Tetration: The operation you were (probably) never taught 

I'm on Mobile and I don't know if this is a bit of trolling or not

Thanks, I love the youtube keyboard commands.

@Hekkaryk Kcalb Damn my mobile only lifestyle

@Hekkaryk Kcalb thanks for the knowledge though.

Dave Marx You can play the video at 25% speed and it’s much easier to pause at the right time.

Monothefox a

Grand Ham!!!

Well... that but much bigger. But that.

One quadrigintillion prefixes for operations later..... g64

I was asking my beaver about how to compute Graham's 64th number, unfortunately he was too busy... ;)

also you can type 3^7600000000000 into wolfram alpha and you get 2.719 × 10^3626121535869

3^3^3^3^3 on wolfram alpha is 10^(10^(10^12.56090264130030)), according to wolfram it has 6.002253567994547×10^3638334640023 digits

@@Kirbykradle Oh wow. When I typed it into WA, it told me 'nope.' :)

Somehow I get a completely different result... 3^^3 = 7.6255975e+12 or 7.625.597.5XX.XXX as you said. 3^^4 on my machine appears to be 4.4342649e+38 sorry for not writing this properly but you get the idea, it is a 4 followed by 38 digits. and I you do 3^^5 I would get the answer (which seems MUCH more logical than the already given answer) of 8.718964e+115 so that's an 8 with 115 digits following it. I mean I would not have fun trying to pronounce 87 novemtrigintillion in a video, but it is definitely still calculateable.

@@NowWeJustWinIt You not doing the exponentiations in the right order: it should be 3^(3^^3)

Interesante. Zapatos antes pantalones tambien? :)

@The Taylor Series, si es estúpido y funciona, no es estúpido :) [eng:] if it's stupid and it works... it ain't stupid

@@Not.Your.Business Verdad, verdad. :) [eng] True, true. :)

Is that a joke lol

I think that's Soviet Russia you're talking about. Also, wouldn't you learn about toutation, one sasquatch, one little bigeddon, one utter oblivion and one BIG FOOT first?

You probably did! Just because others created something too shouldn't rob you of your achievement. If you want, though, there's still some things related to this that need to get invented -- how to use tetration with negative numbers, fractions, decimals, and (goodness forbid) complex numbers. :)

The Taylor Series hmm, thanks for the kind reply. who knows, maybe i could give these a shot in the future :) nice video btw, was very informative and well-made.

same here lol

OMG same

Just like how I invented Holocene Calendar. However, you have a large space to explore unlike me. I failed to breakthrough from that calendar so far.

Yeah, they sound very similar. ;)

I thought he will tell you how to get the right colour during titration. I thought the word is misspelt or something, but then I noticed the sigma and realized it should be related to math.

you get the right colour by pouring the solution along the walls and slowly lmao

I thought it was the titration I've used in making wine.

Do not ration tits, man. Free tits for the people.

Very true! If you wanted to show that for all of them, you could actually use mathematical induction, I think, to show this, but your argument is compelling to me as it is. :)

That makes sense... for higher order operations, the second argument just tells you how many times to repeat the previous operation...since the previous operation only takes in 2 arguments, it stays the same all the way down to addition. Neat find!

That's nice

Tahsin Ahmed as seen in the video 3^^5 is 3^3^3^3^3 which is 5 quantities of 3’s so 2^^2 is simply the same as 2^2 which is 2 quantities of 2’s. I’m not sure how you got 2^^2 equaling 2^2^2 because that’s 3 quantities of 2’s which is equal to 2^^3.

2^^^2 is not 4. Since 2^^2 means raising 2 to the power of 2 twice, or 2^2, then 2^^^2 means tetrating two twice, aka a tower of height 2^2=4. 2^2^2^2= 2^16.

Hahahahaha I know how you feel. :)

All examples of the Ackermann's function

Fr I always thought that tetration was left to right until now

Utterly small in comparison to what other notations can make Have you heard of subcubic graph numbers for example? They get so big that it's basically impossible to know and explain the size with other notations

I know, right? :) Someday!

@@TheTaylorSeries Part 1: Tetration. Part 2: Knuth's up-arrow notation. Part 3: Graham's Number.

@Yndostrul - 3^^5 is just 3↑5; 3^^^5 is 3↑↑5, etc.

Jivan Pal: Where did you get that idea from? A single arrow is just normal exponentiation.

@@Yotanido - Woops, looks like I needed to look up the notation again. I assumed Knuth would've just started with "↑" being tetration since we already use superscripts for exponentiation.

And if the algorithm was run using all the fundamental particles in the visible universe, the inverse tower would still be roughly 5. Quinn, you're gonna need a bigger universe.

Oh wow.

1. Yup! It's called Super Log. I *briefly* mention it in the followup to this video (though I don't go into it other than to mention it). 2. I haven't run into many uses for it; just a cool thing found with this pattern. That doesn't mean that there isn't one, though! 3. That's a good question! We'd have to check for associativity, distributivity, all that good stuff. My guess is that it doesn't have those, as (3^7)^10 != 3^(70) (that is, we lost it with exponentiation). Curious what you can come up with!

@@TheTaylorSeries My first inclination is to start with something like "which is bigger, x^^y or z^^w?" as a good starting point. I'll see what I can figure out :)

@@RobotProctor Right on. :)

At least super log has a few uses. I believe it related to log* which in computer science it can explain theoretical lower limit of the runtime of some types of optimizations that can be recursively optimized with log(n) short-cuts. I believe log* is defined as how many times you need to take log of something before the result is below 2.

@@Carewolf Hm. I haven't heard of using super logs in complexity theory, but I never went crazy deep into it. I would buy that. Do you have any examples?

Yeah, people have been mentioning that. It's so clever, I love it. :)

Yup! It's bazonkers. :)

g1: 3^^3 g2: 3^^^(3^^3 times)^^^3 Repeat until g64

Though of the same thing

I know what you're talking about and you're absolutely right.:) My thing though is that I still don't have any good way of measuring Tree(3), so I've seen. Or at least, that I've been able to make sense of. Do you have any suggested reading for this? :)

The Taylor Series I have learned about Tree (3) and Graham Number thanks to a video-documetary in RU-vid by Numberphile. Interesting fact, the video on Graham Number is taught by Graham itself. Other than those videos, I haven't read much about it, I'm still just a student. But in my opinion they explain it very well. They talk about G64 and then tell you that if you had G64 people each writting one digit of Tree (3) each Planck time since the begining of the universe, the universe would colapse before they can write it down which is a fancy way of saying "We know it's a number, we know it's not infinity, we just can't write it down or even know how big it is exactly". Other than that, I can't help with anything. All I know is that Tree (3) is a Kruskal's Tree and that Kruskal's trees are about joining dots with lines. Last time I tried to read about them was a year ago, maybe now I can understand it better.

Tree (Tree (3))

Yeah, we can go with "higher number" game however long we want. For example, we can use the scheme in this video to define Operation Tree(G64). And then use Tree(64) as both arguments. This game doesn't really make any sense.

Krzysztof Szyszka nah, as I said it's just for fun

And TREE(3)

Wheel Dragon We can wait but Graham number is also too large to comprehend so that won't change anything.

Fun story. I forgot I put that last line in the gag, and there have been like four people to ask me in the comments "what did you say at the party." I intitially thought I was getting random bot comments, because it was so out of left field, and it'd been so long since I wrote that gag. Only now I realize that they were legimately asking questions about the video. So, to clarify, and to answer everyone's question, when I was at the party in question, I said

@@TheTaylorSeries Well done lol

Would the next be Anticip....ation?

True!

The Tree function

@@cubethesquid3919 True.

Rayo’s haha

Well, after Thanos did that [spoiler], I had to find something to do with my time. :)

@@TheTaylorSeries thank you for your amazing knowledge and humor! Greetings

The rabbit hole, it goes very deep. :)

@@TheTaylorSeries :0

trying to imagine using non-integers as the super-exponent in the version of tetration where you swap between using the exponent and the base blows my mind..

@@anto2593 That's interesting! Hm. So, my initial reaction is that you're inventing valid new operations, and you can make an argument that they stem from the same basic operations. But I think those would simply need their own names and be their own independent chains. There's many ways we can expand our understanding with perspective shifts like this, and my take is that we should just name them and explore them. :)

Operation 4! = operation 24.

@Sakkura1 hahaha, didn't think of that.

​@@TheTaylorSeries They are basically all operation 4, and they give different results. The thing is, if order matters for all of these operation 4's, then there'd be infinite versions of operation 5, for each version of operation 4... one could say that the infinite number (N) of versions of the (n)'th operation grows exponentially like this N=∞^(n-3), n => 4.

I think this is the most commonly asked question on this video, which says that there is something here to examine. :) I am going to have to think of a way to answer this well. The irritating but true answer is 'because that's what we discovered when we invented it,' but I don't find it satisfying. Let me see what I can't come up with.

I wrote this earlier in response to a similar variant of this question. I hope you'll forgive me the copy/paste nature of it, but I feel like it's clear. :) It comes down to the nature of the numbers in the operations themselves. When you're multiplying 3 and 5, the 3 and the 5 (called operands in general) are both on equal footing. In multiplication, we can give them more specific names than operands: factors (multiplicand and multiplier have been used, but those really only get meaning when the things you're trying to multiply aren't commutative, like matricies). On the other hand, when you're using exponentiation, the two operands aren't on equal footing; they're not telling you to do the same thing. The base is the number you multiply by itself a number of times, and the exponent is that number of times. Thus, they're no longer on equal footing, so you lose commutivity. And that's really the crux: commutivity is lost when they're not on equal footing.

I talk a *little* bit about that in the next video, but not too deeply. The names of the operations from your line 4: ?; ? are Super-Roots and Super-Logarithms. Not terribly creative, alas, but it works. :) Wikipedia has some documentation on it, but I've yet to really find something awesome that teaches it. As for your questions: A) So far as I know, tetration hasn't been extended past the naturals; it's waiting for an inventor, I think! B) This is basically a duplicate of Graham's up-arrow notation, just with carrots instead of up-arrows. :) They're interpreted in exactly the same way.

Yes i have also thought about this operation which now I came to know is called tetration when I was first introduced to exponentiation

Square root is easily calculated with just addition and division. See: Babylonian methods of square roots. The method converges very quickly.

@@emorag "Square root is easily calculated with just addition and division." Yes, but it requires a storage register, along with iteration, to get common calculator accuracy. My point about not being able to do it easily with just the 4 arithmetic functions, is that it can't be accomplished with 3 or 4 keystrokes; it requires a process of many steps. "The method converges very quickly." Yes, it approximately *doubles* the number of places of precision on each iteration! Finding a square root by the Babylonian method is mathematically equivalent to Newton's Method. The latter, however, can also be applied to any function which, together with its derivative, can be calculated. The Babylonian method is very simple in concept, and works like a charm! You basically just notice that, given x, its sqrt s = √x has the property that x/s = s and that if t is a little greater than s, then x/t < s < t while if t is a little less than s, then x/t > s > t So by taking a guess, t (that isn't way off the mark), and averaging it with x/t, you get closer to s every time. That this process matches up exactly with what Newton's Method does, is a pleasant surprise. As a practical matter of computing very many decimal places, it's still difficult in the end, because the method generates fractions with ever larger terms. And to get n places, you'll generally be doing a division in which both numbers are around ½n places. So if you want 100 places, that's gonna be very tough to do by hand. Unless you have a calculating machine/program that can do arbitrary-length multiple precision. Fred

Interesting. What are the properties of ambi-commutivity and ambi-associativity defined as? That's a new one on me.

*_...ambicommutativity is where either commutativity, or anticommutativity, applies, rather than non commutativity i.e. ab=-ba if a,b are independent imaginary strictly quaternion, e.g. ij=-ji but i(r+i)=(r+i)i... Likewise for ambi-associativity, where anti-associativity applies, not non associativity, for strictly octonion, e.g. (ij)L=-i(jL)..._*

Yep! Check out the next video to see. :)

Counting backwards, dividing, negative powers, negative tetration

Super logarithms is one!

Yes there is, such as super roots or super logarithms, Negative Hyper Operations are just the inverse functions of the Positive Hyper Operations, Ex: 5^(-1)3 = 2, 5^(-2)3 = 5/3 = 1.666..., 5^(-3)3 = cuberoot(5) = 1.709975946676697... , 5^(-4)3 = supercuberoot(5) = 1.78003783883447... where the supercuberoot(5) is the solution for x in this equation. x^x^x=5. Now if you really want to stretch your mind, think about fractional, irrational, transcendental, and complex hyper operations.

I think you guys confuse negative with inverse !

Ian Schimnoski I did too, but did you ever think of meta operations? Like 3 meta 5 would be 3 (operation 5) 3?

But you did invent it, and nobody can take that from you. :)

@@veda-powered I'm working on a math language which I'm hoping will make math (complex calculus trig and even alternate reality logic tables) as easy as child's play, for everyone including beginers, and also, may, I'm hoping, refute Gödel's incompleteness theorems (but I don't actually know Gödel's incompleteness theorems I just need to do me right now, and when I'm done I can look into them)

@@niaschim A good explanation of Godel's Incompleteness Theorem is in The Emperor's New Mind (though the book itself has some weird philosophical elements I don't agree at all with, it explains that well at least). I don't know if it can be refuted per se, but what do I know. :) I wish you best of luck in your language; goodness knows there's enough quirks that may be smoothed out by what we use right now. I don't know if it'll ever remove the need to understand the ideas behind math, but certainly, things could be clearer.

So far as I can tell, no, but that's only based on me Googling it a few times. I keep hoping that someone will find it! Or, better yet -- invent it. :)

@@TheTaylorSeries well if exponents of non integer values work via rooting/exponential combonation (aka the function and it's inverse) the same can be said about tetration logically, so 3^^3 = 7.6×10^12 (7.6×10^12)^^(1/3) = 3 Meaning 2.5^^2.5 = (2.5^^(1/2))^^5 (1.709)^^5 62.943 Boom, defined.

It is defined for all integers, for instance for negative ones?

@@erikjohanson4573 I see what you mean! Tbh I was hoping for a clean and nice "formula", but that's really clever :) Do you mind sharing some interesting results from your Python program? Like idk, 1 ^^ 0.5?

Erik Johanson it’s VERY interesting, thank you!

which is not enough of my 69420D brain

2^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^2 is also easy. It's 4.

Hm. That's an interesting question. I have a feeling that you'd have to be very careful, as the notion of multiplication is more complicated in vectors and matricies -- there's more than one type, after all. Exponentiation would undoubtedly be more intricate, and hyperoperations as well. However, my suspicion is that you are ultimately correct: you could, and I'd be very curious as to what it is! As for quaternions, I think it might be more straightforward; I *think* that they have a straightforward multiplication (albeit computationally challenging), so I imagine extending it would be very natural. Then again, it'd have to be checked, so ... who knows!

​@@TheTaylorSeries Ironically, exponention of a scalar with a matrix as exponent, *is* useful in mathematical physics (rotations in n dimensions, e.g., can be handled this way), and is defined using (you'll never guess!) -- TA-DAAAA -- Taylor's Series! Because TS expansion allows replacing an exponent with a collection of additions and multiplications of that exponent, which are already defined for matrices. If I recall correctly, one downside is that such an infinite series of matrices, doesn't always converge. But in any case, evaluating it, requires discovering the behavior of successive integer powers of a given matrix (also already defined). And that can be problematic. Quaternions - yes, their multiplication is straightforward; it isn't, however, commutative in general. Quaternions can be modeled ("represented") by matrices in several ways, so the problem of applying the various operations to them, reduces to that of applying them to matrices. Division, e.g., is multiplication by the inverse, and for matrices, as for any non-commutative group - there's post- and pre-multiplication, and they are, in general, different. Also to be kept in mind, is that not all matrices *are* invertible. I know of no meaningful way to define a matrix to a matrix power. It might be possible; I don't know. I'd have to think hard about it, or look it up. Failing any such method, would leave no way to define tetration of a matrix by a scalar, let alone by another matrix. As for vectors, there are dot (scalar) product, cross product, and tensor product. Dot product operates on two vectors of equal dimension (nr. of components), and produces a scalar. Cross product also operates on two vectors of equal dimension, n, and produces a tensor of rank n-2. For n=2, the result is a scalar. For n=3, the result is another vector. (This is in common use, and is also called, "vector product.") For n=4, the result is a 2nd-rank tensor (n x n matrix). Etc. Tensor product operates on two vectors which can be the same or different dimension. If the vectors are dimension m and n, respectively, the result is an (m x n) matrix - and one that, even when m=n, and determinant is defined, is always singular (determinant = 0; & thus has no inverse matrix). There's no meaningful inverse to any of these three operations. Fred

@@ffggddss Clearly, what we need to do is to figure out how to write tensors with quaternion dimensions whose elements are permutation cycles and do super-logarithms with those as both the base and argument. ("... so preoccupied if they could they never stopped to think if they should." -- story of my life)

@@TheTaylorSeries "Quaternion dimensions"? Do you mean quaternion elements (components)? "... whose elements are permutation cycles ..."? Could you explain that? I'm unable to make sense of it. I know that cycles (cyclic permutations) are a useful way to analyze permutations in general; that every (finite) permutation can be expressed as a product of cycles; and that permutations have matrix representations, but I don't see how to make cycles into matrix elements. "... as both base and argument." Yes, this needs to be figured out; I'm just not getting any insights about how to do that. But I haven't invested much time toward that yet. Do you have any ideas? Fred

3blue1brown has a great video on quaternions.

And what's weird is that if you want to make something that's perposterously bigger than that ... just go to g_65. That's just monstrously bigger than g_64, even though you only changed one thing by one number. It's crazy.

@@erikjohanson4573 g_g_ ... (g_g_ ... (g_g_ ... (g_g_ ... (g_64) ... g_64) ... g_64) ... g_64). Except, do that nesting g_64 times. Plus 1.

Fair. :)

Ehhh, the sense of the phrase has changed over the last 20-30 years.

@@JivanPal - How's the weather in New Brunswick?

@@jeffjones6951 - I wouldn't know; I'm from the UK. 😉 How's being a prescriptivist? In any case, to cite a dictionary (Wiktionary, specifically): "The sense “raise a question, prompt a question” is more recent and has been proscribed by some commentators, but is now included without comment in some dictionaries. Others suggest that the phrase is hard to understand in any event, and should be avoided, using instead phrases such as “assume the conclusion” (for philosophical sense), "evade the question" for failure to address the question, and “raise the question” or “prompt the question” (for the last sense)." Both the Oxford English Dictionary and Merriam-Webster's list both senses, with MW going so far as to deem the original sense as "formal" and have an entire article on the subject: www.merriam-webster.com/words-at-play/beg-the-question Language is what it is. Also, for what it's worth, Rutgers is still a highly-ranked university (#123 globally in 2016 per the T.H.E. rankings).

@@JivanPal - Touche! Well considered, beautifully penned response. Thank you for that. Link was helpful too. I especially enjoyed the suggestion that people avoid the phrase "beg the question" altogether yet "cultivate an attitude of serene detachment in the face of its use by others." !! Serenity is not one of my strong suits. I agree that English (or any language) is and should remain fluid; words and meanings vary with geography and evolve over time. But when the perversion of a phrase traces to ignorance (like "beg the question" or the US idiom "ax to grind") I remain a purist. Regardless of its street use, i believe that an academician (especially one in the fields of math & logic) has a duty to not perpetuate this particular misuse. Moreover, it's best to avoid cliches entirely in formal writing and speaking. Regarding Rutgers: despite its open-door (>50%!) admission policy the school does comprise some excellent science, math & engineering programs, but (imo) fails miserably in providing those students with a foundation in the liberal arts. This is precisely why I picked on Rutgers! Peace out

Hm. Do you mean like, what is operation -1, or operation 1.5? Those are great questions. Here's a link that kind of explores that: math.stackexchange.com/questions/1227761/example-x-y-and-z-values-for-x-uparrow-alpha-y-z-where-alpha-in-bbb/1241979#1241979 (it's rather technical). The short answer is: It's not yet known, though it's certainly something awesome to think about.

The Taylor Series Thanks, that's really interesting. However, what I was actually asking was how can I 'tetrate' A to B, where B is not a natural number.

Oh! That's also a great question. Put simply, that's called an extension. There are extensions for some functions that we know and hardly think about; if you graph f(x) = 2^x on a computer, for example, you get a smooth curve -- but that means that you are telling your computer to calculate things like 2^1.34824, which is demonstrated by a smooth curve you into which you can zoom infinitely (in principle) far. So, can we extend tetration like this? That has a harder answer. You want a smooth curve that passes through all the points created by the integer values of B, meaning you want it to be continuous and without kinks. Turns out, there's an infinite number of functions that meet these two criteria, which means that we need some way of picking among them. I have no idea how to do that! :) Here's a paper on it I found: web.archive.org/web/20060525195301/ioannis.virtualcomposer2000.com/math/papers/Extensions.pdf But that is SUPER technical, and while I read the first third of it and skimmed the rest, I didn't fully digest all of it. Still, it's really cool.

I'm not sure! I think those may well be open questions.

Well, x^y = 10^(log(x) * y), so I think that x^^y = 10^^(superlog(x) * y)

Just substitute up-arrows for the carrots, and you've got it. :)

SlimThrull MONSTER GROUP

@@TheTaylorSeries Yes

I clearly need to play this game. :)

Ha! Awesome question. I think I'm not ready yet to build up to the idea of a complex-numbered operation; I'd want to figure out fractional and decimal operations first before I went there. That being said, if you think you can figure it out, I'd love to hear!

Its a number

@@TheTaylorSeries Operation 3.5 - Factorial

its a number equal to the square root of negative 1. Its a shame we arn't taught it earlier because it is a very important number perhaps the most important. I say this because we are given the impression that everything builds up from 1 and we do that because it is easy to teach what 1 is and how to abstract layers onto/from that. But then we get to trigonometry and we are taught sine and cosine. I don't know if I am alone in doing this but up until this point we were able to describe the most recently learned material with previous material so I wanted to know the equation that describes the sine wave. I had trouble continuing with trig because I couldn't accept at face value that sine was the relationship between the height of the triangle in relation to the angle of the unit circle, I wanted to describe the sine function with more simple math. But it turns out you cant do that without i. And that is because of the repeating property of sine, how can a function repeat itself it just seems wild, but i also exhibits this strange property. Watch what happens when you multiply it by itself. i = sqrt(-1). i^2 = sqrt(-1) * sqrt(-1) = -1. and again i^3 = -1 * sqrt(-1) = -sqrt(-1) and again i^4 = -sqrt(-1) * sqrt(-1) = 1 and again i^5 = 1 * sqrt(-1) = sqrt(-1) = i and it just keeps going like that in cycles. Put into an equation would look like i^n = i^(n +4) indeed the graph of i^x looks just like cosine(x/Tau) if you only look at the "real" part of the equation. But I think looking at it as imaginary numbers vs real numbers is a very shallow way of looking at it, and it really shows our cultural preference to 1 and its simplicity. But I believe if taught the beauty and power of i at and early age it wouldn't seem imaginary at all. It would just be apparent there isn't a number line but a number plane and to discrimate between one axis or the other by refering to one as real and the other imaginary would look a little silly in retrospect. But I guess I'm going on a tangent. Just know that you were robbed of a complete understanding of numbers that would allow you to conceptualize numbers in such a way that they have an amount and an angle such that positive numbers have an angle of 0 and negative numbers have an angle of 180 and there is a whole range of numbers inbetween that we are missing out on and our current numbering system poorly describes as imaginary when they have very real applications such as trig functions

This. Exactly this.

I seriously considered going into it! I thought, though, that I should probably save that for a future video, as I felt it might have made it overwhelming. I know I so often get caught up in going after the Shiny Amazing Thing that I don't take the time to enjoy the things discovered on the journey there. :)

@@TheTaylorSeries You were also very close to introducing Ackerman's Function. Wonder why that didn't get any press.

@Evi1M4chine But it would have been be fun

@Evi1M4chine Ackerman's function would have been related to the topic of the video because tetration is AN INSTANCE OF ACKERMAN'S FUNCTION.

@@cpuwrite It's just my personal style; I felt it would have been a bridge too far for this video. Someday, I will. :)

Hm. I think that comes out to 10^10^10 = 10^1000000000, whereas a googol is 10^100. So that would be a LOT bigger. But, isn't it amazing how fast it grows?

@@TheTaylorSeries Oh yeah you're right. Gosh this tetration thing is mindblowing me to the max. 😂

@@TheTaylorSeries the growth rate you said about gave me an idea to study these operations with calculus thanks

I think 10^^3 is a googolplex, isn't it?

@@SlimThrull A googolplex is 10^(googol), or just 10^10^100. 10^^3 is 10^10^10

Just assume the 4 numbers to be (x-1),x,(x+1),(x+2) . Mutliply the two numbers x and (x+1) at middle and let it be named A . You'll get A=x^2+x . Similarly multiply the two and edges and you'll get x^2 +x-2 ,which can also be written as A-2 .This might be a helpful algebric proof for you .

Hi! Right now, I'm still getting started with learning how to do all of this stuff, so I don't know that I'm able to collaborate just yet. However, I took a look at your channel; you certainly are doing some very interesting topics, and you've got some good explanations! While I might not be able to collaborate at this time, I'm happy to help if I can. You've got a good head on your shoulders and I'd love to see what you can do.

Well, 9^9^9 would have outright been much larger than 9^81 = 9^9^2. But 9^9^9 = 9^^3. Maybe go 9!^^ ... 9! ...^^9!. You will win. :)

yes!!!! genius!

Indeed, the Ackerman function is nuts. I remember an XKCD comic suggesting that A(g_64, g_64) is outrageously large (and it is), large enough to horrify mathematicians. What's funny, however, is that it's much smaller than ... g_65. Or g_g_64. I feel like this rabbit hole goes very deep ...

My secret plan, revealed!

I looked for it :)

Yeah. I am still looking for away to understand Tree(n) the same way we can understand g_64; that is, algorithmically.

How about G(Graham's number) vs. TREE(3) so I mean instead of going 64 levels like Graham's number you go an actual Graham's number of levels.

centation

I am completely down with using sequences of 'fancy' to describe later operations in this. Thus, we shall call tetration 'fancy fancy fancy fancy counting.' I think? Yes, I feel like I'm okay with this.

You are correct; I was sloppy with how I said it. Thanks for pointing it out. :)

I probably should have used the name the successor function; I didn't have a good reason for omitting it other than counting sounded simplest. Still, it's an oversight I shouldn't have made. Thanks for pointing it out.

Off the top of my head, I'd guess it has something to do with a terrible combination of natural logs. Would have to sit down to try it to see if I couldn't find a pattern. Cool idea though. :)

Well, when they say 'infinitely many degrees of freedom,' they don't usually mean *literally* infinitely. It's that there's so many that keeping track of them in any sensible way would be overwhelming, so just call it infinite and use simpler math to work with it. Counterintuitive, but it works. As for singularities, well, we haven't much observation on them, not directly. So we think we know how it works, and we have some very good descriptions of how it should work (general relativity), but it's possibly one of those things we may never know. Then again, who knows what the future holds. :)

I think the Law says 'no naked singularities', so we have only theoretical/mathematical physicists to take on these questions. Is there a putative 'bridge' between reals and infinites? I'm not seeking a yes/no answer, but I do wonder if others see a problem there. String theory posits 'no infinite densities' and attempts to banish infinities/infinitesimals. I think that IF we could observe 'planckons', (won't happen) they would (1) appear invariant w/respect to relativistic dx or dv thanks to Penrose-Terrell rotation and (b) take a series of photos of a small volume of space, scramble their order, then try to restore their chronological order-- the uncertainty principle would frustrate one's best efforts. Seems that if true these two factoids weaken the motivation for ST and VSL, etc. Thanks for your reply, by the way.

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@@creationisntgood942 Nice to hear from you, So what's Your own Opinion of "Such Math Operations" ?

Alright, I'm with you -- how would it work?

@@TheTaylorSeries Maybe I'll sit down this weekend and figure it out.

I ... know what that is, and will pass, but that is the funniest thing I've read today. :)