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The popular function

f(x) = e^(-1/x^2) when x is nonzero = 0 when x = 0

defines one that is flat at 0 (all derivatives vanish there). My question is: How do I construct a (differentiable) function that is flat at more than one points, say, at 0 and 1, simultaneously.

In general, how do I construct a (differentiable) function that is flat at countably infinitely many points?

I am thinking of this problem in the lines of the Pasting Lemma. As for example, taking different functions flat at the points 0 and 1, and then gluing them. But, doing so might be a compromise on differentiability at the point where the pasting is done. How do I get around this problem?

krianaaa
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  • $e^{-1/x^2}e^{-1/(1-x)^2}$ works for $0$ and $1$. If a function is flat at all rational points it is the zero function. – Kavi Rama Murthy Mar 16 '22 at 11:46
  • @KaviRamaMurthy Thanks. Your example gives a function that is flat at every point between 0 and 1. What if we don't want flatness at some points between 0 and 1, say, in some neighbourhood of 1/ 2? Is there some way to achieve this? – krianaaa Mar 16 '22 at 12:11
  • Kavi Rama Murthy's function is only flat at two points: $0$ and $1$, nowhere else. And the same trick can be immediately extended to any finite number of points. For a countable number of points, you would need to make sure the product didn't converge to $0$ over some unintended range. – Paul Sinclair Mar 16 '22 at 18:46
  • You could find a way in the Wikipedia for smooth bump functions, and here is an example – Joako Apr 10 '25 at 19:28

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