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Tetrolith
Canada
เข้าร่วมเมื่อ 18 มิ.ย. 2023
Beyond Exponentiation: A Tetration Investigation
Repeating addition gives multiplication, repeating multiplication gives exponentiation, but what happens when you repeat exponentiation? That is called tetration, and this video explores some of its properties. More specifically, I examine one possible way to extend the domain of tetration to rational heights.
For further reading:
en.citizendium.org/wiki/tetration
Music - Drifting at 432 Hz | Unicorn Heads
This video is a submission to the Summer of Math Exposition Contest (#SoME3)
For further reading:
en.citizendium.org/wiki/tetration
Music - Drifting at 432 Hz | Unicorn Heads
This video is a submission to the Summer of Math Exposition Contest (#SoME3)
มุมมอง: 111 019
As a complete non mathematician, I have spent absurd amounts of time thinking about tetration It's so elusive, mentally speaking. I find it so hard to keep track of it
I’m finding the 1/2 tetration of 2 right now I found an approximation! It’s about 1.459…
Terribly Wrong. 2x3 is adding 3 to 3 , adding 3 once, right? There is a single + sign, so the operation of addition is done o n c e. „Repeat the operation of 2+2 three times“. Ok, I do that: 2+2, 2+2, 2+2 => 4+4+4 = 12, right? You are a bit sloppy, here: 2x3 = 2 + 2 + 2 , i.e. „add 2 to itself three times“. Really? Ok, three times: 2 (itself) + 2 (once), + 2 (twice), + 2 ( three times), right? Gives 8, right? 3x2 means repeated addition of 2 starting with 0 (zero), i. e. 0 + 2 + 2 + 2 = 6, right? Three + signs, that is three times the operation/addition 2 ( to zero!), right? A n d N o t : 2+ 2 + 2 => „three times“. No, three 2’s but 2 + signs, right? You are Wrong, bc two + signs => two „+“ operations!! NOT THREE, i.e. count the operation signs, not the 2‘s, right? Just said.
Thank you 🙏
This can easily solve tetra roots and tetra logs!?! 👏👏👏👍👍👍👌👌👌
What if we can do (complex number)^^(complex number)???
i^i≈0.2078
@@stickman_lore_official6928 I meant Tetration with complex numbers like i^^i.
@@abdul-muqeet i^^i=0.6552811790031976-0.99917222588218111956172i
Repeated Tetration = ?
Pentation
I kept getting confused as to how 2 to 3 tetrated = 16 then how 2 to 4 tetrated was 65536 but after watching this I gained the proper knowledge on how to preform the function, by simply going down the tower of power I go say 2^2 (for the very top being used to use exponentiation the part of the tower 1 down ) which is 4 then it goes down to next 2 making it 2^4 which is 16 then since no part of the tower remains its just the original value of 2^16 which = 65536, god I love when I finally understand math :D
4 + 2.5 = 4 + (1 + 1 + 0.5) Integers or natural numbers (mostly) 4 × 2.5 = 4 + 4 + (4 × 0.5) Integers or rational numbers (mostly) 4 ^ 2.5 = 4 × 4 × (4 ^ 0.5) Rational or irrational numbers (mostly) 4 ^^ 2.5 = 4 ^ (4 ^ (4 ^^ 0.5)) Irrational or transcendental numbers (mostly)
How big is 4 ^^ 2.5?
@@pentalogue_trialogue But that's 2 ^^ 2.5 I mean 4 ^^ 2.5 Any guesses?
@@XanderTran I was wrong: in fact, the result 4^^2.5 lies between the numbers 256 and 1.341...E154
@@XanderTran Yes, I forgot that the base is four, not two.
@@pentalogue_trialogue Yes. So, what would be the result of 4 ^^ 2.5 if it’s between 256 and 1.341…E154? How would you find the result?
2³=2×2×2=2×4=2+2+2+2+2+2=8
I do thought of repeated exponentiation
After watching some videos, I will try to explain 7 growing levels of making a number bigger. If the explanations aren’t clear, I’m sorry, as I’m only a Gr. 5 student, and I’m only doing this out of boredom. Here are the levels: 1. Succesion 2. Addition 3. Multiplication 4. Exponentiation 5. Tetration 6. Pentation 7. Hexation 1. Succesion is basically adding 1 to the number, which we will set as A, pretty simple. So if A was 1, then Succesion would simply add 1 to it, therefore the equation would be: 1+1, which equals 2. 2. Addition is repeated Succession. It is adding A and B together, which could also be written as adding 1 to A a B amount of times. 3. Multiplication is repeated edition. It is adding A to B, a C amount of times. 4. Exponentiation is repeated Multiplication. It’s multiplying the answer of A times B, a C amount of times. 5. Tetration is repeated Exponentiation. It is exponentiating A to B, a C amount of times. 6. Pentration is repeated Tetration. It is tetrating A to B, a C amount of times. 7. Hexation is repeated Pentration. It is pentrating A to B, a C amount of times. Hopefully that made sense, and keep in mind I’m only in Gr. 5.
I really hope that all the mathetmaticians agree on expanding this marvellous monster operation and getting inspiration from this video! Congrats for this great video! 👏👏👏
could you do ⁻²x with complex numbers?
x^^-2 is -∞, because ln(0)=-∞
20:05 20:05 20:05
Very cool video - well done!
16:14 ln0 or log0 can be intended as -infinity or a expansion for 3d complex numbers since with complex numbers we cant calculate it.
Can you do pentation pls? Idk anything about pentation
In exponentiation positive power shows multiplication and negative power shows division. opposite of positive is negative just like opposite of multiplication is division , so In Tetration positive power shows exponentiation and negative power shows logarithm(log)
I just wrote "Beyond tetration: a pentation investigation" on the search box...
Lore of Beyond Exponentiation: A Tetration Investigation momentum 100
Tetration and also pentation and hexation too are wonderful things. Unfortunately, the numbers they create are way too big
a↑↑b a=2: 2, 4, 16, 65536 a=3: 3, 27, 7625597484987 a=4: 4, 256, 1.1579208924e77 a=5: 5, 3125 a↑↑↑b a=2: 2, 4, 65536 a=3: 3, 7625597484987 a↑↑↑↑b a=2: 2, 4 a=3: 3
A closed form solution of x^x=e is x=e^W(1) ~ 1.76322
5:22 I haven’t watched the video Ye this but I already know a short way. I start with the example x tetrated by 3, this is he same as x to the x to the x. In general x tetrated by n is x to the x n/2 times. now I’ll take the natural log of x tetrated to 3, this is the same as the natural log of x to the x to the x. We multiply the powers to get the natural log of x to the x times x, which is the same as the natural log of x to the x squared. Now a property of logarithms is that the log (including natural log) of x to some power is the number in the exponent section times the log of x. So we fix our equation as x squared times the natural log of x. In our equation we can now say x tetrated to n is the same as x to the n-1 time the natural log of x. And it’s reasonable (and true) to assume this works for all numbers. Therefore we just take e to the power of this formula and we have an equation for tetration.
A linear regression using newtons method would be much quicker and is probably simpler to program. However it's not as fancy as your solution
Why is tetration 2^(2^2) and not (2^2)^2 ? The latter would make more sense to me in the sense of hyperoperations, any insights would be helpful. Thanks!
Exponents are always done first due to the order of operations, so you have to start at the top.
They are two different operations. 2^(2^2) is right tetration while (2^2)^2 is left tetration. Standard tetration was defined to be made on the right because of exponentiation notation 2^2^2 = 2^(2^2).
Left tetration is "boring" because (a^b)^c is the same as a^(bc), so right tetration was chosen.
кто решит дам 1 ТРЛН Рублей: Решай со мной (1+1/x)^x =3 Чему равен Х
This is all fine and dandy, but how do you call repeated tetration?
Pentation, Hexation, Heptation, Octation, Nonation... You can see the pattern.
Pentation
Do yo need to do this complex math? I literally found the 1/2th tetration of 3 with the help of chatgpt.
ChatGPT cannot do this. ChatGPT fails at basic algebra and calculus.
@@denizkirbiyik9221 Well if you ask it directly it won't be able to do it, you just gotta break it down a little.
What value did ChatGPT give you? Using the method in the video I got approximately 1.7068.
Can you do the same video to Pentation ? Like using Real Numbers in Pentation
Can you do tetration tower like ³ ³ 3 Or Pentation 3 ³
@14:46 I don't see how you can get 5a(x) and not 6a(x). If 3a(x) = a(a(a(x))), then 2a(3a(x)) = 2a( a(a(a(x))) ) = a(a(a( a(a(a(x))) ))) = 6a(x)
i mean he’s basing it off a different rule he found earlier in the video, that being ma(na(x))=(m+n)a(x) therefore making 2a(3a(x))=5a(x)
We cant go further to the negative number tetrations ? That was my doubt brother. Also you are absolute genius because you are the only one ive seen do it in this huge platform bro , you deserve more support , thank you man!
Actually, there are 2 ways to tetrate with negative number exponents, using either opposite of exponentiation.
I can't even begin to wonder what a complex tetration would look like
😮
Interesting!
ⁿn or n^^n=n^^^2
Sorry its too much algebra.😕
Why do you set the coefficients of the different powers of x to 1, 1 and 1/2 at 20:33?
I feel like tetration would be more useful in a 4-dimensional universe
Real numbers can be defined with super Logarithm (inverse of Tetration) By definition sLog2 (2^^3) = 3 NOTE: "sLog" is a notation for super Logarithm. Like how Logarithm cancels the base leaving the exponent ex. Log2 (2^3) = 3 super Logarithm does the same with Tetration leaving the super power. We can use super Logarithm to solve non integer super powers since super Logarithm is repeated Logarithm by definition. Let's let sLog2 (16) = 3+x Where 0 ≤ x < 1 (represents a 0 or decimal) sLog2 (2^^3) = sLog2 (2^2^2) => Log2(2^2^2) = 2^2 => Log2(2^2) = 2 =>Log2(2) = 1 At this point we've taken three logs representing our integer part of the solution (given by the fact that the answer is equal to 1). We just take log again for the decimal x (the remainder of 2's that we need.) Log2 (1) = 0 Thus sLog2 (16) = 3+0 = 3 Well let's look at what happens when we go backwards through the same process to see what happens to the remainder. Log2 (Log2 (Log2 (Log2 (16)))) = 0 Log2 (Log2 (Log2 (16))) = 2^0 Log2 (Log2 (16)) = 2^2^0 Log2 (16) = 2^2^2^0 16 = 2^2^2^2^0 = 2^2^2 = 2^^(3+0) The remainder adds an extra '2' to the top of the power tower and the additional 2 is raised to the power of the remainder For 0 ≤ x ≤ 1 By definition sLog a(a^^3+x) => a^a^a^a^x By definition of Tetration a^^3+x = a^a^^2+x = a^a^a^^1+x = a^a^a^a^^x a^a^a^a^^x = a^a^a^a^x a^a^a^^x = a^a^a^x a^a^^x = a^a^x a^^x = a^x by definition for 0 ≤ x ≤ 1 so e^^1/2 = e^1/2 = √e ≈ 1.6487212707
wow, well said. great observation.
what seems to be overlooked a lot is that even if different answers are met with different methods, the answer other people get for e^^1/2 is still very, *_very_* close to sqrt(e).
22:49 you say 1, do you not mean 0?
Someone: My problems get exponentially larger! Someone emo:
My problems get -tetrationally- Pentationally larger
I have entered the Mathematics Realm in TH-cam it appears.
1:11 Can we also do this backwards? i.e. Express addition itself as doing something else repeatedly?
Thats called zeration or hyper0 operator
@@returndislikes6906 So.............basically doing nothing? 😅
@@feynstein1004 its logical operator. it is similar to max function
@@returndislikes6906 Hmm I don't understand
12:23 But... why can you do this???
yknow you could've reused the nested function notation for exp
I initially thought of it as arrow notation rather than the superscript notation Graham’s # is unnecessarily big
It's really small, just ≈ 3{{1}}65
You are so smart. I have conviction that this is very important in math.
What about repeated tetration
Pentation