Find a formula, involving n, the number of e's present in the below function, for the derivative of :

Define

   f(n,x) = e^f(n-1,x)  if n > 0,

          = x           if n = 0.

   Examples:

       f(0,x) = x

       f(1,x) = e^x

       f(2,x) = e^(e^x)

Therefore,

   f'(n,x) = e^[ f(0,x) + ... + f(n-1,x) ]  if n > 0,

           = 1                              if n = 0.

   Examples:

       f'(0,x) = 1

       f'(1,x) = e^x

       f'(2,x) = e^[ x + e^x ]

Proof by mathematical induction on n:

   f'(0,x) = d[x]/dx = 1

   f'(1,x) = d[e^x]/dx = e^x = e^f(0,x)

   Assume that

   f'(k,x) = e^[ f(0,x) + ... + f(k-1,x) ]  for k > 0.

   f'(k+1,x) = d[e^f(k,x)]/dx = e^f(k,x)*d[f(k,x)]/dx
  
             = e^f(k,x)*e^[ f(0,x) + ... + f(k-1,x) ]

             = e^[ f(0,x) + ... + f(k,x) ]

   QED

Posted by Bractals on 2011-02-08 16:17:36