def inv_f(f):
"""
Solve f^(-1)(i_f(r)) to get i_f and i_r using Newton's method
Uses a fixed number of iterations for easy taping and a different
formulation for f>0 and f<0 to ensure convergence in few
iterations. Solution has good accuracy for all values.
"""
# Obtain a good guess first using EKV's simple interpolation function
i1 = 4. * f_simple(f)
i2 = i1
for counter in xrange(4):
# (f > 0) => (i1 >= 3.) The 1e-15 term needed to prevent AD
# library from choking as without it we have log(0) later
sqi1p = np.sqrt(1. + i1 + 1e-15)
sqi11 = sqi1p - 1.
fodfp = (sqi1p - 2. + np.log(sqi11) - f) * 2. * sqi11
i1 = abs(i1 - fodfp)
# f < 0
sqi1 = np.sqrt(1. + i2)
fodfn = (sqi1 - 1. - np.exp(f - sqi1 + 2.)) * 2. * sqi1 \
/ (np.exp(f - sqi1 + 2.) + 1.)
i2 = i2 - fodfn
return ad.condassign(f, i1, i2)
def inv_f(f):
"""
Solve f^(-1)(i_f(r)) to get i_f and i_r using Newton's method
Uses a fixed number of iterations for easy taping and a different
formulation for f>0 and f<0 to ensure convergence in few
iterations. Solution has good accuracy for all values.
"""
# Obtain a good guess first using EKV's simple interpolation function
i1 = 4. * f_simple(f)
i2 = i1
for counter in xrange(4):
# (f > 0) => (i1 >= 3.) The 1e-15 term needed to prevent AD
# library from choking as without it we have log(0) later
sqi1p = np.sqrt(1. + i1 + 1e-15)
sqi11 = sqi1p - 1.
fodfp = (sqi1p - 2. + np.log(sqi11) - f) * 2. * sqi11
i1 = abs(i1 - fodfp)
# f < 0
sqi1 = np.sqrt(1. + i2)
fodfn = (sqi1 - 1. - np.exp(f - sqi1 + 2.)) * 2. * sqi1 \
/ (np.exp(f - sqi1 + 2.) + 1.)
i2 = i2 - fodfn
return ad.condassign(f, i1, i2)
def inv_f1(f):
"""
Approximately solve f^(-1)(i_f(r)) to get i_f and i_r using relaxation
"""
# Obtain a good guess first using EKV's simple interpolation function
i_f = 4. * f_simple(f)
# Use fixed number of iterations for easy taping
for i in xrange(30):
sqi1 = np.sqrt(1. + i_f + 1e-15)
# Uses different formulation for f>0 and f<0 for convergence
i_fnew = ad.condassign(f, (f + 2. - np.log(sqi1 - 1.))**2 - 1.,
(np.exp(f - sqi1 + 2.) + 1.)**2 - 1.)
i_f = .5 * i_f + .5 * i_fnew
return i_f
def inv_f1(f):
"""
Approximately solve f^(-1)(i_f(r)) to get i_f and i_r using relaxation
"""
# Obtain a good guess first using EKV's simple interpolation function
i_f = 4. * f_simple(f)
# Use fixed number of iterations for easy taping
for i in xrange(30):
sqi1 = np.sqrt(1. + i_f + 1e-15)
# Uses different formulation for f>0 and f<0 for convergence
i_fnew = ad.condassign(f,
(f + 2. - np.log(sqi1 - 1.))**2 - 1.,
(np.exp(f - sqi1 + 2.) + 1.)**2 - 1.)
i_f = .5 * i_f + .5 * i_fnew
return i_f
