計(jì)算方法：函數(shù)插值

什么叫插值？溫度oC飽和蒸汽壓kN/m2

焓kJ/kg00.60820101.226242.04202.334683.90304.2474125.69407.3766167.515012.34209.306019.923251.127031.164292.99如何查水在27oC、32.7oC的飽和蒸汽壓和焓?水的物理性質(zhì)函數(shù)關(guān)系：函數(shù)值和自變量的關(guān)系以表格給出，稱列表函數(shù)列表函數(shù)的特點(diǎn)：①自變量與函數(shù)值一一對應(yīng)；②函數(shù)值有很可靠的精確度；③自變量與函數(shù)間的解析表達(dá)式可能不清楚，或者解析表達(dá)式非常復(fù)雜不便于計(jì)算（如為無窮級數(shù)等）；④沒有直接給出未列出點(diǎn)的函數(shù)值，不便于進(jìn)行微分和積分以及計(jì)算機(jī)計(jì)算。1目的：運(yùn)用計(jì)算機(jī)方便地求解函數(shù)值和進(jìn)行微積分等計(jì)算思路：尋找列表函數(shù)的近似解析表達(dá)式(不知道f(x)的函數(shù)表達(dá)形式)

代數(shù)插值問題描述（怎樣進(jìn)行插值）列表函數(shù)例在區(qū)間上連續(xù)設(shè)函數(shù)已知函數(shù)f(x)

在各點(diǎn)(x0,

x1…)上的值方法：建立一個(gè)次數(shù)不超過n的代數(shù)多項(xiàng)式P(x)來近似f(x)溫度x飽和蒸汽壓f(x)x0=00.6087x1=101.2262x2=202.3346x3=304.2474x4=407.3766x5=5012.34x6=6019.923x7=7031.1642用代數(shù)多項(xiàng)式P(x)近似列表函數(shù)y插值法插值方式：全節(jié)點(diǎn)插值——用全部節(jié)點(diǎn)構(gòu)造插值多項(xiàng)式（通過全部節(jié)點(diǎn)）分段插值——只用部分節(jié)點(diǎn)構(gòu)造插值多項(xiàng)式（只通過部分節(jié)點(diǎn)）本節(jié)討論一元n次拉格朗日插值(n+1個(gè)點(diǎn))線性插值（2個(gè)節(jié)點(diǎn)）拋物線插值（3個(gè)節(jié)點(diǎn)）條件：滿足Pn(xi)=yi(i=0,1,…n)Pn(x）稱為函數(shù)f(x)的插值多項(xiàng)式點(diǎn)x0,x1,…xn叫做插值節(jié)點(diǎn)[a，b]為插值區(qū)間。溫度x飽和蒸汽壓f(x)x0=0y0=0.6087x1=10y1=1.2262x2=20y2=2.3346x3=30y3=4.2474x4=40y4=7.3766x5=50y5=12.34x6=60y6=19.923x7=70y7=31.1643拉格朗日(Lagrange)插值由已知的兩點(diǎn)，可寫出直線方程以直線方程作為插值多項(xiàng)式，即下式中的n=1構(gòu)造線性函數(shù)L1(x)，使函數(shù)通過已知的兩點(diǎn)，即滿足線性插值——用直線方程L1(x)近似列表函數(shù)式f(x)需要構(gòu)造一個(gè)直線方程（線性插值多項(xiàng)式）溫度x飽和蒸汽壓f(x)x0=0y0=0.6082x1=10y1=1.2262x2=20y2=2.3346x3=30y3=4.2474x4=40y4=7.3766x5=50y5=12.34x6=60y6=19.923x7=70y7=31.1644由兩點(diǎn)式可看出，L1(x)是由兩個(gè)線性函數(shù)兩點(diǎn)式點(diǎn)斜式的線性組合得到的，線性插值多項(xiàng)式可寫為線性組合系數(shù)線性組合系數(shù)也是線性插值多項(xiàng)式滿足直線方程的兩種構(gòu)成形式,通過點(diǎn)（xk,yk)和(xk+1,yk+1）

滿足以上條件的lk(x)和lk+1(x)稱為線性基本插值多項(xiàng)式或線性插值基函數(shù)，插值多項(xiàng)式可用插值基函數(shù)的線性組合構(gòu)成。5線性插值圖示

幾何意義：用給定兩點(diǎn)的直線y=L1(x)近似替代f(x)。xf(x)xkf(xk+1)f(xk)xk+1yL1(x)溫度x飽和蒸汽壓f(x)x0=0y0=0.6082x1=10y1=1.2262x2=20y2=2.3346x3=30y3=4.2474x4=40y4=7.3766x5=50y5=12.34x6=60y6=19.923x7=70y7=31.1646

二次插值（三點(diǎn)插值或拋物線插值）f(xk+2)xk-1xkxf(xk)f(xk-1)yL1(x)二次插值是用已知的3個(gè)插值節(jié)點(diǎn)，構(gòu)造一個(gè)二次函數(shù)L2(x)并使函數(shù)通過三個(gè)插值節(jié)點(diǎn)構(gòu)造方法：用插值基函數(shù)進(jìn)行線性組合。但此時(shí)因是三個(gè)插值節(jié)點(diǎn)，插值基函數(shù)應(yīng)該是二次函數(shù)。插值基函數(shù)在節(jié)點(diǎn)處應(yīng)滿足條件：xk+17二次插值（三點(diǎn)插值或拋物線插值）如何求滿足條件的二次插值基函數(shù)，分析一次插值基函數(shù)必定含有因子由類推可得二次插值基函數(shù)要滿足條件8利用三個(gè)二次插值基函數(shù)可組合成二次插值多項(xiàng)式二次插值又稱為三點(diǎn)插值或拋物線插值二次插值幾何意義：用通過給定三點(diǎn)的拋物線y=L2(x)近似替代函數(shù)y=f(x)。近似程度通常比線性插值更高。xk-1xkf(xk+2)xf(xk)f(xk-1)yL2(x)f(x)9n次插值線性插值和插值基函數(shù)為插值基函數(shù)滿足lk(xk)

=1,lk(xk+1)

=0lk+1(xk)

=0,lk(xk+1)

=1拋物線插值和插值基函數(shù)為10所以此時(shí)的插值基函數(shù)應(yīng)該滿足用類似的推導(dǎo)方法，可求得n次插值基函數(shù)為利用

n+1個(gè)插值節(jié)點(diǎn)，構(gòu)造一個(gè)n次插值多項(xiàng)式使通過所有n+1個(gè)插值節(jié)點(diǎn)，即滿足(j=0，1,…,n)

將插值基函數(shù)組合可得拉格朗日n次插值多項(xiàng)式拉格朗日插值公式11插值公式通過n+1個(gè)插值節(jié)點(diǎn)，是唯一確定的如果節(jié)點(diǎn)數(shù)有n+1個(gè)，稱為全節(jié)點(diǎn)插值拉格朗日n次插值的幾何意義是否通過的點(diǎn)越多,插值次數(shù)越高越好?12用11點(diǎn)構(gòu)造10次多項(xiàng)式插值的龍格現(xiàn)象某些點(diǎn)插值結(jié)果誤差很大，函數(shù)兩端震蕩加劇在節(jié)點(diǎn)很多的場合，通常不宜采用高次插值分段的低階插值往往效果更好13例：

已知函數(shù)表x0·10·20·30·4

f（x）0·09980·19870·29550·3894

分別用線性插值、2次插值和3次插值求f(x)在x=0.25處的值。解：（1）分段線性插值選取最接近插值點(diǎn)0.25的兩個(gè)插值節(jié)點(diǎn)，求f(x)在x=0.25處的值。由于簡單，可以直接計(jì)算。選取的兩個(gè)插值節(jié)點(diǎn)如下采用由線性插值多項(xiàng)式構(gòu)成的近似多項(xiàng)式代入插值節(jié)點(diǎn)數(shù)據(jù)14C語言：插值計(jì)算水的物性#include<stdio.h>#include<stdlib.h>#include<math.h>#ifndeflagi3#definelagi3#include"lagi3.cpp"#endiffloatwater(floattm1,inta){ staticfloatt[11],mr[6][11]; staticintn=0; floaty; FILE*in; intitm,i; if(n==0){ n=11;if((in=fopen("water.txt","r"))==NULL){printf("Cannotopeninputfile.\n");exit(0);}fscanf(in,"%d",&itm);

do{

while((i=fgetc(in))!=10);/*printf("%c",i);printf("\n");/**/ itm--;}while(itm>0);

for(i=0;i<n;++i){

fscanf(in,"%f",&t[i]);

fscanf(in,"%f",&mr[0][i]);

fscanf(in,"%f",&mr[1][i]);mr[1][i]/=1e3;

fscanf(in,"%f",&mr[2][i]);mr[2][i]/=1e3;

fscanf(in,"%f",&mr[3][i]);

fscanf(in,"%f",&mr[4][i]);

fscanf(in,"%f",&mr[5][i]);mr[5][i]/=1e3;}fclose(in);}#ifndeftiaoshipfor(i=0;i<n;++i){printf("%d%f%f%f\n",i,t[i],mr[0][i],mr[1][i]);}printf("\n");#endif

y=lagi3y(n,t,&mr[a][0],tm1);returny;}#ifndeftiaoshimvoidmain(){floattm1,rho,mu;tm1=15.2;rho=water(tm1,0);mu=water(tm1,2);tm1=25.2;mu=water(tm1,2);rho=water(tm1,0);return;}#endif15lagi3.cpp#include"stdio.h"floatlagi3y(intn,float

a[],float

b[],floatx){ staticinti=0; staticfloatxold=-999.;

intflag,i0,ie; floaty,x1,x2,x3,z1,z2,z3;

if(xold!=x){i0=0;ie=n-3;i=(i0+ie)/2; if(a[i0]<a[ie])flag=1; elseflag=-1;

while(i!=ie){if(x*flag>(a[i+1]+a[i+2])/2*flag)i0=i+1;elseie=i;i=(i0+ie)/2;}//printf("%f%f\n",a[i],x);xold=x;}

x1=a[i];x2=a[i+1];x3=a[i+2];//printf("%f%f%f\n",x1,x2,x3);

z1=(x-x2)/(x1-x2)*(x-x3)/(x1-x3);z2=(x-x1)/(x2-x1)*(x-x3)/(x2-x3);z3=(x-x2)/(x3-x2)*(x-x1)/(x3-x1);