def _eval_simplify(self, ratio, measure):
from sympy.simplify.combsimp import combsimp
# combinatorial function with non-integer arguments is
# automatically passed to gammasimp
expr = combsimp(self)
if measure(expr) <= ratio*measure(self):
return expr
return self
def _eval_simplify(self, ratio, measure, rational, inverse):
from sympy.simplify.combsimp import combsimp
# combinatorial function with non-integer arguments is
# automatically passed to gammasimp
expr = combsimp(self)
if measure(expr) <= ratio * measure(self):
return expr
return self
def _eval_simplify(self, **kwargs):
from sympy.simplify.combsimp import combsimp
# combinatorial function with non-integer arguments is
# automatically passed to gammasimp
expr = combsimp(self)
measure = kwargs['measure']
if measure(expr) <= kwargs['ratio'] * measure(self):
return expr
return self
def test_the_product(m, n):
# g
g = i**3 + 2*i**2 - 3*i
# f = Delta g
f = simplify(g.subs(i, i+1) / g)
# The product
a = m
b = n - 1
P = Product(f, (i, a, b)).doit()
# Test if Product_{m <= i < n} f(i) = g(n) / g(m)
assert combsimp(P / (g.subs(i, n) / g.subs(i, m))) == 1
def _eval_sum_hyper(f, i, a):
""" Returns (res, cond). Sums from a to oo. """
from sympy.functions import hyper
from sympy.simplify import hyperexpand, hypersimp, fraction, simplify
from sympy.polys.polytools import Poly, factor
from sympy.core.numbers import Float
if a != 0:
return _eval_sum_hyper(f.subs(i, i + a), i, 0)
if f.subs(i, 0) == 0:
if simplify(f.subs(i, Dummy("i", integer=True, positive=True))) == 0:
return S.Zero, True
return _eval_sum_hyper(f.subs(i, i + 1), i, 0)
hs = hypersimp(f, i)
if hs is None:
return None
if isinstance(hs, Float):
from sympy.simplify.simplify import nsimplify
hs = nsimplify(hs)
numer, denom = fraction(factor(hs))
top, topl = numer.as_coeff_mul(i)
bot, botl = denom.as_coeff_mul(i)
ab = [top, bot]
factors = [topl, botl]
params = [[], []]
for k in range(2):
for fac in factors[k]:
mul = 1
if fac.is_Pow:
mul = fac.exp
fac = fac.base
if not mul.is_Integer:
return None
p = Poly(fac, i)
if p.degree() != 1:
return None
m, n = p.all_coeffs()
ab[k] *= m ** mul
params[k] += [n / m] * mul
# Add "1" to numerator parameters, to account for implicit n! in
# hypergeometric series.
ap = params[0] + [1]
bq = params[1]
x = ab[0] / ab[1]
h = hyper(ap, bq, x)
f = combsimp(f)
return f.subs(i, 0) * hyperexpand(h), h.convergence_statement
def _eval_sum_hyper(f, i, a):
""" Returns (res, cond). Sums from a to oo. """
from sympy.functions import hyper
from sympy.simplify import hyperexpand, hypersimp, fraction, simplify
from sympy.polys.polytools import Poly, factor
from sympy.core.numbers import Float
if a != 0:
return _eval_sum_hyper(f.subs(i, i + a), i, 0)
if f.subs(i, 0) == 0:
if simplify(f.subs(i, Dummy('i', integer=True, positive=True))) == 0:
return S(0), True
return _eval_sum_hyper(f.subs(i, i + 1), i, 0)
hs = hypersimp(f, i)
if hs is None:
return None
if isinstance(hs, Float):
from sympy.simplify.simplify import nsimplify
hs = nsimplify(hs)
numer, denom = fraction(factor(hs))
top, topl = numer.as_coeff_mul(i)
bot, botl = denom.as_coeff_mul(i)
ab = [top, bot]
factors = [topl, botl]
params = [[], []]
for k in range(2):
for fac in factors[k]:
mul = 1
if fac.is_Pow:
mul = fac.exp
fac = fac.base
if not mul.is_Integer:
return None
p = Poly(fac, i)
if p.degree() != 1:
return None
m, n = p.all_coeffs()
ab[k] *= m**mul
params[k] += [n/m]*mul
# Add "1" to numerator parameters, to account for implicit n! in
# hypergeometric series.
ap = params[0] + [1]
bq = params[1]
x = ab[0]/ab[1]
h = hyper(ap, bq, x)
f = combsimp(f)
return f.subs(i, 0)*hyperexpand(h), h.convergence_statement
def test_the_product(l, n, m):
# Productmand
s = i**3
# First product
a = l
b = n - 1
S1 = Product(s, (i, a, b)).doit()
# Second product
a = l
b = m - 1
S2 = Product(s, (i, a, b)).doit()
# Third product
a = m
b = n - 1
S3 = Product(s, (i, a, b)).doit()
# Test if S1 = S2 * S3 as required
assert combsimp(S1 / (S2 * S3)) == 1
def is_convergent(self):
r"""Checks for the convergence of a Sum.
We divide the study of convergence of infinite sums and products in
two parts.
First Part:
One part is the question whether all the terms are well defined, i.e.,
they are finite in a sum and also non-zero in a product. Zero
is the analogy of (minus) infinity in products as
:math:`e^{-\infty} = 0`.
Second Part:
The second part is the question of convergence after infinities,
and zeros in products, have been omitted assuming that their number
is finite. This means that we only consider the tail of the sum or
product, starting from some point after which all terms are well
defined.
For example, in a sum of the form:
.. math::
\sum_{1 \leq i < \infty} \frac{1}{n^2 + an + b}
where a and b are numbers. The routine will return true, even if there
are infinities in the term sequence (at most two). An analogous
product would be:
.. math::
\prod_{1 \leq i < \infty} e^{\frac{1}{n^2 + an + b}}
This is how convergence is interpreted. It is concerned with what
happens at the limit. Finding the bad terms is another independent
matter.
Note: It is responsibility of user to see that the sum or product
is well defined.
There are various tests employed to check the convergence like
divergence test, root test, integral test, alternating series test,
comparison tests, Dirichlet tests. It returns true if Sum is convergent
and false if divergent and NotImplementedError if it can not be checked.
References
==========
.. [1] https://en.wikipedia.org/wiki/Convergence_tests
Examples
========
>>> from sympy import factorial, S, Sum, Symbol, oo
>>> n = Symbol('n', integer=True)
>>> Sum(n/(n - 1), (n, 4, 7)).is_convergent()
True
>>> Sum(n/(2*n + 1), (n, 1, oo)).is_convergent()
False
>>> Sum(factorial(n)/5**n, (n, 1, oo)).is_convergent()
False
>>> Sum(1/n**(S(6)/5), (n, 1, oo)).is_convergent()
True
See Also
========
Sum.is_absolutely_convergent()
Product.is_convergent()
"""
from sympy import Interval, Integral, log, symbols, simplify
p, q, r = symbols('p q r', cls=Wild)
sym = self.limits[0][0]
lower_limit = self.limits[0][1]
upper_limit = self.limits[0][2]
sequence_term = self.function
if len(sequence_term.free_symbols) > 1:
raise NotImplementedError("convergence checking for more than one symbol "
"containing series is not handled")
if lower_limit.is_finite and upper_limit.is_finite:
return S.true
# transform sym -> -sym and swap the upper_limit = S.Infinity
# and lower_limit = - upper_limit
if lower_limit is S.NegativeInfinity:
if upper_limit is S.Infinity:
return Sum(sequence_term, (sym, 0, S.Infinity)).is_convergent() and \
Sum(sequence_term, (sym, S.NegativeInfinity, 0)).is_convergent()
sequence_term = simplify(sequence_term.xreplace({sym: -sym}))
lower_limit = -upper_limit
upper_limit = S.Infinity
sym_ = Dummy(sym.name, integer=True, positive=True)
sequence_term = sequence_term.xreplace({sym: sym_})
sym = sym_
interval = Interval(lower_limit, upper_limit)
# Piecewise function handle
if sequence_term.is_Piecewise:
for func, cond in sequence_term.args:
# see if it represents something going to oo
if cond == True or cond.as_set().sup is S.Infinity:
s = Sum(func, (sym, lower_limit, upper_limit))
return s.is_convergent()
return S.true
### -------- Divergence test ----------- ###
try:
lim_val = limit_seq(sequence_term, sym)
if lim_val is not None and lim_val.is_zero is False:
return S.false
except NotImplementedError:
pass
try:
lim_val_abs = limit_seq(abs(sequence_term), sym)
if lim_val_abs is not None and lim_val_abs.is_zero is False:
return S.false
except NotImplementedError:
pass
order = O(sequence_term, (sym, S.Infinity))
### --------- p-series test (1/n**p) ---------- ###
p1_series_test = order.expr.match(sym**p)
if p1_series_test is not None:
if p1_series_test[p] < -1:
return S.true
if p1_series_test[p] >= -1:
return S.false
p2_series_test = order.expr.match((1/sym)**p)
if p2_series_test is not None:
if p2_series_test[p] > 1:
return S.true
if p2_series_test[p] <= 1:
return S.false
### ------------- comparison test ------------- ###
# 1/(n**p*log(n)**q*log(log(n))**r) comparison
n_log_test = order.expr.match(1/(sym**p*log(sym)**q*log(log(sym))**r))
if n_log_test is not None:
if (n_log_test[p] > 1 or
(n_log_test[p] == 1 and n_log_test[q] > 1) or
(n_log_test[p] == n_log_test[q] == 1 and n_log_test[r] > 1)):
return S.true
return S.false
### ------------- Limit comparison test -----------###
# (1/n) comparison
try:
lim_comp = limit_seq(sym*sequence_term, sym)
if lim_comp is not None and lim_comp.is_number and lim_comp > 0:
return S.false
except NotImplementedError:
pass
### ----------- ratio test ---------------- ###
next_sequence_term = sequence_term.xreplace({sym: sym + 1})
ratio = combsimp(powsimp(next_sequence_term/sequence_term))
try:
lim_ratio = limit_seq(ratio, sym)
if lim_ratio is not None and lim_ratio.is_number:
if abs(lim_ratio) > 1:
return S.false
if abs(lim_ratio) < 1:
return S.true
except NotImplementedError:
pass
### ----------- root test ---------------- ###
# lim = Limit(abs(sequence_term)**(1/sym), sym, S.Infinity)
try:
lim_evaluated = limit_seq(abs(sequence_term)**(1/sym), sym)
if lim_evaluated is not None and lim_evaluated.is_number:
if lim_evaluated < 1:
return S.true
if lim_evaluated > 1:
return S.false
except NotImplementedError:
pass
### ------------- alternating series test ----------- ###
dict_val = sequence_term.match((-1)**(sym + p)*q)
if not dict_val[p].has(sym) and is_decreasing(dict_val[q], interval):
return S.true
### ------------- integral test -------------- ###
check_interval = None
maxima = solveset(sequence_term.diff(sym), sym, interval)
if not maxima:
check_interval = interval
elif isinstance(maxima, FiniteSet) and maxima.sup.is_number:
check_interval = Interval(maxima.sup, interval.sup)
if (check_interval is not None and
(is_decreasing(sequence_term, check_interval) or
is_decreasing(-sequence_term, check_interval))):
integral_val = Integral(
sequence_term, (sym, lower_limit, upper_limit))
try:
integral_val_evaluated = integral_val.doit()
if integral_val_evaluated.is_number:
return S(integral_val_evaluated.is_finite)
except NotImplementedError:
pass
### ----- Dirichlet and bounded times convergent tests ----- ###
# TODO
#
# Dirichlet_test
# https://en.wikipedia.org/wiki/Dirichlet%27s_test
#
# Bounded times convergent test
# It is based on comparison theorems for series.
# In particular, if the general term of a series can
# be written as a product of two terms a_n and b_n
# and if a_n is bounded and if Sum(b_n) is absolutely
# convergent, then the original series Sum(a_n * b_n)
# is absolutely convergent and so convergent.
#
# The following code can grows like 2**n where n is the
# number of args in order.expr
# Possibly combined with the potentially slow checks
# inside the loop, could make this test extremely slow
# for larger summation expressions.
if order.expr.is_Mul:
args = order.expr.args
argset = set(args)
### -------------- Dirichlet tests -------------- ###
m = Dummy('m', integer=True)
def _dirichlet_test(g_n):
try:
ing_val = limit_seq(Sum(g_n, (sym, interval.inf, m)).doit(), m)
if ing_val is not None and ing_val.is_finite:
return S.true
except NotImplementedError:
pass
### -------- bounded times convergent test ---------###
def _bounded_convergent_test(g1_n, g2_n):
try:
lim_val = limit_seq(g1_n, sym)
if lim_val is not None and (lim_val.is_finite or (
isinstance(lim_val, AccumulationBounds)
and (lim_val.max - lim_val.min).is_finite)):
if Sum(g2_n, (sym, lower_limit, upper_limit)).is_absolutely_convergent():
return S.true
except NotImplementedError:
pass
for n in range(1, len(argset)):
for a_tuple in itertools.combinations(args, n):
b_set = argset - set(a_tuple)
a_n = Mul(*a_tuple)
b_n = Mul(*b_set)
if is_decreasing(a_n, interval):
dirich = _dirichlet_test(b_n)
if dirich is not None:
return dirich
bc_test = _bounded_convergent_test(a_n, b_n)
if bc_test is not None:
return bc_test
_sym = self.limits[0][0]
sequence_term = sequence_term.xreplace({sym: _sym})
raise NotImplementedError("The algorithm to find the Sum convergence of %s "
"is not yet implemented" % (sequence_term))
def is_convergent(self):
r"""Checks for the convergence of a Sum.
We divide the study of convergence of infinite sums and products in
two parts.
First Part:
One part is the question whether all the terms are well defined, i.e.,
they are finite in a sum and also non-zero in a product. Zero
is the analogy of (minus) infinity in products as
:math:`e^{-\infty} = 0`.
Second Part:
The second part is the question of convergence after infinities,
and zeros in products, have been omitted assuming that their number
is finite. This means that we only consider the tail of the sum or
product, starting from some point after which all terms are well
defined.
For example, in a sum of the form:
.. math::
\sum_{1 \leq i < \infty} \frac{1}{n^2 + an + b}
where a and b are numbers. The routine will return true, even if there
are infinities in the term sequence (at most two). An analogous
product would be:
.. math::
\prod_{1 \leq i < \infty} e^{\frac{1}{n^2 + an + b}}
This is how convergence is interpreted. It is concerned with what
happens at the limit. Finding the bad terms is another independent
matter.
Note: It is responsibility of user to see that the sum or product
is well defined.
There are various tests employed to check the convergence like
divergence test, root test, integral test, alternating series test,
comparison tests, Dirichlet tests. It returns true if Sum is convergent
and false if divergent and NotImplementedError if it can not be checked.
References
==========
.. [1] https://en.wikipedia.org/wiki/Convergence_tests
Examples
========
>>> from sympy import factorial, S, Sum, Symbol, oo
>>> n = Symbol('n', integer=True)
>>> Sum(n/(n - 1), (n, 4, 7)).is_convergent()
True
>>> Sum(n/(2*n + 1), (n, 1, oo)).is_convergent()
False
>>> Sum(factorial(n)/5**n, (n, 1, oo)).is_convergent()
False
>>> Sum(1/n**(S(6)/5), (n, 1, oo)).is_convergent()
True
See Also
========
Sum.is_absolutely_convergent()
Product.is_convergent()
"""
from sympy import Interval, Integral, Limit, log, symbols, Ge, Gt, simplify
p, q = symbols('p q', cls=Wild)
sym = self.limits[0][0]
lower_limit = self.limits[0][1]
upper_limit = self.limits[0][2]
sequence_term = self.function
if len(sequence_term.free_symbols) > 1:
raise NotImplementedError("convergence checking for more than one symbol "
"containing series is not handled")
if lower_limit.is_finite and upper_limit.is_finite:
return S.true
# transform sym -> -sym and swap the upper_limit = S.Infinity
# and lower_limit = - upper_limit
if lower_limit is S.NegativeInfinity:
if upper_limit is S.Infinity:
return Sum(sequence_term, (sym, 0, S.Infinity)).is_convergent() and \
Sum(sequence_term, (sym, S.NegativeInfinity, 0)).is_convergent()
sequence_term = simplify(sequence_term.xreplace({sym: -sym}))
lower_limit = -upper_limit
upper_limit = S.Infinity
sym_ = Dummy(sym.name, integer=True, positive=True)
sequence_term = sequence_term.xreplace({sym: sym_})
sym = sym_
interval = Interval(lower_limit, upper_limit)
# Piecewise function handle
if sequence_term.is_Piecewise:
for func, cond in sequence_term.args:
# see if it represents something going to oo
if cond == True or cond.as_set().sup is S.Infinity:
s = Sum(func, (sym, lower_limit, upper_limit))
return s.is_convergent()
return S.true
### -------- Divergence test ----------- ###
try:
lim_val = limit(sequence_term, sym, upper_limit)
if lim_val.is_number and lim_val is not S.Zero:
return S.false
except NotImplementedError:
pass
try:
lim_val_abs = limit(abs(sequence_term), sym, upper_limit)
if lim_val_abs.is_number and lim_val_abs is not S.Zero:
return S.false
except NotImplementedError:
pass
order = O(sequence_term, (sym, S.Infinity))
### ----------- ratio test ---------------- ###
next_sequence_term = sequence_term.xreplace({sym: sym + 1})
ratio = combsimp(powsimp(next_sequence_term/sequence_term))
lim_ratio = limit(ratio, sym, upper_limit)
if lim_ratio.is_number:
if abs(lim_ratio) > 1:
return S.false
if abs(lim_ratio) < 1:
return S.true
### --------- p-series test (1/n**p) ---------- ###
p1_series_test = order.expr.match(sym**p)
if p1_series_test is not None:
if p1_series_test[p] < -1:
return S.true
if p1_series_test[p] >= -1:
return S.false
p2_series_test = order.expr.match((1/sym)**p)
if p2_series_test is not None:
if p2_series_test[p] > 1:
return S.true
if p2_series_test[p] <= 1:
return S.false
### ------------- Limit comparison test -----------###
# (1/n) comparison
try:
lim_comp = limit(sym*sequence_term, sym, S.Infinity)
if lim_comp.is_number and lim_comp > 0:
return S.false
except NotImplementedError:
pass
### ----------- root test ---------------- ###
lim = Limit(abs(sequence_term)**(1/sym), sym, S.Infinity)
lim_evaluated = lim.doit()
if lim_evaluated.is_number:
if lim_evaluated < 1:
return S.true
if lim_evaluated > 1:
return S.false
### ------------- alternating series test ----------- ###
dict_val = sequence_term.match((-1)**(sym + p)*q)
if not dict_val[p].has(sym) and is_decreasing(dict_val[q], interval):
return S.true
### ------------- comparison test ------------- ###
# (1/log(n)**p) comparison
log_test = order.expr.match(1/(log(sym)**p))
if log_test is not None:
return S.false
# (1/(n*log(n)**p)) comparison
log_n_test = order.expr.match(1/(sym*(log(sym))**p))
if log_n_test is not None:
if log_n_test[p] > 1:
return S.true
return S.false
# (1/(n*log(n)*log(log(n))*p)) comparison
log_log_n_test = order.expr.match(1/(sym*(log(sym)*log(log(sym))**p)))
if log_log_n_test is not None:
if log_log_n_test[p] > 1:
return S.true
return S.false
# (1/(n**p*log(n))) comparison
n_log_test = order.expr.match(1/(sym**p*log(sym)))
if n_log_test is not None:
if n_log_test[p] > 1:
return S.true
return S.false
### ------------- integral test -------------- ###
maxima = solveset(sequence_term.diff(sym), sym, interval)
if not maxima:
check_interval = interval
elif isinstance(maxima, FiniteSet) and maxima.sup.is_number:
check_interval = Interval(maxima.sup, interval.sup)
if (
is_decreasing(sequence_term, check_interval) or
is_decreasing(-sequence_term, check_interval)):
integral_val = Integral(
sequence_term, (sym, lower_limit, upper_limit))
try:
integral_val_evaluated = integral_val.doit()
if integral_val_evaluated.is_number:
return S(integral_val_evaluated.is_finite)
except NotImplementedError:
pass
### -------------- Dirichlet tests -------------- ###
if order.expr.is_Mul:
a_n, b_n = order.expr.args[0], order.expr.args[1]
m = Dummy('m', integer=True)
def _dirichlet_test(g_n):
try:
ing_val = limit(Sum(g_n, (sym, interval.inf, m)).doit(), m, S.Infinity)
if ing_val.is_finite:
return S.true
except NotImplementedError:
pass
if is_decreasing(a_n, interval):
dirich1 = _dirichlet_test(b_n)
if dirich1 is not None:
return dirich1
if is_decreasing(b_n, interval):
dirich2 = _dirichlet_test(a_n)
if dirich2 is not None:
return dirich2
_sym = self.limits[0][0]
sequence_term = sequence_term.xreplace({sym: _sym})
raise NotImplementedError("The algorithm to find the Sum convergence of %s "
"is not yet implemented" % (sequence_term))
def simplify(expr, ratio=1.7, measure=count_ops, fu=False):
"""
Simplifies the given expression.
Simplification is not a well defined term and the exact strategies
this function tries can change in the future versions of SymPy. If
your algorithm relies on "simplification" (whatever it is), try to
determine what you need exactly - is it powsimp()?, radsimp()?,
together()?, logcombine()?, or something else? And use this particular
function directly, because those are well defined and thus your algorithm
will be robust.
Nonetheless, especially for interactive use, or when you don't know
anything about the structure of the expression, simplify() tries to apply
intelligent heuristics to make the input expression "simpler". For
example:
>>> from sympy import simplify, cos, sin
>>> from sympy.abc import x, y
>>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2)
>>> a
(x**2 + x)/(x*sin(y)**2 + x*cos(y)**2)
>>> simplify(a)
x + 1
Note that we could have obtained the same result by using specific
simplification functions:
>>> from sympy import trigsimp, cancel
>>> trigsimp(a)
(x**2 + x)/x
>>> cancel(_)
x + 1
In some cases, applying :func:`simplify` may actually result in some more
complicated expression. The default ``ratio=1.7`` prevents more extreme
cases: if (result length)/(input length) > ratio, then input is returned
unmodified. The ``measure`` parameter lets you specify the function used
to determine how complex an expression is. The function should take a
single argument as an expression and return a number such that if
expression ``a`` is more complex than expression ``b``, then
``measure(a) > measure(b)``. The default measure function is
:func:`count_ops`, which returns the total number of operations in the
expression.
For example, if ``ratio=1``, ``simplify`` output can't be longer
than input.
::
>>> from sympy import sqrt, simplify, count_ops, oo
>>> root = 1/(sqrt(2)+3)
Since ``simplify(root)`` would result in a slightly longer expression,
root is returned unchanged instead::
>>> simplify(root, ratio=1) == root
True
If ``ratio=oo``, simplify will be applied anyway::
>>> count_ops(simplify(root, ratio=oo)) > count_ops(root)
True
Note that the shortest expression is not necessary the simplest, so
setting ``ratio`` to 1 may not be a good idea.
Heuristically, the default value ``ratio=1.7`` seems like a reasonable
choice.
You can easily define your own measure function based on what you feel
should represent the "size" or "complexity" of the input expression. Note
that some choices, such as ``lambda expr: len(str(expr))`` may appear to be
good metrics, but have other problems (in this case, the measure function
may slow down simplify too much for very large expressions). If you don't
know what a good metric would be, the default, ``count_ops``, is a good
one.
For example:
>>> from sympy import symbols, log
>>> a, b = symbols('a b', positive=True)
>>> g = log(a) + log(b) + log(a)*log(1/b)
>>> h = simplify(g)
>>> h
log(a*b**(-log(a) + 1))
>>> count_ops(g)
8
>>> count_ops(h)
5
So you can see that ``h`` is simpler than ``g`` using the count_ops metric.
However, we may not like how ``simplify`` (in this case, using
``logcombine``) has created the ``b**(log(1/a) + 1)`` term. A simple way
to reduce this would be to give more weight to powers as operations in
``count_ops``. We can do this by using the ``visual=True`` option:
>>> print(count_ops(g, visual=True))
2*ADD + DIV + 4*LOG + MUL
>>> print(count_ops(h, visual=True))
2*LOG + MUL + POW + SUB
>>> from sympy import Symbol, S
>>> def my_measure(expr):
... POW = Symbol('POW')
... # Discourage powers by giving POW a weight of 10
... count = count_ops(expr, visual=True).subs(POW, 10)
... # Every other operation gets a weight of 1 (the default)
... count = count.replace(Symbol, type(S.One))
... return count
>>> my_measure(g)
8
>>> my_measure(h)
14
>>> 15./8 > 1.7 # 1.7 is the default ratio
True
>>> simplify(g, measure=my_measure)
-log(a)*log(b) + log(a) + log(b)
Note that because ``simplify()`` internally tries many different
simplification strategies and then compares them using the measure
function, we get a completely different result that is still different
from the input expression by doing this.
"""
expr = sympify(expr)
try:
return expr._eval_simplify(ratio=ratio, measure=measure)
except AttributeError:
pass
original_expr = expr = signsimp(expr)
from sympy.simplify.hyperexpand import hyperexpand
from sympy.functions.special.bessel import BesselBase
from sympy import Sum, Product
if not isinstance(expr, Basic) or not expr.args: # XXX: temporary hack
return expr
if not isinstance(expr, (Add, Mul, Pow, ExpBase)):
if isinstance(expr, Function) and hasattr(expr, "inverse"):
if len(expr.args) == 1 and len(expr.args[0].args) == 1 and \
isinstance(expr.args[0], expr.inverse(argindex=1)):
return simplify(expr.args[0].args[0], ratio=ratio,
measure=measure, fu=fu)
return expr.func(*[simplify(x, ratio=ratio, measure=measure, fu=fu)
for x in expr.args])
# TODO: Apply different strategies, considering expression pattern:
# is it a purely rational function? Is there any trigonometric function?...
# See also https://github.com/sympy/sympy/pull/185.
def shorter(*choices):
'''Return the choice that has the fewest ops. In case of a tie,
the expression listed first is selected.'''
if not has_variety(choices):
return choices[0]
return min(choices, key=measure)
expr = bottom_up(expr, lambda w: w.normal())
expr = Mul(*powsimp(expr).as_content_primitive())
_e = cancel(expr)
expr1 = shorter(_e, _mexpand(_e).cancel()) # issue 6829
expr2 = shorter(together(expr, deep=True), together(expr1, deep=True))
if ratio is S.Infinity:
expr = expr2
else:
expr = shorter(expr2, expr1, expr)
if not isinstance(expr, Basic): # XXX: temporary hack
return expr
expr = factor_terms(expr, sign=False)
# hyperexpand automatically only works on hypergeometric terms
expr = hyperexpand(expr)
expr = piecewise_fold(expr)
if expr.has(BesselBase):
expr = besselsimp(expr)
if expr.has(TrigonometricFunction) and not fu or expr.has(
HyperbolicFunction):
expr = trigsimp(expr, deep=True)
if expr.has(log):
expr = shorter(expand_log(expr, deep=True), logcombine(expr))
if expr.has(CombinatorialFunction, gamma):
expr = combsimp(expr)
if expr.has(Sum):
expr = sum_simplify(expr)
if expr.has(Product):
expr = product_simplify(expr)
short = shorter(powsimp(expr, combine='exp', deep=True), powsimp(expr), expr)
short = shorter(short, factor_terms(short), expand_power_exp(expand_mul(short)))
if short.has(TrigonometricFunction, HyperbolicFunction, ExpBase):
short = exptrigsimp(short, simplify=False)
# get rid of hollow 2-arg Mul factorization
hollow_mul = Transform(
lambda x: Mul(*x.args),
lambda x:
x.is_Mul and
len(x.args) == 2 and
x.args[0].is_Number and
x.args[1].is_Add and
x.is_commutative)
expr = short.xreplace(hollow_mul)
numer, denom = expr.as_numer_denom()
if denom.is_Add:
n, d = fraction(radsimp(1/denom, symbolic=False, max_terms=1))
if n is not S.One:
expr = (numer*n).expand()/d
if expr.could_extract_minus_sign():
n, d = fraction(expr)
if d != 0:
expr = signsimp(-n/(-d))
if measure(expr) > ratio*measure(original_expr):
expr = original_expr
return expr
def simplify(expr, ratio=1.7, measure=count_ops, rational=False):
# type: (object, object, object, object) -> object
"""
Simplifies the given expression.
Simplification is not a well defined term and the exact strategies
this function tries can change in the future versions of SymPy. If
your algorithm relies on "simplification" (whatever it is), try to
determine what you need exactly - is it powsimp()?, radsimp()?,
together()?, logcombine()?, or something else? And use this particular
function directly, because those are well defined and thus your algorithm
will be robust.
Nonetheless, especially for interactive use, or when you don't know
anything about the structure of the expression, simplify() tries to apply
intelligent heuristics to make the input expression "simpler". For
example:
>>> from sympy import simplify, cos, sin
>>> from sympy.abc import x, y
>>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2)
>>> a
(x**2 + x)/(x*sin(y)**2 + x*cos(y)**2)
>>> simplify(a)
x + 1
Note that we could have obtained the same result by using specific
simplification functions:
>>> from sympy import trigsimp, cancel
>>> trigsimp(a)
(x**2 + x)/x
>>> cancel(_)
x + 1
In some cases, applying :func:`simplify` may actually result in some more
complicated expression. The default ``ratio=1.7`` prevents more extreme
cases: if (result length)/(input length) > ratio, then input is returned
unmodified. The ``measure`` parameter lets you specify the function used
to determine how complex an expression is. The function should take a
single argument as an expression and return a number such that if
expression ``a`` is more complex than expression ``b``, then
``measure(a) > measure(b)``. The default measure function is
:func:`count_ops`, which returns the total number of operations in the
expression.
For example, if ``ratio=1``, ``simplify`` output can't be longer
than input.
::
>>> from sympy import sqrt, simplify, count_ops, oo
>>> root = 1/(sqrt(2)+3)
Since ``simplify(root)`` would result in a slightly longer expression,
root is returned unchanged instead::
>>> simplify(root, ratio=1) == root
True
If ``ratio=oo``, simplify will be applied anyway::
>>> count_ops(simplify(root, ratio=oo)) > count_ops(root)
True
Note that the shortest expression is not necessary the simplest, so
setting ``ratio`` to 1 may not be a good idea.
Heuristically, the default value ``ratio=1.7`` seems like a reasonable
choice.
You can easily define your own measure function based on what you feel
should represent the "size" or "complexity" of the input expression. Note
that some choices, such as ``lambda expr: len(str(expr))`` may appear to be
good metrics, but have other problems (in this case, the measure function
may slow down simplify too much for very large expressions). If you don't
know what a good metric would be, the default, ``count_ops``, is a good
one.
For example:
>>> from sympy import symbols, log
>>> a, b = symbols('a b', positive=True)
>>> g = log(a) + log(b) + log(a)*log(1/b)
>>> h = simplify(g)
>>> h
log(a*b**(-log(a) + 1))
>>> count_ops(g)
8
>>> count_ops(h)
5
So you can see that ``h`` is simpler than ``g`` using the count_ops metric.
However, we may not like how ``simplify`` (in this case, using
``logcombine``) has created the ``b**(log(1/a) + 1)`` term. A simple way
to reduce this would be to give more weight to powers as operations in
``count_ops``. We can do this by using the ``visual=True`` option:
>>> print(count_ops(g, visual=True))
2*ADD + DIV + 4*LOG + MUL
>>> print(count_ops(h, visual=True))
2*LOG + MUL + POW + SUB
>>> from sympy import Symbol, S
>>> def my_measure(expr):
... POW = Symbol('POW')
... # Discourage powers by giving POW a weight of 10
... count = count_ops(expr, visual=True).subs(POW, 10)
... # Every other operation gets a weight of 1 (the default)
... count = count.replace(Symbol, type(S.One))
... return count
>>> my_measure(g)
8
>>> my_measure(h)
14
>>> 15./8 > 1.7 # 1.7 is the default ratio
True
>>> simplify(g, measure=my_measure)
-log(a)*log(b) + log(a) + log(b)
Note that because ``simplify()`` internally tries many different
simplification strategies and then compares them using the measure
function, we get a completely different result that is still different
from the input expression by doing this.
If rational=True, Floats will be recast as Rationals before simplification.
If rational=None, Floats will be recast as Rationals but the result will
be recast as Floats. If rational=False(default) then nothing will be done
to the Floats.
"""
expr = sympify(expr)
try:
return expr._eval_simplify(ratio=ratio, measure=measure)
except AttributeError:
pass
original_expr = expr = signsimp(expr)
from sympy.simplify.hyperexpand import hyperexpand
from sympy.functions.special.bessel import BesselBase
from sympy import Sum, Product
if not isinstance(expr, Basic) or not expr.args: # XXX: temporary hack
return expr
if not isinstance(expr, (Add, Mul, Pow, ExpBase)):
if isinstance(expr, Function) and hasattr(expr, "inverse"):
if len(expr.args) == 1 and len(expr.args[0].args) == 1 and \
isinstance(expr.args[0], expr.inverse(argindex=1)):
return simplify(expr.args[0].args[0], ratio=ratio,
measure=measure, rational=rational)
return expr.func(*[simplify(x, ratio=ratio, measure=measure, rational=rational)
for x in expr.args])
# TODO: Apply different strategies, considering expression pattern:
# is it a purely rational function? Is there any trigonometric function?...
# See also https://github.com/sympy/sympy/pull/185.
def shorter(*choices):
'''Return the choice that has the fewest ops. In case of a tie,
the expression listed first is selected.'''
if not has_variety(choices):
return choices[0]
return min(choices, key=measure)
# rationalize Floats
floats = False
if rational is not False and expr.has(Float):
floats = True
expr = nsimplify(expr, rational=True)
expr = bottom_up(expr, lambda w: w.normal())
expr = Mul(*powsimp(expr).as_content_primitive())
_e = cancel(expr)
expr1 = shorter(_e, _mexpand(_e).cancel()) # issue 6829
expr2 = shorter(together(expr, deep=True), together(expr1, deep=True))
if ratio is S.Infinity:
expr = expr2
else:
expr = shorter(expr2, expr1, expr)
if not isinstance(expr, Basic): # XXX: temporary hack
return expr
expr = factor_terms(expr, sign=False)
# hyperexpand automatically only works on hypergeometric terms
expr = hyperexpand(expr)
expr = piecewise_fold(expr)
if expr.has(BesselBase):
expr = besselsimp(expr)
if expr.has(TrigonometricFunction, HyperbolicFunction):
expr = trigsimp(expr, deep=True)
if expr.has(log):
expr = shorter(expand_log(expr, deep=True), logcombine(expr))
if expr.has(CombinatorialFunction, gamma):
# expression with gamma functions or non-integer arguments is
# automatically passed to gammasimp
expr = combsimp(expr)
if expr.has(Sum):
expr = sum_simplify(expr)
if expr.has(Product):
expr = product_simplify(expr)
short = shorter(powsimp(expr, combine='exp', deep=True), powsimp(expr), expr)
short = shorter(short, cancel(short))
short = shorter(short, factor_terms(short), expand_power_exp(expand_mul(short)))
if short.has(TrigonometricFunction, HyperbolicFunction, ExpBase):
short = exptrigsimp(short)
# get rid of hollow 2-arg Mul factorization
hollow_mul = Transform(
lambda x: Mul(*x.args),
lambda x:
x.is_Mul and
len(x.args) == 2 and
x.args[0].is_Number and
x.args[1].is_Add and
x.is_commutative)
expr = short.xreplace(hollow_mul)
numer, denom = expr.as_numer_denom()
if denom.is_Add:
n, d = fraction(radsimp(1/denom, symbolic=False, max_terms=1))
if n is not S.One:
expr = (numer*n).expand()/d
if expr.could_extract_minus_sign():
n, d = fraction(expr)
if d != 0:
expr = signsimp(-n/(-d))
if measure(expr) > ratio*measure(original_expr):
expr = original_expr
# restore floats
if floats and rational is None:
expr = nfloat(expr, exponent=False)
return expr
def is_convergent(self):
r"""Checks for the convergence of a Sum.
We divide the study of convergence of infinite sums and products in
two parts.
First Part:
One part is the question whether all the terms are well defined, i.e.,
they are finite in a sum and also non-zero in a product. Zero
is the analogy of (minus) infinity in products as
:math:`e^{-\infty} = 0`.
Second Part:
The second part is the question of convergence after infinities,
and zeros in products, have been omitted assuming that their number
is finite. This means that we only consider the tail of the sum or
product, starting from some point after which all terms are well
defined.
For example, in a sum of the form:
.. math::
\sum_{1 \leq i < \infty} \frac{1}{n^2 + an + b}
where a and b are numbers. The routine will return true, even if there
are infinities in the term sequence (at most two). An analogous
product would be:
.. math::
\prod_{1 \leq i < \infty} e^{\frac{1}{n^2 + an + b}}
This is how convergence is interpreted. It is concerned with what
happens at the limit. Finding the bad terms is another independent
matter.
Note: It is responsibility of user to see that the sum or product
is well defined.
There are various tests employed to check the convergence like
divergence test, root test, integral test, alternating series test,
comparison tests, Dirichlet tests. It returns true if Sum is convergent
and false if divergent and NotImplementedError if it can not be checked.
References
==========
.. [1] https://en.wikipedia.org/wiki/Convergence_tests
Examples
========
>>> from sympy import factorial, S, Sum, Symbol, oo
>>> n = Symbol('n', integer=True)
>>> Sum(n/(n - 1), (n, 4, 7)).is_convergent()
True
>>> Sum(n/(2*n + 1), (n, 1, oo)).is_convergent()
False
>>> Sum(factorial(n)/5**n, (n, 1, oo)).is_convergent()
False
>>> Sum(1/n**(S(6)/5), (n, 1, oo)).is_convergent()
True
See Also
========
Sum.is_absolutely_convergent()
Product.is_convergent()
"""
from sympy import Interval, Integral, Limit, log, symbols, Ge, Gt, simplify
p, q = symbols('p q', cls=Wild)
sym = self.limits[0][0]
lower_limit = self.limits[0][1]
upper_limit = self.limits[0][2]
sequence_term = self.function
if len(sequence_term.free_symbols) > 1:
raise NotImplementedError(
"convergence checking for more than one symbol "
"containing series is not handled")
if lower_limit.is_finite and upper_limit.is_finite:
return S.true
# transform sym -> -sym and swap the upper_limit = S.Infinity
# and lower_limit = - upper_limit
if lower_limit is S.NegativeInfinity:
if upper_limit is S.Infinity:
return Sum(sequence_term, (sym, 0, S.Infinity)).is_convergent() and \
Sum(sequence_term, (sym, S.NegativeInfinity, 0)).is_convergent()
sequence_term = simplify(sequence_term.xreplace({sym: -sym}))
lower_limit = -upper_limit
upper_limit = S.Infinity
sym_ = Dummy(sym.name, integer=True, positive=True)
sequence_term = sequence_term.xreplace({sym: sym_})
sym = sym_
interval = Interval(lower_limit, upper_limit)
# Piecewise function handle
if sequence_term.is_Piecewise:
for func, cond in sequence_term.args:
# see if it represents something going to oo
if cond == True or cond.as_set().sup is S.Infinity:
s = Sum(func, (sym, lower_limit, upper_limit))
return s.is_convergent()
return S.true
### -------- Divergence test ----------- ###
try:
lim_val = limit(sequence_term, sym, upper_limit)
if lim_val.is_number and lim_val is not S.Zero:
return S.false
except NotImplementedError:
pass
try:
lim_val_abs = limit(abs(sequence_term), sym, upper_limit)
if lim_val_abs.is_number and lim_val_abs is not S.Zero:
return S.false
except NotImplementedError:
pass
order = O(sequence_term, (sym, S.Infinity))
### ----------- ratio test ---------------- ###
next_sequence_term = sequence_term.xreplace({sym: sym + 1})
ratio = combsimp(powsimp(next_sequence_term / sequence_term))
lim_ratio = limit(ratio, sym, upper_limit)
if abs(lim_ratio) > 1:
return S.false
if abs(lim_ratio) < 1:
return S.true
### --------- p-series test (1/n**p) ---------- ###
p1_series_test = order.expr.match(sym**p)
if p1_series_test is not None:
if p1_series_test[p] < -1:
return S.true
if p1_series_test[p] >= -1:
return S.false
p2_series_test = order.expr.match((1 / sym)**p)
if p2_series_test is not None:
if p2_series_test[p] > 1:
return S.true
if p2_series_test[p] <= 1:
return S.false
### ----------- root test ---------------- ###
lim = Limit(abs(sequence_term)**(1 / sym), sym, S.Infinity)
lim_evaluated = lim.doit()
if lim_evaluated.is_number:
if lim_evaluated < 1:
return S.true
if lim_evaluated > 1:
return S.false
### ------------- alternating series test ----------- ###
dict_val = sequence_term.match((-1)**(sym + p) * q)
if not dict_val[p].has(sym) and is_decreasing(dict_val[q], interval):
return S.true
### ------------- comparison test ------------- ###
# (1/log(n)**p) comparison
log_test = order.expr.match(1 / (log(sym)**p))
if log_test is not None:
return S.false
# (1/(n*log(n)**p)) comparison
log_n_test = order.expr.match(1 / (sym * (log(sym))**p))
if log_n_test is not None:
if log_n_test[p] > 1:
return S.true
return S.false
# (1/(n*log(n)*log(log(n))*p)) comparison
log_log_n_test = order.expr.match(1 / (sym *
(log(sym) * log(log(sym))**p)))
if log_log_n_test is not None:
if log_log_n_test[p] > 1:
return S.true
return S.false
# (1/(n**p*log(n))) comparison
n_log_test = order.expr.match(1 / (sym**p * log(sym)))
if n_log_test is not None:
if n_log_test[p] > 1:
return S.true
return S.false
### ------------- integral test -------------- ###
maxima = solveset(sequence_term.diff(sym), sym, interval)
if not maxima:
check_interval = interval
elif isinstance(maxima, FiniteSet) and maxima.sup.is_number:
check_interval = Interval(maxima.sup, interval.sup)
if (is_decreasing(sequence_term, check_interval)
or is_decreasing(-sequence_term, check_interval)):
integral_val = Integral(sequence_term,
(sym, lower_limit, upper_limit))
try:
integral_val_evaluated = integral_val.doit()
if integral_val_evaluated.is_number:
return S(integral_val_evaluated.is_finite)
except NotImplementedError:
pass
### -------------- Dirichlet tests -------------- ###
if order.expr.is_Mul:
a_n, b_n = order.expr.args[0], order.expr.args[1]
m = Dummy('m', integer=True)
def _dirichlet_test(g_n):
try:
ing_val = limit(
Sum(g_n, (sym, interval.inf, m)).doit(), m, S.Infinity)
if ing_val.is_finite:
return S.true
except NotImplementedError:
pass
if is_decreasing(a_n, interval):
dirich1 = _dirichlet_test(b_n)
if dirich1 is not None:
return dirich1
if is_decreasing(b_n, interval):
dirich2 = _dirichlet_test(a_n)
if dirich2 is not None:
return dirich2
_sym = self.limits[0][0]
sequence_term = sequence_term.xreplace({sym: _sym})
raise NotImplementedError(
"The algorithm to find the Sum convergence of %s "
"is not yet implemented" % (sequence_term))
def test_combsimp():
assert combsimp(A*B - B*A) == A*B - B*A
def test_issue_14528():
p = symbols("p", integer=True, positive=True)
assert combsimp(binomial(1, p)) == 1 / (factorial(p) * factorial(1 - p))
assert combsimp(factorial(2 - p)) == factorial(2 - p)
def test_issue_6878():
n = symbols('n', integer=True)
assert combsimp(RisingFactorial(
-10, n)) == 3628800 * (-1)**n / factorial(10 - n)
def test_combsimp():
k, m, n = symbols('k m n', integer=True)
assert combsimp(factorial(n)) == factorial(n)
assert combsimp(binomial(n, k)) == binomial(n, k)
assert combsimp(factorial(n) / factorial(n - 3)) == n * (-1 + n) * (-2 + n)
assert combsimp(binomial(n + 1, k + 1) /
binomial(n, k)) == (1 + n) / (1 + k)
assert combsimp(binomial(3*n + 4, n + 1)/binomial(3*n + 1, n)) == \
Rational(3, 2)*((3*n + 2)*(3*n + 4)/((n + 1)*(2*n + 3)))
assert combsimp(factorial(n)**2/factorial(n - 3)) == \
factorial(n)*n*(-1 + n)*(-2 + n)
assert combsimp(factorial(n)*binomial(n + 1, k + 1)/binomial(n, k)) == \
factorial(n + 1)/(1 + k)
assert combsimp(gamma(n + 3)) == factorial(n + 2)
assert combsimp(factorial(x)) == gamma(x + 1)
# issue 9699
assert combsimp((n + 1) * factorial(n)) == factorial(n + 1)
assert combsimp(factorial(n) / n) == factorial(n - 1)
# issue 6658
assert combsimp(binomial(n, n - k)) == binomial(n, k)
# issue 6341, 7135
assert combsimp(factorial(n)/(factorial(k)*factorial(n - k))) == \
binomial(n, k)
assert combsimp(factorial(k)*factorial(n - k)/factorial(n)) == \
1/binomial(n, k)
assert combsimp(factorial(2 * n) / factorial(n)**2) == binomial(2 * n, n)
assert combsimp(
factorial(2 * n) * factorial(k) * factorial(n - k) /
factorial(n)**3) == binomial(2 * n, n) / binomial(n, k)
assert combsimp(factorial(n * (1 + n) - n**2 - n)) == 1
assert combsimp(6*FallingFactorial(-4, n)/factorial(n)) == \
(-1)**n*(n + 1)*(n + 2)*(n + 3)
assert combsimp(6*FallingFactorial(-4, n - 1)/factorial(n - 1)) == \
(-1)**(n - 1)*n*(n + 1)*(n + 2)
assert combsimp(6*FallingFactorial(-4, n - 3)/factorial(n - 3)) == \
(-1)**(n - 3)*n*(n - 1)*(n - 2)
assert combsimp(6*FallingFactorial(-4, -n - 1)/factorial(-n - 1)) == \
-(-1)**(-n - 1)*n*(n - 1)*(n - 2)
assert combsimp(6*RisingFactorial(4, n)/factorial(n)) == \
(n + 1)*(n + 2)*(n + 3)
assert combsimp(6*RisingFactorial(4, n - 1)/factorial(n - 1)) == \
n*(n + 1)*(n + 2)
assert combsimp(6*RisingFactorial(4, n - 3)/factorial(n - 3)) == \
n*(n - 1)*(n - 2)
assert combsimp(6*RisingFactorial(4, -n - 1)/factorial(-n - 1)) == \
-n*(n - 1)*(n - 2)
