# Strange form of Taylor's theorem

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.

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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.

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Thanks. I'll look that up. It's been 25 years since I took calculus so I didn't recognize it. It's appreciated. –  mark leeds Jan 5 '13 at 5:54
thanks paul. when you say it's a parametric version of the same integral, do you mean that the substitution $z = x + \tau(y-x) d \tau$ is being made and that makes the upper and lowers limits change to 0 and 1 respectively. –  mark leeds Jan 5 '13 at 19:34
Exactly! Because $\gamma(\tau) = x+\tau(y-x) ,\ \ \forall \tau \in \left[0,1 \right]$ is the parametric equation of the line which links $x$ to $y$. –  Paul Jan 5 '13 at 23:57