拉格朗日插值公式

拉格朗日插值公式找到一个称为拉格朗日多项式的多项式，该多项式在任意点取特定值。它是函数f(x) 的 n^次多项式表达式。插值法用于在一组离散的已知数据点范围内寻找新的数据点。

拉格朗日插值公式

给定几个实数值 x ₁ , x ₂ , x ₃ , ..., x _n和 y ₁ , y ₂ , y ₃ , ..., y _n并且会有一个实系数满足条件 P(x _i ) = 的多项式 P y _i , ∀i = {1, 2, 3, ..., n} 并且多项式 P 的次数必须小于实数值的计数，即 degree(P) < n。不同阶即 n^阶的拉格朗日插值公式为：

n^阶的拉格朗日插值公式是-

$f(x)=\frac{(x-x_1)(x-x_2)...(x-x_n)}{(x_0-x_1)(x_0-x_2)...(x_0-x_n)}\times y_0+\frac{(x-x_0)(x-x_2)...(x-x_n)}{(x_1-x_0)(x_1-x_2)...(x_1-x_n)}\times y_1+...+\frac{(x-x_0)(x-x_1)...(x-x_n-1)}{(x_n-x_0)(x_n-x_1)...(x_n-x_n-1)}\times y_n$

如果 那么多项式的次数为1，

一^阶多项式的拉格朗日插值公式是-

$f(x)=\frac{(x-x_1)}{(x_0-x_1)}\times y_0+\frac{(x-x_0)}{(x_1-x_0)}\times y_1$

同样对于二阶^多项式，拉格朗日插值公式是-

$f(x)=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\times y_0+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\times y_1+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\times y_2$

拉格朗日定理的证明

Let’s consider a nth degree polynomial of given form

$f(x)=A_0(x-x_1)(x-x_2)(x-x_3)...(x-x_n)+A_1(x-x_0)(x-x_2)(x-x_3)...(x-x_n)+...+A_n(x-x_1)(x-x_2)(x-x_3)...(x-x_{n-1})$

Substitute observations x_i to get A_i

Put x = x₀ then we get A₀ 

f(x₀) = y₀ = A₀(x₀ – x₁)(x₀ – x₂)(x₀ – x₃)…(x₀ – x_n)

A₀ = y₀/(x₀ – x₁)(x₀ – x₂)(x₀ – x₃)…(x₀ – x_n)

By substituting x = x₁ we get A₁

f(x₁) = y₁ = A₁(x₁ – x₀)(x₁ – x₂)(x₁ – x₃)…(x₁ – x_n)

A₁ = y₁/(x₁ – x₀)(x₁ – x₂)(x₁ – x₃)…(x₁ – x_n)

In similar way by substituting x = x_n we get A_n

f(x_n) = y_n = A_n(x_n – x₀)(x_n – x₁)(x_n – x₂)…(x_n – x_n-1)

A_n = y_n/(x_n – x₀)(x_n – x₁)(x_n – x₂)…(x_n – x_n-1)

If we substitute all values of A_i in function f(x) where i = 1, 2, 3, …n then we get Lagrange Interpolation Formula.

$f(x)=\frac{(x-x_1)(x-x_2)...(x-x_n)}{(x_0-x_1)(x_0-x_2)...(x_0-x_n)}\times y_0+\frac{(x-x_0)(x-x_2)...(x-x_n)}{(x_1-x_0)(x_1-x_2)...(x_1-x_n)}\times y_1+...+\frac{(x-x_0)(x-x_1)...(x-x_n-1)}{(x_n-x_0)(x_n-x_1)...(x_n-x_n-1)}\times y_n$

让我们看一些关于拉格朗日插值公式的示例问题。

示例问题

问题 1：对于给定的点集 (1, 2),(3, 4) 求 y 在 x = 2 处的值

解决方案：

Given,

(x₀, y₀) = (1, 2)

(x₁, y₁) = (3, 4)

x = 2

As per the 1^st order Lagrange Interpolation Formula,

$y=\frac{(x-x_1)}{(x_0-x_1)}\times y_0+\frac{(x-x_0)}{(x_1-x_0)}\times y_1$

$=\frac{(2-3)}{(1-3)}\times 2+\frac{(2-1)}{(3-1)}\times 4$

= (-2/-2) + (4/2)

= 1 + 2

y = 3

问题 2：对于给定的点集 (9, 2), (3, 10) 求 y 在 x = 5 处的值

解决方案：

Given,

(x₀, y₀) = (9, 2)

(x₁, y₁) = (3, 10)

x = 5

As per the 1^st order Lagrange Interpolation Formula,

$y=\frac{(x-x_1)}{(x_0-x_1)}\times y_0+\frac{(x-x_0)}{(x_1-x_0)}\times y_1 =\frac{(5-3)}{(9-3)}\times 2+\frac{(5-9)}{(3-9)}\times 10$

= (4/6) + (-40/-6)

= (2/3) + (20/3)

= 22/3

y = 7.33

问题 3：对于给定的点集 (1, 6), (3, 4), (2, 5) 求 y 在 x = 1 处的值

解决方案：

Given,

(x₀, y₀) = (1, 6)

(x₁, y₁) = (3, 4)

(x₂, y₂) = (2, 5)

x = 1

As per the 2^nd order Lagrange Interpolation Formula

$y=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\times y_0+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\times y_1+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\times y_2 =\frac{(1-3)(1-2)}{(1-3)(1-2)}\times 6+\frac{(1-1)(1-2)}{(3-1)(3-2)}\times 4+\frac{(1-1)(1-3)}{(2-1)(2-3)}\times 5 =\frac{(-2)(-1)}{(-2)(-1)}\times 6+\frac{(0)(-1)}{(2)(1)}\times 4+\frac{(0)(-2)}{(1)(-1)}\times 5$

= (12/2) + 0 + 0

y = 6

问题 4：对于给定的点集 (9, 6), (3, 5), (1, 12) 求 y 在 x = 10 处的值

解决方案：

Given,

(x₀, y₀) = (9, 6)

(x₁, y₁) = (3, 5)

(x₂, y₂) = (1, 12)

x = 10

As per the 2^nd order Lagrange Interpolation Formula,

$y=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\times y_0+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\times y_1+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\times y_2 =\frac{(10-3)(10-1)}{(9-3)(9-1)}\times 6+\frac{(10-9)(10-1)}{(3-9)(3-1)}\times 5+\frac{(10-9)(10-3)}{(1-9)(1-3)}\times 12 =\frac{(7)(9)}{(6)(8)}\times 6+\frac{(1)(9)}{(-6)(2)}\times 5+\frac{(1)(7)}{(-8)(-2)}\times 12$

= (63/8) + (-15/4) + (21/4)

= (63-30 + 42)/8

= 75/8

y = 9.375

问题 5：对于给定的点集 (1, 10), (2, 4), (3, 4), (5, 7) 求 y 在 x = 7 处的值

解决方案：

Given,

(x₀, y₀) = (1, 10)

(x₁, y₁) = (2, 4)

(x₂, y₂) = (3, 4)

(x₃, y₃) = (5, 7)

x = 7

As per the 3^rd order Lagrange Interpolation Formula,

$y=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\times y_0+\frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\times y_1+\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\times y_2+\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\times y_3 =\frac{(7-2)(7-3)(7-5)}{(1-2)(1-3)(1-5)}\times 10+\frac{(7-1)(7-3)(7-5)}{(2-1)(2-3)(2-5)}\times 4+\frac{(7-1)(7-2)(7-5)}{(3-1)(3-2)(3-5)}\times 4+\frac{(7-1)(7-2)(7-3)}{(5-1)(5-2)(5-3)}\times 7 =\frac{(5)(4)(2)}{(-1)(-2)(-4)}\times 10+\frac{(6)(4)(2)}{(1)(-1)(-3)}\times 4+\frac{(6)(5)(2)}{(2)(1)(-2)}\times 4+\frac{(6)(5)(4)}{(4)(3)(2)}\times 7$

= -50 + 64 – 60 + 35

= 99 – 110

y = -11

问题 6：对于给定的点集 (5, 12), (6, 13), (7, 14), (8, 15) 求 y 在 x = 10 处的值

解决方案：

Given,

(x₀, y₀) = (5, 12)

(x₁, y₁) = (6, 13)

(x₂, y₂) = (7, 14)

(x₃, y₃) = (8, 15)

x = 10

As per the 3^rd order Lagrange Interpolation Formula

$y=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\times y_0+\frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\times y_1+\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\times y_2+\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\times y_3 =\frac{(10-6)(10-7)(10-8)}{(5-6)(5-7)(5-8)}\times 12+\frac{(10-5)(10-7)(10-8)}{(6-5)(6-7)(6-8)}\times 13+\frac{(10-5)(10-6)(10-8)}{(7-5)(7-6)(7-8)}\times 14+\frac{(10-5)(10-6)(10-7)}{(8-5)(8-6)(8-7)}\times 15 =\frac{(4)(3)(2)}{(-1)(-2)(-3)}\times 12+\frac{(5)(3)(2)}{(1)(-1)(-2)}\times 13+\frac{(5)(4)(2)}{(2)(1)(-1)}\times 14+\frac{(5)(4)(3)}{(3)(2)(1)}\times 15$

= -48 + 195 – 280 + 150

y = 17

问题 7：对于给定的点集 (-2, 5),(1, 7)，在 x = 0 处找到 y 的值

解决方案：

Given,

(x₀, y₀) = (-2, 5)

(x₁, y₁) = (1, 7)

x = 0

As per the 1^st order Lagrange Interpolation Formula

$y=\frac{(x-x_1)}{(x_0-x_1)}\times y_0+\frac{(x-x_0)}{(x_1-x_0)}\times y_1 =\frac{(0-1)}{(-2-1)}\times 5+\frac{(0+2)}{(1+2)}\times 7$

= (5/3) + (14/3)

= 19/3

y = 6.33