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Dec 10, 2016 at 20:55 history edited Fedor Petrov CC BY-SA 3.0
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May 14, 2016 at 14:34 comment added booksee Thank you so much. I added a comment in the end of the post to address some issues when $f$ can be extended analytically.
May 12, 2016 at 19:51 comment added Fedor Petrov It follows from more general en.m.wikipedia.org/wiki/Borel%27s_lemma
May 12, 2016 at 19:38 comment added booksee Sounds right, would you mind giving an example or reference on the claim "Smooth function may have arbitrary sequence of values of derivatives at a point" ?
May 11, 2016 at 5:38 history edited Fedor Petrov CC BY-SA 3.0
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May 9, 2016 at 16:19 comment added booksee I tried to present an example because that allows us to do a bit computation so as to have a first impression of the problem. For sure we can state the problem without examples.
May 9, 2016 at 16:09 comment added booksee Yes, I understand. Alright, since I can't easily cook up an example to my needs, we shall forget about any concrete examples, and assume we have a case where the curve is smooth and $f$ cannot be extended analytically.
May 9, 2016 at 15:12 comment added Fedor Petrov It is again algebraic equation for $\cos t$. Inverse functions for analytic functions are analytic again, right?
May 9, 2016 at 14:56 comment added booksee I've modified the example so it's not trivial to find an analytic function that equals $f$ on the curve.
May 9, 2016 at 6:48 history edited Fedor Petrov CC BY-SA 3.0
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May 8, 2016 at 22:48 comment added booksee Nice answer. I think I made a mistake in my post above because your argument shows that, in this particular setting, $f(z)=\text{Re} z$ coincides with an analytic function on the arc, so $f$ is actually analytic! It seems that I do need to modify the post later so that $f$ is indeed non-analytic.
May 8, 2016 at 20:54 history answered Fedor Petrov CC BY-SA 3.0