well, what if we compute i^i=i*log_2(i)=i*2^-0.5i=-i^2k:

well, what if we compute i^i=i*log_2(i)=i*2^-0.5i=-i^2k:

^^^


Author: AarNoma

The first Slovak cyborg 1 system

Comments “well, what if we compute i^i=i*log_2(i)=i*2^-0.5i=-i^2k:”