04/21/2010, 07:19 PM

I don't have a powerful enough computer to determine this, but I would like someone to tell me that I'm wrong.

I will let the data speak for itself:

e^(1/e) = 1.444...

Let d = 1/e

Set infinity to be some arbitrarily high number, e.g. 9.99e10000000

Take the following sequences:

...^ (1.4444 + d) ^ (1.4444 + d) ^ (1.4444 + d) ^ (1.4444 + d) ^ (1.4444 + d)

...^ (1.4444 + d^2) ^ (1.4444 + d^2) ^ (1.4444 + d^2) ^ (1.4444 + d^2) ^ (1.4444 + d^2)

...^ (1.4444 + d^3) ^ (1.4444 + d^3) ^ (1.4444 + d^3) ^ (1.4444 + d^3) ^ (1.4444 + d^3)

Continue to increase n toward infinity...

...^ (1.4444 + d^n) ^ (1.4444 + d^n) ^ (1.4444 + d^n) ^ (1.4444 + d^n) ^ (1.4444 + d^n)

Each of these sequences reaches "infinity" after the following iterations:

8, 12, 16, 25, 41, 66, 108, 178, 293, 482, 794 when using (d,d^2,d^3,d^4,d^5,d^6,d^7,d^8,d^9,d^10) respectively.

This looks like a geometric series based close to sqrt(e) = (1.645...).

Perhaps the number of iterations to get to "infinity" approaches sqrt(e)???

Bo, I'm awaiting your superior mathematical intellect.

Ryan

I will let the data speak for itself:

e^(1/e) = 1.444...

Let d = 1/e

Set infinity to be some arbitrarily high number, e.g. 9.99e10000000

Take the following sequences:

...^ (1.4444 + d) ^ (1.4444 + d) ^ (1.4444 + d) ^ (1.4444 + d) ^ (1.4444 + d)

...^ (1.4444 + d^2) ^ (1.4444 + d^2) ^ (1.4444 + d^2) ^ (1.4444 + d^2) ^ (1.4444 + d^2)

...^ (1.4444 + d^3) ^ (1.4444 + d^3) ^ (1.4444 + d^3) ^ (1.4444 + d^3) ^ (1.4444 + d^3)

Continue to increase n toward infinity...

...^ (1.4444 + d^n) ^ (1.4444 + d^n) ^ (1.4444 + d^n) ^ (1.4444 + d^n) ^ (1.4444 + d^n)

Each of these sequences reaches "infinity" after the following iterations:

8, 12, 16, 25, 41, 66, 108, 178, 293, 482, 794 when using (d,d^2,d^3,d^4,d^5,d^6,d^7,d^8,d^9,d^10) respectively.

This looks like a geometric series based close to sqrt(e) = (1.645...).

Perhaps the number of iterations to get to "infinity" approaches sqrt(e)???

Bo, I'm awaiting your superior mathematical intellect.

Ryan