I'm no slouch on reasonable pure mathematics, but I can't claim to be an explorer of the outer edges. So I have a question. Aside from perhaps a few instances of simplified notation, are there any applied uses for tetration? Or theoretically, quadration, pentration, sextration, etc? Not that I'm dismissing the field - there is validity in considering the nature of iteration of iteration. I'm just curious to know.
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
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! 👏👏👏
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)
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.
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!
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).
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!
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
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).
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.
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...
