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)

I am talking about your statement "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)." The first sentence doe snot imply the second sentence. In order to avoid shifting of the goalpost, could I ask you to fix the precise statement? You are talking about, say, 3^n for any n, which does not necessarily appear in the computation of a↑↑b. What you stated is "for any natural numbers n and m such that m is not divisible by 10. the last 3 digits of m^n in base 10 coincides with the last 3 digits of m^{n+100}"? Otherwise, please fix a precise statement with full quantifications.

Also, could you understand that even if you are talking only about a↑↑b, your observation of specific values of 2^n does not imply something general? I recommend you the following two:

1. To understand that observing specific values does not imply a general fact.
2. To write a statement in a precise formula with full quantifications.

p-adic 01:53, January 28, 2020 (UTC)

> What you stated is "for any natural numbers n and m such that m is not divisible by 10. the last 3 digits of m^n in base 10 coincides with the last 3 digits of m^{n+100}"?

Yes, although this only holds for n >= 3 if m = 2. In fact, for any m, there can only be four possibilities for the last digit of m^n, (1, 3, 7, and 9 for m coprime to 10, 2, 4, 6, and 8 for m even and not divisible by 5, and the last digit is fixed for m divisible by 5 or 1 mod 5). m^n can only end in 2, 3, 7, or 8 if n is odd, which means that if m^n ends in 3 or 7 and m is 3 mod 4, then m^n is also 3 mod 4 and so the tens digit must be even. If m is 1 mod 4, then m^n is also 1 mod 4 regardless of n, meaning the tens digit must be odd in those cases. On the other hand, if m^n ends in 1 or 9 and neither m+1 or m-1 is divisible by 5, then n must be even, meaning that m^n must be 1 mod 4 since -1^2 = 1, thus the tens place digit must be even. If m-1 is divisible by 5, however, the tens place digit of m^n as n increases cycles through all 10 digits if it doesn't repeat sooner, and so does the hundreds place digit each time you increase n by 10. If m+1 is divisible by 5 however then n must be odd if m^n ends in 9 and thus the tens digit of m^n must be odd. For an even value of m, m^n will always be divisible by 4 for n >= 2, which means the tens place digit of m^n must be even if the final digit is 4 or 8 and odd if the final digit is 2 or 6. In each case, there are possibilities for the last 2 digits that never actually occur. With an even value of m (which is the simplest case, read the previous paragraph), there are 4 possibilities for the units digit and only 5 possibilities for the tens place digit, meaning that the last 2 digits repeat after 20 powers, and there are also 5 possibilities for the hundreds place digit given the last 2 digits, thus the period of the last 3 digits of m^n for all even values of m not divisible by 10 is 20*5 = 100. We can similarly show that the period for the last 4 digits is 500, 2500 for 5 digits, 12500 for 6 digits, and so on.

Proving this for odd values of m is a bit harder. If m == 3 mod 4 and m =/= 1 or -1 mod 5, then m^n == 3 mod 4 if n is odd and 1 mod 4 if n is even. Since 3^2 = 9 and 7^2 = 49, m^n will end in a 9 if n is even and not divisible by 4, or 1 if n is divisible by 4. In both cases, this means that the tens digit must be even, which already rules out a power of m ever ending in 11, 19, 31, 39, 51, 59, 71, 79, 91, or 99. If n is odd, then m^n must be 3 mod 4 and end in either 3 or 7, meaning that the tens digit again must be even, leaving only 20 possibilities for the last 2 digits. m^4 will also be 1 mod 8, since 3^4 = 81 == 1 mod 8 and 7^4 = 2401 == 1 mod 8. This means that m^n mod 8 repeats with a period of 4, which evenly divides 20, thus the first value of n for which the last 2 digits of m^n repeat is also 1 mod 8, leaving only five choices for the hundreds digit given the two digits to the right, giving 100 possibilities for 3 digits. In the case where m == 1 mod 5, this is already proven since there are 10 possibilities for the last 2 digits, and the period can be multiplied by no more than 10 if we add another digit. Lastly, if m == 1 mod 4, m^n is always 1 mod 4 regardless of n, meaning the tens digit must be odd if the last digit is 3 or 7, and must be even if the last digit is 1 or 9. 1^2 is just 1, and 5^2 = 25 == 1 mod 8, meaning if m is 1 mod 4, then m^2 and thus m^n for all even values of n must be 1 mod 8. We have already proved that the last 2 digits repeat after 20 and that m^20 == 1 mod 8. There are only 5 possibilities for the last 3 digits that are both congruent to 1 mod 8 and 25, meaning the last 3 digits once again repeat after 100 powers. Note that simply proving that the last 3 digits repeat after 100 powers is already enough to prove that the recursive modular exponentiation method for tetration works in base 10, since the period can be multiplied by no more than 10 if we add another digit. Allam948736 (talk) 02:21, January 28, 2020 (UTC)

> Yes, although this only holds for n >= 3 if m = 2.

Well, what are you going to verify then? You understand that what you stated is wrong, while you are continuing explaining the classification of the possibilities.

> Note that simply proving that the last 3 digits repeat after 100 powers is already enough to prove that the recursive modular exponentiation method for tetration works in base 10, since the period can be multiplied by no more than 10 if we add another

As you already noticed that it does not hold for the case n < 3, it is not provable... What precise statement are you going to show? Please write down full statements before trying to show them.

p-adic 09:39, January 28, 2020 (UTC)

It only doesn't hold for n < 3 if m is congruent to 2 mod 4, because m^n is always divisible by 8 for n >= 3 if m is even, but if m is congruent to 2 mod 4, then neither m nor m^2 are divisible by 8. If m == 4 mod 8 then m is obviously not divisible by 8, but m^2 is a multiple of 8, meaning in those cases it starts working at n = 2. Allam948736 (talk) 15:38, January 28, 2020 (UTC)

It might be related (a link from the last digits of Graham's number in Wikipedia.) Triakula (talk) 15:50, January 28, 2020 (UTC)

@Allam948736

> It only doesn't hold ... because'

Again, you are wrong in this logic. What you explained is that the statement does not hold in that case. It does not imply that it holds otherwise. What you are stating is for any natural numbers n and m satisfying the following, the last 3 decimal digits of m^n coincides with those of m^{n+100}, right? Please fix a statement by yourself, instead of saying "it does not hold if..." or "I assume ...".

1. m is not divisible by 10.
2. n≧3 if m is congruent to 2 mod 4.
3. n≧2 if m is congruent to 4 mod 8.

@Triakula

We know it, because we were talking about it so many times. Allam948736 wrote wrong arguments again and again on last digits of powers without fixing precise statements or showing explicit proofs. There were so many wrong descriptions on last digits in this wiki written by Allam948736 without sources. That is why I require Allam948736 to fix a precise statement before explaining how it is believed to be correct.

p-adic 22:48, January 28, 2020 (UTC)

By the recursive modular exponentiation method to compute the last \(d\) decimal digits of \(^yx\), I mean this:

\(a_1 = x\) \(a_n = x^{a_{n-1}} mod 10^d\) repeated until we reach \(a_y\)

I have proven that it doesn't work if we replace the 10 with another base (say, 11), but it works in base 10! The last 3 decimal digits of m^n always coincide with those of m^(n+100) if m is odd or divisible by 8, and for even numbers not divisible by 8, the values of 1 and 2 are just special cases. This is already enough to prove that the above method works (already explained why). Allam948736 (talk) 03:25, January 30, 2020 (UTC)

> I have proven that it doesn't work if we replace the 10 with another base (say, 11), but it works in base 10!

Although you are talking as if it were your own great discovery, I told you so many times that it does not necessarily hold when the base is not 10 (while you first stated by yourself that it worked for any non-prime bases) and I have explicitly written down the proof for cases including base 10.

The last 3 decimal digits of m^n always coincide with those of m^(n+100) if m is odd or divisible by 8, and for even numbers not divisible by 8, the values of 1 and 2 are just special cases.

You fixed your statement. Why couldn't you fix your statements before you are aked so many times?

>This is already enough to prove that the above method works (already explained why).

No, the observation of the behaviour of the last 3 digits does not imply the validity of the recursive modular exponentiation method. Should I repeat to say "observing lower values itself does not gives a proof" again? How many time should I say? (I conjecture that you will change your statement by saying "it is obvious that d is always assumed to be 3" or something like that, although you intensionally skipped the quantification of d, which I required so many times.)

p-adic 10:58, January 30, 2020 (UTC)

The reason why proving that the period of the last 3 digits is 100 is enough to verify that the recursive modular exponentiation method works in base 10 is because the period can be multiplied by no more than 10 if we add another digit. Allam948736 (talk) 15:50, January 30, 2020 (UTC)

> because the period can be multiplied by no more than 10 if we add another digit.

You have not verified it. Moreover, there is more general a method applicable to wider bases in the article. Why are you trying to prove it in the case where the base is 10? Just you could not understand the article, right?

p-adic 22:50, January 30, 2020 (UTC)

No, I understood the article. Also, could someone actually calculate the last 20 decimal digits of sufficiently high power towers of 2, 3, 4, and 5 using the method described in the article and see if the results are the same as those found here ? Allam948736 (talk) 02:00, January 31, 2020 (UTC)

If you understood the articke, then why would you try to "prove" the statement for the special case as if it were a novel idea?

p-adic 10:52, January 31, 2020 (UTC)

The reason I'm trying to prove the statement for the special case b = 10 is simply because base 10 is the most commonly used. My earlier results found here  were in base 10. Allam948736 (talk) 15:46, January 31, 2020 (UTC)

I used the following method in Wolfram Mathematica: \(s_1 = a\), \(s_n = a^{s_{n-1} \mod \phi(10^d)} \mod 2\times10^d\), and I obtained ...98,615,075,353,432,948,736 as the last 20 decimal digits of a sufficiently high power tower of 2s, and ...04,575,627,262,464,195,387 as the last 20 decimal digits of a sufficiently high power tower of 3s, which are the same results I got using the basic recursive modular exponentiation method which were found here. Allam948736 (talk) 16:15, February 10, 2020 (UTC)

It does not mean that your method generally works for any tetration... Why couldn't you understand it although I repeat to tell you that coincidence at a specific value does not mean that the algorithm always works?

p-adic 23:37, February 10, 2020 (UTC)

I tried it for other values too, and got the same results: for a base of 4, ...22,302,555,290,411,728,896, for a base of 5, ...17,493,152,618,408,203,125. I already explained why the correct method should coincide with the basic recursive modular exponentiation method. This is because the period of the last d decimal digits of \(m^n\) evenly divides 10^d iff d >= 2 for all values of m not divisible by 10. Allam948736 (talk) 01:34, February 11, 2020 (UTC)

I said that checking finitely many specific values does not give a proof and I explained why your explanation is not a proof. How many time could I repeat the same comments?

p-adic 01:47, February 11, 2020 (UTC)

But can it even be proven generally? What if in "computing" many last digits we're always relying only on guesses? Triakula (talk) 08:59, February 11, 2020 (UTC)

It is not necessarily provable in a reasonable length. If it is not proved, then it is just good not to write down it in the article as if it were a verified fact. I am requiring him proofs because he added so many guesses without proofs to articles and stated that they were correct without showing proofs or explicit algorithms. However, once I disproved that segmental informations of his secret method yield contradiction, he changed the statement and started to state as if he had noticed the issue by himself. Now this is the time when he states as if he verified that his method works without showing explicit proofs. He just lists correct equalities, which are not proofs. It is something like writing 1+1=2, 2+3=5, and so on and stating "therefore Fermat's last theorem was proved by me". Listing correct equalities does nothing bad, but does not yield a proof. I explained this so many times, but he always ignores the requirement. Nevertheless, he added his guess without proofs again. I would like to know when he realises the importance of a proof.

p-adic 09:40, February 11, 2020 (UTC)