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Start with $i=\sqrt{-1}$.

This will be $a_1$.

$a_2$ will be $i^i$.

$a_3$ will be $i^{i^{i}}$.

$\vdots$

etc.

In Knuth up-arrow notation:

$$a_n=i\uparrow\uparrow n$$

And, amazingly, you can evaluate $\lim_{n\to\infty}a_n=\lim_{n\to\infty}i\uparrow\uparrow n=e^{-W(-\ln(i))}\approx0.4383+0.3606i$.

You can check this, it does indeed converge to this value.

In fact, I decided to make a graph of $a_n$ to show that it converges. (y axis is imaginary part, x axis is real part.)

And, to little astonishment, I quickly noticed that there is an apparent pattern to the graph.

Commonly, we define $x\uparrow\uparrow0=1$, which I have included in the graph.

So the pattern seems very obvious. It follows a curved path that converges onto the point that was given above.

And, if you connect the dots, starting with the first point (given on the left as the first point) and trace a nice line to the second, third, and so fourth numbers, you will find an interesting spiral. I thought that at first, this spiral was writable as an equation, but apparently, there are a few implications.

You will notice that the blue dots are way closer to the converging point and that the red and black dots are a little closer. So whatever equation you can come up with should account that $a_{3n}$ is closest to the number you are trying to converge to.

I want (so desperately) to see if anyone can come up with an equation that allows the computation of $a_{0.5}$ that satisfies $$i^{a_{0.5}}=a_{1.5}$$a well known identity you can find on the Wikipedia.

At first glance of the graph I went on to think that perhaps, just perhaps, I (or you) could find a formula that allows us to define $i\uparrow\uparrow 0.5$.

If you are familiar with De'Moivres formula, it is a formula that allows us to perform compute

$$\sqrt{i}$$

with relative ease. It was derived when De'Moivre noticed an interesting pattern to $(a+bi)^n$. He proceeded to write his formula concerning the distance from zero and the angle from the positive real axis.

So I must tell you that I wish for the same to occur with $i\uparrow\uparrow n$. Perhaps the answer lies in using a different coordinate system. Perhaps the answer lies in calculating the distance one of the points on one of the lines (black, red, or blue) is from the converging spot and the adding in the angle at which the next point changes.

My progress on determining such a formula has gone nowhere. The most I can say is that $a_n$ is probably not chaotic and does indeed converge in a way that is most certainly not random.

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  • $\begingroup$ See aso math.stackexchange.com/a/432339/442 $\endgroup$ – GEdgar Feb 2 '16 at 23:12
  • $\begingroup$ @GEdgar Yes, thank you, but unlike your graph, I wish to make it continuous without hard corners. A "smooth" curve. $\endgroup$ – Simply Beautiful Art Feb 2 '16 at 23:14
  • $\begingroup$ So your sequence is $i\uparrow\uparrow n$, also written as ${}^ni$, and you're looking for a "smooth" extension to real heights ${}^xi$. Wikipedia has some discussion about it, though it begins with the statement that "At this time there is no commonly accepted solution to the general problem of extending tetration to the real or complex values of $n$." $\endgroup$ – Rahul Feb 3 '16 at 4:44
  • 1
    $\begingroup$ But see this previous answer by mike4ty4. $\endgroup$ – Rahul Feb 3 '16 at 4:48
  • $\begingroup$ @Rahul That was a wonderful answer! I will see how it goes here. Thank you! $\endgroup$ – Simply Beautiful Art Feb 3 '16 at 22:35
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You can find a non-trivial interpolation for the fractional iteration-height when you write down the consecutive iterates in log-polar-form (with center at the fixpoint). The nearer you come to the final fixpoint the log of the distance as well as the angle come nearer and nearer to a linear relation with the index and this suggests an obvious method of interpolation for fractional iteration-heights.

I found it interesting that that type of interpolation agrees well with the solution, which you would find via the method invented by E. Schröder in the late 19 century. Although this log-polar/Schröder-interpolation gives a straightforward solution, there seems to be a better one (better in what sense? - too difficult to discuss it here) in the spirit of Kneser's analytic solution for the fractional interpolation of the $\exp()$-function (implementations available by tetrationforum). The latter can seemingly be approximated by a simple (but computationally much involved) procedure involving matrix-diagonalization and computing fractional powers of that matrix.

You can find an introductory comparision of that mentioned methods (however for a different base for the exponentiation) in this small essay of mine


Here is an image for an interpolation to fractional heights starting at $z_0=1$ going to $z_1=î,z_2 \approx 0.2078,...,z_\infty \approx 0.438+0.361 î$ using the Schröder-mechanism. For instance for the half-iterate we find by this method $z_{0.5} \approx 1.1667+0.734 î$. The grey dotted line indicates the integer-iterates (should be the same as that of @GEdgar)

image



I got Sh. Levenstein's Pari/GP-program "fatou.gp" (from the tetrationforum) for the (extended) Kneser-method working. Here is a comparision of the orbits produced by the two methods. For instance, the half-iterates differ even visually:

image2

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  • $\begingroup$ Wow, an article of yours? It is wonderful that people are spreading knowledge through more physical means nowadays. $\endgroup$ – Simply Beautiful Art Feb 4 '16 at 21:47
  • $\begingroup$ Thanks Gottfried. You might want to look at how sexp(z-0.1*I) looks too. As $\Re(z)$ goes to $-\infty$, it goes to the repelling fixed point, yet as $\Re(z)$ goes to $+\infty$ is is still going to the attracting fixed point. This is the big difference between the Schroeder function solution, and the Kneser solution. $\endgroup$ – Sheldon L Feb 6 '16 at 20:22
  • $\begingroup$ Hi Sheldon - thanks, that is an interesting information for me (possibly I'll start understanding that method once :-) ) . As the question of fixpoints are of relevance now I've tried to look systematically at the fixpoints to base (i) and observe, that we do not only have on simple attracting fixpoint for the log but also a set of 3 periodic points attracting for the log. Is such a property in any way relevant for the (generalized) Kneser-ansatz? (See a picture in the tetration-forum math.eretrandre.org/tetrationforum/… ) $\endgroup$ – Gottfried Helms Feb 7 '16 at 4:43
  • $\begingroup$ This looks like something you'll want to see. math.stackexchange.com/questions/2288488/… $\endgroup$ – Simply Beautiful Art May 20 '17 at 11:56
  • $\begingroup$ yepp, @SimplyBeautifulArt, I've already seen your new thread, thanks for the link. It's a nice picture, indeed. (Perhaps I'll come one day back to my pen&paper and complete the little compilation of simple tetration-approaches - I didn't consider this specific one so far...) $\endgroup$ – Gottfried Helms May 20 '17 at 13:45
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As Gottfried hints, there is yet another solution to $^{0.5}i \approx 1.07571355731 + 0.873217399108i$

I will use this question to describe a unique Abel function for $f(z)=i^z$. I wrote a pari-gp complex base tetration program available for download at math.eretrandre.org. The results posted here were generated with that program. I will use this question about base(i) to show that if there is a solution of this type, than it has to be unique. This tetration can be regarded as an extension of Kneser's solution for real bases>$\exp(\frac{1}{e})$, to tetration for complex bases. So what is "this type" of complex base slog/abel function solution?

The answer is this Abel function involves both primary fixed points. The op points out the attracting fixed point, $l_1 \approx 0.438282936727 + 0.360592471871i\;$. There is also a repelling fixed point $l_2 \approx -1.86174307501 - 0.410799968836i\;$. Henryk Trapmann's uniqueness criteria
says if you can make a sickle between the two fixed points, bounded on one side by a defined curve f(z), and bounded on the other side by $i^{f(z)}$. For sexp base(i), we can choose f(z) as a straight line between the primary fixed points. Henryk's proof says if there is a one to one analytic mapping between the sickle, and the Abel function, excluding the two fixed points, and if the derivative of the Abel function is never zero, than it is unique to an additive constant. The additive constant is uniquely determined by the requirement that Tetration have the slog(1)=abel(1)=0.

Here is a picture of the sickle, and $\alpha(z)$ or the abel/slog on the sickel. You can see the one-to-one mapping between the two fixed points, extending between $-\Im \infty$ and $+\Im \infty$. The mapping between the straight line, and f(z) are always be definition exactly one cycle apart, since $\alpha(f(z))=\alpha(z)+1$. I also filled in vertical grid lines for sexp(z+0.25), sexp(z+0.5) and sexp(z+0.75). The two graphs are colored identical to allow visual verification of the one to one mapping. Because $\exp_i(z)$ is well defined, the sexp(z) function can be extended to the right over the entire complex plane, and extended to the left except for logarithmic branch singularities. So this slog on a sickle defines sexp(z) base(i) for the entire complex plane! Henryk Trapmann's uniqueness proof generates a mapping function between this solution and the other purported solution. Since both functions are analytic on the strip, it turns out both the mapping function and its inverse have to be entire, which can only be the case if the two slogs are the same except for an additive constant.

Near the attracting fixed point, the function approaches arbitrarily closely to the attracting fixed point Abel/Schroeder function, and near the repelling fixed point, the function approaches the repelling fixed point Abel/Schroeder function.

slog and abel function baes(i) on a sickel

sexp base i in the complex plane, grids are 1 unit apart. You can see the logarithmic singularity at z=-2.

sexp base i on the complex plane

The Abel function Taylor series was computed using the following form: $$\alpha(z)=\frac{\ln(z-l_1)}{\ln(\lambda_1)} + \frac{\ln(z-l_2)}{\ln(\lambda_2)} + p(z)$$

$\lambda_1$ and $\lambda_2$ are the multipliers at the two fixed points, $l_1$ and $l_2$, $$i^{l_1+z} = l_1 + \lambda_1 \cdot z + a_2 \cdot z^2 + a_3 \cdot z^3...$$

It turns out $p(z)$ has a relatively mild singularity at each of the two fixed points when this form is used for the Abel/slog function. For example, $p(z)$ and its derivative are both continuous and differentiable at both of the two fixed points, although the 3rd and higher derivatives are not continuous, since the periodicity at the two fixed points is less than 3.

The pari-gp fatou.gp complex base sexp program would be used as follows:

\r fatou.gp
setmaxconvergence();  /* base i is poorly behaved */
sexpinit(i);
sexp(0.5)
1.07571355731392 + 0.873217399108003*I

Here are numerical values for $l_1, l_2, r_1=\frac{1}{\ln(\lambda_1)}$, $r_2=\frac{1}{\ln(\lambda_1)}$, and p(z), and equation for the slogestimation. The radius of convergence for p(z) is $|\frac{l1-l2}{2}|$, centered between the fixed points.

l1 = 0.4382829367270321116269751636 + 0.3605924718713854859529405269*I;
l2 = -1.861743075013160391397055791 - 0.4107999688363923093542478071*I;
r1 = -0.02244005259030164710115539234 - 0.4414842544742195824980579384*I;    
r2 = 0.3613567874856575121871741974 + 0.4459440823588587557573111438*I;
slogest(z) = {
  z = r1*(log(I*(z-l1))-Pi*I/2) + r2*(log(-I*(z-l2))+Pi*I/2) + 
  subst(p,x,(z-0.5*(l1+l2)));
  return(z);
}
{p= -0.06582860911769610907611153624 - 0.6391834058813427803550150237*I
+x^ 1* ( 0.0004701290774740458290098771596 - 0.04537158729375693129580356342*I)
+x^ 2* (-0.003324372336079859782821095201 + 0.001495132937745569349230811243*I)
+x^ 3* ( 0.0007980787520098490845820065316 - 0.001533441799004958947560304185*I)
+x^ 4* (-0.001108786744422696031980666816 - 2.731877902187453470989686831 E-6*I)
+x^ 5* ( 0.0001798802115603965459766944797 + 0.0001776744851391085901363383617*I)
+x^ 6* (-0.0001598048157256642978352955851 - 3.381203527058705270044424867 E-5*I)
+x^ 7* ( 4.834500417029476499351747515 E-5 + 7.971199385246578717457250724 E-5*I)
+x^ 8* (-2.079867322054674351760533351 E-5 - 1.406842037326640069256998532 E-5*I)
+x^ 9* ( 1.690770367738385075341590185 E-5 + 2.135309134452918173269411762 E-5*I)
+x^10* (-3.353412252728033524441156034 E-6 - 5.845155821267264231283805042 E-6*I)
+x^11* ( 6.565965239846111713090140941 E-6 + 4.769875342842561685863158675 E-6*I)
+x^12* (-1.296330399893039321872277846 E-6 - 2.456868593278006094540299988 E-6*I)
+x^13* ( 2.533916981224509417637955994 E-6 + 7.218102304226136498196092124 E-7*I)
+x^14* (-8.409501999009543726430092781 E-7 - 9.279879295518162345796637972 E-7*I)
+x^15* ( 9.275250588492317644336514121 E-7 - 9.631817386499723279878826279 E-8*I)
+x^16* (-5.215083989292973029369039510 E-7 - 2.615530503953161606154084492 E-7*I)
+x^17* ( 3.116111111914753868488936298 E-7 - 1.665350480228628392912034933 E-7*I)
+x^18* (-2.781400382567721610094378621 E-7 - 1.112607415413251118507915520 E-8*I)
+x^19* ( 9.051635326999330520247332230 E-8 - 1.085008978701155103767460830 E-7*I)
+x^20* (-1.238964597578335282733301968 E-7 + 5.485260567253507938012071652 E-8*I)
+x^21* ( 1.846113879795222761048581419 E-8 - 5.632856406052825347059503708 E-8*I)
+x^22* (-4.197980145789475821721904427 E-8 + 5.240770851948157536559348001 E-8*I)
+x^23* (-1.428548543349343274791836858 E-9 - 2.602689861858463106421605234 E-8*I)
+x^24* (-6.065810598994532136326922961 E-9 + 3.328188440463381773510055778 E-8*I)
+x^25* (-4.925408783783587354755417128 E-9 - 1.098256950809547844459017995 E-8*I)
+x^26* ( 5.571041113925408468110396754 E-9 + 1.638316780708641282846896470 E-8*I)
+x^27* (-4.149193648847472629362625045 E-9 - 4.116268895249720851777701930 E-9*I)
+x^28* ( 6.721271351954440168744328856 E-9 + 5.947866395141685553477517779 E-9*I)
+x^29* (-2.739160795070694522203609350 E-9 - 1.172354292086004770247804721 E-9*I)
+x^30* ( 4.623071483414304725202549852 E-9 + 9.232228309095999063811309141 E-10*I)
+x^31* (-1.585766089923197553788716462 E-9 - 1.651307950491239271118345156 E-11*I)
+x^32* ( 2.363704675846105632188520360 E-9 - 8.296748095830550218237145087 E-10*I)
+x^33* (-8.032209583204614846555211647 E-10 + 3.448796470634182196661522301 E-10*I)
+x^34* ( 8.586697889632180390697042972 E-10 - 1.035571885972986467540525699 E-9*I)
+x^35* (-3.283321823970989260857161769 E-10 + 3.700931769620798039214959724 E-10*I)
+x^36* ( 1.060749698244576767546142485 E-10 - 7.216733126248904197113800569 E-10*I)
+x^37* (-7.435780239422554167112328213 E-11 + 2.755791031783421936772787526 E-10*I)
+x^38* (-1.572500505206983217824447542 E-10 - 3.667364964073739957874509082 E-10*I)
+x^39* ( 3.623731852785295824889239864 E-11 + 1.629657324418246834695914694 E-10*I)
+x^40* (-1.802976354364813915048318195 E-10 - 1.261880617212203404824890625 E-10*I) }
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  • $\begingroup$ In tetration Abel's equation only extends tetration where the base is $\large a=e^{\frac{1}{e}}$. For hyperbolic fixed points as $a=i$ Schroeder's equation must be used. See Complex Dynamics with Carleson & Gamelin, the chapter Fixed Points and Conjugations. $\endgroup$ – Daniel Geisler Feb 6 '16 at 17:40
  • $\begingroup$ @DanielGeisler I am not aware of any treatment of the Abel equation on a sickle (using both fixed points) in Carleson & Gamelin. While Kneser's method only works for real bases>$e^{1/e}$, this method uniquely extends it to complex bases, including complex bases where one of the fixed points is indifferent or attracting. $\endgroup$ – Sheldon L Feb 6 '16 at 17:44
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Consider the functions $$f_n(z) = \frac{\sin \pi z}{z - n}$$ for $n\in\Bbb Z$. Note that $f_n(k) = 0$ if $n \ne k$ and $f_n(n) = 1$.

Define $$f(z) := \sum_{n=-\infty}^{\infty} a_{|n|}f_n(z) = \sin \pi z \sum_{n=-\infty}^{\infty} \frac{a_{|n|}}{z - n} = \sin \pi z\left(\frac{a_0} z + 2z\sum_{n=1}^{\infty} \frac{a_n}{z^2 - n^2}\right)$$

(for the appropriate definition of the double-infinite sum). Since the $a_n$ converge, they are bounded, so the last sum converges for all non-integer $z$, with simple poles that the integers, which are cancelled out by the multiplication of $\sin \pi z$, leaving $f(n) = a_{|n|}$ for all $n \in \Bbb Z$.

You can define $a_z = f(z)$ for arbitrary $z$.

However, this is just one possible function. For any entire function $h$, the function $g(z) = f(z) + h(z)\sin \pi z$ also satisfies $g(n) = a_{|n|}$ for $n\in \Bbb Z$, but generally has different values from $f$ off of $\Bbb Z$.

There are other choices as well. Maybe you want to define $a_n$ for negatives by the inverse recursion formula $a_n = -\frac {2\log a_{n+1}}{\pi}$. A similar construction can be done, though one has to take more care to ensure convergence (the reason I went with an even function for $f$).

So without some extra criteria, there is not a well-defined answer. Your formula $f(z + 1) = i^{f(z)}$ may resolve the issue, but I haven't worked that out yet.

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  • $\begingroup$ Interesting, I will look at this. $\endgroup$ – Simply Beautiful Art Feb 3 '16 at 22:46
  • $\begingroup$ @Simple Art - but it falls well short of an answer. Fortunately Rahul's link does a much better job. $\endgroup$ – Paul Sinclair Feb 3 '16 at 23:24
  • $\begingroup$ True, but your answer is well thought out and a very nice attempt. I enjoy watching creativity blossom. $\endgroup$ – Simply Beautiful Art Feb 3 '16 at 23:26
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The number $ ^\infty i \approx 0.4383+0.3606i$ where $ i^{^\infty i} = {^\infty i}$ is a hyperbolic fixed point. Let $\epsilon$ be a very small number such that $\epsilon^2 \approx 0$. Then

$\large ^\infty i + \epsilon \rightarrow i^{^\infty i+ \epsilon} = {^\infty i} \times i^\epsilon = {^\infty i} \times e^{Ln(i) \epsilon} = {^\infty i}(1 + Ln(i) \epsilon) = {^\infty i} + Ln(^\infty i) \epsilon$

$\large ^\infty i + \epsilon \rightarrow {^\infty i} + Ln(^\infty i) \epsilon$

The number $ Ln(^\infty i) \approx Ln(0.4383+0.3606i) \approx -0.566386 +0.688444 i$ is called the multiplier in dynamics.

In polar coordinates $ Ln(^\infty i), r\approx 0.891487 ; \theta \approx 129.444°$. This describes the triangles in the plot and why they exist. The logarithm of the multiplier is called the Lyapunov exponent which is $-0.114865 +2.25923 i$ Because the Lyapunov exponent's x value is negative the system has a hyperbolic attractor at the fixed point.

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  • $\begingroup$ We have two simple fixpoints: 0.4383+0.3606i (attracting for i^z) , -1.86-0.41i (attracting for $\log _i(z)$) , and three periodic points : -1.14 + 0.71i, 1.65-0.19i, -0.07 - 0.32i (attracting for $\log _i(z)$) $\endgroup$ – Gottfried Helms Feb 7 '16 at 13:18
  • $\begingroup$ @GottfriedHelms Since logarithms are infinitely multivalued taking the same branch repeatedly in $log_i(z)$ gives one of an infinite number of attractive fixed points. This is why I consider tetration to typically be infinitely multivalued. In numerical simulations I have correctly estimated the multiplier of a neighboring fixed point. $\endgroup$ – Daniel Geisler Feb 8 '16 at 1:13
  • $\begingroup$ Just to avoid misunderstanding: the three additional periodic points do not occur because of using a different branch in the logarithm. While the above fixpoint $ ^\infty i$ is approached by iterated exponentiation, the other fixpoint (resp the periodic points) are approached by repeated logarithmizing (always the principal value was taken!) So in short $ ^{- \infty} i $ arrives at $-1.86-0.41$ when started from a value in the green area and one of the periodic points when started from the red,magenta or seagreen area in the picture go.helms-net.de/math/tetdocs/bilder/BATBaseI_3Log.png $\endgroup$ – Gottfried Helms Feb 8 '16 at 6:19
  • $\begingroup$ I tried a similar picture for the fixpoint with which you work in your answer ($^\infty i$) - unfortunately there are regions of complex starting values where the iteration escapes (at least temporarily) to gigantic values and I don't know whether from that regions they even might diverge to infinity. The blue colored area in the picture go.helms-net.de/math/tetdocs/bilder/BATBaseI_3Exp.png indicate initial values where the iteration goes towards the fixpoint and from the grey area it is unknown to me so far where iterations go. Some examples looked at individually manually converged. $\endgroup$ – Gottfried Helms Feb 8 '16 at 6:26

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