clock()
n=100
def f(x):
f = zeros([n])
for i in arange(0,n-1,1):
f[i] = (3 + 2*x[i])*x[i] — x[i-1] — 2*x[i+1] — 2
f [0] = (3 + 2*x[0] )*x[0] — 2*x[1] — 3
f[n-1] = (3 + 2*x[n-1] )*x[n-1] — x[n-2] — 4
return f
x0 =zeros([n])
sol = optimize.<\/p>\n

root(f,x0, method='krylov')
print('Solution:
', sol.x)
print('Krylov method iteration = ',sol.nit)
print('Optimize root time', round(time.<\/p>\n

clock()-ti,3), 'seconds') \u0422\u043e\u043b\u044c\u043a\u043e \u043e\u0434\u0438\u043d \u0438\u0437 \u043c\u0435\u0442\u043e\u0434\u043e\u0432, \u043f\u0440\u0438\u0432\u0435\u0434\u0435\u043d\u043d\u044b\u0445 \u0432 \u0434\u043e\u043a\u0443\u043c\u0435\u043d\u0442\u0430\u0446\u0438\u0438 [3] \u043f\u0440\u043e\u0448\u0451\u043b \u0442\u0435\u0441\u0442\u0438\u0440\u043e\u0432\u0430\u043d\u0438\u0435 \u043f\u043e \u0440\u0435\u0437\u0443\u043b\u044c\u0442\u0430\u0442\u0443 \u0440\u0435\u0448\u0435\u043d\u0438\u044f \u043c\u043e\u0434\u0435\u043b\u044c\u043d\u043e\u0439 \u0444\u0443\u043d\u043a\u0446\u0438\u0438, \u044d\u0442\u043e \u043c\u0435\u0442\u043e\u0434 'krylov'<\/b>. \u0420\u0435\u0448\u0435\u043d\u0438\u0435 \u0434\u043b\u044f n=100: Solution: [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. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.<\/p>\n

\n
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. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.] Krylov method iteration = 4219 Optimize root time 7.239 seconds: \u0420\u0435\u0448\u0435\u043d\u0438\u0435 \u0434\u043b\u044f n=200<\/b>Solution: [1.00000018 0.99999972 0.99999985 1.00000001 0.99999992 1.00000049 0.99999998 0.99999992 0.99999991 1.<\/p>\n<\/p><\/div>\n
00000001 1.00000013 1.00000002 0.9999997 0.99999987 1.00000005 0.99999978 1.0000002 1.00000012 1.00000023 1.00000017 0.99999979 1.00000012 1.00000026 0.99999987 1.00000014 0.99999979 0.99999988 1.00000046 1.00000064 1.00000007 1.00000049 1.00000005 1.00000032 1.00000031 1.00000028 0.99999992 1.0000003 1.0000001 0.99999971 1.00000023 1.00000039 1.0000003 1.<\/p>\n
00000013 0.9999999 0.99999993 0.99999996 1.00000008 1.00000016 1.00000034 1.00000004 0.99999993 0.99999987 0.99999969 0.99999985 0.99999981 1.00000051 1.0000004 1.00000035 0.9999998 1.00000065 1.00000061 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.<\/p>\n
0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.<\/p>\n
\n
0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.0000006 1.00000059 1.00000056 1.00000047 1.00000016 1.00000018 0.99999988 1.00000061 1.00000002 1.00000033 1.00000034 1.0000004 1.00000046 1.00000009 1.00000024 1.00000017 1.00000014 1.00000054 1.00000006 0.99999964 0.99999968 1.00000005 1.00000049 1.<\/p>\n<\/blockquote>\n
0000005 1.00000028 1.00000029 1.00000027 1.00000027 0.9999998 1.00000005 0.99999974 0.99999978 0.99999988 1.00000015 1.00000007 1.00000005 0.99999973 1.00000006 0.99999995 1.00000021 1.00000031 1.00000058 1.00000023 1.00000023 1.00000044 0.99999985 0.99999948 0.99999977 0.99999991 0.99999974 0.99999978 0.99999983 1.0000002 1.00000016 1.00000008 1.00000013 1.<\/p>\n
00000007 0.99999989 0.99999959 1.00000029 1.0000003 0.99999972 1.00000003 0.99999967 0.99999977 1.00000017 1.00000005 1.00000029 1.00000034 0.99999997 0.99999989 0.99999945 0.99999985 0.99999994 0.99999972 1.00000029 1.00000016] Krylov method iteration = 9178 Optimize root time 23.397 seconds \u0412\u044b\u0432\u043e\u0434:<\/b> \u0421 \u0443\u0432\u0435\u043b\u0438\u0447\u0435\u043d\u0438\u0435\u043c \u0447\u0438\u0441\u043b\u0430 \u0443\u0440\u0430\u0432\u043d\u0435\u043d\u0438\u0439 \u0432\u0434\u0432\u043e\u0435 \u0437\u0430\u043c\u0435\u0442\u043d\u043e \u043f\u043e\u044f\u0432\u043b\u0435\u043d\u0438\u0435 \u043e\u0448\u0438\u0431\u043e\u043a \u0432 \u0440\u0435\u0448\u0435\u043d\u0438\u0438. \u041f\u0440\u0438 \u0434\u0430\u043b\u044c\u043d\u0435\u0439\u0448\u0435\u043c \u0443\u0432\u0435\u043b\u0438\u0447\u0435\u043d\u0438\u0438 n \u0440\u0435\u0448\u0435\u043d\u0438\u0435 \u0441\u0442\u0430\u043d\u043e\u0432\u0438\u0442\u0441\u044f \u043d\u0435 \u043f\u0440\u0438\u0435\u043c\u043b\u0435\u043c\u044b\u043c, \u0447\u0442\u043e \u0432\u043e\u0437\u043c\u043e\u0436\u043d\u043e \u0438\u0437-\u0437\u0430 \u0430\u0432\u0442\u043e\u043c\u0430\u0442\u0438\u0447\u0435\u0441\u043a\u043e\u0439 \u0430\u0434\u0430\u043f\u0442\u0430\u0446\u0438\u0438 \u043a \u0448\u0430\u0433\u0443, \u044d\u0442\u0430 \u0436\u0435 \u043f\u0440\u0438\u0447\u0438\u043d\u0430 \u0440\u0435\u0437\u043a\u043e\u0433\u043e \u043f\u0430\u0434\u0435\u043d\u0438\u044f \u0431\u044b\u0441\u0442\u0440\u043e\u0434\u0435\u0439\u0441\u0442\u0432\u0438\u044f. \u041d\u043e \u044d\u0442\u043e \u0442\u043e\u043b\u044c\u043a\u043e \u043c\u043e\u0451 \u043f\u0440\u0435\u0434\u043f\u043e\u043b\u043e\u0436\u0435\u043d\u0438\u0435. <\/p>\n
\u041f\u0440\u043e\u0433\u0440\u0430\u043c\u043c\u0430 \u0434\u043b\u044f \u0442\u0435\u0441\u0442\u0438\u0440\u043e\u0432\u0430\u043d\u0438\u044f \u043d\u0430 \u043c\u043e\u0434\u0435\u043b\u044c\u043d\u043e\u0439 \u0444\u0443\u043d\u043a\u0446\u0438\u0438 c \u0440\u0435\u0437\u0443\u043b\u044c\u0442\u0430\u0442\u0430\u043c\u0438 \u0440\u0435\u0448\u0435\u043d\u0438\u044f \u0441\u0438\u0441\u0442\u0435\u043c\u044b \u0430\u043b\u0433\u0435\u0431\u0440\u0430\u0438\u0447\u0435\u0441\u043a\u0438\u0445 \u043d\u0435\u043b\u0438\u043d\u0435\u0439\u043d\u044b\u0445 \u0443\u0440\u0430\u0432\u043d\u0435\u043d\u0438\u0439 \u0441 \u043f\u043e\u043c\u043e\u0449\u044c\u044e \u043f\u0440\u043e\u0433\u0440\u0430\u043c\u043c\u044b \u043d\u0430\u043f\u0438\u0441\u0430\u043d\u043d\u043e\u0439 \u043d\u0430 Python 3 \u0441 \u0443\u0447\u0451\u0442\u043e\u043c \u0441\u043e\u043e\u0442\u043d\u043e\u0448\u0435\u043d\u0438\u0439 (1)-(8) \u0434\u043b\u044f \u043e\u0442\u044b\u0441\u043a\u0430\u043d\u0438\u044f \u043a\u043e\u0440\u043d\u0435\u0439 \u043f\u043e \u043c\u043e\u0434\u0438\u0444\u0438\u0446\u0438\u0440\u043e\u0432\u0430\u043d\u043d\u043e\u043c\u0443 \u043c\u0435\u0442\u043e\u0434\u0443 \u041d\u044c\u044e\u0442\u043e\u043d\u0430<\/h3>\n
\u041f\u0440\u043e\u0433\u0440\u0430\u043c\u043c\u0430 \u043e\u0442\u044b\u0441\u043a\u0430\u043d\u0438\u044f \u043a\u043e\u0440\u043d\u0435\u0439 \u043f\u043e \u043c\u043e\u0434\u0438\u0444\u0438\u0446\u0438\u0440\u043e\u0432\u0430\u043d\u043d\u043e\u043c\u0443 \u043c\u0435\u0442\u043e\u0434\u0443 \u041d\u044c\u044e\u0442\u043e\u043d\u0430<\/b>from numpy import*
import time
ti = time.clock()
def jacobian(f, x):
h = 1.<\/p>\n
0e-4
n = len(x)
Jac = zeros([n,n])
f0 = f(x)
for i in arange(0,n,1):
tt = x[i]
x[i] = tt + h
f1= f(x)
x[i] = tt
Jac [:,i] = (f1 — f0)\/h
return Jac, f0
def newton(f, x, tol=1.<\/p>\n
0e-9):
iterMax = 50
for i in range(iterMax):
Jac, fO = jacobian(f, x)
if sqrt(dot(fO, fO) \/ len(x)) < tol: return x, i dx = linalg.<\/p>\n
solve(Jac, fO)
x = x — dx
print («Too many iterations for the Newton method»)
n=100
def f(x):
f = zeros([n])
for i in arange(0,n-1,1):
f[i] = (3 + 2*x[i])*x[i] — x[i-1] — 2*x[i+1] — 2
f [0] = (3 + 2*x[0] )*x[0] — 2*x[1] — 3
f[n-1] = (3 + 2*x[n-1] )*x[n-1] — x[n-2] — 4
return f
x0 =zeros([n])
x, iter = newton(f, x0)
print ('Solution:
', x)
print ('Newton iteration = ', iter)
print('Newton method time', round(time.clock()-ti,3), 'seconds') \u0420\u0435\u0448\u0435\u043d\u0438\u0435 \u0434\u043b\u044f n=100: Solution: [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. 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. 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. 1. 1. 1. 1.] Newton iteration = 13 Newton method time 0.496 seconds \u0420\u0435\u0448\u0435\u043d\u0438\u0435 \u0434\u043b\u044f n=200: Solution: [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. 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. 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. 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. 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. 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. 1. 1. 1. 1. 1. 1. 1. 1.] Newton iteration = 14 Newton method time 1.869 seconds \u0427\u0442\u043e\u0431\u044b \u0443\u0431\u0435\u0434\u0438\u0442\u044c\u0441\u044f \u0432 \u0442\u043e\u043c, \u0447\u0442\u043e \u043f\u0440\u043e\u0433\u0440\u0430\u043c\u043c\u0430 \u0434\u0435\u0439\u0441\u0442\u0432\u0438\u0442\u0435\u043b\u044c\u043d\u043e \u0440\u0435\u0448\u0430\u0435\u0442 \u0441\u0438\u0441\u0442\u0435\u043c\u0443, \u043f\u0435\u0440\u0435\u043f\u0438\u0448\u0435\u043c \u043c\u043e\u0434\u0435\u043b\u044c\u043d\u0443\u044e \u0444\u0443\u043d\u043a\u0446\u0438\u044e \u0434\u043b\u044f \u0443\u0445\u043e\u0434\u0430 \u043e\u0442 \u043a\u043e\u0440\u043d\u044f \u0441\u043e \u0437\u043d\u0430\u0447\u0435\u043d\u0438\u0435\u043c 1 \u0432 \u0432\u0438\u0434\u0435: n=10
def f(x):
f = zeros([n])
for i in arange(0,n-1,1):
f[i] = (3 + 2*x[i])*x[i]*sin([i]) — x[i-1] — 2*x[i+1] — 2+e**-x[i]
f [0] = (3 + 2*x[0] )*x[0] — 2*x[1] — 3
f[n-1] = (3 + 2*x[n-1] )*x[n-1] — x[n-2] — 4
return f \u041f\u043e\u043b\u0443\u0447\u0438\u043c: Solution: [ 0.96472166 0.87777036 0.48175823 -0.26190496 -0.63693762 0.49232062 -1.31649896 0.6865098 0.89609091 0.98509235] Newton iteration = 16 Newton method time 0.046 seconds <\/p>\n
<\/p>\n