Muhmmad al-Xorazmiy nomidagi tatu qarshi filiali tt va kt fakulteti akt 11 21 Guruh talabasi

Muhmmad al-Xorazmiy nomidagi TATU Qarshi filiali
TT va KT fakulteti AKT 11_21 Guruh talabasi
Sherxonov Sherdorning Algortimlarni loyihalash fanidan

1-amaliy ishi

Bajardi:. Sherxonov.Sh

Qabul qildi:. Samandarova.Sh

Mavzu:
Amaliy ishini bajarish tartibi. Laboratoriya ishini bajarishda quyidagi tartibga amal qiling: 1. Guruh jurnalidagi nomerga ko‘ra o‘z variantingizni aniqlang 2. Masalani yechish uchun algoritm va dastur quring. 3. Kichik hajmdagi ma’lumotlar uchun dasturning to‘g‘ri ishlayotganligiga ishonch hosil qiling. 4. Bajarilgan ishlar haqida hisobot tayyorlang. Amaliy topshiriqlari variantlari Quyidagi tenglamalar: 1. Ildizlarning qisqa atrofini analitik yoki grafik usulda aniqlang. 2. Aniqlangan oraliqda ildizning taqribiy qiymatini 0.01 aniqlikda iteratsiya usuli bilan hisoblang 3. Aniqlangan oraliqda ildizni ketma-ket yaqinlashish va ikkiga bo‘lish usuli bilan taqribiy hisoblang

N=16
1) tg(0.3x+0.4)=0
2) x3+4x-6=0
1. tg(0.3x+0.4)=0

Sinx(0,3x+0.4)=0
Cos(0,3x+0,4)≠0



0.3x+0,4=πк
0,3x+0.4≠π/2+ πк

0.3x=-0.4+πк

X=4/3+πк/0.3

0.3x+0.4≠π/2+π

X=-0.4/0.3+ π/0.6+ πк/0.3

Dastur kodi:

#include
#include

double equation(double x) {

return tan(0.3 * x + 0.4);
}

int main() {

double x = -10.0; // Starting value for x
double epsilon = 0.0001; // Tolerance for convergence
int maxIterations = 100; // Maximum number of iterations

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

double result = equation(x);
if (fabs(result) < epsilon) {
std::cout << "Solution found: x = " << x << std::endl;
return 0;
}

// Update x for the next iteration

x += 0.1; // You can adjust the step size according to your needs
}

std::cout << "No solution found within the given iterations." << std::endl;

return 0;
}

N atija:

Misol:2

X³ +4x-6=0

Dastur kodi:

#include
#include

double f(double x) {

return x * x * x + 4 * x - 6;
}

double f_derivative(double x) {

return 3 * x * x + 4;
}

double newton_raphson(double x0, double epsilon) {

double x1 = x0 - f(x0) / f_derivative(x0);
while (fabs(x1 - x0) > epsilon) {
x0 = x1;
x1 = x0 - f(x0) / f_derivative(x0);
}
return x1;
}

int main() {

double x0 = 1.0; // Boshlang'ich taxminiy qiymat
double epsilon = 0.0001; // Epsilon qiymati (qanday qatorda yaxlitlashni aniqlash)
double root = newton_raphson(x0, epsilon);
std::cout << "Tenglama yechimi: x = " << root << std::endl;
return 0;
}
Natija: