1、摘要求函數在給定區間上的定積分，在微積分學中已給出了許多計算方法，但是，在實際問題計算中，往往僅給出函數在一些離散點的值，它的解析表達式沒有明顯的給出，或者，雖然給出解析表達式，但卻很難求得其原函數。這時我們可以通過數值方法求出函數積分的近似值。當然再用近似值代替真實值時，誤差精度是我們需要考慮因素，但是除了誤差精度以外，還可以用代數精度來判斷其精度的高低。已知n+1點的Newton-Cotes型積分公式，當n為奇數時，其代數精度為n；當n為偶數時，其代數精度達到n+1。若對隨機選取的n+1個節點作插值型積分公式也僅有n次代數精度。如何選取適當的節點，能使代數精度提高？Gauss型積分公式可是

2、實現這一點，但是Gauss型求積公式，需要被積函數滿足的條件是正交，這一條件比較苛刻。因此本實驗將針對三種常用的Gauss型積分公式進行討論并編程實現。關鍵詞：Newton-Cotes型積分公式 正交多項式 代數精度1、實驗目的1) 通過本次實驗體會并學習Gauss型積分公式，在解決如何取節點能提高代數精度這一問題中的思想方法。2) 通過對Gauss型積分公式的三種常見類型進行編程實現，提高自己的編程能力。3) 用實驗報告的形式展現，提高自己在寫論文方面的能力。2、算法流程下面介紹三種常見的Gauss型積分公式1) 高斯-勒讓德（Gauss-Legendre）積分公式勒讓德（Legendre）

3、多項式如下定義的多項式Lnx=12nn!dndxn(x2-1)n,x-1,1,n=0,1,2稱作勒讓德多項式。由于(x2-1)n是2n次多項式，所以Lnx是n次多項式，其最高次冪的系數An與多項式12nn!dndxnx2n=12nn!2n(2n-1)(2n-2)(n+1)xn的系數相同。也就是說n次勒讓德多項式具有正交性即勒讓德多項式Lnx是在-1,1上帶x=1的n次正交多項式，而且Lm,Ln=-11Lm(x)Ln(x)dx=0, mn22n+1, m=n 這時Gauss型積分公式的節點就取為上述多項式Lnx的零點，相應的Gauss型積分公式為-11f(x)dxk=1nAkf(xk)此積分公式

4、即成為高斯-勒讓德積分公式。其中Gauss-Legendre求積公式的系數Ak=-11xnxx-xk'nxdx=-11xLnxx-xkL'nxdx其中k的取值范圍為k=1,2,nGauss點和系數不容易計算，但是在實際計算中精度要求不是很高，所以給出如下表所示的部分Gauss點xk和系數Ak，在實際應用中只需查表即可。nxAnxA1026±0.9324695142±0.6612093865±1.23861918160.1713244920.3607615730.4679139342±0.577350269217±0.949107

5、9123±0.7415311856±0.405845151400.1294849660.2797053910.3818300500.4179591833±0.7745966692000.55555555560.88888888894±0.8611363116±0.33998104360.34785484510.65214515498±0.9602898565±0.7966664774±0.5255324099±01012285360.2223810340.3137066450.3

6、626837835±0.9061798459±0.538469310100.23692688510.47862867050.56888888892) 高斯-拉蓋爾（Gauss-Laguerre）積分公式拉蓋爾（Laguere）多項式Lnx=exdndxn(xne-x),0x<+,n=0,1,2稱為拉蓋爾多項式。其首項系數為(-1)n,且具有性質：正交性，在區間0,+上關于權函數x=e-x正交，而且Lm,Ln=0e-xLm(x)Ln(x)dx=0, mn(n!)2, m=n 積分區間為0,+，權函數為x=e-x的Gauss型積分公式稱為高斯-拉蓋爾積分公式，其中Gaus

7、s點為拉蓋爾多項式Lnx的零點，高斯-拉蓋爾積分公式為0e-xf(x)dxk=1nAkf(xk)同樣高斯-拉蓋爾積分公式的Gauss點和求積系數如下表所示：nxAnxA20.58578643763.41421356240.85355339050.146446609450.26356031971.41340305913.59642577107.085810005812.64080084420.52175561050.39866681100.07594244970.00361175870.000023370030.41577455672.29428036026.28994508290.711093

8、00990.278517733540.32254768961.74576110114.53662029699.39507091230.60315410430.35741869240.03888790850.000539294760.2228466041199273632605.77514356919.837467418315.98287398060.45896467930.41700083070.11337338200.01039919750.00026101720.00000089853) 高斯-埃爾米特（Gauss-Hermite）積分公式埃爾米特（Hermite

9、）多項式Hnx=(-1)nex2dne-x2dxn,-<x<+,n=0,1,2被稱作埃爾米特多項式，其首項系數為2n，具有性質如下正交性，在區間-,+上關于權函數e-x2正交，而且Hm,Hn=-+e-x2Hm(x)Hn(x)dx=0, mn2nn!, m=n 積分區間為-,+，權函數為x=e-x2的Gauss型積分公式稱為Gauss-Hermite積分公式，其Gauss點就是Hermite正交多項式Hnx的零點。Gauss-Hermite求積公式為-e-x2f(x)dxk=1nAkf(xk)同樣高斯-埃爾米特積分公式的Gauss點和求積系數如下表所示：nxAnxA2±0.

10、70710678110.88622692557±0.4360774119±1.3358490704±2.35030497360.7246295952000453000993±1.224744871400.29540897511.81635900064±0.5246476232±1.65068012380.80491409000.08131283548±0.8162878828±1.6735516287±2.651961356300.42560725260.05451558280.

11、00097178120.81026461755±0.9585724646±2.02018287040.39361932310.01995324210.94530872040.56888888893、算法實例1) 用3點Gauss型求積公式計算-11cosxdx解：根據積分限可以知道應該用Gauss-Legendre積分公式，具體程序如下所示#include <iostream>#include <math.h>using namespace std; const int M(10);void main()int i=0;int n=0;int m=0

12、;int sign=0;double sum=0;double xM=0;double AM=0;double x1=0;double x2=-0.57735502692,0.57735502692;double x3=-0.77459666920,0.77459666920,0;double x4=-0.8611363116,0.8611363116,-0.3399810436,0.3399810436;double x5=-0.9061798459,0.9061798459,-0.53846931010,0.53846931010,0;double x6=-0.9324695142,0.9

13、324695142,-0.6612093865,0.6612093865,-1.2386191816,1.2386191816;double x7=-0.9491079123,0.9491079123,-0.7415311856,0.7415311856,-0.40584515140,0.40584515140,0;double x8=-0.9602898565,0.9602898565,-0.7966664774,0.7966664774,-0.5255324099,0.5255324099,-0.1834346425,0.1834346425;double A1=2;double A2=1

14、;double A3=0.5555555556,0.8888888889;double A4=0.3478548451,0.6521451549;double A5=0.2369268851,0.4786286705,0.5688888889;double A6=0.1713244924,0.3607615730,0.4679139346;double A7=0.1294849662,0.2797053915,0.3818300505,0.4179591834;double A8=0.1012285363,0.2223810345,0.3137066459,0.3626837834;cout&

15、lt;<"請輸入節點個數"<<endl;cin>>n;switch(n)case 1:for(i=0;i<n;i+)xi=x1i;Ai=A1i;break;case 2:for(i=0;i<n;i+)xi=x2i;for(i=0;i<n;i+)Ai=A2i/2;break;case 3:for(i=0;i<n;i+)xi=x3i;for(i=0;i<n;i+)Ai=A3i/2;break;case 4:for(i=0;i<n;i+)xi=x4i;for(i=0;i<n;i+)Ai=A4i/2;break

16、;case 5:for(i=0;i<n;i+)xi=x5i;for(i=0;i<n;i+)Ai=A5i/2;break;case 6:for(i=0;i<n;i+)xi=x6i;for(i=0;i<n;i+)Ai=A6i/2;break;case 7:for(i=0;i<n;i+)xi=x7i;for(i=0;i<n;i+)Ai=A7i/2;break;case 8:for(i=0;i<n;i+)xi=x8i;for(i=0;i<n;i+)Ai=A8i/2;break;default:cout<<"輸入出錯，請從新輸入！&q

17、uot;<<endl;break;for(i=0;i<n;i+)sum=sum+Ai*cos(xi);cout<<sum;運行結果：2) 用兩點Gauss型求積公式計算積分0e-10xsinxdx解：根據積分限可以知道應該用Gauss-Laguerre積分公式，具體程序如下所示#include "stdafx.h"#include <iostream>#include <math.h>using namespace std;const int M(10);void main()int i=0;int n=0;int m=

18、0;int sign=0;double sum=0;double xM=0;double AM=0;double x2=0.5857864376,3.4142135624;double x3=0.4157745567,2.2942803602,6.2899450829;double x4=0.3225476896,1.7457611011,4.5366202969,9.3950709123;double x5=0.2635603197,1.4134030591,3.5964257710,7.0858100058,12.6408008442;double x6=0.2228466041,1.18

19、89321016,2.9927363260,5.7751435691,9.8374674183,15.9828739806;double A2=0.8535533905,0.1464466094;double A3=0.7110930099,0.2785177335,0.0103892565;double A4=0.6031541043,0.3574186924,0.0388879085,0.0005392947;double A5=0.5217556105,0.3986668110,0.0759424497,0.0036117587,0.0000233700;double A6=0.4589

20、646793,0.4170008307,0.1133733820,0.0103991975,0.0002610172,0.0000008985;cout<<"請輸入節點個數"<<endl;cin>>n;switch(n)case 2:for(i=0;i<n;i+)xi=x2i;for(i=0;i<n;i+)Ai=A2i;break;case 3:for(i=0;i<n;i+)xi=x3i;for(i=0;i<n;i+)Ai=A3i;break;case 4:for(i=0;i<n;i+)xi=x4i;for(

21、i=0;i<n;i+)Ai=A4i;break;case 5:for(i=0;i<n;i+)xi=x5i;for(i=0;i<n;i+)Ai=A5i;break;case 6:for(i=0;i<n;i+)xi=x6i;for(i=0;i<n;i+)Ai=A6i;break;default:cout<<"輸入出錯，請從新輸入！"<<endl;break;for(i=0;i<n;i+)sum=sum+Ai*sin(xi)*exp(-9*xi);cout<<sum;運行結果：3) 用兩點Gauss型求積公式

22、計算積分-e-x2cosxdx解：根據積分限可以知道應該用Gauss-Hermite積分公式，具體程序如下所示#include "stdafx.h"#include <iostream>#include <math.h>using namespace std;const int M(10);void main()int i=0;int n=0;int m=0;int sign=0;double sum=0;double xM=0;double AM=0;double x2=-0.7071067811,0.7071067811;double x3=-1

23、.2247448714,1.2247448714,0;double x4=-0.5246476232,0.5246476232,-1.6506801238,1.6506801238;double x5=-0.9585724646,0.9585724646,-2.0201828704,2.0201828704;double x6=-0.4360774119,0.4360774119,-1.3358490704,1.3358490704,-2.3506049736,2.3506049736;double x7=-0.8162878828,0.8162878828,-1.6735516287,1.6735516287,-2.65196135630,2.65196135630,0;double A2=0.8862269255;double A3=0.2954089751,1.8163590006;double A4=0.8049140900,0.0813128354;double A5=0.3936193231,0.0199532421,0.9453087204,0.5688888889;double A6=0.7246295952,0.157067