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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.2012-07-04
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    @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).2012-07-04
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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$2012-07-04
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    what is $\beta$?2012-07-04
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    @copper.hat: it is one of 0.45,0.475,0.5,0.525,0.55. Again, not my choice :(2012-07-04
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    @user1963: but t is somewhere near $\frac12$?2012-07-04
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    @robjohn: yes, $\beta$ is always in the vicinity of 0.5. The passage in the paper is around equation (3.10) pp12. "Then, the atypical curves were the composition of 5 amplitude functions generated by (3.9) and the inverse of these warping functions". (3.9) is $\sin(6\pi x)/(6\pi x)$. Does that help? I had no problem replicating the other settings in this paper, but this one i'm stuck :(.2012-07-04
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    Close to $x=0$ we have $f(x) \simeq 2x^\beta$ so the inverse is $\simeq (x/2)^{1/\beta}$. Close to $x=1$ we have $f(x)\simeq x^\beta$ so the inverse here is $\simeq x^{1/\beta}$. This gives a rough approximation to the inverse.2016-03-10

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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.2012-07-04