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.)