本文整理汇总了Python中sympy.besseli函数的典型用法代码示例。如果您正苦于以下问题:Python besseli函数的具体用法?Python besseli怎么用?Python besseli使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了besseli函数的20个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Python代码示例。
示例1: test_diff
def test_diff():
assert besselj(n, z).diff(z) == besselj(n - 1, z)/2 - besselj(n + 1, z)/2
assert bessely(n, z).diff(z) == bessely(n - 1, z)/2 - bessely(n + 1, z)/2
assert besseli(n, z).diff(z) == besseli(n - 1, z)/2 + besseli(n + 1, z)/2
assert besselk(n, z).diff(z) == -besselk(n - 1, z)/2 - besselk(n + 1, z)/2
assert hankel1(n, z).diff(z) == hankel1(n - 1, z)/2 - hankel1(n + 1, z)/2
assert hankel2(n, z).diff(z) == hankel2(n - 1, z)/2 - hankel2(n + 1, z)/2
开发者ID:Abhityagi16,项目名称:sympy,代码行数:7,代码来源:test_bessel.py
示例2: test_airybiprime
def test_airybiprime():
z = Symbol('z', real=False)
t = Symbol('t', negative=True)
p = Symbol('p', positive=True)
assert isinstance(airybiprime(z), airybiprime)
assert airybiprime(0) == 3**(S(1)/6)/gamma(S(1)/3)
assert airybiprime(oo) == oo
assert airybiprime(-oo) == 0
assert diff(airybiprime(z), z) == z*airybi(z)
assert series(airybiprime(z), z, 0, 3) == (
3**(S(1)/6)/gamma(S(1)/3) + 3**(S(5)/6)*z**2/(6*gamma(S(2)/3)) + O(z**3))
assert airybiprime(z).rewrite(hyper) == (
3**(S(5)/6)*z**2*hyper((), (S(5)/3,), z**S(3)/9)/(6*gamma(S(2)/3)) +
3**(S(1)/6)*hyper((), (S(1)/3,), z**S(3)/9)/gamma(S(1)/3))
assert isinstance(airybiprime(z).rewrite(besselj), airybiprime)
assert airyai(t).rewrite(besselj) == (
sqrt(-t)*(besselj(-S(1)/3, 2*(-t)**(S(3)/2)/3) +
besselj(S(1)/3, 2*(-t)**(S(3)/2)/3))/3)
assert airybiprime(z).rewrite(besseli) == (
sqrt(3)*(z**2*besseli(S(2)/3, 2*z**(S(3)/2)/3)/(z**(S(3)/2))**(S(2)/3) +
(z**(S(3)/2))**(S(2)/3)*besseli(-S(2)/3, 2*z**(S(3)/2)/3))/3)
assert airybiprime(p).rewrite(besseli) == (
sqrt(3)*p*(besseli(-S(2)/3, 2*p**(S(3)/2)/3) + besseli(S(2)/3, 2*p**(S(3)/2)/3))/3)
assert expand_func(airybiprime(2*(3*z**5)**(S(1)/3))) == (
sqrt(3)*(z**(S(5)/3)/(z**5)**(S(1)/3) - 1)*airyaiprime(2*3**(S(1)/3)*z**(S(5)/3))/2 +
(z**(S(5)/3)/(z**5)**(S(1)/3) + 1)*airybiprime(2*3**(S(1)/3)*z**(S(5)/3))/2)
开发者ID:KonstantinTogoi,项目名称:sympy,代码行数:33,代码来源:test_bessel.py
示例3: test_airyai
def test_airyai():
z = Symbol('z', real=False)
t = Symbol('t', negative=True)
p = Symbol('p', positive=True)
assert isinstance(airyai(z), airyai)
assert airyai(0) == 3**(S(1)/3)/(3*gamma(S(2)/3))
assert airyai(oo) == 0
assert airyai(-oo) == 0
assert diff(airyai(z), z) == airyaiprime(z)
assert series(airyai(z), z, 0, 3) == (
3**(S(5)/6)*gamma(S(1)/3)/(6*pi) - 3**(S(1)/6)*z*gamma(S(2)/3)/(2*pi) + O(z**3))
assert airyai(z).rewrite(hyper) == (
-3**(S(2)/3)*z*hyper((), (S(4)/3,), z**S(3)/9)/(3*gamma(S(1)/3)) +
3**(S(1)/3)*hyper((), (S(2)/3,), z**S(3)/9)/(3*gamma(S(2)/3)))
assert isinstance(airyai(z).rewrite(besselj), airyai)
assert airyai(t).rewrite(besselj) == (
sqrt(-t)*(besselj(-S(1)/3, 2*(-t)**(S(3)/2)/3) +
besselj(S(1)/3, 2*(-t)**(S(3)/2)/3))/3)
assert airyai(z).rewrite(besseli) == (
-z*besseli(S(1)/3, 2*z**(S(3)/2)/3)/(3*(z**(S(3)/2))**(S(1)/3)) +
(z**(S(3)/2))**(S(1)/3)*besseli(-S(1)/3, 2*z**(S(3)/2)/3)/3)
assert airyai(p).rewrite(besseli) == (
sqrt(p)*(besseli(-S(1)/3, 2*p**(S(3)/2)/3) -
besseli(S(1)/3, 2*p**(S(3)/2)/3))/3)
assert expand_func(airyai(2*(3*z**5)**(S(1)/3))) == (
-sqrt(3)*(-1 + (z**5)**(S(1)/3)/z**(S(5)/3))*airybi(2*3**(S(1)/3)*z**(S(5)/3))/6 +
(1 + (z**5)**(S(1)/3)/z**(S(5)/3))*airyai(2*3**(S(1)/3)*z**(S(5)/3))/2)
开发者ID:KonstantinTogoi,项目名称:sympy,代码行数:34,代码来源:test_bessel.py
示例4: test_hyperexpand_bases
def test_hyperexpand_bases():
assert (
hyperexpand(hyper([2], [a], z))
== a + z ** (-a + 1) * (-a ** 2 + 3 * a + z * (a - 1) - 2) * exp(z) * lowergamma(a - 1, z) - 1
)
# TODO [a+1, a-S.Half], [2*a]
assert hyperexpand(hyper([1, 2], [3], z)) == -2 / z - 2 * log(exp_polar(-I * pi) * z + 1) / z ** 2
assert hyperexpand(hyper([S.Half, 2], [S(3) / 2], z)) == -1 / (2 * z - 2) + log((sqrt(z) + 1) / (-sqrt(z) + 1)) / (
4 * sqrt(z)
)
assert hyperexpand(hyper([S(1) / 2, S(1) / 2], [S(5) / 2], z)) == (-3 * z + 3) / 4 / (z * sqrt(-z + 1)) + (
6 * z - 3
) * asin(sqrt(z)) / (4 * z ** (S(3) / 2))
assert hyperexpand(hyper([1, 2], [S(3) / 2], z)) == -1 / (2 * z - 2) - asin(sqrt(z)) / (
sqrt(z) * (2 * z - 2) * sqrt(-z + 1)
)
assert hyperexpand(hyper([-S.Half - 1, 1, 2], [S.Half, 3], z)) == sqrt(z) * (6 * z / 7 - S(6) / 5) * atanh(
sqrt(z)
) + (-30 * z ** 2 + 32 * z - 6) / 35 / z - 6 * log(-z + 1) / (35 * z ** 2)
assert hyperexpand(hyper([1 + S.Half, 1, 1], [2, 2], z)) == -4 * log(sqrt(-z + 1) / 2 + S(1) / 2) / z
# TODO hyperexpand(hyper([a], [2*a + 1], z))
# TODO [S.Half, a], [S(3)/2, a+1]
assert hyperexpand(hyper([2], [b, 1], z)) == z ** (-b / 2 + S(1) / 2) * besseli(b - 1, 2 * sqrt(z)) * gamma(
b
) + z ** (-b / 2 + 1) * besseli(b, 2 * sqrt(z)) * gamma(b)
开发者ID:kendhia,项目名称:sympy,代码行数:25,代码来源:test_hyperexpand.py
示例5: test_rewrite
def test_rewrite():
from sympy import polar_lift, exp, I
assert besselj(n, z).rewrite(jn) == sqrt(2*z/pi)*jn(n - S(1)/2, z)
assert bessely(n, z).rewrite(yn) == sqrt(2*z/pi)*yn(n - S(1)/2, z)
assert besseli(n, z).rewrite(besselj) == \
exp(-I*n*pi/2)*besselj(n, polar_lift(I)*z)
assert besselj(n, z).rewrite(besseli) == \
exp(I*n*pi/2)*besseli(n, polar_lift(-I)*z)
nu = randcplx()
assert tn(besselj(nu, z), besselj(nu, z).rewrite(besseli), z)
assert tn(besseli(nu, z), besseli(nu, z).rewrite(besselj), z)
开发者ID:Abhityagi16,项目名称:sympy,代码行数:11,代码来源:test_bessel.py
示例6: test_bessel_eval
def test_bessel_eval():
from sympy import I
assert besselj(-4, z) == besselj(4, z)
assert besselj(-3, z) == -besselj(3, z)
assert bessely(-2, z) == bessely(2, z)
assert bessely(-1, z) == -bessely(1, z)
assert besselj(0, -z) == besselj(0, z)
assert besselj(1, -z) == -besselj(1, z)
assert besseli(0, I*z) == besselj(0, z)
assert besseli(1, I*z) == I*besselj(1, z)
assert besselj(3, I*z) == -I*besseli(3, z)
开发者ID:Abhityagi16,项目名称:sympy,代码行数:14,代码来源:test_bessel.py
示例7: test_branching
def test_branching():
from sympy import exp_polar, polar_lift, Symbol, I, exp
assert besselj(polar_lift(k), x) == besselj(k, x)
assert besseli(polar_lift(k), x) == besseli(k, x)
n = Symbol('n', integer=True)
assert besselj(n, exp_polar(2*pi*I)*x) == besselj(n, x)
assert besselj(n, polar_lift(x)) == besselj(n, x)
assert besseli(n, exp_polar(2*pi*I)*x) == besseli(n, x)
assert besseli(n, polar_lift(x)) == besseli(n, x)
def tn(func, s):
from random import uniform
c = uniform(1, 5)
expr = func(s, c*exp_polar(I*pi)) - func(s, c*exp_polar(-I*pi))
eps = 1e-15
expr2 = func(s + eps, -c + eps*I) - func(s + eps, -c - eps*I)
return abs(expr.n() - expr2.n()).n() < 1e-10
nu = Symbol('nu')
assert besselj(nu, exp_polar(2*pi*I)*x) == exp(2*pi*I*nu)*besselj(nu, x)
assert besseli(nu, exp_polar(2*pi*I)*x) == exp(2*pi*I*nu)*besseli(nu, x)
assert tn(besselj, 2)
assert tn(besselj, pi)
assert tn(besselj, I)
assert tn(besseli, 2)
assert tn(besseli, pi)
assert tn(besseli, I)
开发者ID:Abhityagi16,项目名称:sympy,代码行数:28,代码来源:test_bessel.py
示例8: test_bessel
def test_bessel():
from sympy import (besselj, Heaviside, besseli, polar_lift, exp_polar,
powdenest)
assert simplify(integrate(besselj(a, z)*besselj(b, z)/z, (z, 0, oo),
meijerg=True, conds='none')) == \
2*sin(pi*a/2 - pi*b/2)/(pi*(a - b)*(a + b))
assert simplify(integrate(besselj(a, z)*besselj(a, z)/z, (z, 0, oo),
meijerg=True, conds='none')) == 1/(2*a)
# TODO more orthogonality integrals
# TODO there is some improvement possible here:
# - the result can be simplified to besselj(y, z))
assert powdenest(simplify(integrate(sin(z*x)*(x**2-1)**(-(y+S(1)/2)),
(x, 1, oo), meijerg=True, conds='none')
*2/((z/2)**y*sqrt(pi)*gamma(S(1)/2-y))),
polar=True) == \
exp(-I*pi*y/2)*besseli(y, z*exp_polar(I*pi/2))
# Werner Rosenheinrich
# SOME INDEFINITE INTEGRALS OF BESSEL FUNCTIONS
assert integrate(x*besselj(0, x), x, meijerg=True) == x*besselj(1, x)
assert integrate(x*besseli(0, x), x, meijerg=True) == x*besseli(1, x)
# TODO can do higher powers, but come out as high order ... should they be
# reduced to order 0, 1?
assert integrate(besselj(1, x), x, meijerg=True) == -besselj(0, x)
assert integrate(besselj(1, x)**2/x, x, meijerg=True) == \
-(besselj(0, x)**2 + besselj(1, x)**2)/2
# TODO more besseli when tables are extended or recursive mellin works
assert integrate(besselj(0, x)**2/x**2, x, meijerg=True) == \
-2*x*besselj(0, x)**2 - 2*x*besselj(1, x)**2 \
+ 2*besselj(0, x)*besselj(1, x) - besselj(0, x)**2/x
assert integrate(besselj(0, x)*besselj(1, x), x, meijerg=True) == \
-besselj(0, x)**2/2
assert integrate(x**2*besselj(0, x)*besselj(1, x), x, meijerg=True) == \
x**2*besselj(1, x)**2/2
assert integrate(besselj(0, x)*besselj(1, x)/x, x, meijerg=True) == \
(x*besselj(0, x)**2 + x*besselj(1, x)**2 - \
besselj(0, x)*besselj(1, x))
# TODO how does besselj(0, a*x)*besselj(0, b*x) work?
# TODO how does besselj(0, x)**2*besselj(1, x)**2 work?
# TODO sin(x)*besselj(0, x) etc come out a mess
# TODO can x*log(x)*besselj(0, x) be done?
# TODO how does besselj(1, x)*besselj(0, x+a) work?
# TODO more indefinite integrals when struve functions etc are implemented
# test a substitution
assert integrate(besselj(1, x**2)*x, x, meijerg=True) == \
-besselj(0, x**2)/2
开发者ID:amakelov,项目名称:sympy,代码行数:50,代码来源:test_meijerint.py
示例9: test_chi_noncentral
def test_chi_noncentral():
k = Symbol("k", integer=True)
l = Symbol("l")
X = ChiNoncentral("x", k, l)
assert density(X)(x) == (x**k*l*(x*l)**(-k/2)*
exp(-x**2/2 - l**2/2)*besseli(k/2 - 1, x*l))
开发者ID:vprusso,项目名称:sympy,代码行数:7,代码来源:test_continuous_rv.py
示例10: test_meijerg_eval
def test_meijerg_eval():
from sympy import besseli, exp_polar
from sympy.abc import l
a = randcplx()
arg = x*exp_polar(k*pi*I)
expr1 = pi*meijerg([[], [(a + 1)/2]], [[a/2], [-a/2, (a + 1)/2]], arg**2/4)
expr2 = besseli(a, arg)
# Test that the two expressions agree for all arguments.
for x_ in [0.5, 1.5]:
for k_ in [0.0, 0.1, 0.3, 0.5, 0.8, 1, 5.751, 15.3]:
assert abs((expr1 - expr2).n(subs={x: x_, k: k_})) < 1e-10
assert abs((expr1 - expr2).n(subs={x: x_, k: -k_})) < 1e-10
# Test continuity independently
eps = 1e-13
expr2 = expr1.subs(k, l)
for x_ in [0.5, 1.5]:
for k_ in [0.5, S(1)/3, 0.25, 0.75, S(2)/3, 1.0, 1.5]:
assert abs((expr1 - expr2).n(
subs={x: x_, k: k_ + eps, l: k_ - eps})) < 1e-10
assert abs((expr1 - expr2).n(
subs={x: x_, k: -k_ + eps, l: -k_ - eps})) < 1e-10
expr = (meijerg(((0.5,), ()), ((0.5, 0, 0.5), ()), exp_polar(-I*pi)/4)
+ meijerg(((0.5,), ()), ((0.5, 0, 0.5), ()), exp_polar(I*pi)/4)) \
/(2*sqrt(pi))
assert (expr - pi/exp(1)).n(chop=True) == 0
开发者ID:KonstantinTogoi,项目名称:sympy,代码行数:28,代码来源:test_hyper.py
示例11: test_bessel
def test_bessel():
from sympy import besselj, besseli
assert simplify(integrate(besselj(a, z) * besselj(b, z) / z, (z, 0, oo), meijerg=True, conds="none")) == 2 * sin(
pi * (a / 2 - b / 2)
) / (pi * (a - b) * (a + b))
assert simplify(integrate(besselj(a, z) * besselj(a, z) / z, (z, 0, oo), meijerg=True, conds="none")) == 1 / (2 * a)
# TODO more orthogonality integrals
assert simplify(
integrate(sin(z * x) * (x ** 2 - 1) ** (-(y + S(1) / 2)), (x, 1, oo), meijerg=True, conds="none")
* 2
/ ((z / 2) ** y * sqrt(pi) * gamma(S(1) / 2 - y))
) == besselj(y, z)
# Werner Rosenheinrich
# SOME INDEFINITE INTEGRALS OF BESSEL FUNCTIONS
assert integrate(x * besselj(0, x), x, meijerg=True) == x * besselj(1, x)
assert integrate(x * besseli(0, x), x, meijerg=True) == x * besseli(1, x)
# TODO can do higher powers, but come out as high order ... should they be
# reduced to order 0, 1?
assert integrate(besselj(1, x), x, meijerg=True) == -besselj(0, x)
assert integrate(besselj(1, x) ** 2 / x, x, meijerg=True) == -(besselj(0, x) ** 2 + besselj(1, x) ** 2) / 2
# TODO more besseli when tables are extended or recursive mellin works
assert (
integrate(besselj(0, x) ** 2 / x ** 2, x, meijerg=True)
== -2 * x * besselj(0, x) ** 2
- 2 * x * besselj(1, x) ** 2
+ 2 * besselj(0, x) * besselj(1, x)
- besselj(0, x) ** 2 / x
)
assert integrate(besselj(0, x) * besselj(1, x), x, meijerg=True) == -besselj(0, x) ** 2 / 2
assert integrate(x ** 2 * besselj(0, x) * besselj(1, x), x, meijerg=True) == x ** 2 * besselj(1, x) ** 2 / 2
assert integrate(besselj(0, x) * besselj(1, x) / x, x, meijerg=True) == (
x * besselj(0, x) ** 2 + x * besselj(1, x) ** 2 - besselj(0, x) * besselj(1, x)
)
# TODO how does besselj(0, a*x)*besselj(0, b*x) work?
# TODO how does besselj(0, x)**2*besselj(1, x)**2 work?
# TODO sin(x)*besselj(0, x) etc come out a mess
# TODO can x*log(x)*besselj(0, x) be done?
# TODO how does besselj(1, x)*besselj(0, x+a) work?
# TODO more indefinite integrals when struve functions etc are implemented
# test a substitution
assert integrate(besselj(1, x ** 2) * x, x, meijerg=True) == -besselj(0, x ** 2) / 2
开发者ID:Carreau,项目名称:sympy,代码行数:47,代码来源:test_meijerint.py
示例12: test_bessel_rand
def test_bessel_rand():
assert td(besselj(randcplx(), z), z)
assert td(bessely(randcplx(), z), z)
assert td(besseli(randcplx(), z), z)
assert td(besselk(randcplx(), z), z)
assert td(hankel1(randcplx(), z), z)
assert td(hankel2(randcplx(), z), z)
assert td(jn(randcplx(), z), z)
assert td(yn(randcplx(), z), z)
开发者ID:Abhityagi16,项目名称:sympy,代码行数:9,代码来源:test_bessel.py
示例13: test_rewrite
def test_rewrite():
from sympy import polar_lift, exp, I
assert besselj(n, z).rewrite(jn) == sqrt(2*z/pi)*jn(n - S(1)/2, z)
assert bessely(n, z).rewrite(yn) == sqrt(2*z/pi)*yn(n - S(1)/2, z)
assert besseli(n, z).rewrite(besselj) == \
exp(-I*n*pi/2)*besselj(n, polar_lift(I)*z)
assert besselj(n, z).rewrite(besseli) == \
exp(I*n*pi/2)*besseli(n, polar_lift(-I)*z)
nu = randcplx()
assert tn(besselj(nu, z), besselj(nu, z).rewrite(besseli), z)
assert tn(besselj(nu, z), besselj(nu, z).rewrite(bessely), z)
assert tn(besseli(nu, z), besseli(nu, z).rewrite(besselj), z)
assert tn(besseli(nu, z), besseli(nu, z).rewrite(bessely), z)
assert tn(bessely(nu, z), bessely(nu, z).rewrite(besselj), z)
assert tn(bessely(nu, z), bessely(nu, z).rewrite(besseli), z)
assert tn(besselk(nu, z), besselk(nu, z).rewrite(besselj), z)
assert tn(besselk(nu, z), besselk(nu, z).rewrite(besseli), z)
assert tn(besselk(nu, z), besselk(nu, z).rewrite(bessely), z)
# check that a rewrite was triggered, when the order is set to a generic
# symbol 'nu'
assert yn(nu, z) != yn(nu, z).rewrite(jn)
assert hn1(nu, z) != hn1(nu, z).rewrite(jn)
assert hn2(nu, z) != hn2(nu, z).rewrite(jn)
assert jn(nu, z) != jn(nu, z).rewrite(yn)
assert hn1(nu, z) != hn1(nu, z).rewrite(yn)
assert hn2(nu, z) != hn2(nu, z).rewrite(yn)
# rewriting spherical bessel functions (SBFs) w.r.t. besselj, bessely is
# not allowed if a generic symbol 'nu' is used as the order of the SBFs
# to avoid inconsistencies (the order of bessel[jy] is allowed to be
# complex-valued, whereas SBFs are defined only for integer orders)
order = nu
for f in (besselj, bessely):
assert hn1(order, z) == hn1(order, z).rewrite(f)
assert hn2(order, z) == hn2(order, z).rewrite(f)
assert jn(order, z).rewrite(besselj) == sqrt(2)*sqrt(pi)*sqrt(1/z)*besselj(order + S(1)/2, z)/2
assert jn(order, z).rewrite(bessely) == (-1)**nu*sqrt(2)*sqrt(pi)*sqrt(1/z)*bessely(-order - S(1)/2, z)/2
# for integral orders rewriting SBFs w.r.t bessel[jy] is allowed
N = Symbol('n', integer=True)
ri = randint(-11, 10)
for order in (ri, N):
for f in (besselj, bessely):
assert yn(order, z) != yn(order, z).rewrite(f)
assert jn(order, z) != jn(order, z).rewrite(f)
assert hn1(order, z) != hn1(order, z).rewrite(f)
assert hn2(order, z) != hn2(order, z).rewrite(f)
for func, refunc in product((yn, jn, hn1, hn2),
(jn, yn, besselj, bessely)):
assert tn(func(ri, z), func(ri, z).rewrite(refunc), z)
开发者ID:KonstantinTogoi,项目名称:sympy,代码行数:59,代码来源:test_bessel.py
示例14: test_bessel_eval
def test_bessel_eval():
from sympy import I, Symbol
n, m, k = Symbol('n', integer=True), Symbol('m'), Symbol('k', integer=True, zero=False)
for f in [besselj, besseli]:
assert f(0, 0) == S.One
assert f(2.1, 0) == S.Zero
assert f(-3, 0) == S.Zero
assert f(-10.2, 0) == S.ComplexInfinity
assert f(1 + 3*I, 0) == S.Zero
assert f(-3 + I, 0) == S.ComplexInfinity
assert f(-2*I, 0) == S.NaN
assert f(n, 0) != S.One and f(n, 0) != S.Zero
assert f(m, 0) != S.One and f(m, 0) != S.Zero
assert f(k, 0) == S.Zero
assert bessely(0, 0) == S.NegativeInfinity
assert besselk(0, 0) == S.Infinity
for f in [bessely, besselk]:
assert f(1 + I, 0) == S.ComplexInfinity
assert f(I, 0) == S.NaN
for f in [besselj, bessely]:
assert f(m, S.Infinity) == S.Zero
assert f(m, S.NegativeInfinity) == S.Zero
for f in [besseli, besselk]:
assert f(m, I*S.Infinity) == S.Zero
assert f(m, I*S.NegativeInfinity) == S.Zero
for f in [besseli, besselk]:
assert f(-4, z) == f(4, z)
assert f(-3, z) == f(3, z)
assert f(-n, z) == f(n, z)
assert f(-m, z) != f(m, z)
for f in [besselj, bessely]:
assert f(-4, z) == f(4, z)
assert f(-3, z) == -f(3, z)
assert f(-n, z) == (-1)**n*f(n, z)
assert f(-m, z) != (-1)**m*f(m, z)
for f in [besselj, besseli]:
assert f(m, -z) == (-z)**m*z**(-m)*f(m, z)
assert besseli(2, -z) == besseli(2, z)
assert besseli(3, -z) == -besseli(3, z)
assert besselj(0, -z) == besselj(0, z)
assert besselj(1, -z) == -besselj(1, z)
assert besseli(0, I*z) == besselj(0, z)
assert besseli(1, I*z) == I*besselj(1, z)
assert besselj(3, I*z) == -I*besseli(3, z)
开发者ID:KonstantinTogoi,项目名称:sympy,代码行数:54,代码来源:test_bessel.py
示例15: test_hyperexpand_bases
def test_hyperexpand_bases():
assert hyperexpand(hyper([2], [a], z)) == \
a + z**(-a + 1)*(-a**2 + 3*a + z*(a - 1) - 2)*exp(z)*lowergamma(a - 1, z) - 1
# TODO [a+1, a-S.Half], [2*a]
assert hyperexpand(hyper([1, 2], [3], z)) == -2/z - 2*log(-z + 1)/z**2
assert hyperexpand(hyper([S.Half, 2], [S(3)/2], z)) == \
-1/(2*z - 2) + log((z**(S(1)/2) + 1)/(-z**(S(1)/2) + 1))/(4*z**(S(1)/2))
assert hyperexpand(hyper([S(1)/2, S(1)/2], [S(5)/2], z)) == \
(-3*z + 3)/(4*z*(-z + 1)**(S(1)/2)) \
+ (6*z - 3)*asin(z**(S(1)/2))/(4*z**(S(3)/2))
assert hyperexpand(hyper([1, 2], [S(3)/2], z)) == -1/(2*z - 2) \
- asin(z**(S(1)/2))/(z**(S(1)/2)*(2*z - 2)*(-z + 1)**(S(1)/2))
assert hyperexpand(hyper([-S.Half - 1, 1, 2], [S.Half, 3], z)) == \
z**(S(1)/2)*(6*z/7 - S(6)/5)*atanh(z**(S(1)/2)) \
+ (-30*z**2 + 32*z - 6)/(35*z) - 6*log(-z + 1)/(35*z**2)
assert hyperexpand(hyper([1+S.Half, 1, 1], [2, 2], z)) == \
-4*log((-z + 1)**(S(1)/2)/2 + S(1)/2)/z
# TODO hyperexpand(hyper([a], [2*a + 1], z))
# TODO [S.Half, a], [S(3)/2, a+1]
assert hyperexpand(hyper([2], [b, 1], z)) == \
z**(-b/2 + S(1)/2)*besseli(b - 1, 2*z**(S(1)/2))*gamma(b) \
+ z**(-b/2 + 1)*besseli(b, 2*z**(S(1)/2))*gamma(b)
开发者ID:TeddyBoomer,项目名称:wxgeometrie,代码行数:22,代码来源:test_hyperexpand.py
示例16: test_inverse_laplace_transform
def test_inverse_laplace_transform():
from sympy import sinh, cosh, besselj, besseli, simplify, factor_terms
ILT = inverse_laplace_transform
a, b, c, = symbols("a b c", positive=True, finite=True)
t = symbols("t")
def simp_hyp(expr):
return factor_terms(expand_mul(expr)).rewrite(sin)
# just test inverses of all of the above
assert ILT(1 / s, s, t) == Heaviside(t)
assert ILT(1 / s ** 2, s, t) == t * Heaviside(t)
assert ILT(1 / s ** 5, s, t) == t ** 4 * Heaviside(t) / 24
assert ILT(exp(-a * s) / s, s, t) == Heaviside(t - a)
assert ILT(exp(-a * s) / (s + b), s, t) == exp(b * (a - t)) * Heaviside(-a + t)
assert ILT(a / (s ** 2 + a ** 2), s, t) == sin(a * t) * Heaviside(t)
assert ILT(s / (s ** 2 + a ** 2), s, t) == cos(a * t) * Heaviside(t)
# TODO is there a way around simp_hyp?
assert simp_hyp(ILT(a / (s ** 2 - a ** 2), s, t)) == sinh(a * t) * Heaviside(t)
assert simp_hyp(ILT(s / (s ** 2 - a ** 2), s, t)) == cosh(a * t) * Heaviside(t)
assert ILT(a / ((s + b) ** 2 + a ** 2), s, t) == exp(-b * t) * sin(a * t) * Heaviside(t)
assert ILT((s + b) / ((s + b) ** 2 + a ** 2), s, t) == exp(-b * t) * cos(a * t) * Heaviside(t)
# TODO sinh/cosh shifted come out a mess. also delayed trig is a mess
# TODO should this simplify further?
assert ILT(exp(-a * s) / s ** b, s, t) == (t - a) ** (b - 1) * Heaviside(t - a) / gamma(b)
assert ILT(exp(-a * s) / sqrt(1 + s ** 2), s, t) == Heaviside(t - a) * besselj(
0, a - t
) # note: besselj(0, x) is even
# XXX ILT turns these branch factor into trig functions ...
assert simplify(
ILT(a ** b * (s + sqrt(s ** 2 - a ** 2)) ** (-b) / sqrt(s ** 2 - a ** 2), s, t).rewrite(exp)
) == Heaviside(t) * besseli(b, a * t)
assert ILT(a ** b * (s + sqrt(s ** 2 + a ** 2)) ** (-b) / sqrt(s ** 2 + a ** 2), s, t).rewrite(exp) == Heaviside(
t
) * besselj(b, a * t)
assert ILT(1 / (s * sqrt(s + 1)), s, t) == Heaviside(t) * erf(sqrt(t))
# TODO can we make erf(t) work?
assert ILT(1 / (s ** 2 * (s ** 2 + 1)), s, t) == (t - sin(t)) * Heaviside(t)
assert ILT((s * eye(2) - Matrix([[1, 0], [0, 2]])).inv(), s, t) == Matrix(
[[exp(t) * Heaviside(t), 0], [0, exp(2 * t) * Heaviside(t)]]
)
开发者ID:whimsy-Pan,项目名称:sympy,代码行数:47,代码来源:test_transforms.py
示例17: test_besselsimp
def test_besselsimp():
from sympy import besselj, besseli, besselk, bessely, jn, yn, exp_polar, cosh
assert besselsimp(exp(-I*pi*y/2)*besseli(y, z*exp_polar(I*pi/2))) == \
besselj(y, z)
assert besselsimp(exp(-I*pi*a/2)*besseli(a, 2*sqrt(x)*exp_polar(I*pi/2))) == \
besselj(a, 2*sqrt(x))
assert besselsimp(sqrt(2)*sqrt(pi)*x**(S(1)/4)*exp(I*pi/4)*exp(-I*pi*a/2) * \
besseli(-S(1)/2, sqrt(x)*exp_polar(I*pi/2)) * \
besseli(a, sqrt(x)*exp_polar(I*pi/2))/2) == \
besselj(a, sqrt(x)) * cos(sqrt(x))
assert besselsimp(besseli(S(-1)/2, z)) == sqrt(2)*cosh(z)/(sqrt(pi)*sqrt(z))
assert besselsimp(besseli(a, z*exp_polar(-I*pi/2))) == exp(-I*pi*a/2)*besselj(a, z)
开发者ID:marshall2389,项目名称:sympy,代码行数:12,代码来源:test_simplify.py
示例18: test_inverse_laplace_transform
def test_inverse_laplace_transform():
from sympy import (expand, sinh, cosh, besselj, besseli, exp_polar,
unpolarify, simplify)
ILT = inverse_laplace_transform
a, b, c, = symbols('a b c', positive=True)
t = symbols('t')
def simp_hyp(expr):
return expand(expand(expr).rewrite(sin))
# just test inverses of all of the above
assert ILT(1/s, s, t) == Heaviside(t)
assert ILT(1/s**2, s, t) == t*Heaviside(t)
assert ILT(1/s**5, s, t) == t**4*Heaviside(t)/24
assert ILT(exp(-a*s)/s, s, t) == Heaviside(t - a)
assert ILT(exp(-a*s)/(s + b), s, t) == exp(b*(a - t))*Heaviside(-a + t)
assert ILT(a/(s**2 + a**2), s, t) == sin(a*t)*Heaviside(t)
assert ILT(s/(s**2 + a**2), s, t) == cos(a*t)*Heaviside(t)
# TODO is there a way around simp_hyp?
assert simp_hyp(ILT(a/(s**2 - a**2), s, t)) == sinh(a*t)*Heaviside(t)
assert simp_hyp(ILT(s/(s**2 - a**2), s, t)) == cosh(a*t)*Heaviside(t)
assert ILT(a/((s + b)**2 + a**2), s, t) == exp(-b*t)*sin(a*t)*Heaviside(t)
assert ILT(
(s + b)/((s + b)**2 + a**2), s, t) == exp(-b*t)*cos(a*t)*Heaviside(t)
# TODO sinh/cosh shifted come out a mess. also delayed trig is a mess
# TODO should this simplify further?
assert ILT(exp(-a*s)/s**b, s, t) == \
(t - a)**(b - 1)*Heaviside(t - a)/gamma(b)
assert ILT(exp(-a*s)/sqrt(1 + s**2), s, t) == \
Heaviside(t - a)*besselj(0, a - t) # note: besselj(0, x) is even
# XXX ILT turns these branch factor into trig functions ...
assert simplify(ILT(a**b*(s + sqrt(s**2 - a**2))**(-b)/sqrt(s**2 - a**2),
s, t).rewrite(exp)) == \
Heaviside(t)*besseli(b, a*t)
assert ILT(a**b*(s + sqrt(s**2 + a**2))**(-b)/sqrt(s**2 + a**2),
s, t).rewrite(exp) == \
Heaviside(t)*besselj(b, a*t)
assert ILT(1/(s*sqrt(s + 1)), s, t) == Heaviside(t)*erf(sqrt(t))
开发者ID:Acebulf,项目名称:sympy,代码行数:41,代码来源:test_transforms.py
示例19: hopkins
def hopkins(u):
return sm.besseli(1,x) * sm.exp(-(x/2)**2 / L)
开发者ID:XNShen,项目名称:turbulence_pdfs,代码行数:2,代码来源:sympy_moments.py
示例20: test_mellin_transform_bessel
def test_mellin_transform_bessel():
from sympy import Max, Min, hyper, meijerg
MT = mellin_transform
# 8.4.19
assert MT(besselj(a, 2*sqrt(x)), x, s) == \
(gamma(a/2 + s)/gamma(a/2 - s + 1), (-re(a)/2, S(3)/4), True)
assert MT(sin(sqrt(x))*besselj(a, sqrt(x)), x, s) == \
(2**a*gamma(S(1)/2 - 2*s)*gamma((a+1)/2 + s) \
/ (gamma(1 - s- a/2)*gamma(1 + a - 2*s)),
(-(re(a) + 1)/2, S(1)/4), True)
# TODO why does this 2**(a+2)/4 not cancel?
assert MT(cos(sqrt(x))*besselj(a, sqrt(x)), x, s) == \
(2**(a+2)*gamma(a/2 + s)*gamma(S(1)/2 - 2*s)
/ (gamma(S(1)/2 - s - a/2)*gamma(a - 2*s + 1)) / 4,
(-re(a)/2, S(1)/4), True)
assert MT(besselj(a, sqrt(x))**2, x, s) == \
(gamma(a + s)*gamma(S(1)/2 - s)
/ (sqrt(pi)*gamma(1 - s)*gamma(1 + a - s)),
(-re(a), S(1)/2), True)
assert MT(besselj(a, sqrt(x))*besselj(-a, sqrt(x)), x, s) == \
(gamma(s)*gamma(S(1)/2 - s)
/ (sqrt(pi)*gamma(1 - a - s)*gamma(1 + a - s)),
(0, S(1)/2), True)
# NOTE: prudnikov gives the strip below as (1/2 - re(a), 1). As far as
# I can see this is wrong (since besselj(z) ~ 1/sqrt(z) for z large)
assert MT(besselj(a - 1, sqrt(x))*besselj(a, sqrt(x)), x, s) == \
(gamma(1-s)*gamma(a + s - S(1)/2)
/ (sqrt(pi)*gamma(S(3)/2 - s)*gamma(a - s + S(1)/2)),
(S(1)/2 - re(a), S(1)/2), True)
assert MT(besselj(a, sqrt(x))*besselj(b, sqrt(x)), x, s) == \
(2**(2*s)*gamma(1 - 2*s)*gamma((a+b)/2 + s)
/ (gamma(1 - s + (b-a)/2)*gamma(1 - s + (a-b)/2)
*gamma( 1 - s + (a+b)/2)),
(-(re(a) + re(b))/2, S(1)/2), True)
assert MT(besselj(a, sqrt(x))**2 + besselj(-a, sqrt(x))**2, x, s)[1:] == \
((Max(re(a), -re(a)), S(1)/2), True)
# Section 8.4.20
assert MT(bessely(a, 2*sqrt(x)), x, s) == \
(-cos(pi*a/2 - pi*s)*gamma(s - a/2)*gamma(s + a/2)/pi,
(Max(-re(a)/2, re(a)/2), S(3)/4), True)
assert MT(sin(sqrt(x))*bessely(a, sqrt(x)), x, s) == \
(-2**(2*s)*sin(pi*a/2 - pi*s)*gamma(S(1)/2 - 2*s)
* gamma((1-a)/2 + s)*gamma((1+a)/2 + s)
/ (sqrt(pi)*gamma(1 - s - a/2)*gamma(1 - s + a/2)),
(Max(-(re(a) + 1)/2, (re(a) - 1)/2), S(1)/4), True)
assert MT(cos(sqrt(x))*bessely(a, sqrt(x)), x, s) == \
(-2**(2*s)*cos(pi*a/2 - pi*s)*gamma(s - a/2)*gamma(s + a/2)*gamma(S(1)/2 - 2*s)
/ (sqrt(pi)*gamma(S(1)/2 - s - a/2)*gamma(S(1)/2 - s + a/2)),
(Max(-re(a)/2, re(a)/2), S(1)/4), True)
assert MT(besselj(a, sqrt(x))*bessely(a, sqrt(x)), x, s) == \
(-cos(pi*s)*gamma(s)*gamma(a + s)*gamma(S(1)/2 - s)
/ (pi**S('3/2')*gamma(1 + a - s)),
(Max(-re(a), 0), S(1)/2), True)
assert MT(besselj(a, sqrt(x))*bessely(b, sqrt(x)), x, s) == \
(-2**(2*s)*cos(pi*a/2 - pi*b/2 + pi*s)*gamma(1 - 2*s)
* gamma(a/2 - b/2 + s)*gamma(a/2 + b/2 + s)
/ (pi*gamma(a/2 - b/2 - s + 1)*gamma(a/2 + b/2 - s + 1)),
(Max((-re(a) + re(b))/2, (-re(a) - re(b))/2), S(1)/2), True)
# NOTE bessely(a, sqrt(x))**2 and bessely(a, sqrt(x))*bessely(b, sqrt(x))
# are a mess (no matter what way you look at it ...)
assert MT(bessely(a, sqrt(x))**2, x, s)[1:] == \
((Max(-re(a), 0, re(a)), S(1)/2), True)
# Section 8.4.22
# TODO we can't do any of these (delicate cancellation)
# Section 8.4.23
assert MT(besselk(a, 2*sqrt(x)), x, s) == \
(gamma(s - a/2)*gamma(s + a/2)/2, (Max(-re(a)/2, re(a)/2), oo), True)
assert MT(besselj(a, 2*sqrt(2*sqrt(x)))*besselk(a, 2*sqrt(2*sqrt(x))), x, s) == \
(4**(-s)*gamma(2*s)*gamma(a/2 + s)/gamma(a/2 - s + 1)/2,
(Max(-re(a)/2, 0), oo), True)
# TODO bessely(a, x)*besselk(a, x) is a mess
assert MT(besseli(a, sqrt(x))*besselk(a, sqrt(x)), x, s) == \
(gamma(s)*gamma(a + s)*gamma(-s + S(1)/2)/(2*sqrt(pi)*gamma(a - s + 1)),
(Max(-re(a), 0), S(1)/2), True)
assert MT(besseli(b, sqrt(x))*besselk(a, sqrt(x)), x, s) == \
(2**(2*s - 1)*gamma(-2*s + 1)*gamma(-a/2 + b/2 + s)*gamma(a/2 + b/2 + s) \
/(gamma(-a/2 + b/2 - s + 1)*gamma(a/2 + b/2 - s + 1)),
(Max(-re(a)/2 - re(b)/2, re(a)/2 - re(b)/2), S(1)/2), True)
# TODO products of besselk are a mess
# TODO this can be simplified considerably (although I have no idea how)
mt = MT(exp(-x/2)*besselk(a, x/2), x, s)
assert not mt[0].has(meijerg, hyper)
assert mt[1:] == ((Max(-re(a), re(a)), oo), True)
开发者ID:goodok,项目名称:sympy,代码行数:88,代码来源:test_transforms.py
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