0
$\begingroup$

As the question asks I want to verify that $\beta(t) = (t^{3},|t^{3}|)$ is $C^{2}$ as well as the rank of its derivative $D\beta(0)$ rank.

My main concern with regards to verifying that the function was $C^{2}$ had to do with treating the absolute value bars in the second component. The first component is straight forward. The only idea that came to me was treating it with first principals in the one dimensional case as such I did the following:

$$ \lim_{h \to \infty} \frac{\beta(a+h) - \beta(a)}{h} = \lim_{h \to \infty} \frac{|(a+h)^{3}| - |a^{3}|}{|h|} \leq \lim_{h \to \infty} \frac{|a^{3}| +|a^2||h| + |a||h^2|+|h^3| - |a^3|}{|h|} \\ = \lim_{h \to \infty}\frac{|h|(|a^{2}| + |a||h|+|h^2|}{|h|} = |a^2|$$

This first instance would establish that it is $C^{1}$, then I would have to perform the same process over again to establish that this component is $C^{2}$

The issue of determining the rank of the matrix $D\beta(0)$ I find a bit more challenging. Since I could only find the derivative using first principals would it mean that the matrix $D\beta(t)$ is the following:

$$D\beta(t) = \left( \begin{array}{ccc} 3t^{2} \\ |t^2| \\ \end{array} \right )$$

I don't feel comfortable with this result and then how would I proceed to find the rank?

$\endgroup$
1
$\begingroup$

To check that $\beta$ is $C^2$, it suffices to check that each component is $C^2$. Now, the function $f(t)=|t|^3$ is $C^2$, the only issue is at $t=0$, where you can just compute $f'(0)$ by definition (with the limit, as you've done).

The process above gives you $D\beta (0)=(0 \ 0)^T$.

$\endgroup$
2
  • $\begingroup$ And since $D\beta (0)=(0 \ 0)^T$ this implies that the rank would be $0$? $\endgroup$ – dc3rd Jan 7 '20 at 2:53
  • $\begingroup$ @dc3rd Yes, exactly $\endgroup$ – Reveillark Jan 7 '20 at 4:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.