Protip: youtube actually has a frame-advance feature. You can use , and . to skip forward and backward by individual frames in a video. Super useful for the one-frame gag youtubers like to use all the time.
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?
I actually thought about this at school, but didn't found anything about this operations anywhere (I had got no internet back then), so I just assumed it was of no interest. What I later realized, however, is that the number 2 yields the the same value for all orders: 2 + 2 = 4 2 * 2 = 4 2 ^ 2 = 4 2 ^^ 2 = 4 ...
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!
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.
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.
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.
I had heard of tetration before, but even after looking it up I didn't understand it. After watching this video I do, and I understand just how utterly insane the numbers you can calculate with it (and pentration etc) is.
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
@@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.
since i was a kid i knew there was possible and had to be a "next level" operation a.k.a more and bigger hyperoperations than the potentiation or exponentiation. i was so wondered when i saw there was a theory behind it and that it had it's own symboles (i had to invent ones when i didn't knew the most used ones) Thank you for explaining this to more people!
Many computer scientists call this the “tower function,” while the “super log” is referred to as the inverse-tower function. Some complex algorithms can be shown to take place in inverse-tower of n time. Note that if such an algorithm was run using all of the atoms in the universe as n, the inverse tower would be roughly 5
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.
1. Is there a log equivalent for tetration? 2. What sort of problems does tetration help solve? 3. What algebraic properties does tetration have? Commutation is out of the question, but if these numbers are there and we can't count them in our universe it'd be interesting to work with them in a different way.
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 :)
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?
I thought it was fascinating how you explained all those operations. Some of the schools I know just fail really hard to even explain multiplication, they just do some examples and tell "roll with it". Exponentiation sometimes is just "grab a calc, do this". With the really easy and didactic way you showed, I feel I don't need a calc anymore lmao. Amazing video
WHAT YOU SAID?? In all seriousness though, I loved this video. With your sense of humor and dedication to the number language, I can easily see your channel blowing up soon! Keep it up!!
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.
Look at the definition of Grahams number. It uses exactly this mechanism and creates a mind-bogglingly huge number (each iteration takes the following number as the grade of operation done to 3 and 3)
Well this is amazing, even if you go to college, there are somethings that remain secrets, that shows how wonderful and big humankind knowledge has become so far. Great video The Taylor Series!!!
✓ - Liked!! Right at the beginning - the calculator you show, I call a 5-function calculator, because in addition to the 4 arithmetic functions, it has a square-root. I don't consider % a separate function, because it's really just a multiply or divide, together with a 2-digit shift. There are lots of both kinds of calculator, and √x is next to impossible to compute with just +, -, x, and ÷, so I regard the distinction as important. Tetration - I didn't have a name for it, but I toyed with the idea in Jr. High. As I'm sure lots of kids did who, like me, were math enthusiasts. Yes, inverting each of those operations, is a natural followup question, and of course, that's a known procedure for opns. 0 - 3. 0: counting down (instead of up) 1: subtraction 2: division 3: root extraction; logarithms (this one has two inverses because, unlike opns. 0-2, it isn't commutative) 4: ?; ? (there will again be two, since tetration is also not commutative) Also of interest: A) How can each operation be carried out with non-integers? Is there an obvious way to apply the answer to this for exponentiation, to opns 4 and up? B) What connection, if any, is there between these and Graham's up-arrow notation (see: Graham's Number)? Thanks for doing fun stuff! Fred
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.
@@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
Well, then there's Trees and Graham Numbers like G64 and Tree (3). G64 is mibdbogingly huge and just explaining how vastly huge it is would take a good teacher about ten minutes to half an hour. Then theres Tree (3). Let's just explain it this way; G64 is closer to zero than it is to Tree (3). So much closer in fact that if Tree(3) was 100000000, G64 would be almost equal to zero, it would be hard to understand even the number of digits Tree (3) is bigger than G64. But here's a bigger number, Tree (4), and yet again, Tree (3) is closer to zero than it is to Tree (4). And now, using these operations we can think of a stupidly large number, so much it wouldn't even be worth using it except for giving curious nerds like me a hardon. That number is Tree (G64) Fun fact, I think G64 is like operation 64 or 64...Oh no, wait sorry, G63 is the number of ^ symbols you need to use to get to G64, and G62 is the number of ^ used to get to G63 and so on. As you might guess, just trying to call that Operation N and explaining the value of N would take quite a lot of time. Just look it up, these numbers are absolutely huge and interesting to study, and Tree and G could be thought of as operations.
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 TH-cam 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.
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.
4:17 You can use logarithms to get that in powers of 10... But I was lazy, so I used Wolfram Alpha and get 10^10^10^12.5609026413003 and it has 6.002253567994547×10^3638334640023 digits... The last few digits are 9660355387...
Alternatively, you might generalise based on the pattern x*y = exp(log(x)+log(y)). So, the next higher operation is defined as the exp of the previous operation applied to the logs. Then you don't lose commutativity, but you end up with a different series of operations. Exponentiation is replaced by x^y = exp(log(x)*log(y)), etc.
As a computer science major, I was taught in college that these operations are described using something called Ackerman's Function. It is written A(x, y, z) Where x is the type of operation and y and, if needed, z are the arguments. Counting up to 5 would be A(1, 5) = 5. 1 for counting and 5 for what you're counting up to. No z argument needed here. Adding 3 and 4 would be A(2, 3, 4) = 7 -- 2 for addition and 3 and 4 for what's being added. Multiplication would be A(2, 3, 5) = 15 -- 3 for multiplication and 3 and 5 for what's being multiplied. A(4, y, z) is, of course, tetration.
First time I've seen one of your videos and you've immediately gained another subscriber. Great quality work. Let's hope your subscription rate increases exponentially, initially at least!
3:47 you can actually use , and . to go frame by frame. Incidentally, I managed to pause on the message without doing it, but normally I'd just go frame by frame to see a message like this.
When I was younger, before I even knew about powers, after learning about multiplication, I thought of this very thing only I didn’t have a good name for it, later I thought about meta functions like 3 meta 5 would be 3 (operation 5) 3 and so on. Great video, I haven’t seen anyone explain this on TH-cam until now, and i’m glad i’m not the only one to come up with this.
But... that is just one version of operation 4! There's many different thinkable opperation 4's! since order does not matter in addition, there is only one version of multiplication. And since order does not matter in multiplication, there is only one version of exponentiation. But order matters in exponentiation! Look at this. 2^^5 means: 1) Take 2^2 2) Use the result as the *exponent* of another calculation with the previous base: 2^x, 3) Repeat step 2 until you have writen the number 2, 5 times. Does that not seem arbitrary? Why not put 2^2 into the *base* of a calculation with the previous exponent? that operation would look like this: 2^^5 = ((((2^2)^2)^2)^2), and it fulfills the criteria of unpacking into repeated exponentiation. If we are allowed to mix it up, there is an infinite amout of versions of operation 4. We can use the *exponent* in every second calculation, and the *base* in the rest, or we can use the exponent only in every third, or forth or... All of these infinitely many operations unpack into repeated exponentiation, don't they? I think that's why tetration is not that popular. It is not an operation, but an idea, containing infinitely many operations. And that's just because order in exponentiation matters, unlike in all the previous operations, where a similar, infinite amout of ways to do the operation, all yield the same result. Which btw, i find insanely fascinating.
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. :)
@@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.
Is there a nice intuitive reason why the different operations behave so differently regarding their netral element and commutativity? The neutral element for addition is 0, for higher operations it is 1. And while addition and multiplication are still commutative, exponentiation and higher order operations are not. Why don't our operations behave in a more uniform way, since their definition is quite uniform (the (n+1)-th operation is just repeatedly doing the n-th operation).
Ah and then there's Graham's number defined as the 64th level of the following iterative process: On each level put 3 ^^^^....^^^3 where the number of arrows is the value of the previous level. Start at level 1 with 3^^^^3 already which is unimaginably big!
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.
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
An obvious generalisation is to make the first number (0 for counting, 1 for addition, etc.) a parameter for the function. This is what Ackermann did, and led to the so-called Ackermann function, which originally had 3 parameters, but 2 parameter versions have been devised too. An inverse to (one variation of) the Ackermann function even shows up in some applications in Computer Science, the standard union/find algorithm performs at time complexity of O(n alpha(n)), where 'alpha' is that inverse function. Of course, for all practical purposes alpha(n) is like a constant.
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.
what are the derivatives and integrals of x^^n and a^^x, and are these functions analytic. Also is it worth it to create operators that do more complicated things than addition and multiplication (we have the general forms with sums and products, but I'm talking about making shorthand notations and memorizing results like we do with times tables)
You're right about that...this is something I never learned in school. Yesterday, I asked myself, is there a term for x^x? 🤔 Little did I know, this was a just a "key", leading to a whole new realm of mathematics...Tetration! Very mind-boggling! Maybe aliens in higher dimensions use this type of math.👽😁
Norman Wildberger has nice 10 - 20 minute videos about very large numbers, and tetration as you called it, comes up in MF17, MF173, MF176, MF179 (Math Foundations)... very interesting
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.
3:47 But who's counting? *Us! Because we're doing tetration, and that's operation 4, and counting is operation 0, and so if you dig deep enough you'll find counting, though I really wouldn't suggest counting to 5^^3. You might be there for a very long time. Also, congrats on timing the pause button press to read this episode's silly pause gag. Some people hate explaining their jokes, yeah, but not me. You see, this one time, I was at a party with friends,and I said,*
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
I'm sad because when I was younger, I independantly discovered tetration, and until I saw this: I thought it was some cool new out-there idea of mine which I would have to battle to have recognized... ...ahh well, these things happen😆
@@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.
*_...no, I think you want to get into a discussion of operation rules of order, commutativity, associativity... 3^2 ≢ 2^3, is-not-commutative (nor ambi-commutative like quaternions), 3^3^3 is-not-associative (3^3)^3 ≡ 3^(3*3) ≢ 3^(3^3) (nor ambi-associative like octonions)..._*
*_...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)..._*
Much more interesting are fractional tetrations. So, what about 2^^0.5? Or 3.4^^1.243? Or even pi^^e? In common case tatration is none elementary function. And here is fair question: is there formula for tetration, in a kind of infinity converging sum or multiplication, or some integral form like Euler integral for gamma function? What features and laws the tetration has? All of that is pretty interesting...
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.
@@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?
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. :)
I once did what this cat is talking about, operation "unpacking", by implementing Ackermann's function in Pascal and printing out its intermediate results. It was very slow.
in seventh grade the schools gave us our own graphing calculator, and i used to put 2^2^2^2^2... until my calculator overflowed, or did sqrt of a number until it became a mess, and nested trig functions. it was fun making the digits so long
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 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
Wonder what the cube superroot (analogous to division/rooting for tetration) of a googol is.... An unfathomably large number being used by an unfathomably-small-ifying function should make a number we can work with, right?
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?
Why does exponentiation lose its commutative propety? What is unique about operation 3 that causes this property to get removed? Is there any similar property that gets removed when you go from addition to multiplication?
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.
At 2:52 "This begs the question..." No! That is a term of logic (type of fallacy). Look it up. Just say "RAISES the question." Otherwise people will think you graduated from Rutgers
@@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
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 ...
Hello! I like your content and I also have started a math channel, I was wondering if you want to collab for a video or something? I surely do. Let me know!!
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.
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.
NEAT. Many times in solving various equations one runs across "tetration". It just falls out of the works, sometimes. But throughout 10 Years of Ph.D. level engineering math studies I NEVER heard the word tetration used to describe the operation.
You know, there's actually a notation of creating new operations. They're called extended operators. So, how they work is that a{c}b is a{c-1}a{c-1}a...{c-1}a, with b copies of a. Here, let me explain. 3{2}3 would be 3^^3. 3{5}3 would be 3^^^^^3. We can then make nesting within operators, such as 3{3{3}3}3, which is 3{{1}}3. a{{1}}b is a{a{a{...{a}...}a}a}a with b a's from the center out. You can also do a{{2}}b, and so on, creating more and more curly brackets.
+The Taylor Series Are there intermediate operations? (E.g., Operation 'one half'?) Also, could operations corresponding to complex numbering make any sort of sense? Cheers.
Thought about this many times when I was young. never really put much thought into it because I couldn't think of any real life applications for anything that could produce such large numbers.
Can you interpolate/generalize from Integer counting of operation levels to Rational, Real or Complex numbers as operation levels? Maybe you can use 2+2=2×2=2^2=2^^2=2^^^2=...=4 as a kind of fix point for the "lifting" (to the next operation level) functor L('+', 1) = '×' between operations, i.e. between two-dimensional functions. In your video you're using the ("repetition") functor R(f,r) between one-dimensional functions. If c(x)=x+1 is the counting (i.e. incrementing) function, then R(c, x)='+x' , with neutral element 0, i.e. R(c, 0)(x)=x and R(c, x)(0)=x. For the next level we get R('+x', y)='+(x*y)'. We have to apply this function to the neutral element to get the definition of '*' R(R(c,x),y)(0)=x*y. Maybe it's a good start to consider functions p(x)=x+x and m(x)=x×x in functor equation S(p, 1)=m. Another candidate for fix point consideration is 1. 1×1=1^1=1^^1=1 But unfortunately 1+1=2 is an oddball here.
I am not sure! It is an interesting question, but I feel like it has more than one answer. I suspect that there's a such thing as quotient steps -- like, there might be an operator halfway between adding and multiplication. But I am not sure about real numbers, and not even sure how the notion would extend to the complex numbers. But! That is, to some extent, merely a comment on my perception, not the reality of the situation. It may well be possible!
So, here we go. About three years ago I was pondering about arithmetical and geometrical progressions, and I realized the operation pattern, so I thought "I could make infinite types of progressions using this," and so I did. For that I created two operations, I called the first one cycle, and the second one theorericalling. My notation for them was a little circle superscript with the number inside for cycle and a rectangle for theoreticalling. Using those operations I created "exponential progression" and "cyclical progression", and kind of proved to myself that I could create infinite of those by creating "new" operations and using the pattern on the formula. There is just one big difference that makes it different from tetration, though: Instead of exponatiating the exponent, I exponentiated the base. So when I started watching this video, I thought I knew what is was going to talk about when you began to explain the concept, but then I was surprised, which makes me wonder: does the mathematical field have anything like my "cycle", or am I just ignorant and don't know a fundamental flaw with it?
I always have enjoyed the vastly increasing of combinational powers of progressing through operations, that grows so fast it leaves plaid streaks like the ships traveling at ludicrous speed in the movie Spaceballs 😀 But how does one write out the 5th operation, the multiple tetratations?
I found that if you multiply any two numbers that are next to each other in order, like 5 and 6 you will get the result minus 2 for the numbers in the edges, 4 and 7 in this scenario. This works for any four rational numbers that are one number away from each other. In this example 5*6=4*7+2. (try other numbers too). 7*8=6*9+2 If you can tell me why this is so or prove it algebraically I'm all ears. This has been bugging me for over a year, trying to link addition to multiplication and why the number 2. Thanks for whoever read all of this and gave it a try...
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 .
Hey Taylor Series, Loved the video, I had never heard about this operation and it certainly makes sense for it to be the next step. I would like to point out that when you described exponentiation you said "3 times itself 5 times" which is not equal to 3^5. Instead it should be something like the product of 5 threes. The former actually gives you 3^6.
Protip: youtube actually has a frame-advance feature. You can use , and . to skip forward and backward by individual frames in a video. Super useful for the one-frame gag youtubers like to use all the time.
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.
And that's how you get Graham's number.
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... ;)
Here in Spain first we learn about Graham's number, then tetration, exponentiation, multiplication, addition and finally counting
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?
I actually thought about this at school, but didn't found anything about this operations anywhere (I had got no internet back then), so I just assumed it was of no interest.
What I later realized, however, is that the number 2 yields the the same value for all orders:
2 + 2 = 4
2 * 2 = 4
2 ^ 2 = 4
2 ^^ 2 = 4
...
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.
3^7.6 trillion is trivial to calculate! what are you talking about?
It's 10 in base 3^7.6 trillion. ezpz
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)
at first i thought this was like titration, a chemistry thing, when i saw the title
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.
I had heard of tetration before, but even after looking it up I didn't understand it. After watching this video I do, and I understand just how utterly insane the numbers you can calculate with it (and pentration etc) is.
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
3^^5, also known as 3↑↑5
The end of this video would have so nicely flown into Knuth's arrows...
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.
since i was a kid i knew there was possible and had to be a "next level" operation a.k.a more and bigger hyperoperations than the potentiation or exponentiation. i was so wondered when i saw there was a theory behind it and that it had it's own symboles (i had to invent ones when i didn't knew the most used ones) Thank you for explaining this to more people!
Oh wow.
Woa, insanely good production value! Good luck on this channel
Many computer scientists call this the “tower function,” while the “super log” is referred to as the inverse-tower function. Some complex algorithms can be shown to take place in inverse-tower of n time. Note that if such an algorithm was run using all of the atoms in the universe as n, the inverse tower would be roughly 5
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.
0:39 my last brain cell in math exam
Hahahahaha I know how you feel. :)
1. Is there a log equivalent for tetration?
2. What sort of problems does tetration help solve?
3. What algebraic properties does tetration have? Commutation is out of the question, but if these numbers are there and we can't count them in our universe it'd be interesting to work with them in a different way.
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?
3:25
I commend you man. I can see your dedication to the subject you love!
I thought it was fascinating how you explained all those operations. Some of the schools I know just fail really hard to even explain multiplication, they just do some examples and tell "roll with it". Exponentiation sometimes is just "grab a calc, do this". With the really easy and didactic way you showed, I feel I don't need a calc anymore lmao.
Amazing video
Was on TH-cam too distracted to fall asleep.
This helped alot, thank you
Counting, addition, multiplication, exponentiation, tetration, pentation, hexation, etc.…
WHAT YOU SAID??
In all seriousness though, I loved this video. With your sense of humor and dedication to the number language, I can easily see your channel blowing up soon!
Keep it up!!
and as a 12 year old i thought i invented it myself hahah
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.
MR TAYLOR!! I was in Wineberg’s class next door to you a couple years ago. It’s so cool to see you on TH-cam!
Wow the quality of this video is amazing, you deserve so many more subscribers
It's past 1 am, I have no idea how I got here, and holy goodness, I freaking loved it.
This is a great intro to Knuth’s Up Arrow Notation!
Though of the same thing
16 500 subscribers is a rather small number for this type of quality content
Look at the definition of Grahams number. It uses exactly this mechanism and creates a mind-bogglingly huge number (each iteration takes the following number as the grade of operation done to 3 and 3)
Yup! It's bazonkers. :)
g1: 3^^3
g2: 3^^^(3^^3 times)^^^3
Repeat until g64
Well this is amazing, even if you go to college, there are somethings that remain secrets, that shows how wonderful and big humankind knowledge has become so far. Great video The Taylor Series!!!
✓ - Liked!!
Right at the beginning - the calculator you show, I call a 5-function calculator, because in addition to the 4 arithmetic functions, it has a square-root. I don't consider % a separate function, because it's really just a multiply or divide, together with a 2-digit shift.
There are lots of both kinds of calculator, and √x is next to impossible to compute with just +, -, x, and ÷, so I regard the distinction as important.
Tetration - I didn't have a name for it, but I toyed with the idea in Jr. High. As I'm sure lots of kids did who, like me, were math enthusiasts.
Yes, inverting each of those operations, is a natural followup question, and of course, that's a known procedure for opns. 0 - 3.
0: counting down (instead of up)
1: subtraction
2: division
3: root extraction; logarithms (this one has two inverses because, unlike opns. 0-2, it isn't commutative)
4: ?; ? (there will again be two, since tetration is also not commutative)
Also of interest:
A) How can each operation be carried out with non-integers?
Is there an obvious way to apply the answer to this for exponentiation, to opns 4 and up?
B) What connection, if any, is there between these and Graham's up-arrow notation (see: Graham's Number)?
Thanks for doing fun stuff!
Fred
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
Omg I finally found a channel that shares my geekiness. I always think about this things before bed. Maybe that's why I don't get enough sleep!
Well, then there's Trees and Graham Numbers like G64 and Tree (3).
G64 is mibdbogingly huge and just explaining how vastly huge it is would take a good teacher about ten minutes to half an hour. Then theres Tree (3). Let's just explain it this way; G64 is closer to zero than it is to Tree (3). So much closer in fact that if Tree(3) was 100000000, G64 would be almost equal to zero, it would be hard to understand even the number of digits Tree (3) is bigger than G64.
But here's a bigger number, Tree (4), and yet again, Tree (3) is closer to zero than it is to Tree (4).
And now, using these operations we can think of a stupidly large number, so much it wouldn't even be worth using it except for giving curious nerds like me a hardon.
That number is Tree (G64)
Fun fact, I think G64 is like operation 64 or 64...Oh no, wait sorry, G63 is the number of ^ symbols you need to use to get to G64, and G62 is the number of ^ used to get to G63 and so on. As you might guess, just trying to call that Operation N and explaining the value of N would take quite a lot of time.
Just look it up, these numbers are absolutely huge and interesting to study, and Tree and G could be thought of as operations.
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 TH-cam 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
according to wolfram alpha the answer to 3^^5 is 6.002253567994547*10^3638334640023
"Is there something that grows quicker"
Check out the busy beaver function!
True!
The Tree function
@@cubethesquid3919 True.
Rayo’s haha
you are EXCELLENT at explaining sir
4:17 You can use logarithms to get that in powers of 10... But I was lazy, so I used Wolfram Alpha and get 10^10^10^12.5609026413003 and it has
6.002253567994547×10^3638334640023 digits... The last few digits are 9660355387...
Yeah, people have been mentioning that. It's so clever, I love it. :)
I'm sorry, why dont you have more subscribers??? I'll be sticking around for a long time, keep up the awesome work!
Type "CTRL + SHIFT + I" on this window (if you're on a computer). Go to the javascript console and try this operation. You get infinity!
I'm glad TH-cam added this video in my recommendations. Your vids are really well done! :D
Wow Thor is teaching Maths! Greetings, nice video, awesome maths.
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
Here's a cool identity - the "pseudo-distributive property" of tetration over exponentiation:
(a^^b) ^ (a^^c) = (a^^(c+1)) ^ (a^^(b-1))
Alternatively, you might generalise based on the pattern x*y = exp(log(x)+log(y)). So, the next higher operation is defined as the exp of the previous operation applied to the logs. Then you don't lose commutativity, but you end up with a different series of operations. Exponentiation is replaced by x^y = exp(log(x)*log(y)), etc.
As a computer science major, I was taught in college that these operations are described using something called Ackerman's Function.
It is written A(x, y, z)
Where x is the type of operation and y and, if needed, z are the arguments.
Counting up to 5 would be A(1, 5) = 5. 1 for counting and 5 for what you're counting up to. No z argument needed here.
Adding 3 and 4 would be A(2, 3, 4) = 7 -- 2 for addition and 3 and 4 for what's being added.
Multiplication would be A(2, 3, 5) = 15 -- 3 for multiplication and 3 and 5 for what's being multiplied.
A(4, y, z) is, of course, tetration.
“Too large to comprehend” Just wait until these guys find out about Graham’s number ;)
And TREE(3)
Wheel Dragon We can wait but Graham number is also too large to comprehend so that won't change anything.
First time I've seen one of your videos and you've immediately gained another subscriber. Great quality work. Let's hope your subscription rate increases exponentially, initially at least!
3:47 you can actually use , and . to go frame by frame. Incidentally, I managed to pause on the message without doing it, but normally I'd just go frame by frame to see a message like this.
Tetration (and the notation used) makes very large numbers more "pleasing to the eye".
0:05 that's one beautiful calculator!
Wonderful episode. Thanks, Professor!
When I was younger, before I even knew about powers, after learning about multiplication, I thought of this very thing only I didn’t have a good name for it, later I thought about meta functions like 3 meta 5 would be 3 (operation 5) 3 and so on. Great video, I haven’t seen anyone explain this on TH-cam until now, and i’m glad i’m not the only one to come up with this.
Was really looking forward to hear "Ackerman function" in the video :)
But... that is just one version of operation 4! There's many different thinkable opperation 4's! since order does not matter in addition, there is only one version of multiplication. And since order does not matter in multiplication, there is only one version of exponentiation. But order matters in exponentiation!
Look at this. 2^^5 means:
1) Take 2^2
2) Use the result as the *exponent* of another calculation with the previous base: 2^x,
3) Repeat step 2 until you have writen the number 2, 5 times.
Does that not seem arbitrary? Why not put 2^2 into the *base* of a calculation with the previous exponent? that operation would look like this: 2^^5 = ((((2^2)^2)^2)^2), and it fulfills the criteria of unpacking into repeated exponentiation.
If we are allowed to mix it up, there is an infinite amout of versions of operation 4. We can use the *exponent* in every second calculation, and the *base* in the rest, or we can use the exponent only in every third, or forth or...
All of these infinitely many operations unpack into repeated exponentiation, don't they? I think that's why tetration is not that popular. It is not an operation, but an idea, containing infinitely many operations.
And that's just because order in exponentiation matters, unlike in all the previous operations, where a similar, infinite amout of ways to do the operation, all yield the same result. Which btw, i find insanely fascinating.
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.
Is there a nice intuitive reason why the different operations behave so differently regarding their netral element and commutativity? The neutral element for addition is 0, for higher operations it is 1. And while addition and multiplication are still commutative, exponentiation and higher order operations are not. Why don't our operations behave in a more uniform way, since their definition is quite uniform (the (n+1)-th operation is just repeatedly doing the n-th operation).
Hm. That is a deep question. I feel like there is, but I don't have it in my brain right now. I'll think on this.
Ah and then there's Graham's number defined as the 64th level of the following iterative process: On each level put 3 ^^^^....^^^3 where the number of arrows is the value of the previous level. Start at level 1 with 3^^^^3 already which is unimaginably big!
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.
Doing all those commas manually. What a mad lad...
They were refugee Oxford commas that nobody else wanted. Figured I should treat them kindly.
Humm... could we define a similar operation to vectors or matrices? Or maybe to quaternions?
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.
An obvious generalisation is to make the first number (0 for counting, 1 for addition, etc.) a parameter for the function. This is what Ackermann did, and led to the so-called Ackermann function, which originally had 3 parameters, but 2 parameter versions have been devised too. An inverse to (one variation of) the Ackermann function even shows up in some applications in Computer Science, the standard union/find algorithm performs at time complexity of O(n alpha(n)), where 'alpha' is that inverse function. Of course, for all practical purposes alpha(n) is like a constant.
How would you apply this function on negative or non-integer values in the height parameter?
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.
what are the derivatives and integrals of x^^n and a^^x, and are these functions analytic. Also is it worth it to create operators that do more complicated things than addition and multiplication (we have the general forms with sums and products, but I'm talking about making shorthand notations and memorizing results like we do with times tables)
You're right about that...this is something I never learned in school. Yesterday, I asked myself, is there a term for x^x? 🤔 Little did I know, this was a just a "key", leading to a whole new realm of mathematics...Tetration! Very mind-boggling! Maybe aliens in higher dimensions use this type of math.👽😁
which is not enough of my 69420D brain
Hello! Can you also tetrate with 1 and 0? Example: 5^^1 and 5^^0. Is this possibile?
this is 27..... ok got that
this is 7 trillion...... well
this is something absurdly large.... Jeez
The rabbit hole, it goes very deep. :)
@@TheTaylorSeries :0
Norman Wildberger has nice 10 - 20 minute videos about very large numbers, and tetration as you called it, comes up in MF17, MF173, MF176, MF179 (Math Foundations)... very interesting
Is there a negative operation ?
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 !
So glad I was recommended this, this is awesome! Subscribed, keep it up :)
3:47
But who's counting?
*Us! Because we're doing tetration, and that's operation 4, and counting is operation 0, and so if you dig deep enough you'll find counting, though I really wouldn't suggest counting to 5^^3. You might be there for a very long time. Also, congrats on timing the pause button press to read this episode's silly pause gag. Some people hate explaining their jokes, yeah, but not me. You see, this one time, I was at a party with friends,and I said,*
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
The Ackermann function is some generalized form of countng,addition,multiplication,exponentiation,tetration, and so on.
I'm sad because when I was younger, I independantly discovered tetration, and until I saw this: I thought it was some cool new out-there idea of mine which I would have to battle to have recognized... ...ahh well, these things happen😆
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.
Nice to know that im not the only one who thought about what would come after exponentiation.
Hexation 😂😂 i will name it hesitation instead
Would the next be Anticip....ation?
Your way of explaining things is great, wonderful presentation
*_...no, I think you want to get into a discussion of operation rules of order, commutativity, associativity... 3^2 ≢ 2^3, is-not-commutative (nor ambi-commutative like quaternions), 3^3^3 is-not-associative (3^3)^3 ≡ 3^(3*3) ≢ 3^(3^3) (nor ambi-associative like octonions)..._*
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)..._*
Much more interesting are fractional tetrations. So, what about 2^^0.5? Or 3.4^^1.243? Or even pi^^e? In common case tatration is none elementary function. And here is fair question: is there formula for tetration, in a kind of infinity converging sum or multiplication, or some integral form like Euler integral for gamma function? What features and laws the tetration has? All of that is pretty interesting...
Is Tetration defined for non-integers?
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!
Did that high-pitched whining **ping... ping... ping** in the music hurt anyone else's ears?
You were so close to introduce Grahm's number
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. :)
I once did what this cat is talking about, operation "unpacking", by implementing Ackermann's function in Pascal and printing out its intermediate results.
It was very slow.
So Tetration is almost Arrow Notation?
Just substitute up-arrows for the carrots, and you've got it. :)
SlimThrull MONSTER GROUP
@@TheTaylorSeries Yes
in seventh grade the schools gave us our own graphing calculator, and i used to put 2^2^2^2^2... until my calculator overflowed, or did sqrt of a number until it became a mess, and nested trig functions. it was fun making the digits so long
What's operation i?
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
Wonder what the cube superroot (analogous to division/rooting for tetration) of a googol is.... An unfathomably large number being used by an unfathomably-small-ifying function should make a number we can work with, right?
So a googol will be 10^^3?
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
Why does exponentiation lose its commutative propety? What is unique about operation 3 that causes this property to get removed? Is there any similar property that gets removed when you go from addition to multiplication?
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.
At 2:52 "This begs the question..." No! That is a term of logic (type of fallacy). Look it up. Just say "RAISES the question." Otherwise people will think you graduated from Rutgers
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
I didn't recall the term "tetration", but definitely remember the Ackermann function.
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 ...
Hello! I like your content and I also have started a math channel, I was wondering if you want to collab for a video or something? I surely do. Let me know!!
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.
I didn't know this was considered a whole separate operation, I've just been considering it fancy exponentiation.
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.
Gah! You were so, so close to explaining Graham's Number!!!
NEAT. Many times in solving various equations one runs across "tetration". It just falls out of the works, sometimes. But throughout 10 Years of Ph.D. level engineering math studies I NEVER heard the word tetration used to describe the operation.
You know, there's actually a notation of creating new operations. They're called extended operators. So, how they work is that a{c}b is a{c-1}a{c-1}a...{c-1}a, with b copies of a. Here, let me explain. 3{2}3 would be 3^^3. 3{5}3 would be 3^^^^^3. We can then make nesting within operators, such as 3{3{3}3}3, which is 3{{1}}3. a{{1}}b is a{a{a{...{a}...}a}a}a with b a's from the center out. You can also do a{{2}}b, and so on, creating more and more curly brackets.
What a FUN video. Disappointed that it was so short. :-)
+The Taylor Series
Are there intermediate operations? (E.g., Operation 'one half'?)
Also, could operations corresponding to complex numbering make any sort of sense?
Cheers.
Thought about this many times when I was young. never really put much thought into it because I couldn't think of any real life applications for anything that could produce such large numbers.
Can you interpolate/generalize from Integer counting of operation levels to Rational, Real or Complex numbers as operation levels?
Maybe you can use 2+2=2×2=2^2=2^^2=2^^^2=...=4 as a kind of fix point for the "lifting" (to the next operation level) functor L('+', 1) = '×' between operations, i.e. between two-dimensional functions.
In your video you're using the ("repetition") functor R(f,r) between one-dimensional functions.
If c(x)=x+1 is the counting (i.e. incrementing) function, then R(c, x)='+x' , with neutral element 0, i.e. R(c, 0)(x)=x and R(c, x)(0)=x.
For the next level we get R('+x', y)='+(x*y)'. We have to apply this function to the neutral element to get the definition of '*' R(R(c,x),y)(0)=x*y.
Maybe it's a good start to consider functions p(x)=x+x and m(x)=x×x in functor equation S(p, 1)=m.
Another candidate for fix point consideration is 1.
1×1=1^1=1^^1=1 But unfortunately 1+1=2 is an oddball here.
I am not sure! It is an interesting question, but I feel like it has more than one answer. I suspect that there's a such thing as quotient steps -- like, there might be an operator halfway between adding and multiplication. But I am not sure about real numbers, and not even sure how the notion would extend to the complex numbers.
But! That is, to some extent, merely a comment on my perception, not the reality of the situation. It may well be possible!
I stopped doing my math to watch this, nobody thought I wasn't doing my school.
So, here we go. About three years ago I was pondering about arithmetical and geometrical progressions, and I realized the operation pattern, so I thought "I could make infinite types of progressions using this," and so I did. For that I created two operations, I called the first one cycle, and the second one theorericalling. My notation for them was a little circle superscript with the number inside for cycle and a rectangle for theoreticalling. Using those operations I created "exponential progression" and "cyclical progression", and kind of proved to myself that I could create infinite of those by creating "new" operations and using the pattern on the formula. There is just one big difference that makes it different from tetration, though: Instead of exponatiating the exponent, I exponentiated the base. So when I started watching this video, I thought I knew what is was going to talk about when you began to explain the concept, but then I was surprised, which makes me wonder: does the mathematical field have anything like my "cycle", or am I just ignorant and don't know a fundamental flaw with it?
I really liked this video that you made, thank you.
Just found your channel and really well made videos, surprised when you had so few subs.
I always have enjoyed the vastly increasing of combinational powers of progressing through operations, that grows so fast it leaves plaid streaks like the ships traveling at ludicrous speed in the movie Spaceballs 😀
But how does one write out the 5th operation, the multiple tetratations?
I found that if you multiply any two numbers that are next to each other in order, like 5 and 6 you will get the result minus 2 for the numbers in the edges, 4 and 7 in this scenario. This works for any four rational numbers that are one number away from each other. In this example 5*6=4*7+2. (try other numbers too).
7*8=6*9+2
If you can tell me why this is so or prove it algebraically I'm all ears. This has been bugging me for over a year, trying to link addition to multiplication and why the number 2. Thanks for whoever read all of this and gave it a try...
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 .
3:47 - I recently learned a neat trick for stuff like that. Use comma and period to step back and forth a single frame. QED
Hey Taylor Series,
Loved the video, I had never heard about this operation and it certainly makes sense for it to be the next step.
I would like to point out that when you described exponentiation you said "3 times itself 5 times" which is not equal to 3^5. Instead it should be something like the product of 5 threes. The former actually gives you 3^6.
You are correct; I was sloppy with how I said it. Thanks for pointing it out. :)
i knew there had to be a term for this kind of operation, thank you