Given data points $(x_i,y_i)\in \mathbb{R}^m\times \mathbb{R}^n$ with $n>m$ satisfying $y_i=f (x_i)$ with a sufficiently smooth injective unknown function $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$ and a value $y \in Im (f)$, I want to find an estimate for $x\in \mathbb{R}^m$ s.t. $f (x)=y$. Does a problem of this sort have a name. Is there any literature available? I am thankful for any suggestions.
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$\begingroup$ Is $f$ known? Or are there known bounds on its derivatives? $\endgroup$– Robert IsraelCommented Sep 20, 2017 at 7:17
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$\begingroup$ $f$ is not known. We dont have explicit bounds on $f$ but we know that it is supposed to be a "nice" function with small derivatives. Since this is an applied math problem I could not state a mathematically rigorous problem and there is probably not a unique solution to it. I am looking more towards ways to approach such kinds of problems. $\endgroup$– user35593Commented Sep 20, 2017 at 8:06
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$\begingroup$ Isn't this essentially interpolation on $f^{-1}$? (if $f$ is injective --- otherwise there is no unique solution.) $\endgroup$– Federico PoloniCommented Sep 20, 2017 at 8:14
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$\begingroup$ @Federico Poloni: basically yes, it is also why I called it inverse interpolation. However f^{-1} is defined on Im(f) which is an $m$-dimensional submanifold of $R^n$. So we have to do interpolation on a (unknown) manifold. Furthermore we know that $f$ has small derivatives but not of $f^{-1}$. $\endgroup$– user100927Commented Sep 20, 2017 at 8:23
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1 Answer
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After having calculated an "explicit" interpolating function $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$, satisfying $y_i=f (x_i)$, you can calculate the local inverse via evaluation of the implicit taylor series or some other series expansion of implicit functions.
If you are however satisfied with very pragmatic solution, you can chose a data point $(x_i,y_i)$ that minimizes $\|y-y_i\|$ and then via the method of steepest ascend or descent identify the "nearest" $x$, for which $f(x)=y$ w.r.t. the interpolating function $f$; that should also disambiguate the calculation.
If $f$ is smooth, it can be assumed to be (piecewise) polynomial and thus the gradients that are used in the method of steepest ascend or descent can be easily calculated.
answered Sep 20, 2017 at 8:43