Minuit gives Nan?

ROOT
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1

Trying to fit some data to a function that I call Ffitr but when running the code below I only get “nan” to all values. The problem lies within the chiaf if you remove if from the return sum the code works. I suspect that it as something to do with the log part of Ffitr but then I think that python should display “invalid value encountered in log”. What else could be the reason for minuit settting all parameters to nan?

P.s. sorry that my code is a bit hard to read.

import numpy as np
from iminuit import Minuit

#Parameters to fit to:
Zsrgi=np.array([1.47, 1.5, 1.54, 1.58])

Za=np.array([ 0.9468,  0.9632,  0.9707,  0.9756])

af=np.array([ 0.0575465 ,  0.05547301,  0.04983955,  0.05096624,  0.04737787,
        0.04716958,  0.04579672,  0.0454464 ,  0.0437895 ,  0.04202845,
        0.04717754,  0.04625286,  0.04245786,  0.04105158,  0.04052182,
        0.04077226,  0.03857616,  0.03776707,  0.03453072,  0.0414683 ,
        0.0395172 ,  0.03501315,  0.03553733,  0.03483842,  0.03207193,
        0.03173218,  0.03206222,  0.03104359,  0.03046799,  0.02715095,
        0.0273168 ,  0.02622413])

amudr=np.array([ 0.01737619,  0.01377279,  0.00808912,  0.00970136,  0.00580544,
        0.00496463,  0.00397211,  0.00383197,  0.0019517 ,  0.00102585,
        0.01227267,  0.01208733,  0.00673   ,  0.00537933,  0.00521267,
        0.00450733,  0.00328733,  0.002288  ,  0.000658  ,  0.00950714,
        0.00923442,  0.00537273,  0.00533831,  0.00534351,  0.0022987 ,
        0.00224545,  0.00126688,  0.00588165,  0.00582215,  0.00267405,
        0.00266899,  0.0013943 ])

a=np.array([ 0.0904,  0.0755,  0.0647,  0.0552])

#coefficients
afij=np.array([  1.09440000e+00,  -1.00000000e+00,  -4.73000000e-02,
        -1.90580000e+00,  -1.25000000e+00,  -2.44535000e+02,
        -1.54989000e+01,  -9.39460000e+00,  -3.45830000e+00])


def Ffitr(X,s,pp,k,B=2.61,Fc=92.8,mu=770): #fit function
    temp1 = (2*B*X)/(4*pi*Fc)**2
    temp2 = temp1*(afij[0]+afij[1]*np.log((2*B*X)/mu**2))
    temp3 = temp1**2*(afij[2]+afij[3]*np.log((2*B*X)/mu**2)+\
                   afij[4]*(np.log((2*B*X)/mu**2))**2)
    temp4 = temp1**3*(afij[5]+afij[6]*np.log((2*B*X)/mu**2)+\
                      afij[7]*(np.log((2*B*X)/mu**2))**2+\
                      afij[8]*(np.log((2*B*X)/mu**2))**3)
    return Fc*pp*(1+k*s)*(1+temp2+temp3+temp4)


def chiwork(a0,a1,a2,a3,b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13,b14,b15,b16,b17,b18,b19,b20,b21,b22,b23,b24,b25,b26,b27,b28,b29,b30,b31,\
            z0,z1,z2,z3,za0,za1,za2,za3,\
            am0,am1,am2,am3,am4,am5,am6,am7,am8,am9,am10,am11,am12,am13,am14,am15,am16,am17,am18,am19,am20,am21,am22,am23,am24,am25,am26,am27,am28,am29,am30,am31,\
            k,y1): #chisquare

    ana=np.array([a0,a1,a2,a3])
    zs=np.array([z0,z1,z2,z3])
    za=np.array([za0,za1,za2,za3])
    am=np.array([am0,am1,am2,am3,am4,am5,am6,am7,am8,am9,am10,am11,am12,am13,am14,am15,am16,am17,am18,am19,am20,am21,am22,am23,am24,am25,am26,am27,am28,am29,am30,am31])
    bss=np.array([b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13,b14,b15,b16,b17,b18,b19,b20,b21,b22,b23,b24,b25,b26,b27,b28,b29,b30,b31])
    betalat=[0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,3]
    chia=sum((ana-a)**2)+sum((bss*a[betalat]/hc-amssr)**2)
    chiz=sum((zs-1/Zsrgi)**2)+sum((za-Za)**2)
    chiamd=sum((amudr-(1+y1*ana[betalat]**2)*ana[betalat]*am*zs[betalat]/hc)**2)
    chiaf=sum((af-ana[betalat]/hc*Ffitr(am,bss**2-mssph**2,ana[betalat]/za[betalat],k))**2)
    #chiaB=sum(((ampir**2/(2*amudr))-ampm)**2)
    return chia+chiz+chiamd+chiaf
#minuit part
mc=Minuit(chiwork)

fminc,paramc=mc.migrad()
2

I suggest to print the values of all the intermediate steps. If you pipe nan or infinity into a calculation by Minuit, it cannot do more than return nan.

3

How would I print all the intermediate steps? If I add print(am) in the function it just prints all the am’s which are all nan.

4

In Ffitr:

print(temp1, temp2, temp3, temp4, Fc*pp*(1+k*s)*(1+temp2+temp3+temp4))?

In chi2work:

print(chia, chiz, .....)

And indeed, the code is a bit hard to read.

Further, I cannot believe that the function you try to optimise has so many parameters. Be aware that Minuit will try to measure everything that you pass into chiwork. Data and constants should come from outside of the function passed to Minuit. This is an example from the IMinuit tutorials:

def least_squares(a, b):
    yvar = 0.01
    return sum((data_y - line(data_x, a, b)) ** 2 / yvar)

Minuit will only measure a and b here.

5

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