Quadratic Taylor series using epsilon expression Hello I have a question about Taylor series. While studying Mathematical Statistics written by Hogg, I found this expression below (page 314).
$$m(t)=m(0)+m'(0)t+\frac{m''(\epsilon)t^2}{2}$$
where $0<\epsilon<t$
Could you tell me how $\epsilon$ can be used?
As expression above and Taylor series, the equality below is right.
$$\frac{m''(\epsilon)t^2}{2} = \frac{m''(0)t^2}{2}+\frac{m'''(0)t^3}{3!} + \frac{m''''(0)t^4}{4!} + \dots $$
Why?
 A: As commented, the expression is one form of Taylor's theorem with an exact expression for the remainder.  It is derived from the mean value theorem as follows:
Suppose $f(x)$ is $n$ times differentiable in the closed interval $[0,a]$ (at the ends of the interval we suppose $f$ has $n$ left or right one-sided derivatives as applicable).  Then for any $t, 0 < t \leqslant a$ there exists $h$ (dependent on $t$) with $0 < h < t$ such that
\begin{align}
f(t) = f(0) + f'(0) t + f''(0) \frac{t^2}{2!} + \cdots + f^{(n-1)}(0)\frac{t^{n-1}}{(n-1)!}+f^{(n)}(h)\frac{t^n}{n!}.
\end{align}
The formula you quote is the case where $n=2$ and the interval is  $[0,t]$.
It can be used where $f$ is twice differentiable in an interval containing $[0,t]$ (allowing one-sided derivatives at the end points) to obtain a linear approximation of the form $f(t) \bumpeq f(0)+f'(0)t$ where the error is then given by $f''(\varepsilon)t^2/2$.  The error term may not immediately be of much use since $\varepsilon$ is not explicitly revealed but, for example, if $f''$ is known to be bounded then you can obtain a worst case estimate of the error size.
Higher order Taylor series give approximations for $f$ as a quadratic or cubic and so on, each with an associated error expression.
Taylor's theorem with this form of final term can be derived by defining the function,
$$
g(x) = f(t)-\sum_{k=0}^{n-1} \frac{(t-x)^kf^{(k)}(x)}{k!}
$$
which is differentiable on $[0,t]$ with $g(t)=0$, and then applying the mean value theorem to the function,
$$\phi(x) = g(x) -\frac{t-x}{t}g(0)$$ so
constructed top make $\phi(0) = \phi(t) = 0$.
