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Solve \(g(1+{g}^{-1}(a))=f(1+{f}^{-1}(1+f(-1+{f}^{-1}(b))))\) for \(f(1+{f}^{-1}(b))\)

 May 2, 2019
 #1
avatar+68 
+2

First, I need to find f –1(x), g –1(x), and ( f o g)–1(x):

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Inverting  f (x):

Inverting g(x):

Finding the composed function:   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

Inverting the composed function:

Now I'll compose the inverses of  f(x) and g(x) to find the formula for (g–1 o f –1)(x):

Note that the inverse of the composition (( f o g)–1(x)) gives the same result as does the composition of the inverses ((g–1 o f –1)(x)). So I would conclude that

f (x) = 2x – 1 
    y = 2x – 1 
    y + 1 = 2x 
    (y + 1)/2 = x 
    (x + 1)/2 = y 
    (x + 1)/2 =  f –1(x)

g(x) = (1/2)x + 4 
    y = (1/2)x + 4 
    y – 4 = (1/2)x 
    2(y – 4) = x 
    2y – 8 = x 
    2x – 8 = y 
    2x – 8 = g –1(x) 

( f o g)(x) = f (g(x)) = f ((1/2)x + 4) 
    = 2((1/2)x + 4) – 1 
    = x + 8 – 1 
    = x + 7

( f o g)(x) = x + 7 
    y = x + 7 
    y – 7 = x 
    x – 7 = y 
    x – 7 = ( f o g)–1(x)

(g–1 o f –1)(x) = g–1( f –1(x)) 
    = g–1( (x + 1)/2 ) 
    = 2( (x + 1)/2 ) – 8 
    = (x + 1) – 8 
    = x – 7 = (g–1 o f –1)(x)

( f o g)–1(x) = (g–1 o f –1)(x)

While it is beyond the scope of this lesson to prove the above equality, I can tell you that this equality is indeed always true, assuming that the inverses and compositions exist — that is, assuming there aren't any problems with the domains and ranges and such.

 May 2, 2019
 #2
avatar+68 
+1

did you get the right answer??

 May 2, 2019
 #3
avatar+65 
+1

I actually don't know the answer (or even how to go about solving it), so I asked the question to see if anyone else could figure it out. However, if it is solved, then the results could be groundbreaking.

creepercraft97T3  May 2, 2019
 #4
avatar
+1

groundbreaking how?

Guest May 2, 2019
 #5
avatar+65 
+2

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.

creepercraft97T3  May 2, 2019
 #6
avatar
+1

Hyperoperations are binary functions

Guest May 2, 2019
 #8
avatar+65 
+1

They could also be trinary functions, if you want the base to also change around.

creepercraft97T3  May 6, 2019
 #9
avatar
+1

Yes but then there is only one such function.

 

Besides you were referring to some hyperoperations as f(x) despite the fact that none of them are unary.

Guest May 6, 2019
edited by Guest  May 6, 2019
 #7
avatar+7499 
+2

Presumably, both f(x) and g(x) has inverse functions. 

\(g(1+{g}^{-1}(a))=f(1+{f}^{-1}(1+f(-1+{f}^{-1}(b))))\\ f^{-1}(g(1+{g}^{-1}(a))) = 1+{f}^{-1}(1+f(-1+{f}^{-1}(b)))\\ f(f^{-1}(g(1+{g}^{-1}(a)))-1) = 1+f(-1+{f}^{-1}(b))\\ f(-1+f^{-1}(b)) = f(f^{-1}(g(1+{g}^{-1}(a)))-1)-1\)

But I can't solve for f(1 + f^{-1}(b))

 May 2, 2019

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