import numpy as np

from scipy.linalg import solve

rows, cols = (2**5, 2**5)

A = [[np.random.uniform(0,2,None) for i in range(cols)] for j in range(rows)]

print('Matrix A:')

for row in A:

print(row)

B = [[A[j][i] for j in range(len(A))] for i in range(len(A[0]))]

print('Matrix B:')

for row in B:

print(row)

AB = [[0 for i in range(cols)] for j in range(rows)]

for i in range(len(A)):

for j in range(len(B[0])):

for k in range(len(B)):

#resulted matrix

AB[i][j] += A[i][k] * B[k][j]

print('')

print('Matrix AB:')

for row in AB:

print(row)

def normalize(x):

fac = abs(x).max()

x_n = x / x.max()

return fac, x_n

x = np.array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])

for i in range(8):

x = np.dot(A, x)

lambda1, x = normalize(x)

print('Eigenvalue:', lambda1)

print('Eigenvector:', x)

import matplotlib.pyplot as plt

h1 = 10**4

x1 = np.arange(0, 1, h1)

y1method1 = [1]

y1method2 = [1]

def f(c, x, y): #this is the function f(x)

return (-c + x + y + (2 * x * y))

def bisectionMethod (A, B, x, y):

return bisectionMethod(a, c, x, y) #eliminates the right side of x using recursion

for i in range(0, len(x1)-1):

y1method1.append(y1method1[i] + h1 * bisectionMethod(-10000000, 10000000,x1[i], y1method1[i]))

#print(y1method1[i])

for i in range(0, len(x1)-1):

k1 = bisectionMethod(-10000, 10000, x1[i], y1method2[i])

k2 = bisectionMethod(-10000, 10000, x1[i] + 1/2*h1, y1method2[i] + 1/2*h1*k1)

k3 = bisectionMethod(-10000, 10000, x1[i] + 1/2*h1, y1method2[i] + 1/2*h1*k2)

k4 = bisectionMethod(-10000, 10000, x1[i] + h1, y1method2[i] + h1*k3)

y1method2.append(y1method2[i] + 1/6*h1*(k1 + 2*k2 + 2*k3 + k4))

plt.plot(x1, y1method1, label = '10^5, euler')

plt.plot(x1, y1method2, label = '10^5, RK4')

plt.legend()

plt.show()