实验一--三次样条插值的三弯矩法

实验一 三次样条插值的三弯矩法

一、实验目的

用三次样条插值的三弯矩法,编制第一与第二种边界条件的程序.已知数据i x ,()i i y f x =,0,,i n =及边界条件(

)(),0,,1,2k j j y x j n k ==,编程计算)(x f 的三

次样条插值函数)(x S .具体要求为:输出用追赶法解出的弯矩向量0[,,]n M M M =及()(),0,,,0,1,2k i S t i m k ==的值;画出)(x S y =的图形,图形中描出插值点(,)i i x y 及(,())i i t S t 分别用‘o ’和‘*’标记.

二、实验原理

（1）由(x k ?y k )(k =0,1,2,…,n ),按公式

S k ‘’(x )=M k?1x k ?x k +M k x ?x k?1k

,?k =x k ?x k?1 λk =

?k+1?k +?k+1,μk =?k ?k +?k+1, d k =6

?k +?k+1(y k+1?y k ?k+1?y k?y k?1?k ) d 0=6λ0?1(y 1?y 0

?1?y 0′)+2(1?λ0)y 0′′,d n =6μn

?n (y n ′?y n ?y n?1

?n )+2(1?μn )y n

′′ 求出?k ,λk ,μk ,d k .

（2）由边界条件确定λ0，d 0和λn ,d n 。

（3）用追赶法解线性方程组

[

2 λ0μ1 2 λ1 μ1 2 ? λ2?? μn?22λn?2μn?1 2 λn?1 μn 2 ] [ M 0M 1M 2?M n?2M n?1M n ] =[ d 0d 1d 2?d n?2d n?1d n ] 求出M 0,M 1,M 2,…,M n

（4）由

S k (x )=M k?1(x k ?x )36?k +M k (x ?x k?1)36?k +x k ?x ?k [y k?1?M k?16

?k 2]+x ?x k?1?k [y k ?M k 6?k

2] 确定S k (x ),得到S(x).

三、实验结果

所用数据:

x=[-5,-3.8884,-3.0142,-2.3216,-1.7433,-1.3771,-1.1318,-0.86821,-0.58598,-0.4314,-

0.0293,0.4314,0.58598,0.86821,1.1318,1.3771,1.7433,2.1242,2.3216,3.01 42,3.8884,5];

y=[0.03846,0.06204,0.09915,0.1565,0.2476,0.3453,0.4384,0.5702,0.7444, 0.8431,0.9991,0.8431,0.7444,0.5702,0.4384,0.3453,0.2476,0.1814,0.1565 ,0.09915,0.06204,0.03846];

d2s1=0.0084;d2sn=0.0084;%第二种边界条件

t=[-4.166,-3.233,-2.495,-1.888,-1.469,-1.193,-0.9341,-0.6565,-0.47,-0.1298,0.3162,0.5473,0.7977,1.066,1.316,1.652,2.029,2.272,2.841,3.67];

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