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RE: Intervals of Tetration - UVIR - 10/06/2007
bo198214 Wrote:UVIR Wrote:I *think* (but I am not absolutely sure about it) that any method which defines \( {^y}x=f(x,y) \) for reasonable \( f(x,y) \), for \( y\in [0,1] \) could conceivably be extended to a \( C^{\infty} \)
I dont get this. If I define it on \( (0,1) \) and I demand that \( {^{x+1}b}=b^{^xb} \) then the function is already determined for all \( x>0 \) there is no place left for further manipulations (i.e. to make it differentiable).
Andrew's approach was to choose it as an analytic function on \( (0,1) \) and he determines the coefficients of the seriesexpansion at 0 by demanding that it is \( C^{\infty} \) (or even analytic). However this approach works computably only on the \( \text{slog} \), the inverse of \( {^xb} \).
I'll do one example to force the function to be \( C^2 \) at y=1 so you can see what I mean for the function I used: \( f(x,y)=x^y \). All calculations were done with Maple, so hopefully there are no mistakes.
Left of y=1:
\( f(x,y)=x^y\Rightarrow\\
D^2 f(x,y)=x^y*ln(x)^2\Rightarrow\\
lim_{y\to 1^-}D^2 f(x,y)=x*ln(x)^2 \)
For the other side (right of y=1):
\(
D^2 x^{f(x,1-y)}=x^{x^{1-y}}*(x^{1-y})^2*ln(x)^4+x^{x^{1-y}}*x^{1-y}*ln(x)^3\Rightarrow\\
lim_{y\to 1^+}D^2 f(x,y)=x*ln(x)^4+x*ln(x)^3 \)
Solving the equation:
\( x*ln(x)^2=x*ln(x)^4+x*ln(x)^3 \) numerically, we get:
\( x=1.855276959 \)
In other words, the tetration function defined by:
\( f(1.855276959,y)=1.855276959^y \) in (0,1) and by \( ^{y+1}1.855276959=1.855276959^{({^y}1.855276959)} \) is \( C^2 \) at y=1, assuming the way it was constructed by joining at the naturals with frac{y} as with my first construction.
Similar calculations show that using \( f(1.148776058,y) \) the corresponding function is \( C^6 \) at y=1, etc. I think one can go as high as one wants, provided a solution for x exists. It looks as if a solution \( x>1 \) exists for any even order derivative.
What do you guys think?
RE: Intervals of Tetration - bo198214 - 10/06/2007
UVIR Wrote:In other words, the tetration function defined by:
\( f(1.855276959,y)=1.855276959^y \) in (0,1) and by \( ^{y+1}1.855276959=1.855276959^{({^y}1.855276959)} \) is \( C^2 \) at y=1, assuming the way it was constructed by joining at the naturals with frac{y} as with my first construction.
Similar calculations show that using \( f(1.148776058,y) \) the corresponding function is \( C^6 \) at y=1, etc. I think one can go as high as one wants, provided a solution for x exists. It looks as if a solution \( x>1 \) exists for any even order derivative.
What do you guys think?
I didnt verify it directly but you dont need only the second derivative or only the 6th or only the \( n \)th derivative to be continuous for \( x\mapsto f(b,x) \) being in \( C^n \) but also all previous derivations. This means you get an equation system of at least \( n \) equations, but in only one variable.
Did you prove that in your case also the first or the first till fifth derivative is continuous?
RE: Intervals of Tetration - UVIR - 10/07/2007
bo198214 Wrote:I didnt verify it directly but you dont need only the second derivative or only the 6th or only the \( n \)th derivative to be continuous for \( x\mapsto f(b,x) \) being in \( C^n \) but also all previous derivations. This means you get an equation system of at least \( n \) equations, but in only one variable.
Did you prove that in your case also the first or the first till fifth derivative is continuous?
I checked it and it doesn't appear to work. Forcing each condition, \( C^2 \), \( C^3 \),...\( C^n \) gives separate and distinct solutions in x, so forcing all conditions simultaneously would give an overdetermined system and would thus be impossible. Back to the drawing board I guess.