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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
    Is it $-\sqrt{2}$, or $\sqrt{-2}$?2011-12-07
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    @user7530. Is $-\sqrt{2}$, sorry, I edited.2011-12-07
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    Taking the principal value of the exponentials ... 8 steps is **very** close to zero. Around $10^{-45}$. Then 9 steps is close to 1, and then at 10 starts over again. So if you are doing this numerically, you had better take many more than 45 places in your approximations!2011-12-07
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    I'm guessing the whole thing approaches a cycle of size 9.2011-12-07
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    @GEdgar, unfortunally the computer couldn't calculated much more =// . If you can generate some other values, please post here ^^.2011-12-07
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    The iteration looks pretty chaotic. Try `Graphics[{AbsolutePointSize[2], Line[Composition[Through, {Re, Im}] /@ Drop[NestList[((-Sqrt[2])^#) &, N[-Sqrt[2], 100], 600], 150]]}, Axes -> True, PlotRange -> {{-5/2, 5/2}, {-5/2, 5/2}}]`2011-12-08
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    Raising a negative number (or any complex number that is not a positive real) to a non-integer power is not a well-defined thing to do. The basic example is $(-1)^{1/2}$. Is that $i$ or $-i$? It just gets worse with irrational powers. $-\sqrt{2}^{-\sqrt{2}}$ could be interpreted as $\sqrt{2}\exp(-\pi i\sqrt{2})$, $\sqrt{2}\exp(-3\pi i\sqrt{2})$, $\sqrt{2}\exp(-5\pi i\sqrt{2})$, etc. and all of these are different. You might decide to enforce that all arguments always be reduced to values in $(-\pi,\pi]$, but that would be unnatural and give discontinuous operations.2011-12-08
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    Using the branches of the *lambert-w* we get *~2* for the first negative branch (http://www.wolframalpha.com/input/?i=-W%28-1%2C-log%28-sqrt%282%29%29%29%2Flog%28-sqrt%282%29%29) and *~4* for the second negative branch (http://www.wolframalpha.com/input/?i=-W%28-2%2C-log%28-sqrt%282%29%29%29%2Flog%28-sqrt%282%29%29). Both fixpoints are repelling, so we approximate them only by the iteration of the inverse operation $\small x_{j+1}=(\log(x_j)+k*2 \pi i)/\log(b) $ where $\small b=-\sqrt2$ for *k=-1* and *k=-2*2011-12-08
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    Very interesting plot @J.M.2011-12-09
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    Yes, you're right @alex.jordan. But as I explain in the question, maybe this tetration looks like something like [$i^i=e^{-\frac{\pi }{2}+2k\pi} \quad (k \in \mathbb{Z})$](http://en.wikipedia.org/wiki/Imaginary_unit).2011-12-09
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    @GottfriedHelms, lol. Interesting thing, could this show the number converge to $2$ too? Maybe it converges to $2$ and several another ones. (I'm not afilliated to this $k=-1$ and $k=-2$, so I need see this a little).2011-12-09
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    @GarouDan: :-) No, the "attracting" one is the 9-period-orbit. But for the inverse operation (the log) and the multiple branches of the log there are infinitely many fixpoints, such that $\small \log(x_0)/log(b)=x_0$ which may be called the k=0'th fixpoint (at x_0). The other *k*'th fixpoints are defined by $ \small (\log(x_k)+ k \cdot 2 \pi i )/log(b)=x_k$ and are "attracting" for that log-operations (try the iteration with some initial value). These *k* define the branches for the loag as well as for the lambert-w-function. This is all more obvious if the other direction of indexing is used.2011-12-09
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    @GarouDan I think that's my point. If you raise a general complex number to a general complex power (as in $i^i$) there are infinitely many interpretations of the result ($\ldots\exp(-\pi/2), \exp(3\pi/2), \ldots$), and none of them are canonical. $(-\sqrt2)^{-\sqrt{2}}$ has the same problem. How do you choose which representative from $\{\sqrt2^{-\sqrt2}\left(\cos(\sqrt{2}(\pi+2\pi k))-i\sin(\sqrt{2}(\pi+2\pi k))\right)\}$ to continue working with? Because they are all different.2011-12-10
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    Did you mean $$(-\sqrt{2})^{ (-\sqrt{2})^{\cdots}} $$ or $$-\left( \sqrt{2}^{- \left(\sqrt{2}^{-\left( \cdots \right)} \right)} \right)$$2013-02-24
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    First option Hurkyl.2013-02-24
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    @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