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I was dreaming about uniqueness for a halfiterate 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.
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This minimal length idea lead to conjecture L
if f is minimal length then so is f^^n.
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To support conjecture L it might be a good exercise to prove that exp(x) is the shortest lenght solution to the halfiterate of exp(exp(x)).
regards
tommy1729
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05/08/2014, 12:57 AM
(This post was last modified: 05/08/2014, 09:54 AM by hixidom.)
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 rediscovered 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 closedform 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 doublepost! 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).
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(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
Gottfried Helms, Kassel
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05/08/2014, 07:31 AM
(This post was last modified: 05/08/2014, 07:31 AM by hixidom.)
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)
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(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
Gottfried Helms, Kassel
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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.
Your derivation in the other thread is really cool, by the way.
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(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.
~ Jay Daniel Fox
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05/08/2014, 07:24 PM
(This post was last modified: 05/08/2014, 07:24 PM by jaydfox.)
(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:
\( W(x^2)=bx\\
\frac{W(x^2)}{x}=b \)
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.
~ Jay Daniel Fox
