用三次樣條插值求斜率
三次樣條插值的matlabt程序function x=followup(a,b,c,d)n=length(d);" u+ M0 d `: l; G, B) y
a(1)=0;2 }- D7 s" m$ K0 E+ o3 h
%“追”的過(guò)程' y! b8 ]& J9 h9 D
L(1)=b(1);
y(1)=d(1)/L(1);
u(1)=c(1)/L(1);
for i=2
n-1)9 p; K1 d7 N0 t" G8 ]% e, t6 A' |6 YL(i)=b(i)-a(i)*u(i-1);6 Y- d' O( Z6 t" R" ]9 o! L) |4 O
y(i)=(d(i)-y(i-1)*a(i))/L(i);
u(i)=c(i)/L(i);; E- ?' O1 ~3 k$ |/ M
end
L(n)=b(n)-a(n)*u(n-1);; y7 v" E% p. z0 A, R& l& c
y(n)=(d(n)-y(n-1)*a(n))/L(n);0 `# y5 o- C9 t1 W! O
%“趕”的過(guò)程/ A) o% N2 {; k
x(n)=y(n);
for i=(n-1):-1:15 L& z3 k: Z! @6 ?
x(i)=y(i)-u(i)*x(i+1);! ^1 j/ L5 b8 |
end7 e' a& \" V" c3 Y$ S
4 L- U2 C8 @+ f l P, C* g
function[s,y0]=spline3 (x,y,x0) `* z4 S% M" u/ b) w, P
%x,y為數(shù)表x0為插值點(diǎn)s表示插值函數(shù)y0為x0對(duì)應(yīng)的插值函數(shù)值
syms t |: M0 I& b+ p
n=length(x);
%得出n+ T; c+ a7 D, R. J9 T* j2 n. X! c7 |$ Y* s
for i=1:n-1;
h(i)=x(i+1)-x(i);( b) _' b/ @* E: R' {# ~' L
end1 S; {+ r- S" r% d" L: ^$ K
for i=2:n-1;# N! w- L8 g1 {! G7 H: u, P
lamda(i)=h(i)/(h(i-1)+h(i));( Q0 t4 v+ X. ]8 B
miu(i)=1-lamda(i);
g(i)=3*(lamda(i)*((y(i)-y(i-1))/h(i-1))+miu(i)*((y(i+1)-y(i))/h(i)));- R+ t( f* s9 I* g5 t* @2 F/ I
end
g(1)=3*((y(2)-y(1))/h(1));7 {1 d+ k4 L( V7 }8 v/ t+ X. H% M
g(n)=3*((y(n)-y(n-1))/h(n-1));
%前邊求出lamda miu和g從而可以確定系數(shù)矩陣 m) v' E j& F5 P
miu(1)=1;8 F. \4 M& Q9 x% f
miu(4)=0;( p" F* i4 G1 U/ D) L" I
lamda(n)=1;
lamda(1)=0;
%根據(jù)第二邊界條件補(bǔ)充兩個(gè)lamda和miu的值
for i=1:n
beta(i)=2;6 } f' W( Q" I/ t+ b
end
m=followup(lamda,beta,miu,g)
%解出m的值從而可確定st st為各段的插值多項(xiàng)式. S }8 c: K* K
for i=1:n-1- s+ e8 F, q# [% c
st(i)=(t-x(i+1))^2*(h(i)+2*(t-x(i)))*y(i)/(h(i)^3)...% d" {' B' {6 }& j& x$ M+ f. l
+(t-x(i))^2*(h(i)+2*(x(i+1)-t))*y(i+1)/(h(i)^3)...
+(t-x(i))^2*(t-x(i+1))*m(i+1)/(h(i)^2)..., h9 y4 ~8 n* R* @" m8 |0 O
+(t-x(i+1))^2*(t-x(i))*m(i)/(h(i)^2);: h' K8 _. O% B, W) `* ~
end
%得到插值的結(jié)果各段的t的表達(dá)式
%接下來(lái)要將插值點(diǎn)x0代入首先確定x0所在的插值區(qū)間% J# e3 G( v" t7 C5 K, s
for i=1:n-12 c6 J" l! }+ _' a# O5 z& }$ A
if (x(i)>x0)
in=i;
end
end
s=st(in);& E" `# |( {/ W' m) m0 Q4 {
s=expand(s);) M' |& u. i6 u( P$ z+ F
s=collect(s,'t');
y0=subs(s,'t',x0)" |) w, U, f8 a9 S
%s是插值多項(xiàng)式y(tǒng)0是插值點(diǎn)的函數(shù)值$ d" H/ N" J/ w7 M
( z9 p Q6 h6 L
在matlab中輸入
x=[1 2 4 5]；
y=[1 3 4 2]；
spline3(x,y,2)
會(huì)得到各點(diǎn)的斜率0 `, M) f! X& e" d' z
1 G* Y( Y4 f( j/ Q' F c
& F3 t5 q$ d1 H9 N" I2 u