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OK, I think I have a path to a solution. I have changed the differential equation so that it has zeros and this code prints the values on either side of where the function crosses zero. I can do something similar for the extrema. So I guess I am now asking if there is a function call that does this, or if there is a smarter way. Thanks, Wayne

var('t y z') P=desolve_system_rk4([z,-z^2*(-2+t/(1+t))-y],[y,z],ics=[0.1,10,2],ivar=t,end_points=100) Q=[ [t1,y1] for t1,y1,z1 in P] line(Q).show() print len(Q) [a1,b1]=Q[0] for i in range(1,len(Q)): [a2,b2]=Q[i] if(((b1<0.)and(b2>0.)) or ((b1>0.)and(b2<0.))): print a1,b1,a2,b2 a1=a2 b1=b2

OK, I think I have a path to a solution. I have changed the differential equation so that it has zeros and this code prints the values on either side of where the function crosses zero. I can do something similar for the extrema. So I guess I am now asking if there is a function call that does this, or if there is a smarter way. Thanks, Wayne

var('t y z') P=desolve_system_rk4([z,-z^2*(-2+t/(1+t))-y],[y,z],ics=[0.1,10,2],ivar=t,end_points=100) z')

P=desolve_system_rk4([z,-z^2*(-2+t/(1+t))-y],[y,z],ics=[0.1,10,2],ivar=t,end_points=100)

Q=[ [t1,y1] for t1,y1,z1 in P] line(Q).show() P]

line(Q).show()

print len(Q) [a1,b1]=Q[0] len(Q)

[a1,b1]=Q[0]

for i in range(1,len(Q)): range(1,len(Q)):

[a2,b2]=Q[i]
  if(((b1<0.)and(b2>0.)) or ((b1>0.)and(b2<0.))):
      print a1,b1,a2,b2
  a1=a2
    b1=b2
 b1=b2

OK, I think I have a path to a solution. I have changed the differential equation so that it has zeros and this code prints the values on either side of where the function crosses zero. I can do something similar for the extrema. So I guess I am now asking if there is a function call that does this, or if there is a smarter way. Thanks, Wayne

var('t y z')

P=desolve_system_rk4([z,-z^2*(-2+t/(1+t))-y],[y,z],ics=[0.1,10,2],ivar=t,end_points=100)

Q=[ [t1,y1] for t1,y1,z1 in P]

line(Q).show()

print len(Q)

[a1,b1]=Q[0]

for i in range(1,len(Q)):

[a2,b2]=Q[i]

if(((b1<0.)and(b2>0.)) or ((b1>0.)and(b2<0.))):

    print a1,b1,a2,b2

a1=a2

b1=b2

OK, I think I have a path to a solution. I have changed the differential equation so that it has zeros and this code prints the values on either side of where the function crosses zero. I can do something similar for the extrema. So I guess I am now asking if there is a function call that does this, or if there is a smarter way. Thanks, Wayne

var('t y z')
z')
 P=desolve_system_rk4([z,-z^2*(-2+t/(1+t))-y],[y,z],ics=[0.1,10,2],ivar=t,end_points=100)
 
P=desolve_system_rk4([z,-z^2*(-2+t/(1+t))-y],[y,z],ics=[0.1,10,2],ivar=t,end_points=100)
 
Q=[ [t1,y1] for t1,y1,z1 in P]P]
 line(Q).show()
 
line(Q).show()print len(Q)
 [a1,b1]=Q[0]
 
print len(Q)
 
[a1,b1]=Q[0]
 
for i in range(1,len(Q)):range(1,len(Q)):
  
 [a2,b2]=Q[i]

 if(((b1<0.)and(b2>0.)) or ((b1>0.)and(b2<0.))):

     print a1,b1,a2,b2

 a1=a2

 b1=b2