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Consider the sequence $x_{n+1} = 2x_{n}-\frac{1}{x_n},n\geq0 $.

  1. For $x_{0} = 0,87$ we have

$$ \begin{aligned} X(1) &\approx 0,590574712643678 \\ X(2) &\approx -0,512116436915835\\ X(3) &\approx 0,928448055572567\\ X(4) &\approx 0,779829931731029\\ X(5) &\approx 0,277328986805082\\ X(6)&\approx -3,05116776974367\\ X(7)&\approx -5,77459217187698\\ X(8)&\approx -11,37601194123\\ X(9)&\approx-22,6641196146756\\ X(10)&\approx -45,2841166242379\\ X(11)&\approx -90,5461504504763\\ X(12)&\approx -181,081256809127\\ X(13)&\approx -362,156991235554\\ X(14)&\approx -724,311221237654\\ X(15)&\approx -1448,62106185332\\ X(16)&\approx -2897,24143339498\\ X(17)&\approx -5794,48252163406\\ X(18)&\approx -11588,9648706901\\ X(19)&\approx -23177,9296550913\\ X(20)&\approx -46355,8592670381\\ X(21)&\approx -92711,718512504\\\ X(22) &\approx -185423,437014222\\ X(23) &\approx -370846,874023051\\ X(24) &\approx -741693,748043405\\ X(25) &\approx -1483387,49608546 \end{aligned} \begin{aligned} X(26) &\approx -2966774,99217025\\ X(27) &\approx -5933549,98434016\\ X(28) &\approx -11867099,9686802\\ X(29) &\approx -23734199,9373602\\ X(30) &\approx -47468399,8747204\\ X(31) &\approx -94936799,7494408\\ X(32) &\approx -189873599,498882\\ X(33) &\approx -379747198,997763\\ X(34) &\approx -759494397,995526\\ X(35) &\approx -1518988795,99105\\ X(36) &\approx -3037977591,98211\\ X(37) &\approx -6075955183,96421\\ X(38) &\approx -12151910367,9284\\ X(39) &\approx -24303820735,8568\\ X(40) &\approx -48607641471,7137\\ X(41) &\approx -97215282943,4274\\ X(42) &\approx -194430565886,855\\ X(43) &\approx -388861131773,709\\ X(44) &\approx -777722263547,419\\ X(45) &\approx -1555444527094,84\\ X(46) &\approx -3110889054189,68\\ X(47) &\approx -6221778108379,35\\ X(48) &\approx -12443556216758,7\\ X(49) &\approx -24887112433517,4\\ X(50) &\approx -49774224867034,8 \end{aligned} $$

and for $x_{0} = 0,88$ we have

$$ \begin{aligned} X(1) &\approx 0,623636363636364\\ X(2) &\approx -0,356225815001326\\ X(3) &\approx 2,09475648880333\\ X(4) &\approx 3,71213052034011\\ X(5) &\approx 7,15487396107172\\ X(6) &\approx 14,1699830562016\\ X(7) &\approx 28,2693943978123\\ X(8) &\approx 56,503414850717\\ X(9) &\approx 112,989131656057\\ X(10) &\approx 225,969412903359\\ X(11) &\approx 451,934400429021\\ X(12) &\approx 903,866588147526\\ X(13) &\approx 1807,73206993709\\ X(14) &\approx 3615,46358669485\\ X(15) &\approx 7230,9268968\\ X(16) &\approx 14461,8536553051\\ X(17) &\approx 28923,7072414629\\ X(18) &\approx 57847,414448352\\ X(19) &\approx 115694,828879417\\ X(20) &\approx 231389,657750191\\ X(21) &\approx 462779,31549606\\ X(22) &\approx 925558,630989959\\ X(23) &\approx 1851117,26197884\\ X(24) &\approx 3702234,52395714\\ X(25) &\approx 7404469,047914 \end{aligned} \begin{aligned} X(26) &\approx 14808938,0958279\\ X(27) &\approx 29617876,1916557\\ X(28) &\approx 59235752,3833113\\ X(29) &\approx 118471504,766623\\ X(30) &\approx 236943009,533245\\ X(31) &\approx 473886019,06649\\ X(32) &\approx 947772038,13298\\ X(33) &\approx 1895544076,26596\\ X(34) &\approx 3791088152,53192\\ X(35) &\approx 7582176305,06384\\ X(36) &\approx 15164352610,1277\\ X(37) &\approx 30328705220,2554\\ X(38) &\approx 60657410440,5108\\ X(39) &\approx 121314820881,022\\ X(40) &\approx 242629641762,043\\ X(41) &\approx 485259283524,086\\ X(42) &\approx 970518567048,172\\ X(43) &\approx 1941037134096,34\\ X(44) &\approx 3882074268192,69\\ X(45) &\approx 7764148536385,38\\ X(46) &\approx 15528297072770,8\\ X(47) &\approx 31056594145541,5\\ X(48) &\approx 62113188291083\\ X(49) &\approx 124226376582166\\ X(50) &\approx 248452753164332 \end{aligned} $$

What explains this?

Also, is it possible to determine the number $q$ located between $0.87$ and $0.88$ in the radical change of behavior occurs this sequence?

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    $\begingroup$ what is there to explain ? $\endgroup$ – mercio May 4 '13 at 13:40
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    $\begingroup$ This is called chaos, or The Butterfly Effect. $\endgroup$ – Ma Ming May 4 '13 at 13:42
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    $\begingroup$ sensitive dependence on initial conditions. $\endgroup$ – Cheerful Parsnip May 4 '13 at 13:45
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    $\begingroup$ DO NOT close this question. I think this is a valid question. $\endgroup$ – user17762 May 4 '13 at 14:37
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    $\begingroup$ Speaking as someone who voted to close this question, I would vote to reopen it if the OP edited it to be explicit about what s/he considered bizarre, rather than making us all guess. $\endgroup$ – Micah May 4 '13 at 14:51
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The answers of mercio and Mark Bennet explain how to find the antecedents of $1$, $0$, and $-1$. This answer may help to visualize the structure of the set.

animated schematic

The horizontal axis represents $x_0$. The black horizontal line segments at height $1$ indicate intervals on which the limiting value of $x_n$ is $+\infty,$ while the black horizontal line segments at height $-1$ indicate intervals on which the limiting value of $x_n$ is $-\infty.$ The colored dots at heights $\pm1$ indicate values of $x_0$ for which the fixed points $\pm1$ are reached, while the colored dots on the horizontal axis represent values of $x_0$ for which the fixed point $\infty$ is reached (by first reaching $0$). Redder colors indicate fixed points that are reached in a small number of iterations; bluer colors indicate fixed points that are reached in a larger number of iterations. The image is animated. If the animation has stopped, it may be necessary to reload the page.

Some data:

$x_0$ reaching fixed points $(1,\infty,-1)$ after $1$ step: $$\left( \begin{array}{ccc} -0.5 & 0. & 0.5 \\ \end{array} \right)$$

$x_0$ reaching fixed points $(1,\infty,-1)$ after $2$ steps: $$\left( \begin{array}{ccc} -0.84307 & -0.707107 & -0.59307 \\ 0.59307 & 0.707107 & 0.84307 \\ \end{array} \right)$$

$x_0$ reaching fixed points $(1,\infty,-1)$ after $3$ steps: $$\left( \begin{array}{ccc} -0.948618 & -0.905646 & -0.870752 \\ -0.574217 & -0.552092 & -0.527083 \\ 0.527083 & 0.552092 & 0.574217 \\ 0.870752 & 0.905646 & 0.948618 \\ \end{array} \right)$$

$x_0$ reaching fixed points $(1,\infty,-1)$ after $4$ steps: $$\left( \begin{array}{ccc} -0.982971 & -0.968882 & -0.957545 \\ -0.865086 & -0.858475 & -0.851051 \\ -0.587509 & -0.582428 & -0.577977 \\ -0.522169 & -0.516059 & -0.508662 \\ 0.508662 & 0.516059 & 0.522169 \\ 0.577977 & 0.582428 & 0.587509 \\ 0.851051 & 0.858475 & 0.865086 \\ 0.957545 & 0.968882 & 0.982971 \\ \end{array} \right)$$

$x_0$ reaching fixed points $(1,\infty,-1)$ after $5$ steps: $$\left( \begin{array}{ccc} -0.994334 & -0.989663 & -0.985915 \\ -0.955713 & -0.953578 & -0.951185 \\ -0.869077 & -0.86755 & -0.866214 \\ -0.849598 & -0.847795 & -0.845616 \\ -0.591285 & -0.589765 & -0.588514 \\ -0.577225 & -0.576336 & -0.575323 \\ -0.52566 & -0.524341 & -0.52317 \\ -0.507143 & -0.505222 & -0.502849 \\ 0.502849 & 0.505222 & 0.507143 \\ 0.52317 & 0.524341 & 0.52566 \\ 0.575323 & 0.576336 & 0.577225 \\ 0.588514 & 0.589765 & 0.591285 \\ 0.845616 & 0.847795 & 0.849598 \\ 0.866214 & 0.86755 & 0.869077 \\ 0.951185 & 0.953578 & 0.955713 \\ 0.985915 & 0.989663 & 0.994334 \\ \end{array} \right)$$

Explanation: Taking the first row of the last table, $$\begin{pmatrix} -0.994334 & -0.989663 & -0.985915\end{pmatrix},$$ as an example,

  • $x_0=-0.994334$ implies $x_n=1$ for $n\ge5;$
  • $x_0=-0.989663$ implies $x_n=\infty$ for $n\ge5$ since $x_4=0;$
  • $x_0=-0.985915$ implies $x_n=-1$ for $n\ge5;$
  • for $x_0\in(-0.994334,-0.989663),$ we have $\displaystyle\lim_{n\to\infty}x_n=+\infty;$
  • for $x_0\in(-0.989663,-0.985915),$ we have $\displaystyle\lim_{n\to\infty}x_n=-\infty.$

The decimal numbers in this discussion should, of course, be replaced by the algebraic numbers they approximate.

Added: Here's a better answer than the one I gave in the comments (now deleted) to your question about how one knows that the sequence converges to $+\infty$ for $x_0\in(-0.994334,-0.989663).$ In addition, I will try to clarify what $+\infty$ and $-\infty$ are. I should also say that nearly everything there is to know about this problem is well-covered in mercio's answer. Since it took me a while to understand where all of his statements come from, I provide some of the details I was able to fill in, in case they're helpful to you or others.

It is a little bit neater to work with the real line extended by the point at infinity. One can do this formally by working on the projective line whose points are $(a,b),$ with $a,$ $b$ not both zero, and with the equivalence relation $(a,b)\sim(\lambda a,\lambda b)$ for any non-zero $\lambda.$ Then the real number $x$ is identified with $(x,1)$ and $\infty$ is identified with $(0,1).$ Arithmetic works as follows: $$ \begin{aligned} (a,b)\cdot(c,d)&=(ac,bd)\\ \frac{1}{(a,b)}&=(b,a)\\ (a,b)+(c,d)&=(ad+bc,bd)\\ -(a,b)&=(-a,b). \end{aligned} $$ The map $x\mapsto2x-\frac{1}{x}$ becomes $(a,b)\mapsto(2a^2-b^2,ab).$ This formulation allows us to say things like $$0\leftrightarrow(0,1)\mapsto(-1,0)\sim(1,0)\leftrightarrow\infty,$$ and $$\infty\leftrightarrow(1,0)\mapsto(2,0)\sim(1,0)\leftrightarrow\infty.$$

In this system there is no distinction between the point $\infty$ and the point $-\infty.$ Nevertheless, it turns out that if the sequence $x_0,$ $x_1,$ $x_2,\ldots$ converges to $\infty,$ then it does so either through positive values only or through negative values only. In no case does a sequence that converges to $\infty$ alternate infinitely many times between positive and negative. For this reason, it makes sense to make statements like "the sequence converges to $+\infty$" or "the sequence converges to $-\infty$" as shorthand. The only other way that sequence can converge to $\infty$ is for it to map there directly. This happens either if $x_0=\infty$ or if one of $x_0,$ $x_1,$ $x_2,\ldots$ equals $0.$ From now on, I will drop the formal language of ordered pairs. I hope it will be clear what I mean when I refer to $\infty,$ $+\infty,$ and $-\infty.$

To verify the statement above that a sequence converging to $\infty$ does so either through positive values only or through negative values only, compute the first and second differences: $$\begin{aligned}&x_{n+1}-x_n=x_n-\frac{1}{x_n}\\ &(x_{n+2}-x_{n+1})-(x_{n+1}-x_n)=x_{n+1}-\frac{1}{x_{n+1}}-x_n+\frac{1}{x_n}=x_n-\frac{1}{x_{n+1}}. \end{aligned}$$ If $x_n>1,$ the first and second differences are both positive, so the sequence $x_n,$ $x_{n+1},$ $x_{n+2},\ldots$ is monotonically increasing towards $+\infty.$ Since the map $x\mapsto2x-\frac{1}{x}$ is an odd function, we also conclude that if $x_n<1$ then $x_n,$ $x_{n+1},$ $x_{n+2},\ldots$ is monotonically decreasing towards $-\infty.$

Regarding the projective line as a circle, the map $f:x\mapsto2x-\frac{1}{x}$ is continuous and covers the circle exactly twice. Specifically, $f$ is increasing on each of the intervals $(-\infty,0)$ and $(0,+\infty)$ and maps both of them onto the interval $(-\infty,+\infty)$.

To understand the full set of points $x_0$ that generate sequences converging to $\infty,$ we determine the set of points $S_n$ that have not reached $(1,+\infty)\cup\{\infty\}\cup(-\infty,-1)$ after $n$ iterations of $f,$ with $n=0,1,2,\ldots.$ Equivalently, we determine the set of points $x_0$ such that $x_n\in[-1,1].$ Clearly $$S_0=[-1,1].$$ Given any point $x,$ the preimages of $x$ are $$ \begin{aligned} p_-(x)&=\begin{cases}\frac{1}{4}\left(x-\sqrt{x^2+8}\right) & \text{$x$ real,}\\ 0 & x=\infty,\end{cases}\\ p_+(x)&=\begin{cases}\frac{1}{4}\left(x+\sqrt{x^2+8}\right) & \text{$x$ real,}\\ \infty & x=\infty.\end{cases} \end{aligned} $$ Observe that $p_-$ is increasing and maps $(-\infty,\infty)$ onto $(-\infty,0),$ and that $p_+$ is increasing and maps $(-\infty,\infty)$ onto $(0,\infty).$ The image of the interval $[-1,1]$ under $p_-$ is $\left[-1,-\frac{1}{2}\right].$ (This image lies within $[-1,1],$ as it must, since, under the map $f,$ a point outside of $[-1,1]$ maps to a point outside of $[-1,1].$) Similarly, the image of the interval $[-1,1]$ under $p_+$ is $\left[\frac{1}{2},1\right].$ Therefore $$ S_1=p_-(S_0)\cup p_+(S_0)=\left[-1,-\frac{1}{2}\right]\cup\left[\frac{1}{2},1\right]. $$ In a similar manner, $$ \begin{aligned} S_2=&p_-(S_1)\cup p_+(S_1)\\ =&p_-\left(\left[-1,-\frac{1}{2}\right]\right)\cup p_-\left(\left[\frac{1}{2},1\right]\right)\cup p_+\left(\left[-1,-\frac{1}{2}\right]\right)\cup p_+\left(\left[\frac{1}{2},1\right]\right)\\ =&[-1,-0.84307]\cup[-0.59307,-0.5]\cup[0.5,0.59307]\cup[0.84307,1]. \end{aligned} $$ The approximations $0.59307$ and $0.84307$ actually stand for the algebraic numbers $$\frac{1}{8}\left(-1+\sqrt{33}\right)\approx0.59307,\qquad\frac{1}{8}\left(1+\sqrt{33}\right)\approx0.84307.$$ Continuing in the same way, applying $p_-$ and $p_+$ to each of the four intervals making up $S_2,$ we get $$ \begin{aligned} S_3=&[-1,-0.94862]\cup[-0.87075,-0.84307]\cup[-0.59307,-0.57422]\cup[-0.52708,-0.5]\\ &\cup[0.5,0.52708]\cup[0.57422,0.59307]\cup[0.84307,0.87075]\cup[0.94862,1]. \end{aligned} $$ The intervals with negative endpoints come from the application of $p_-,$ while those with positive endpoints come from the application of $p_+.$ Likewise, $$ \begin{aligned} S_4=&[-1, -0.98297]\cup[-0.95754, -0.94862]\\ &\cup[-0.87075, -0.86509]\cup[-0.85105, -0.84307]\\ &\cup[-0.59307, -0.58751]\cup[-0.57798, -0.57422]\\ &\cup[-0.52708, -0.52217]\cup[-0.50866, -0.5]\\ &\cup[0.5, 0.50866]\cup[0.52217,0.52708]\\ &\cup[0.57422, 0.57798]\cup[0.58751, 0.59307]\\ &\cup[0.84307, 0.85105]\cup[0.86509, 0.87075]\\ &\cup[0.94862, 0.95754]\cup[0.98297, 1] \end{aligned} $$ and $$ \begin{aligned} S_5=&[-1, -0.99433]\cup[-0.98592, -0.98297]\\ &\cup[-0.95754, -0.95571]\cup[-0.95119, -0.94862]\\ &\cup[-0.87075, -0.86908]\cup[-0.86621, -0.86509]\\ &\cup[-0.85105, -0.84960]\cup[-0.84562, -0.84307]\\ &\cup[-0.59307, -0.59128]\cup[-0.58851, -0.58751]\\ &\cup[-0.57798, -0.57722]\cup[-0.57532, -0.57422]\\ &\cup[-0.52708, -0.52566]\cup[-0.52317, -0.52217]\\ &\cup[-0.50866, -0.50714]\cup[-0.50285, -0.5]\\ &\cup[0.5, 0.50285]\cup[0.50714, 0.50866]\\ &\cup[0.52217, 0.52317]\cup[0.52566, 0.52708]\\ &\cup[0.57422, 0.57532]\cup[0.57722, 0.57798]\\ &\cup[0.58751, 0.58851]\cup[0.59128, 0.59307]\\ &\cup[0.84307, 0.84562]\cup[0.84960, 0.85105]\\ &\cup[0.86509, 0.86621]\cup[0.86908, 0.87075]\\ &\cup[0.94862, 0.95119]\cup[0.95571, 0.95754]\\ &\cup[0.98297, 0.98592]\cup[0.99433, 1]. \end{aligned} $$

To make the foregoing more explicit, we parameterize intervals and their endpoints by sign sequences. Let $\sigma_0\sigma_1\sigma_2,\ldots$ be a sequence of signs: $\sigma_j\in\{-,+\},$ $j\in\{0,1,2,\ldots\}.$ Define $$ \begin{aligned} L()&=-1\\ R()&=1\\ I()&=[L(),R()]=[-1,1], \end{aligned} $$ where the empty argument denotes the zero length sequence; for $\ell>0,$ define $$ \begin{aligned} L\left(\sigma_0\sigma_1\ldots\sigma_{\ell-1}\right)&=p_{\sigma_0}\left(L\left(\sigma_1\sigma_2\ldots\sigma_{\ell-1}\right)\right)\\ R\left(\sigma_0\sigma_1\ldots\sigma_{\ell-1}\right)&=p_{\sigma_0}\left(R\left(\sigma_1\sigma_2\ldots\sigma_{\ell-1}\right)\right)\\ I\left(\sigma_0\sigma_1\ldots\sigma_{\ell-1}\right)&=\left[L\left(\sigma_0\sigma_1\ldots\sigma_{\ell-1}\right),R\left(\sigma_0\sigma_1\ldots\sigma_{\ell-1}\right)\right]. \end{aligned} $$ Let $Q_\ell=\{\sigma=\sigma_0\sigma_1\ldots\sigma_{\ell-1}\vert \sigma_j\in\{-,+\},\ j\in\{0,1,\ldots,\ell-1\}\}$ be the set of all sign sequences of length $\ell.$ Then $$ S_\ell=\bigcup_{\sigma\in Q_\ell} I\left(\sigma_0\sigma_1\ldots\sigma_{\ell-1}\right). $$ Observe that if sign sequences, $\sigma=\sigma_0\sigma_1\ldots\sigma_{\ell-1}$ and $\tau=\tau_0\tau_1\ldots\tau_{\ell-1}$ are compared lexicographically, then for $\sigma<\tau,$ we have $L(\sigma)<L(\tau)$ and $R(\sigma)<R(\tau).$ Furthermore, we have $L(\sigma)<R(\sigma).$ These statements are proved inductively using the property that $p_-$ and $p_+$ are increasing and $L()<R().$

Note that the interval $S_0=I$ is of length $2,$ and that the sum of the lengths of the two intervals $I_-$ and $I_+$ making up $S_1$ is $1.$ Moreover, the sum of the lengths of the intervals making up $S_n$ is $2^{1-n}.$ That is, the total length of $S_\ell$ is halved with every iteration. Turning things around, the total length of the intervals containing points in $[-1,1]$ that wind up in $(1,\infty)\cup\{\infty\}\cup(-\infty,-1)$ after $n$ applications of $f$ is $2\left(1-2^{-n}\right),$ which means that almost all points in $[-1,1]$ converge to $\infty.$

To see why this is true, observe that each set $S_n,$ $n>0,$ is the union of pairs of intervals of the form $[-b,-a]\cup[a,b]$ where $b>a>0.$ Specifically, such a pair is formed by a pair of complementary sign sequences: $$ a=L(\sigma),\quad b=R(\sigma),\quad -b=L(\overline\sigma),\quad -a=R(\overline\sigma), $$ with $\sigma_0=+$ and where $\overline\sigma$ denotes the sequence obtained by negating every element of $\sigma.$ Under $p_-,$ this pair maps to $$ \left[\frac{1}{4}\left(-b-\sqrt{b^2+8}\right),\frac{1}{4}\left(-a-\sqrt{a^2+8}\right)\right]\cup\left[\frac{1}{4}\left(a-\sqrt{a^2+8}\right),\frac{1}{4}\left(b-\sqrt{b^2+8}\right)\right], $$ which has total length $\frac{1}{2}(b-a).$ Similarly, the image of the pair under $p_+$ has length $\frac{1}{2}(b-a).$ Since $S_{n+1}$ is the union of the images of $S_n$ under $p_-$ and $p_+,$ the pair $[-b,-a]\cup[a,b],$ of total length $2(b-a),$ maps to intervals of total length $b-a$ in $S_{n+1}.$

We have understood the process of producing $S_{n+1}$ from $S_n$ as one of applying the maps $p_-$ and $p_+$ to $S_n,$ and then taking the union of the results. It can also be understood as a process of deleting an open interval from the middle region of each of the closed intervals making up $S_n.$ Observe that, for $\sigma=\sigma_0\sigma_1\ldots\sigma_{n-1},$ we have $$ \begin{aligned} L(\sigma--\ldots-)&=L(\sigma),\\ R(\sigma++\ldots+)&=R(\sigma). \end{aligned} $$ The first of these follows by noting that $$L(\sigma)=p_{\sigma_0}(p_{\sigma_1}(\ldots p_{\sigma_{n-1}}(-1)\ldots)),$$ and then observing that $p_-(-1)=-1.$ The second follows similarly from $$R(\sigma)=p_{\sigma_0}(p_{\sigma_1}(\ldots p_{\sigma_{n-1}}(1)\ldots))$$ and $p_+(1)=1.$ Therefore, the interval $I(\sigma)=[L(\sigma),R(\sigma)]$ in $S_n$ becomes two intervals $$ I(\sigma-)\cup I(\sigma+)=[L(\sigma-),R(\sigma-)]\cup[L(\sigma+),R(\sigma+)] $$ in $S_{n+1}.$ The outermost endpoints stay the same: $L(\sigma-)=L(\sigma),$ $R(\sigma+)=R(\sigma).$ Two new interior endpoints are created by removing the open interval $(R(\sigma-),L(\sigma+)).$ This viewpoint of the process of creating $S_{n+1}$ from $S_n$ makes it clear that $S_{n+1}\subset S_n.$

The points in the removed open interval, $(R(\sigma-),L(\sigma+)),$ are points that map outside of $[-1,1]$ under $f^{\circ(n+1)}.$ There is some point $B(\sigma-)\in(R(\sigma-),L(\sigma+))$ such that $f^{\circ(n+1)}$ maps $(R(\sigma-),B(\sigma-))$ to $(1,\infty)$ and $(B(\sigma-),L(\sigma+))$ to $(-\infty,-1).$ The point $B(\sigma)$ is the preimage of $\infty,$ defined by $$ \begin{aligned} B()&=\infty,\\ B\left(\sigma_0\sigma_1\ldots\sigma_{\ell-1}\right)&=p_{\sigma_0}\left(B\left(\sigma_1\sigma_2\ldots\sigma_{\ell-1}\right)\right). \end{aligned} $$ We can therefore say that points in $(-0.994334,-0.989663)$ converge to $+\infty$ because $$ -0.994334\approx R(-----),\quad -0.989663\approx B(-----), $$ and hence $f^{\circ5}$ maps points in the interval to $(1,\infty).$ Likewise, points in $(-0.989663,-0.985915)$ converge to $-\infty$ because $-0.985915\approx L(----+)$ and hence $f^{\circ5}$ maps point in this interval to $(-\infty,-1).$

The set of real numbers that do not converge to $\infty$ is $$ S=S_0\cap S_1\cap S_2\cap\ldots. $$ In comments to his answer, mercio states that there is a one-to-one correspondence between points of $S$ and semi-infinite sign sequences $\sigma=\sigma_0\sigma_1\sigma_2\ldots.$ To see why this is so, observe that $$ I(\sigma_0)\supset I(\sigma_0\sigma_1)\supset I(\sigma_0\sigma_1\sigma_2)\supset\ldots. $$ Hence the sequence of intervals associated with finite initial subsequences of $\sigma$ is a set of nested intervals. By Cantor's Intersection Theorem, the intersection of a set of nested closed intervals whose diameter tends to $0$ contains a single point. Call this point $x(\sigma).$

The image of the interval $I(\sigma_0\sigma_1\ldots\sigma_{\ell-1})$ under $f$ is $I(\sigma_1\sigma_2\ldots\sigma_{\ell-1})$ since $f(p_{\sigma_0}(x))=x.$ It follows that $$f(x(\sigma_0\sigma_1\sigma_2\ldots))=x(\sigma_1\sigma_2\sigma_3\ldots).$$ This implies that sign sequences with period $\ell$ are associated with points that have orbits of period $\ell$ under $f.$ (Proof: Let $\sigma$ have period $\ell:$ $\sigma=\sigma_0\sigma_1\ldots\sigma_{\ell-1}\sigma.$ Then $$f^{\circ\ell}(x(\sigma))=f^{\circ\ell}(x(\sigma_0\sigma_1\ldots\sigma_{\ell-1}\sigma)=x(\sigma).$$ Furthermore, $f^{\circ j}(x(\sigma))\ne x(\sigma)$ for $j<\ell$ since the map $\sigma\mapsto x(\sigma)$ is one-to-one.)

You ask whether there are non-periodic orbits that do not converge to $\infty.$ The comments of mercio imply yes. In fact, any non-periodic sign sequence $\sigma$ produces an $x(\sigma)$ that has such a non-periodic orbit.

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  • $\begingroup$ How can you justify that to the limit of the sequence $x_0\in (-0,994334,-0,989663)$ is infinite. What is $+\infty$ and it is $-\infty$? $\endgroup$ – medicu Aug 31 '14 at 6:22
  • $\begingroup$ My previous comments in response to your question were misleading, so I've deleted them, and expanded my answer instead. In particular, there was nothing non-rigourous about the statement that points in $(-0.994334,-0.989663)$ converge to $+\infty$ as I had stated in my comment. (I'd forgotten how I carried out the original calculations back in May 2013. $\endgroup$ – Will Orrick Sep 11 '14 at 17:00
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Extending the function $f : x \mapsto 2x- \frac 1x$ to the circle $\Bbb R \cup \{\infty\}$ by defining $f(0) = f(\infty) = \infty$, $f$ is a $2$-to-$1$ function with $3$ fixpoints, one attractive ($\infty$), and two repulsive ($1,-1$).

So, for most starting reals, the generated sequence will "converge" to $\infty$. Since $f$ is continuous, the set of reals that generate a sequence diverging to $+ \infty$ is an open subset, as well as the set of reals that generate a sequence diverging to $- \infty$.

those two open sets "come in contact" at the real numbers that generate a sequence eventually going to $0$ and then that stays on $\infty$. Here, you can plainly see that there is an $x \in (0.87 ; 0.88)$ such that $f^{6}(x) = 0$.

To find the precise value of that $x$, you need to compute the successive antecedents of $0$ by $f$ that lay in-between the two sequences.

$f^{-1}(y) = \frac{y \pm \sqrt {y^2+8}}4$, so the first one is $\frac {\sqrt 2}2$ (betwwen the two $X(5)$ values), then $\frac {\sqrt 2 + \sqrt {34}}8$ (between the two $X(4)$ values), and you can go on until you reach the incriminating $x \in (0.87;0.88)$.

Also, the set of all the antecedents of $0$ has accumulation points (starting with $1$ and $-1$, but they could be more), so it's very possible to find many incriminating values between the two initial values, by going back long enough in the antecedents tree.

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  • $\begingroup$ Indeed, there are many incriminating values between $0.87$ and $0.88$. They all seem to be located between $0.8702$ and $0.8708$. (The point with $f^6(x) = 0$ is $x \approx 0.870427$.) $\endgroup$ – TMM May 4 '13 at 17:36
  • $\begingroup$ @TMM : I think the closure of the antecedents of $0$ (or any $x \in [-1;1]$) is a Cantor set (independent of $x$). $\endgroup$ – mercio May 4 '13 at 17:48
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    $\begingroup$ @TMM : $x_2=-1$ for $x_0=(1+\sqrt{33})/8\approx0.84307,$ while $x_3=1$ for $x_0=\left(\sqrt{2(273-\sqrt{33})}+\sqrt{33}-1\right)/32\approx0.870752.$ The limiting value of $x_n$ changes from $+\infty$ to $-\infty$ infinitely often between these two values. The least critical value greater than $0.87$ is $x_0\approx0.870214,$ at which point $x_6=-1,$ so one can also say that the limiting value of $x_n$ changes from $+\infty$ to $-\infty$ infinitely often between $\approx0.870214$ and $\approx0.870752.$ $\endgroup$ – Will Orrick May 5 '13 at 10:59
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    $\begingroup$ The set of starting points that don't generate a sequence that converge to $\infty$ is a Cantor subset $S$ of $[-1;1]$. $S$ has uncountably many points and is fixed by $f$. For $x \in S$ you can attach a sequence of signs $\phi(x) = (sign(f^{\circ n}(x)))_{n \in \Bbb N}$, and $\phi$ is actually a bijection from $S$ to $\{+,-\}^\Bbb N$ (which can be made continuous with the adequate topology on those sequences). We clearly have $\phi \circ f \circ \phi^{-1} ((s_n)_{n \in \Bbb N}) = (s_{n+1})_{n \in \Bbb N}$. $\endgroup$ – mercio Aug 31 '14 at 14:07
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    $\begingroup$ Now via this $\phi$ the points in the Cantor set that generate a periodic sequence correspond exactly to the periodic sign sequences. There are a lot of them, and none of them is attractive. $\endgroup$ – mercio Aug 31 '14 at 14:07
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The recurrence has fixed points at 1 and -1.

If $x_n \lt -1$ then the sequence becomes increasingly negative.

If $x_n \gt 1$ then the sequence becomes increasingly positive.

The sequence crashes if $x_n=0$

The behaviour outside the interval $(-1,1)$ is clear, so we don't need long strings of values to explain it.

The way to understand what is going on is to see that the interval $(-1,1)$ can be separated into subintervals in which the values exhibit the same behaviour. The behaviour changes when we pass through a critical point - and these are just the points at which the sequence eventually terminates at $0$ or $\pm1$.

So, for example, if $x_n \in (-0.5,0)$ we find that $x_{n+1} \gt 1$ and $x_n \in (0,0.5)$ gives $x_{n+1} \lt -1$. The behaviour changes as we pass through $0$.

We note that if $x_n=\pm0.5$ then $x_{n+1}=\mp1$, so these are critical points. Also if $x_n=\pm \frac {\sqrt 2} 2$ then $x_{n+1}=0$.

If $x=a$ is a critical point, we get the next one(s) - the point(s) which map onto $a$ - by solving $2x-\cfrac 1 x=a$, this is equivalent to $$2x^2-ax-1=0$$ with solutions $$\frac {a\pm\sqrt{a^2+8}}4$$

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  • $\begingroup$ Thank you! I noticed the same thing myself. I have tried to describe the set of "bad" but I could not. $\endgroup$ – medicu May 4 '13 at 17:10
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If you make the transformation $y_n=\sqrt{2}x_n$, then the equation is transformed to $$y_{n+1}=2\left(y_n-\frac{1}{y_n}\right),\ n\geq 0$$

So, basically we're talking about the following kind of map $$f(x)=\lambda\left(x+\frac{a}{x}\right),\ a,x\in \mathbb{R},\ \lambda >0$$ For $a<0$, this function has fixed points at $x={}_{-}^+\sqrt{\frac{\lambda a}{1-\lambda}},\infty$. ${}_{-}^+\sqrt{-a}$ are both attracting for $0<\lambda<1$ and repelling for $\lambda>1$. $\infty$ is attracting for $\lambda<1$. This mapping exhibits rich dynamics as good as the famous logistic map.

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