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Let $f:\Bbb{R}\to\Bbb{R}$ such that

$$f(x) = \begin{cases} c_n & \text{if }x=\frac1n\text{ for some }n\in\mathbb N \\ 0 & \text{elsewhere} \\ \end{cases}$$
where $c_n$ is a given sequence. Find the condition on the sequence $c_n$ such that $f'(0)$ exists.

Here is what I got, I'm actually not sure only about the last part, when I choose the condition for my $N$.

So now, be definition of derivative, $f'(0)$ defined when $$\lim_{x\to0}{f(x)-f(0)\over x-0}=\lim_{x\to0}{f(x)\over x}$$ Now, if $x={1\over n}$ then ${f(x)\over x} = \begin{cases} nc_n & \text{for n =1,2,3...} \\ 0 & \text{elsewhere} \\ \end{cases} $

We have $$f'(0)=\lim_{x\to0}{f(x)\over x}=\lim_{n\to\infty}{f(1/n)\over 1/n}=lim_{n\to\infty}{c_n\over 1/n}=\lim_{n\to\infty}{nc_n}.$$

Now, $f'(0)$ exists only when $\lim_{n\to\infty}{nc_n}=0$. So this is our condition. Let $\epsilon>0$, need to find an $N>0$ such that $$|nc_n-0|<\epsilon \text{ for }n>N$$ $\rightarrow |c_n|<{\epsilon\over |n|}$

Let $N={\epsilon\over |n|}$ so that we have $$|nc_n|<|n|{\epsilon\over |n|}=\epsilon$$

So $\lim_{n\to\infty}nc_n=0$ hence, $f$ is differentiable at $0$. And we have it (?)

What do you think?

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I think your accept rate is painfully low. I'd enhance it at once. –  DonAntonio Nov 16 '12 at 19:58
    
what do you mean by accept rate? –  Akaichan Nov 16 '12 at 20:00
    
Read it under your nick,@user45593. You have accpet rate of 25%, which means you've only accepted as "the best answer" about one quarter of the questions you've asked, and this could mean you don't really like the answers you get here. –  DonAntonio Nov 16 '12 at 20:02
    
Oh, ok, thanks. And I do like the answer I get here. And sometimes, I answered question myself also, just like this one. –  Akaichan Nov 16 '12 at 20:04
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You "wrote something wrong" at some point : when you write : when $x = 1/n$, $f(x) = \{ nc_n,$ if $x = 1/n$ for $n = 1,2, \dots$. It should be written $c_n$ instead of $nc_n$. And you could've stopped at "So this is our condition." The rest is irrelevant. But the argument is fine –  Patrick Da Silva Nov 16 '12 at 20:07
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1 Answer 1

Most books define differentiable functions on an interval $ (a, b) $. But we can define the derivative of a function $ f: A \to \mathbb {R} $ at a point $ c \in A $ if is $ c $ is an accumulation point ( or cluster point ) of $ A $.

Then $f^\prime(c)$ is unic number that have the propert:

$$ \forall \epsilon >0, \exists \delta=\delta(c,\epsilon)>0\quad \mbox{ such that } x\in A \mbox{ e }0<|c-x|<\delta \implies \left|\frac{f(x)-f(c)}{x-c}- f^\prime(c) \right|<\epsilon. $$

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