數(shù)值計(jì)算方法上機(jī)實(shí)習(xí)

1、實(shí)習(xí)報(bào)告課程名稱：數(shù)值計(jì)算方法上機(jī)學(xué)院：電氣信息工程學(xué)院姓名：李勇學(xué)號(hào)：年12月30日數(shù)值計(jì)算方法上機(jī)習(xí)題TOCo1-5hz153101.用Newton迭代法求解方程x-3x_1=0在2.0附近的實(shí)根，2源程序：#include#includemain()doublex,f,f1;scanf(%lf,&x);運(yùn)行結(jié)果：r=1.879385輸入：for(;x=x-f/f1)f=x*x*x-3*x-1;f1=3*x*x-3;if(fabs(f)0.5e-5)break;printf(r=%lfn,x);getch();return0;2.02.設(shè)AX=B-5791011681091(1)A=710

2、87B=1I5765_1I1一f2x1+x2_3x3=5(2)丿3x,+6x2+2x3=9彳Xr+5x2_3x3=5試用高斯-約旦消去法求方程組的解。(1)源程序：#include#includemain()floata55=0,0,0,0,0,0,5,7,9,10,0,6,8,10,9,0,7,10,8,7,0,5,7,6,5,b5=0,1,1,1,1,s,t,am;inti,j,l,k;clrscr();for(k=1;k=4;k+)am=fabs(akk);l=k;for(i=k;iam)am=fabs(aik);l=i;if(alk=0)printf(qiyijuzhen.);brea

3、k;if(l!=k)for(j=k;j=4;j+)t=akj;akj=alj;alj=t;s=bl;bl=bk;bk=s;for(j=k+1;j=4;j+)akj=akj/akk;bk=bk/akk;for(i=1;i=4;i+)if(i=k)continue;bi=bi-aik*bk;for(j=k+1;j=4;j+)aij=aij-aik*akj;printf(x1=%fn,b1);printf(x2=%fn,b2);printf(x3=%fn,b3);printf(x4=%fn,b4);運(yùn)行結(jié)果：x1=19.999809x2=-11.999882x3=-4.999953x4=2.9999

4、72(2)源程序：#include#includemain()floata44=0,0,0,0,0,2,1,-3,0,3,6,2,0,4,5,-3,b4=0,-5,9,5,s,t,am;inti,j,l,k;clrscr();for(k=1;k=3;k+)am=fabs(akk);l=k;for(i=k;iam)am=fabs(aik);l=i;if(alk=0)printf(qiyijuzhen.);break;if(l!=k)for(j=k;j=3;j+)t=akj;akj=alj;alj=t;s=bl;bl=bk;bk=s;for(j=k+1;j=3;j+)akj=akj/akk;bk=

5、bk/akk;for(i=1;i=3;i+)if(i=k)continue;bi=bi-aik*bk;for(j=k+1;j=3;j+)aij=aij-aik*akj;printf(x1=%fn,b1);printf(x2=%fn,b2);printf(x3=%fn,b3);運(yùn)行結(jié)果：x1=-11.000001x2=8.000000 x3=-3.0000003已知函數(shù)y二sin(x)的數(shù)表Xi0.320.340.360.38Sinxi0.31456660.33348710.35227420.3709205試用拉格朗日插值法求x=0.3324時(shí)的函數(shù)值。源程序：#includevoidmain(

6、)floatx4=0.32,0.34,0.36,0.38,y4=0.3145666,0.3334871,0.3522742,0.3709205;floatu,v,l;inti,j;clrscr();scanf(%f,&u);v=0;for(j=0;j=3;j+)l=1;for(i=0;i=3;i+)if(i=j)continue;l=l*(u-xi)/(xj-xi);v=v+l*yj;printf(u=%fnv=%f,u,v);運(yùn)行結(jié)果：u=0.332400v=0.3263134.用復(fù)化Simpson方法求積分源程序：要求誤差不超過10#include#includevoidmain()int

7、i,n=2;floats;floatf(float);for(i=0;i=2;i+)s=simpson(f,0,1,n);printf(s(%d)=%fn,n,s);n*=2;getch();floatsimpson(float(*f)(float),floata,floatb,intn)intk;floats,s1,s2=0.0;floath=(b-a)/n;s1=f(a+h/2);for(k=1;k=n_1;k+)s1+=f(a+k*h+h/2);s2+=f(a+k*h);s=h/6*(f(a)+4*s1+2*s2+f(b);returns;floatf(floatx)return5/(1

8、+x*x);運(yùn)行結(jié)果：S(2)=3.926961S(4)=3.926991S(8)=3.9269915.用阿達(dá)姆斯預(yù)測(cè)-校正法求解八cos(sin(x)(0*1的初值問題。y(0)=0分別取步長(zhǎng)h=0.05和h=0.2，分析其結(jié)果，保留5位有效數(shù)字源程序：#include#includedoublef(doublex)returncos(sin(x);main()doubleh,x,y=0,y2;clrscr();scanf(%lf,&h);printf(h=%f”,h);for(x=0;x=1.001;x=x+h)y2=y+h/24.0*(9*f(x+h)+19*f(x)-5*f(x-h)+

9、f(x-2*h);y=y2;printf(%f%lfn,y2);運(yùn)行結(jié)果：h=0.200000 x=0.000000y=0.198720 x=0.200000y=0.389827x=0.400000y=0.567151x=0.600000y=0.727046x=0.800000y=0.868845x=1.000000y=0.994710運(yùn)行結(jié)果：h=0.050000 x=0.000000y=0.049979x=0.050000y=0.099834x=0.100000y=0.149441x=0.150000y=0.198680 x=0.200000y=0.247436x=0.250000y=0.

10、295600 x=0.300000y=0.343069x=0.350000y=0.389749x=0.400000y=0.435555x=0.450000y=0.480414x=0.500000y=0.524262x=0.550000y=0.567047x=0.600000y=0.608729x=0.650000y=0.649279x=0.700000y=0.688682x=0.750000y=0.726934x=0.800000y=0.764041x=0.850000y=0.800023x=0.900000y=0.834910 x=0.950000y=0.868740 x=1.000000y=0.901564為期四天的數(shù)值計(jì)算方法上機(jī)實(shí)習(xí)已經(jīng)接近尾聲，盡管時(shí)間很短，卻讓我獲益良多。首先，做什么事都不可能一蹴而就，必要的準(zhǔn)備和持之不懈的努力，以及敢于面對(duì)失敗，愈挫愈勇的精神。編程問題是非常繁瑣的，要想正確的編出程序，必須要在充分了解C語言程序的基礎(chǔ)上透徹的分析題目的的要求和目的，在進(jìn)行程序的編程，然而在編程的過程中不可避免的要面臨諸多的困難和阻礙，想要一下子就變出正確的程序幾乎不可能，面對(duì)錯(cuò)誤也不要失去耐心，尋根糾結(jié)，發(fā)現(xiàn)并改正錯(cuò)誤，最終取得成功。其次，我們要注意要虛心求教，人并不是萬能的，我們總有一些知識(shí)漏點(diǎn)，或者一