I found the following theorem, but I don't understand it and was unable to prove it. Is it true? Is there a proof for it?
Theorem: Let $f : R^d → R$ be such that $f$ is twice-differentiable and
has continuous derivatives in an open ball $B$ around the point $x ∈ R^d$. Then for any small enough $∆x ∈ R^d$ such that
$x + ∆x$ is also contained in the ball $B$, we have the following:
$$
f(x + \Delta x) = f(x) + \Delta x^T\nabla f|_x + \frac{1}{2}(\Delta x)^T (\nabla^2f|_w)(\Delta x)
$$
Where $(\nabla^2f|_w)$ is the Hessian of $f$ evaluated at a point $w ∈ R^d$
that lies on the line connecting $x$ and $x + ∆x$
I understand that this is a second-order Taylor expansion of $f$ about $x$, and I understand why it is in this form.
But, I don't get is why the Hessian can be evaluated at the point $w$ rather than at $x$. If it is a Taylor expansion about $x$, shouldn't all derivatives be evaluated at $x$? Why is this expansion valid?
For reference, this is where I found the theorem: https://www.cs.princeton.edu/courses/archive/fall18/cos597G/lecnotes/lecture3.pdf
On page 2.
Best Answer
The short answer is, that the part with the Hessian of $f$ is an error correcting term. Thus the theorem states that if $f$ is twice differentiable you can approximate your function by an (affine) linear function (your first two terms with zero- and first order differentials) and the error of that approximation evaluated at any point $x+\Delta x$ is given by the Hessian of $f$ evaluated at a certain point on the line between $x$ and $x+\Delta x$.
You will find the same idea in the one-dimensional statement about Taylor series. Sometimes this will be refered to as Lagrangian form (of the remainder).