5 ms·
No, that bit works. If 2=x^x^x^x^x^... then we can raise x to the power of both sides, giving x^2=x^(x^x^x^x^...) But the RHS is equal to x^x^x^x^... and hen
by RiderOfGiraffes 16y ago
No, that bit works. If 2=x^x^x^x^x^... then we can raise x to the power of both sides, giving
x^2=x^(x^x^x^x^...)
But the RHS is equal to x^x^x^x^... and hence equal to 2. Thus x^2=2.
- dedward 16y agoIm' still thinking the original equation is not solvable = 2=x^x^x^x^x^x... is invalid.
- RiderOfGiraffes 16y agoBut (purely for brevity) using r=sqrt(2) we find that the sequence r, r^r, r^r^r, r^r^r^r, ... converges to 2. To my mind that makes it perfectly reasonable to say that 2=x^x^x^... has a solution with x=sqrt(2).
- pbhjpbhj 16y agoI think implicit in the parent is that the equation is not solvable for ∀x, whilst you're saying ∃x where x converges toward a "solution" of 2 = x^^\inf.