[MSE] f(f(z)) = z , f(exp(z)) = exp(f(z))

[tommy1729]

Ok this is very sketchy, I ignore branches and validity and locality and complex analysis and such so take with a grain of salt.

f f = id
g h = h g = id

f = g( 1 - h(x) )

so

f(exp(x)) = exp(f(x))

g( 1 - h exp ) = exp g(1 - h)

g( 1 - h exp g ) = exp g(1 - x)

g( 1 - h exp g(1-x) ) = exp(g(x))

1 - x = x if x = 1/2

g( 1 - h(exp(g(1/2))) ) = exp(g(1/2))

exp(g(1/2)) = Q

g(1 - h(Q)) = Q

f(Q) = Q


so

f(exp(Q)) = exp(f(Q))

f(exp(Q)) = exp(Q)

so exp(Q) is another fixpoint for f ?

also

g(1 - h(Q)) = Q , g(h(Q)) = Q

so 

1 - h(Q) = h(Q)

so h(Q) = 1/2 = h(exp(g(1/2))) = 1/2

but also

h(g(1/2)) = 1/2.

so maybe h(exp(x)) = h(x) ?
But then g( h(exp(x)) ) = g(h(x)) = x and = (gh) exp = exp(x) 
contradiction.

I need a drink ...


regards

tommy1729