Might results which show the same result for tetration as for exponentiation be of any use (like in the range from 2 to e^(1/e))?

Question

I experimented with this and found 9 numbers which have the same height and exponent and show nearly the same result for tetration and exponentiation.

Might this be of any use or worth looking at it any further?

Starting with 2 here's the list:

b² = ²b ... b = 2
b³ = ³b ... b ~ 1.825767861009766
$b^4 = {{^4}b}$ ... b ~ 1.72232999999934
$b^5 = {{^5}b}$ ... b~ 1.6554299999989017
$b⁶ = {{^6}b}$ ... b ~ 1.6096399999986017
$b⁷ = {{^7}b}$ ... b ~ 1.5769699999983877
$b⁸ = {{^8}b}$ ... b ~ 1.55289999999823
$b⁹ = {{^9}b}$ ... b ~ 1.5347099999981109

Finding more numbers slowed my computer down considerably and/or would break the floating point arithmetic of JavaScript. Probably 15 digits of precision ain't enough to find more.

Do you know Pari/GP? It is easily programmable; and you can compute with thousands of digits precision. It comes out as an interpreter, so you have a very simple dialog-system but with ability to program loops and to use transcendental functions (for instance Lambert-W-functions). Google for Pari/GP and university of Bordeaux. — Gottfried Helms, Nov 28 '23 at 12:21
@Tyma Gaidash - didn't try it analytically. For an approx see my answer; perhaps a determination can be found in the tetration-forum (https://tetrationforum.org), some people played around with very similar questions and maybe it is somehow this "generalized Euler number" (phhh... little colorful name btw.). — Gottfried Helms, Nov 28 '23 at 13:09
@TymaGaidash - ps.: I think the limit is $\lambda=\exp(\exp(-1))$ btw. — Gottfried Helms, Nov 28 '23 at 13:45
@ТymaGaidash - there is a constant $c$ which allows incredibly accuracy in prognose of the value $b$ for some height $h \in {100...100.000}$. Let $c=\exp(\text{LambertW}(\log(45)))$ (such that $c^c=45$ then $ b_h \approx c^2/h^2 + \lambda$. Perhaps there is a powerseries on $h$ in the background with large constant term... Don't have more analysis at hand... — Gottfried Helms, Nov 30 '23 at 08:07
@GottfriedHelms I think it's accuracte enough to try it with real heights. I played around with ²0.5 for fun and then with 8.5 to calculate the height for 0.5 and the result was between the linear and the quadratic approximation. I even thought of changing the formula to ask for the base. Congrats on finding that so quickly! — Matthias Liszt, Dec 01 '23 at 08:29
Matthias - sounds interesting. It would be good you could write a bit about your interpolation procedure to fractional heights (there exist many approaches, you know...) Unfortunately I shall not be able to go much deeper in this myself, (loosing energy due to ageing) and at best can do "here a sip, there a sip..." - - - Plus: I didn't refer to your question "is there any use of this equality for something in number-theory", so I might do it here: do not see such a sideffect so far, and have no idea about one. But might you be lucky and find such! :-) — Gottfried Helms, Dec 01 '23 at 08:46
@GottfriedHelms $b^x$ ~ ${^{x}b}$ for x >= 0 and x<=1 is mentioned as linear approximation on wikipedia and that quadratic is the same for base e. Your formula - I thought of it as (ssrt(h)^2)/(h^2)+(e^(1/e)) - gets more accurate for higher heights. The integer value can be removed with repeated logarithms of the right base. So you got ${^{10000.5}b}$ you got ${^{0.5}b}$. — Matthias Liszt, Dec 01 '23 at 09:58
Ah, thank you, that sounds good. I'll see in the afternoon. Thx! — Gottfried Helms, Dec 01 '23 at 10:02

Gottfried Helms · Answer 1 · 2023-12-01T09:02:34.320

I've recalculated your problem using Pari/GP up to tetration-"height" $h=50$ (update:$h=1.000.000$).
The used internal precision was 200 dec digits. It needed 2 secs to compute the list of first 50 entries. (The error using the found $b$ is less than 1e-180).

Here is an excerpt of the solutions for your problem $\;^hb = b^h $ (or: $ \;^{h-1}b = h$ or: $\;^{h-2}b\cdot \log(b) = \log(h)$):

      h   b
      -+----------------------
      2 2
      3 1.825455022924830
      4 1.722191912772396
      5 1.655390902316565
      6 1.609627746636388
      7 1.576964928603568
      8 1.552901631749337
      9 1.534710270835191
     10 1.520655935607996
     11 1.509593123816016
     12 1.500742202664287
     13 1.493558789695041
     14 1.487654237606995
     15 1.482745596652516
     ...    ...
     48 1.448984047079777
     49 1.448814814881550
     50 1.448655327023237
    ... ...
    199 1.444929953048032
    200 1.444927354272319
    ...  ...
    499 1.444709831304874
    500 1.444709663947240
    ...  ...
   1000 1.444678333141238
    ...  ...
  10000 1.444667965901445
    ...  ...
 100000 1.444667862058820
1000000 1.444667861020257
  ...  ...
  ...  ...
 e^1/e  1.444667861009766

To illustrate the weirdness of that approximations, here we see the base $b=b_{100}$ as found for tetration-height $h=100$ and $h=1000$:

Example 1 ($b_{100} \approx 1.445692829534854 $): $$ \small \begin{array} {llll} a_1&=\;^{100} b_{100} &\approx& 1.017659983991973 E16 \\ a_2&={b_{100}}^{100} & \approx &1.017659983991973 E16 \\ a_1-a_2 & & &= 5.389496515861541 E-190 \end{array}$$ _{Computations done with $200$ dec digits precision}

or even more:

Example 2 ($b_{1000} \approx 1.444678333141238$): $$ \small \begin{array} {llll} a_1&=\;^{1000} b_{1000} &\approx& 5.904177641015545 E159 \\ a_2&={b_{1000}}^{1000} & \approx &5.904177641015545 E159 \\ a_1-a_2 & & &= -2.113731367714944 E-639 \end{array}$$ _{Computations done with $800$ dec digits precision}

The program in Pari/GP is (for the first 50 or so entries; for isolated values of heights of $h=1000$ or $h=1.000.000$ one needs a more sophisticated procedure to avoid numerical overflows):

default(realprecision,200) \\ this is my default
\ procedure for iterated exponentiation for heights h 
tetr(b,h)=my(a=b);for(k=2,h,a=b^a);return(a)
\ actually computing...
vb=vectorv(50); vb[1]=1; vb[2]=2; \ this vector gets the set of solutions
   \\ compute the solutions for heights 3 to 50:
   for(h=3,50, vb[h]=solve(b=vb[h-1],1.4,tetr(b,h-2)*log(b)-log(h)))


\ printout
printp(Mat(vb))

update: Here is a table for further use, say, interpolations or to find some better estimation formula for $b$ depending on $h$ than that of simply using the guessed constant $c \approx 3.23...$ (see earlier comment) as given for $w_2$. The shown guess is already good: it allows to really approximate $b$ even for high $h$ in more reasonable computing time and enabled me to get an estimate for $h=1.000.000$:
Table 2: $$\small \begin{array} {r|r} h & b & w_1=1/(b-\lambda) & w_2=w_1^{0.5} \cdot c \approx h\\ \hline 10 & 1.520655935607996 & 13.15995970798423 & 11.74997695258401 \\ 20 & 1.467343859391106 & 44.09949159384668 & 21.50931235098078 \\ 50 & 1.448655327023237 & 250.7858365743097 & 51.29337051500824 \\ 100 & 1.445692829534854 & 975.6397152922338 & 101.1706337147851 \\ 200 & 1.444927354272319 & 3853.664600623257 & 201.0697286191011 \\ 500 & 1.444709663947240 & 23921.76388643725 & 500.9637885955461 \\ 1000 & 1.444678333141238 & 95491.54368949162 & 1000.903463303749 \\ 2000 & 1.444670481514372 & 381605.8929592410 & 2000.862622116834 \\ 5000 & 1.444668280509221 & 2383793.325683243 & 5000.852378528801 \\ 10000 & 1.444667965901445 & 9533644.761419189 & 10000.90306047686 \\ 20000 & 1.444667887234663 & 38131704.16346160 & 20001.05216061220 \\ 50000 & 1.444667865205928 & 238313029.9051765 & 50001.56863347198 \\ 100000 & 1.444667862058820 & 953239447.1432710 & 100002.4725468315 \\ 200000 & 1.444667861272031 & 3812933641.379400 & 200004.3117840303 \\ 500000 & 1.444667861051729 & 23830749177.08709 & 500009.8763919837 \\ 1000000 & 1.444667861020257 & 95322887666.87609 & 1000019.180814657 \end{array}$$

I multiplied the constant with $(1−(b-λ))^{0.5}$ and then subtracted 1 from the result and even for higher heights like 2 or 3 it was about 7 % off. Here's the full formula $((1/d)^{0.5})c((1-d)^{0.5})-1$ with d being (b-λ) and c being 3.238990951 and λ being $e^{(1/e)}$. — Matthias Liszt, Dec 05 '23 at 08:45
@MatthiasLiszt - yes, a couple of tests ($h$ up to $100,000$) gave a better estimate to the true value for $b$ with your formula. For $h=1000$ I got estimate $b_{He}- b \approx -1.8E-8$ and $b_{Li} -b \approx 2.13E-9$ , for $h=10,000$ I got $b_{He}-b \approx -1.89E-11$ and $b_{Li} -b \approx 2.04E-12$ and for $h=100,000$ I got $b_{He}-b \approx -5.18E-14$ and $b_{Li} -b \approx -3.08E-14$ Didn't try larger $h$ . Nice... — Gottfried Helms, Dec 05 '23 at 12:46
Yesterday I noticed a difference of 1.2 rather than 1 so for unknown reasons it might be $λ^{0.5}$ (rather than subtracting 1)... However the formula is not exact enough for heights between 1 and 2. I suppose it comes close to e for heights like 1.00001 when common approximations are used for tetration with real heights. $2.041^{0.9}$ ~ 1.9 , $2.704^{0.01}$ ~ 1.01 just by playing around — Matthias Liszt, Dec 06 '23 at 08:25
actually all the JS code I used for calculating and testing can now be found on https://github.com/MatthiasLiszt/tetrationalPlayground --- of interest might be my method of calculating the super square root using double logarithms — Matthias Liszt, Dec 18 '23 at 10:05

Might results which show the same result for tetration as for exponentiation be of any use (like in the range from 2 to e^(1/e))?

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