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
Best Answer
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?