Відео

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)

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?

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 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!

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?

Do yo need to do this complex math? I literally found the 1/2th tetration of 3 with the help of chatgpt.

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)

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!

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

22:49 you say 1, do you not mean 0?

Someone: My problems get exponentially larger! Someone emo:

I have entered the Mathematics Realm in UA-cam it appears.

1:11 Can we also do this backwards? i.e. Express addition itself as doing something else repeatedly?

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

You are so smart. I have conviction that this is very important in math.

What about repeated tetration

googology flashbacks

The methods you have employed in this exploration are really quite primitive. Lambert's numbers, first off, have no generalized exponentiation manifold that can be used to complete the same analysis you have here postulated. Furthermore, Euler's grid-based derivative theorems have no tangible sequential proof that can be reasonably extrapolated to iterate f(x)=e^$ combinatorically. Functions of tetration are not numerically ordinal AT ALL. No one past the 2nd grade can make the claims that you have, unless they hit their head and the brain cells containing all mention of Fermat's Polynomial Truncated Formula are instantly destroyed. ALSO, nice partial-numerical truncation function bypass at 8:45, I know you tried to slip that past unsuspecting viewers but I caught it.

Great Video. Hope you post more, but I do wonder how one would generalize to the irrational numbers.

Well.. I can't think of anyone more suited to explore the next hyper-operation... I guess "pentatration".... Which sounds risky. I honestly dread the thought...

I'll have what he's having.

Tetrisation

Is using 3 parameters to the "exp" function (a superscript, a subscript, and a regular parameter) standard? I didn't know what this meant and the internet didn't help. Also, is there any way that non-integer hyperoperators (eg the 1.5th hyperoperator between addition and multiplication) make sense? :)