# [Numpy-discussion] NaN (Not a Number) occurs in calculation of complex number for Bessel functions

Happyman bahtiyor_zohidov@mail...
Fri Dec 21 09:14:45 CST 2012

```I think you advised about the code which is the same appearance.
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Problem is not here Sir....
I will give you exactly what I was talking about. I have ready codes already(It would be kind of you if you checked the following codes, may be):
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## Bessel function of the first kind
# mathematical form: Jn(x)--> n=arg1, x=arg2
# returns --> Jn(x) value in a complex form
------------------------------
def bes_1(arg1,arg2):
nu=arg1+0.5                          # nu=n+0.5: Jn(x) --> Jnu(x)
return sqrt(pi/(2*arg2))*np.round(jv(nu,arg2),5)      # jv(nu,arg2)--> from 'numpy.special' in PYTHON
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## Bessel function of the second kind
# mathematical form: Yn(x)--> n=arg1, x=arg2
# returns --> Yn(x) value in a complex form
---------------------------------
def bes_2 ( arg1, arg2 ):
nu = arg1 + 0.5                        # nu=n+0.5: Yn(x)--> Ynu(x)
return sqrt ( pi / ( 2 * arg2 ) ) * np.round ( yv ( nu , arg2 ) , 5)       # yv(nu,arg2)--> from 'numpy.special' in PYTHON
--------------------------------- -------------------------------------------------------------------------------------------------------
## Hankel function of the first kind
# mathematical form: Hn(x)= Jn(x)+Yn(x)j
# returns --> Hn(x) value in a complex form
---------------------------------
def hank_1 ( arg1, arg2 ):
return bes_1 ( arg1 , arg2 ) + bes_2 ( arg1 , arg2 ) * 1.0j            # Hn(x)= Jn(x)+Yn(x)j
-------------------------------------------------------------------------------------------------------------------------------------------
## Bessel function of the first kind derivative
# mathematical form: d(z*jn(z))=z*jn-1(z)-n*jn(z) where, z=x or m*x
---------------------------------
def bes_der1 ( arg1 , arg2 ):
return arg2 * bes_1 ( arg1 - 1, arg2 ) - arg1 * bes_1 ( arg1 , arg2 )
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## Hankel function of the first kind derivative
# mathematical form: d(z*hankeln(z))=z*hankeln-1(z)-n*hankeln(z) where, z=x or m*x
def hank_der1 ( arg1 , arg2 ):
return arg2 * hank_1 ( arg1 - 1, arg2 ) - arg1 * hank_1( arg1, arg2 )
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FOR MY CASE: m = 2513.2741228718346 + 201.0619298974676j
x = 502.6548245743669
def F(m,x):
nmax = x + 2.0 + 4.0 * x ** ( 1.0 / 3.0 )          #    nmax= gives 536.0 as expected value
nstop = np.round( nmax )
n = np.arange ( 0.0 ,nstop, dtype = float)
z = m * x
m2 = m * m

val1 = m2 * bes_1 ( en , z ) * bes_der1 ( en, x)
val2 = bes_1 ( en , x ) * bes_der1 ( en , z )
val3 = m2 * bes_1 ( en , z ) * hank_der1 ( en , x )
val4 = hank_1 ( en , x ) * bes_der1 ( en , z )
an = ( val1 - val2 ) / ( val3 - val4 )
val5 = bes_1 ( en , z ) * bes_der1 ( en, x )
val6 = bes_1 ( en , x ) * bes_der1 ( en, z )
val7 = bes_1 ( en , z ) * hank_der1 ( en, x )
val8 = hank_1 ( en , x ) * bes_der1 ( en , z )
bn = ( val5 - val6 ) / ( val7 - val8 )
return an, bn !!!!  PROBLEM IS RETURNING THE an, bn at a given value which I showed before F(m,x) function

WHAT IS WRONG WITH THIS??????

Пятница, 21 декабря 2012, 13:59  от Pauli Virtanen <pav@iki.fi>:
>Dag Sverre Seljebotn <d.s.seljebotn <at> astro.uio.no> writes:
>[clip]
>> Do you have an implemention of the Bessel functions that work as you
>> wish in C or Fortran? If so, that could be wrapped and called from Python.
>
>For spherical Bessel functions it's possible to also use the relation
>to Bessel functions, which have a better implementation (AMOS) in Scipy:
>
>import numpy as np
>from scipy.special import jv
>
>def sph_jn(n, z):
>    return jv(n + 0.5, z) * np.sqrt(np.pi / (2 * z))
>
>print sph_jn(536, 2513.2741228718346 + 201.0619298974676j)
># (2.5167666386507171e+81+3.3576357192536334e+81j)
>
>
>This should solve the problem.
>
>--
>Pauli Virtanen
>
>_______________________________________________
>NumPy-Discussion mailing list
>NumPy-Discussion@scipy.org
>http://mail.scipy.org/mailman/listinfo/numpy-discussion

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