==========================================================================
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):
------------------------------------------------------------------------------------------------------------------------------
## 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
-------------------------------------------------------------------------------------------------------------------------------- 

## 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 )
--------------------------------------------------------------------------------------------------------------------------------------------- 

## 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 )
--------------------------------------------------------------------------------------------------------------------------------------------- 
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

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

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