exp(x) x f ' (exp(x)) = exp(f(x)) f(x)

[tommy1729]

I considered and wrote about this before but I felt like mentioning it again.

I want to approximate the semi-exp(x) by lower bound functions for it,  by giving equations for defining that lower bound function.

Ofcourse this depends on the type of tetration we pick, or maybe not.
But lets ignore that for now.

Lower bound principle :

if g(x) is strictly increasing then 

D [exp( g(x) x )] >= D [exp( v x )]

where v is the value of g at x.

SO

D [exp( g(x) x )]  > v exp( v x ).

So

D [exp( g(x) x )]  > g(x) exp( v x ).

**

Now semi-exp(x) = exp( semi-log(x) )

and

exp( semi-log(x) ) = exp( semi-log(x)/x  x )

let v be the value of semi-log(x)/x at x.

Then 

D [semi-exp(x)] = D [exp( semi-log(x) )] = D[exp( semi-log(x)/x x )] >  semi-log(x)/x * exp(semi-log(x)/x x)

So 

D [semi-exp(x)] = D [exp( semi-log(x) )] > semi-log(x)/x * semi-exp(x)

or equivalent :

D [semi-log( exp(x) )] > semi-log(x)/x * exp( semi-log(x) )

LET f(x) = semi-log(x)


Lets use the chain rule on the left hand.

f ' ( exp(x) ) exp(x) > exp( f(x) ) f(x) / x

so

exp(x)  x  f ' ( exp(x) ) > exp( f(x) )  f(x)

Since we are talking about lower bounds and equations for it , we replace > with = ;

exp(x)  x  f ' ( exp(x) ) = exp( f(x) )  f(x)

and we try to solve that equation.

Or at least get an asymptotic solution for it , for x large and positive.

This is the equation from the title btw.

***

Notice we cannot replace exp( f(x) ) with f( exp(x) ) and vice versa , in that final equation because that would lead us to 

f ' (exp(x)) exp(x) = f(exp(x)) f(x)/x

integrating

f( exp(x) ) = integral exp(f(x)) f(x)/x

which would imply that

f(x)/x = f ' (x)

which is not what we want !

We have build many methods of solving exotic equations here over the years, so maybe this differential like equation can be somewhat solved as well ?



regards

tommy1729