练习：已知连续函数f(x)在x=-1,0,2,3的值分别是-4，-1,0,3，用牛顿插值求f(1.5)的近似值。

牛顿插值法的核心是通过构造差商表，利用已知数据点逐步构建插值多项式。本题中，已知4个数据点，需计算到三阶差商，构造三次牛顿插值多项式。关键步骤包括：

1. 计算各阶差商，注意相邻节点间距；
2. 构造牛顿多项式，基函数形式为$(x+1), (x+1)x, (x+1)x(x-2)$；
3. 代入$x=1.5$，逐项计算求和。

计算差商

一阶差商

$\begin{aligned}f[x_0, x_1] &= \frac{f(x_1)-f(x_0)}{x_1 - x_0} = \frac{-1 - (-4)}{0 - (-1)} = 3, \\f[x_1, x_2] &= \frac{f(x_2)-f(x_1)}{x_2 - x_1} = \frac{0 - (-1)}{2 - 0} = \frac{1}{2}, \\f[x_2, x_3] &= \frac{f(x_3)-f(x_2)}{x_3 - x_2} = \frac{3 - 0}{3 - 2} = 3.\end{aligned}$

二阶差商

$\begin{aligned}f[x_0, x_1, x_2] &= \frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0} = \frac{\frac{1}{2} - 3}{2 - (-1)} = -\frac{5}{6}, \\f[x_1, x_2, x_3] &= \frac{f[x_2, x_3] - f[x_1, x_2]}{x_3 - x_1} = \frac{3 - \frac{1}{2}}{3 - 0} = \frac{5}{6}.\end{aligned}$

三阶差商

$f[x_0, x_1, x_2, x_3] = \frac{f[x_1, x_2, x_3] - f[x_0, x_1, x_2]}{x_3 - x_0} = \frac{\frac{5}{6} - \left(-\frac{5}{6}\right)}{3 - (-1)} = \frac{5}{12}.$

构造牛顿插值多项式

$N_3(x) = f(x_0) + f[x_0, x_1](x - x_0) + f[x_0, x_1, x_2](x - x_0)(x - x_1) + f[x_0, x_1, x_2, x_3](x - x_0)(x - x_1)(x - x_2).$
代入数据：
$N_3(x) = -4 + 3(x+1) - \frac{5}{6}(x+1)x + \frac{5}{12}(x+1)x(x-2).$

代入$x=1.5$

逐项计算：
$\begin{aligned}3(x+1) &= 3 \times (1.5 + 1) = 7.5, \\-\frac{5}{6}(x+1)x &= -\frac{5}{6} \times 2.5 \times 1.5 = -3.125, \\\frac{5}{12}(x+1)x(x-2) &= \frac{5}{12} \times 2.5 \times 1.5 \times (-0.5) = -0.78125.\end{aligned}$
总和：
$N_3(1.5) = -4 + 7.5 - 3.125 - 0.78125 = -0.40625.$