数值微分
\[\begin{align} f(x) &= p_n(x)+R_n(x) \\ &= \sum_{k=0}^{n}f(x_k)L_k(x) + \frac{f^{(n+1)}(\xi(x))}{(n+1)!}\prod_{k=0}^n(x-x_k) \\ \end{align}\]于是我们可以得到：
\[\begin{align} f'(x_j) &= \sum_{k=0}^{n}f(x_k)L_k'(x_j) \\ &+ \frac{f^{(n+1)}(\xi(x_j))}{(n+1)!} \end{align}\]end-point
\[\begin{align} f'(x_0) &= \frac{-3f(x_0)+4f(x_0+h)-f(x_0+2h)}{2h} \\ &+ \frac{h^2}{3}f^{(3)}(\xi) \end{align}\]mid-point
\[\begin{align} f'(x_0) &= \frac{f(x_0+h)-f(x_0-h)}{2h} \\ &- \frac{h^2}{6}f^{(3)}(\xi) \end{align}\] \[\begin{align} f''(x_0) &= \frac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} \\ &- \frac{h^2}{12}f^{(4)}(\xi) \end{align}\]数值积分
\(\begin{align} \int_b^a f(x)\,\mathrm{d}x &= \sum_{k=0}^nf(x_k)\int_b^aL_k(x)\,\mathrm{d}x \\ &+ \frac{1}{(n+1)!}\int_b^af^{(n+1)}(\xi(x))\prod_{k=0}^n(x-x_k)\,\mathrm{d}x \\ &= \sum_{k=0}^na_kf(x_k)+E(f) \end{align}\)
Simpson’s Rule
\(\begin{align} \int_b^af(x)\mathrm{d}x &= \frac{b-a}{6}\left(f(a)+4f(\frac{a+b}{2})+f(b)\right) \\ &- \frac{(b-a)^5}{2880}f^{(4)}(\xi) \end{align}\)
等距数值积分
如果\(x_0=a\)，\(x_a=\frac{a+b}{2}\)，\(x_2=b\)，\(h=\frac{b-a}{2}\)。
\[\begin{align} a_0 &= \int_{x_0}^{x_2}\frac{(x-x_1)(x-x_2)}{2h^2}\,\mathrm{d}x &= \frac13h \\ a_1 &= \int_{x_0}^{x_2}\frac{(x-x_0)(x-x_2)}{2h^2}\,\mathrm{d}x &= \frac43h \\ a_2 &= \int_{x_0}^{x_2}\frac{(x-x_0)(x-x_1)}{2h^2}\,\mathrm{d}x &= \frac13h \\ \end{align}\]于是我们得到
\[\begin{align} \int_b^af(x)\mathrm{d}x &= \frac{h}{3}\left(f(a)+4f(\frac{a+b}{2})+f(b)\right) \\ &+ \frac16\int_b^a(x-x_0)(x-x_1)(x-x_2)f'''(\xi)\mathrm{d}x \end{align}\]再由泰勒展式，可以比较得到最后的余项为：
\[\begin{align} -\frac{1}{36}h^5f^{(4)}(\xi_1)+\frac{1}{60}h^5f^{(4)}(\xi_2) \end{align}\]The degree of accuracy, or precision
使得 \(x^k\) 精确的最大的整数。
Newton-Cotes 公式
\(x_k=a+kh\), for \(h=\frac{b-a}{2}\) and \(k=0:n\)。
\[\int_b^af(x)\mathrm{d}x\approx\sum_{k=0}^na_kf(x_k)\]其中
\[a_k=\int_b^aL_k(x)\mathrm{d}x.\]