Tetration Forum - Uniqueness of half-iterate of exp(x) ?

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I was dreaming about uniqueness for a half-iterate of exp(x).
And what other properties it might have.

Let f ( f (x) ) = exp(x) where f (x) and f ' (x) are continu and increasing for all real x.

Define L1 as the limit for x -> +oo :

L1 = lim L2(x) / (a(x)+b(x))

where a(x) = x + f (x) , b(x) = sqrt(f (x)^2 + x^2)

and L2(x) = integral sqrt ( 1 + f ' (x)^2) dx where the integral goes from 0 to x.

The conjectured uniqueness is the lowest possible value for L1.

Thus let L be the lowest possible value for L1.

Then L = L1 and L has a unique function f (x) associated with it.

This minimal length idea lead to conjecture L

if f is minimal length then so is f^^n.

To support conjecture L it might be a good exercise to prove that exp(x) is the shortest lenght solution to the half-iterate of exp(exp(x)).

regards

tommy1729

f(x) can be found in terms of the Lambert W function. I haven't been able to find this anywhere else so I'm not sure if I discovered it (highly unlikely, I know) or re-discovered it.

Anyways...
Let's say f is of the following form,
$f(x)=e^{bx}$

Then the original definition becomes
$f(f(x))=e^{be^{bx}}=e^x$

Taking natural log of both sides,
$be^{bx}=x\\\Rightarrow bxe^{bx}=x^2\\\Rightarrow W(x^2)=bx\\\Rightarrow f(x)=e^{W(x^2)}$

which, by definition, can also be written as
$f(x)=x^2/W(x^2)$

Perhaps someone can verify that I'm using the W function properly, as I'm rather new to applying it. There isn't a closed-form expression for W as far as I know, so perhaps writing f(x) in this way doesn't actually answer any of your questions.

EDIT: So I just noticed that I posted this somewhere else on the forum several months back. Sorry about the double-post! I completely for got about it, but I guess I never did find out why it doesn't work (if that is indeed the case).

(05/08/2014, 12:57 AM)hixidom Wrote: [ -> ]which, by definition, can also be written as
$f(x)=x^2/W(x^2)$

Hmm, I get the following using Pari/GP:

Code:
x0=1.0

 %1876 = 1.00000000000

x1=exp(x0)

 %1877 = 2.71828182846

x05 = x0^2/LW(x0^2)

 %1878 = 1.76322283435

 x05^2/LW(x05^2)

 %1879 = 2.91030427217

Perhaps some cofactor could improve the formula?

Gottfried

Or maybe it's just plain wrong. A test using Matlab suggests that the f(x) I provided is not the solution. That's so weird; It seemed airtight to me, but there must be some pathology in my derivation. Hopefully someone with a better understanding of the W function can spot it for me.

Code:
x=(.5:.01:3);

f=@(x)exp(lambertw(x.^2));

for i=1:numel(x);

    ffx(i)=f(f(x(i)));

end

figure(1);

clf();

plot(x,exp(x),x,ffx)

(05/08/2014, 07:31 AM)hixidom Wrote: [ -> ]Or maybe it's just plain wrong. A test using Matlab suggests that the f(x) I provided is not the solution. That's so weird; It seemed airtight to me, but there must be some pathology in my derivation. Hopefully someone with a better understanding of the W function can spot it for me.

Code:
x=(.5:.01:3); f=@(x)exp(lambertw(x.^2)); for i=1:numel(x); ffx(i)=f(f(x(i))); end figure(1); clf(); plot(x,exp(x),x,ffx)

Well, I meant that something possibly similar like the cofactor in my older posting http://math.eretrandre.org/tetrationforu...hp?tid=785 which I didn't see myself and did not spoil the formula but was needed to explain/correct the result. Perhaps something like this can be introduced into your formula as well to make it working...

Gottfried

Perhaps, but why would it be necessary? Anyways, my solution seems to be off by more than a cofactor. I've tried several other functions as cofactors as well, but none seem to produce a satisfactory result.
[attachment=1054]

Your derivation in the other thread is really cool, by the way.

(05/08/2014, 07:31 AM)hixidom Wrote: [ -> ]Or maybe it's just plain wrong. A test using Matlab suggests that the f(x) I provided is not the solution. That's so weird; It seemed airtight to me, but there must be some pathology in my derivation. Hopefully someone with a better understanding of the W function can spot it for me.

I think the problem is that b is not a constant. It is a function of x. That's where the math goes wrong:

(05/08/2014, 12:57 AM)hixidom Wrote: [ -> ]Let's say f is of the following form,
$f(x)=e^{bx}$

Then the original definition becomes
$f(f(x))=e^{be^{bx}}=e^x$

The first equation should be written:
$f(x)=e^{b(x)x}$

Then your second equation becomes:
$f(f(x))=e^{b(x)e^{b(x)x}}=e^x$

But this is incorrect. The correct expansion is thus:
$f(f(x))=e^{b(e^{b(x)x})e^{b(x)x}}=e^x$

From there, the rest of the derivation is wrong.

BTW, it looked suspicious to me at first, because I knew b wasn't constant, but otherwise the derivation looked good, so I couldn't spot the error right away.

(05/08/2014, 07:04 PM)jaydfox Wrote: [ -> ]I think the problem is that b is not a constant. It is a function of x. That's where the math goes wrong:

The first equation should be written:
$f(x)=e^{b(x)x}$

Hmm, now that I think about it, you really can treat b as a constant. From your derivation, one of the steps is:

(05/08/2014, 12:57 AM)hixidom Wrote: [ -> ] $W(x^2)=bx$

Given a value of x, we can solve for b:

$W(x^2)=bx\\<br /> \frac{W(x^2)}{x}=b$

For example, if we let x = 1, then we find b=0.567143290409784
Then we validate:

let x = 1
let b = 0.567143290409784
exp(b*exp(b*x)) = 2.718... = exp(1)

As another example, let x=3. Then we find that b=0.559672139928533

Putting it together, exp(b*exp(b*x))=20.0855369231877 = exp(3)

So it works. The problem is, the value of b will vary with each value of x. If we make b constant, then it will only work for a small set of values of x (possibly a single value of x). Otherwise, we have to make b a function of x, and then the derivation does not work, as I previously pointed out.

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