Intervals of Tetration UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 10/06/2007, 05:26 PM 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? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/06/2007, 07:26 PM 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? UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 10/07/2007, 06:43 PM 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. « Next Oldest | Next Newest »