F(f(x))=exp(exp(x)).

[Gottfried]

Try this one:

Code:

g(x) =2*( 1  -0.693147180560*(x/2-1) - 1.32551766575*(x/2-1)^2 - 2.50705892304*(x/2-1)^3 - 4.77252348338*x^4
- 9.15698933257*x^5 - 17.7013067056*x^6 - 34.4511806278*x^7 - 67.4604160110*x^8 - 132.822317065*x^9
- 262.806118782*x^10 - 522.322180952*x^11 - 1042.32389137*x^12 - 2087.72385956*x^13 - 4195.78438654*x^14
- 8458.67594200*x^15 - 17101.4712999*x^16 - 34666.5488624*x^17 - 70444.6162583*x^18 - 143472.156426*x^19  
- 292819.543812*x^20 - 598799.786851*x^21 - 1226745.99525*x^22 - 2517482.69947*x^23 - 5174508.71132*x^24  
- 10651690.6480*x^25 - 21957071.7238*x^26 - 45321021.1391*x^27 - 93661312.1666*x^28 - 193787001.856*x^29  
- 401387548.839*x^30 - 832244988.655*x^31) + O(x^32)

With a convergence range of about \(1/2 \) this should give the "dual function" (as I once christened this) for \( f(x) = b^x \) with \(b= \sqrt 2 \) to the effect that \(g(g(x))=f(f(x)) \) but \( g(x) \ne f(x) \)

A short inspection, (using Euler-summing of the series when (not-too-much) out of convergence) gave
  \[ g(0) = 2.46791405115 \qquad g(g(0)) = \sqrt 2   \]
while, of course,
  \[ f(0) = 1 \qquad f(f(0)) = \sqrt 2   \]
Note, that the value \(2.46791405115 \) is exactly the value in my picture (recently posted here) of imaginary iteration of \(b^x\) where \( b= \sqrt 2\) and \( f(0)=1 \) is iterated with imaginary height .

The power series of the function \(g(x) \) was created using the Schroeder-procedere as mentioned in my first answer with the help of diagonalizing the Carlemanmatrix for \(t(x) = 2^x-1 \).              

Gottfried

[JmsNxn]

Note, that the value \(2.46791405115 \) is exactly the value in my picture (recently posted here) of imaginary iteration of \(b^x\) where \( b= \sqrt 2\) and \( f(0)=1 \) is iterated with imaginary height .

Don't you just love the parallel of \(\sqrt{-1} = i\)!

[Gottfried]

Note, that the value \(2.46791405115 \) is exactly the value in my picture (recently posted here) of imaginary iteration of \(b^x\) where \( b= \sqrt 2\) and \( f(0)=1 \) is iterated with imaginary height .

Don't you just love the parallel of \(\sqrt{-1} = i\)!

Yeppo !