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Suppose I have the function


with zeroes $y=0$ (2x), $y=1$, $y=3/2$, which I only need on the part of the domain $0\le y\le 1$.

Is there a transformation $y\rightarrow y'$, such that

  • $f(y')$ is even
  • The transformation is invertible on the specified domain
  • The transformation of the derivative $\frac{d^2f}{dy^2}$ still looks 'nice'
  • $0$ and $1$ get mapped to $y'_0$ and $-y'_0$ or vice versa.

I am aware that my third requirement is a bit vague, but I don't know a better formulation for now.

I tried some transformations that first maps $3/2$ and $1$ to the same value, but it will not be invertible for sure.

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up vote 1 down vote accepted

You have $f(y)=2y^2(y-1)(y-\frac 32)$, with a double root followed by two single roots. You won't be able to get it even with an affine transformation $(y'=ay+b)$ or any order preserving transformation because of that pattern. For an even differentiable function the roots have to be symmetric around zero, including multiplicity. If you don't respect order, it becomes harder to have the transformation invertible, and the odds of it being 'nice' are slim.

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I was a bit afraid this would be the answer. I will accept your answer later on, if no other answer appear. Thanks. – Bernhard Oct 15 '12 at 15:21

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