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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

  • 0
    No we did not yet2012-10-17

2 Answers 2

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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.

  • 0
    @user43418 No it is not $h$ for h>0.2012-10-18
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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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    @EuYu I did not know that was a common term, thanks (+$1$)2012-10-17