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Let $\{x_n\}, \{y_n\}$ be two sequences of real numbers with $\{x_n\}$ consisting of distinct numbers such that $x_n\to\infty$. Does there exist a smooth function $f:\mathbb{R}\to\mathbb{R}$ such that $f(x_i)=y_i$ for all $i$? I know there are ways to prove this for finitely many points $x_i$, but here we have infinitely many, so I don't know how to construct such a function.

Aadi Rane
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  • Try looking at this question – Egor Larionov Oct 15 '24 at 13:00
  • That question's answer gives a complex function $f:\mathbb{C}\to\mathbb{C}$, which is not what I'm looking for. – Aadi Rane Oct 15 '24 at 13:24
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    Unless you impose some extra assumption I don't think it's possible. For example, if $x_n \to 0$ and $y_n \to \infty$ then $f(x)$ cannot be continuous at $0$. – George Giapitzakis Oct 15 '24 at 13:46
  • The linked question covers the broader case $f:\mathbb{C}\to\mathbb{C}$ so it includes your $f:\mathbb{R}\to\mathbb{R}$ case assuming $|x_n| \to \infty$. – Steven Clark Oct 15 '24 at 16:41
  • But doesn't $f:\mathbb{C}\to\mathbb{C}$ give a possibility that a real number gets mapped onto a complex number? – Aadi Rane Oct 15 '24 at 19:37
  • The reals are a subset of the complexes, and I don't see any restriction in the linked question's answer that rules out $z_n$ and $w_n$ (your $x_n$ and $y_n$) both being real. The answer seems to suggest $z_n$ and $w_n$ can be arbitrary (except $|z_n|\to\infty$) which would include your restriction to the reals. – Steven Clark Oct 16 '24 at 00:08

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