def test_arithmetic_sums():
assert summation(1, (n, a, b)) == b - a + 1
assert Sum(nan, (n, a, b)) is nan
assert Sum(x, (n, a, a)).doit() == x
assert Sum(x, (x, a, a)).doit() == a
assert Sum(x, (n, 1, a)).doit() == a * x
lo, hi = 1, 2
s1 = Sum(n, (n, lo, hi))
s2 = Sum(n, (n, hi, lo))
assert s1 != s2
assert s1.doit() == 3 and s2.doit() == 0
lo, hi = x, x + 1
s1 = Sum(n, (n, lo, hi))
s2 = Sum(n, (n, hi, lo))
assert s1 != s2
assert s1.doit() == 2 * x + 1 and s2.doit() == 0
assert Sum(Integral(x, (x, 1, y)) + x, (x, 1, 2)).doit() == \
y**2 + 2
assert summation(1, (n, 1, 10)) == 10
assert summation(2 * n, (n, 0, 10**10)) == 100000000010000000000
assert summation(4*n*m, (n, a, 1), (m, 1, d)).expand() == \
2*d + 2*d**2 + a*d + a*d**2 - d*a**2 - a**2*d**2
assert summation(cos(n), (n, -2, 1)) == cos(-2) + cos(-1) + cos(0) + cos(1)
assert summation(cos(n), (n, x, x + 2)) == cos(x) + cos(x + 1) + cos(x + 2)
assert isinstance(summation(cos(n), (n, x, x + Rational(1, 2))), Sum)
assert summation(k, (k, 0, oo)) == oo
def test_arithmetic_sums():
assert summation(1, (n, a, b)) == b - a + 1
assert Sum(nan, (n, a, b)) is nan
assert Sum(x, (n, a, a)).doit() == x
assert Sum(x, (x, a, a)).doit() == a
assert Sum(x, (n, 1, a)).doit() == a*x
lo, hi = 1, 2
s1 = Sum(n, (n, lo, hi))
s2 = Sum(n, (n, hi, lo))
assert s1 != s2
assert s1.doit() == 3 and s2.doit() == 0
lo, hi = x, x + 1
s1 = Sum(n, (n, lo, hi))
s2 = Sum(n, (n, hi, lo))
assert s1 != s2
assert s1.doit() == 2*x + 1 and s2.doit() == 0
assert Sum(Integral(x, (x, 1, y)) + x, (x, 1, 2)).doit() == \
y**2 + 2
assert summation(1, (n, 1, 10)) == 10
assert summation(2*n, (n, 0, 10**10)) == 100000000010000000000
assert summation(4*n*m, (n, a, 1), (m, 1, d)).expand() == \
2*d + 2*d**2 + a*d + a*d**2 - d*a**2 - a**2*d**2
assert summation(cos(n), (n, -2, 1)) == cos(-2) + cos(-1) + cos(0) + cos(1)
assert summation(cos(n), (n, x, x + 2)) == cos(x) + cos(x + 1) + cos(x + 2)
assert isinstance(summation(cos(n), (n, x, x + Rational(1, 2))), Sum)
assert summation(k, (k, 0, oo)) == oo
def test_Sum_doit():
f = Function('f')
assert Sum(n*Integral(a**2), (n, 0, 2)).doit() == a**3
assert Sum(n*Integral(a**2), (n, 0, 2)).doit(deep=False) == \
3*Integral(a**2)
assert summation(n*Integral(a**2), (n, 0, 2)) == 3*Integral(a**2)
# test nested sum evaluation
s = Sum(Sum(Sum(2, (z, 1, n + 1)), (y, x + 1, n)), (x, 1, n))
assert 0 == (s.doit() - n*(n + 1)*(n - 1)).factor()
assert Sum(Sum(KroneckerDelta(m, n), (m, 1, 3)), (n, 1, 3)).doit() == 3
assert Sum(Sum(KroneckerDelta(k, m), (m, 1, 3)), (n, 1, 3)).doit() == \
3*Piecewise((1, And(Integer(1) <= k, k <= 3)), (0, True))
assert Sum(f(n)*Sum(KroneckerDelta(m, n), (m, 0, oo)), (n, 1, 3)).doit() == \
f(1) + f(2) + f(3)
assert Sum(f(n)*Sum(KroneckerDelta(m, n), (m, 0, oo)), (n, 1, oo)).doit() == \
Sum(Piecewise((f(n), And(Le(0, n), n < oo)), (0, True)), (n, 1, oo))
l = Symbol('l', integer=True, positive=True)
assert Sum(f(l)*Sum(KroneckerDelta(m, l), (m, 0, oo)), (l, 1, oo)).doit() == \
Sum(f(l), (l, 1, oo))
# issue sympy/sympy#2597
nmax = symbols('N', integer=True, positive=True)
pw = Piecewise((1, And(Integer(1) <= n, n <= nmax)), (0, True))
assert Sum(pw, (n, 1, nmax)).doit() == Sum(pw, (n, 1, nmax))
def test_Sum_doit():
f = Function('f')
assert Sum(n * Integral(a**2), (n, 0, 2)).doit() == a**3
assert Sum(n*Integral(a**2), (n, 0, 2)).doit(deep=False) == \
3*Integral(a**2)
assert summation(n * Integral(a**2), (n, 0, 2)) == 3 * Integral(a**2)
# test nested sum evaluation
s = Sum(Sum(Sum(2, (z, 1, n + 1)), (y, x + 1, n)), (x, 1, n))
assert 0 == (s.doit() - n * (n + 1) * (n - 1)).factor()
assert Sum(Sum(KroneckerDelta(m, n), (m, 1, 3)), (n, 1, 3)).doit() == 3
assert Sum(Sum(KroneckerDelta(k, m), (m, 1, 3)), (n, 1, 3)).doit() == \
3*Piecewise((1, And(Integer(1) <= k, k <= 3)), (0, True))
assert Sum(f(n)*Sum(KroneckerDelta(m, n), (m, 0, oo)), (n, 1, 3)).doit() == \
f(1) + f(2) + f(3)
assert Sum(f(n)*Sum(KroneckerDelta(m, n), (m, 0, oo)), (n, 1, oo)).doit() == \
Sum(Piecewise((f(n), And(Le(0, n), n < oo)), (0, True)), (n, 1, oo))
l = Symbol('l', integer=True, positive=True)
assert Sum(f(l)*Sum(KroneckerDelta(m, l), (m, 0, oo)), (l, 1, oo)).doit() == \
Sum(f(l), (l, 1, oo))
# issue sympy/sympy#2597
nmax = symbols('N', integer=True, positive=True)
pw = Piecewise((1, And(Integer(1) <= n, n <= nmax)), (0, True))
assert Sum(pw, (n, 1, nmax)).doit() == Sum(pw, (n, 1, nmax))
def test_sympyissue_2787():
n, k = symbols('n k', positive=True, integer=True)
p = symbols('p', positive=True)
binomial_dist = binomial(n, k) * p**k * (1 - p)**(n - k)
s = Sum(binomial_dist * k, (k, 0, n))
res = s.doit().simplify()
assert res == Piecewise(
(n * p, And(Or(-n + 1 < 0, Ne(p / (p - 1), 1)), p / Abs(p - 1) <= 1)),
(Sum(k * p**k * (-p + 1)**(-k) * (-p + 1)**n * binomial(n, k),
(k, 0, n)), True))
def test_sympyissue_2787():
n, k = symbols('n k', positive=True, integer=True)
p = symbols('p', positive=True)
binomial_dist = binomial(n, k)*p**k*(1 - p)**(n - k)
s = Sum(binomial_dist*k, (k, 0, n))
res = s.doit().simplify()
assert res == Piecewise(
(n*p, And(Or(-n + 1 < 0, Ne(p/(p - 1), 1)), p/Abs(p - 1) <= 1)),
(Sum(k*p**k*(-p + 1)**(-k)*(-p + 1)**n*binomial(n, k), (k, 0, n)),
True))
def test_evalf_sum():
assert Sum(n, (n, 1, 2)).evalf() == 3.
assert Sum(n, (n, 1, 2)).doit().evalf() == 3.
# the next test should return instantly
assert Sum(1/n, (n, 1, 2)).evalf() == 1.5
# issue 8219
assert Sum(E/factorial(n), (n, 0, oo)).evalf() == (E*E).evalf()
# issue 8254
assert Sum(2**n*n/factorial(n), (n, 0, oo)).evalf() == (2*E*E).evalf()
# issue 8411
s = Sum(1/x**2, (x, 100, oo))
assert s.n() == s.doit().n()
def test_evalf_sum():
assert Sum(n, (n, 1, 2)).evalf() == 3.
assert Sum(I*n, (n, 1, 2)).evalf() == 3.*I
assert Sum(n, (n, 1, 2)).doit().evalf() == 3.
# the next test should return instantly
assert Sum(1/n, (n, 1, 2)).evalf() == 1.5
# issue sympy/sympy#8219
assert Sum(E/factorial(n), (n, 0, oo)).evalf() == (E*E).evalf()
# issue sympy/sympy#8254
assert Sum(2**n*n/factorial(n), (n, 0, oo)).evalf() == (2*E*E).evalf()
# issue sympy/sympy#8411
s = Sum(1/x**2, (x, 100, oo))
assert s.evalf() == s.doit().evalf()
def test_conjugate_transpose():
A, B = symbols("A B", commutative=False)
p = Sum(A*B**n, (n, 1, 3))
assert p.adjoint().doit() == p.doit().adjoint()
assert p.conjugate().doit() == p.doit().conjugate()
assert p.transpose().doit() == p.doit().transpose()
def check_exact(f, a, b, m, n):
A = Sum(f, (k, a, b))
s, e = A.euler_maclaurin(m, n)
assert (e == 0) and (s.expand() == A.doit())
def test_karr_convention():
# Test the Karr summation convention that we want to hold.
# See his paper "Summation in Finite Terms" for a detailed
# reasoning why we really want exactly this definition.
# The convention is described on page 309 and essentially
# in section 1.4, definition 3:
#
# \sum_{m <= i < n} f(i) 'has the obvious meaning' for m < n
# \sum_{m <= i < n} f(i) = 0 for m = n
# \sum_{m <= i < n} f(i) = - \sum_{n <= i < m} f(i) for m > n
#
# It is important to note that he defines all sums with
# the upper limit being *exclusive*.
# In contrast, diofant and the usual mathematical notation has:
#
# sum_{i = a}^b f(i) = f(a) + f(a+1) + ... + f(b-1) + f(b)
#
# with the upper limit *inclusive*. So translating between
# the two we find that:
#
# \sum_{m <= i < n} f(i) = \sum_{i = m}^{n-1} f(i)
#
# where we intentionally used two different ways to typeset the
# sum and its limits.
i = Symbol("i", integer=True)
k = Symbol("k", integer=True)
j = Symbol("j", integer=True)
# A simple example with a concrete summand and symbolic limits.
# The normal sum: m = k and n = k + j and therefore m < n:
m = k
n = k + j
a = m
b = n - 1
S1 = Sum(i**2, (i, a, b)).doit()
# The reversed sum: m = k + j and n = k and therefore m > n:
m = k + j
n = k
a = m
b = n - 1
S2 = Sum(i**2, (i, a, b)).doit()
assert simplify(S1 + S2) == 0
# Test the empty sum: m = k and n = k and therefore m = n:
m = k
n = k
a = m
b = n - 1
Sz = Sum(i**2, (i, a, b)).doit()
assert Sz == 0
# Another example this time with an unspecified summand and
# numeric limits. (We can not do both tests in the same example.)
f = Function("f")
# The normal sum with m < n:
m = 2
n = 11
a = m
b = n - 1
S1 = Sum(f(i), (i, a, b)).doit()
# The reversed sum with m > n:
m = 11
n = 2
a = m
b = n - 1
S2 = Sum(f(i), (i, a, b)).doit()
assert simplify(S1 + S2) == 0
# Test the empty sum with m = n:
m = 5
n = 5
a = m
b = n - 1
Sz = Sum(f(i), (i, a, b)).doit()
assert Sz == 0
e = Piecewise((exp(-i), Mod(i, 2) > 0), (0, True))
s = Sum(e, (i, 0, 11))
assert s.evalf(3) == s.doit().evalf(3)
def test_karr_convention():
# Test the Karr summation convention that we want to hold.
# See his paper "Summation in Finite Terms" for a detailed
# reasoning why we really want exactly this definition.
# The convention is described on page 309 and essentially
# in section 1.4, definition 3:
#
# \sum_{m <= i < n} f(i) 'has the obvious meaning' for m < n
# \sum_{m <= i < n} f(i) = 0 for m = n
# \sum_{m <= i < n} f(i) = - \sum_{n <= i < m} f(i) for m > n
#
# It is important to note that he defines all sums with
# the upper limit being *exclusive*.
# In contrast, diofant and the usual mathematical notation has:
#
# sum_{i = a}^b f(i) = f(a) + f(a+1) + ... + f(b-1) + f(b)
#
# with the upper limit *inclusive*. So translating between
# the two we find that:
#
# \sum_{m <= i < n} f(i) = \sum_{i = m}^{n-1} f(i)
#
# where we intentionally used two different ways to typeset the
# sum and its limits.
i = Symbol("i", integer=True)
k = Symbol("k", integer=True)
j = Symbol("j", integer=True)
# A simple example with a concrete summand and symbolic limits.
# The normal sum: m = k and n = k + j and therefore m < n:
m = k
n = k + j
a = m
b = n - 1
S1 = Sum(i**2, (i, a, b)).doit()
# The reversed sum: m = k + j and n = k and therefore m > n:
m = k + j
n = k
a = m
b = n - 1
S2 = Sum(i**2, (i, a, b)).doit()
assert simplify(S1 + S2) == 0
# Test the empty sum: m = k and n = k and therefore m = n:
m = k
n = k
a = m
b = n - 1
Sz = Sum(i**2, (i, a, b)).doit()
assert Sz == 0
# Another example this time with an unspecified summand and
# numeric limits. (We can not do both tests in the same example.)
f = Function("f")
# The normal sum with m < n:
m = 2
n = 11
a = m
b = n - 1
S1 = Sum(f(i), (i, a, b)).doit()
# The reversed sum with m > n:
m = 11
n = 2
a = m
b = n - 1
S2 = Sum(f(i), (i, a, b)).doit()
assert simplify(S1 + S2) == 0
# Test the empty sum with m = n:
m = 5
n = 5
a = m
b = n - 1
Sz = Sum(f(i), (i, a, b)).doit()
assert Sz == 0
e = Piecewise((exp(-i), Mod(i, 2) > 0), (0, True))
s = Sum(e, (i, 0, 11))
assert s.evalf(3) == s.doit().evalf(3)
def test_sympyissue_15943():
s = Sum(binomial(n, k) * factorial(n - k), (k, 0, n))
assert s.doit().simplify() == E * (gamma(n + 1) - lowergamma(n + 1, 1))
def test_sympyissue_8016():
n, m = symbols('n, m', integer=True, positive=True)
k = symbols('k', integer=True)
s = Sum(binomial(m, k) * binomial(m, n - k) * (-1)**k, (k, 0, n))
assert s.doit().simplify() == \
cos(pi*n/2)*gamma(m + 1)/gamma(n/2 + 1)/gamma(m - n/2 + 1)
def test_conjugate_transpose():
A, B = symbols("A B", commutative=False)
p = Sum(A * B**n, (n, 1, 3))
assert p.adjoint().doit() == p.doit().adjoint()
assert p.conjugate().doit() == p.doit().conjugate()
assert p.transpose().doit() == p.doit().transpose()
def check_exact(f, a, b, m, n):
A = Sum(f, (k, a, b))
s, e = A.euler_maclaurin(m, n)
assert (e == 0) and (s.expand() == A.doit())
def test_sympyissue_8016():
n, m = symbols('n, m', integer=True, positive=True)
k = symbols('k', integer=True)
s = Sum(binomial(m, k)*binomial(m, n-k)*(-1)**k, (k, 0, n))
assert s.doit().simplify() == \
cos(pi*n/2)*gamma(m + 1)/gamma(n/2 + 1)/gamma(m - n/2 + 1)
def test_karr_convention():
# A simple example with a concrete summand and symbolic limits.
# The normal sum: m = k and n = k + j and therefore m < n:
m = k
n = k + j
a = m
b = n - 1
S1 = summation(i**2, (i, a, b))
# The reversed sum: m = k + j and n = k and therefore m > n:
m = k + j
n = k
a = m
b = n - 1
S2 = summation(i**2, (i, a, b))
assert simplify(S1 + S2) == 0
# Test the empty sum: m = k and n = k and therefore m = n:
m = k
n = k
a = m
b = n - 1
Sz = summation(i**2, (i, a, b))
assert Sz == 0
# Another example this time with an unspecified summand and
# numeric limits. (We can not do both tests in the same example.)
f = Function('f')
# The normal sum with m < n:
m = 2
n = 11
a = m
b = n - 1
S1 = summation(f(i), (i, a, b))
# The reversed sum with m > n:
m = 11
n = 2
a = m
b = n - 1
S2 = summation(f(i), (i, a, b))
assert simplify(S1 + S2) == 0
# Test the empty sum with m = n:
m = 5
n = 5
a = m
b = n - 1
Sz = summation(f(i), (i, a, b))
assert Sz == 0
e = Piecewise((exp(-i), Mod(i, 2) > 0), (0, True))
s = Sum(e, (i, 0, 11))
assert s.evalf(3) == s.doit().evalf(3)
