Resolution of the asin() problem

Let's take the example function



In [1]:

    
f(x) = asin(x)/x - pi/3









    Out[1]:





f (generic function with 1 method)

It has two symmetrical roots, -1/2 and 1/2.

The zero-finding algorithms will encounter the roots as long as we start with the intervals fitting in the domain of arcsin:



In [2]:

    
using KrawczykMethod









    



Syntax: krawczyk(function, Interval(lo, hi), precision [default is 64])



In [3]:

    
krawczyk(f, Interval(-0.9, 0.9))









    



Unique zero in Interval(-4.99999999999999468768e-01 with 64 bits of precision,-4.99999999999999462507e-01 with 64 bits of precision)
Maybe a zero in Interval(-3.12198595744902126454e-15 with 64 bits of precision,-2.32258541169002204847e-15 with 64 bits of precision)
Maybe a zero in Interval(-2.32258541169002204885e-15 with 64 bits of precision,-1.52318486593102283268e-15 with 64 bits of precision)
Maybe a zero in Interval(-1.52318486593102283287e-15 with 64 bits of precision,-7.23784320172023617034e-16 with 64 bits of precision)
Maybe a zero in Interval(-7.23784320172023617226e-16 with 64 bits of precision,7.56162255869755988157e-17 with 64 bits of precision)
Maybe a zero in Interval(7.56162255869755988037e-17 with 64 bits of precision,8.75016771345974814737e-16 with 64 bits of precision)
Maybe a zero in Interval(8.75016771345974814689e-16 with 64 bits of precision,1.67441731710497403067e-15 with 64 bits of precision)
Maybe a zero in Interval(1.67441731710497403048e-15 with 64 bits of precision,2.47381786286397324665e-15 with 64 bits of precision)
Maybe a zero in Interval(2.47381786286397324646e-15 with 64 bits of precision,3.27321840862297246234e-15 with 64 bits of precision)
Unique zero in Interval(4.99999999999999461721e-01 with 64 bits of precision,4.99999999999999468741e-01 with 64 bits of precision)






    Out[3]:





10-element Array{Any,1}:
 {Interval(-4.99999999999999468768e-01 with 64 bits of precision,-4.99999999999999462507e-01 with 64 bits of precision),:unique}  
 {Interval(-3.12198595744902126454e-15 with 64 bits of precision,-2.32258541169002204847e-15 with 64 bits of precision),:possible}
 {Interval(-2.32258541169002204885e-15 with 64 bits of precision,-1.52318486593102283268e-15 with 64 bits of precision),:possible}
 {Interval(-1.52318486593102283287e-15 with 64 bits of precision,-7.23784320172023617034e-16 with 64 bits of precision),:possible}
 {Interval(-7.23784320172023617226e-16 with 64 bits of precision,7.56162255869755988157e-17 with 64 bits of precision),:possible} 
 {Interval(7.56162255869755988037e-17 with 64 bits of precision,8.75016771345974814737e-16 with 64 bits of precision),:possible}  
 {Interval(8.75016771345974814689e-16 with 64 bits of precision,1.67441731710497403067e-15 with 64 bits of precision),:possible}  
 {Interval(1.67441731710497403048e-15 with 64 bits of precision,2.47381786286397324665e-15 with 64 bits of precision),:possible}  
 {Interval(2.47381786286397324646e-15 with 64 bits of precision,3.27321840862297246234e-15 with 64 bits of precision),:possible}  
 {Interval(4.99999999999999461721e-01 with 64 bits of precision,4.99999999999999468741e-01 with 64 bits of precision),:unique}



In [4]:

    
using NewtonMethod









    



Syntax: newton(function, Interval(lo, hi), precision [default is 64])



In [5]:

    
newton(f, Interval(-0.9, 0.9))









    



Function calls: 13






    Out[5]:





2-element Array{Interval,1}:
 Interval(-4.99999999999999321696e-01 with 64 bits of precision,-4.99999999999999319094e-01 with 64 bits of precision)
 Interval(4.99999999999999319094e-01 with 64 bits of precision,4.99999999999999321181e-01 with 64 bits of precision)

But if we set the interval which goes outside of the domain, we get errors.



In [6]:

    
krawczyk(f, Interval(-2, 2))









    



DomainError
while loading In[6], in expression starting on line 1

 in asin at mpfr.jl:573
 in asin at /home/kriukov/main/work/mexico/unam/interval-methods/modules/IntervalArithmetic.jl:507
 in f at In[1]:1
 in differentiate at /home/kriukov/main/work/mexico/unam/interval-methods/modules/AutoDiff.jl:101
 in krawczyk_internal at /home/kriukov/main/work/mexico/unam/interval-methods/modules/KrawczykMethod.jl:37
 in krawczyk at /home/kriukov/main/work/mexico/unam/interval-methods/modules/KrawczykMethod.jl:64
 in krawczyk at /home/kriukov/main/work/mexico/unam/interval-methods/modules/KrawczykMethod.jl:25



In [7]:

    
newton(f, Interval(-2, 2))









    



DomainError
while loading In[7], in expression starting on line 1

 in asin at mpfr.jl:573
 in asin at /home/kriukov/main/work/mexico/unam/interval-methods/modules/IntervalArithmetic.jl:507
 in f at In[1]:1
 in differentiate at /home/kriukov/main/work/mexico/unam/interval-methods/modules/AutoDiff.jl:101
 in N at /home/kriukov/main/work/mexico/unam/interval-methods/modules/NewtonMethod.jl:16
 in newton at /home/kriukov/main/work/mexico/unam/interval-methods/modules/NewtonMethod.jl:33
 in newton at /home/kriukov/main/work/mexico/unam/interval-methods/modules/NewtonMethod.jl:13



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