Strange form of Taylor's theorem

Question

I am reading a proof in a convex optimization book and at the beginning of the proof it says the following:

for any $x, y$ in $\mathbb{R}^n$, we have $f'(y) = f'(x) + \int_{0}^{1} f''\left( x + \tau(y-x)\right)d~\tau$.

I think this is some form of taylor expansion but I don't understand it. thanks for any enlightenment.

Answer 1

Actually, it's the fundametal theorem of calculus (which corresponds also to a first order Taylor's expantion) applied to function $F=f^\prime$ wich must be derivable over $\left[ x,y \right]$ so f must be two times derivable.

$$ f^{'}(y) - f^{'}(x) = F(y) - F(x) = \int_{x}^{y} F'( z) d~z = \int_{0}^{1} f^{''}( x + \tau(y-x)) d~\tau$$

Note that \begin{align} \int_{0}^{1} f^{''}( x + \tau(y-x)) d~\tau = \int_{x}^{y} f^{''}( z) d~z \end{align} It's a parametric verstion of the same integral.