K kicma Jan 2013 50 1 Croatia Jan 5, 2016 #1 Solve (w^k+n)^w where k € w\{0} and n € w. I managed to show that w^w <= (w^k+n)^w < w^(w^2).
C chiro MHF Helper Sep 2012 6,608 1,263 Australia Jan 5, 2016 #2 Hey kicma. What constraints can you have on your ordinal? Are you just looking at w^(f(x)) where x is some whole positive number?
Hey kicma. What constraints can you have on your ordinal? Are you just looking at w^(f(x)) where x is some whole positive number?
H HallsofIvy MHF Helper Apr 2005 20,249 7,909 Jan 6, 2016 #4 What do you mean by "solve" here? To calculate the given value?
K kicma Jan 2013 50 1 Croatia Jan 27, 2016 #6 I'll post solution here if anyone needs in future. w^w=w^(k*w)=(w^k)^w<=(w^k+n)^w<=(w^k+w^k)^w<=((w^k)*w)^w=(w^(k+1))^w=w^((k+1)*w)=w^w
I'll post solution here if anyone needs in future. w^w=w^(k*w)=(w^k)^w<=(w^k+n)^w<=(w^k+w^k)^w<=((w^k)*w)^w=(w^(k+1))^w=w^((k+1)*w)=w^w