Batasheva Durdonaning

Muhammad al-Xorazmiy nomidagi Toshkent axborot texnologiyalari universiteti Televizion
texnologiyalar/Audiovizual texnologiyalari
fakulteti 512-20 guruh talabasi

Batasheva Durdonaning

Algaritmlarni loyihalash fanidan

1-amaliy ishi

Bajardi: Batasheva Durdona.

Tekshirdi: Begimov O’ktam.
4-variant:

1. 
   
   

#include

#include
#define f(x) log(pow(x,2)+3)*pow(sin(x+1),2)
using namespace std;
int main()
{
float lower, upper, integration=0.0, stepSize, k;
int i, subInterval;

cout<<"Enter lower limit of integration: ";

cin>>lower;
cout<<"Enter upper limit of integration: ";
cin>>upper;
cout<<"Enter number of sub intervals: ";
cin>>subInterval;

stepSize = (upper - lower)/subInterval;

integration = f(lower) + f(upper);

for(i=1; i<= subInterval-1; i++)
{
k = lower + i*stepSize;
integration = integration + 2 * (f(k));
}

integration = integration * stepSize/2;

cout<< endl<<"Required value of integration is: "<< integration;

return 0;

}


Dastur Natijasi

2.

#include
#include

using namespace std;

#define MAX_ITER 1000000

double func(double x)

{
return x-cos(x);
}

void regulaFalsi(double a, double b)

{
if(func(a)*func(b) >= 0)
{
cout << " You have not assumed right a and b\n";
return;
}
double c = a;
for(int i = 0; i < MAX_ITER; i ++)
{
c = (a*func(b) - b*func(a)) / (func(b)-func(a));
if(func(c) == 0)
break;
else if(func(c)*func(a) < 0)
b = c;
else
a = c;
}
cout << "The value of root is : " << c;
}

int main()

{
double a = 0, b = 1;
regulaFalsi(a, b);

return 0;
}


b)
#include
#include

using namespace std;

#define MAX_ITER 1000000

double func(double x)

{
return x*x*x+3*x-1;
}

void regulaFalsi(double a, double b)

{
if(func(a)*func(b) >= 0)
{
cout << " You have not assumed right a and b\n";
return;
}
double c = a;
for(int i = 0; i < MAX_ITER; i ++)
{
c = (a*func(b) - b*func(a)) / (func(b)-func(a));
if(func(c) == 0)
break;
else if(func(c)*func(a) < 0)
b = c;
else
a = c;
}
cout << "The value of root is : " << c;
}

int main()

{
double a = 0, b = 1;
regulaFalsi(a, b);

return 0;
}


3-savol
[Forwarded from Durov]
from cmath import log
import numpy as np
def newton(f,Df,x0,epsilon,max_iter):
xn = x0
for n in range(0,max_iter):
fxn = f(xn)
if abs(fxn) < epsilon:
print(f'{n} ta iteratsiyadan song topilgan javoblar: ')
return xn
Dfxn = Df(xn)
if Dfxn == 0:
print('Xosila nol, yechim topilmadi.')
return None
xn = xn - fxn/Dfxn
print('Iteratsiya maximal dan oshib ketdi, yechim topilmadi')
return None
p = lambda x: x - np.cos(x)
Dp = lambda x: 1 + np.sin(x)
approx = newton(p,Dp,1,1e-3,10)
print(approx)
'''
p = lambda x: x**3 + 3*x - 1
Dp = lambda x: 3*x**2 + 3
approx = newton(p,Dp,1,1e-3,10)
print(approx)
'''
Dastur NAtijasi