Taylor’s Theorem
Let f and its n-derivative be continuous on [a,b] and differentiable on (a,b). Let x_0 \in [a,b]. For each x \in [a,b] such that x \not = x_0. Then there exists c between x,x_0 such that
\begin{align*} f(x) &= f(x_0) + f^\prime(x_0)(x-x_0) + \frac{f^{\prime\prime}(x_0)}{2!}(x-x_0)^2 + \cdots \\ &+ \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}. \end{align*}
We can think of the final term as measuring the failure of approximation given by the first n+1 terms.
Fix x \in [a,b] such that x_0 \not = x. Let M be the unique solution to
f(x) = f(x_0) + \cdots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n + M(x-x_0)^{n+1}.
Define
F(t) = f(t) + f^\prime(t)(x-t) + \cdots + \frac{f^{(n)}(x_0)}{n!}(x-t)^n + M(x-t)^{n+1}.
Our assumuptions imply F(t) and its n-derivatives are continuous on [a,b] and differentiable on (a,b). Note that F(x) = f(x). Our choice of M implies that F(x_0) = f(x). By the MVT, there exists c between x,x_0 such that
F^\prime(c) = \frac{F(x)-F(x_0)}{x-x_0} = 0
since F(x) = F(x_0) = f(x). We can calculate F^\prime(t), and find that all terms except the last two will cancel out. Thus
F^\prime(t) =\frac{f^{(n+1)}(t)}{n!}(x-t)^n - M(n+1)(x-t)^n.
Thus, F^\prime(c) = 0 implies
F^\prime(c) = \frac{f^{(n+1)}(c)}{n!}(x-t)^n - M(n+1)(x-c)^n = 0
M = \frac{f^{(n+1)}(c)}{(n+1)!}.
Therefore we have proved the Taylor series is an accurate representation.