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In the context of neural networks and cryptography, I would like to approximate some activation functions. However, I need to approximate them into polynomial forms for my purpose.

It seems that chebyshev polynomial approximation is the more suited one for my case, but while I read in some papers that we can enlarge the approximation for a bigger interval, all the algorithms i've seen throughout the web only mention the case $[-1,1]$...

So how do we compute this process to get a polynomial approximation of a function given an interval $[a,b]$ (the interval can be considered as $[-a,a]$ if it is easier)

If some of you have a clue to this, I would be very grateful! Thank you

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  • $\begingroup$ Let $\phi: [-1, 1] \to [a, b]$ be the bijection given by $x \mapsto \frac{b - a}{2}x + \frac{b + a}{2}$. It is simply an affine transformation given by scaling and translation. Then $f \circ \phi$ would be a $[0, 1] \to \mathbb{R}$ function. Now apply your approximation, you get some $[0, 1] \to \mathbb{R}$ approximation function, call it $\overline{f \circ \phi}$. Finally, applying $\phi^{-1}$ gives $\phi^{-1} \circ \overline{f \circ \phi}$ which is a $[a, b] \to \mathbb{R}$ approximation function. Does it work for you? $\endgroup$ – Alex Vong Dec 8 '18 at 7:17
  • $\begingroup$ I haven't tried yet but it should definitely work! Thank you very much! $\endgroup$ – Robin S. Dec 8 '18 at 8:32
  • $\begingroup$ Glad it helped! $\endgroup$ – Alex Vong Dec 8 '18 at 9:21
  • $\begingroup$ I actually have an issue (sorry i didn't mention it earlier...) I want to approximate the sigmoid function, but on the interval [-8,8] roughly.. Given what you told me, the bijection is only x maps to 8x, but when I compute the approximation polynomial to sigmoid(8x), and then apply the inverse of our scaling, I basically get the approximation polynomial of the usual sigmoid (hence effective only really close to 1)... Did I miss something? $\endgroup$ – Robin S. Dec 14 '18 at 7:07
  • $\begingroup$ I understand what's wrong now. I wrote $\phi^{-1} \circ \overline{f \circ \phi}$ but in fact it should be $\overline{f \circ \phi} \circ \phi^{-1}$. Does the approximation work now? $\endgroup$ – Alex Vong Dec 14 '18 at 13:19

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