I know that Taylor series for $f(x)$ at $x_0$ is $$ f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f^{(2)}(x_0)}{2!}(x-x_0)^2 + \frac{f^{(3)}(x_0)}{3!}(x-x_0)^3 + \frac{f^{(4)}(x_0)}{4!}(x-x_0)^4 + \cdots $$ where $f^{(k)}$ implies the $k$-th derivative of $f$.
Sometimes, we shorten the equation above using big-$O$ or small-$o$ notation, which are:
Let $g$ be a real-valued function defined in some neighborhood of $0\in\Bbb{R}$, with $g(x)\ne0$ if $x\ne0$. Let $f:\Omega\to\Bbb{R}$ be defined in a domain $\Omega\subset\Bbb{R}$ that include $0$. Then,
- $f(x)=O(g(x))$ to mean that the quotient $\displaystyle\frac{|f(x)|}{|g(x)|}$ is bounded near $0$; that is, there exist numbers $K\gt0$ and $\delta\gt0$ such that if $|x|\lt\delta$, $x\in\Omega$, then $\displaystyle\frac{|f(x)|}{|g(x)|}\le K$.
- $f(x)=o(g(x))$ to mean that $$\lim_{x\to0,x\in\Omega} \frac{|f(x)|}{|g(x)|}=0$$
Then, $$ f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f^{(2)}(x_0)}{2!}(x-x_0)^2 + O(|x-x_0|^3)\\ or\\ f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f^{(2)}(x_0)}{2!}(x-x_0)^2 + o(|x-x_0|^2) $$
However, the textbook I am reading expresses Taylor series expansion of $f$ at $x_0$ as $$ f(x)=f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x^*)}{2}(x-x_0)^2 $$ where $x^*$ lies between $x_0$ and $x$.
Please let me understand why this expression is just. Thank you for reading my question.