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let $f:\mathbb{R} \to \mathbb{R}\ $ be infinitely times differentiable.such that $f(\frac{1}{n})=\frac{1}{n} \forall n \in\mathbb\ {N}$. then find $f(0),f^{k}(0)\ \forall\ k\in \mathbb{N} $.

it is easy to see because of continuity of $f$ ,$f(0)=0$. Now apply taylor formula for $f$ in $[0,x]$.we get $f(x)=xf'(0)+\frac{x^2}{2}f''(t)$ for some $t$ in $(0,x)$.since $f'' $ is bounded in $[0,x ]$ and putting $x=\frac{1}{n}$ in above expression and taking limit as $n\to \infty$. i get $f'(0)=1$.similiary taking other higher order taylor expenssion i get $f^{n}(0)=0$.

is this explaination right.please help.is there any better method to do this ?

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Here's a way that uses the same idea but is a bit cleaner: let $h(x) = f(x) - x$. Then $h$ is infinitely differentiable. Suppose that there is a minimal $n \in \Bbb{N}$ such that $h^{(n)}(0) \neq 0$. For $a$ in some open interval $I$ containing $0$, we may then write $h(a) = h^{(n)}(0)a^n + O(|a|^{n+1})$. This is clearly impossible since it would imply that $h(1/m) \neq 0$ for $m \in \Bbb{N}$ sufficiently large: as $a_m = 1/m$ becomes very small, the $h^{(n)}(0)a_m^n$ term (which is nonzero by assumption) dominates the error term, so they cannot cancel each other. Therefore we have a contradiction and all derivatives of $h$ vanish at zero, so by linearity of differentiation all but the first derivative of $f$ vanish there.

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  • $\begingroup$ It isn't much clear why that would imply $h(1/n) \neq 0$ for $n$ sufficiently large. Could you please elaborate on that? $\endgroup$ – user384138 Jan 13 '17 at 14:44
  • $\begingroup$ Sure, I'll expand in the body of the answer. $\endgroup$ – Vik78 Jan 13 '17 at 14:46
  • $\begingroup$ To be even more explicit, let $c, d$ be nonzero constants, and note that as $a$ goes to zero, $|\frac{da^{n+1}}{ca^n}|$ goes to zero. Therefore for $a$ sufficiently close to zero $|ca^n| > |da^{n+1}|$, which is exactly the property I used in my answer: the dominant term in the Taylor expansion is $ca^n$ for some nonzero $c$, and for $a$ sufficiently close to zero the error term is bounded above in absolute value by $|da^{n+1}|$ for a constant $d$ independent of $a$. Therefore for $a$ small enough the error term cannot cancel the main term, and $f(a) \neq 0$. $\endgroup$ – Vik78 Jan 13 '17 at 14:57
  • $\begingroup$ Oh, it is much more clear as such. Thank you! (not the OP, by the way). +1 $\endgroup$ – user384138 Jan 13 '17 at 15:02

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