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 Tetration extension for bases between 1 and eta dantheman163 Junior Fellow Posts: 13 Threads: 3 Joined: Oct 2009 11/05/2009, 03:00 AM (This post was last modified: 11/05/2009, 09:14 PM by dantheman163.) I believe I have found an analytic extension of tetration for bases 1 < b <= e^(1/e). This is based on the assumption (1) The function y=b^^x is a smooth, monotonic concave down function Conjecture: If assumption (1) is true then ${}^x b = \lim_{k\to \infty} (log_{b}^{ok}(x({}^k b- {}^{(k-1)} b)+{}^k b) )$ for $-1 \le x\le 0$ Some properties: This formula converges rapidly for values of b that are closer one. For base eta it converges to b^^x for all x but this is not true for the other bases. Interestingly for b= sqrt(2) and x=1 it seems to be converging to the super square root of 2 I will try to post a proof in the next couple of days I just need some time to type it up. Thanks bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/05/2009, 01:44 PM (This post was last modified: 11/05/2009, 01:47 PM by bo198214.) Hey Dan, thank you for this contribution. The formula you mention is the inversion of Lévy's formula (which is generally applicable to functions with f'(p)=1 for a fixed point p of f). See e.g. the tetration methods draft, formula 2.26 (where formula 2.27 is called Lévy's formula). The formula is known to give the regular tetration for $b=e^{1/e}$. However it is new to apply it to $1. So I am really curious about your proof. PS: for writing formulas in this forum please have a look at this post. dantheman163 Junior Fellow Posts: 13 Threads: 3 Joined: Oct 2009 11/05/2009, 11:53 PM (This post was last modified: 11/06/2009, 12:24 AM by dantheman163.) I'm not sure how rigorous this is but here it is Proof: Assumption (1)The function $f(x)={}^x b$ is a smooth, monotonic concave down function Based on (1) we can establish (2)Any line can only pass through f(x) a maximum of 2 times (3)The intermediate value theorem holds for the entire domain of f(x) Take the region R bounded on the x axis by x=-1 and x=0 and bounded on the y axis by y=f(-1) and y=f(0) ( y=0 and y=1). Because of (3) every value, x, has a corresponding value, f(x), on the interval. If we take a point (x,y) in R and assume that it is on the curve f(x) we can then use the relation $f(x+1)=b^{f(x)}$ and obtain the new point $(x',y')$ which equals $(x+1,b^y)$. Applying this repeatedly we obtain the point $(x+k,{Exp}_b ^k (y))$. We can now establish (4)The point (x,y) in the region R is on the curve f(x) if $(x+k,{Exp}_b ^k (y))$ is not on the secant line that touches the curve at 2 other known points of f(x) for any value of k. Now we will find the equation of the secant line that touches f(x) at 2 consecutive known points. Using the point slope formula we find the equation to be $g(x)= ({}^k b-{}^{(k-1)} b)(x-k)+{}^k b$. We must also note that if a point (x,y) is above the curve in the region R then $(x+k,{Exp}_b ^k (y))$ is above the curve for any value of k. we shall now extend (4) to say (5)The point (x,y) in the region R is on or above the curve f(x) if $g(x+k) < {Exp}_b ^k (y)$for any value of k Finally if we take the limit as k approaches infinity we will find that the slope of the secant line approaches zero and therefore follows f(x) exactly because f(x) has an asymptote as x goes towards infinity. Therefore $g(x+k) = {Exp}_b ^k (y)$ for infinity large values of k Now solving for y and taking the limit as k approaches infinity we obtain the desired result: ${}^x b = \lim_{k\to \infty} (log_{b}^{ok}(x({}^k b- {}^{(k-1)} b)+{}^k b) )$ for $-1 \le x\le 0$ q.e.d bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/07/2009, 09:31 AM Let me rephrase in my words: We consider the linear functions $g_k$ on (k-1,k) determined by $g_k(k-1)=f(k-1)={^{ k-1}b}$ and $g_k(k)=f(k)={^k b}$, they are given by: $g_k(x) = ({^k b} - {^{ k-1}}b)(x-k) + {^k b}$. As f is concave $f(x+k) \ge g_k(x+k)$ for $x\in (-1,0)$, and as $\log_b$ is strictly increasing we have also $f(x) \ge \log_b^{\circ k} g_k(x+k)$. On the other hand we know that $f(x+k)=\exp_b^{\circ k}(f(x))\uparrow a$ for $k\to\infty$, hence $a > f(x+k) > g_k(x+k)$ and $g_k(x+k)\uparrow a$ for $k\to\infty$. We have now $f(x+k)-g_k(x+k)\downarrow 0$ for $k\to\infty$. But I think that does not directly show the convergence $f(x) - \log_b^{\circ k} g_k(x+k)\downarrow 0$. Any ideas? bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/07/2009, 05:11 PM (11/07/2009, 09:31 AM)bo198214 Wrote: We have now $f(x+k)-g_k(x+k)\downarrow 0$ for $k\to\infty$. But I think that does not directly show the convergence $f(x) - \log_b^{\circ k} g_k(x+k)\downarrow 0$. Any ideas? Actually I think one can show that $\log_b^{\circ k} g_k(x+k)$ is strictly increasing with $k$ because $\log_b g_k(x+k)$ is concave and hence $g_{k-1}(x+k-1) < \log_b g_k(x+k)$ and thatswhy $\log_b^{\circ k-1} g_{k-1}(x+k-1)<\log_b^{\circ k} g_{k}(x+k)$ also it is bounded and hence must have a limit. But now there is still the question why the limit $f(x)$ indeed satisfies $f(x+1)=b^{f(x)}$? bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/07/2009, 08:12 PM (11/07/2009, 05:11 PM)bo198214 Wrote: But now there is still the question why the limit $f(x)$ indeed satisfies $f(x+1)=b^{f(x)}$? Actually it is not the case but we can obtain something very similar. I just read in [1], p. 31, th. 10, that we have the following limit for a function $f(x)=f_1 x+f_2x^2+f_3 x^3 + ...$ with $0: $\lim_{n\to\infty} \frac{f^{\circ n}(t)-f^{\circ n}(\theta)}{f^{\circ n}(t)-f^{\circ n+1}(t)}=w\frac{1-f_1^w}{1-f_1}$ for $\theta = f^{\circ w}(t)$. If we invert the formula we get $f^{\circ w}(t) = \lim_{n\to\infty} f^{\circ -n}\left(w\frac{1-f_1^w}{1-f_1}\left(f^{\circ n+1}(t)-f^{\circ n}(t)\right)+f^{\circ n}(t)\right)$ In our case though we dont have f(0)=0 but there is some fixed point $z_f$ of $f$, $f(z_f)=z_f$. In this case however the formula is quite similar, the only change is that $f_1=f'(z_f)$. [1] Ecalle: Theorie des invariants holomorphes dantheman163 Junior Fellow Posts: 13 Threads: 3 Joined: Oct 2009 11/07/2009, 11:30 PM If we set $f(x+1)=b^{f(x)}$ which is to say $\lim_{k\to \infty} (log_{b}^{ok}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (log_{b}^{o(k-1)}(x({}^k b- {}^{(k-1)} b)+{}^k b) )$ then reduce it to $\lim_{k\to \infty} (log_{b}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (x({}^k b- {}^{(k-1)} b)+{}^k b)$ Then just strait up plug in infinity for k we get $log_{b} {}^\infty b = {}^\infty b$ which is the same as ${}^\infty b = {}^\infty b$ This is really weird because if i do $\lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) )$ for $b= sqrt2$ i get about 1.558 which is substantially larger then $sqrt2$ Can anyone else confirm that $\lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \approx 1.558$ for $b= sqrt2$ ? bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/08/2009, 02:44 PM (This post was last modified: 11/08/2009, 02:47 PM by bo198214.) (11/07/2009, 11:30 PM)dantheman163 Wrote: If we set $f(x+1)=b^{f(x)}$ which is to say $\lim_{k\to \infty} (log_{b}^{ok}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (log_{b}^{o(k-1)}(x({}^k b- {}^{(k-1)} b)+{}^k b) )$ then reduce it to $\lim_{k\to \infty} (log_{b}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (x({}^k b- {}^{(k-1)} b)+{}^k b)$ Then just strait up plug in infinity for k we get $log_{b} {}^\infty b = {}^\infty b$ which is the same as ${}^\infty b = {}^\infty b$ Slowly, slowly. The first line is what you want to show. You show from the first line something true, but you can also show something true starting from something wrong; so thats not sufficient. Also it seems as if you confuse limit equality with sequence equality. Lets have a look at the inverse function $g=f^{-1}$, $g(x)=\lim_{k\to\infty} \frac{\exp_b^{\circ k}(x)-{^k b}}{{^k b}-({^{k-1} b})}$ it should satisfy $g(b^x)=g(x)+1$. Then lets compute $g(b^x)-g(x)= \lim_{k\to\infty}\frac{\exp_b^{\circ k+1}(x)-{^k b}}{{^k b}-({^{k-1} b})} - \frac{\exp_b^{\circ k}(x)-{^k b}}{{^k b}-({^{k-1} b})} =\lim_{k\to\infty} \frac{\exp_b^{\circ k+1}(x) - \exp_b^{\circ k}(x)}{{^k b}-({^{k-1} b})}$ Take for example $x=1$ then the right side converges to the derivative of $b^x$ at the fixed point; and not to 1 as it should be. This is the reason why the formula is only valid for functions that have derivative 1 at the fixed point, e.g. $e^{x/e}$, i.e. $b=e^{1/e}$. Quote:This is really weird because if i do $\lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) )$ for $b= sqrt2$ i get about 1.558 which is substantially larger then $sqrt2$ Can anyone else confirm that $\lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \approx 1.558$ for $b= sqrt2$ ? Try the same with $b=e^{1/e}$ and it will work; but for no other base; except you use the modified formula I described before. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/12/2009, 07:11 PM Oops... you may have noticed this just got a "1 star" rating. I just happened to accidentally click the mouse on the "rate" thing, sorry dantheman163 Junior Fellow Posts: 13 Threads: 3 Joined: Oct 2009 12/15/2009, 01:01 AM (This post was last modified: 12/15/2009, 01:02 AM by dantheman163.) Upon closer study i think i have found a formula that actually works. ${}^ x b = \lim_{k\to \infty} \log_b ^k({}^ k b (ln(b){}^ \infty b)^x-{}^ \infty b(ln(b){}^ \infty b)^x+{}^ \infty b)$ Also i have noticed that this can be more generalized to say, if $f(x)=b^x$ then $f^n(x)= \lim_{k\to \infty} \log_b ^k( Exp_b^k(x) (ln(b){}^ \infty b)^n-{}^ \infty b(ln(b){}^ \infty b)^n+{}^ \infty b)$ numerical evidence shows this to be true. some plots ${}^x sqrt2$     half iterite of $sqrt2^x$     The graph falls apart at x=4 because $Exp_b^k(4.01)$ diverges as k goes to infinity. thanks. Edit: sorry for the huge pictures « Next Oldest | Next Newest »

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