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 Operator similar to tetration rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 05/27/2008, 03:57 AM Daniel Geisler referred me to this forum, and I'm surprised how active a forum is on a topic like tetration I have a question about an operator that is similar to a tetration that I'm trying to locate. I'm a bit new to this topic so forgive the poor terminology. An infinite tetriation with exponent (1/2) could be written as a recurrence relation, I believe: F(n) = square_root( F(n-1) ) F(0) =.5 I'm trying to locate the mathematical operator that describes something very similar, but just multiplying a negative through each iteration (see example). F(n) = square_root( -1 * F(n-1) ) Please let me know if you could help me out with this. I think this has some very interesting properties. Any guidance would be appreciated. Thanks. -Ryan G. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/27/2008, 08:09 AM (This post was last modified: 05/29/2008, 01:57 PM by bo198214.) rsgerard Wrote:Daniel Geisler referred me to this forum, and I'm surprised how active a forum is on a topic like tetration Quote:I have a question about an operator that is similar to a tetration that I'm trying to locate. I'm a bit new to this topic so forgive the poor terminology. An infinite tetriation with exponent (1/2) could be written as a recurrence relation, I believe: F(n) = square_root( F(n-1) ) F(0) =.5 Thats not infinite tetration. If you expand this, you get $F(0)=1/2$ $F(1)=(1/2)^{1/2}$ $F(2)=\left((1/2)^{1/2}\right)^{1/2}=(1/2)^{1/4}$ $F(3)=\left((1/2)^{1/4}\right)^{1/2}=(1/2)^{1/8}$ $F(n)=(1/2)^{2^{-n}}$ $\lim_{n\to\infty} F(n)=1$ I would guess what you mean is (at least this resembles infinite tetration): $F(0)=1$ $F(1)=1/2$ $F(2)=(1/2)^{1/2}$ $F(3)=(1/2)^{(1/2)^{1/2}}$ $F(n+1)=(1/2)^{F(n)}$ yes, no? Quote:I'm trying to locate the mathematical operator that describes something very similar, but just multiplying a negative through each iteration (see example). F(n) = square_root( -1 * F(n-1) ) Ok, taking your original definition a bit more generally: $F(n+1) = (c*F(n))^\alpha$ $F(1)=c^{\alpha} F(0)^{\alpha}$ $F(2)=\left(c c^{\alpha} F(0)^{\alpha}\right)^{\alpha}=c^{\alpha+\alpha^2} F(0)^{\alpha^2}$ $F(3)=\left(c c^{\alpha+\alpha^2} F(0)^{\alpha^2}\right)^\alpha= c^{\alpha+\alpha^2+\alpha^3} F(0)^{\alpha^3}$ $F(n)=c^{\alpha+\alpha^2+\dots+\alpha^n} F(0)^{\alpha^n}$ Now we know that for $0<\alpha<1$ $\lim_{n\to\infty} 1+\alpha+\alpha^2+\dots+\alpha^n = 1/(1-\alpha)$ hence for $F(0)>0$ $\lim_{n\to\infty} F(n) = c^{\alpha/(1-\alpha)}$ in our case $\alpha=1/2$ and $c=-1$ we get $\lim_{n\to\infty} F(n) = (-1)^{(1/2)/(1/2)} = -1$ however the derivation is somewhat sloppy as the exponential laws are generally not applicable for complex numbers (as painfully observed by Gottfried ). So the above should be considered only valid for $c>0$. Indeed one sees that the the sequence $F(n)$ has no limit, but oscillates between two values, i.e. has two limit points. The task for the reader is to determine these both points rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 05/28/2008, 07:05 PM bo198214, Quote:Thats not infinite tetration. If you expand this, you get $F(0)=1/2$ $F(1)=(1/2)^{1/2}$ $F(2)=\left((1/2)^{1/2}\right)^{1/2}=(1/2)^{1/4}$ $F(3)=\left((1/2)^{1/4}\right)^{1/2}=(1/2)^{1/8}$ $F(n)=(1/2)^{2^{-n}}$ $\lim_{n\to\infty} F(n)=1$ I would guess what you mean is (at least this resembles infinite tetration): $F(0)=1$ $F(1)=1/2$ $F(2)=(1/2)^(1/2)$ $F(3)=(1/2)^{(1/2)^{1/2}}$ $F(n+1)=(1/2)^{F(n)}$ yes, no? You are completely correct, the recurrence relation I suggested does not evaluate the exponent from "highest to lowest". I agree that the rewritten expression accurately describes a tetration. Quote:Ok, taking your original definition a bit more generally: $F(n+1) = (c*F(n))^\alpha$ $F(1)=c^{\alpha} F(0)^{\alpha}$ $F(2)=\left(c c^{\alpha} F(0)^{\alpha}\right)^{\alpha}=c^{\alpha+\alpha^2} F(0)^{\alpha^2}$ $F(3)=\left(c c^{\alpha+\alpha^2} F(0)^{\alpha^2}\right)^\alpha= c^{\alpha+\alpha^2+\alpha^3} F(0)^{\alpha^3}$ $F(n)=c^{\alpha+\alpha^2+\dots+\alpha^n} F(0)^{\alpha^n}$ Now we know that for $0<\alpha<1$ $\lim_{n\to\infty} 1+\alpha+\alpha^2+\dots+\alpha^n = 1/(1-\alpha)$ hence for $F(0)>0$ $\lim_{n\to\infty} F(n) = c^{\alpha/(1-\alpha)}$ in our case $\alpha=1/2$ and $c=-1$ we get $\lim_{n\to\infty} F(n) = (-1)^{(1/2)/(1/2)} = -1$ however the derivation is somewhat sloppy as the exponential laws are generally applicable for complex numbers (as painfully observed by Gottfried ). So the above should be considered only valid for $c>0$. Indeed one sees that the the sequence $F(n)$ has no limit, but oscillates between two values, i.e. has two limit points. The task for the reader is to determine these both points First, thanks a lot for expanding the math out here. This is very helpful to see. When I attempt to evaluate this recurrence, I see the result oscillate between 2 values when the principle root is used and -1 otherwise. So, I get the following solutions: $\lim_{n\to\infty} F(n) = -1, .5 \pm \sqrt{3}/2 i$ I believe these are the roots for $x^3=-1$ My two specific questions are: 1. Is this above correct? 2. Has something similar ever been shown in a more general case: For example: $F(n)=\sqrt[3]{-1 * F(n-1)}$ $\lim_{n\to\infty} F(n)$ seems to be the roots for $x^4=-1$ ??? Sorry for being a bit off topic to tetrations but this operation performed an infinite number of times relates to taking the 1/(n+1) exponent. Please let me know if this is on the right track, or have I derailed somewhere Thanks. -Ryan bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/29/2008, 03:41 PM (This post was last modified: 05/29/2008, 03:46 PM by bo198214.) rsgerard Wrote:When I attempt to evaluate this recurrence, I see the result oscillate between 2 values when the principle root is used and -1 otherwise. So, I get the following solutions: $\lim_{n\to\infty} F(n) = -1, .5 \pm \sqrt{3}/2 i$ I believe these are the roots for $x^3=-1$ My two specific questions are: 1. Is this above correct? 2. Has something similar ever been shown in a more general case: For example: $F(n)=\sqrt[3]{-1 * F(n-1)}$ $\lim_{n\to\infty} F(n)$ seems to be the roots for $x^4=-1$ ??? First the roots of $x^n=-1=e^{i\pi+2\pi i k}$ are $e^{\frac{2 \pi i }{n}k+i\frac{\pi}{n}}=\cos\left(k\frac{2\pi}{n}+\frac{\pi}{n}\right)+i\sin\left(k\frac{2\pi}{n}+\frac{\pi}{n}\right)$, $k=0\dots n-1$. So the roots of $x^3=-1$ are at $\alpha=\frac{\pi}{3}$, $\frac{2\pi}{3}+\frac{\pi}{3}=\pi$ and $\frac{4\pi}{3}+\frac{\pi}{3}=\frac{5\pi}{3}\equiv -\frac{\pi}{3}$: If we look up the values $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$, $\cos(\frac{\pi}{3})=\frac{1}{2}$ we indeed get that the roots of $x^3=-1$ are $\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$ and $-1$. To tackle the general question we first have to know how a root at the complex domain is defined. More precisely which root $v$ is chosen from the $n$ possible roots of $v^n=w$. The general definition of the power in the complex domain is: $z^\alpha = e^{\log(z)\alpha}$, where $\log(z)$ is the standard branch of the logarithm, i.e. is chosen such that the $-\pi< \Im(\log(z))\le\pi$. The oddity is that $\log(z)$ has a jump (not continuous) at the negative real axis. If we approach -1 from the upper plane we get $\log(z)\to i\pi$ and if we approach -1 from the lower plane we get $\log(z)\to -i\pi$. Both values are $2\pi i$ apart and if we repeat winding around 0 we get all the other branches of the logarithm $\log(z)+2\pi i k$. So how applies this to the case of roots? $z^{1/n} = e^{\log(z)/n} = e^{(\ln(|z|)+i\arg(z))/n}=\sqrt[n]{|z|}e^{i\frac{\arg(z)}{n}}$, where again $-\pi< \arg(z)\le \pi$ is chosen. Taking the n-th root divides the argument/angle by n, in the way the angle is chosen it moves the point towards the positive real axis, like a scissor. If we however consider $(-z)^{1/n}$ (for simplicity for $|z|=1$) we first mirror $z$ at 0 and then divide the angle by $n$ towards the positive real axis. Mirroring at 0 means either to add $\pi$ for $-\pi<\arg(z)\le 0$ or to subtract $\pi$ for $0<\arg(z)\le \pi$. Say we start with a value $z_0$ in the upper halfplane, i.e. $0< \arg(z) <\pi$, then $z_1=\sqrt[n]{-z}$, is a point in the lower halfplane, $z_2$ is again in the upper halfplane and so on: $0<\alpha_0<\pi$ $\alpha_1=\frac{\alpha_0-\pi}{n}$ $\alpha_2=\frac{\alpha_1+\pi}{n}=\frac{\frac{\alpha_0-\pi}{n}+\pi}{n}$ $\alpha_3=\frac{\alpha_2-\pi}{n}=\frac{\frac{\alpha_1+\pi}{n}-\pi}{n}$ Generally $\alpha_{2m} = \frac{\frac{\alpha_{2(m-1)}-\pi}{n}+\pi}{n}$ with $0<\alpha_{2m}<\pi$ and $\alpha_{2m+1} = \frac{\frac{\alpha_{2(m-1)+1}+\pi}{n}-\pi}{n}$ with $-\pi<\alpha_{2m+1}<0$. We dont know yet whether the sequences $\alpha_{2m}$ and $\alpha_{2m+1}$ have a limit, but if they have then the limit $\alpha$ for $\alpha_{2m}$ must satisfy: $\alpha=\frac{\frac{\alpha-\pi}{n}+\pi}{n}$ hence $n^2\alpha = \alpha - \pi + \pi n$ $(n^2-1)\alpha=(n-1)\pi$, via $n^2-1=(n+1)(n-1)$: $\alpha=\frac{\pi}{n+1}$ Similarly the limit $\beta$ of $\alpha_{2m+1}$ would be $\beta = -\frac{\pi}{n+1}$. To be really complete one have to show that $\alpha_{2m}$ is strictly increasing for a starting value $\alpha_0<\frac{\pi}{n+1}$ and striclty decreasing for a starting value $\alpha_0>\frac{\pi}{n+1}$, because then the existence of the limit is guarantied. Ok, conclusion, if we compare above the roots of $-1$ we see that the two limit points of $z_{m+1}=\sqrt[n]{-z_m}$ are exactly the both roots of $z^{n+1}=-1$ that are nearest the positive real axis ($k=0$ and $k=n-1$). (That $|z_m|\to 1$ was also clear.) rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 05/29/2008, 10:34 PM bo198214, Thanks for going above and beyond and actually going through the details on how this function behaves. I find it particularly interesting that an infinite number of square roots can relate to the cube root function and so on. Any suggestions on where to read up on this more? I really haven't found too many places to learn about sequences with roots. Also, have you come across this particular sequence in any of your other work? Thanks again for the thorough response. -Ryan bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/30/2008, 06:11 AM rsgerard Wrote:I find it particularly interesting that an infinite number of square roots can relate to the cube root function and so on.Yes, thats really surprising. Quote:Any suggestions on where to read up on this more? I really haven't found too many places to learn about sequences with roots. Also, have you come across this particular sequence in any of your other work? No, sorry, I dont know about any references. Though I think there must be some, perhaps in a more general context. « Next Oldest | Next Newest »

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