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 Uniqueness of Ansus' extended sum superfunction bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 09/05/2009, 01:26 PM (This post was last modified: 09/05/2009, 01:27 PM by bo198214.) First let me recapitulate the method from my point of view. We start with as usual with a function $f$ and want to obtain a superfunction $\sigma$ of it, i.e. a function that satisfies (1) $\sigma(z+1)=f(\sigma(z))$. Now by differentiating once we get $\sigma'(z+1)=\sigma'(z) f'(\sigma(z))$ expanding twice: $\sigma'(z+2)=\sigma'(z) f'(\sigma(z)) f'(\sigma(z+1))$ and generally by induction: $\sigma'(z+n)=\sigma'(z) \prod_{k=0}^{n-1} f'(\sigma(z+k))$ making $z=n$ our new variable: $\frac{\sigma'(z_0+z)}{\sigma'(z_0)}= \prod_{k=0}^{z-1} f'(\sigma(z_0+k))$ To extend the product to non-integer $z$ we have a look at the sum (2) $\ln\sigma'(z_0+z)-\ln\sigma'(z_0)= \sum_{k=0}^{z-1} \ln f'(\sigma(z_0+k))$ Extending the sum to non-integer boundaries The sum can extended to non-integer values $z$ via Faulhaber's formula which provides the sum for monomials: $\sum_{k=n_0}^{n} k^p = \frac{\varphi_{p+1}(n+1) - \varphi_{p+1}(n_0)}{p+1}$, where $\varphi_{p}(x) = \sum_{k=0}^{p} \left(p\\k\right) B_k x^{p-k}$ is the $p$-th Bernoulli polynomial with $B_k$ the Bernoulli numbers. The sum operator is linear for natural boundaries so one can extend Faulhaber's formula to linear combinations of monomials (also called polynomials ), and finally to powerseries. So for non-integer $z$ we define for $f(z)=\sum_{n=0}^\infty f_n z^n$ the extended sum $\sum_{\zeta = z_0}^{z} f(\zeta) := \sum_{n=0}^\infty f_n \frac{\varphi_{n+1}(z+1)-\varphi_{n+1}(z_0)}{n+1}$. To simplify the matter we can define the antidifference or indefinite sum: $\Delta^{-1}[f](z) :=\sum_{n=0}^\infty f_n \frac{\varphi_{n+1}(z)}{n+1}$ and get $\sum_{\zeta = z_0}^{z} f(\zeta) = \Delta^{-1}[f](z+1) - \Delta^{-1}[f](z_0)$ Extended sum and shifts We hope that the extended sum behaves like a normal sum with integer boundaries, e.g. we want to have (3) $\sum_{\zeta=a}^b f(\zeta+c) = \sum_{\zeta=a+c}^{b+c} f(\zeta)$. We show it for polynomials. Let $p$ be a polynomial of degree $N$, then we know that (3) is satisfied for all integers $c$ and both sides are polynomials of maximal degree $N+1$ in $c$. But equality at $N+1$ different points already implies the equality of both polynomials. Hence not only for integer $c$ but for every $c$ the equality holds. This than carries over to powerseries (as limit of a polynomial sequence) if no convergence issues arise. This property lets us formulate (2) in a much nicer way by applying (3) with $c=z_0$ and afterwards setting $z:=z+z_0$: (4) $\ln\sigma'(z)- \ln\sigma'(z_0)=\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\zeta))$ Uniqueness of extended sum superfunction Sketch of proof: Let $\rho$ beside $\sigma$ be another at $z_0$ analytic super-exponential (1) that satisfies (2). We further assume $\sigma$ and $\rho$ to be invertible at $z_0$ and $\sigma(z_0)=\rho(z_0)$. We set $\delta=\sigma^{-1}\circ \rho$ in a vicinity of $z_0$, so $\rho(z)=\sigma(\delta(z))$ there. $\delta(z_0)=z_0$ and by (1) $\delta(z+n)=\delta(z)+n$ so $\delta(z_0+n)=z_0+n$ for integer $n$. So put $\rho$ into (4), with $\rho'(z)=\sigma'(\delta(z)) \delta'(z)$: $\ln\sigma'(\delta(z))- \ln\sigma'(z_0)+\ln\delta'(z) =\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\delta(\zeta)))$ the left difference can be replaced with (4), knowing $\delta(z_0)=z_0$: (5) $\sum_{\zeta=\delta(z_0)}^{\delta(z)-1} \ln f'(\sigma(\zeta))+\ln\delta'(z) =\sum_{\zeta=z_0}^{z-1} \ln f'(\sigma(\delta(\zeta)))$ Now consider the forward difference $\Delta (h\circ \delta) (z) = h(\delta(z+1))-h(\delta(z)) = h(\delta(z)+1)-h(\delta(z)) = \Delta(h)\circ \delta(z)$. Hence $\Delta^{-1} (h\circ \delta) = \Delta^{-1}(h) \circ \delta$ for any $h$ up to perhaps a constant, so $\sum_{\zeta=z_0}^{z-1} h(\delta(\zeta)) = \sum_{\zeta=\delta(z_0)}^{\delta(z)-1} h(\zeta)$. Now we apply this to (5): $\ln\delta'(z) =0$ $\delta'(z)=1$ $\delta(z)=a z + b$ together with $\delta(z_0)=z_0$ and $\delta(z_0+1)=z_0+1$ this gives $a=1$ and $b=0$, so $\delta(z)=z$ and $\sigma(z)=\rho(z)$. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 09/06/2009, 04:56 AM (This post was last modified: 09/06/2009, 05:16 AM by Base-Acid Tetration.) nice proof! Does this prove the uniqueness for all tetration? i.e. that all tetrations are equal where they are invertible, since they are supposed to be analytic, equal to 1 at 0, and is a superfunction of exponentiation, ex hypothesis. bo198214 Wrote:if no convergence issues arise. I am confused on exactly what the problem you tetration guys are trying to solve with tetration. I am guessing that it may be convergence or holomorphicity. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 09/06/2009, 06:54 AM (09/06/2009, 04:56 AM)Tetratophile Wrote: Does this prove the uniqueness for all tetration? i.e. that all tetrations are equal where they are invertible, since they are supposed to be analytic, equal to 1 at 0, and is a superfunction of exponentiation, ex hypothesis. No, this is particularly about Ansus' construction, i.e. all tetrations that satisfy (4) are equal. There are so far no numerical testing on the other tetrations, I will see what I can do about this in the next time. Quote:bo198214 Wrote:if no convergence issues arise. I am confused on exactly what the problem you tetration guys are trying to solve with tetration. I am guessing that it may be convergence or holomorphicity. Most methods compute coefficients of powerseries as limits of some computations. To be on solid ground we need to prove that: 1. The limit for each coefficient exists. 2. The resulting series has non-zero convergence radius. Nothing of that is proved neither for the matrix power method, the intuitive method nor Ansus' extended sum method. Though numerically everything looks good. 3. Also it would be nice to prove that the corresponding super-exponential is holomorphic on C without (-oo,-2]. Which also looks good numerically. tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/06/2009, 08:08 PM (09/06/2009, 06:54 AM)bo198214 Wrote: 3. Also it would be nice to prove that the corresponding super-exponential is holomorphic on C without (-oo,-2]. Which also looks good numerically. but I dont think it is. i will give arguments when i find the time... tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 10/25/2013, 11:27 PM (This post was last modified: 10/25/2013, 11:29 PM by tommy1729.) @Bo In post 1 for your proof I doubt the statements concerning the forward difference. Could you elaborate on that ? I prefer no use of triangles and clear brackets to avoid confusion. Afterall this is continuum summation and the hard subject called tetration. I doubt the uniqueness. regards tommy1729 « Next Oldest | Next Newest »

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