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Finding extrema and zeros in lists

asked 2018-01-11 06:27:25 -0600

Wayne gravatar image

updated 2018-01-12 07:51:30 -0600

tmonteil gravatar image

I create a list Q (below) of pairs of numbers [t1,z1], I want the pairs (maybe just a print of them?) at the local extrema (of z1) and where z1 crosses zero. Is there a simple way to do this?

var('t y z')
P=desolve_system_rk4([z,0.005*exp(-y/8000)*z^2-10],[y,z],ics=[0,32000,0],ivar=t,end_points=400)
Q=[ [t1,z1] for t1,y1,z1 in P]

Thanks, Wayne

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answered 2018-01-12 17:15:13 -0600

dan_fulea gravatar image

updated 2018-01-13 06:17:58 -0600

The code:

var('t y z')
P = desolve_system_rk4( [ z, 0.005 * exp(-y/8000) * z^2 - 10]
                        , [y,z]
                        , ics = [ 0, 32000, 0 ]
                        , ivar=t
                        , end_points=400 )
Q = [ (t1,z1) for t1,y1,z1 in P ]

for k in xrange( 1, len(Q)-1 ):
    k1, k2, k3 = k-1, k, k+1
    (t1,z1), (t2,z2), (t3,z3) = Q[k1], Q[k2], Q[k3]

    if z1 <= z2 and z2 >= z3:
        print "( %s, %s ) is a local MAX" % ( t2, z2 )
    if z1 >= z2 and z2 <= z3:
        print "( %s, %s ) is a local MIN" % ( t2, z2 )
    if z1 <= 0 and z2 >=0:
        print "ZERO BETWEEN ( %s, %s ) and ( %s, %s )" % ( t1, z1, t2, z2 )
    if z1 >= 0 and z2 <=0:
        print "ZERO BETWEEN ( %s, %s ) and ( %s, %s )" % ( t1, z1, t2, z2 )    # EDITED HERE, sorry

finds...

ZERO BETWEEN ( 0, 0 ) and ( 0.1, -0.9999969473920776 )
( 37.8, -231.6456047650853 ) is a local MIN

but misses of course a possible zero at the end of the interval. Bur looking at

sage: Q[-2:]
[(399.9, -36.4642141232692), (400.0, -36.4558355793491)]

we see there is no zero at the other end of the interval.

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Thanks very much!!

Wayne gravatar imageWayne ( 2018-01-15 15:41:22 -0600 )edit
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answered 2018-01-12 20:37:35 -0600

calc314 gravatar image

Another option is to use splines to approximate the curve. Here is some code inspired by https://stackoverflow.com/questions/3...

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=[ [float(t1),float(z1)] for t1,y1,z1 in P]
x_points=([p[0] for p in Q])
y_points=([p[1] for p in Q])
tmp = interpolate.splrep(x_points, y_points)
f=lambda x: interpolate.splev(x, tmp)

You could then find a zero using:

find_root(f,12,20)

Also, you can compute the derivative using:

g=lambda x: interpolate.splev(x, tmp,der=1)

Then, find potential extrema using find_root.

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Both of your solutions are very helpful! Thanks much!

Wayne gravatar imageWayne ( 2018-01-15 15:44:03 -0600 )edit
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answered 2018-01-11 12:07:44 -0600

Wayne gravatar image

updated 2018-01-12 20:12:12 -0600

calc314 gravatar image

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
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Asked: 2018-01-11 06:27:25 -0600

Seen: 67 times

Last updated: Jan 13