Berilgan tenglamaning taqribiy yechimini ε>0 aniqlikda urinmalar

1. 
   Labaratoriya ishi
   

1.2 Berilgan tenglamaning taqribiy yechimini ε>0 aniqlikda urinmalar
(Nyuton) va vatarlar usullarida hisoblang. Aniqlikka erishganlik sharti
sifatida |xn+1-xn|<ε tengsizlikdan foydalaning. Boshlang‘ich x0ϵ(a;b) olinadi.
Natijada ildiz taqribiy qiymati va zarur bo‘lgan qadamlar soni chiqarilsin.
a)

#Urinmalar usulida
#y=(x-1)**2-e**x/2
def f(a):
return (a-1)**2-e**a/2
e=2.71
def f2(a):
return 2*(a-1)-e**a/2
e=2.71
ep=0.0001
i=0
x=[0,1]
x[0]=1
x[1+i]=x[i]-f(x[i])/f2(x[i])
while abs(x[i+1]-x[i])>=ep:
i+=1
x.append(x[i])
x[1+i]=x[i]-f(x[i])/f2(x[i])
print('X=',x[i])
print(f(x[i]))

b)

#x**3-0.1*x**2+0.4*x+2
#Vatarlar usuli
def f(x):
return x**3-0.1*x**2+0.4*x+2
a=-2
b=-1
i=0
c=[-2,-1]
c[0]=-2
e=0.0001
while abs(c[i+1]-c[i])>=e:
c.append(c[i])
c[i+1]=c[i]-f(c[i])/(f(b)-f(c[i]))*(b-c[i])
i+=1
print('x=',c[i-1])
print(f(c[i-1]))