Regarding Taylor polynomial of 2nd order,
$$ f(x+ p)=f(x)+\nabla f(x + tp)^Tp $$
I have the following question: Why is there a transpose?
I have to do a optimization course and unfortunately my last linear algebra course is already a while back.
Regarding Taylor polynomial of 2nd order,
$$ f(x+ p)=f(x)+\nabla f(x + tp)^Tp $$
I have the following question: Why is there a transpose?
I have to do a optimization course and unfortunately my last linear algebra course is already a while back.
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I assume $x$ is an $n \times 1$ vector and $f$ is a scalar function.
Then $\nabla f(x_1)$ is also an $n \times 1$ vector. A scalar product $\langle x,y \rangle = x^Ty$. That's the reason for the first transpose.
$\nabla^2 f(x_1)$ is an $n \times n$ matrix. In the end, you need a scalar, so here you transpose the first occurence of $(x-x_1)$.