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I do know that if you use tetration the equation would look like this. $$^3x=-1$$

You could then theoretically use the super-root function to solve the equation, but I do not know how to use the super-root function nor do I know of any website online that could calculate that equation.

Mathematica also doesn't help.

Mathematica

So how would you attempt to solve it?

Also, in case anyone is interested, here is the graph for $$z = \Re[(x+iy)^{\Large (x+iy)^{(x+iy)}}]$$

Real tetration

And $$z = \Im[(x+iy)^{\Large (x+iy)^{(x+iy)}} ]$$

enter image description here

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    $\begingroup$ Well $-1$ is a trivial one $\endgroup$ – Renato Faraone Aug 30 '15 at 19:09
  • $\begingroup$ Oh yeah, I forgot about that. But I'm assuming there are more answers? $\endgroup$ – Paul Aug 30 '15 at 19:12
  • $\begingroup$ Are you only looking for real solutions? $\endgroup$ – Tintarn Aug 30 '15 at 19:12
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    $\begingroup$ I mean x^ (x^x) $\endgroup$ – Paul Aug 30 '15 at 19:36
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    $\begingroup$ What does $x^{x^x}$ even mean for general complex $x$? Even something like $i^i$ is not well-defined. It should be the same as $\exp(i\ln(i))$, but $\ln$ has many branches and $\ln(i)$ can reasonably be interpreted as both $\frac{\pi}{2}i$ and $\frac{-3\pi}{2}i$ just to get started. Then $i^i$ could mean both $\exp(-\pi/2)$ and $\exp(3\pi/2)$, just to get started. Maybe for some complex $x$, one of the many branch values of $x^{x^x}$ is equal to $-1$. But writing $x^{x^x}=-1$, using the equals sign, when working with general complex $x$ is going to lead to confusion. $\endgroup$ – alex.jordan Aug 31 '15 at 8:34
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Let's define $$f(z)=\;^3z+1$$ and for an inital evaluation/rootfinder $$g(z) = \log\log(f(z)-1)-\log(\log(-1)) \qquad \text{ or }\\ g(z)=\log\log(z)+z \log(z)-\log(\log(-1))$$ .

When doing root-finding with the Newtonalgorithm, then using $f(z)$ we'll easily run into numerical under- or overflows, so I tried first the rootfinding using $g(z)$. This gave for the range $z=\pm 10 \pm 10î$ the roots $x_0,x_1,x_2$ so initially $3$ solutions.

        x:     roots for g(z)                                
 ------------------------------------------  initial findings
  x0 = -1                                    (pure red colour)
  x1 = -0.158908751582 + 0.0968231909176*I   (pure blue colour)
  x2 =  2.03426954187 + 0.678025662373*I     (pure green colour)
 ------------------------------------------ update: additional findings
  x_3 =  2.21022616044 + 2.14322152216*I
  x_4 =  2.57448299040 + 3.39212026316*I
  x_5 =  2.93597198855 + 4.49306256310*I
  x_6 =  3.27738123699 + 5.51072853255*I
  x_7 =  3.60013285730 + 6.47345617876*I
  x_8 =  3.90713751281 + 7.39619042452*I
  x_9 =  4.20091744993 + 8.28794173821*I
  x_10 =  4.48346951212 + 9.15465399776*I
  x_11 =  4.75636133031 + 10.0005052039*I

by checking the range -10-10i to 10+10i in steps of 1/20 with 200 decimal digits internal precision.

These are so far "principal" solutions, where "principal" means, we do not ponder the various branches of the complex logarithm.

The Pari/GP-routines are (in principle, much improved to draw the picture)

fmt(200,12) \\ user routine to set internal precision(200 digits)
            \\ and displayed digits
lIPi=log(I*Pi)
myfun(x)=local(lx=log(x));log(lx)+lx*x
mydev(x) =local(h=1e-12); (myfun(x+h/2)- myfun(x-h/2))/h

{mynewton(root,z=lIPi) =local(err);
      for(k=1,150, 
            err= precision((myfun(root)-z)/mydev(root),200);
            root = root-err;
            if(abs(err)<1e-100,return(root));
          );
       return([err,root]);}              
  \\ --------------------------------------------------

 {for(r=-10,10,for(c=-10,10, z0= c -r*I;
        if(z0==-1,print(z0);next());
        if(z0==0 | z0==1 ,print(z0," fixpoint!");next());
        print([z0,mynewton(z0)]);
        );}


Here is a plot of the array of the complex initial values $z$ from $-10+10î \ldots 10-10î$ leading to the solutions $x_0,x_1,x_2,... x_{11}$ for roots-finding on $g(z)$ and some cases did not converge.
The pure blue colour mark the area, for which $x_1$ is attracting for the Newton-iteration, the pure green colour mark the area, for which $x_2$ is attracting, the pure red colour where $x_0=-1$ is attracting. The other roots have modified/mixed colours. The shading shows roughly the number of iterations needed, the less iterations the lighter is the colour.
The iteration had to exclude the fixpoints 1,0,-1 to avoid infinite number of iterations.
* root-finding for $g(z)$ *

bild1

(There are some spurious dots, not visible in the picture, these are coordinates where the Newton-iteration did not sufficiently converge)

Here are the locations of the first 12 found roots so far. It suggests, we'll find infinitely many...

bild2

update 2020 After we see that the roots occur nearly in line when $\text{real}(x_k)>2.5$ I took then another simpler search routine to fill the gaps, where the initial value in the newton-algorithm for $\;^3 x_k - (-1) \to 0$ inserted for $x_{k+1}$ is guessed as linear continuation from $x_{k-1}$ to $x_k$ : $x_{k+1, \text{init}} = x_k + 0.97 (x_k-x_{k-1})$.
I got now

        x                                     
 ------------------------------------------  
x_0 = -1                                    
x_1 = -0.158908751582 + 0.0968231909176*I   
x_2 =  2.03426954187 + 0.678025662373*I     
x_3 =  2.21022616044 + 2.14322152216*I
x_4 =  2.57448299040 + 3.39212026316*I
x_5 =  2.93597198855 + 4.49306256310*I
x_6 =  3.27738123699 + 5.51072853255*I
x_7 =  3.60013285730 + 6.47345617876*I
x_8 =  3.90713751281 + 7.39619042452*I
x_9 =  4.20091744993 + 8.28794173821*I
x_10 =  4.48346951212 + 9.15465399776*I
x_11 =  4.75636133031 + 10.0005052039*I
  ...               ...
x_{k+1} ~ x_k + (x_k-x_{k-1})*0.96  as initial value for the Newton 
                                      algorithm on xk^xk^xk - (-1) => 0

Of all of this likely infinitely many roots their complex conjugates are as well roots.

update 2020 jan (2)
A finer analysis detects more complex roots aside of the indicated location in the previous picture. It seems that there are further roots rather arbitrarily scattered in the right half plane. I the following picture I just find a second linear looking region (green dots) and a third region (blue dots) and some more scattered roots but which I didn't document here.
Here is the improved picture: picture

update 2020 jan (3)
I improved again the search-routine for the complex roots of $\;^3x=-1$ and found a lot of new roots by scanning the neighbourhood of two known roots from the first version. I just stepped at a given root $x_k + 0î ... 1î$ in steps of $1/1000$ and took this values as initial values for the Newton-rootfinder. This gave a lot (about $200$) of new roots which I inserted in the picture. I'm not sure, whether I should conjecture again some structure in this scatterplot; the only thing that surprises me that for all that new roots the locus of the first found ones (documented in the first and second picture, here in red and pink) give somehow an "outer" boundary for all that randomly(?) scattered set of roots (in blue color).
picture

update 2020 feb (1) The rough structure in the scattering of the roots for $f(z)$ made me suspecting that the roots might lie somehow on dotted lines and looking at the imaginary and the real parts of $f(z)$ separately the resp roots might be connected by continuous lines. This seems to be true; I show three pictures. This pictures show the neighbourhood of a known zero $z_0 \approx 5.277 + 11.641î$ with $\pm$ one unit extension.
First picture $real(f(z))$ : green color indicates negative values, red color positive values. Small absolute values dark, large absolute values light. Where neighboured dots have alternating sign I draw white points, indicating continuous lines of zero values:

picture5

second picture $imag(f(z))$. Again we find continuous lines with zero value:
picture6

Now the overlay shows discrete locations for the complex zeros of $f(z)$ as met in the recent investigation; they are just on the points, where the white lines from the real and from the complex images intersect:

picture7

Interesting the further structure of the roots-locations: that slightly rotated rectangular shape towards the south-east of the picture.
But I think I'll go not further in this matter -with that findings someone interested and talented might dive deeper in this and give bigger pictures and/or more insight.

(data can be shared on request; the excel-sheet has the data in about six digits precision which are enough to reconstruct better approximations easily using them as initial values for the Newton-rootfinder, for instance with Pari/GP and arbitrary precision)

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  • $\begingroup$ Thanks for the algorithm. And also, if we did care about the other branches of the complex logarithm, how would you calculate them easily (given the principle solutions)? $\endgroup$ – Paul Aug 31 '15 at 14:31
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    $\begingroup$ @Paul: I think myfun(x)=local(lx=(log(x)+6*2*Pi*I));(log(lx)+6*2*Pi*I) + lx*x ; no modification of the mydev() is needed (the branch is handled directly at the call of the logarithm, sorry: I had an error in my previous comment here). But anyway - no guarantee.... $\endgroup$ – Gottfried Helms Sep 1 '15 at 2:36
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    $\begingroup$ This is beautiful. I would upvote again if I could. $\endgroup$ – Paul Nov 11 '15 at 16:54
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    $\begingroup$ The picture is more than beautiful - you shouldn't delete it - but obviously corresponds to a different iteration. And it should also be symmetrical wrt the real axis, shouldn't it? $\endgroup$ – Wolfgang Feb 2 at 21:34
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    $\begingroup$ Yeah, the last pictures look much more as what one should be expect for such an infinite sequence of roots with supposedly a "regular" pattern (in terms of monotonicity, monotonicity of neighboring differences etc.), unlike e.g. the zeta zeros. $\endgroup$ – Wolfgang Feb 7 at 11:51
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Clearly $x=-1$ is a solution. Here I'll prove that it's the only real solution, complex solutions are a different matter.

Given $z,\alpha \in \mathbb{C}$ , we have $$z^{\alpha} = \exp(\alpha [\log |z| + (\arg z)i])$$ So let $x = re^{i\theta} \in \mathbb{C}$. Then $$\begin{align} x^{x^x} &= \exp(x^x[\log r + \theta i]) \\ &= \exp\big( \exp(x\log r +\theta i)[\log r + \theta i]\big)\\ &= \exp\big( r^x(\cos \theta + i \sin \theta)(\log r + \theta i) \big)\\ &= \underbrace{\exp\big( r^x(\cos \theta \log r - \theta \sin \theta) \big)}_{\in \mathbb{R}^+} \cdot \exp \big( r^x(\sin \theta \log r + \theta\cos \theta)i \big) \end{align}$$

Thus $\arg x^{x^x} = r^x(\sin \theta \log r + \theta \cos \theta)$.

For example, if $x \in \mathbb{R}$ and $x<0$ then $\arg x^{x^x} = -(-x)^x\pi$. So if we're going to have $x^{x^x}=-1$ and $x \in \mathbb{R}$ then certainly we need $x<0$ so that $\arg x^{x^x} = \pi$. Hence if $x<0$ and $x^{x^x}=-1$ then we have $$-(-x)^x = -1$$ which is equivalent to $(-x)^{(-x)}=1$. As the only solution to $y^y=1$ with $y>0$ is $y=1$, this means that $x=-1$ is the only real solution.

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  • $\begingroup$ I've spotted an error in my answer; my claim that $\exp \left( r^x (\cos \theta \log r - \theta \sin \theta) \right) \in \mathbb{R}^+$ is only true if $x \in \mathbb{R}$. For the purposes of finding real solutions this is fine, but it means the formula for $\arg x^{x^x}$ is off. I don't have the time to fix this now, but if I remember later then I'll do it. $\endgroup$ – Clive Newstead Aug 30 '15 at 21:03
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    $\begingroup$ I think it's simpler than this if you focus on reals. For reals, $x^{x^x}$ is only defined when $x$ is nonnegative or is a odd negative integer. For most negative $x$, it isn't even defined. For $x^{x^x}$ to be $-1$, negative, $x$ would have to be one of those odd negative integers. But for odd positive $n$, $(-n)^{(-n)^{-n}}=-1\implies(-n)^{1/(-n)^n}=-1\implies(-n)^{1/-n^n}=-1 \implies -n=(-1)^{-n^n}=-1$. $\endgroup$ – alex.jordan Aug 31 '15 at 8:19
  • $\begingroup$ Or perhaps you are allowing $x^{x^x}$ to be defined for negative $x$ by giving complex meaning to the $x^x$, which would be a fair thing to do. $\endgroup$ – alex.jordan Aug 31 '15 at 8:21
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I've just entered in this problem again and propose now to use the power series for the inversion $ \;^3 W(x)= \text{reverse}(x \cdot \exp(x \cdot \exp(x)))$ using the Lagrange series-inversion. You'll get a series with a very limited radius of convergence; however it seems to be finite and not really zero. But the signs of the coefficients alternate, so you can apply Euler-summation or similar tools at them.

Then let us define $x$ as unknown and $u=\log(x)$ its logarithm, and $y=x^{x^x} = -1 $ the known value and $v=\log(y)$ its logarithm.

Then $u = \;^3W(v)$ (in the range of convergence) and $x=\exp(u)$ .

Using Euler-summation (of complex order and 128 terms for the partial series) I arrive at
$\qquad u=0.762831989634 + 0.321812259776î \qquad$ and
$\qquad x=2.03425805694 + 0.678225493699î \qquad$. (In my older post I gave
$\qquad x=2.03426954187 + 0.678025662373î \qquad$ by Newton-approximation).

The check gives $x^{x^x}=-0.998626839391 + 0.0000476837419237î$ which is by $0.00137 + 0.000047î$ apart.

I think this way is in principle viable, however one needs then better convergence-acceleration / summation tools. And possibly it is a meaningful starting-point for the classical Newton-approximation.

A longer treatize showing more details can be found in my webspace

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In addition to $x=-1$, there are complex roots. These can be found numerically. For example,

fsolve(x^(x^x) = -1, complex, avoid = {x = -1});

$$- 0.1589087516+ 0.09682319092\,i $$

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  • $\begingroup$ What language did you write the computation in? $\endgroup$ – Paul Aug 30 '15 at 19:25
  • $\begingroup$ f[u_, v_] := u^v; NSolve[f[ u, f[u, u]] = -1, u] can be modified in Mma? $\endgroup$ – Narasimham Aug 30 '15 at 19:29
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    $\begingroup$ Unfortunately the so called super-cube root doesn't have a closed form yet, even in terms of Lambert's W and generalized Lambert's functions. So if you want numerical approximations use the Newton's method. @Paul $\endgroup$ – Renato Faraone Aug 30 '15 at 19:54
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    $\begingroup$ @RenatoFaraone: we can have a taylor-series $g(x)=(x+1)^{(x+1)^{(x+1)}}-1$ and then try to evaluate $f(x)=g(x-1)+1$. Because the formal powerseries of $g(x)$ is invertible we can also have $f^{-1}(x) = g^{-1}(x-1)+1$. I don't know about the ranges of convergence of all these power series and functions. But since the signs of the coefficients are seemingly roughly alternating, one can possibly extend the range of convergence to a range of summability using Cesaro-,Euler- or Borel-summation. $\endgroup$ – Gottfried Helms Aug 31 '15 at 9:32
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    $\begingroup$ @renatoFaraone: I don't think we can do it with a finite composition of common /elementary functions. Maybe there is some iterative scheme like it is with the LambertW-implementation, for instance as in wikipedia/rosetta-code... What I'd really like were if one could Fourier-decompose the set of coefficients of $g^{-1}(x)$. Perhaps we get some Fourierseries or Theta-series from this. $\endgroup$ – Gottfried Helms Aug 31 '15 at 10:20

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