test for convergence

i need to test for convergence for x^x^x^..., that converges to 2 when x=sqrt(2)

where at it, does anyone know how to solve for x exactly when x^x^x=2?

if x^y = x^x^x^... then y=x^x^x... and thus you can write x^2=2, x being sqrt(2)

i don't think you can solve x^x^x=2

Quote:

---

if x^y = x^x^x^... then y=x^x^x... and thus you can write x^2=2, x being sqrt(2)

---

that only works if x^x^x... indeed converges to 2. try this:

x^x^x^...=4

you will get x=sqrt(2) also.

yeah true, and there's isn't much else I can say, but it looks like it converges to 2, not 4 :p

Ok, this is strange...

If x^x^x...=k,
Then k^k is still x^x^x...
Then k^k=k???
:confused:

according to the order of operations,

3^4^5^3^4^5=3^(4^(5^(3^(4^5))))

and not (3^4^5)^(3^4^5)

i have solved x^x^x=2

Sprite will be the next Fermat.

did i mention that according to the rule i mentioned above,

if x^x^x...=k,

k^k then is not x^x^x...

Yes, I assumed that from what you said.