Taylor series of upx function
#22
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-integers
It 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).
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Messages In This Thread
Taylor series of upx function - by Zagreus - 09/07/2008, 10:56 AM
RE: Taylor series of upx function - by bo198214 - 09/10/2008, 01:56 PM
RE: Taylor series of upx function - by andydude - 10/23/2008, 10:16 PM
RE: Taylor series of upx function - by Kouznetsov - 11/16/2008, 01:40 PM
RE: Taylor series of upx function - by bo198214 - 11/16/2008, 07:17 PM
RE: Taylor series of upx function - by Kouznetsov - 11/17/2008, 04:11 AM
RE: Taylor series of upx function - by bo198214 - 11/18/2008, 11:49 AM
RE: Taylor series of upx function - by Kouznetsov - 12/02/2008, 12:03 PM
RE: Taylor series of upx function - by Kouznetsov - 12/05/2008, 12:30 AM

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