0
$\begingroup$

enter image description here

How to get formula (2) by formula(1)?

$\endgroup$
1
$\begingroup$

You have to take the derivative with respect to the vector $x$ and set it equal to zero. For a constant matrix $A$, the derivative of $A^Tx$ is $A$, and the derivative of $\frac12 x^TA^Tx=Ax$. So taking the derivative of $(1)$ gives

$$\frac{\partial D}{\partial x}+\frac{\partial^2D}{\partial x^2}x\tag{1}$$

Setting $(1)$ equal to zero results in $(2)$.

$\endgroup$
  • $\begingroup$ $\frac12 x^TA^Tx$ is a constant,$Ax$ is a vector? $\endgroup$ – zychen Aug 31 '18 at 8:57
  • $\begingroup$ you say "a constant matrix A ",do you mean $(\frac{\partial D}{\partial x})^T$ is a constant and $(\frac{\partial^2D}{\partial x^2})^T$ also is a constant? $\endgroup$ – zychen Aug 31 '18 at 9:08
  • $\begingroup$ @zychen: Yes, to both questions, but I would say $\frac12x^TA^Tx$ is a scalar, which is probably what you meant. So $(1)$ is a scalar, and $(2)$ is a vector. And the derivatives of $D$ are evaluated at $x=0$, that's why they're constant. $\endgroup$ – Matt L. Aug 31 '18 at 10:52

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.