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$\begingroup$

Problem:

How we can strictly prove $(-\sqrt{2})^{(-\sqrt{2})^{(-\sqrt{2})^\ldots}}$ can't be 2?

Can $(-\sqrt{2})^{(-\sqrt{2})^{(-\sqrt{2})^\ldots}}$ have the value expressed by complex numbers? (See below, in calculation details), like $i^i=e^{-\frac{\pi }{2}+2k\pi} \quad (k \in \mathbb{Z})$?

Details:

In this my other question, Are the solutions of $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$ correct?, I show how to find the possible solutions for $x$ in the equation. $\sqrt{2}$ is one root and this is ok, but, what about $-\sqrt{2}$? Maybe it isn't a root, but how to strictly prove this? The prove given in these question (as we can read in the comments) looks wrong, so I'm searching for a correct one.

Calculation details

This was posted in the other question (as a answer, but the truth is just a huge comment, as I explained) and I think is pertinent to here too.

" Looks like $-\sqrt{2}$ isn't a solution for the equation, but I'm not sure. Looks like too, the power tower of a number should converge only on a specific interval ($[e^{−e},e^{1/e}]$).

But using Mathematica and the ProductLog function (wich the Lambert $W(z)$ function) we find some strange things:

Using $h(z)=z^{z^{z^{\ldots}}}=-\frac{W(-\log (z))}{\log (z)}$ (h[z_]:=(-ProductLog[-Log[z]])/Log[z])

Calculating the power tower to $\sqrt{2}$ we have N[h[Sqrt[2]], 10]=2.000000000

And the power tower to $-\sqrt{2}$ we have N[h[-Sqrt[2]], 10]=0.2513502988 + 0.3162499180 I

Calculating explicity, by iteration

${-\sqrt{2}},{(-\sqrt{2})}^{({-\sqrt{2}})},{(-\sqrt{2})}^{({-\sqrt{2})}^{\ldots}}$ we have

Table[N[Re[PowerTower[-Sqrt[2], i]], 30] + I*N[Im[PowerTower[-Sqrt[2], i]], 5], {i, 1, 15}] // TableForm

-1.41421356237309504880168872421 -0.163093997943414854921937604558+0.59044 I  0.140921295793052749536215801866-0.044791 I  1.10008630700672531426983704055+0.50079 I -0.268168781568546776692908102136-0.14235 I  0.894980750563013739735614892750-1.1090 I -33.5835630157562847787187418023+29.118 I  6.49187847255812829134661655850*10^-46-1.5181*10^-45 I  1.00000000000000000000000000000+1.5134*10^-45 I -1.41421356237309504880168872421-2.2930*10^-44 I -0.163093997943414854921937604558+0.59044 I  0.140921295793052749536215801866-0.044791 I  1.10008630700672531426983704055+0.50079 I -0.268168781568546776692908102136-0.14235 I  0.894980750563013739735614892750-1.1090 I 

Ploting the real and imaginary part of the function $h$, we have:

To the real part:

Plot[Re[N[h[x]], {x, -2, 0}, Epilog -> {PointSize[0.01], Point[{-Sqrt[2], N[Re[h[-Sqrt[2]]]]}]}]

Real part of the tower of -Sqrt[2]

and to the imaginary part:

Plot[Im[N[h[x]], {x, -2, 0}, Epilog -> {PointSize[0.01], Point[{-Sqrt[2], N[Im[h[-Sqrt[2]]]]}]}]

Imaginary part of the tower of -Sqrt[2]

So looks like the function converges, but, unfortunally not to $2$.

"

  • 0
    @GarouDan how can we figure about some initial value of $a$ and $b$ where $a+bi$ is your $(-\sqrt{2})^{ (-\sqrt{2})^{\cdots}}$?2014-05-25

4 Answers 4

8

This is the list of the absolute values of the iterates of $\small b=-\sqrt{2} $ . Because it has a very obvious 9-periodic pattern at the beginning I separated the trajectory in vertical columns of only 9 subsequent iterates. (computed with Pari/GP, internal precision 1200 decimal digits)

$ \Tiny \begin{array} {rrrr} 1.4142136 & 1.4142136 & 1.4142136 & 1.4142136 & 1.4142136 \\ 0.61254733 & 0.61254733 & 0.61254733 & 0.61254733 & 0.61254733 \\ 0.14786831 & 0.14786831 & 0.14786831 & 0.14786831 & 0.14786831 \\ 1.2087108 & 1.2087108 & 1.2087108 & 1.2087108 & 1.2087108 \\ 0.30360689 & 0.30360689 & 0.30360689 & 0.30360689 & 0.30360689 \\ 1.4251197 & 1.4251197 & 1.4251197 & 1.4251197 & 1.4251197 \\ 44.448758 & 44.448758 & 44.448758 & 44.448758 & 44.448758 \\ 1.6510623E-45 & 1.6510623E-45 & 1.6510623E-45 & 1.6510623E-45 & 1.6510623E-45 \\ 1.0000000 & 1.0000000 & 1.0000000 & 1.0000000 & 1.0000000 \end{array} $ And here are the first-order differences between subsequent columns. We see, that the differences vanish quickly so this indicates strongly a convergence to a 9-period-fixpoint structure for the lenghtes: $ \Tiny \begin{array} {rrrr} 1.4142136 & -4.2758815E-45 & 1.5415076E-84 & -3.5904308E-124 & 7.0561983E-164 \\ 0.61254733 & 4.5034171E-44 & -6.6081983E-84 & 8.9967024E-124 & -1.1106139E-163 \\ 0.14786831 & -2.0529932E-44 & 2.8802422E-84 & -3.7054357E-124 & 4.1945575E-164 \\ 1.2087108 & -1.4713104E-44 & -9.1472657E-85 & 6.2615464E-124 & -1.6552948E-163 \\ 0.30360689 & 7.6680831E-44 & -1.1071058E-83 & 1.4777436E-123 & -1.7723010E-163 \\ 1.4251197 & 1.7072251E-43 & -3.7497694E-83 & 7.1363187E-123 & -1.2382368E-162 \\ 44.448758 & 4.3205739E-41 & -7.3721684E-81 & 1.1721448E-120 & -1.7432962E-160 \\ 1.6510623E-45 & -2.4711650E-85 & 3.4434076E-125 & -4.3898227E-165 & 4.8945927E-205 \\ 1.0000000 & -6.8798480E-85 & 8.6346349E-125 & -9.3717918E-165 & 7.3761858E-205 \end{array} $


This is the same with the arg (angle in the complex plane) of the iterates: $ \Tiny \begin{array} {rrrr} 3.1415927 & -3.1415927 & -3.1415927 & -3.1415927 & -3.1415927 \\ 1.8403024 & 1.8403024 & 1.8403024 & 1.8403024 & 1.8403024 \\ -0.30774541 & -0.30774541 & -0.30774541 & -0.30774541 & -0.30774541 \\ 0.42719397 & 0.42719397 & 0.42719397 & 0.42719397 & 0.42719397 \\ -2.6536012 & -2.6536012 & -2.6536012 & -2.6536012 & -2.6536012 \\ -0.89181076 & -0.89181076 & -0.89181076 & -0.89181076 & -0.89181076 \\ 2.4273000 & 2.4273000 & 2.4273000 & 2.4273000 & 2.4273000 \\ -1.1666935 & -1.1666935 & -1.1666935 & -1.1666935 & -1.1666935 \\ 1.5133579E-45 & 1.5133579E-45 & 1.5133579E-45 & 1.5133579E-45 & 1.5133579E-45 \end{array} $ and the differences $\Tiny \begin{array} {rrrr} 3.1415927 & -6.2831853 & -2.3079194E-84 & 3.0257404E-124 & -3.5304957E-164 \\ 1.8403024 & 5.4860495E-45 & -3.7116105E-84 & 9.7966678E-124 & -2.0437302E-163 \\ -0.30774541 & -3.3111505E-44 & 1.0414472E-83 & -2.3245643E-123 & 4.4644292E-163 \\ 0.42719397 & -6.5587742E-44 & 1.0295176E-83 & -1.5111378E-123 & 2.0580665E-163 \\ -2.6536012 & 3.4000673E-44 & -1.5018863E-83 & 3.6815608E-123 & -7.4238709E-163 \\ -0.89181076 & -2.1319630E-43 & 2.7199457E-83 & -3.0364896E-123 & 2.5759836E-163 \\ 2.4273000 & -5.1815970E-43 & 3.9336883E-83 & 6.3329419E-124 & -1.1315814E-162 \\ -1.1666935 & -3.9316194E-41 & 1.1769005E-80 & -2.5814484E-120 & 4.9090189E-160 \\ 1.5133579E-45 & -4.2285808E-85 & 9.0338626E-125 & -1.6915902E-164 & 2.9002867E-204 \end{array} $ Again we see, that they vanish quickly and since the angles and the lengthes (absolute values) seem to approximate 9 values its a good hypothese, that this is 9-periodic fixpoints.

The mean of the fixpoints is then approximately $\small m_\infty = -3.5881168+3.2125197*I $

  • 0
    @GarouDan The way I feel with this guy's post. Man, lengthy and rich.2016-12-22
3

From Maple: Let $x[1] = -\sqrt{2}$ and $x[n+1] = (-\sqrt{2})^{x[n]}$. Some numerical results using the principal value for the exponentials.

x[  1] = -1.414213562373095048801688724209698078569671875376948073176679737990732     +0.000000000000000000000000000000000000000000000000000000000000000000000 i  x[  2] = -0.163093997943414854921937604558017962499793965727869386822402042154175         +0.590435919538534820623115985186637206577707103759559066773060672459998 i  x[  3] = 0.140921295793052749536215801866147804169595017211147358356211625506612         -0.044790898341731398052941928863585335872205418214802335628821017645623 i  x[  4] = 1.100086307006725314269837040545062800076092484472530604917053566055528         +0.500791346054036352743920991367697779027338760297890701264222550336166 i  x[  5] = -0.268168781568546776692908102135608983545959166888483749205191974796848         -0.142346920553878866119517412480250679122455204231360202301605215735912 i  x[  6] = 0.894980750563013739735614892750496547408631351459698268792989443758413         -1.109042679617501408669836777920686914041610008570543149982061855745641 i  x[  7] = -33.583563015756284778718741802267673776552912769626319413100735640780956         +29.117630558936738895309311011584496358235358548332430845036010574467116 i  x[  8] = 0.000000000000000000000000000000000000000000000649187847255812829134662         -0.000000000000000000000000000000000000000000001518078385198129994507332 i  x[  9] = 1.000000000000000000000000000000000000000000004994175265501475052691346         +0.000000000000000000000000000000000000000000001513357895454095189313419 i  x[ 10] = -1.414213562373095048801688724209698078569671871101066587094934996112240         -0.000000000000000000000000000000000000000000022930276777769681861169385 i  x[ 11] = -0.163093997943414854921937604558017962499793980957618715410603002901938         +0.590435919538534820623115985186637206577707146273368937624772962839578 i 

and evidence of periodicity...

x[  1] = -1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641573         +0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 i x[ 10] = -1.414213562373095048801688724209698078569671871101066587094934996112240314574971797998806072494141208         -0.000000000000000000000000000000000000000000022930276777769681861169384867946447483376541158075718190 i x[ 19] = -1.414213562373095048801688724209698078569671871101066587094934996112240314574971798000347580082307702         -0.000000000000000000000000000000000000000000022930276777769681861169384867946447483373277267119819928 i x[ 28] = -1.414213562373095048801688724209698078569671871101066587094934996112240314574971798000347580082307702         -0.000000000000000000000000000000000000000000022930276777769681861169384867946447483373277267119819928 i 
  • 0
    `GEdgar`, I'll take a look and return. Thx.2011-12-09
3

This is more of an extended comment, but I'll leave it here and maybe flesh it out later to a full answer if I can figure out more from this.

I managed to confirm the 9-cycle for the iteration $x_{k+1}=(-\sqrt2)^{x_k}$ by using Floyd's cycle finding algorithm.

Here's a Mathematica implementation (adapted from here):

Options[CycleFixedPoint] = {SameTest -> SameQ, CycleTest -> SameQ,                              IterationCount -> False};  CycleFixedPoint[f_, start_, opts___] :=     Block[{nth, twonth, counter, result, value, sameTest, cycleTest,           iterationCount},           {sameTest, cycleTest, iterationCount} =          {SameTest, CycleTest, IterationCount} /. {opts} /.           Options[CycleFixedPoint];           nth = f[start]; twonth = f[f[start]]; counter = 1;           While[(! sameTest[nth, twonth]) && (counter < $IterationLimit),                 nth = f[nth]; twonth = f[f[twonth]]; ++counter;]           If[counter == $IterationLimit, result = nth,             (* else *)             value = f[nth];             result = Prepend[Flatten[Last[                              Reap[While[! cycleTest[value, nth],                                         Sow[value]; value = f[value];                                   ]]                              ], 1], nth]             ];           If[TrueQ[iterationCount], {result, counter}, result]          ] 

Using 120 significant digits, I executed Short[AbsoluteTiming[{cycle, count} = CycleFixedPoint[((-Sqrt[2])^#) &, N[-Sqrt[2], 120], IterationCount -> True]]] to find the 9-cycle. Here is the cycle I found to fifty digits:

-1.41421356237309504880168872420969807856967187110107  -0.16309399794341485492193760455801796249979398095762 +  0.59043591953853482062311598518663720657770714627337 i   0.14092129579305274953621580186614780416959499616264 -   0.04479089834173139805294192886358533587220541666218 i   1.10008630700672531426983704054506280007609250392744 +   0.50079134605403635274392099136769777902733868204980 i  -0.26816878156854677669290810213560898354595922977896 -   0.14234692055387886611951741248025067912245524930130 i   0.8949807505630137397356148927504965474086312222303 -   1.1090426796175014086698367779206869140416103322351 i  -33.583563015756284778718741802267673776552930326435 +   29.117630558936738895309311011584496358235404253331 i   6.491878472558128291346616558500415541158598093491 * 10^(-46) -   1.5180783851981299945073316151972156311102228546016 * 10^(-45) i   1.00000000000000000000000000000000000000000000499418 +  1.5133578954540951893134194357192492465553697834027 * 10^(-45) i 

It does look as if the eighth member isn't zero; a plot of the curves $\Re((-\sqrt 2)^{x+iy})=0$ and $\Im((-\sqrt 2)^{x+iy})=0$ in the vicinity of the seventh member of the cycle gives a strong hint:

is that a zero I see?

(An analytical proof of this will still be needed, though.)

Here are plots of the complex plane trajectory. Since the seventh member of the cycle is a bit larger in magnitude than the others, I have also thrown in a close-up:

power tower trajectory


This paper by Galidakis seems to give some conditions for a power tower iteration (see page 990 for instance), but I seem to have some difficulty applying his criterion for determining the cycle length of the power tower iteration. It involves a multivariate generalization of the Lambert function which Mathematica doesn't have, so I'll need to study this a bit more and see if Galidakis's results can be adapted for this situation.

  • 0
    This is something curious. Define the function $\small it9(x) = for(k=1,9,x=\exp(bl*x));return(x) \qquad \text{ where } bl=\log(-sqrt(2)) $. Then, using Pari/GP, one can ask for the power series of $\small it9(x-1) $ . We get $ \displaystyle it9(x-1) = (0.140921295793 - 0.0447908983417*I) + (0.E-1821552 + 0.E-1821552*I)*x + \ldots $ where all following coefficients are machine-epsilon. Surprisingly the constant is the 4'th or 5'th iterate of $\small x_{k+1}=exp(bl*x_{k})$ beginning at $\small x_0=0 $. Perhaps this helps for focusing the analytical reasoning?2011-12-10
2

Real solutions: $\enspace x\in\{2;4\}$

Complex solutions: $\enspace\displaystyle x=\frac{W_n(-i\pi - \ln\sqrt{2})}{-i\pi - \ln \sqrt{2}} \enspace$ for $\enspace n\in\mathbb{Z}\enspace$ where $\,W_n\,$ is the analytic

continuation of the product log function (see Lambert W-function and it’s branches,

e.g. https://cs.uwaterloo.ca/research/tr/1993/03/W.pdf , chapter 4).