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I'm trying to replicate a simulation study in a paper. For that i would need the inverse of this function:

$f(x)=x^{\beta}\left((x-1)^6+1\right), x\in[0,1]$

Plugging this unto Maxima returns:

solve(y=x^(beta)*((x-1)^6+1),x);
   beta        beta + 3                                                                            
[x     = y - x        ]

which doesn't help me one bit. The fact that it's used in this paper leads me to think that the inverse of this function is probably well known. It's just not known to me.

Alternatively, is there a general numerical strategy to find the inverse of a function for a grid of values of x?

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I am afraid you cannot hope for an elementary function. –  Siminore Jul 4 '12 at 8:46
    
@Siminore: Thanks, that's already a great step forward. Could you elaborate a bit on why (i could change the simulation setting i guess, but i will have to justify). –  user1963 Jul 4 '12 at 8:47
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It's very hard to prove rigorously that the inverse (if it exists) cannot be expressed by elementary functions. Most of the times, it is something you feel :-) In general, it is impossible to solve explicitly $P(x)Q(x)=y$ w.r.t. $x$ –  Siminore Jul 4 '12 at 8:50
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what is $\beta$? –  copper.hat Jul 4 '12 at 8:51
    
@copper.hat: it is one of 0.45,0.475,0.5,0.525,0.55. Again, not my choice :( –  user1963 Jul 4 '12 at 8:52
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1 Answer 1

up vote 2 down vote accepted

As noted in the comments, there is no closed form for the inverse. There are numerical techniques for finding very good approximations. One of the simplest and best is called Newton's Method - you will have no trouble finding it.

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For the non-math inclined, we look for $min_x(f(x)−y)^2$ for say a grid a values of y between $\max(f(x))$ and $\min(f(x))$ using a Newton-Raphson algorithm. –  user1963 Jul 4 '12 at 10:04
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