I´ve tried to calculate symbolic limit in sympy but got an error message.
import sympy as smp
from sympy import *
n,x=smp.symbols('n x')
limit(simplify(integrate(exp(-x)*exp(-smp.I*n*x),x).args[0][0]),x,smp.oo)
I`ve got following error message:
---------------------------------------------------------------------------
NotImplementedError Traceback (most recent call last)
Cell In[26], line 1
----> 1 limit(exp(-n*x),x,smp.oo)
File ~\anaconda3\Lib\site-packages\sympy\series\limits.py:64, in limit(e, z, z0, dir)
13 def limit(e, z, z0, dir="+"):
14 """Computes the limit of ``e(z)`` at the point ``z0``.
15
16 Parameters
(...)
61 limit_seq : returns the limit of a sequence.
62 """
---> 64 return Limit(e, z, z0, dir).doit(deep=False)
File ~\anaconda3\Lib\site-packages\sympy\series\limits.py:375, in Limit.doit(self, **hints)
372 l = None
374 try:
--> 375 r = gruntz(e, z, z0, dir)
376 if r is S.NaN or l is S.NaN:
377 raise PoleError()
File ~\anaconda3\Lib\site-packages\sympy\series\gruntz.py:732, in gruntz(e, z, z0, dir)
729 else:
730 raise NotImplementedError("dir must be '+' or '-'")
--> 732 r = limitinf(e0, z)
734 # This is a bit of a heuristic for nice results... we always rewrite
735 # tractable functions in terms of familiar intractable ones.
736 # It might be nicer to rewrite the exactly to what they were initially,
737 # but that would take some work to implement.
738 return r.rewrite('intractable', deep=True)
File ~\anaconda3\Lib\site-packages\sympy\core\cache.py:72, in __cacheit.<locals>.func_wrapper.<locals>.wrapper(*args, **kwargs)
69 @wraps(func)
70 def wrapper(*args, **kwargs):
71 try:
---> 72 retval = cfunc(*args, **kwargs)
73 except TypeError as e:
74 if not e.args or not e.args[0].startswith('unhashable type:'):
File ~\anaconda3\Lib\site-packages\sympy\series\gruntz.py:452, in limitinf(e, x)
450 c0, e0 = mrv_leadterm(e.min, x)
451 else:
--> 452 c0, e0 = mrv_leadterm(e, x)
453 sig = sign(e0, x)
454 if sig == 1:
File ~\anaconda3\Lib\site-packages\sympy\core\cache.py:72, in __cacheit.<locals>.func_wrapper.<locals>.wrapper(*args, **kwargs)
69 @wraps(func)
70 def wrapper(*args, **kwargs):
71 try:
---> 72 retval = cfunc(*args, **kwargs)
73 except TypeError as e:
74 if not e.args or not e.args[0].startswith('unhashable type:'):
File ~\anaconda3\Lib\site-packages\sympy\series\gruntz.py:554, in mrv_leadterm(e, x)
546 #
547 # The positive dummy, w, is used here so log(w*2) etc. will expand;
548 # a unique dummy is needed in this algorithm
(...)
551 # improved, or just find limits of Re and Im components separately.
552 #
553 w = Dummy("w", positive=True)
--> 554 f, logw = rewrite(exps, Omega, x, w)
555 try:
556 lt = f.leadterm(w, logx=logw)
File ~\anaconda3\Lib\site-packages\sympy\series\gruntz.py:647, in rewrite(e, Omega, x, wsym)
645 sig = sign(g.exp, x)
646 if sig != 1 and sig != -1 and not sig.has(AccumBounds):
--> 647 raise NotImplementedError('Result depends on the sign of %s' % sig)
648 if sig == 1:
649 wsym = 1/wsym # if g goes to oo, substitute 1/w
NotImplementedError: Result depends on the sign of -sign(n)
.........................................................................................................................
In mathematics result should be 0
Can anybody explain me output and fix that?
Setting n and Im to be a positive number when creating the symbol will resolve this issue.
import sympy as smp
from sympy import *
x=symbols('x')
#Im is symbol for imaginary unit to avoid overwriting smp.I
n,Im=symbols('n Im',positive=True)
limit(simplify(integrate(exp(-x)*exp(-Im*n*x),x)),x,smp.oo) # 0