+65 Solve \(g(1+{g}^{-1}(a))=f(1+{f}^{-1}(1+f(-1+{f}^{-1}(b))))\) for \(f(1+{f}^{-1}(b))\)
+82 First, I need to find f –1(x), g –1(x), and ( f o g)–1(x):
Advertisement
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.
+65 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.
+65 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.
+65 They could also be trinary functions, if you want the base to also change around.
+9519 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))
