Talk: Tetration

Here, it uses what looks like circle notation - can circle notation accept parameters? If it can, this should be mentioned on Circle notation. {{subst:1x|~~}}{{subst:1x|~~}}

Circle notation, an extension to Steinhaus-Moser Notation, is entirely different from Rob Munafo's hyper-operators. Followed by 100 zeroes (talk | contribs) 21:01, 7 March 2009 (UTC)

I've been trying for a while to decode Geisler's site, but the math is giving me migraines -.- Is anyone interested in perusing that? FB100Z • talk • contribs 03:09, April 29, 2012 (UTC)

Small examples only?

Well, I think that only giving small examples doesn't tell you how large the larger numbers are generated by tetration, so it won't be helpful. Large examples tell you how this function is important in googology. Cloudy176 (talk) 10:38, November 18, 2012 (UTC)

Okay, you're right. Some examples reaching into scientific notation and beyond would be nice, but don't let's fill the page with seas of digits. FB100Z • talk • contribs 20:56, November 18, 2012 (UTC)

Real tetration

It's don't seem to me that \(^{1/x}a\) = \(\sqrt[x]{a}_s\).

Just see that the infinith-super root of 2 = \(\sqrt{2}\), but \(^{0}a\) = 1, and 1/0 = infinity. Ikosarakt1 (talk) 13:11, January 13, 2013 (UTC)

What's about defining \(a \uparrow\uparrow (1/b)\) as \(\sqrt[a \uparrow\uparrow (b-1)](a)\). That's reasonable, since for b=1 it gives a, and as \(b \rightarrow \infty\) it tends to 0.

Last digits convergence

We know very well that for integers \(n\), the last digits of power towers of \(n\) converge, making it easy to predict the last digits of a very large power tower (such as a chained arrow or BEAF array). For example, a power tower of threes converges to ...04575627262464195387, the final digits of Graham's number.

There's a page on the InterNet about "oology," which explores infinities recreationally. (Perhaps coincidentally, the page is from the André Joyce Fan Club, from which the term "googology" first originated.) It has silly things like turning fractions backwards: \(1/3 = 0.333\ldots\), so \(3\backslash 1\) ("3 conquered by 1," as in "divide and conquer") is \(\ldots 333\). Although clearly this is recreational mathematics and somewhat tongue-in-cheek, these look a lot like the notation for digit convergence in tetration. This can give us funny-looking equations such \(^\infty3 = \ldots 04575627262464195387\). FB100Z • talk • contribs 02:08, February 17, 2013 (UTC)

If I take the conquer operator for granted, \(x/1\) is the reversal of \(1\backslash x\). So I guess we have \(1\backslash ^\infty3 = 0.78359146426272657540\ldots\) okay, what the hell did I just write FB100Z • talk • contribs 02:12, February 17, 2013 (UTC)

I've typed a simple-math book about integer tetration, exploring its p-adic convergence properties. There are also  pseudo-convergence properties involving the figures next to the rightmost ones. The caos on the left fight against the order on the right of the number and there are some rules to point out this feature. In short, it's easy to predict the digit on the left of the leftmost frozen digit (without calculating tetration or using modular arithmetics and cycles), but it's very complex to predict other digits, even if you have already found their mutual constraints.

SPIqr (talk) 02:14, February 19, 2013 (UTC)

In popular culture

I found a rare occurrence of tetration humor: http://www.smbc-comics.com/?id=1666 (click on the red button below the comic) FB100Z • talk • contribs 06:04, September 12, 2013 (UTC)

What about Knuth Paper-Stack Notation? -- ☁ I want more clouds! ⛅ 07:17, September 12, 2013 (UTC)

True, true. Still, the SMBC comic may be the world's first sex joke with tetration in it. FB100Z • talk • contribs 07:57, September 12, 2013 (UTC)

And then they went on to make one with pentation and hexation in it. ArtismScrub (talk) 02:38, December 1, 2017 (UTC) 

"natural" tetration

is there any tetrational analog to the function \(f(x) = e^x\)? that is, is there a constant \(\nu\) such that \(f(x) = {}^x\nu\) is a "natural" function? FB100Z • talk • contribs 08:58, September 24, 2013 (UTC)

We have to define tetration for real heights before answering this question. Ikosarakt1 (talk ^ contribs) 15:22, September 24, 2013 (UTC)

Since e is the infinite sum of the factorial reciprocals, I would say that your \(\nu\) might possibly be the infinite sum of the expofactorial reciprocals. Then again, since ^ba is not defined for non-integer b, this is meaningless. ^{—Preceding unsigned comment added by ArtismScrub (talk • contribs) 03:52, January 24, 2018 (UTC)}

Definition for ordinals

I think it's natural to define w^^(1+a) = w^(w^^a), but not w^^(a+1) = w^(w^^a). For finite "a", it doesn't matter, but for a = w there is the fact that w^^(1+w) = w^(w^^w) = w^(e(0)) = e(0) = w^^w. That's consistent with the fact that 1+w = w.

For w^^(a+1), we should use Saibian's definition I believe. So w^^(a+1) = {w^^a}^{w}. Don't be confused with something like (w^^a)^w, look how the proper variant works for a = 4: w^^(4+1) = {w^^4}^{w} = {w^w^w^w}^{w} = w^w^w^w^w, but not (w^w^w^w)^w. Ikosarakt1 (talk ^ contribs) 07:56, July 27, 2014 (UTC)

Nontrivial solutions

in the natural numbers to a^^b = c^^d? Are there numbers expressible as a tetration in more than one way? —Preceding unsigned comment added by 160.94.192.193‎ (talk • contribs)

I think I have proven that there are none, but I'm not enirely sure. Full proof with all details would be quite lengthly. LittlePeng9 (talk) 15:59, August 6, 2014 (UTC)

Could you say what your approach was? —Preceding unsigned comment added by 160.94.192.193‎ (talk • contribs)

Suppose that we had a solution. Then we would have a^(a^^(b-1))=c^(c^^(d-1)). Every solution of p^q=r^s has p and r being integer powers of some number k. So we have a=k^l, c=k^m. From non-triviality k isn't 1. Now by laws of exponentiation we can get l*(k^l)^^(b-2)=m*(k^m)^^(d-2). If we assume l>=m then we get l/m is an integer and a power of k. By further manipulation we get that l must be divisible by m*(k^m)^^(d-2), which leads to the contradiction. I used induction to show this, and I'm not entirely sure if this will work. LittlePeng9 (talk) 18:21, August 6, 2014 (UTC)

Actually, it is l*(k^l)^^(b-1) = m*(k^m)^^(d-1). Continuing, let l = m*k^n. Then (m*k^n)((k^(m*k^n))^^(b-1)) = m*(k^m)^^(d-1), and therefore n + (m*k^n)*((k^(m*k^n))^^(b-2)) = m*((k^m)^^(d-2)). Thus n = m*((k^m)^^(d-2)) - (m*k^n)*((k^(m*k^n))^^(b-2)) = m*k^u - m*k^v, where we have v >= n and therefore u >= n. So k^n divides n, which is impossible.

This is of course assuming a,b,c,d > 1 and a != c. Deedlit11 (talk) 03:23, August 7, 2014 (UTC)

***

I attempted to create a redirect there, but this was blocked by a spam filter. --84.61.216.30 21:36, February 24, 2018 (UTC)

Can somebody else create the redirect from *** to tetration? --84.61.216.30 21:49, February 24, 2018 (UTC)

I can't create the redirect as well. I suggest reporting this to w:c:vstf:Report:Spam filter problems. -- ☁ I want more clouds! ⛅ 08:01, February 25, 2018 (UTC)

And what about the redirect from / to Extended Cascading-E Notation#Further extensions? --84.61.216.30 13:37, February 25, 2018 (UTC)

Last digits 2

In December, I was at the center of a whole debate on here about the computation of last digits of tetrations. It turns out that the original description on this article was indeed wrong, as the method works in base 10 but fails in most other bases. Observe:

\(7 \uparrow\uparrow 2\) = 823,543 == 985 mod 11^3

\(7 \uparrow\uparrow 3\) = \(7^{823543}\) == \(7^{985}\) mod \(11^3\) = 857

\(7 \uparrow\uparrow 4\) = \(7^{7^{823543}}\) == \(7^{857}\) mod \(11^3\) = 378

but this is wrong, because \(\phi(1331)\) (using Euler's totient function) is 1,210, which doesn't evenly divide into 1331. Using the correct method, we have:

\(7 \uparrow\uparrow 2\) = 823,543 == 985 mod 11^3

\(7 \uparrow\uparrow 3\) = \(7^{823543}\) == \(7^{823543 mod 1210}\) or \(7^{743}\) mod \(11^3\) = 332

\(7 \uparrow\uparrow 4\) = \(7^{7^{823543}}\) == \(7^{7^{823543} mod 1210}\) or \(7^{453}\) mod \(11^3\) = 24

However, using Mathematica, I observed that 3^15597484987 ends with the same last 10 digits as 3^5597484987 (...6100739387) and thus \(3 \uparrow\uparrow 4\) also ends in ...6100739387. Similarly, 3^16100739387 ends with the same last 10 digits as 3^6100739387, confirming that the last 10 digits of \(3 \uparrow\uparrow 5\) are indeed ...9660355387. This is because the period of the last d decimal digits of a^b for integers a and b is always a divisor of 10^d for d >= 2 (but not for d = 1).

\(2^{36}\) = 68,719,476,736

\(2^{136}\) = 87,112,285,931,760,246,646,623,899,502,532,662,132,736

This means that 2^...36 = ...736 regardless of the digits before the first ...36, so the last 2 digits of n indeed determine the last 3 digits of 2^n (the same goes with 3^n, 4^n, 5^n, etc). Allam948736 (talk) 20:47, January 27, 2020 (UTC)

No. It has a counterexample. Finding a pattern in specific value can never be a proof, as I pointed out so many time. Please understand the meaning even if you do not find a counterexample.

p-adic 23:30, January 27, 2020 (UTC)

In the case of "2^...36 always ends with ...736", multiplying by \(2^{100}\) in the example I gave clearly has no effect on the last 3 digits, meaning that the last digits will always be ...736 no matter how many times you multiply by \(2^{100}\). I was talking specifically about the last decimal digits of \(^ba\) when a is not a multiple of 10.  Allam948736 (talk) 01:15, January 28, 2020 (UTC)