f(f(x))=exp(x) and extensions to noninteger numbers of f.

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mollwollfumble
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f(f(x))=exp(x) and extensions to noninteger numbers of f.

Postby mollwollfumble » Thu Dec 29, 2011 3:48 am UTC

Let's use the notation
[imath]e_1=\exp(x), e_2=\exp(\exp(x)), e_n=\exp(e_{n-1})[/imath].
Let's extend that
[imath]e_{1/2}(e_{1/2}(x))=\exp(x), e_{1/3}(e_{1/3}(e_{1/3}(x)))=\exp(x)[/imath] etc.
And in this way generate
[imath]e_{n/m}(x)[/imath] where [imath]n[/imath] and [imath]m[/imath] are positive integers.
Let [imath]n/m[/imath] tend to real number [imath]a[/imath] to get [imath]e_a(x)[/imath].

Can it be proved that [imath]e_a(e_b(x))=e_{a+b}(x)[/imath]?

Does it then make sense to say [imath]\log(x)=e_{-1}(x)[/imath] and [imath]\log(\exp(x))=e_{-1}e_1=e_0=1[/imath] and for [imath]e_{-a}(x)[/imath] to be defined by [imath]e_a(e_{-a}(x))=1[/imath] when [imath]a\ne 0[/imath]?

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jestingrabbit
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Re: f(f(x))=exp(x) and extensions to noninteger numbers of f

Postby jestingrabbit » Thu Dec 29, 2011 11:40 am UTC

mollwollfumble wrote:Let's extend that
[imath]e_{1/2}(e_{1/2}(x))=\exp(x), e_{1/3}(e_{1/3}(e_{1/3}(x)))=\exp(x)[/imath] etc.


If you want to exted like this, you need to demonstrate that such functions actually exist. Its not at all clear to me that this is the case. For instance, an initial question that needs answering would be "what is the domain and range of [math]e_{1/2}?[/math]"

You probably also want them to have nice nice properties, like continuity, convexity or concavity, and differentiability. The strongest such condition that log and exp both share is that they are infinitely differentiable and locally equal to their power series. Assuming that we get statements like

[math]\exp(x) = \frac{d}{dx} exp(x) = \frac{d}{dx} e_{1/2}(e_{1/2}(x)) = (e_{1/2})'(x) (e_{1/2})'(e_{1/2}(x))[/math]

If you work those sorts of statements, then you might be able to find a power series representation for [math]e_{1/2}[/math], but you at least need to demonstrate that functions like this must exist if you want to say anything more about them.
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NathanielJ
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Re: f(f(x))=exp(x) and extensions to noninteger numbers of f

Postby NathanielJ » Thu Dec 29, 2011 8:24 pm UTC

An article about the function e1/2 can be found here, and the fact that such an article exists suggests that the function exists. Unfortunately, however, I don't read German, so that's about all I can say.
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Re: f(f(x))=exp(x) and extensions to noninteger numbers of f

Postby mfb » Thu Dec 29, 2011 11:24 pm UTC

The abstract states that the author is able to construct e1/2, which is analytic (can be approximated with a power series) and defined on R->R.
But it is not so trivial that this is possible, so I would be careful with a general e1/n.

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Re: f(f(x))=exp(x) and extensions to noninteger numbers of f

Postby Qaanol » Fri Dec 30, 2011 3:47 am UTC

NathanielJ wrote:An article about the function e1/2 can be found here, and the fact that such an article exists suggests that the function exists. Unfortunately, however, I don't read German, so that's about all I can say.

Thanks for the link, how did you find it without reading German? I don’t speak German either unfortunately. That site has a link to download the article as a PDF, which has the text as an image. I uploaded it to Google Docs, which has an OCR functionality. It also has translation capability, although machine translations are notoriously poor.

Here is the Google Docs generated OCR version in German, and here is the Google auto-translated version in “English”. Quite a lot of things do not show up properly in the OCR version, and more yet were mistranslated, but this could at least be a start toward understanding.

The equations in particular did not all show up as they ought, so I recommend viewing or downloading the original German version from the linked site, to see what the equations are meant to be. I have not read through the auto-translated version nor made any attempt at comprehending it yet, so I cannot vouch for it one way or another.
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mfb
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Re: f(f(x))=exp(x) and extensions to noninteger numbers of f

Postby mfb » Fri Dec 30, 2011 12:56 pm UTC

I can try to give the idea of the paper:

After working a bit with the functional equations (see the equations), the author searches for stable points of the exponential function, that means a (complex) c with e^c = c. In §3, you can see the complex plane with two lines, corresponding to solutions with respect to one (real) degree of freedom. From the equations, he concludes that for each interval 2pi*k < Im(c) < 2pi*k + pi/2, there is one unique c., where negative Im(c) correspond to the complex conjugated numbers of the solutions with positive Im(c).
On the next page, he restricts the analysis to c~=0.318+1.337 i.
He gets a function [imath]\chi(z)[/imath] (eq 11) which is defined and analytic for Im(z)>=0 except at {0,1,e,e^e,e^e^e,...}. That is summarized in theorem 6 ("Satz 6"), and after that some properties are discussed.
Then he does a lot of magic with curves and complex analysis on them and finds a function for equation 16 within some curve, which is then extended to an analytic function.
Theorem 9: [imath]\Psi(z)[/imath] is in the area [the 3 symbols] (except z=c) and the real axis regular. On the real axis, the derivative is positive and the function reaches all positive [real] values. In [2 symbols] (without z=c) and on the real axis, it satisfies the functional equation (17).

(§6) Now define [imath]\Phi(z)=\Psi^{-1}(z)[\Psi(z)+1/2][/imath], it is the solution of the initial problem.

tomtom2357
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tetration

Postby tomtom2357 » Sat Dec 31, 2011 7:00 am UTC

If there was a function for tetration, t(x)=e^t(x-1), then we could solve the functional equation f(t(x))=t(x+0.5) to get your function. I am working on a way to make a taylor series for t(x) then extend it to f(x). The main problem is that the equations to solve for the coefficients gets very complicated, very fast.

Edit: This is the understatement of the century. Even the quadratic approximation involves complex terms.
Last edited by tomtom2357 on Sat Dec 31, 2011 8:12 am UTC, edited 2 times in total.
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mollwollfumble
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Re: f(f(x))=exp(x) and extensions to noninteger numbers of f

Postby mollwollfumble » Sat Dec 31, 2011 7:02 am UTC

> If you want to extend like this, you need to demonstrate that such functions actually exist. Its not at all clear to me that this is the case.

Thanks for that. The problem is not so much whether such functions exist, they do, but whether they can be made unique by extra assumptions. To show that they exist, consider the following:

Any number of solutions to f(f(...f(x)))=exp(x) can be generated using the axiom of choice. Suppose that there are n occurrences of f in the above. Set x_0=x and choose x_1=f(x_0), x_2=f(x_1) up to x_{n-1} by any way that you like to choose. Then x_n=f(x_{n-1}) is fixed to equal exp(x). Then also x_{n+1}=f(x_n)=exp(x_1) and in general x_{n+m}=exp(m). This is also done backwards. x_{-1} = ln(x_{n-1}) etc.

Assuming the sensible choice of having x_0<x_1<\ldots<x_n, we carry out the same procedure for all x between x_0 and x_1 and this automatically generates the whole function.

But to make it unique it is necessary to add extra conditions. A good one seems to be that the derivative to all orders exists and is monotonic.

Image
This image shows numerical calculations of e_1=exp(x), e_{1/2}, e_{1/4} and e_{1/phi} where phi is the golden ratio.

The numerical simulation for this image was done using the above method of guessing x_1, x_2, etc and then adjusting them until the numerical first and second derivatives were monotonic.
Last edited by mollwollfumble on Sat Dec 31, 2011 7:15 am UTC, edited 1 time in total.

tomtom2357
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Re: f(f(x))=exp(x) and extensions to noninteger numbers of f

Postby tomtom2357 » Sat Dec 31, 2011 7:09 am UTC

The function that my formula would generate would (probably) be the unique function that is infinitely differentiable that satisfies the above requirements (especially the monotonic condition).
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Desiato
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Re: f(f(x))=exp(x) and extensions to noninteger numbers of f

Postby Desiato » Sun Jan 08, 2012 1:26 am UTC

mollwollfumble wrote:The problem is not so much whether such functions exist, they do, but whether they can be made unique by extra assumptions. To show that they exist, consider the following:

Any number of solutions to f(f(...f(x)))=exp(x) can be generated using the axiom of choice. Suppose that there are n occurrences of f in the above. Set x_0=x and choose x_1=f(x_0), x_2=f(x_1) up to x_{n-1} by any way that you like to choose. Then x_n=f(x_{n-1}) is fixed to equal exp(x). Then also x_{n+1}=f(x_n)=exp(x_1) and in general x_{n+m}=exp(m). This is also done backwards. x_{-1} = ln(x_{n-1}) etc.

Assuming the sensible choice of having x_0<x_1<\ldots<x_n, we carry out the same procedure for all x between x_0 and x_1 and this automatically generates the whole function.

Typo: You mean "... in general x_{n*m}=exp(x_m)"?

Nitpick that's easily fixable: You need to require all x_i to be different (which you later add by "sensibly" chosing x_0<...<x_n); otherwise, you no longer would have a function.

There's something more that I find subtly nontrivial about things you omitted for brevity, but I'll need to think more before I can phrase that clearly.

Edit: Changed n+m to n*m in index of x

Edit the second: Okay, so I realize what irks me, or rather I think I'm now able to phrase that coherently. Your use of axiom of choice is at least non-obvious and at worst non-valid. I've spent considerable time trying to turn your sketch into something solid, and I fail. Basically, the weakness is that AoC doesn't allow you to do an arbitrary construction an uncountable number of times. It allows you one (or many, but only as many as you can actually write down in a proof) choice simultanously. But you need to avoid collisions, so you can't chose all necessary images at once.

I'll open another post with a more general question related to this.

Edit the third: Really, I just let myself be confused by not realizing this is just transfinite recursion. So the above construction (while I still think it could be phrased a bit more clearly) is correct.


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