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 Operator similar to tetration 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.) « Next Oldest | Next Newest »

 Messages In This Thread Operator similar to tetration - by rsgerard - 05/27/2008, 03:57 AM RE: Operator similar to tetration - by bo198214 - 05/27/2008, 08:09 AM RE: Operator similar to tetration - by rsgerard - 05/28/2008, 07:05 PM RE: Operator similar to tetration - by bo198214 - 05/29/2008, 03:41 PM RE: Operator similar to tetration - by rsgerard - 05/29/2008, 10:34 PM RE: Operator similar to tetration - by bo198214 - 05/30/2008, 06:11 AM

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