Exponential factorial

[mike3]

Hi.

I heard of a function called "exponential factorial" that works like this:

\( EF(1) = 1 \)
\( EF(2) = 2^1 = 2 \)
\( EF(3) = 3^{2^1} = 9 \)
\( EF(4) = 4^{3^{2^1}} = 262144 \)
\( EF(5) = 5^{4^{3^{2^1}}} = 5^{262144} \)
...
\( EF(n) = n^{EF(n-1)} \)

As one can see it is similar to tetration in that it involves a power tower, but it is not defined by iteration but by a different type of recurrence, similar to the factorial. Could there be a way to derive a smooth/analytic extension for this like there is with the factorial and gamma function and like how extensions have been proposed for tetration?

[Base-Acid Tetration]

Andy has some info atthis post. How could we confirm that EF(0)=euler-mascheroni constant?

[mike3]

Andy has some info atthis post. How could we confirm that EF(0)=euler-mascheroni constant?

I wouldn't know but I'm still trying to figure out how precisely he got those coefficients. It would seem a little good to be true wouldn't it, given how exotic this function is?

[andydude]

I wouldn't know but I'm still trying to figure out how precisely he got those coefficients. It would seem a little good to be true wouldn't it, given how exotic this function is?

Essentially, by the method of undetermined coefficients.

Assume that the solution is a polynomial, say of the form \( EF(x) = a_0 + a_1x + a_2 x^2 \), and substitute that in the functional equation gives
\( \log_{x+1}(a_0 + a_1(x+1) + a_2(x+1)^2) = a_0 + a_1x + a_2 x^2 \)
and expanding the left-hand-side of this equation about x=1, and truncating the result at 3 or so terms gives an approximately equal equation \( O(x^3) \). Since this equation should hold for all \( 1 \le x \le 2 \) (by assumption*), you can match each coefficient of x on each side to form a system of 3 equations in 3 unknowns \( (a_0, a_1, a_2) \). Even though this system of algebraic equations is simpler (in a way) than the single functional equation, they are not linear in the unknowns, so use your favorite nonlinear system method, and solve for \( (a_0, a_1, a_2) \).

Does this help?

Andrew Robbins

* Assuming the functional equation holds for all \( 0 \le x \le 1 \) seems to lead to a contradiction, because \( EF(0) = EF'(1) \) and yet \( 0^x = 0 \) for all x, so you can't solve for x such that the functional equation holds.