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An infinite exponentiation is something which is raised to a power which is something raised to a power ad infinitum> Suppose
This equation might seem a puzzlement as to whether it has any solution other than the obvious one of a=1 and G=1. However a little manipulation turns it into a seemiongly trivial problem. The manipulation is to take the G-th root of both sides giving
Now if we want a value of a that gives G as a solution we need only take the G-th root of G and we have the answer. For example, for G=2, a=2½=√2. Thus
Furthermore since (½)² = 1/4
To verify these relations consider an iterative scheme of the form
The results of the first 20 iterations are:
However everything is not as simple as the preceding. For one thing G1/C does not have a single valued inverse.
Thus 41/4=21/2 so the infinite exponentiation of 41/4 does not converge to 4, instead it converges to 2.
Also the infinite exponentiation of the cube root of 3 ; i.e.,
should give 3 as a result but the iteration scheme does not converge to 3, instead it converges to a value of 2.478.... which is a lower value of G such that G1/G is also equal to the cube root of 3.
The function G1/G reaches its maximum for G=e, the base of the natural logarithms 2.71828.. At that value of G, G1/G is equal to 1.44466786100977…, call it ζ. This is the maximum value of a for which infinite exponentiation has a finite value and
For details see Maximum.
Thus the maximum value an infinite exponentiation can converge to is e and the maximum base for the infinite exponentiation is ζ=1.44466786100977 so this is why the procedure worked for G=2 and G=1/2 but not for G=3.
(To be continued.)
The maximum of the function is found by finding the value of G such that the derivative is equal to zero. The derivative is found for a function in which its argument variable appears in more than one place is to get the derivative for the variable in each place it appears treating it as a constant in the other places.
For a(G)=G1/G suppose the function is represented as G11/G2, then
Setting this equal to zero
The maximum value of the function G1/G is then e1/e.
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