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 Uniqueness of half-iterate of exp(x) ? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/21/2014, 11:55 PM 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. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/26/2014, 01:26 PM This minimal length idea lead to conjecture L if f is minimal length then so is f^^n. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/27/2014, 11:23 PM 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 hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 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 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). Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 05/08/2014, 05:28 AM (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 hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 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) Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 05/08/2014, 01:49 PM (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 hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 05/08/2014, 04:28 PM 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. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 05/08/2014, 07:04 PM (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 jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 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 « Next Oldest | Next Newest »

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