In this video, I explained the cases in which a tetration can equal an exponentiation. I also showed the other forms of writing hyperoperations and how to say them
Your way of learning is amazing. You explain everything clearly, and you're always smiling, I would love to have you as my prof I'm only in the equivalent of high school but I understand (almost) everything !
Loving your videos. Your teaching style is unique, and very enjoyable. I hope that your clue about "many arrows" is going to be a Graham's Number video. I've yet to see one that I fully understood but I think if anyone can make it clear, it will be you! I also wanted to ask: you showed that 0 and 1 are solutions to 3^^x=3^x but is there an algebraic solution to this type of question? e.g. can we solve something like 2^^x=9^x?
Thank you. I tried an algebraic solution but it was not helpful. Looking for a better way. I would consider superlogarithm and super roots. I hope I can.
well for that example, yes. x=0 is a solution, and if you graph it, I suspect there's a second value in-between -1 and 0 (something like x~-0.843), and a third between 3 and 4 (something like x~3.62), but these would require a proper definition for tetration of real numbers to properly represent.
@@PrimeNewtons I've showed tetration to my math teacher. She has never heard of it. Math theory and just big math I think is the most fun thing to learn about. Thank you!
i feel like including x{2}y (brace notation) and {x,y,2} (array notation) couldve been other good ways to show how tetration was written. I personally really like brace notation for when the arrows get to be a lot, and i like array notation a lot.
Before watching, is tetration even... properly defined for non-integer values? I remember a video from SoME that went quite in depth about it, I'll have to rewatch it, but til then, I know x=0 and x=1 are two solutions to this. as n↑0=1 and n↑↑0=1, similarly n↑1=n and n↑↑1=n. Perhaps there are other solutions that come up when you properly define tetration in the reals. 4:36 nice fix lol
Yes it is. For a^x = a^^x 0 ≤ x ≤ 1 We can better understand using 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. We can use super Logarithm to solve non integer super powers since super Logarithm is repeated Logarithm by definition until the result is less than 1 Let's let sLog2 (16) = 3+x Where 0 ≤ x < 1 (represents a 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 to see what happens to the remainder of 2's that we need. Log2 (1) = 0 Thus sLog2 (16) = 3+0 = 3 Well look what happens when we go backwards through the same process 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 example take sLog2 (20) = 3+x Log2 (Log2 (Log2 (Log2 (20)))) = 0.1088761602 Log2 (Log2 (Log2 (20))) = 2^0.1088761602 Log2 (Log2 (20)) = 2^2^ 0.1088761602 Log2 (20) = 2^2^2^0.1088761602 20 = 2^2^2^2^0.1088761602 = 2^^3.1088761602 So sLog2 (20) = 3.1088761602 meaning 2^^3.1088761602 = 20
And I'd one and zero will give you always a solution for any natural base. The interesting thing are bases between e^(1/e) and 2 --- probably even between e^(1/e) and e where you get a solution which is not just 0 and 1.
Damn it, got me on the first question. I saw the formula and the first question and said to myself with full confidence, "NO" and the very moment he started saying what the two answers were, I realized what an idiot I am. I should have known better, lol.
One question you didn't answer is "What is the practical use of tetration?" I would have imagined that 3††4 is a very large number. But my calculator says that 3††4 is not a number.
while it's evident that 0 and 1 are answers to the equation, I see no proof that there's no other solutions. only a good explanation of tetration. by the way, there's still another way to write tetration. using conway chain arrow. instead of 3↑↑4 you can use 3→4→2 I prefer this method because, for example, instead of 3↑↑↑↑↑↑↑↑↑4 you can use 3→4→9 (a bit more easy to read even if nobody in his right mind would try to compute it)
The equation 3^^x = 3^x actually has infinite solutions We can understand better 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 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
