class TimeSum:
def setup(self):
self.expr = Sum(
Ts(int32ASU2L, ((), ((0, 1, -I / 2), (1, 0, I / 2)), ()), ext12Pi, int12Pi)
* Ts(int42ASU2L, ((), ((0, 1, -I / 2), (1, 0, I / 2)), ()), ext22Pi, int22Pi)
* Ts(int52ASU2L, ((), ((0, 1, -I / 2), (1, 0, I / 2)), ()), int12Pi, ext32Pi)
* Ts(int62ASU2L, ((), ((0, 1, -I / 2), (1, 0, I / 2)), ()), int22Pi, ext42Pi)
* KroneckerDelta(int32ASU2L, int42ASU2L)
* KroneckerDelta(int52ASU2L, int62ASU2L),
(int12Pi, 1, 2),
(int22Pi, 1, 2),
(exgg12SU2L, intgg11SU2L, intgg11SU2L),
(ext12Pi, ext11Pi, ext11Pi),
(exgg22SU2L, intgg11SU2L, intgg11SU2L),
(ext22Pi, ext31Pi, ext31Pi),
(exgg32SU2L, intgg21SU2L, intgg21SU2L),
(ext32Pi, ext21Pi, ext21Pi),
(exgg42SU2L, intgg21SU2L, intgg21SU2L),
(ext42Pi, ext41Pi, ext41Pi),
(int32ASU2L, 0, 2),
(int42ASU2L, 0, 2),
(int52ASU2L, 0, 2),
(int62ASU2L, 0, 2),
)
def time_doit(self):
self.expr.doit()
def test_arithmetic_sums():
assert summation(1, (n, a, b)) == b - a + 1
assert Sum(S.NaN, (n, a, b)) is S.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() == s2.doit() == 3
lo, hi = x, x + 1
s1 = Sum(n, (n, lo, hi))
s2 = Sum(n, (n, hi, lo))
assert s1 != s2
assert s1.doit() == s2.doit() == 2*x + 1
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 + S.Half)), Sum)
def test_Sum_doit():
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(S(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), n >= 0), (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 2597
nmax = symbols('N', integer=True, positive=True)
pw = Piecewise((1, And(S(1) <= n, n <= nmax)), (0, True))
assert Sum(pw, (n, 1, nmax)).doit() == Sum(pw, (n, 1, nmax))
def test_issue_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, -n + 1 >= 0),
Or(-n + 1 < 0, Ne(p/(p - 1), 1)), p*Abs(1/(p - 1)) <= 1)),
(Sum(k*p**k*(-p + 1)**(-k)*(-p + 1)**n*binomial(n, k), (k, 0, n)), True))
def test_issue_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, p/Abs(p - 1) <= 1),
((-p + 1)**n*Sum(k*p**k*(-p + 1)**(-k)*binomial(n, k), (k, 0, n)),
True))
def test_sho():
n, m = symbols('n,m')
h_n = Bd(n) * B(n) * (n + Rational(1, 2))
H = Sum(h_n, (n, 0, 5))
o = H.doit(deep=False)
b = FixedBosonicBasis(2, 6)
m = matrix_rep(o, b)
# We need to double check these energy values to make sure that they
# are correct and have the proper degeneracies!
diag = [1, 2, 3, 3, 4, 5, 4, 5, 6, 7, 5, 6, 7, 8, 9, 6, 7, 8, 9, 10, 11]
for i in range(len(diag)):
assert diag[i] == m[i, i]
def test_issue_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, -n + 1 >= 0), Or(-n + 1 < 0, Ne(p / (p - 1), 1)),
p * Abs(1 / (p - 1)) <= 1)),
(Sum(k * p**k * (-p + 1)**(-k) * (-p + 1)**n * binomial(n, k),
(k, 0, n)), True))
def test_sho():
n, m = symbols('n m')
h_n = Bd(n)*B(n)*(n + Rational(1, 2))
H = Sum(h_n, (n, 0, 5))
o = H.doit(deep = False)
b = FixedBosonicBasis(2, 6)
m = matrix_rep(o, b)
# We need to double check these energy values to make sure that they
# are correct and have the proper degeneracies!
diag = [1, 2, 3, 3, 4, 5, 4, 5, 6, 7, 5, 6, 7, 8, 9, 6, 7, 8, 9, 10, 11]
for i in range(len(diag)):
assert diag[i] == m[i, i]
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(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_issue_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, p/Abs(p - 1) <= 1),
((-p + 1)**n*Sum(k*p**k*(-p + 1)**(-k)*binomial(n, k), (k, 0, n)),
True))
# Issue #17165: make sure that another simplify does not change/increase
# the result
assert res == res.simplify()
def _marginalise(self, expr, rv, evaluate):
if isinstance(rv.pspace.distribution, SingleFiniteDistribution):
rv_dens = rv.pspace.distribution.pmf(rv)
else:
rv_dens = rv.pspace.distribution.pdf(rv)
rv_dom = rv.pspace.domain.set
if rv.pspace.is_Discrete or rv.pspace.is_Finite:
expr = Sum(expr * rv_dens, (rv, rv_dom._inf, rv_dom._sup))
else:
expr = Integral(expr * rv_dens, (rv, rv_dom._inf, rv_dom._sup))
if evaluate:
return expr.doit()
return expr
def test_Identity():
A = MatrixSymbol('A', n, m)
i, j = symbols('i j')
In = Identity(n)
Im = Identity(m)
assert A*Im == A
assert In*A == A
assert transpose(In) == In
assert In.inverse() == In
assert In.conjugate() == In
assert In[i, j] != 0
assert Sum(In[i, j], (i, 0, n-1), (j, 0, n-1)).subs(n,3).doit() == 3
assert Sum(Sum(In[i, j], (i, 0, n-1)), (j, 0, n-1)).subs(n,3).doit() == 3
# If range exceeds the limit `(0, n-1)`, do not remove `Piecewise`:
expr = Sum(In[i, j], (i, 0, n-1))
assert expr.doit() == 1
expr = Sum(In[i, j], (i, 0, n-2))
assert expr.doit().dummy_eq(
Piecewise(
(1, (j >= 0) & (j <= n-2)),
(0, True)
)
)
expr = Sum(In[i, j], (i, 1, n-1))
assert expr.doit().dummy_eq(
Piecewise(
(1, (j >= 1) & (j <= n-1)),
(0, True)
)
)
assert Identity(3).as_explicit() == ImmutableMatrix.eye(3)
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()
# Integer assumes finite
assert Sum(KroneckerDelta(x, y), (x, -oo, oo)).doit() == Piecewise(
(1, And(-oo <= y, y < oo)), (0, True))
assert Sum(KroneckerDelta(m, n), (m, -oo, oo)).doit() == 1
assert Sum(m * KroneckerDelta(x, y), (x, -oo, oo)).doit() == Piecewise(
(m, And(-oo <= y, y < oo)), (0, True))
assert Sum(x * KroneckerDelta(m, n), (m, -oo, oo)).doit() == x
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(S(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(f(n), (n, 1, oo))
# issue 2597
nmax = symbols('N', integer=True, positive=True)
pw = Piecewise((1, And(S(1) <= n, n <= nmax)), (0, True))
assert Sum(pw, (n, 1, nmax)).doit() == Sum(
Piecewise((1, nmax >= n), (0, True)), (n, 1, nmax))
q, s = symbols('q, s')
assert summation(1 / n**(2 * s), (n, 1, oo)) == Piecewise(
(zeta(2 * s), 2 * s > 1), (Sum(n**(-2 * s), (n, 1, oo)), True))
assert summation(1 / (n + 1)**s, (n, 0, oo)) == Piecewise(
(zeta(s), s > 1), (Sum((n + 1)**(-s), (n, 0, oo)), True))
assert summation(1 / (n + q)**s, (n, 0, oo)) == Piecewise(
(zeta(s, q), And(q > 0, s > 1)), (Sum((n + q)**(-s),
(n, 0, oo)), True))
assert summation(1 / (n + q)**s, (n, q, oo)) == Piecewise(
(zeta(s, 2 * q), And(2 * q > 0, s > 1)), (Sum((n + q)**(-s),
(n, q, oo)), True))
assert summation(1 / n**2, (n, 1, oo)) == zeta(2)
assert summation(1 / n**s, (n, 0, oo)) == Sum(n**(-s), (n, 0, 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(KroneckerDelta(m, n), (m, -oo, oo)).doit() == Piecewise((1, And(-oo < n, n < oo)), (0, True))
assert Sum(x*KroneckerDelta(m, n), (m, -oo, oo)).doit() == Piecewise((x, And(-oo < n, n < oo)), (0, True))
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(S(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 2597
nmax = symbols('N', integer=True, positive=True)
pw = Piecewise((1, And(S(1) <= n, n <= nmax)), (0, True))
assert Sum(pw, (n, 1, nmax)).doit() == Sum(pw, (n, 1, nmax))
q, s = symbols('q, s')
assert summation(1/n**(2*s), (n, 1, oo)) == Piecewise((zeta(2*s), 2*s > 1),
(Sum(n**(-2*s), (n, 1, oo)), True))
assert summation(1/(n+1)**s, (n, 0, oo)) == Piecewise((zeta(s), s > 1),
(Sum((n + 1)**(-s), (n, 0, oo)), True))
assert summation(1/(n+q)**s, (n, 0, oo)) == Piecewise(
(zeta(s, q), And(q > 0, s > 1)),
(Sum((n + q)**(-s), (n, 0, oo)), True))
assert summation(1/(n+q)**s, (n, q, oo)) == Piecewise(
(zeta(s, 2*q), And(2*q > 0, s > 1)),
(Sum((n + q)**(-s), (n, q, oo)), True))
assert summation(1/n**2, (n, 1, oo)) == zeta(2)
assert summation(1/n**s, (n, 0, oo)) == Sum(n**(-s), (n, 0, oo))
def test_Sum_doit():
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(S(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), n >= 0), (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))
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, sympy 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.n(3) == s.doit().n(3)
S k 1 n = sum i**k {i=0->n} = 1/(k+1) (sum C(k+1, i) B[i] n**(k+1-i) {i} - B[k+1])
'''
import sympy
from sympy.abc import k, n, z, j, i
import sympy.abc
# ff = sympy.functions.combinatorial.factorials.FallingFactorial
# rf = RisingFactorial
from sympy import Sum, ff, rf, roots, binomial as C, \
symbols, factor, factorial, Product, floor, gcd_terms, gcdex
a, b, k, n = symbols('a b k n', positive=True, integer=True)
s = Sum(i**k * z**i, (i, 0, n-1))
ans = s.doit()
q = Sum(i**k * (1-z)**i, (i, 0, n-1))
u = s * (z-1)**(k+1) - n**k*z**n*(z-1)**k
'''
t = ans.euler_maclaurin()
(0**k/2 + z**(n - 1)*(n - 1)**k/2 + Integral(_x**k*z**_x, (_x, 0, n - 1)),
Abs(-oo*0**k*k - 0**k*log(z)/12 + k*z**(n - 1)*(n - 1)**k/(12*(n - 1)) + z**(n - 1)*(n - 1)**k*log(z)/12))
'''
if 0:
t = lambda _k = 0: ans.subs(k, _k).doit()
print(ans)
for i in range(5):
print('S(k={k}, z, n) = {s}'.format(k=i, s=t(i)))
elif 0:
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, sympy 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.n(3) == s.doit().n(3)
def test_BernoulliProcess():
B = BernoulliProcess("B", p=0.6, success=1, failure=0)
assert B.state_space == FiniteSet(0, 1)
assert B.index_set == S.Naturals0
assert B.success == 1
assert B.failure == 0
X = BernoulliProcess("X", p=Rational(1, 3), success='H', failure='T')
assert X.state_space == FiniteSet('H', 'T')
H, T = symbols("H,T")
assert E(X[1] + X[2] * X[3]
) == H**2 / 9 + 4 * H * T / 9 + H / 3 + 4 * T**2 / 9 + 2 * T / 3
t, x = symbols('t, x', positive=True, integer=True)
assert isinstance(B[t], RandomIndexedSymbol)
raises(ValueError,
lambda: BernoulliProcess("X", p=1.1, success=1, failure=0))
raises(NotImplementedError, lambda: B(t))
raises(IndexError, lambda: B[-3])
assert B.joint_distribution(B[3], B[9]) == JointDistributionHandmade(
Lambda(
(B[3], B[9]),
Piecewise((0.6, Eq(B[3], 1)), (0.4, Eq(B[3], 0)),
(0, True)) * Piecewise((0.6, Eq(B[9], 1)),
(0.4, Eq(B[9], 0)), (0, True))))
assert B.joint_distribution(2, B[4]) == JointDistributionHandmade(
Lambda(
(B[2], B[4]),
Piecewise((0.6, Eq(B[2], 1)), (0.4, Eq(B[2], 0)),
(0, True)) * Piecewise((0.6, Eq(B[4], 1)),
(0.4, Eq(B[4], 0)), (0, True))))
# Test for the sum distribution of Bernoulli Process RVs
Y = B[1] + B[2] + B[3]
assert P(Eq(Y, 0)).round(2) == Float(0.06, 1)
assert P(Eq(Y, 2)).round(2) == Float(0.43, 2)
assert P(Eq(Y, 4)).round(2) == 0
assert P(Gt(Y, 1)).round(2) == Float(0.65, 2)
# Test for independency of each Random Indexed variable
assert P(Eq(B[1], 0) & Eq(B[2], 1) & Eq(B[3], 0)
& Eq(B[4], 1)).round(2) == Float(0.06, 1)
assert E(2 * B[1] + B[2]).round(2) == Float(1.80, 3)
assert E(2 * B[1] + B[2] + 5).round(2) == Float(6.80, 3)
assert E(B[2] * B[4] + B[10]).round(2) == Float(0.96, 2)
assert E(B[2] > 0, Eq(B[1], 1) & Eq(B[2], 1)).round(2) == Float(0.60, 2)
assert E(B[1]) == 0.6
assert P(B[1] > 0).round(2) == Float(0.60, 2)
assert P(B[1] < 1).round(2) == Float(0.40, 2)
assert P(B[1] > 0, B[2] <= 1).round(2) == Float(0.60, 2)
assert P(B[12] * B[5] > 0).round(2) == Float(0.36, 2)
assert P(B[12] * B[5] > 0, B[4] < 1).round(2) == Float(0.36, 2)
assert P(Eq(B[2], 1), B[2] > 0) == 1
assert P(Eq(B[5], 3)) == 0
assert P(Eq(B[1], 1), B[1] < 0) == 0
assert P(B[2] > 0, Eq(B[2], 1)) == 1
assert P(B[2] < 0, Eq(B[2], 1)) == 0
assert P(B[2] > 0, B[2] == 7) == 0
assert P(B[5] > 0, B[5]) == BernoulliDistribution(0.6, 0, 1)
raises(ValueError, lambda: P(3))
raises(ValueError, lambda: P(B[3] > 0, 3))
# test issue 19456
expr = Sum(B[t], (t, 0, 4))
expr2 = Sum(B[t], (t, 1, 3))
expr3 = Sum(B[t]**2, (t, 1, 3))
assert expr.doit() == B[0] + B[1] + B[2] + B[3] + B[4]
assert expr2.doit() == Y
assert expr3.doit() == B[1]**2 + B[2]**2 + B[3]**2
assert B[2 * t].free_symbols == {B[2 * t], t}
assert B[4].free_symbols == {B[4]}
assert B[x * t].free_symbols == {B[x * t], x, t}
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 sum_progression(a_n, end):
r'''\sum^{end}_{n=1} a_n
'''
expr = Sum(a_n, (n, 1, end))
return expr.doit()
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 test_issue_8016():
k = Symbol('k', integer=True)
n, m = symbols('n, m', integer=True, positive=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)
