Well that ended abruptly lol

@brainandbodytraining

11 ай бұрын

"I'm going to answer you guys this" *ends the video*

@HuTaoEvil

8 ай бұрын

​@@brainandbodytrainingask* not answer

@brainandbodytraining

8 ай бұрын

@@HuTaoEvil thank you for correcting me. You're absolutely right

@tioa.p.1058

8 ай бұрын

​okkkkkkkk

@tioa.p.1058

8 ай бұрын

​

The dual black & red pen handling is awesome!

@carollane8694

6 ай бұрын

I too am just as impressed by this as his mathematical knowledge

@milansunar143

2 ай бұрын

Truly ❤

The final question is clearly no. The LHS is 16*65536 which is just over 1 million. The RHS is clearly way bigger because just 5 2’s in the stack already gives a number with 19 729 digits. On another point, another way to write tetration is to use the up-arrow notation. a tetrated to b is written as a↑↑b (2 arrows) meaning we stack a in a power tower b times. I like this notation more personally as we can actually generalise this to more up arrows. Because exponentiation can also be written as a^b = a↑b (1 arrow). So to have more up arrows, we just repeat the previous level.

@bobbylarue6704

11 ай бұрын

I looked into it and 65536 is 2^16 which would be 2^2^2^2^2 because when you get exponents being multiplied to the power, you multiply them together. For instance, 2^2^3 would be 2^6. I believe he made a mistake because otherwise 2^2^2 would be 2^8. It seems like a simplified way of doing exponential multiplication. I could be wrong though. Edit:I found out my mistake, because I was going up instead of down the chain. I just watched another video and found my mistake.

@Ninja20704

11 ай бұрын

@@bobbylarue6704 a power tower with no brackets means you start from the top and work downwards. So 2^2^2^2 (4 2’s) = 2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536 The law of exponent does not apply because thats when we have (a^b)^c, when what we have is a^(b^c). So theres no simplification we can do besides just manually working it out

@SuryaBudimansyah

8 ай бұрын

RIP Ronald Graham

@devinanderson7615

8 ай бұрын

I did a little research and saw it was ≈18 quintillion

@Bruh-bk6yo

8 ай бұрын

​@@SuryaBudimansyah Graham's number still bigger bruh.

Wow, it's so fast to understand tetration, thank you!

For tetrations to be added, like when multiplying two powers of the same base, ex: (3^5)×(3^4)=(3^9) To achieve this in tetratiin we would have to raise the tetratiin to the power of the tetratiin, ex: (2^^3)^(2^^4)=(2^^7) This can be further generalized for the nth-tration(don't know the general term for tetratiins or pentrations etc) because when you use the nth-tration operation with itself the power/titration multiplies. Ex: ((3^4)^5)=(3^20) And also can be generalized for when you add the power/titration by using the (n-1)th-tration Ex: (4^2)×(4^5)=(4^7) I think this can be further generalized for (n+/-x)th-tration being used simultaenously, such as powers used with pentration, or powers(tritration, I think) used with multiplication. Please excuse the likely incomprehensible jargon I've said as I am neither an expert in this nor am I awake enough to be typing this.

@chaosinsurgency884

11 ай бұрын

Someone please bring some pentrations into the mix and other higher order operations.

@Ninja20704

11 ай бұрын

I don’t think that what you said about taking powers of tetration towers is correct. (2^^3)^(2^^4)=(2^2^2)^(65536) =(2^4)^65536 =2^262,144 =2^2^18 Which is quite clearly not 2^^7. The problem with coming up with rules is that unlike addition or multiplication, exponentiation is neither commutative nor asscociative, so it would be much harder to come up with rules.

@SuperEMT6957

8 ай бұрын

@@Ninja20704 2 to the hyper power of 3 time 2 to the hyper power of 4 is written as: 2^4 * 2^16=2^20=1,048,576 OR 16 * 65,536= 1,048,576 This is so much fun! 😀

@alexeynezhdanov2362

7 ай бұрын

Nope. If you write (2^^3)^(2^^4)==2^^7 then the cat on his t-shirt cries even more. Tetration must be done from the up down and you break that order.

@SYAgencies0379

7 ай бұрын

That why, scientific says, Womankind has 2 xx ,our marker to indentifier that we multiply everything we do. ❤

Master of using two markers at a time! Beautiful.

I am now more confused. 2 to the third tetration is 16, but to fourth is 65536? Shouldn’t that be third tetration is 256? First 2x2=4 Second 4x4=16 Third 16x16=256 Fourth 256x256=65536 I am 50 and never needed more than basic algebra since I left high school so I have forgotten everything😢

@davidknight9709

4 ай бұрын

OK wait.. what I need to do is look at the stack of exponents right? Third is 2x2x2=16 Fourth is 2 multiplied against itself 16 times. Ok. I feel like I can outthink my 6 year old for now. 😊

@user-mg5jd9nf4l

2 ай бұрын

To three its 2^2^2 so 2^(2×2)=2^4= 2×2×2×2=16. Vut to four its 2^2^2^2 so 2^2^4=2^16 so 2×2×2×2×2×2×2... with a 2 being doubled 16 times. The difference is like folding a paper 4 times and folding it 16 times

@collinparham350

Ай бұрын

no. 2 to the third tetration is 2 to the 16th power.

Should WE use the "left exponent" notation, or Knuth arrows notation ? Like 2 ↑↑4 = 2^2^2^2=65536

Just wait for pentation to arrive

Can you define a continuous (extension) tetration function on IR with base e such that it is compatable with other operations of power , multiplication ,..etc.

Thank you for the video.

Do it simply with 2 and 2^2. Just the first and second tetrations. 2*4=8 and that does not equal what the third retraction is. 2^2^2 is 2^4 which is 16. 16 does not equal 8. Therefore a^2 * b^2 does not equal a+b^2. I wish I could notate that better but I’m unsure how to do that in comments on my phone.

So good I could not understand it better'

Great teacher,❤❤

"Which if I remember correctly is 65,536." let me double check that real quick: 2^^4 is 2^2^2^2 so 2^16. I don't know off hand what that is, but I can use exponent properties to break it down into something more manageable because unless I'm mistaken, (2^4)^4 is equal to 16^4, which by using exponent properties is equal to 16^2*16^2. Now for the hard part: calculating 256^2. That would be 256*6+256*50+256*200, so 1536+12,800+51,200=65,536. Checks out.

I'm wondering if you can generalize the height of the power tower (tetration) to any real or complex number.

@tobybartels8426

11 ай бұрын

There's a pretty good discussion in the English Wikipedia article for Tetration, under Extensions. (Short answer: Even generalizing the base can be tricky, since you have to make branch cuts; generalizing the exponent is _really_ hard. Except unlike with the other operations, you can often get interesting results with infinite exponents!)

@ryanman0083

8 ай бұрын

Real numbers yes you can Take slog a(z) = b+x --> a^^(b+x) = z NOTE: "slog" is a common notation for super logarithm, defined as repeated Logarithm where z is not an integer hyper power of a, that means b=Z and 0 Super Log by definition is repeated log until the 0 ≤ answer ≤ 1. (ex: slog2(16) --> Log2(16)=4 --> Log2(4)=2 --> Log2(2)=1: for 2^^x = n, x=(T-1)+r where T is one less than the total logs performed and r is the remainder of the last log. For 2^^x = 16 --> x=(3-1)+1 = 3 thus 2^^3 = 16) By definition of Tetration, a^^(b+x) = a^a^^(b-1+x)^...(b copies)...^a^^x By definition of Super Log, a^^(b+x) = a^a^...(b copies)...^a^x (per the definition above) Both towers by definition are equal to z so they must equal eachother a^^(b+x) --> a^a^^(b-1+x)^...(b copies)...^a^^x = a^a^...(b copies)...^a^x Repeat Log a() on both sides b times to cancel the towers and we get a^^x = a^x By definition, a^^1 = Log a(a^a) = a = a^1 and, a^^0 = Log a(a) = 1 = a^0 Thus we have our Extension of Real tetration powers in the positive direction R+ --> a^^x = a^x For 0 ≤ x ≤ 1 We can use this to define Negative real Tetration powers for 0 ≤ x ≤ 1 a^^x = Log a(a^^x+1) a^^(-x) = Log a(a^^-x+1) = Log a(a^^1-x) Given 0 ≤ x ≤ 1 is true, 0 ≤ (1-x) ≤ 1 is also true, thus a^^(1-x) = a^(1-x) Log a(a^^1-x) = Log a(a^1-x) = (1-x)Log a(a) = 1-x Now we have our two extentions R+ --> a^^x = a^x, For 0 ≤ x ≤ 1 R- --> a^^(-x) = 1-x, For 0 ≤ x ≤ 1, a≠0,1 (Log base 1 and 0 are undefined) I don't know of any extensions for complex hyper powers

I'd never guess that ³2 × ⁴2 = ⁷2; there's no pattern to suggest that. Sure, we have 2³ × 2⁴ = 2⁷, but that's just one level. We don't have 3•2 × 4•2 = 7•2, so clearly this rule only works for exponentiation, not multiplication or tetration. What I _would_ guess is ³2 ^ ⁴2 = ⁷2. This fits a pattern: 3•2 + 4•2 = 7•2, then 2³ × 2⁴ = 2⁷, so why not ³2 ^ ⁴2 = ⁷2? If you write both sides out as power towers, they even have the same number of 2s in the tower. But this isn't true either! Ultimately, this is because exponentiation (unlike addition and multiplication) isn't associative. So (2^2^2)^(2^2^2^2) isn't the same as 2^2^2^2^2^2^2; the parentheses matter.

@argonwheatbelly637

8 ай бұрын

Exponentiation is right-associative. And that makes all the difference. 😊

@tobybartels8426

8 ай бұрын

@@argonwheatbelly637 : Yes, that's right. Although notice that associativity is a property that an operation might or might not have; multiplication has it, and exponentiation doesn't, and there's nothing that we can do about that. But right associativity is a convention about the order of operations, which we can take or leave as we like; it's more convenient for exponentiation, that's all.

@ktejakrishna9620

8 ай бұрын

How did you get to write that tetration format?

@tobybartels8426

8 ай бұрын

@@ktejakrishna9620 : You mean the superscript 3 in ³2? On my phone keyboard, I can press and hold the 3 and that option will pop up. But if you can't get it on your keyboard, you can google something like "unicode superscript three" one of the first few hits should be a page about the Unicode character, where you can copy it from.

just asking, what's the inverse function of tetration? like log to exp, or how would a "root" would work?

@titan7789

7 ай бұрын

The two inverses of tetration are called super-root and super-logarithm

2 to the hyper power of 3 time 2 to the hyper power of 4 2^4 * 2^16=2^20=1,048,576 Is that correct? This is so much fun! 😀

I didn't come further than high school math, but I find this to be very interesting. However, I've seen some videos in which it is said that notations are done differently with upward arrows, because there are higher levels of hyperoperations. About your question at the end of the video: To me it's obvious that ³2•⁴2 can't be equal to ⁷2, because you're multiplying two 2's somewhere inbetween, which means that you're not exponentiating 2 seven times consecutively. The answer to this question is 1,048,576, while the number ⁷2 is much, much bigger. I suspect that you'll have to calculate (³2) to the power of (⁴2) to get to ⁷2, because you're adding 4 more times of exponentiation with 2 to the first 3 times (I think that the parentheses are very important here, because the whole number must be involved, and otherwise you would probably just involve the base without the exponent directly). But I could be wrong, because the problem with exponentiation is that you can't just swap the base and the exponent to get the same result the same way as you could swap numbers with multiplication to get the same result. However, I honestly don't know if you can just add the hyperpowers to get ⁷2 if you calculate (³2) to the power of (⁴2), because (like I said) using (⁴2) as the base and (³2) as the exponent instead would probably give you another answer. In fact, I think you can't even get ⁷2 with either calculations, because you must probably calculate everything between parentheses first, even if it is before hyperoperations. So I'm not sure how all of this could work in a way similar to multiplying exponentiations (probably not at all). And now that I think of it while typing all of this, ⁷2 is calculated exponent by exponent from the top down, so even parentheses don't work in this case. Anyway, multiplying ³2 by ⁴2 doesn't get you even close to ⁷2. And as for ⁷2: ⁷2=2^(2^(2^(2^(2^(2²))))) ⁷2=2^(2^(2^(2^(2⁴) ⁷2=2^(2^(2^(2¹⁶) ⁷2=2^(2^(2⁶⁵⁵³⁶) And from here it doesn't make any sense to go on, because I can't even calculate this with any divice available to me. This number is just insanely high already, so '³2•⁴2 is nowhere near equal to ⁷2' is a very firm understatement. 😄

2 with any hyperoperation to 2 is 4.

Next we could use Knuth up arrow notation and generalise these binary operations a whole lot more...

Can you have a hyper power and a normal exponent on the same number or variable? If you can how do you evaluate to get the proper value?

@megotsnodex

6 ай бұрын

Ah yes, the class PTEMDAS

Had many calc courses at Duke University, and of getting my MA in International finance, had trig, econometrics, stats, and taught math at one of the top private schools in the US and at two universities. Yet, I cannot recall ever teaching or working with tetration, that in spell check comes up as a wrongful word! Yes, the 80's are a bit fuzzy, and apparently beer is not a great study aid, hence my recall might be faulty. However is tetration being taught in math curriculum now? Goodness, I would just write 2 to the 4th power as that, instead of using tetration. Perhaps there is some practical usage in science with exceptionally large or small numbers, but...I wonder if this is taught at any secondary or university math curricula?

On what occasion would anyone use tetrations?

@josephmalone253

4 ай бұрын

When writing a tower of powers in a shorter neater method. Instead of simply stating a number or exponent using tetration keeps track of how many times a base was raised to a power. This could be useful for simplifying some equations or knowing how many times something was doubled not just the end value. In finance and chemistry the doubling time is important (assuming a constant rate of change). You can think of it as a notational trick.

I personally developed a method to calculate fractional tetration, and even complex tetration thanks to someone. My method can also be extended to calculate fractional (and complex) iterations of functions under certain circumstances, and I demonstrated it! So yeah, tetration is among my favorite topics in math, as well as iterated functions, and extending definitions.

@Georgebushdidit

11 ай бұрын

Can you share the method?

@fantiscious

11 ай бұрын

well dont leave us hangin, whats the method

@jordan4835

11 ай бұрын

@@fantisciousright😂

@pierre8235

11 ай бұрын

@@fantiscious Wait a sec, rn I'm busy but I'll manage to later explain it without math symbols.

@pierre8235

11 ай бұрын

@@Georgebushdidit Wait a sec, rn I'm busy but I'll manage to later explain it without math symbols.

Can you express e Outside its polynomial series definition so i can properly relate it to π ahs arclength S=r theta

Time to ask this question from my classmates 😁

Would 2 terraced to the 3 and then all of that raised to the power of 2 tetrated to the 4th equal 2 tetrated to the 7

There’s also pentation and hexation.

@raymax4960

3 ай бұрын

And more, but we dont go that far tho

@EyeSooGuy

3 ай бұрын

@@raymax4960 yup. Heptation, octation, etc etc.

Clear 😊

Wow, nice ❤❤❤

Finally i find a teacher who knows what he is doing😂😂

Nothing is visible from 2:50 (m:s) to rest of the video, that is, tetration part, due to placement of subtitles.

Is tetration defined for all real numbers? Like does 2^^(√2) have a value? ( "^^" means hyperpower )

@Ninja20704

11 ай бұрын

For now, i dont think so. We would need to come up with alternate definitions just like normal exponentiation.

@ryanman0083

8 ай бұрын

Take slog a(z) = b+x --> a^^(b+x) = z NOTE: "slog" is a common notation for super logarithm, defined as repeated Logarithm where z is not an integer hyper power of a, that means b=Z and 0 Super Log by definition is repeated log until the 0 ≤ answer ≤ 1. (ex: slog2(16) --> Log2(16)=4 --> Log2(4)=2 --> Log2(2)=1: for 2^^x = n, x=(T-1)+r where T is one less than the total logs performed and r is the remainder of the last log. For 2^^x = 16 --> x=(3-1)+1 = 3 thus 2^^3 = 16) By definition of Tetration, a^^(b+x) = a^a^^(b-1+x)^...(b copies)...^a^^x By definition of Super Log, a^^(b+x) = a^a^...(b copies)...^a^x (per the definition above) Both towers by definition are equal to z so they must equal eachother a^^(b+x) --> a^a^^(b-1+x)^...(b copies)...^a^^x = a^a^...(b copies)...^a^x Repeat Log a() on both sides b times to cancel the towers and we get a^^x = a^x By definition, a^^1 = Log a(a^a) = a = a^1 and, a^^0 = Log a(a) = 1 = a^0 Thus we have our Extension of Real Tetration powers in the positive direction R+ --> a^^x = a^x For 0 ≤ x ≤ 1 So 2^^√2 = 2^2^√(2)-1 ≈ 2.5185128141

@Zyrkoon

8 ай бұрын

I dont think it is possible to define "a tower of √2 numbers stacked"

@ryanman0083

8 ай бұрын

@@Zyrkoon Well you won't be able to exactly define it since it's impossible to precisely define an irrational number, but it does have an exact form it can be written as x^^√2 = x^x^(√(2)-1) Using more and more decimals of √2 will allow you to be more accurate, but like with any irrational it will never truly be 100%

@AlbertTheGamer-gk7sn

5 ай бұрын

​@@ryanman0083 Just like how all BEAF (and other googolism functions using the Fast-Growing Hierarchy) only accepts nonnegative integers, and extending the domains to the reals would be very hectic, as in order to extend a function's domain from the rationals to the reals, you must define a function that is infinitely differentiable throughout its domain, like the Gamma function for the factorial function, and the diGamma function for the harmonic sequence.

I'm going to guess "no" simply because you started the question with the word "unfortunately".

Yo! When this patch came out?

bro thanks

Kind of curious if there is a notation for instead of going to 2^2^2, going with 2^3^4? For the Power Tower is it always the same number?

@Ninja20704

11 ай бұрын

For tetration it’s always the same number throughout. If you want other numbers you just have to write them.

@danielbickford3458

11 ай бұрын

@@Ninja20704 good to know. So it sounds like that if you have a bunch of different numbers raised to each other then you need to find a way to convert them all into the same number. Not sure how you'd go about doing that, probably something involving logarithms.

@Ninja20704

11 ай бұрын

@@danielbickford3458 i do know about something called the exponential factorial. Its very similar to the normal factorial, except we exponentiate instead of multiply. The notation is n$, which means n$=n^(n-1)^(n-2)^…^3^2^1

@danielbickford3458

11 ай бұрын

@@Ninja20704 nifty

@tobybartels8426

11 ай бұрын

One level down, there's notation for repeated multiplication where you keep increasing the factor, which is to put a bar above the exponent. (Actually there are several notations with this, and several names too, but this is the most extendable.) So 2³ with a bar above the 3 (like how Ē has a bar above the E) means 2×3×4. (You can put the bar below the exponent if you want the factors to decrease instead.) So you could adapt that and use the same bar here, writing 2^3^4 as ³2 with a bar above the 3.

Does titration have any application to physics or chemistry?

@citizenwolf8720

7 ай бұрын

Titration is used all the time in chemical reactions. But perhaps you meant to ask about tetration?

Can anyone explain me how (x+y)²=x²+y² Because if I am correct i remember that in my book it was written that x²+y²=x²•2xy•y²

@DarkvsLight69

3 ай бұрын

(x+y)² is not equal to x²+y² (x+y)² = x² + 2xy + y² x² + y² = (x+y)² - 2xy

is tetration the same of tower power?

@The123Adrian

11 ай бұрын

Yes

³2 ⁴2=³2 2^(³2) which puts us in Lambert W territory so no simple rule here.

I think the rules of exponentiation do not apply with respect to tetration and the reason is that the left-sided exponent is not to indicate the sum of three left-sided exponents with the value of one, and the term repeated exponentiation gives actually already evidence to this cause the base with respect to the upper exponent in every tetration is thus changing when becoming the exponent itself till you reach the base... ...by the way 2tothehyperpowerof 3 times 2tothehyperpowerof 4 is 1048576 thus 2tothepowerof 20 and this is calculated on the spot in my head but 2tothehyperpowerof 7 is somewhat unrealistic as it implies to be able to calculate 2tothepowerof 65536 and you wouldn't still be through.... ...just to give you a notion of how grand the number is iuxtapositioned to 2tothepowerof20... ...by the way qua complete induction with a reductio ad absurdum ( proof by contradiction of the antithesis... ...in this case assuming it be possible... ) you can proof this arithmetically... Le p'tit Daniel, if I got anything wrong just give me a note with respect to my one... ...at least the binomials do I get right cause I know the triangle according to Pascal by heart

2x2 is 4, 4x4 is 16, 16x16 is 256. Can someone explain why 2 hyperpower 4 (pardon the jargon) isn't 256?

@SuperEMT6957

8 ай бұрын

2 to the hyper power of 4 = 2 raised to the power of 16 = 2^16= 65,536 Let me explain. It’s a little confusing🧐. A). 2 to the hyper power of 4, when written out in equation form, will have a TOTAL of four number twos stacked on top of each other (a “tetration tower”). With the base 2 counting as the first one in the series then the exponent 2 written as a superscript to itself three times. 2^[^2((^2)(^2))] B). When calculating tetrations you have too work from the top of the tower back down to the base number. In other words, calculate the exponents first. Starting with the ultimate exponent; power of 2 raised to the power of 2 = 2^2= 4 (your new exponent is 4). Continue down towards the base; the power of 2 raised to the power of 4 = 2^4= 16 (your new and final exponent is 16, in this specific example). C). Finally, multiple the base of 2 to the power of 16= 2^16= 65,536 I hope this helps to clarify 😊. Math is fun!😀

@ronjones1414

8 ай бұрын

@SuperEMT6957 perfect, I went from the bottom up instead of the top down. I wonder if you could put a "ghost" number under the bottom exponent and have it work. I'll have to play with it.

Number 1 is everywhere‼️

what happens if u tetrate by 0?

How do you write (2^3)^2 ?

@abdulrahmanbinsaeed7317

6 ай бұрын

2^6 or 8^2

I want to know how many tetrations of 2 markers this guy can write with at the same time

Why is the text not appearing?

Division is the inverse of multiplication, and logarithms are the inverse of exponentiation. So then, what's the inverse of tetration?

@kerrybarneyiii1202

3 ай бұрын

I believe it’s just stacked Logarithms? Lol I don’t have a calculator right now to do any checking, but I assume it would be taking the log of the log of something. log(log(log(x))) would be the inverse of x^^3 That’s my best guess!

2×2×2 VS 2^2^2 vel 8 VS 2^16

Your Tshirt tells us the reality of majority students 😂, Awesome

How to make seven even Me: 7 pentated to the 2nd

2^4 + 2^16 16 + 65536 < 2^1024. So 2 ^ 1024 is a number so big that even calculators couldn't display!

12^2

tetration doesn’t have large implications or practical uses. So it’s not taught. It’s just a small case of exponent. U can create more such cases of your own.

When did the multiplication sign become a dot? 3.2 is not 3x2 or 3*2. It's 3 1/5

@kerrybarneyiii1202

3 ай бұрын

Putting the “dot” higher up than a decimal is multiplication. Don’t remember why, don’t really care since I always use an asterisk, but we used the dot in high school.

How to read tetration notation like ³2 ??

@hlaingthazin7339

4 ай бұрын

Third titration of 2

Can we go over that multiplication part again? It's adding? You never went over adding!

what about pentation

What comes after tetration?

@SalutLunar

11 ай бұрын

It's called pentation. And that's followed by hexation.

@trihgtwo.Se2

11 ай бұрын

​@@SalutLunarand 10-ation

@eleSDSU

9 ай бұрын

Usually a nap.

@scmtuk3662

3 ай бұрын

​@@trihgtwo.Se2 you forgot heptation, octation, and enneation (although some people call it nonation, but this is Greek, not Latin).

@trihgtwo.Se2

3 ай бұрын

okay sorry decation

Imagine a Millinillion tetrated to a Millinillion to Millinillion💀

I would say, by how exponentially more powerful tetration is the ^3 (2) + ^4 (2) = ^12 (2)

Isn't 2 hyper power4 256?

Now tell us about pentration and infinity multiplication series 😂

= 2^64?

I guess this stuff isn't taught at school probably because practical applications probably are fairly limited - unlike those of ordinary exponentiation and multiplication. Unless you get deeper into some subfields perhaps.

It will be cool to say I earn 2 to the hyper-power of 4 annually :)

that should be "3 reduce to the impotence of 2" 😂😂😂

Answer is 2^20. It can't be written in perfect tetration form

Hi

Isn’t the main reason it isn’t taught in school is because it can form numbers with about 10 billion digits

2↑↑7 = 2^2^2^65536

No, if adding, yes, multiplied, no.

3^^4

teaching system in Turkey teaching us tetration in 5th grade

and pentation

The 2 tetrated to 3 * 2 tetrated to 4 is similar to 2^3+2^4 But if 2 tetrated to 3 is raised to 2 retracted to 4 it is simply (2^2^2)^(2^2^2^2)=2^2^2^2^2^2^2= 2 tetrated to 7

²2=2²=2•2=2+2

16

8

I think ⁵2

Pentatation is the repetition of tetrations

I knew

It's just going to break everything very quickly when the numbers get higher..

its not, right?

If tetration is so *basic* , why isn't it taught, and what are its major applications? That's what makes math basic. I think the word you want is *simple* , because it is indeed easy to understand.

@RyanSmith-lg1cn

4 ай бұрын

It can generate numbers with 10 billion digits and no one is writing all thay

@farmergiles1065

4 ай бұрын

@@RyanSmith-lg1cn There's nothing complex about 10 billion digits. It's just too long to manage any other way. The idea is still simple.

@kerrybarneyiii1202

3 ай бұрын

Simple, yes. Useful? Not really at all.

@farmergiles1065

3 ай бұрын

@@kerrybarneyiii1202 I have to agree. It's part of what I was saying about it being simple, but not basic.

If you change that last problem to exponentiation instead of multiplication, it'll be true.

@tobybartels8426

11 ай бұрын

It seems more reasonable that way, but it still won't work; exponentiation isn't associative, and (2^2^2)^(2^2^2^2) ≠ 2^2^2°2^2^2^2.

@chitlitlah

11 ай бұрын

@@tobybartels8426 Yeah, now that you write it out, it does look wrong. Mea culpa.

I don't see well because the sentence cover....

Answer to the final question is...not even close!

@firebladetenn6633

7 ай бұрын

If I'm not mistaken, the first part of the question is (2^16)*16, and the second part of that equation is 2^128.

I don't like this notation, I prefer Knuth's up-arrow notation it's simply written 2^^3.

t-shirt: =1/4

@sepehrhaghverdi8977

12 күн бұрын

fact: square root of 2↑↑infinity=2

Not serious !!!

@Scp-4419

7 ай бұрын

What do you mean? 😅

The answer is 2 billion

No need to re-invent the wheel.

@Scp-4419

7 ай бұрын

What do you want to say exactly 💯🤔

tu tudetu tudetu tudetu

Tetration is not taught in school because it has no meaning!

tu t’es trompé

It's not taught in school or even college (engineering) because it's pretty much useless.

@PRStudios208

21 күн бұрын

Its only useless because we think in 2d instead of the actual 3d world we inhabit.

@gdevelek

21 күн бұрын

@@PRStudios208 No. It's useless because in the 3d world we inhabit, it doesn't come up in any physics laws, i.e. it doesn't apply anywhere in nature. It's entirely made up.