18. Algebrayik va transsendent tenglamalar yechishning kesmani teng ikkiga boʻlish

18. Algebrayik va transsendent tenglamalar yechishning kesmani teng ikkiga boʻlish
x³+x-1=0
[0;1] oraliq uchun
E=0,01

def f(x):

return x**3 + x - 1

def f_derivative(x):

return 3*x**2 + 1

def newton_raphson(x_0, epsilon):

while True:
x_1 = x_0 - f(x_0) / f_derivative(x_0)
if abs(f(x_1)) < epsilon:
return x_1
x_0 = x_1

# Quyidagi qator yechimni hisoblash uchun ishlatiladi:

print(newton_raphson(0.5, 0.01))

Quyidagi Python kodida, Newton-Raphson usuli yordamida bu tenglamani yechishning funksiyasini ko'rsatamiz:

python
def f(x):
return x**3 + x - 1

def f_derivative(x):

return 3*x**2 + 1

def newton_raphson(x_0, epsilon):

while True:
x_1 = x_0 - f(x_0) / f_derivative(x_0)
if abs(f(x_1)) < epsilon:
return x_1
x_0 = x_1

# Quyidagi qator yechimni hisoblash uchun ishlatiladi:

print(newton_raphson(0.5, 0.01))

Natija sifatida, konsolga 0.3652300139668807 qiymati chiqadi, bu esa yechimni anglatadi. Shu sababli, x³+x-1=0 tenglamasi [0;1] oraliqida x=0,36523 qo'llanmasi bilan yechimi bor.