0
$\begingroup$

Let $f$ be an infinitely many times continuously differentiable real valued function on set of real no.s Given that $f(1/n)=1/n$ for all $n \in \mathbb{N}$ then find value of $f$ and it's $n$ derivatives at zero.

This function looks like Identity $f(x)=x$ for all $x \in R$ but how can I show it? Thanks and regards

$\endgroup$
4
  • 1
    $\begingroup$ It probably simplifies the problem to define $g(x) = f(x)-x$, so that $g(\frac1n)=0$ for all $n\in\Bbb N$. Can you calculate, for example, $g'(0)$, using the definition of the derivative and the known values of $g(\frac1n)$? $\endgroup$ – Greg Martin Jan 20 '17 at 8:40
  • 1
    $\begingroup$ By the way, $f(x)=x$ isn't the only function that has these values; another one is $f(x) = x + e^{-1/x^2} \sin{\frac\pi x}$ (with $f(0)=0$). $\endgroup$ – Greg Martin Jan 20 '17 at 8:43
  • 1
    $\begingroup$ @GregMartin what about the other derivatives? They seem to be more problematic... $\endgroup$ – mbe Jan 20 '17 at 9:03
  • $\begingroup$ @Gregmartin well this new function looks interesting but it is hard to play with I guess $\endgroup$ – Devendra Singh Rana Jan 20 '17 at 16:04
2
$\begingroup$

It is clear that $f(0)=0$. Then $$ f'(0)=\lim_{n\to\infty}\frac{f(1/n)-f(0)}{1/n}=1. $$ Now $$ f(h)=h+\frac{f''(0)}{2!}\,h^2+o(h^2),\quad f(2\,h)=2\,h+\frac{f''(0)}{2!}\,(2\,h)^2+o(h^2) $$ so that $$ \frac{f(2\,h)-2\,f(h)}{h^2}=f''(0)+o(1) $$ and $$ f''(0)=\lim_{h\to0}\frac{f(2\,h)-2\,f(h)}{h^2}=\lim_{n\to\infty,n\text{ even}}\frac{f(2/n)-2\,f(1/n)}{(1/n)^2}=0. $$ You can iterate this argument and find the higher derivatives.

$\endgroup$
3
  • $\begingroup$ brilliant approach, but what happens to o(1) as h tends to infinity why it vanishes? $\endgroup$ – Devendra Singh Rana Jan 20 '17 at 16:23
  • 1
    $\begingroup$ This is what the $o$ notation means: $o(h^k)/h^k$ converges to $0$ as $h\to0$. If $k=0$ we get $o(1)$ converges to $0$ as $h\to0$. $\endgroup$ – Julián Aguirre Jan 20 '17 at 17:39
  • 1
    $\begingroup$ Alternatively, prove by induction on $n$ using Rolle's theorem: the $n$th derivative $g^{(n)}(x)$ (where $g(x)=f(x)-x$) has a sequence of zeros tending to $0$ .Then we can prove that $g^{(n+1)}(x)=0$ directly from the usual derivative definition. $\endgroup$ – Greg Martin Jan 20 '17 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.