Пример #1
0
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()
Пример #2
0
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()
Пример #3
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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)
Пример #4
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)