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

[tommy1729]

Yes yes I know this is all far from formal.

And all iterations of exp or ln commute compositionally with exp(x) and ln(x).
So that might create issues.

Im just explaining the main idea.

It is thus not formally confirmed that one of the solutions is smaller than semi-exp.

And the amount of solutions and their structure is also unclear.

Then again those are good reasons to investigate not ?


regards

tommy1729

[JmsNxn]

Hey, Tommy. I looked at this before; and found a numerical procedure that looks similar to this.

We start by using differential-bullet notation. So we start by setting:

\[
\int_a^x g(s,z)\,ds\bullet z = h(x,z)\\
\]

Then \(h\) is the unique solution to the equation:

\[
\begin{align}
h'(x,z) &= g(x,h(x,z))\\
h(a,z) &= z\\
\end{align}
\]

When taking the half iterate \(f(f(x)) = \exp(x)\); we know that:

\[
f'(x) = \frac{\exp(x)}{f'(f(x))}\\
\]

Let's try to sequentially solve this equation for \(|x| < \delta\)--by writing:

\[
f_{n+1}'(x) = \frac{\exp(x)}{f_n'(f_{n+1}(x))}\\
\]

This would mean that:

\[
f_{n+1}(x) = \int_0^x \frac{\exp(s)}{f_{n}'(z)}\,ds\bullet z\\
\]

If you choose \(|x| < \delta\)--this becomes very close to a linear operation; as:

\[
f_{n+1}(x) = z + \int_0^x \frac{\exp(s)}{f_{n}'(z)}\,ds + O(x^2)\\
\]

Which satisfies the proper initial condition:

\[
\begin{align}
f_{n+1}(0) &= z\\
f_{n+1}'(0) &= \frac{1}{f_n'(z)}\\
\end{align}
\]

I would always start this iteration with \(f_0(x) = z+x\).

I could never prove it converges; but it did look like it was converging for certain values of \(z\). But it was fairly uneconomical-and I never wrote good code for it. What this would converge to is a function:

\[
\begin{align}
f(0) = z\\
f(f(x)) = \exp(x)\\
\end{align}
\]

So what you choose here is the appropriate value of \(z\), which is somewhere in \((-\infty,0)\). And it would only converge for specific values of \(z\); which I could never figure out... Plus my hacked together mock proof of the convergence of this sequence could only get smooth convergence; never holomorphy.

For example, if you take \(z = \text{tet}_K(\frac{1}{2} + \text{slog}_K(0)) = \text{tet}_K(-\frac{1}{2})\), for Kneser's tetration, it did look like it was was converging. I wasn't detailed enough to tell if it actually converged to Kneser though...

I eventually abandoned this approach when I started with the beta method--as that seemed more promising. But maybe it's time to crack this open again .

---

EDIT: I should add the description of differential bullet product:

\[
\int_a^b g(s,z)\,ds\bullet z = \Omega_{j=1}^M \left(z+ g(s_j,z)\Delta s_j \right)\bullet z\\
\]

Where omega is composition in z; and \(s_j\) are the sample points of \(s_j \in [a,b]\) and \(\Delta s_j\) are the difference in samples.... This is just a fancier, more modern way of writing Euler's method for first order differential equations. Or, Euler's method.

[tommy1729]

When taking the half iterate \(f(f(x)) = \exp(x)\); we know that:

\[
f'(x) = \frac{\exp(x)}{f'(f(x))}\\
\]

Let's try to sequentially solve this equation for \(|x| < \delta\)--by writing:

\[
f_{n+1}'(x) = \frac{\exp(x)}{f_n'(f_{n+1}(x))}\\
\]

I think everyone here considered this at least once.

The problem here is you get even more complicated integrals and differential type equations at every step , especially if you want a closed form.
But even numerical.

Maybe you wanted to write 

 

\[
f_{n+1}'(x) = \frac{\exp(x)}{f_n'(f_{n-1}(x))}\\
\]

That seems easier because your method depends on earlier functions and not present or future ones.

On the other hand that might give worse solutions or worse convergeance.

Another thing I do not like is that those integrals are still asymptotics to closed form functions , so you are getting very slowly towards an exotic function.

I think I called it pade-tommy method or so in the past :

( inspired by pade approximations )

\[
f_{n+1}'(x) = \frac{2\exp(x)}{(f_n'(f_{n-1}(x)) + f_n'(f_{n-2}(x))} \mod x^{2n+4}\\
\]

IN mod x^(2n+4)

I guess you can see the pade analogue.

So we get increasing degree polynomials that should converge to the semi-exp(x).
We set the initial guess so that f_w(0) is between 0 and 1 ( because id(z) and exp(z) is between 0 and 1 ).


regards

tommy1729

[JmsNxn]

Well, yes and no. I wanted:

\[
f'_{n+1}(x) = \frac{\exp(x)}{f'_n(f_{n+1}(x))}\\
\]

Because this is a first order ODE; and is perfectly solvable. It's simply an equation:

\[
y'_{n+1} = g_n(x,y_{n+1})\\
\]

Which is really easy to solve for locally. Once you have \(f_n\), you get \(g_n\); this completely determines \(f_{n+1}(x)\) and ensures differentiability. It produces an exotic function yes; but it's approximating a half-exponential, which is the most exotic function I can think of.

It is more so a numerical procedure than a closed form expression, I'll give you that.

[tommy1729]

Well, yes and no. I wanted:

\[
f'_{n+1}(x) = \frac{\exp(x)}{f'_n(f_{n+1}(x))}\\
\]

Because this is a first order ODE; and is perfectly solvable. It's simply an equation:

\[
y'_{n+1} = g_n(x,y_{n+1})\\
\]

Which is really easy to solve for locally. Once you have \(f_n\), you get \(g_n\); this completely determines \(f_{n+1}(x)\) and ensures differentiability. It produces an exotic function yes; but it's approximating a half-exponential, which is the most exotic function I can think of.

It is more so a numerical procedure than a closed form expression, I'll give you that.

I started a new thread to get perhaps better initial guess by using sums like f_0(x) = exp(ln(x)^2) + exp(exp(ln(ln(x))^2))/10 + ... and such ideas.

see

https://math.eretrandre.org/tetrationfor...p?tid=1707



and then perhaps you could use half-plane integral representations or series representations
( meaning series or integrals valid on a half plane , like Laplace or Dirichlet etc ) to solve the recursively defined differential equation.


regards

tommy1729

[tommy1729]

Well, yes and no. I wanted:

\[
f'_{n+1}(x) = \frac{\exp(x)}{f'_n(f_{n+1}(x))}\\
\]

Because this is a first order ODE; and is perfectly solvable. It's simply an equation:

\[
y'_{n+1} = g_n(x,y_{n+1})\\
\]

Which is really easy to solve for locally. Once you have \(f_n\), you get \(g_n\); this completely determines \(f_{n+1}(x)\) and ensures differentiability. It produces an exotic function yes; but it's approximating a half-exponential, which is the most exotic function I can think of.

It is more so a numerical procedure than a closed form expression, I'll give you that.

I started a new thread to get perhaps better initial guess by using sums like f_0(x) = exp(ln(x)^2) + exp(exp(ln(ln(x))^2))/10 + ... and such ideas.

see

https://math.eretrandre.org/tetrationfor...p?tid=1707



and then perhaps you could use half-plane integral representations or series representations
( meaning series or integrals valid on a half plane , like Laplace or Dirichlet etc ) to solve the recursively defined differential equation.


regards

tommy1729

I assume that would converge to the solution given by the 2sinh method.

regards

tommy1729

[tommy1729]

A simple idea comes to mind for x > 1 :

semi-exp(x) is locally about equal to

exp^[v]( c ln^[v](x) )

where v is log*(x) 

see 

https://en.wikipedia.org/wiki/Iterated_logarithm

and c is some constant between 1 and 2 , probably sqrt(2).



regards

tommy1729

[tommy1729]

A simple idea comes to mind for x > 1 :

semi-exp(x) is locally about equal to

exp^[v]( c ln^[v](x) )

where v is log*(x) 

see 

https://en.wikipedia.org/wiki/Iterated_logarithm

and c is some constant between 1 and 2 , probably sqrt(2).



regards

tommy1729

maybe ...

confusing lol