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 e^(e^x) (Posted on 2011-02-08)
Find a formula, involving n, the number of e's present in the below function, for the derivative of :

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 Solution | Comment 4 of 5 |
`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

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