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I have a question regarding the following exercise.

Let $f(x)= \begin{cases} x^n \mbox{ for } x \geq 0\\ 0 \mbox{ for } x<0 \end{cases}$

Show that the iterated derivatives $f^{(1)}$ through $f^{(n-1)}$ exist at all real numbers x, but the n-th iterated derivative at 0 does not.

I was able to prove that the n-th iterated derivative at 0 does not exist: at right of 0, its value is n! and left of 0, its value is 0.

Since the derivative left and right of 0 is different, the n-th iterated derivative is therefore non differentiable in 0.

But can someone show me how to prove that the first to the n-1 derivatives exist?

Thank you

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Have you been taught the epsilon-delta definition of a differentiation? If so, just try to plug it in, using $\displaystyle \frac{x^n-y^n}{x-y} = \sum_{i=0}^{n-1}x^iy^{n-1-i}$ –  Lord_Farin Oct 17 '12 at 19:10
    
No we did not yet –  user43418 Oct 17 '12 at 19:10
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2 Answers

up vote 1 down vote accepted

Consider the quotient $$\frac{f^{(n-1)}(h+0)-f^{(n-1)}(0)}{h}.$$

  • What if $h\to0$ from below (that is $h<0$)?
  • What if $h\to0$ from above (that is $h>0$)?

If you wish you might start with $n=1$.

Edit For $n=1$ we should look at $$\frac{f^{(0)}(h+0)-f^{(0)}(0)}{h}=\frac{f(h)-f(0)}{h},\tag{1}$$ with $$f(x)=\begin{cases} x \text{ for } x \geq 0\\ 0 \text{ for } x<0 \end{cases}\tag{2}$$ Now, substitute (2) into (1). What happens when $h\to0$ for $h<0$ and what happens when $h\to0$ for $h>0$?

Now go for $n=2$, $n=3$, etc. until you see and can explain the general picture.

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I don't understand –  user43418 Oct 17 '12 at 20:32
    
@user43418 Is the edit easier? –  AD. Oct 17 '12 at 21:48
    
If h>0, the quotient equals h and its limit is therefore 0 If h<0, the quotient equals 0 Since the limit left and right of 0 is the same, then $f^{(0)}$ exists. –  user43418 Oct 17 '12 at 22:24
    
And now I suppose it is true for n-2 terms and show that it is true for n-1 right ? –  user43418 Oct 17 '12 at 22:24
    
@user43418 No it is not $h$ for $h>0$. –  AD. Oct 18 '12 at 5:25
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You would do something very similar. The $k$th derivative would be given by $$f^{(k)}(x) = \begin{cases} (n)_kx^{n-k} \mbox{ for } x > 0\\ 0 \mbox{ for } x<0 \end{cases}$$ where $(n)_k = n(n-1)\cdots (n-k+1)$ is the falling factorial. There's no issues away from $0$, but as $x\rightarrow 0$ from the left and the right, do the limits coincide?

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yes I think they do –  user43418 Oct 17 '12 at 19:32
    
I think they do too. But can you prove that they do? I would probably use Newton's quotient explicitly to show that the limits coincide. –  EuYu Oct 17 '12 at 19:33
    
Can you prove it for me. I just did 10 problems of derivatives and this is my last question. Please :( –  user43418 Oct 17 '12 at 19:34
1  
That isn't how this site works. I'm sorry. I'd be happy to provide assistance if you show me some progress though. –  EuYu Oct 17 '12 at 19:35
    
So I have to show it to be true for n=1 and then for n terms by induction ? –  user43418 Oct 17 '12 at 20:33
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