O’zbekiston respublikasi axborot texnologiyalari

Dastur kodi

#include

using namespace std;

int main ()

{

float x[]={-0.2,-0.1,0,0.1,0.2,0.3,0.4,0.5,0.6,0.7};

float y[]={4.7936,6.9441,9.0370,10.9361,12.7296,14.3601,15.8096,17.0625,18.1056,18.9281};

float s[50], b[10], p=1, X=-0.07, sum=y[0];

int m=0, a=9;

cout<<"\n"<<"Nyuton interpalatsion ko'pxadini tuzish va uni yechish"<<"\n\n";

cout<<"ayirmalarda chiqqan birinchi natijalar "<<"\n\n";

for (int i=0;i<9;i++,m++)

{

for (int j=0;j

{

s[j]=(y[j+1]-y[j])/(x[j+1+m]-x[j]);

y[j]=s[j];

}

b[i]=s[0]; a--;

cout<

}

for (int i=0;i<9;i++)

{

for (int j=0;j<10;j++)

{

p=p*(X-x[j]);

}

sum=sum+p*b[i];

}

cout<<"\n"<<"Nyuton interpalatsion ko'pxadining qiymati = "<

cout<<" Umarov Otabek !!!";

return 0;

}




6-Laboratoriya ishi

MAVZU : APPROKSIMATSIYA MASALASINI YECHISHDA ENG KICHIK KVADRATLAR USULI. CHIZIQLI VA KVADRATIK MODELLAR
Berilgan jadvaldan k = 2(n–1) formula orqali n=15 guruh jurnalidagitartibnomerimdan boshlab 10 tа qiymat ko’chiriboldim

k=28. Jadvaldagi variantga deb olindi, ya’ni va variant jadvali i=0,1,2,…,9 ko’rinishida belgilanadi. Jadval qiymatlaridagisistematik bartaraf qilib bo’lmas xatoliklartartibi tartibida ekanligima’lum deb funksiyaning ; ; ;nuqtalaridagi qiymatlarni aniqladim.

Berilgan x qiymatlari (-0.2,-0.1, 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7)

y qiymatlari (4.7936,6.9441,9.037,10.9361,12.7296,14.3601,15.8096,17.0625,18.1056,18.9281)

endi esa hisoblashlar;

berilgan x larning o’rtacha qiymati = 0.25

berilgan y larning o’rtacha qiymati = 12.8706

x*y larning o’rtacha qiymati = 4.52626

x kvadratlarning o’rtacha qiymati = 0.145

a= formula orqali a ning qiymati a = 15.8618

b= formula orqali b = 8.90518 qiymatlar hosil bo’ldi.

Endi esa shartga larning qiymatini topamiz

= -0.07;

0.38= 0.18;

= 0.52;
Endi bu uchta nuqtadagi qiymatlarni quyidagi dastur orqali ko’rib chiqamiz:

Dastur kodi:

#include

using namespace std;

int main(){

float x[]={-0.2,-0.1,0,0.1,0.2,0.3,0.4,0.5,0.6,0.7};

float y[]={4.7936,6.9441,9.037,10.9361,12.7296,14.3601,15.8096,17.0625,18.1056,18.9281};

float delx=0,dely=0,delxy=0,delx2=0,a,b,a1,a2,a3,X,y1,y2,y3;

a1=x[0]+0.13;

a2=x[0]+0.38;

a3=x[0]+0.72;

cout<<"x^2"<<"\t"<<"x*y"<<"\t"<<"\n";

for(int i=0;i<10;i++) {

cout<


for(int i=0;i<10;i++) {

delx+=x[i]/10;

dely+=y[i]/10;

delxy+=x[i]*y[i]/10;

delx2+=pow(x[i],2)/10; }

cout<<"\n"<<"\n"<<"x ning o'rtacha qiymati = "<

cout<<"\n"<<"\n"<<"y ning o'rtacha qiymati = "<

cout<<"\n"<<"\n"<<"ko'paytmaning o'rtacha qiymati = "<

cout<<"\n"<<"\n"<<"x^2 ning o'rtacha qiymati = "<

a=(delxy-delx*dely)/(delx2-pow(delx,2));

cout<<"\n"<<"\n"<<" a = "<

b=dely-a*delx;

cout<<"\n"<<" b = "<

y1=X*a1+b;

y2=X*a2+b;

y3=X*a3+b;

cout<<"\n"<<"\n"<<"funksiyaning a1 a2 a3 nuqtalardagi taqribiy qiymati"<<"\n";

cout<<"\n"<<"y1 = "<

cout<<"Umarov Otabek!!!";

return 0;

}

Dasturning ko’rinishi:





7-Laboratoriya ishi

Mavzu: ANIQ INTEGRALLARNI TAQRIBIY HISOBLASHDA SIMPSON VA MONTE-KARLO USULLARI

15.

[0;1]

10

160

Quyida ushbu misolni C++ tilida dasturini ko’rib chiqamiz:

#include

using namespace std;

int main()

{

float x[10], y[10], S=0, S1=0, S2=0, S3;

double h;

float n=10, a=0, b=1;

h = (b-a)/n;

cout << " Aniq integrallarni taqribiy hisoblash" << "\n\n";

cout << " x " << "\t" << " y " << "\n";

for (int i=0; i<=10; i++ )

{

x[i] = a+i*h;

y[i] = cos(x[i]*x[i]+1)*pow(x[i]*x[i]+5*x[i]+8,1/3);

cout << x[i]<< "\t" << y[i] << "\n";

}

for (int i=0; i<10; i++ )

{

S+= y[i];

}

cout << "\n" << " chap to'g'ri to'rtburchaklar formulasi bo'yicha = " << h*S << "\n\n";

for (int i=1; i<=10; i++ )

{

S1+= y[i];

}

cout << " o'ng to'g'ri to'rtburchaklar formulasi bo'yicha = " << h*S1 << "\n\n";

for (int i=1; i<10; i++ )

{

S2+= y[i];

}

cout << " Trapetsiya usulida = "<< h*((1+y[10])/2+S2) << "\n\n";

cout << " Simpson usulida = ";

S3 = h/3*(y[0]+y[10]+4*(y[1]+y[3]+y[5]+y[7]+y[9])+2*(y[2]+y[4]+y[6]+y[8]));

cout << S3 << "\n\n";

cout << " Umarov Otabek!!!";

return 0;

}