >> So what. ? >> >> > e is an approximation... and your idea is not general for any n. e is not an approximation, it is a symbolic name for an exact transcendental number which cannot be specified exactly by any finite number of decimal places. Your comment about "n" is irrelevant, since Euler's Identity e**(i*pi)=-1 has nothing to do with "n". But in case you actually meant "i", again you are mistaken. i is a symbolic name for an exact number. -- Steven -- https://mail.python.org/mailman/listinfo/python-list
