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 Exponential factorial mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/02/2009, 02:55 AM 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 Fellow Posts: 94 Threads: 15 Joined: Apr 2009 10/02/2009, 03:40 AM Andy has some info atthis post. How could we confirm that EF(0)=euler-mascheroni constant? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/06/2009, 12:05 AM (10/02/2009, 03:40 AM)Base-Acid Tetration Wrote: 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 Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/07/2009, 02:04 AM (10/06/2009, 12:05 AM)mike3 Wrote: 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. « Next Oldest | Next Newest »

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