creepercraft97T3

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 #5
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The hyperoperations are basically a sequence that extends the 3 basic functions everyone is familiar with.

Addition is hyperoperation 1

Multiplication is hyperoperation 2

Exponentiation is hyperoperation 3

Tetration is hyperoperation 4

and it continues on forever (although hyperoperations greater than or equal to 4 have problems with their domains).

 

Let f(x) be the 'n'th hyperoperation. You can get from hyperoperation 'n' to hyperoperation 'n+1' by setting the 'x' of f(x) to a certain value and then iterating f(x) 'x' times. That takes a lot of work, but for going from hyperoperation 'n' to hyperoperation 'n-1', you just need to use the formula \(f(1+{f}^{-1}(x))\), where \(f^{-1}(x)\) is the functional inverse of \(f(x)\). Using this, you can find out how to go from hyperoperation 'n' to hyperoperation 'n-2' or other values. Basically, you can go from hyperoperation 'n' to hyperoperation 'n-v', where 'v' is an integer.

 

Hyperoperation 'n-2' from hyperoperation 'n': \(f(1+f^{-1}(1+f(-1+f^{-1}(b))))\)

 

I then remembered something I did about a year ago. I set up this equation \(f(a)=f^2(b)\) and solved for \(f(b)\), and I got \(f^{0.5}(a)=f(b)\). Then I did it for the 3rd iteration, then the 4th, and so on. In other words, I learned how to get \(f^{\frac1n}(x)\) from \(f^n(x)\). So I figured that I could apply this to hyperoperations.


Basically, if you set up this equation \(g(1+g^{-1}(a))=f(1+f^{-1}(1+f(-1+f^{-1}(b))))\) ('n-1' and 'n-2' equations equal to each other) and manage to solve the right side for \(f(1+{f}^{-1}(x))\) (the 'n-1' equation), the left side will, consequently, turn from an 'n-1' equation into an 'n-0.5' equation. This would give us the ability to define fractional hyperoperations (and not just for 'n-0.5'), and if a useful, usable pattern is found, can lead to the generalization of all hyperoperations.

May 2, 2019