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 Change of base formula for Tetration bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/15/2007, 01:42 PM (This post was last modified: 08/15/2007, 01:47 PM by bo198214.) jaydfox Wrote:Well, most of real and complex analysis would fall apart if limiting cases were not sufficient to provide proofs! If you ever looked into an analysis book you would know that there are quite rigorous proofs for convergence. You merely telled something about an $\epsilon$ that goes rapidly to 0. Neither is clear whether this $\epsilon$ is inside or outside the parenthesis nor why this would imply the convergence of the sequence. And about what limiting cases do you speak? A proof could perhaps look like this: We want to show that the sequence $t_n:=\log_a^{\circ n}(\exp_b^{\circ n}(y))$ converges. By the law $\log_a(b^wr)=w\cdot\log_a(b) + \log_a( r)=w\left(\log_a(b)+\frac{ \log_a ( r ) }{w}\right)$ we inductively construct a supplemental sequence by $r_{n,m}=\log_a(b)+\frac{\log_a ( r_{n,m-1} ) }{\exp_b^{\circ n-m}(y)}$ and $r_{n,0}=1$. This sequence is exactly chosen such that $t_n=\log_a^{\circ n-m}\left(\exp_b^{\circ n-m}(y)r_{n,m})$ particularly $t_n=yr_{n,n}$. Now it is clear by looking at the derivative of $\log_a$ that $\log_a(x+\delta)<\log_a(x)+\delta$ for each $x>1,\delta>0$. If we repeatedly apply this to the formula of $r_n$, while assuming that $1 and hence $\log_a(b)>1$, we get $r_{n,m}=\log_a(b)+\frac{\log_a ( r_{n,m-1} ) }{\exp_b^{\circ n-m}(y)}\\ \le \log_a(b)+\log_a(\log_a(b))\left(\frac{1}{\exp_b^{\circ n-m}(y)}+\frac{1}{\exp_b^{\circ n-m}(y)\exp_b^{\circ n-m+1}(y)}+\dots+\frac{1}{\exp_b^{\circ n-m}(y)\dots \exp_b^{n-2}(y)}\right)$ and further for $n\ge 1$ $t_n\le y\left(\log_a(b)+(\log_a(\log_a(b))\sum_{k=0}^{n-2}\frac{1}{\prod_{j=0}^k \exp_b^{\circ j}(y)}\right)$ Now is $\prod_{j=0}^k \exp_b^{\circ j}(y)\ge y^{k+1}$, because $b^x\ge x$ for $b\ge\eta$. But we know that the series $\sum_{k=0}^\infty \frac{1}{y^{k+1}$ converges for $y>1$ and hence is the sequence $t_n$ bounded from above. An induction shows that $r_{n,n}$ is increasing. We show $r_{n,m}\ge r_{n-1,m-1}$ (for arbitrary $n, n\ge m$) by induction over m. Induction base: $m=0$ $r_{n,1}=\log_b(a)\ge 1 =r_{n-1,0)$ and for the induction step show it for $m\mapsto m+1$: From the assumption follows by monotone increase of $\log_a$: $\log_a ( r_{n,m} ) \ge \log_a(r_{n-1,m-1})$ then $\frac{\log_a ( r_{n,m} ) }{\exp_b^{\circ n-(m+1)}(y)}\ge \frac{\log_a ( r_{n-1,m-1} ) }{\exp_b^{\circ (n-1)-m}(y)}$ which yields $r_{n,m+1}\ge r_{n-1,m}$, the induction assertion. So particularely $t_n=yr_{n,n}$ is increasing and bounded from above (for $y>0$) and so has a limit, given that $b>a$ and that $b>\eta$. Theorem. The sequence $\left(\log_a^{\circ n}({}^{x+n}b)\right)_n$ converges if $b>a>1$, $b>\eta$ (and ${}^xb>0$). bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/15/2007, 09:00 PM (This post was last modified: 08/15/2007, 09:02 PM by bo198214.) jaydfox Wrote:Essentially, if k is the number of terms at which we truncate the series expansion, then there is a non-zero radius for which the series is initially convergent (i.e., the root-test for terms 1 through k would all be less than 1). I dont understand a word. If you truncate the series, it is a polynomial and a polynomial is defined on the whole complex numbers, i.e. infinite radius of convergence. Quote:Regardless, if the proof has already been shown, then combined with my change of base formula, we now have a unique solution to tetration of bases greater than eta. As it appears to me your change of base formula works merely for base $b$ greater than $\eta$ and $a. But thats exactly the wrong direction. We need to change the base from $b\le\eta$ to $a>\eta$. Even if we had a proper change of base formula we need to check that it is consistent for change of bases smaller than $\eta$, i.e. that it transforms the already known unique solution for base $b\le\eta$ into the already known unique solution for base $a\le\eta$. If we had a change of base formula then we anyway dont need the converging solution for $b=\eta$ (via $e^x-1$), because we could simply compute that solution from a smaller base. Quote:By the way, for the reference to Ecalle, where can I get a copy, and more importantly, is there an English translation available? Not sure, I lent it from the library. There seems no translation into english. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/15/2007, 09:19 PM bo198214 Wrote:Quote:Regardless, if the proof has already been shown, then combined with my change of base formula, we now have a unique solution to tetration of bases greater than eta. As it appears to me your change of base formula works merely for base $b$ greater than $\eta$ and $a.How do you figure that it only works for a-2$ I guess. However if we can show analyticity it suffices to define it in a small vicinity of some point $x_0$ because from there we can analytically extend it. However the question then is what is $x_0$ and what $\varepsilon>0$ has the vicinity. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/16/2007, 05:51 PM (This post was last modified: 08/16/2007, 06:00 PM by jaydfox.) Hey, it's a change of base formula for **tetration**. Some people only define tetration for integers $n \ge 1$, others for integers $n \ge 0$, others for integers $n \ge -1$. Some people define it for real numbers $x \ge 0$, or $x \ge -1$ or $x > -2$. Some people might define tetration over all reals, or over the complex numbers. I prefer reals greater than -2 myself, but your mileage may vary. For real numbers, my formula is valid, assuming you define tetration as iterated exponentials, with negative iterations equivalent to logarithms. I don't claim validity over complex numbers, mainly because the formula's very basis for convergence is the increasing modulus of the successive tetrations, and complex tetrations can cause the modulus to decrease, which invalidates the limit. For real x, use whatever domain you use for tetration. If *you* define tetration as valid for x>-2, what did you expect to happen when you put in x <= -2? The problem isn't with my formula, it's with your indecision on what the valid domain for x is. Use whatever domain you consider valid. Moving along: There are two main base conversion formulae for exponentiation for base a, given an exact solution for base b: $ \begin{eqnarray} a^x & = & b^{\log_b(a^x)} \\ \vspace{5} \\ a^x & = & b^{\log_b(a)\times x} \\ \end{eqnarray}$ The first is a trivial restatement of the definition of log_b(z) as the inverse function of b^z. The second displays some "fundamental truth" about exponentiation that isn't obvious from looking solely at the first formula. It allows you to solve for arbitrary exponentiations of base a, having no knowledge of how to do so explicitly, but having knowledge of how to exponentiate base b, along with knowledge of the constant log_b(a). There are two main change of base formulae for tetration: $ \begin{eqnarray} {}^x a & = & {}^{\text{slog}_b({}^x a)} b \\ \vspace{5} \\ {\Large ^{\normalsize x} a} & = & {\Large \lim_{n \to \infty}\log_a^{\circ n}\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)} \end{eqnarray}$ In either case, if it makes you feel better, you can explicitly state that x is a real > -2, or x is an integer >= -1, or whatever. And I've already stated that a and b should be greater than eta, though as a tool for fractionally iterating logarithms, it has applications with bases between 1 and eta. The first formula, again, is a trivial restatement of the definition of slog_b(x) as the inverse of b^^x. The second displays some fundamental truth about the relationship of tetration in various bases, which isn't at all obvious by looking solely at the first formula, and it also allows us solve tetration for base a when we have no knowledge of how to do so explicitly, so long as we know how to do so with base b, and we know the value of the constant mu_b(a). ~ Jay Daniel Fox bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/16/2007, 06:40 PM (This post was last modified: 08/16/2007, 06:43 PM by bo198214.) Sorry Jay we dont speak about the same thing. You ask me why I assume $a. I answered: because your formula does not converge for some x if we allow $a>b$. You answered: that I didnt pay attention to $\mu_b(a)$. I answered: there are despite values of x for which it doesnt converge. And I can not relate your current answer to this problem. To express it in formulas and numbers: Your definition: For a given tetration ${}^xb$ for base $b$ compute another tetration for base $a$ by the formula ${}^xa = \lim_{n\to\infty} \log_a^{\circ n}({}^{x+b+\mu_b(a)}b)$ where $\mu_b(a)$ is chosen accordingly that ${}^1a=a$. This formula is equivalent to ${}^xa = \lim_{n\to\infty} \log_a^{\circ n}(\exp_b^{\circ n}({}^{x+\mu_b(a)}b)$ By definition ${}^0 b=1$, so if I chose $x=-\mu_b(a)$ then ${}^{x+\mu_b(a)}b=1$. If I now compute the sequence $t_n = \lim_{n\to\infty} \log_a^{\circ n}(\exp_b^{\circ n}(1.0))$ say for bases $a=3.0>\eta$ and $b=1.5>\eta$ I get for $n=1,2,3$: $.3690702464, -.5382273450, -.9460372733+2.859600867*I$ For me this means your formula gives no result in this case. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/16/2007, 09:21 PM Posts are missing. ~ Jay Daniel Fox « Next Oldest | Next Newest »

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