s
maxstp
20
iprint
2
q
#
#  Choose problem 1
#
a
1
2
x
2,2
i
p
x
3,3
i
p
x
100,1
i
p
x
1,100
i
p
x
0.4,0.68267
i
p
x
0.4,0.68268
i
p
#
# Switch to problem 2.
#
a
2
i
p
x
1000,1000,1000,1000
i
p
x
-1000,0,0,0
i
p
#
# Switch to Newton method with central finite difference jacobian.
#
m
f
c
#
# Choose problem 1
#
a
1
2
x
2,2
i
p
x
3,3
i
p
x
100,1
i
p
x
1,100
i
p
x
0.4,0.68267
i
p
x
0.4,0.68268
i
p
#
# Switch to problem 2..
#
a
2
i
p
x
1000,1000,1000,1000
i
p
x
-1000,0,0,0
i
p
#
# Now compare the performance of the finite difference jacobians for
# a tiny h, and a large h=0.01.  
#
# The first run uses the default h, which should be roughly 1.0e-16
# for double precision
#
s
abserr
0
q

x
0.01, 0.01, 0.01, 0.01
i
p
#
# Now we reset h to 0.01 and try again.  Since the difference
# between the iterates and the true solution itself is of
# order 0.01, we expect poor performance.
#
s
difjac
0.01
q
x
0.01, 0.01, 0.01, 0.01
i
p
#
# Switch to Broyden method, with initial matrix the identity
#
m
b
i
#
# Choose problem 1
#
a
1
2
x
2,2
i
p
x
3,3
i
p
x
100,1
i
p
x
1,100
i
p
x
0.4,0.68267
i
p
x
0.4,0.68268
i
p
#
# Switch to problem 2...
#
a
2
i
p
x
1000,1000, 1000, 1000
i
p
x
-1000,0, 0, 0
i
p
#
# Switch to Broyden method, with initial matrix the jacobian
#
m
b
j
#
# Choose problem 1
#
a
1
2
x
2,2
i
p
x
3,3
i
p
x
100,1
i
p
x
1,100
i
p
x
0.4,0.68267
i
p
x
0.4,0.68268
i
p
#
# Switch to problem 2....
#
a
2
i
p
x
1000,1000, 1000, 1000
i
p
x
-1000,0, 0, 0
i
p
#
# That's all!
#
q