Taylor's theorem

Statement

Let $k \ge 1$, let $I$ be an open interval, let $a \in I$, and let $f \colon I \to \mathbb{R}$ be $k$ times differentiable at $a$. Let $P_k$ be the Taylor polynomial of degree $k$ of $f$ at $a$. Then there is a function $h_k \colon I \to \mathbb{R}$ with $h_k(x) \to 0$ as $x \to a$ and $$f(x) = P_k(x) + h_k(x) (x - a)^k \quad \text{for all } x \in I.$$

Remarks

In little o notation, the conclusion is the Peano form $f(x) - P_k(x) = o((x - a)^k)$. Under stronger hypotheses, the remainder has explicit forms: the Lagrange form, the Cauchy form, and the integral form. For $k = 1$, it says that $f(x) = f(a) + f'(a)(x - a) + o(x - a)$.

Explanations

L1 human

Near $a$, a function is close to its Taylor polynomial. The error is small even compared with $(x - a)^k$.