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I am looking for a homotopy between $y=x$ to $y = {\rm step}(x-1)$ (to the limit; a very steep sigmoid is also ok). More formally, I look for a family $y_\epsilon \in C^1(A,A)$ where $A = [0,\infty)$ $\epsilon \in [0, 1)$ such that $y_0 = y$ $\lim_{\epsilon \rightarrow 1} y_\epsilon = {\rm step}(x-1)$ Any hint is appreciated.

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    @copper.hat: You can equivalently think of a homotopy of cost function $J_\epsilon$, with solutions $y_\epsilon$. But it's easier to think directly in terms of $y_\epsilon$.2012-05-07

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How about: $ y_\epsilon(x) = (1-\epsilon) x + \epsilon\left( \frac{1}{2} + \frac{1}{\pi} \arctan\left(\frac{x-1}{1-\epsilon}\right)\right) $ Clearly $y_0(x) = x$ and $ \lim_{\epsilon \uparrow 1} y_\epsilon(x) = \lim_{\epsilon \uparrow 1} \left( \frac{1}{2} + \frac{1}{\pi}\arctan\left(\frac{x-1}{1-\epsilon}\right)\right) = \cases{1 & x > 1 \\ 1/2 & x=1 \\ 0 & x<1} $

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    Terrific, thanks! This will do. First time I check this website and already love it. Thanks guys! Stefano2012-05-07