$f:\mathbb{R} \to \mathbb{R}$ st $f(\frac{1}{2^n})=0 \forall n \in \mathbb{N}$ , show $f'(0)=f''(0)=0$

Question

Let$f:\mathbb{R} \to \mathbb{R}$ be twice differentiable function such that $f(\frac{1}{2^n})=0 \forall n \in \mathbb{N}$. Show that $f'(0)=f''(0)=0$

Applying mean value theorem on $(\frac{1}{2^n},\frac{1}{2^{n-1}})$ we see that $\exists c_n$ st $f'(c_n)=\frac{f(\frac{1}{2^{n-1}})-f(\frac{1}{2^{n}})}{\frac{1}{2^{n-1}}-\frac{1}{2^{n}}}$ , thus $f'(c_n) =0$ $\implies$ $\lim_{n\to \infty}f'(c_n)=0 \implies f'(0)=0$

But I can't use similar argument for $f''$ since it is not known whether it is continuous or not!

$f$ is not defined at $0$, therefore $f'(0)$ and $f''(0)$ do not exist. — HorizonsMaths, May 20 '14 at 14:24
@Mercy ? Why isn't $;f;$ defined in zero? How do you know that? It is given it is a twice differentiable function...(on the whole of $;\Bbb R;$ , according to the given info), so why do you say it is not defined at zero? — DonAntonio, May 20 '14 at 14:26
Yes @Mercy, I can see there's a difference between the question's title and the question's body... — DonAntonio, May 20 '14 at 14:28

Daniel Fischer · Accepted Answer · 2014-06-18T15:06:34.210

7

Hint: For any sequence $x_n \to 0$ with $x_n \neq 0$ for all $n$, you have

$$f''(0) = \lim_{n\to\infty} \frac{f'(x_n)-f'(0)}{x_n}.$$

Picking the right sequence yields the result quickly.

For example the sequence $(c_n)$ of zeros of $f'$, with $c_n \in \left(\frac{1}{2^n},\frac{1}{2^{n-1}}\right)$ whose existence was established in the question. Then each of the quotients is $0$, hence so is the limit.

edited Jun 18 '14 at 15:06

answered May 20 '14 at 14:35

Daniel Fischer

206,697

okay so choosing $x_n= \frac{1}{2^n}$ will work, BTW you are replacing the usual $h$ in definition of derivative by sequence which is going to zero to re write the expression for derivative so is it obvious kind of thing or it requires a proof? – Mathronaut May 20 '14 at 14:39
No, $x_n = \frac{1}{2^n}$ works for $f'(0)$, for $f''(0)$ you should take something else. By definition of the derivative, the derivative is the limit of the difference quotients for all sequences converging to $0$, so we can just take any convenient sequence to determine the limit. – Daniel Fischer May 20 '14 at 14:43
oops I see $x_n =\frac{1}{2^n}$ will work for $f'(0)$ – Mathronaut May 20 '14 at 14:45
which sequence will work for $f''(0)$? – Mathronaut May 20 '14 at 15:08
One with $f'(x_n) = 0$ would be nice. – Daniel Fischer May 20 '14 at 15:15
how do we know $\exists$ such a sequence? – Mathronaut May 20 '14 at 15:17
2

You have such a sequence in your question. It is called $c_n$ there. – Daniel Fischer May 20 '14 at 15:19
oh I see! I was thinking of some example – Mathronaut May 20 '14 at 15:21
Let us continue this discussion in chat. – Mathronaut May 20 '14 at 15:22

Ted Shifrin · Answer 2 · 2014-05-20T14:56:22.203

1

An alternative argument is to write down the second-order Taylor polynomial $P$ of $f$ at $0$ (which, given the hypotheses, we know exists). It has the property that

$$\lim_{x\to 0} \frac{f(x)-P(x)}{x^2}=0. \tag{$\star$}$$

(Set $P(x)=f(0)+f'(0)x+\frac12 f''(0)x^2$ and use ($\star$) when we set $x=2^{-n}$ to show $f'(0)=f''(0)=0$.)

edited May 20 '14 at 14:56

answered May 20 '14 at 14:41

Ted Shifrin

115,160

$f:\mathbb{R} \to \mathbb{R}$ st $f(\frac{1}{2^n})=0 \forall n \in \mathbb{N}$ , show $f'(0)=f''(0)=0$

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