數(shù)值分析實(shí)驗(yàn)課

實(shí)驗(yàn)報(bào)告求解方程迭代法實(shí)驗(yàn)（數(shù)值計(jì)算方法，2學(xué)時(shí)）一實(shí)驗(yàn)?zāi)康恼莆辗匠痰ㄋ枷耄瑫?huì)用Aitken法、牛頓法、弦截法求解方程。二實(shí)驗(yàn)內(nèi)容1用下列方法求方程X3-3X-1=0在％=2附近的根，要求準(zhǔn)確到四位有效數(shù)字。（1）牛頓法。（2）單點(diǎn)弦截法（應(yīng)數(shù)）。（3）雙點(diǎn)弦截法（信科）。2用Aitken法求方程x3-x2-1=0在x°=1.5附近的根，精度要求為￡=10-4。三實(shí)驗(yàn)步驟（算法）與結(jié)果1（1）實(shí)驗(yàn)步驟：輸出孔C語(yǔ)言代碼:#include"stdio.h”#include"math.h”

#defineNUM100intmain(){doublex[NUM];doublep,f;inti;x[0]=2;p=pow(10,-4);printf("thenumberis%5.4f\n",x[0]);for(i=1;i<20;i++){x[i]=x[i-1]-(pow(x[i-1],3)-3*x[i-1]-1)/(3*pow(x[i-1],2)-3);if(x[i]>x[i-1])f=x[i]-x[i-1];elsef=x[i-1]-x[i];printf("thenumberis%5.4f\n”,x[i]);if(f<p){break;}}getch();return0;}運(yùn)算結(jié)果：*?QMJ5BFskA￡G「\DeMktQE逑隹方析作業(yè)'受途果匕上⑴Nlewtar^eKe'ilnumboF±?2.0300il我皿人觸卜Isilnumberis1=8795numbei*is1.87^4

(2)C語(yǔ)言代碼：#include"stdio.h”#include"math.h”#defineNUM100intmain(){doublex[NUM];doublep,q,f;inti;x[0]=2;x[1]=1.5;f=pow(x[0],3)-3*x[0]-1;p=pow(10,-4);printf("thenumberis%5.4f\n”,x[0]);for(i=1;i<10;i++){x[i+1]=(x[0]*(pow(x[i],3)-3*x[i]-1)-x[i]*f)/(pow(x[i],3)-3*x[i]-1-f);if(x[i]>x[i-1])

q=x[i]-x[i-1];elseq=x[i-1]-x[i];printf("thenumberis%5.4f\n”,x[i+1]);if(q<p){break;}}getch();return0;}運(yùn)算步驟：C:\User&\Acer\De如叩\數(shù) 作業(yè)\實(shí)發(fā)誤11住J5ecantexetliethetlieLliethethetlicnunbermmbernLinberriLinbernunbernunbernLinbarLSISLSLSLSLSZ.MUUt)1.84001.S7601.87911.87V41.87941.8794,口回?將曲=響解■外=機(jī)心由兌出擔(dān)，城代兩次盼三個(gè)相牝的迭代廿沃u，H-zi-Kn)fl=Ol'￡f5h=f『+Il'￡f5輸出七C語(yǔ)言代碼:#include"stdio.h”#include"math.h”#defineNUM100intmain(){doublex[NUM];doubley[NUM];doublez[NUM];doublep,f;inti;x[0]=1.5;y[1]=pow(x[0],3)-pow(x[0],2)+x[0]-1;z[1]=pow(y[1],3)-pow(y[1],2)+y[1]-1;p=pow(10,-4);for(i=1;i<10;i++){x[i]=(x[i-1]*z[i]-pow(y[i],2))/(x[i-1]-2*y[i]+z[i]);y[i+1]=pow(x[i],3)-pow(x[i],2)+x[i]-1;z[i+1]=pow(y[i+1],3)-pow(y[i+1],2)+y[i+1]-1;printf("thenumberis%5.4f\n”,x[i]);if(x[i]>x[i-1])f=x[i]-x[i-1];elsef=x[i-1]-x[i];if(f<p){break;}}getch();return0;}運(yùn)算結(jié)果：四實(shí)驗(yàn)收獲與教師評(píng)語(yǔ)收獲：學(xué)會(huì)通過(guò)編程，利用Aitken,Newton,Secant三種迭代解法，計(jì)算方程的解，有實(shí)際應(yīng)用