In [1]:

    

from __future__ import division

import numpy as np
import scipy.linalg as lin
import pylab as pl

from locore.identifiability import crit_l1_analysis
from locore.operators import finite_diff_1d


    

In [2]:

    

n = 100


    

In [3]:

    

A = np.eye(n)


    

In [4]:

    

D = finite_diff_1d(n)


    

In [5]:

    

x = np.ones((n,1))
x[n//2:n//2+10] = 2


    

In [6]:

    

crit_l1_analysis(D, A, x)


    

    
Out[6]:





(0.97999999999999987,
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