User:Wkkim11

The square root of 2 is also the only real number other than 1 whose infinite tetrate is equal to its square.


 * $$\sqrt{2}^ {\sqrt{2}^ {\sqrt{2}^ {\ \cdot^ {\cdot^ \cdot}}}} = 2.$$

However, the square root of 2 is also the real number other than 1 whose infinite tetrate of the square root of 2 is equal to 4


 * $$\sqrt{2}^ {\sqrt{2}^ {\sqrt{2}^ {\ \cdot^ {\cdot^ \cdot}}}} = 4$$

so that the infinite tetrate of the square root of 2 should diverge.