Is there any continuous function $\phi$ such that $\phi(\cos x) = \sin x$ over $[0,2\pi)$? If so, could you give me an example?

I stumbled across this problem after trying to train a single layer neural network to do the same thing as my purported continuous function. By the Universal Approximation Theorem I figured that if I can't train the neural net (training error is very high) it means there isn't a continuous function there to approximate the neural net towards. NN are dense in the space of continuous functions. Any thoughts?

EDIT: If it's relevant, I trained a radial basis function neural net.

  • $\begingroup$ What's the purpose of the subscript $0$ ? $\endgroup$ – Yves Daoust Oct 28 '18 at 13:44
  • $\begingroup$ Are you asking about the existence of a continuous function $\phi$ such that $\phi(\cos x)=\sin x$ over $[0,2\pi)$ ? $\endgroup$ – Yves Daoust Oct 28 '18 at 13:45
  • $\begingroup$ There is no purpose for the subscript 0. Yes. That's my question, @YvesDaoust $\endgroup$ – asd11 Oct 28 '18 at 13:51

There cannot be such a function because $\phi(\cos(2\pi-x))=\phi(\cos x)$ while $\sin(2\pi-x)=-\sin x$.


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