I like these better.
[(([(x^1/2)x]^1/2))x]^1/2.... and so on equals 777. What is x?
Hm, I'm not sure whether the operation order is well-defined or not:
1) Let's define the following sequence:
a_0 = x^1/2
a_n = (x * a_(n-1))^1/2
a_0 = x^1/2, a_1 = ((x^1/2)x)^1/2, a_2 = ((((x^1/2)x)^1/2)x)^1/2, ...
If a_n converges to a then:
a_n = (x * a_(n-1))^1/2 -> a = (x*a)^1/2 -> x = a
Since we want a = 777, we can take x = 777
Let's prove it converges:
a_0 = x^1/2 > 0 and if a_(n-1) > 0 then a_n = (x * a_(n-1))^1/2 > 0 so a_n > 0 for all n.
a_n - a_(n-1) = (x * a_(n-1))^1/2 - (x * a_(n-2))^1/2 = (x*a_(n-1) - x*a_(n-2))/((x * a_(n-1))^1/2 + (x * a_(n-2))^1/2) ->
-> sign (a_n - a_(n-1)) = sign (a_(n-1) - a_(n-2))
And since a_1 - a_0 = ((777^1/2)777)^1/2 - 777^1/2 > 0 -> a_n increases
a_0 = x^1/2 < 777. Let's assume a_(n-1) < 777 -> x*a_(n-1) < 777² -> a_n = (x*a_(n-1))^1/2 < 777 so a_n < 777 for all n.
Since a_n increases and is bounded -> converges, QED.
2) We could also define the following sequence:
a_0 = 1
a_n = x*(a_(n-1))^1/2
a_0 = 1, a_1 = x, a_2 = (x^1/2)x, a_3 = (((x^1/2)x)^1/2)x, a_4 = (((((x^1/2)x)^1/2)x)^1/2)x, ...
If a_n converges to a then:
a_n = x*(a_(n-1))^1/2 -> a = x*a^1/2 -> x = a^1/2
Since we want a = 777, we can take x = 777^1/2
It can be proven that converges in a similar way.
So the value of x depends on the order you do the operations...
?????