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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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 possible solution | Comment 3 of 5 |

let f(n,x) be defined recursively as
f(0,x)=x
f(n,x)=e^f(n-1,x)
now we want to find df/dx
now
d/dx f(n,x) = d/dx e^f(n-1,x)
f'(n-1,x) * e^f(n-1,x)
thus if we define g(n,x) = d/dx f(n,x)
then
g(n,x) = g(n-1,x) * e^f(n-1,x)
with g(0,x)=1=e^(0)
so now I will prove by induction on n that
g(n,x) = e^( sum f(t,x) for t=0 to n-1 )
now for n=0 this holds
assume it holds for n=k
g(k+1,x) from above is equal to
g(k,x)*e^f(k,x)
g(k,x) = e^(sum f(t,x) for t=0 to k-1)
thus
g(k+1,x) = e^(sum f(t,x) for t=0 to k)
thus by the induction hypothesis it holds for all n>=0

so in simplier terms, the derivative is equal to e to the power of the sum of all the smaller towers.

 Posted by Daniel on 2011-02-08 15:59:41

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