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Let $f,g\in C^{\infty}(\mathbb{R})$ such that \begin{equation*} f(1/n)=1, g(1/n)=\frac{n}{1-n^2} \end{equation*} for $n=2,3,4,\dots$

What are the possibile values of $f(\pi)$ ?

Do you have any suggestion (I do not want the solution) how to start?

The only thing I can say is that, by continuity, $f(0)=1$, but actually I do not understand how this information can help me, since the exercise requires the value of $f$ in $\pi$.

Thanks in advance for any advice!

P.S. : I copied the entire text of the exercise (which has other two points), but the question is only about the first points of it, which does not involve the function $g$.

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    $\begingroup$ You can't conclude anything about $f(\pi)$ with that information. – I suggest that you remove the facts about $g$, which are irrelevant here. $\endgroup$ Commented Aug 8 at 8:48
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    $\begingroup$ If the context were complex differentiable functions, you could. But with $C^{\infty}(\mathbb{R})$, not. $\endgroup$ Commented Aug 8 at 8:49
  • $\begingroup$ Are you sure they are not entire functions? $\endgroup$ Commented Aug 8 at 10:05
  • $\begingroup$ Unfortunately, $\displaystyle e^{-1/x^2}\sin\frac\pi{x}$ (and $0$ for $x=0$) is $C^\infty$, and $0$ for $x=1/n$, so there‘s no uniqueness. $\endgroup$ Commented Aug 8 at 10:06
  • $\begingroup$ Unfortunately, the text of the exercise says only that $f$ (and $g$) is a smooth function, but real...I do not think the context is that of complex analysis. If they were complex functions, what could I say about $f(\pi)$ ? $\endgroup$ Commented Aug 8 at 10:20

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We are given a function $f\in C^\infty(\mathbb{R})$ satisfying

$ f\!\Bigl(\tfrac1n\Bigr)=1, \qquad n=2,3,4,\dots $

and we wish to determine the possible values of $f(\pi)$.

Claim. The value $f(\pi)$ may be chosen arbitrarily in $\mathbb{R}$

Argument. Since the points $\{1/n : n\ge2\}$ accumulate only at $0$, continuity forces $ f(0)=\lim_{n\to\infty}f\!\Bigl(\tfrac1n\Bigr)=1, $ but there is no restriction at any other point of $\mathbb{R}$. In particular $\pi$ is not a limit point of $\{1/n\}$, so one can prescribe any desired value $A$ at $x=\pi$ without violating smoothness.

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