Find a formula, involving
n, the number of e's present in the below function, for the derivative of :
let f(n,x) be defined recursively as
f(0,x)=x
f(n,x)=e^f(n1,x)
now we want to find df/dx
now
d/dx f(n,x) = d/dx e^f(n1,x)
f'(n1,x) * e^f(n1,x)
thus if we define g(n,x) = d/dx f(n,x)
then
g(n,x) = g(n1,x) * e^f(n1,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 n1 )
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 k1)
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 20110208 15:59:41 