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+--- Thread: Taylor series of upx function (/showthread.php?tid=199)
RE: Taylor series of upx function - Kouznetsov - 03/03/2009
Ansus Wrote:Thank you.
What did you find about simmetricity?
If y=J[x], then, the symmetry with respect to line x=-y implies that
-x=J[-y], and then
-x=J[-J[x]].
Please, plot J[-J[x]]+1 versus x; if symmetry, then, it should be zero.
You may just add the line
PJ = Plot[{J[-J[z]] + z}, {z, -1.999, 1.9}, PlotRange -> All];
to the coide, and get the graphic below:
The deviation is small, of order of a percent. Could you confirm this result?
RE: Taylor series of upx function - bo198214 - 03/03/2009
Ansus Wrote:Anyway the deviation should go to zero with higher z even if the functions differ because they have common limit.
Does it decline with increased precision?
Guys, this really pisses me off.
I gave a demonstration here that no smooth function \( f \) can satisfy the three criteria
\( f(0)=1 \)
\( f(x+1)=\exp_{\sqrt{2}}(f(x)) \)
\( f(-f(x))=-x \).
Ansus can you please tell me why you still insist on that claim?!
It was your idea that by those 3 properties (which are (1),(2), and (7)/(10) in your original post) the function \( f \) is already defined on a dense set!
You Wrote:andydude Wrote:Even if 7 and 10 are true, which I think they are, it seems that it does not (by itself) give solutions for all real numbers, it only gives solutions for specific non-integersIt gives solutions for infinite quantity of real numbers in each interval (and for each given number you can find as close solution as you wish).
Simply because you can use (1) property of tetration, you can project all solutions for non-integer numbers onto interval, say, (0;1). Since there is infinite quantity of non-integer solutions, you will fill this interval with solutions as dense as you want.
I just carried out your idea (after some misunderstandings in the beginning).
I confirmed that it defines \( f \) on a dense set.
But it also turned out that \( f \) is fractal instead of smooth.
Andrew even announced that fractal nature before me.
Quote:I have done some computations with (H2) and (10) and after about 8 iterations of these constraints, then you can see that the points start to "overlap" in some places, indicating that if this does form
a function defined over reals \( x>-2 \), then it will be like the Weierstrass function, although it may not even be continuous! I have attached a zoomed graph about this.
So what reason do you have against all evidence to claim that there is a smooth \( \text{sexp}_{\sqrt{2}} \)?
What do you doubt about the demonstrations?
Why dont you just verify your own idea with mathematica?
These some iterations can never imply so big error terms that would explain the fractal look; but even for that reason I increased the precision to 3000 bits with no change of the picture.
Of course it was a beautiful idea to have a symmetric \( \text{sexp}_{\sqrt{2}} \). It even would have been a uniqueness criterion. But if the mathematical reality is different, you can not just ignore it.
So please do your homework and verify yourself the graph which is defined by your properties (1),(2) and (7).
RE: Taylor series of upx function - sheldonison - 03/05/2009
Kouznetsov Wrote:There is EXACT Tailor series, if we insist that upx(z^*) = upx(z)^*
and upx(z) is holomorphic outside the part of the negative part of axis,
id est, holomorphic everywhere except \( z\le 2 \).
Then the solution is unique.
You can calculate so many derivatives as you like in any regular point, and, in particular, at z=0. The algorithm of evaluation is described in
http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp99.pdf
Ok so far. My higher math skills aren't as good as I'd like them to be, so I'm not able to follow Dimitrii's paper. I'm looking at the pdf paper, and in section 2.1, Dimitrii writes:
Quote:(1.4) F(z + 1) = exp(F(z))Shortly after thereafter, Dimitrii generates the value L for the critical point, but by then, I was already lost. I think the key is that in the complex plane, exp and ln, equations (1.4) and (2.1) are no longer inverses of each other, or perhaps more likely, that f(z) is an approximation function, which may not hold when adding increments of 2pi*i. Do you need the taylor series for f(z) to generate L? Is the key adding increments of 2*pi*i to the taylor series expansion for approximations of F(z)? Or does that require the contour integrals, in section (4)?
....
(2.1) ln(F(z + 1)) = F(z)
....
In this paper I assume that logarithm ln is single-valued function, that is analytic at the complex plane with cut along the negative part of the real axis. Then, all the solutions of equation (2.1) are solutions of (1.4), but it is not obvious, that a solution F of (1.4) satisfies also (2.1). One could add term 2\( \pi \)i to the right hand side of equation (2.1), and the solution of the resulting equation will be also solution of (1.4). Searching for the simple tetration, I begin with analytic solutions, which satisfy also equation (2.1).
- Sheldon