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 The (forgotten) left hyper operations bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 04/26/2008, 08:27 PM (This post was last modified: 04/26/2008, 08:53 PM by bo198214.) When one starts considering tetration then one usually has the choice between chosing $x[4](n+1)=x^{x[4]n}$ or $x[4'](n+1)=(x[4]n)^x$. Usually people always prefer the first right bracketed law, because the left bracketed law is too simple: $x[4']n=x^{x^{n-1}}$, assuming $x[4']1=x$. However as I described here it is more appropriate to assume $x[4']0=x$ getting the even more simple $x[4']n=x^{x^n}$. Though this is a trivial start, it does not remain trivial when considering left pentation: $b[5']0=b$ $b[5'](n+1)=(b[5']n)[4']b=(b[5']n)^{(b[5']n)^b}$ For example $x[5']1=x^{x^x}$, $x[5']2= \left( {x}^{{x}^{x}} \right) ^{ \left( {x}^{{x}^{x}} \right) ^{x}}=x^{x^x x^{xx^x}}=x^{x^{x+x^{x+1}}$ So there is no reason to neglect the left hyper operation sequence by being too trivial! But before proceding something about extension of the left tetration to the reals. There is one quite obvious way for an extension: We can describe it by $b[4']n= f^{\circ n}(b)$ where $f(x)=x^b$. Then we even have the desired $b[4']0=f^{\circ 0}(b)=b$. Now $x^b$ has the fixed point 1 and this is the only analytic fixed point there. So of course we consider regular iteration at this fixed point which indeed yields $f^{\circ t}(x)=x^{b^t}$ (without proof) so $b[4']t=b^{b^t}$. We do similar for left pentation, let $f(x)=x^{x^b}$ then $b[5']n=f^{\circ n}(b)$. Now $f(x)$ has again a fixed point at 1 and we can apply regular iteration there. Generally: $b[k+1']t={f_k}^{\circ t}(b)$ where $f_k(x)=x[k']b$. We can show by induction that $f_k(1)=1$ for $k\ge 3$ and that $f_k'(1)=1$ for $k\ge 4$. To obtain real values we maybe must assume that $b\ge 1$ and $f_k(x)$ defined only on $x\ge 1$ for $k\ge 4$. I think it also can be shown that 1 is the only real fixed point of $f_k$, $k\ge 4$. So the situation here is quite similar to the one of balanced tetration. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 04/26/2008, 09:00 PM Some pictures of left tetration     red, green, yellow, blue corresponds to x[4']0, x[4']1, x[4']2, x[4']3 and left pentation     red, green, yellow, blue corresponds to x[5']0, x[5']1, x[5']2, x[5']3 The conjecture of course is that $\lim_{x\to 0} x[2k]t = 1$, $t>0$. $\lim_{x\to 0} x[2k+1]t = 0$ Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 04/27/2008, 07:17 AM I was wondering why left and balanced are not being talked about here too much since they obviously give different results when iterated. There is also a possibility to make a mix of them as well on a way, changing bracketing in the middle of sequences, or any place, or changing it continuously. Probably(?) the problem is that there is fast multiplying of different results and not satisfactoy simple way to describe the various infinite trees or paths arising. My intuitive guess is that there either already exists in mathematics way to describe relations between different values obtained by different bracketing,(trees? tree generating series?) or , if not, especially in continuous extension of iterations, such has to be found, which is probably as very difficult task. But You know that I am not so good in details, so excuse me for having such general views. Ivars « Next Oldest | Next Newest »

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