→ theoculus :再改一下就變 Lagrange 了 08/06 22:56
推 cacud :高中的插值多項式~ 計算方便的假設方式 08/06 23:06
→ recorriendo :其實就是Lagrange 中間過程那幾個式子裡a b c d的係 08/07 01:37
→ recorriendo :樹就是Lagrange的分母 08/07 01:37
※ 引述《iddee ()》之銘言：
: 設 f(x) 是三次多項式且滿足
: f(1981) = 1
: f(1982) = 9
: f(1983) = 8
: f(1984) = 5
: 求 f(1985)
這題還可以這樣作
命 f(x) = a(x-1981)(x-1982)(x-1983)+b(x-1981)(x-1982)(x-1984)
+c(x-1981)(x-1983)(x-1984)+d(x-1982)(x-1983)(x-1984)
所以 1 = f(1981) = -6d, d = -1/6
9 = f(1982) = 2c, c = 9/2
8 = f(1983) = -2b, b = -4
5 = f(1984) = 6a, a = 5/6
=> f(1985) = (5/6)*24 + (-4)*12 + (9/2)*8 + (-1/6)*6
= 20 - 48 + 36 - 1 = 7.
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