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

  • 1
    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 $g(\frac1n)$? – 2017-01-20
  • 1
    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 $f(0)=0$). – 2017-01-20
  • 1
    @GregMartin what about the other derivatives? They seem to be more problematic... – 2017-01-20
  • 0
    @Gregmartin well this new function looks interesting but it is hard to play with I guess – 2017-01-20

1 Answers 1

2

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.

  • 0
    brilliant approach, but what happens to o(1) as h tends to infinity why it vanishes? – 2017-01-20
  • 1
    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 $h\to0$. – 2017-01-20
  • 1
    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. – 2017-01-20