def probability(self, condition, **kwargs):
cond_inv = False
if isinstance(condition, Ne):
condition = Eq(condition.args[0], condition.args[1])
cond_inv = True
expr = condition.lhs - condition.rhs
rvs = random_symbols(expr)
z = Dummy('z', real=True, Finite=True)
dens = self.compute_density(expr)
if any([pspace(rv).is_Continuous for rv in rvs]):
from sympy.stats.crv import (ContinuousDistributionHandmade,
SingleContinuousPSpace)
if expr in self.values:
# Marginalize all other random symbols out of the density
randomsymbols = tuple(set(self.values) - frozenset([expr]))
symbols = tuple(rs.symbol for rs in randomsymbols)
pdf = self.domain.integrate(self.pdf, symbols, **kwargs)
return Lambda(expr.symbol, pdf)
dens = ContinuousDistributionHandmade(dens)
space = SingleContinuousPSpace(z, dens)
result = space.probability(condition.__class__(space.value, 0))
else:
from sympy.stats.drv import (DiscreteDistributionHandmade,
SingleDiscretePSpace)
dens = DiscreteDistributionHandmade(dens)
space = SingleDiscretePSpace(z, dens)
result = space.probability(condition.__class__(space.value, 0))
return result if not cond_inv else S.One - result
def probability(self, condition, **kwargs):
z = Dummy('z', real=True, finite=True)
cond_inv = False
if isinstance(condition, Ne):
condition = Eq(condition.args[0], condition.args[1])
cond_inv = True
# Univariate case can be handled by where
try:
domain = self.where(condition)
rv = [rv for rv in self.values if rv.symbol == domain.symbol][0]
# Integrate out all other random variables
pdf = self.compute_density(rv, **kwargs)
# return S.Zero if `domain` is empty set
if domain.set is S.EmptySet or isinstance(domain.set, FiniteSet):
return S.Zero if not cond_inv else S.One
if isinstance(domain.set, Union):
return sum(
Integral(pdf(z), (z, subset), **kwargs) for subset in
domain.set.args if isinstance(subset, Interval))
# Integrate out the last variable over the special domain
return Integral(pdf(z), (z, domain.set), **kwargs)
# Other cases can be turned into univariate case
# by computing a density handled by density computation
except NotImplementedError:
from sympy.stats.rv import density
expr = condition.lhs - condition.rhs
dens = density(expr, **kwargs)
if not isinstance(dens, ContinuousDistribution):
dens = ContinuousDistributionHandmade(dens)
# Turn problem into univariate case
space = SingleContinuousPSpace(z, dens)
result = space.probability(condition.__class__(space.value, 0))
return result if not cond_inv else S.One - result
def solver():
vals = {i: base_eq(i) for i in range(1, 12)}
coefficients = a, b, c, d, e, f, g, h, i, j, k = symbols(','.join('abcdefghijk'))
(x, n), lhs = symbols('x n'), a
answer = 1
for i, element in enumerate(coefficients[1:]):
lhs = lhs * n + element
equation = Eq(x, lhs)
results = solve([equation.subs(dict(n=t, x=vals[t])) for
t in range(1, i+3)])
results[n] = i + 3
if element != k:
answer += equation.subs(results).args[1]
return answer
def solve_it(eqn, knowns, x):
LHS, RHS = eqn['sympy'].split('=')
e = Eq(sympify(LHS), sympify(RHS))
voi = Symbol(x['symbol']) # upgrade sympy to 0.7.6 if this throws a unicode error
vals = dict()
for k in knowns:
vals[Symbol(k['symbol'])] = k['value']
for c in eqn['constants']:
vals[Symbol(c['symbol'])] = c['value']
answers = solve(e.subs(vals),voi)
answer = answers[0]
return answer.evalf()
def probability(self, condition):
complement = isinstance(condition, Ne)
if complement:
condition = Eq(condition.args[0], condition.args[1])
try:
_domain = self.where(condition).set
if condition == False or _domain is S.EmptySet:
return S.Zero
if condition == True or _domain == self.domain.set:
return S.One
prob = self.eval_prob(_domain)
except NotImplementedError:
from sympy.stats.rv import density
expr = condition.lhs - condition.rhs
dens = density(expr)
if not isinstance(dens, DiscreteDistribution):
dens = DiscreteDistributionHandmade(dens)
z = Dummy('z', real = True)
space = SingleDiscretePSpace(z, dens)
prob = space.probability(condition.__class__(space.value, 0))
if (prob == None):
prob = Probability(condition)
return prob if not complement else S.One - prob
def test_issue_9849():
assert ConditionSet(x, Eq(x, x), S.Naturals) == S.Naturals
assert ConditionSet(x, Eq(Abs(sin(x)), -1), S.Naturals) == S.EmptySet
#!/usr/bin/python
from sympy import latex
from sympy import solve
from sympy import symbols
from sympy import Eq
from sympy import Rational
a, d, t, v, v0 = symbols('a d t v v0')
kin1 = Eq(d, v0 * t + Rational(1,2) * a * t**2)
kin2 = Eq(v, v0 + a * t)
print ""
print "The basic equations for one-dimensional motion are:"
print ""
print " {}".format(latex(kin1))
print " {}".format(latex(kin2))
print ""
print "How long does it take an object released from rest to fall five meters?"
print ""
values = dict()
values[d] = -5
values[v0] = 0
values[a] = -9.81
print "The parameters for this problem are:"
print ""
for param, value in values.iteritems():
def test_Equality_rewrite_as_Add():
eq = Eq(x + y, y - x)
assert eq.rewrite(Add) == 2*x
assert eq.rewrite(Add, evaluate=None).args == (x, x, y, -y)
assert eq.rewrite(Add, evaluate=False).args == (x, y, x, -y)
def given(expr, condition=None, **kwargs):
r""" Conditional Random Expression
From a random expression and a condition on that expression creates a new
probability space from the condition and returns the same expression on that
conditional probability space.
Examples
========
>>> from sympy.stats import given, density, Die
>>> X = Die('X', 6)
>>> Y = given(X, X > 3)
>>> density(Y).dict
{4: 1/3, 5: 1/3, 6: 1/3}
Following convention, if the condition is a random symbol then that symbol
is considered fixed.
>>> from sympy.stats import Normal
>>> from sympy import pprint
>>> from sympy.abc import z
>>> X = Normal('X', 0, 1)
>>> Y = Normal('Y', 0, 1)
>>> pprint(density(X + Y, Y)(z), use_unicode=False)
2
-(-Y + z)
-----------
___ 2
\/ 2 *e
------------------
____
2*\/ pi
"""
if not is_random(condition) or pspace_independent(expr, condition):
return expr
if isinstance(condition, RandomSymbol):
condition = Eq(condition, condition.symbol)
condsymbols = random_symbols(condition)
if (isinstance(condition, Equality) and len(condsymbols) == 1
and not isinstance(pspace(expr).domain, ConditionalDomain)):
rv = tuple(condsymbols)[0]
results = solveset(condition, rv)
if isinstance(results, Intersection) and S.Reals in results.args:
results = list(results.args[1])
sums = 0
for res in results:
temp = expr.subs(rv, res)
if temp == True:
return True
if temp != False:
# XXX: This seems nonsensical but preserves existing behaviour
# after the change that Relational is no longer a subclass of
# Expr. Here expr is sometimes Relational and sometimes Expr
# but we are trying to add them with +=. This needs to be
# fixed somehow.
if sums == 0 and isinstance(expr, Relational):
sums = expr.subs(rv, res)
else:
sums += expr.subs(rv, res)
if sums == 0:
return False
return sums
# Get full probability space of both the expression and the condition
fullspace = pspace(Tuple(expr, condition))
# Build new space given the condition
space = fullspace.conditional_space(condition, **kwargs)
# Dictionary to swap out RandomSymbols in expr with new RandomSymbols
# That point to the new conditional space
swapdict = rs_swap(fullspace.values, space.values)
# Swap random variables in the expression
expr = expr.xreplace(swapdict)
return expr
def test_rsolve():
f = y(n + 2) - y(n + 1) - y(n)
h = sqrt(5)*(S.Half + S.Half*sqrt(5))**n \
- sqrt(5)*(S.Half - S.Half*sqrt(5))**n
assert rsolve(f, y(n)) in [
C0 * (S.Half - S.Half * sqrt(5))**n + C1 *
(S.Half + S.Half * sqrt(5))**n,
C1 * (S.Half - S.Half * sqrt(5))**n + C0 *
(S.Half + S.Half * sqrt(5))**n,
]
assert rsolve(f, y(n), [0, 5]) == h
assert rsolve(f, y(n), {0: 0, 1: 5}) == h
assert rsolve(f, y(n), {y(0): 0, y(1): 5}) == h
assert rsolve(y(n) - y(n - 1) - y(n - 2), y(n), [0, 5]) == h
assert rsolve(Eq(y(n), y(n - 1) + y(n - 2)), y(n), [0, 5]) == h
assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0
f = (n - 1) * y(n + 2) - (n**2 + 3 * n - 2) * y(n + 1) + 2 * n * (n +
1) * y(n)
g = C1 * factorial(n) + C0 * 2**n
h = -3 * factorial(n) + 3 * 2**n
assert rsolve(f, y(n)) == g
assert rsolve(f, y(n), []) == g
assert rsolve(f, y(n), {}) == g
assert rsolve(f, y(n), [0, 3]) == h
assert rsolve(f, y(n), {0: 0, 1: 3}) == h
assert rsolve(f, y(n), {y(0): 0, y(1): 3}) == h
assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0
f = y(n) - y(n - 1) - 2
assert rsolve(f, y(n), {y(0): 0}) == 2 * n
assert rsolve(f, y(n), {y(0): 1}) == 2 * n + 1
assert rsolve(f, y(n), {y(0): 0, y(1): 1}) is None
assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0
f = 3 * y(n - 1) - y(n) - 1
assert rsolve(f, y(n), {y(0): 0}) == -3**n / 2 + S.Half
assert rsolve(f, y(n), {y(0): 1}) == 3**n / 2 + S.Half
assert rsolve(f, y(n), {y(0): 2}) == 3 * 3**n / 2 + S.Half
assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0
f = y(n) - 1 / n * y(n - 1)
assert rsolve(f, y(n)) == C0 / factorial(n)
assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0
f = y(n) - 1 / n * y(n - 1) - 1
assert rsolve(f, y(n)) is None
f = 2 * y(n - 1) + (1 - n) * y(n) / n
assert rsolve(f, y(n), {y(1): 1}) == 2**(n - 1) * n
assert rsolve(f, y(n), {y(1): 2}) == 2**(n - 1) * n * 2
assert rsolve(f, y(n), {y(1): 3}) == 2**(n - 1) * n * 3
assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0
f = (n - 1) * (n - 2) * y(n + 2) - (n + 1) * (n + 2) * y(n)
assert rsolve(f, y(n), {y(3): 6, y(4): 24}) == n * (n - 1) * (n - 2)
assert rsolve(f, y(n), {
y(3): 6,
y(4): -24
}) == -n * (n - 1) * (n - 2) * (-1)**(n)
assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0
assert rsolve(Eq(y(n + 1), a * y(n)), y(n), {y(1): a}).simplify() == a**n
assert rsolve(y(n) - a*y(n-2),y(n), \
{y(1): sqrt(a)*(a + b), y(2): a*(a - b)}).simplify() == \
a**(n/2)*(-(-1)**n*b + a)
f = (-16 * n**2 + 32 * n - 12) * y(n - 1) + (4 * n**2 - 12 * n + 9) * y(n)
assert expand_func(rsolve(f, y(n), \
{y(1): binomial(2*n + 1, 3)}).rewrite(gamma)).simplify() == \
2**(2*n)*n*(2*n - 1)*(4*n**2 - 1)/12
assert (rsolve(y(n) + a * (y(n + 1) + y(n - 1)) / 2, y(n)) -
(C0 * ((sqrt(-a**2 + 1) - 1) / a)**n + C1 *
((-sqrt(-a**2 + 1) - 1) / a)**n)).simplify() == 0
def test_conditional_eq():
E = Exponential('E', 1)
assert P(Eq(E, 1), Eq(E, 1)) == 1
assert P(Eq(E, 1), Eq(E, 2)) == 0
assert P(E > 1, Eq(E, 2)) == 1
assert P(E < 1, Eq(E, 2)) == 0
def test_dice_bayes():
X, Y, Z = Die('X', 6), Die('Y', 6), Die('Z', 6)
BayesTest(X > 3, X + Y < 5)
BayesTest(Eq(X - Y, Z), Z > Y)
BayesTest(X > 3, X > 2)
def test_tsolve():
a, b = symbols('a,b')
x, y, z = symbols('x,y,z')
assert solve(exp(x)-3, x) == [log(3)]
assert solve((a*x+b)*(exp(x)-3), x) == [-b/a, log(3)]
assert solve(cos(x)-y, x) == [acos(y)]
assert solve(2*cos(x)-y,x)== [acos(y/2)]
raises(NotImplementedError, "solve(Eq(cos(x), sin(x)), x)")
assert solve(exp(x) + exp(-x) - y, x, simplified=False) == [
log((y/2 + sqrt(y**2 - 4)/2)**2)/2,
log((y/2 - sqrt(y**2 - 4)/2)**2)/2]
assert solve(exp(x)-3, x) == [log(3)]
assert solve(Eq(exp(x), 3), x) == [log(3)]
assert solve(log(x)-3, x) == [exp(3)]
assert solve(sqrt(3*x)-4, x) == [Rational(16,3)]
assert solve(3**(x+2), x) == [zoo]
assert solve(3**(2-x), x) == [zoo]
assert solve(4*3**(5*x+2)-7, x) == [(-2*log(3) - 2*log(2) + log(7))/(5*log(3))]
assert solve(x+2**x, x) == [-LambertW(log(2))/log(2)]
assert solve(3*x+5+2**(-5*x+3), x) in [
[-((25*log(2) - 3*LambertW(-10240*2**(Rational(1, 3))*log(2)/3))/(15*log(2)))],
[Rational(-5, 3) + LambertW(log(2**(-10240*2**(Rational(1, 3))/3)))/(5*log(2))],
[-Rational(5,3) + LambertW(-10240*2**Rational(1,3)*log(2)/3)/(5*log(2))],
[(-25*log(2) + 3*LambertW(-10240*2**(Rational(1, 3))*log(2)/3))/(15*log(2))],
[-((25*log(2) - 3*LambertW(-10240*2**(Rational(1, 3))*log(2)/3)))/(15*log(2))],
[-(25*log(2) - 3*LambertW(log(2**(-10240*2**Rational(1, 3)/3))))/(15*log(2))],
[(25*log(2) - 3*LambertW(log(2**(-10240*2**Rational(1, 3)/3))))/(-15*log(2))]
]
assert solve(5*x-1+3*exp(2-7*x), x) == \
[Rational(1,5) + LambertW(-21*exp(Rational(3, 5))/5)/7]
assert solve(2*x+5+log(3*x-2), x) == \
[Rational(2,3) + LambertW(2*exp(-Rational(19, 3))/3)/2]
assert solve(3*x+log(4*x), x) == [LambertW(Rational(3,4))/3]
assert solve((2*x+8)*(8+exp(x)), x) == [-4, log(8) + pi*I]
assert solve(2*exp(3*x+4)-3, x) in [
[Rational(-4, 3) + log(2**Rational(2, 3)*3**Rational(1, 3)/2)],
[-Rational(4, 3)+log(Rational(3, 2))/3],
[Rational(-4, 3) - log(2)/3 + log(3)/3],
]
assert solve(2*log(3*x+4)-3, x) == [(exp(Rational(3,2))-4)/3]
assert solve(exp(x)+1, x) == [pi*I]
assert solve(x**2 - 2**x, x) == [2]
assert solve(x**3 - 3**x, x) == [-3*LambertW(-log(3)/3)/log(3)]
A = -7*2**Rational(4, 5)*6**Rational(1, 5)*log(7)/10
B = -7*3**Rational(1, 5)*log(7)/5
result = solve(2*(3*x+4)**5 - 6*7**(3*x+9), x)
assert len(result) == 1 and expand(result[0]) in [
Rational(-4, 3) - 5/log(7)/3*LambertW(A),
Rational(-4, 3) - 5/log(7)/3*LambertW(B),
]
assert solve(z*cos(x)-y, x) == [acos(y/z)]
assert solve(z*cos(2*x)-y, x) == [acos(y/z)/2]
assert solve(z*cos(sin(x))-y, x) == [asin(acos(y/z))]
assert solve(z*cos(x), x) == [acos(0)]
# issue #1409
assert solve(y - b*x/(a+x), x) in [[-a*y/(y - b)], [a*y/(b - y)]]
assert solve(y - b*exp(a/x), x) == [a/log(y/b)]
# issue #1408
assert solve(y-b/(1+a*x), x) in [[(b - y)/(a*y)], [-((y - b)/(a*y))]]
# issue #1407
assert solve(y-a*x**b , x) == [(y/a)**(1/b)]
# issue #1406
assert solve(z**x - y, x) == [log(y)/log(z)]
# issue #1405
assert solve(2**x - 10, x) == [log(10)/log(2)]
from sympy.functions import *
from sympy.interactive import init_printing
from sympy.physics.units import Quantity
from sympy.physics.units import convert_to
from utils import *
init_printing()
hp_to_watts = 745.7
W0 = Quantity("W0")
W = Quantity("W")
m = Quantity("m")
v = Quantity("v")
g = Quantity("g")
theta = symbols("theta")
set_quantity(W0, un.power, 85 * hp_to_watts * un.watts)
set_quantity(W, un.power, 20 * hp_to_watts * un.watts)
set_quantity(m, un.mass, 1200 * un.kg)
set_quantity(v, un.velocity, 48 * un.km / un.hour)
set_quantity(g, un.acceleration, 9.81 * un.m / un.s**2)
eq = Eq(sin(theta), (W0 - W) / (m * g * v))
pretty_print(eq)
theta = solve(eq, theta)[1]
theta = convert_to(theta, un.watts).n()
pretty_print(convert_to(theta, un.degrees).n())
# Program 02g: A second order ODE.
from sympy import symbols, dsolve, Function, Eq, sin
t = symbols('t')
I = symbols('I', cls=Function)
sol = dsolve(Eq(I(t).diff(t, t) + 5 * I(t).diff(t) + 6 * I(t), 10 * sin(t)),
I(t))
print(sol)
def test_evalf_relational():
assert Eq(x / 5, y / 10).evalf() == Eq(0.2 * x, 0.1 * y)
def test_Intersection_as_relational():
x = Symbol('x')
assert (Intersection(Interval(0, 1), FiniteSet(2),
evaluate=False).as_relational(x)
== And(And(Le(0, x), Le(x, 1)), Eq(x, 2)))
def test_Union_as_relational():
x = Symbol('x')
assert (Interval(0, 1) + FiniteSet(2)).as_relational(x) == \
Or(And(Le(0, x), Le(x, 1)), Eq(x, 2))
assert (Interval(0, 1, True, True) + FiniteSet(1)).as_relational(x) == \
And(Lt(0, x), Le(x, 1))
def test_Finite_as_relational():
x = Symbol('x')
y = Symbol('y')
assert FiniteSet(1, 2).as_relational(x) == Or(Eq(x, 1), Eq(x, 2))
assert FiniteSet(y, -5).as_relational(x) == Or(Eq(x, y), Eq(x, -5))
def test_solve_trig_abs():
assert solveset(Eq(sin(Abs(x)), 1), x, domain=S.Reals) == \
Union(ImageSet(Lambda(n, n*pi + (-1)**n*pi/2), S.Naturals0),
ImageSet(Lambda(n, -n*pi - (-1)**n*pi/2), S.Naturals0))
def test_random_parameters_given():
mu = Normal('mu', 2, 3)
meas = Normal('T', mu, 1)
assert given(meas, Eq(mu, 5)) == Normal('T', 5, 1)
def test_equality_subs2():
f = Function("f")
x = abc.x
eq = Eq(f(x)**2, 16)
assert bool(eq.subs(f(x), 3)) == False
assert bool(eq.subs(f(x), 4)) == True
def test_conjugate_priors():
mu = Normal('mu', 2, 3)
x = Normal('x', mu, 1)
assert isinstance(simplify(density(mu, Eq(x, y), evaluate=False)(z)),
Integral)
def __eq__(self, other):
return Eq(self.symbol, _param_to_symbol(other))
def test_tsolve():
assert solve(exp(x) - 3, x) == [log(3)]
assert set(solve((a * x + b) * (exp(x) - 3), x)) == set([-b / a, log(3)])
assert solve(cos(x) - y, x) == [acos(y)]
assert solve(2 * cos(x) - y, x) == [acos(y / 2)]
assert solve(Eq(cos(x), sin(x)), x) == [-3 * pi / 4, pi / 4]
assert set(solve(exp(x) + exp(-x) - y, x)) == set([
log(y / 2 - sqrt(y**2 - 4) / 2),
log(y / 2 + sqrt(y**2 - 4) / 2),
])
assert solve(exp(x) - 3, x) == [log(3)]
assert solve(Eq(exp(x), 3), x) == [log(3)]
assert solve(log(x) - 3, x) == [exp(3)]
assert solve(sqrt(3 * x) - 4, x) == [Rational(16, 3)]
assert solve(3**(x + 2), x) == []
assert solve(3**(2 - x), x) == []
assert solve(x + 2**x, x) == [-LambertW(log(2)) / log(2)]
assert solve(3 * x + 5 + 2**(-5 * x + 3), x) in [
[
-((25 * log(2) - 3 * LambertW(-10240 * 2**
(Rational(1, 3)) * log(2) / 3)) /
(15 * log(2)))
],
[
Rational(-5, 3) +
LambertW(log(2**(-10240 * 2**(Rational(1, 3)) / 3))) / (5 * log(2))
],
[
-Rational(5, 3) +
LambertW(-10240 * 2**Rational(1, 3) * log(2) / 3) / (5 * log(2))
],
[(-25 * log(2) +
3 * LambertW(-10240 * 2**(Rational(1, 3)) * log(2) / 3)) /
(15 * log(2))],
[
-((25 * log(2) - 3 * LambertW(-10240 * 2**
(Rational(1, 3)) * log(2) / 3))) /
(15 * log(2))
],
[
-(25 * log(2) -
3 * LambertW(log(2**(-10240 * 2**Rational(1, 3) / 3)))) /
(15 * log(2))
],
[(25 * log(2) - 3 * LambertW(log(2**(-10240 * 2**Rational(1, 3) / 3))))
/ (-15 * log(2))]
]
assert solve(5*x-1+3*exp(2-7*x), x) == \
[Rational(1,5) + LambertW(-21*exp(Rational(3, 5))/5)/7]
assert solve(2*x+5+log(3*x-2), x) == \
[Rational(2,3) + LambertW(2*exp(-Rational(19, 3))/3)/2]
assert solve(3 * x + log(4 * x), x) == [LambertW(Rational(3, 4)) / 3]
assert set(solve((2 * x + 8) * (8 + exp(x)),
x)) == set([S(-4), log(8) + pi * I])
eq = 2 * exp(3 * x + 4) - 3
ans = solve(eq, x)
assert len(ans) == 3 and all(eq.subs(x, a).n(chop=True) == 0 for a in ans)
assert solve(2 * log(3 * x + 4) - 3, x) == [(exp(Rational(3, 2)) - 4) / 3]
assert solve(exp(x) + 1, x) == [pi * I]
assert solve(x**2 - 2**x, x) == [2]
assert solve(x**3 - 3**x, x) == [-3 * LambertW(-log(3) / 3) / log(3)]
A = -7 * 2**Rational(4, 5) * 6**Rational(1, 5) * log(7) / 10
B = -7 * 3**Rational(1, 5) * log(7) / 5
result = solve(2 * (3 * x + 4)**5 - 6 * 7**(3 * x + 9), x)
assert len(result) == 1 and expand(result[0]) in [
Rational(-4, 3) - 5 / log(7) / 3 * LambertW(A),
Rational(-4, 3) - 5 / log(7) / 3 * LambertW(B),
]
assert solve(z * cos(x) - y, x) == [acos(y / z)]
assert solve(z * cos(2 * x) - y, x) == [acos(y / z) / 2]
assert solve(z * cos(sin(x)) - y, x) == [asin(acos(y / z))]
assert solve(z * cos(x), x) == [acos(0)]
# issue #1409
assert solve(y - b * x / (a + x), x) in [[-a * y / (y - b)],
[a * y / (b - y)]]
assert solve(y - b * exp(a / x), x) == [a / log(y / b)]
# issue #1408
assert solve(y - b / (1 + a * x), x) in [[(b - y) / (a * y)],
[-((y - b) / (a * y))]]
# issue #1407
assert solve(y - a * x**b, x) == [(y / a)**(1 / b)]
# issue #1406
assert solve(z**x - y, x) == [log(y) / log(z)]
# issue #1405
assert solve(2**x - 10, x) == [log(10) / log(2)]
def test_given():
X = Die('X', 6)
density(X, X > 5) == {S(6): S(1)}
where(X > 2, X > 5).as_boolean() == Eq(X.symbol, 6)
sample(X, X > 5) == 6
#!/usr/bin/python
from sympy import latex
from sympy import solve
from sympy import symbols
from sympy import Eq
from sympy import Rational
P, V, n, R, T = symbols('P V n R T')
igl = Eq(P * V, n * R * T)
values = {
R : 8.3144621,
T : 273,
P : 101300,
n : 1
}
answers = solve(igl.subs(values),V)
answer = answers[0]
print ""
print "The ideal gas law is ",latex(igl)
print ""
print "Where the gas constant R is {} in SI units".format(values[R])
print "Standard temperature is {} Kelvin, or 0 degrees Celsius".format(values[T])
print "Standard pressure is {} pascals, or 1 atm".format(values[P])
print ""
print "So, at standard temperature and pressure, one mole of an ideal gas"
def system_of_linear_equations_solver(sub):
#NOTE: tests may fail due to sets being randomly ordered, but the answers are correct
r"""System of Linear Equations Checker/Solver.
Checks whether a given string is a system of linear equations in two variables,
and if so, returns an explanation of how to solve it.
Parameters
----------
sub : str
The submitted expression, as a math string, to be passed to SymPy.
Returns
-------
explanation:
False if unable to parse as linear system,
A worked thorugh $\LaTeX$ explanation otherwise.
Examples
--------
>>> system_of_linear_equations_solver("")
False
>>> system_of_linear_equations_solver("something abstract")
False
>>> system_of_linear_equations_solver("x+1")
False
>>> print(system_of_linear_equations_solver("x=1"))
False
>>> print(system_of_linear_equations_solver("a=1,b=0"))
False
>>> print(system_of_linear_equations_solver("a=b+1,0=a-b"))
False
>>> print(system_of_linear_equations_solver("x+y=0,2x-y=9,2x+y=3"))
False
>>> system_of_linear_equations_solver("x**2+y=1,xy=1")
False
>>> print(system_of_linear_equations_solver("a+2b = 1,a-b=3"))
Let's solve the system of equations:
\[
\begin{align*}
a + 2 b&=1\\
a - b&=3
\end{align*}
\]
First we multiply each equation by the coefficient of $b$ in the other equations.
This way they will all have the same coefficient for $b$.
\begin{align*}
-1\left(a + 2 b\right)&=-1\left(1\right)\\
- a - 2 b&=-1
\end{align*}
\begin{align*}
2\left(a - b\right)&=2\left(3\right)\\
2 a - 2 b&=6
\end{align*}
If we subtract the second equation from the first, we get:
\begin{align*}
- 3 a&=-7
\end{align*}
Since this is now an equation in one variable, we can solve it like we would
any other linear equation.
\begin{align*}
\end{align*}
Let's solve the equation:
\[
- 3 a = -7
\]
We have just one term on the left:
The variable $a$ with coefficient $-3$.
Divide both sides by $-3$:
\begin{align*}
\frac{ - 3 a }{ -3 } &=
\frac{ -7 }{ -3 } \\
a &= \frac{7}{3}
\end{align*}
The equation is in the form $a = \frac{7}{3}$;
That is, the value of $a$ is $\frac{7}{3}$.\begin{align*}
\end{align*}
Plugging in $\frac{7}{3}$ for $a$ in the first equation, we can now
solve for $b$ like we would any other linear equation.
\begin{align*}
\end{align*}
Let's solve the equation:
\[
2 b + \frac{7}{3} = 1
\]
First, we subtract 7/3 from both sides:
\begin{align*}
(2 b + \frac{7}{3})-(7/3) &= 1-(7/3) \\
2 b &= - \frac{4}{3}
\end{align*}
We have just one term on the left:
The variable $b$ with coefficient $2$.
Divide both sides by $2$:
\begin{align*}
\frac{ 2 b }{ 2 } &=
\frac{ - \frac{4}{3} }{ 2 } \\
b &= - \frac{2}{3}
\end{align*}
The equation is in the form $b = - \frac{2}{3}$;
That is, the value of $b$ is $- \frac{2}{3}$.\begin{align*}
\end{align*}
Thus we have found the pair of values which satisfy the two linear equations:
$b=- \frac{2}{3}$ and $a=\frac{7}{3}$.
<BLANKLINE>
>>> print(system_of_linear_equations_solver("3x+2/3y=5/2,5x-y=2"))
Let's solve the system of equations:
\[
\begin{align*}
3 x + \frac{2 y}{3}&=\frac{5}{2}\\
5 x - y&=2
\end{align*}
\]
First we multiply each equation by the coefficient of $y$ in the other equations.
This way they will all have the same coefficient for $y$.
\begin{align*}
-1\left(3 x + \frac{2 y}{3}\right)&=-1\left(\frac{5}{2}\right)\\
- 3 x - \frac{2 y}{3}&=- \frac{5}{2}
\end{align*}
\begin{align*}
\frac{2}{3}\left(5 x - y\right)&=\frac{2}{3}\left(2\right)\\
\frac{10 x}{3} - \frac{2 y}{3}&=\frac{4}{3}
\end{align*}
If we subtract the second equation from the first, we get:
\begin{align*}
- \frac{19 x}{3}&=- \frac{23}{6}
\end{align*}
Since this is now an equation in one variable, we can solve it like we would
any other linear equation.
\begin{align*}
\end{align*}
Let's solve the equation:
\[
- \frac{19 x}{3} = - \frac{23}{6}
\]
We have just one term on the left:
The variable $x$ with coefficient $- \frac{19}{3}$.
Divide both sides by $- \frac{19}{3}$:
\begin{align*}
\frac{ - \frac{19 x}{3} }{ - \frac{19}{3} } &=
\frac{ - \frac{23}{6} }{ - \frac{19}{3} } \\
x &= \frac{23}{38}
\end{align*}
The equation is in the form $x = \frac{23}{38}$;
That is, the value of $x$ is $\frac{23}{38}$.\begin{align*}
\end{align*}
Plugging in $\frac{23}{38}$ for $x$ in the first equation, we can now
solve for $y$ like we would any other linear equation.
\begin{align*}
\end{align*}
Let's solve the equation:
\[
\frac{2 y}{3} + \frac{69}{38} = \frac{5}{2}
\]
First, we subtract 69/38 from both sides:
\begin{align*}
(\frac{2 y}{3} + \frac{69}{38})-(69/38) &= \frac{5}{2}-(69/38) \\
\frac{2 y}{3} &= \frac{13}{19}
\end{align*}
We have just one term on the left:
The variable $y$ with coefficient $\frac{2}{3}$.
Divide both sides by $\frac{2}{3}$:
\begin{align*}
\frac{ \frac{2 y}{3} }{ \frac{2}{3} } &=
\frac{ \frac{13}{19} }{ \frac{2}{3} } \\
y &= \frac{39}{38}
\end{align*}
The equation is in the form $y = \frac{39}{38}$;
That is, the value of $y$ is $\frac{39}{38}$.\begin{align*}
\end{align*}
Thus we have found the pair of values which satisfy the two linear equations:
$y=\frac{39}{38}$ and $x=\frac{23}{38}$.
<BLANKLINE>
>>> print(system_of_linear_equations_solver("3a+2b=5,a+2b=3"))
Let's solve the system of equations:
\[
\begin{align*}
3 a + 2 b&=5\\
a + 2 b&=3
\end{align*}
\]
If we subtract the second equation from the first, we get:
\begin{align*}
4 a&=4
\end{align*}
Since this is now an equation in one variable, we can solve it like we would
any other linear equation.
\begin{align*}
\end{align*}
Let's solve the equation:
\[
4 a = 4
\]
We have just one term on the left:
The variable $a$ with coefficient $4$.
Divide both sides by $4$:
\begin{align*}
\frac{ 4 a }{ 4 } &=
\frac{ 4 }{ 4 } \\
a &= 1
\end{align*}
The equation is in the form $a = 1$;
That is, the value of $a$ is $1$.\begin{align*}
\end{align*}
Plugging in $1$ for $a$ in the first equation, we can now
solve for $b$ like we would any other linear equation.
\begin{align*}
\end{align*}
Let's solve the equation:
\[
2 b + 3 = 5
\]
First, we subtract 3 from both sides:
\begin{align*}
(2 b + 3)-(3) &= 5-(3) \\
2 b &= 2
\end{align*}
We have just one term on the left:
The variable $b$ with coefficient $2$.
Divide both sides by $2$:
\begin{align*}
\frac{ 2 b }{ 2 } &=
\frac{ 2 }{ 2 } \\
b &= 1
\end{align*}
The equation is in the form $b = 1$;
That is, the value of $b$ is $1$.\begin{align*}
\end{align*}
Thus we have found the pair of values which satisfy the two linear equations:
$b=1$ and $a=1$.
<BLANKLINE>
>>> print(system_of_linear_equations_solver("x+y=5,x+y=3"))
Let's solve the system of equations:
\[
\begin{align*}
x + y&=5\\
x + y&=3
\end{align*}
\]
If we subtract the second equation from the first, we get:
\begin{align*}
0&=2
\end{align*}
But these values aren't equal, so there are no solutions to this system of equations.
<BLANKLINE>
>>> print(system_of_linear_equations_solver("x+y=5,2x+2y=10"))
Let's solve the system of equations:
\[
\begin{align*}
x + y&=5\\
2 x + 2 y&=10
\end{align*}
\]
First we multiply each equation by the coefficient of $y$ in the other equations.
This way they will all have the same coefficient for $y$.
\begin{align*}
2\left(x + y\right)&=2\left(5\right)\\
2 x + 2 y&=10
\end{align*}
If we subtract the second equation from the first, we get:
\begin{align*}
0&=0
\end{align*}
In this case, both variables ended up cancelling out since the two
equations were scalar multiples of each other. Thus there are an
infinite number of solutions to this system of equations.
<BLANKLINE>
"""
#separate the equations at the comma
eqns = sub.split(",")
if len(eqns) != 2:
return False
# Check if SymPy can parse the expressions as equations
try:
expr = []
for i in eqns:
expr.append(
parse_expr(i,
transformations=(*standard_transformations,
implicit_multiplication,
convert_equals_signs)))
except:
return False
# Verify the structure of the equations
# Check if the expressions are in 2 variables
variables = set()
for i in expr:
variables.update(i.free_symbols)
if len(i.free_symbols) != 2:
return False
if len(variables) != 2:
return False
x, y, = variables
# Check if they are linear equations
for i in expr:
if not isinstance(i, Eq):
return False
if not i.rhs.is_constant():
return False
if not i.lhs.diff(x).is_constant():
return False
if not i.lhs.diff(y).is_constant():
return False
# Now that we know the structure of the equations,
# we can turn them into a worked-through solution.
lhs = []
rhs = []
coeff = []
for i in expr:
lhs.append(i.lhs)
rhs.append(i.rhs)
coeff.append((lhs[-1].coeff(x), lhs[-1].coeff(y)))
explanation = dedent("""\
Let's solve the system of equations:
\\[
\\begin{{align*}}
{left1}&={right1}\\\\
{left2}&={right2}
\\end{{align*}}
\\]
""".format(left1=latex(lhs[0]),
left2=latex(lhs[1]),
right1=latex(rhs[0]),
right2=latex(rhs[1])))
new_coeff = [1]
new_coeff.extend(new_coeff[-1] * i[0] for i in coeff)
new_coeff = new_coeff[-1]
new_lhs = [lhs[i] * new_coeff / coeff[i][0] for i in range(len(coeff))]
new_rhs = [rhs[i] * new_coeff / coeff[i][0] for i in range(len(coeff))]
if any(i[0] != coeff[0][0] for i in coeff):
explanation += dedent("""\
First we multiply each equation by the coefficient of ${variable}$ in the other equations.
This way they will all have the same coefficient for ${variable}$.
""".format(variable=latex(x)))
if coeff[1][0] != 1:
explanation += dedent("""\
\\begin{{align*}}
{coeff1}\\left({left1}\\right)&={coeff1}\\left({right1}\\right)\\\\
{new_left1}&={new_right1}
\\end{{align*}}
""".format(variable=latex(x),
coeff1=latex(new_coeff / coeff[0][0]),
left1=latex(lhs[0]),
new_left1=latex(new_lhs[0]),
right1=latex(rhs[0]),
new_right1=latex(new_rhs[0])))
if coeff[0][0] != 1:
explanation += dedent("""\
\\begin{{align*}}
{coeff2}\\left({left2}\\right)&={coeff2}\\left({right2}\\right)\\\\
{new_left2}&={new_right2}
\\end{{align*}}
""".format(variable=latex(x),
coeff2=latex(new_coeff / coeff[1][0]),
left2=latex(lhs[1]),
new_left2=latex(new_lhs[1]),
right2=latex(rhs[1]),
new_right2=latex(new_rhs[1])))
combined_lhs = new_lhs[0] - new_lhs[1]
combined_rhs = new_rhs[0] - new_rhs[1]
explanation += dedent("""\
If we subtract the second equation from the first, we get:
\\begin{{align*}}
{left}&={right}
\\end{{align*}}
""".format(left=latex(combined_lhs), right=latex(combined_rhs)))
if combined_lhs.is_constant():
if (combined_lhs - combined_rhs).is_zero:
explanation += dedent("""\
In this case, both variables ended up cancelling out since the two
equations were scalar multiples of each other. Thus there are an
infinite number of solutions to this system of equations.
""")
else:
explanation += dedent("""\
But these values aren't equal, so there are no solutions to this system of equations.
""")
return explanation
y_explanation, y_value = linear_logic(Eq(combined_lhs, combined_rhs), y)
x_explanation, x_value = zip(
*[linear_logic(i.subs(y, y_value), x) for i in expr])
explanation += dedent("""\
Since this is now an equation in one variable, we can solve it like we would
any other linear equation.
\\begin{{align*}}
\\end{{align*}}
""".format())
explanation += y_explanation
explanation += dedent("""\
\\begin{{align*}}
\\end{{align*}}
Plugging in ${value}$ for ${variable}$ in the first equation, we can now
solve for ${other}$ like we would any other linear equation.
\\begin{{align*}}
\\end{{align*}}
""".format(value=latex(y_value), variable=latex(y), other=latex(x)))
explanation += x_explanation[0]
explanation += dedent("""\
\\begin{{align*}}
\\end{{align*}}
Thus we have found the pair of values which satisfy the two linear equations:
${var1}={val1}$ and ${var2}={val2}$.
""".format(var1=latex(x),
var2=latex(y),
val1=latex(x_value[0]),
val2=latex(y_value)))
assert all(
i == x_value[0] for i in x_value
), "Deduced value for {x} should be equal for all expressions".format(x=x)
return explanation
from sympy import symbols, Eq, Derivative, Integral, log, solve, exp, ccode
globals().update(symbs) # see common.py: x, Y, Z, k_f, ...
eqs = []
# rate of x
rate_expr = k_f * (Y - x) * (Z - x)
rate_eq = Eq(Derivative(x, t), rate_expr)
eqs.append(rate_eq)
integrand = k_f / rate_expr
inte_eq_lhs = Integral(integrand.subs({x: chi}), (chi, 0, x))
inte_eq_rhs = Integral(k_f, (tau, 0, t))
inte_eq = Eq(inte_eq_lhs, inte_eq_rhs)
eqs.append(inte_eq)
expl_in_x_eq = inte_eq.doit(manual=True).simplify()
eqs.append(expl_in_x_eq)
expl_in_t_eq = Eq(x, solve(expl_in_x_eq, x)[0])
alt_expl_in_t = Y * (1 - exp(k_f * t * (Z - Y))) / (Y / Z - exp(k_f * t * (Z - Y)))
assert (alt_expl_in_t - expl_in_t_eq.rhs).simplify() == 0
alt_expl_in_t_eq = Eq(x, alt_expl_in_t)
eqs.append(alt_expl_in_t_eq)
def main():
# GENERATES WITHOUT ARGUMENTS: irrev_binary_1.tex irrev_binary_2.tex irrev_binary_rate.tex irrev_binary_k_b.c irrev_binary_K_eq.c
def test_distribution_over_equality():
assert Product(Eq(x*2, f(x)), (x, 1, 3)).doit() == Eq(48, f(1)*f(2)*f(3))
assert Sum(Eq(f(x), x**2), (x, 0, y)) == \
Eq(Sum(f(x), (x, 0, y)), Sum(x**2, (x, 0, y)))
def set_free_surface(self, d, b, side, algo='robertsson'):
"""
set free surface boundary condition to boundary d, at index b
:param indices: list of indices, e.g. [t,x,y,z] for 3D
:param d: direction of the boundary surface normal
:param b: location of the boundary (index)
:param algo: which algorithm to use to compute ghost cells
algo == 'robertsson' [1]: setting all velocities at ghost cells to zero
algo == 'levander' [2]: only valid for 4th spatial order. using 2nd order FD approximation for velocities
side: lower boundary (0) or upper boundary (1)
e.g. set_free_surface([t,x,y,z],1,2,0)
set y-z plane at x=2 to be lower free surface
ghost cells are calculated using reflection of stress fields
store the code to populate ghost cells to self.bc
[1] Robertsson, Johan OA. "A numerical free-surface condition for elastic/viscoelastic finite-difference modeling in the presence of topography." Geophysics 61.6 (1996): 1921-1934.
[2] Levander, Alan R. "Fourth-order finite-difference P-SV seismograms." Geophysics 53.11 (1988): 1425-1436.
"""
idx = list(self.indices)
if d not in self.direction:
if (not algo == 'levander') or (not self.direction[0] == self.direction[1]):
# shear stress, e.g. Tyz no need to recalculate at x boundary (only depends on dV/dz and dW/dy)
self.bc[d][side] = []
return
else:
# normal stress, need to recalcuate Tyy, Tzz at x boundary
expr = self.dt
derivatives = get_all_objects(expr, DDerivative)
for deriv in derivatives:
if deriv.var == idx[d]:
# replacing dx at x boundary with dy, dz terms
expr2 = self.sfields[d].dt
deriv_0 = deriv
deriv_sub = solve(expr2, deriv)[0]
break
expr = expr.subs(deriv_0, deriv_sub)
derivatives = get_all_objects(expr, DDerivative)
# substitution dictionary
dict1 = {}
for deriv in derivatives:
dict1[deriv] = deriv.fd[4]
expr = expr.subs(dict1)
eq = Eq(self.d[0][1].fd[2], expr)
eq = eq.subs(idx[d], b)
t = idx[0]
idx[0] = t+hf
idx[d] = b
# eq = eq.subs(t, t+hf)
# idx[0] = t+1
# idx[d] = b
# solve for Txx(t+1/2)
lhs = self[idx]
rhs = solve(eq, lhs)[0]
rhs = self.align(rhs)
# change t+1/2 to t+1
lhs = lhs.subs(t, t+hf)
rhs = rhs.subs(t, t+hf)
eq2 = Eq(lhs, rhs)
self.bc[d][side] = [eq2]
return
# use anti-symmetry to ensure stress at boundary=0
# this applies for all algorithms
idx = list(self.indices) # ghost cell
idx2 = list(self.indices) # cell inside domain
if not self.staggered[d]:
# if not staggered, assign T[d]=0, assign T[d-1]=-T[d+1]
idx[d] = b
idx2[d] = b
eq1 = Eq(self[idx])
else:
# if staggered, assign T[d-1/2]=T[d+1/2], assign T[d-3/2]=T[d+3/2]
idx[d] = b - (1-side)
idx2[d] = idx[d] + (-1)**side
eq1 = Eq(self[idx], -self[idx2])
eq1 = eq1.subs(idx[0], idx[0]+1)
self.bc[d][side] = [eq1]
for depth in range(self.order[d]/2-1):
# populate ghost cells
idx[d] -= (-1)**side
idx2[d] += (-1)**side
eq = Eq(self[idx], -self[idx2])
# change t to t+1
eq = eq.subs(idx[0], idx[0]+1)
self.bc[d][side].append(eq)
def test_issue_10555():
f = Function('f')
assert solveset(f(x) - pi/2, x, S.Reals) == \
ConditionSet(x, Eq(2*f(x) - pi, 0), S.Reals)
def test_conditionset_equality():
''' Checking equality of different representations of ConditionSet'''
assert solveset(Eq(tan(x), y), x) == ConditionSet(x, Eq(tan(x), y),
S.Complexes)
def test_unrad():
s = symbols('s', cls=Dummy)
# checkers to deal with possibility of answer coming
# back with a sign change (cf issue 2104)
def check(rv, ans):
rv, ans = list(rv), list(ans)
rv[0] = rv[0].expand()
ans[0] = ans[0].expand()
return rv[0] in [ans[0], -ans[0]] and rv[1:] == ans[1:]
def s_check(rv, ans):
# get the dummy
rv = list(rv)
d = rv[0].atoms(Dummy)
reps = zip(d, [s] * len(d))
# replace s with this dummy
rv = (rv[0].subs(reps).expand(), [
(p[0].subs(reps), p[1].subs(reps)) for p in rv[1]
], [a.subs(reps) for a in rv[2]])
ans = (ans[0].subs(reps).expand(), [
(p[0].subs(reps), p[1].subs(reps)) for p in ans[1]
], [a.subs(reps) for a in ans[2]])
return str(rv[0]) in [str(ans[0]), str(-ans[0])] and \
str(rv[1:]) == str(ans[1:])
assert check(unrad(sqrt(x)), (x, [], []))
assert check(unrad(sqrt(x) + 1), (x - 1, [], []))
assert s_check(unrad(sqrt(x) + x**Rational(1, 3) + 2),
(2 + s**2 + s**3, [(s, x - s**6)], []))
assert check(unrad(sqrt(x) * x**Rational(1, 3) + 2), (x**5 - 64, [], []))
assert check(unrad(sqrt(x) + (x + 1)**Rational(1, 3)),
(x**3 - (x + 1)**2, [], []))
assert check(unrad(sqrt(x) + sqrt(x + 1) + sqrt(2 * x)),
(-2 * sqrt(2) * x - 2 * x + 1, [], []))
assert check(unrad(sqrt(x) + sqrt(x + 1) + 2), (16 * x - 9, [], []))
assert check(unrad(sqrt(x) + sqrt(x + 1) + sqrt(1 - x)),
(-4 * x + 5 * x**2, [], []))
assert check(unrad(a * sqrt(x) + b * sqrt(x) + c * sqrt(y) + d * sqrt(y)),
((a * sqrt(x) + b * sqrt(x))**2 -
(c * sqrt(y) + d * sqrt(y))**2, [], []))
assert check(unrad(sqrt(x) + sqrt(1 - x)), (2 * x - 1, [], []))
assert check(unrad(sqrt(x) + sqrt(1 - x) - 3),
(36 * x + (2 * x - 10)**2 - 36, [], []))
assert check(unrad(sqrt(x) + sqrt(1 - x) + sqrt(2 + x)),
(-5 * x**2 + 2 * x - 1, [], []))
assert check(
unrad(sqrt(x) + sqrt(1 - x) + sqrt(2 + x) - 3),
(-25 * x**4 - 376 * x**3 - 1256 * x**2 + 2272 * x - 784, [], []))
assert check(unrad(sqrt(x) + sqrt(1 - x) + sqrt(2 + x) - sqrt(1 - 2 * x)),
(-41 * x**4 - 40 * x**3 - 232 * x**2 + 160 * x - 16, [], []))
assert check(unrad(sqrt(x) + sqrt(x + 1)), (S(1), [], []))
eq = sqrt(x) + sqrt(x + 1) + sqrt(1 - sqrt(x))
assert check(unrad(eq), (16 * x**3 - 9 * x**2, [], []))
assert solve(eq, check=False) == [0, S(9) / 16]
assert solve(eq) == []
# but this one really does have those solutions
assert solve(sqrt(x) - sqrt(x + 1) + sqrt(1 - sqrt(x))) == [0, S(9) / 16]
'''real_root changes the value of the result if the solution is
simplified; `a` in the text below is the root that is not 4/5:
>>> eq
sqrt(x) + sqrt(-x + 1) + sqrt(x + 1) - 6*sqrt(5)/5
>>> eq.subs(x, a).n()
-0.e-123 + 0.e-127*I
>>> real_root(eq.subs(x, a)).n()
-0.e-123 + 0.e-127*I
>>> (eq.subs(x,simplify(a))).n()
-0.e-126
>>> real_root(eq.subs(x, simplify(a))).n()
0.194825975605452 + 2.15093623885838*I
>>> sqrt(x).subs(x, real_root(a)).n()
0.809823827278194 - 0.e-25*I
>>> sqrt(x).subs(x, (a)).n()
0.809823827278194 - 0.e-25*I
>>> sqrt(x).subs(x, simplify(a)).n()
0.809823827278194 - 5.32999467690853e-25*I
>>> sqrt(x).subs(x, real_root(simplify(a))).n()
0.49864610868139 + 1.44572604257047*I
'''
eq = (sqrt(x) + sqrt(x + 1) + sqrt(1 - x) - 6 * sqrt(5) / 5)
ra = S('''-1484/375 - 4*(-1/2 + sqrt(3)*I/2)*(-12459439/52734375 +
114*sqrt(12657)/78125)**(1/3) - 172564/(140625*(-1/2 +
sqrt(3)*I/2)*(-12459439/52734375 + 114*sqrt(12657)/78125)**(1/3))''')
rb = S(4) / 5
ans = solve(sqrt(x) + sqrt(x + 1) + sqrt(1 - x) - 6 * sqrt(5) / 5)
assert all(abs(eq.subs(x, i).n()) < 1e-10 for i in (ra, rb)) and \
len(ans) == 2 and \
sorted([i.n(chop=True) for i in ans]) == \
sorted([i.n(chop=True) for i in (ra, rb)])
ans = solve(sqrt(x) + sqrt(x + 1) - \
sqrt(1 - x) - sqrt(2 + x))
assert len(ans) == 1 and NS(ans[0])[:4] == '0.73'
# the fence optimization problem
# http://code.google.com/p/sympy/issues/detail?id=1694#c159
F = Symbol('F')
eq = F - (2 * x + 2 * y + sqrt(x**2 + y**2))
X = solve(eq, x, hint='minimal')[0]
Y = solve((x * y).subs(x, X).diff(y), y, simplify=False, minimal=True)
ans = 2 * F / 7 - sqrt(2) * F / 14
assert any((a - ans).expand().is_zero for a in Y)
raises(ValueError,
lambda: unrad(sqrt(x) + sqrt(x + 1) + sqrt(1 - sqrt(x)) + 3))
raises(ValueError,
lambda: unrad(sqrt(x) + (x + 1)**Rational(1, 3) + 2 * sqrt(y)))
# same as last but consider only y
assert check(unrad(sqrt(x) + (x + 1)**Rational(1, 3) + 2 * sqrt(y), y),
(4 * y - (sqrt(x) + (x + 1)**(S(1) / 3))**2, [], []))
assert check(unrad(sqrt(x / (1 - x)) + (x + 1)**Rational(1, 3)),
(x**3 / (-x + 1)**3 - (x + 1)**2, [], [(-x + 1)**3]))
# same as last but consider only y; no y-containing denominators now
assert s_check(unrad(sqrt(x / (1 - x)) + 2 * sqrt(y), y),
(x / (-x + 1) - 4 * y, [], []))
assert check(unrad(sqrt(x) * sqrt(1 - x) + 2, x),
(x * (-x + 1) - 4, [], []))
# http://tutorial.math.lamar.edu/
# Classes/Alg/SolveRadicalEqns.aspx#Solve_Rad_Ex2_a
assert solve(Eq(x, sqrt(x + 6))) == [3]
assert solve(Eq(x + sqrt(x - 4), 4)) == [4]
assert solve(Eq(1, x + sqrt(2 * x - 3))) == []
assert solve(Eq(sqrt(5 * x + 6) - 2, x)) == [-1, 2]
assert solve(Eq(sqrt(2 * x - 1) - sqrt(x - 4), 2)) == [5, 13]
assert solve(Eq(sqrt(x + 7) + 2, sqrt(3 - x))) == [-6]
# http://www.purplemath.com/modules/solverad.htm
assert solve((2 * x - 5)**Rational(1, 3) - 3) == [16]
assert solve((x**3 - 3 * x**2)**Rational(1, 3) + 1 - x) == []
assert solve(x + 1 - (x**4 + 4*x**3 - x)**Rational(1, 4)) == \
[-S(1)/2, -S(1)/3]
assert solve(sqrt(2 * x**2 - 7) - (3 - x)) == [-8, 2]
assert solve(sqrt(2 * x + 9) - sqrt(x + 1) - sqrt(x + 4)) == [0]
assert solve(sqrt(x + 4) + sqrt(2 * x - 1) - 3 * sqrt(x - 1)) == [5]
assert solve(sqrt(x) * sqrt(x - 7) - 12) == [16]
assert solve(sqrt(x - 3) + sqrt(x) - 3) == [4]
assert solve(sqrt(9 * x**2 + 4) - (3 * x + 2)) == [0]
assert solve(sqrt(x) - 2 - 5) == [49]
assert solve(sqrt(x - 3) - sqrt(x) - 3) == []
assert solve(sqrt(x - 1) - x + 7) == [10]
assert solve(sqrt(x - 2) - 5) == [27]
assert solve(sqrt(17 * x - sqrt(x**2 - 5)) - 7) == [3]
assert solve(sqrt(x) - sqrt(x - 1) + sqrt(sqrt(x))) == []
# don't posify the expession in unrad and use _mexpand
z = sqrt(2 * x + 1) / sqrt(x) - sqrt(2 + 1 / x)
p = posify(z)[0]
assert solve(p) == []
assert solve(z) == []
assert solve(z + 6 * I) == [-S(1) / 11]
assert solve(p + 6 * I) == []
def plot_implicit(expr,
x_var=None,
y_var=None,
adaptive=True,
depth=0,
points=300,
line_color="blue",
show=True,
**kwargs):
"""A plot function to plot implicit equations / inequalities.
Arguments
=========
- ``expr`` : The equation / inequality that is to be plotted.
- ``x_var`` (optional) : symbol to plot on x-axis or tuple giving symbol
and range as ``(symbol, xmin, xmax)``
- ``y_var`` (optional) : symbol to plot on y-axis or tuple giving symbol
and range as ``(symbol, ymin, ymax)``
If neither ``x_var`` nor ``y_var`` are given then the free symbols in the
expression will be assigned in the order they are sorted.
The following keyword arguments can also be used:
- ``adaptive`` Boolean. The default value is set to True. It has to be
set to False if you want to use a mesh grid.
- ``depth`` integer. The depth of recursion for adaptive mesh grid.
Default value is 0. Takes value in the range (0, 4).
- ``points`` integer. The number of points if adaptive mesh grid is not
used. Default value is 300.
- ``show`` Boolean. Default value is True. If set to False, the plot will
not be shown. See ``Plot`` for further information.
- ``title`` string. The title for the plot.
- ``xlabel`` string. The label for the x-axis
- ``ylabel`` string. The label for the y-axis
Aesthetics options:
- ``line_color``: float or string. Specifies the color for the plot.
See ``Plot`` to see how to set color for the plots.
Default value is "Blue"
plot_implicit, by default, uses interval arithmetic to plot functions. If
the expression cannot be plotted using interval arithmetic, it defaults to
a generating a contour using a mesh grid of fixed number of points. By
setting adaptive to False, you can force plot_implicit to use the mesh
grid. The mesh grid method can be effective when adaptive plotting using
interval arithmetic, fails to plot with small line width.
Examples
========
Plot expressions:
.. plot::
:context: reset
:format: doctest
:include-source: True
>>> from sympy import plot_implicit, cos, sin, symbols, Eq, And
>>> x, y = symbols('x y')
Without any ranges for the symbols in the expression:
.. plot::
:context: close-figs
:format: doctest
:include-source: True
>>> p1 = plot_implicit(Eq(x**2 + y**2, 5))
With the range for the symbols:
.. plot::
:context: close-figs
:format: doctest
:include-source: True
>>> p2 = plot_implicit(
... Eq(x**2 + y**2, 3), (x, -3, 3), (y, -3, 3))
With depth of recursion as argument:
.. plot::
:context: close-figs
:format: doctest
:include-source: True
>>> p3 = plot_implicit(
... Eq(x**2 + y**2, 5), (x, -4, 4), (y, -4, 4), depth = 2)
Using mesh grid and not using adaptive meshing:
.. plot::
:context: close-figs
:format: doctest
:include-source: True
>>> p4 = plot_implicit(
... Eq(x**2 + y**2, 5), (x, -5, 5), (y, -2, 2),
... adaptive=False)
Using mesh grid without using adaptive meshing with number of points
specified:
.. plot::
:context: close-figs
:format: doctest
:include-source: True
>>> p5 = plot_implicit(
... Eq(x**2 + y**2, 5), (x, -5, 5), (y, -2, 2),
... adaptive=False, points=400)
Plotting regions:
.. plot::
:context: close-figs
:format: doctest
:include-source: True
>>> p6 = plot_implicit(y > x**2)
Plotting Using boolean conjunctions:
.. plot::
:context: close-figs
:format: doctest
:include-source: True
>>> p7 = plot_implicit(And(y > x, y > -x))
When plotting an expression with a single variable (y - 1, for example),
specify the x or the y variable explicitly:
.. plot::
:context: close-figs
:format: doctest
:include-source: True
>>> p8 = plot_implicit(y - 1, y_var=y)
>>> p9 = plot_implicit(x - 1, x_var=x)
"""
has_equality = False # Represents whether the expression contains an Equality,
#GreaterThan or LessThan
def arg_expand(bool_expr):
"""
Recursively expands the arguments of an Boolean Function
"""
for arg in bool_expr.args:
if isinstance(arg, BooleanFunction):
arg_expand(arg)
elif isinstance(arg, Relational):
arg_list.append(arg)
arg_list = []
if isinstance(expr, BooleanFunction):
arg_expand(expr)
#Check whether there is an equality in the expression provided.
if any(
isinstance(e, (Equality, GreaterThan, LessThan))
for e in arg_list):
has_equality = True
elif not isinstance(expr, Relational):
expr = Eq(expr, 0)
has_equality = True
elif isinstance(expr, (Equality, GreaterThan, LessThan)):
has_equality = True
xyvar = [i for i in (x_var, y_var) if i is not None]
free_symbols = expr.free_symbols
range_symbols = Tuple(*flatten(xyvar)).free_symbols
undeclared = free_symbols - range_symbols
if len(free_symbols & range_symbols) > 2:
raise NotImplementedError("Implicit plotting is not implemented for "
"more than 2 variables")
#Create default ranges if the range is not provided.
default_range = Tuple(-5, 5)
def _range_tuple(s):
if isinstance(s, Symbol):
return Tuple(s) + default_range
if len(s) == 3:
return Tuple(*s)
raise ValueError('symbol or `(symbol, min, max)` expected but got %s' %
s)
if len(xyvar) == 0:
xyvar = list(_sort_gens(free_symbols))
var_start_end_x = _range_tuple(xyvar[0])
x = var_start_end_x[0]
if len(xyvar) != 2:
if x in undeclared or not undeclared:
xyvar.append(Dummy('f(%s)' % x.name))
else:
xyvar.append(undeclared.pop())
var_start_end_y = _range_tuple(xyvar[1])
#Check whether the depth is greater than 4 or less than 0.
if depth > 4:
depth = 4
elif depth < 0:
depth = 0
series_argument = ImplicitSeries(expr, var_start_end_x, var_start_end_y,
has_equality, adaptive, depth, points,
line_color)
#set the x and y limits
kwargs['xlim'] = tuple(float(x) for x in var_start_end_x[1:])
kwargs['ylim'] = tuple(float(y) for y in var_start_end_y[1:])
# set the x and y labels
kwargs.setdefault('xlabel', var_start_end_x[0].name)
kwargs.setdefault('ylabel', var_start_end_y[0].name)
p = Plot(series_argument, **kwargs)
if show:
p.show()
return p
def solve_constants(eq, ics, d_ics):
udiff = Eq(d_ics[0][1], eq.rhs.diff(t))
system = [eq.subs(ics), udiff.subs(t, 0)]
consts = solve(system, [C1, C2])
return eq.subs(consts)
def test_dice():
# TODO: Make iid method!
X, Y, Z = Die('X', 6), Die('Y', 6), Die('Z', 6)
a, b, t, p = symbols('a b t p')
assert E(X) == 3 + S.Half
assert variance(X) == Rational(35, 12)
assert E(X + Y) == 7
assert E(X + X) == 7
assert E(a * X + b) == a * E(X) + b
assert variance(X + Y) == variance(X) + variance(Y) == cmoment(X + Y, 2)
assert variance(X + X) == 4 * variance(X) == cmoment(X + X, 2)
assert cmoment(X, 0) == 1
assert cmoment(4 * X, 3) == 64 * cmoment(X, 3)
assert covariance(X, Y) is S.Zero
assert covariance(X, X + Y) == variance(X)
assert density(Eq(cos(X * S.Pi), 1))[True] == S.Half
assert correlation(X, Y) == 0
assert correlation(X, Y) == correlation(Y, X)
assert smoment(X + Y, 3) == skewness(X + Y)
assert smoment(X + Y, 4) == kurtosis(X + Y)
assert smoment(X, 0) == 1
assert P(X > 3) == S.Half
assert P(2 * X > 6) == S.Half
assert P(X > Y) == Rational(5, 12)
assert P(Eq(X, Y)) == P(Eq(X, 1))
assert E(X, X > 3) == 5 == moment(X, 1, 0, X > 3)
assert E(X, Y > 3) == E(X) == moment(X, 1, 0, Y > 3)
assert E(X + Y, Eq(X, Y)) == E(2 * X)
assert moment(X, 0) == 1
assert moment(5 * X, 2) == 25 * moment(X, 2)
assert quantile(X)(p) == Piecewise((nan, (p > 1) | (p < 0)),\
(S.One, p <= Rational(1, 6)), (S(2), p <= Rational(1, 3)), (S(3), p <= S.Half),\
(S(4), p <= Rational(2, 3)), (S(5), p <= Rational(5, 6)), (S(6), p <= 1))
assert P(X > 3, X > 3) is S.One
assert P(X > Y, Eq(Y, 6)) is S.Zero
assert P(Eq(X + Y, 12)) == Rational(1, 36)
assert P(Eq(X + Y, 12), Eq(X, 6)) == Rational(1, 6)
assert density(X + Y) == density(Y + Z) != density(X + X)
d = density(2 * X + Y**Z)
assert d[S(22)] == Rational(1, 108) and d[S(4100)] == Rational(
1, 216) and S(3130) not in d
assert pspace(X).domain.as_boolean() == Or(
*[Eq(X.symbol, i) for i in [1, 2, 3, 4, 5, 6]])
assert where(X > 3).set == FiniteSet(4, 5, 6)
assert characteristic_function(X)(t) == exp(6 * I * t) / 6 + exp(
5 * I * t) / 6 + exp(4 * I * t) / 6 + exp(3 * I * t) / 6 + exp(
2 * I * t) / 6 + exp(I * t) / 6
assert moment_generating_function(X)(
t) == exp(6 * t) / 6 + exp(5 * t) / 6 + exp(4 * t) / 6 + exp(
3 * t) / 6 + exp(2 * t) / 6 + exp(t) / 6
assert median(X) == FiniteSet(3, 4)
D = Die('D', 7)
assert median(D) == FiniteSet(4)
# Bayes test for die
BayesTest(X > 3, X + Y < 5)
BayesTest(Eq(X - Y, Z), Z > Y)
BayesTest(X > 3, X > 2)
# arg test for die
raises(ValueError, lambda: Die('X', -1)) # issue 8105: negative sides.
raises(ValueError, lambda: Die('X', 0))
raises(ValueError, lambda: Die('X', 1.5)) # issue 8103: non integer sides.
# symbolic test for die
n, k = symbols('n, k', positive=True)
D = Die('D', n)
dens = density(D).dict
assert dens == Density(DieDistribution(n))
assert set(dens.subs(n, 4).doit().keys()) == {1, 2, 3, 4}
assert set(dens.subs(n, 4).doit().values()) == {Rational(1, 4)}
k = Dummy('k', integer=True)
assert E(D).dummy_eq(Sum(Piecewise((k / n, k <= n), (0, True)), (k, 1, n)))
assert variance(D).subs(n, 6).doit() == Rational(35, 12)
ki = Dummy('ki')
cumuf = cdf(D)(k)
assert cumuf.dummy_eq(
Sum(Piecewise((1 / n, (ki >= 1) & (ki <= n)), (0, True)), (ki, 1, k)))
assert cumuf.subs({n: 6, k: 2}).doit() == Rational(1, 3)
t = Dummy('t')
cf = characteristic_function(D)(t)
assert cf.dummy_eq(
Sum(Piecewise((exp(ki * I * t) / n, (ki >= 1) & (ki <= n)), (0, True)),
(ki, 1, n)))
assert cf.subs(
n,
3).doit() == exp(3 * I * t) / 3 + exp(2 * I * t) / 3 + exp(I * t) / 3
mgf = moment_generating_function(D)(t)
assert mgf.dummy_eq(
Sum(Piecewise((exp(ki * t) / n, (ki >= 1) & (ki <= n)), (0, True)),
(ki, 1, n)))
assert mgf.subs(n,
3).doit() == exp(3 * t) / 3 + exp(2 * t) / 3 + exp(t) / 3
def set_free_surface(self, d, b, side, algo='robertsson'):
"""
- set free surface boundary condition to boundary d, at index b
:param d: direction of the boundary surface normal
:param b: location of the boundary (index)
:param side: lower boundary (0) or upper boundary (1)
:param algo: which algorithm to use to compute ghost cells
algo == 'robertsson' [1]: setting all velocities at ghost cells to zero
algo == 'levander' [2]: only valid for 4th spatial order. using 2nd order FD approximation for velocities
- e.g. set_free_surface([t,x,y,z],1,2,0)
set y-z plane at x=2 to be lower free surface
- ghost cells are calculated using reflection of stress fields
- store the symbolic equations to populate ghost cells in self.bc
[1] Robertsson, Johan OA. "A numerical free-surface condition for elastic/viscoelastic finite-difference modeling in the presence of topography." Geophysics 61.6 (1996): 1921-1934.
[2] Levander, Alan R. "Fourth-order finite-difference P-SV seismograms." Geophysics 53.11 (1988): 1425-1436.
"""
if algo == 'levander':
# use this stress field to solve for ghost cell expression
field = self.sfields[d]
expr = field.dt
idx = list(self.indices)
# create substituion dictionary
dict1 = {}
derivatives = get_all_objects(expr, DDerivative)
for deriv in derivatives:
# using 2nd order approximation
dict1[deriv] = deriv.fd[2]
expr = expr.subs(dict1)
if self.staggered[d]:
# if staggered, solve ghost cell using T'[b]=0 (e.g. W at z surface, using Tzz)
eq = Eq(expr)
shift = hf
t = b - hf # real boundary location
else:
# if not staggered, solve ghost cell using T'[b-1/2]=T'[b+1/2] (e.g. U at z surface, using Txz)
eq = Eq(expr.subs(idx[d], idx[d]-hf),
expr.subs(idx[d], idx[d]+hf))
shift = 1
t = b
idx[d] -= ((-1)**side)*shift
lhs = self[idx]
rhs = solve(eq, lhs)[0]
lhs = lhs.subs(self.indices[d], t)
rhs = self.align(rhs.subs(self.indices[d], t))
# change ti to t+1
lhs = lhs.subs(idx[0], idx[0]+1)
rhs = rhs.subs(idx[0], idx[0]+1)
self.bc[d][side] = [Eq(lhs, rhs)]
elif algo == 'robertsson':
idx = list(self.indices)
if self.staggered[d]:
# e.g. W at z boundary
idx[d] = b - (1-side)
else:
# e.g. U at z boundary
idx[d] = b - (-1)**side
eq = Eq(self[idx])
eq = eq.subs(idx[0], idx[0]+1)
self.bc[d][side] = [eq]
# populate all ghost cells
for depth in range(self.order[d]/2-1):
idx[d] -= (-1)**side
eq = Eq(self[idx])
eq = eq.subs(idx[0], idx[0]+1)
self.bc[d][side].append(eq)
else:
raise ValueError('Unknown boundary condition algorithm')
def circular_random_walk():
fig, axs = plt.subplots()
plt.xlabel("Position - x")
plt.ylabel("Position - y")
Radius = 100
step_addition_A = 0
step_addition_B = 0
#####
##Choosing initial points
r_A = np.random.uniform(0, Radius)
r_B = np.random.uniform(0, Radius)
thet_A = np.random.uniform(0, 2 * np.pi)
thet_B = np.random.uniform(0, 2 * np.pi)
x_A = r_A * np.cos(thet_A)
y_A = r_A * np.sin(thet_A)
x_B = r_B * np.cos(thet_B)
y_B = r_B * np.sin(thet_B)
####
randomWalk_x_A = [x_A]
randomWalk_y_A = [y_A]
randomWalk_x_B = [x_B]
randomWalk_y_B = [y_B]
flag_A_ref = False
flag_B_ref = False
steps = 0
counter = 0
while True:
angle_A = np.random.uniform(0, 2 * np.pi, size=1)
angle_B = np.random.uniform(0, 2 * np.pi, size=1)
distance_A = np.random.uniform(0, 1, size=1)
distance_B = np.random.uniform(0, 1, size=1)
x_coordinate_A = distance_A[0] * np.cos(angle_A[0])
y_coordinate_A = distance_A[0] * np.sin(angle_A[0])
x_coordinate_B = distance_B[0] * np.cos(angle_B[0])
y_coordinate_B = distance_B[0] * np.sin(angle_B[0])
new_x_A = randomWalk_x_A[-1] + x_coordinate_A
new_y_A = randomWalk_y_A[-1] + y_coordinate_A
new_x_B = randomWalk_x_B[-1] + x_coordinate_B
new_y_B = randomWalk_y_B[-1] + y_coordinate_B
distance_from_origin_A = ((new_x_A)**2 + (new_y_A)**2)**(0.5)
distance_from_origin_B = ((new_x_B)**2 + (new_y_B)**2)**(0.5)
while distance_from_origin_A >= (Radius):
new_x_A = '%.3f' % (new_x_A)
new_y_A = '%.3f' % (new_y_A)
old_x_A = '%.3f' % (randomWalk_x_A[-1])
old_y_A = '%.3f' % (randomWalk_y_A[-1])
p1, p2 = Point(old_x_A, old_y_A), Point(new_x_A, new_y_A)
if p1.equals(p2):
randomWalk_x_A.append(old_x_A)
randomWalk_y_A.append(old_y_A)
flag_A_ref = True
break
line = Line(p1, p2)
a, b, c = line.coefficients
x, y = symbols('x,y')
eq1 = Eq((x)**2 + (y)**2 - (Radius)**2, 0, evaluate=False)
eq2 = Eq((a * x) + (b * y) + c, 0, evaluate=False)
sol = solve([eq1, eq2], [x, y])
#find the point on the circumference where our line will touch, call it circpoint
sol1 = (sol[0][0], sol[0][1])
sol2 = (sol[1][0], sol[1][1])
d1 = ((p2[0] - sol1[0])**2 + (p2[1] - sol1[1])**2)**(0.5)
d2 = ((p2[0] - sol2[0])**2 + (p2[1] - sol2[1])**2)**(0.5)
if d1 <= d2:
circ_point = sol[0]
else:
circ_point = sol[1]
circ_point = Point(circ_point[0], circ_point[1])
#we know the overflow amount, call that overflowed
overflowed = p2.distance(circ_point)
#get the vector from our curr_pos to circpoint, call it incident
incident = np.array([circ_point[0] - p1[0], circ_point[1] - p1[1]])
#circpoint to origin vector
normal = np.array([circ_point[0], circ_point[1]])
#-find the angle between incident and normal using dot product wali equation, call that theta
if p1.equals(circ_point):
randomWalk_x_A.append(old_x_A)
randomWalk_y_A.append(old_y_A)
flag_A_ref = True
break
dot = (circ_point[0] * incident[0]) + (circ_point[1] * incident[1])
magnitude1 = math.sqrt(circ_point[0]**2 + circ_point[1]**2)
magnitude2 = math.sqrt(incident[0]**2 + incident[1]**2)
ang = dot / (magnitude1 * magnitude2)
main_ang = math.acos(ang)
Theta = main_ang
if p2[0] < 0 and p2[1] > 0: #second
Theta = -Theta
elif p2[0] > 0 and p2[1] > 0: #first
if incident[1] >= incident[0]:
Theta = -Theta
elif p2[0] > 0 and p2[1] < 0:
if incident[1] > -incident[0]: #fourth
Theta = -Theta
elif p2[0] < 0 and p2[1] < 0:
if incident[1] <= incident[0]:
Theta = -Theta
#-form a 2x2 rotation matrix, uska format dekh lo aram se mile ga
rotation_matrix = np.array([[np.cos(Theta), -np.sin(Theta)],
[np.sin(Theta),
np.cos(Theta)]])
#-multiply the rotation matrix by the normal vector to get another 2x1 vector, call it reflected
reflected = np.dot(rotation_matrix, normal)
reflected *= -1
#-find the unit vector in the direction of reflected, call it ref_unit
ref_unit = reflected / math.sqrt(reflected[0]**2 + reflected[1]**2)
#-scalar multiply ref_unit with overflow
ref_point = ref_unit * overflowed
#-us vector ke components ko add to circpoint ke coordinates to get the final coordinates
final = circ_point + ref_point
#-add both circ point (optional but will look better) and final coord to the arrays
randomWalk_x_A.append(circ_point[0])
randomWalk_y_A.append(circ_point[1])
distance_from_origin = math.sqrt((final[0])**2 + (final[1])**2)
if distance_from_origin >= Radius:
flag_A_ref = True
break
else:
randomWalk_x_A.append(final[0])
randomWalk_y_A.append(final[1])
step_addition_A += 1
flag_A_ref = True
break
while distance_from_origin_B >= (Radius):
new_x_B = '%.3f' % (new_x_B)
new_y_B = '%.3f' % (new_y_B)
old_x_B = '%.3f' % (randomWalk_x_B[-1])
old_y_B = '%.3f' % (randomWalk_y_B[-1])
p1, p2 = Point(old_x_B, old_y_B), Point(new_x_B, new_y_B)
if p1.equals(p2):
randomWalk_x_B.append(old_x_B)
randomWalk_y_B.append(old_y_B)
flag_B_ref = True
break
line = Line(p1, p2)
a, b, c = line.coefficients
x, y = symbols('x,y')
eq1 = Eq((x)**2 + (y)**2 - (Radius)**2, 0, evaluate=False)
eq2 = Eq((a * x) + (b * y) + c, 0, evaluate=False)
sol = solve([eq1, eq2], [x, y])
#find the point on the circumference where our line will touch, call it circpoint
sol1 = (sol[0][0], sol[0][1])
sol2 = (sol[1][0], sol[1][1])
d1 = ((p2[0] - sol1[0])**2 + (p2[1] - sol1[1])**2)**(0.5)
d2 = ((p2[0] - sol2[0])**2 + (p2[1] - sol2[1])**2)**(0.5)
if d1 <= d2:
circ_point = sol[0]
else:
circ_point = sol[1]
circ_point = Point(circ_point[0], circ_point[1])
#we know the overflow amount, call that overflowed
overflowed = p2.distance(circ_point)
#get the vector from our curr_pos to circpoint, call it incident
incident = np.array([circ_point[0] - p1[0], circ_point[1] - p1[1]])
#circpoint to origin vector
normal = np.array([circ_point[0], circ_point[1]])
#-find the angle between incident and normal using dot product wali equation, call that theta
if p1.equals(circ_point):
randomWalk_x_B.append(old_x_B)
randomWalk_y_B.append(old_y_B)
flag_B_ref = True
break
dot = (circ_point[0] * incident[0]) + (circ_point[1] * incident[1])
magnitude1 = math.sqrt(circ_point[0]**2 + circ_point[1]**2)
magnitude2 = math.sqrt(incident[0]**2 + incident[1]**2)
ang = dot / (magnitude1 * magnitude2)
main_ang = math.acos(ang)
Theta = main_ang
if p2[0] < 0 and p2[1] > 0: #second
Theta = -Theta
elif p2[0] > 0 and p2[1] > 0: #first
if incident[1] >= incident[0]:
Theta = -Theta
elif p2[0] > 0 and p2[1] < 0:
if incident[1] > -incident[0]: #fourth
Theta = -Theta
elif p2[0] < 0 and p2[1] < 0:
if incident[1] <= incident[0]:
Theta = -Theta
#-form a 2x2 rotation matrix, uska format dekh lo aram se mile ga
rotation_matrix = np.array([[np.cos(Theta), -np.sin(Theta)],
[np.sin(Theta),
np.cos(Theta)]])
#-multiply the rotation matrix by the normal vector to get another 2x1 vector, call it reflected
reflected = np.dot(rotation_matrix, normal)
reflected *= -1
#-find the unit vector in the direction of reflected, call it ref_unit
ref_unit = reflected / math.sqrt(reflected[0]**2 + reflected[1]**2)
#-scalar multiply ref_unit with overflow
ref_point = ref_unit * overflowed
#-us vector ke components ko add to circpoint ke coordinates to get the final coordinates
final = circ_point + ref_point
#-add both circ point (optional but will look better) and final coord to the arrays
randomWalk_x_B.append(circ_point[0])
randomWalk_y_B.append(circ_point[1])
distance_from_origin = ((final[0])**2 + (final[1])**2)**(0.5)
if distance_from_origin >= Radius:
flag_B_ref = True
break
else:
randomWalk_x_B.append(final[0])
randomWalk_y_B.append(final[1])
step_addition_B += 1
flag_B_ref = True
break
if flag_A_ref:
pass
else:
randomWalk_x_A.append(new_x_A)
randomWalk_y_A.append(new_y_A)
if flag_B_ref:
pass
else:
randomWalk_x_B.append(new_x_B)
randomWalk_y_B.append(new_y_B)
flag_A_ref = False
flag_B_ref = False
steps += 1
curr_A = (randomWalk_x_A[-1], randomWalk_y_A[-1])
curr_B = (randomWalk_x_B[-1], randomWalk_y_B[-1])
dist_between_particles = ((curr_B[0] - curr_A[0])**2 +
(curr_B[1] - curr_A[1])**2)**(0.5)
counter += 1
if counter % 1000 == 0:
print(counter)
if dist_between_particles <= 1:
break
#return steps
size_A = steps + step_addition_A
size_B = steps + step_addition_B
########################################################
#UNCOMMENT FOR ANIMATION
# xdata, ydata = [], []
# ln, = plt.plot(xdata,ydata)
# def init():
# #Radius = 5.64189584 when area of circle = 100 unit^2
# axs.set_xlim(-Radius,Radius)
# axs.set_ylim(-Radius,Radius)
# ln.set_data([], [])
# return ln,
# def update(frame):
# xdata.append(randomWalk_x[int(frame)])
# ydata.append(randomWalk_y[int(frame)])
# ln.set_data(xdata, ydata)
# return ln,
# ani = FuncAnimation(fig, update, frames=np.linspace(0, size, size+1, endpoint=True), interval = 20,
# init_func=init, blit=True, repeat=False)
#ANIMATION CODE
#######################################################
circle1 = plt.Circle((0, 0), Radius, color='r', alpha=0.2)
plt.gcf().gca().add_artist(circle1)
plt.plot(randomWalk_x_A, randomWalk_y_A)
plt.plot(randomWalk_x_B, randomWalk_y_B)
plt.show()
def test_equality_subs2():
f = Function('f')
eq = Eq(f(x)**2, 16)
assert bool(eq.subs(f(x), 3)) is False
assert bool(eq.subs(f(x), 4)) is True
def test_diophantine():
assert check_solutions((x - y) * (y - z) * (z - x))
assert check_solutions((x - y) * (x**2 + y**2 - z**2))
assert check_solutions((x - 3 * y + 7 * z) * (x**2 + y**2 - z**2))
assert check_solutions(x**2 - 3 * y**2 - 1)
assert check_solutions(y**2 + 7 * x * y)
assert check_solutions(x**2 - 3 * x * y + y**2)
assert check_solutions(z * (x**2 - y**2 - 15))
assert check_solutions(x * (2 * y - 2 * z + 5))
assert check_solutions((x**2 - 3 * y**2 - 1) * (x**2 - y**2 - 15))
assert check_solutions((x**2 - 3 * y**2 - 1) * (y - 7 * z))
assert check_solutions((x**2 + y**2 - z**2) * (x - 7 * y - 3 * z + 4 * w))
# Following test case caused problems in parametric representation
# But this can be solved by factoring out y.
# No need to use methods for ternary quadratic equations.
assert check_solutions(y**2 - 7 * x * y + 4 * y * z)
assert check_solutions(x**2 - 2 * x + 1)
assert diophantine(x - y) == diophantine(Eq(x, y))
# 18196
eq = x**4 + y**4 - 97
assert diophantine(eq, permute=True) == diophantine(-eq, permute=True)
assert diophantine(3 * x * pi - 2 * y * pi) == {(2 * t_0, 3 * t_0)}
eq = x**2 + y**2 + z**2 - 14
base_sol = {(1, 2, 3)}
assert diophantine(eq) == base_sol
complete_soln = set(signed_permutations(base_sol.pop()))
assert diophantine(eq, permute=True) == complete_soln
assert diophantine(x**2 + x * Rational(15, 14) - 3) == set()
# test issue 11049
eq = 92 * x**2 - 99 * y**2 - z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(9, 7, 51)}
assert diophantine(eq) == {
(891 * p**2 + 9 * q**2, -693 * p**2 - 102 * p * q + 7 * q**2,
5049 * p**2 - 1386 * p * q - 51 * q**2)
}
eq = 2 * x**2 + 2 * y**2 - z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(1, 1, 2)}
assert diophantine(eq) == {(2 * p**2 - q**2, -2 * p**2 + 4 * p * q - q**2,
4 * p**2 - 4 * p * q + 2 * q**2)}
eq = 411 * x**2 + 57 * y**2 - 221 * z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(2021, 2645, 3066)}
assert diophantine(eq) == \
{(115197*p**2 - 446641*q**2, -150765*p**2 + 1355172*p*q -
584545*q**2, 174762*p**2 - 301530*p*q + 677586*q**2)}
eq = 573 * x**2 + 267 * y**2 - 984 * z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(49, 233, 127)}
assert diophantine(eq) == \
{(4361*p**2 - 16072*q**2, -20737*p**2 + 83312*p*q - 76424*q**2,
11303*p**2 - 41474*p*q + 41656*q**2)}
# this produces factors during reconstruction
eq = x**2 + 3 * y**2 - 12 * z**2
coeff = eq.as_coefficients_dict()
assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \
{(0, 2, 1)}
assert diophantine(eq) == \
{(24*p*q, 2*p**2 - 24*q**2, p**2 + 12*q**2)}
# solvers have not been written for every type
raises(NotImplementedError, lambda: diophantine(x * y**2 + 1))
# rational expressions
assert diophantine(1 / x) == set()
assert diophantine(1 / x + 1 / y - S.Half) == {(6, 3), (-2, 1), (4, 4),
(1, -2), (3, 6)}
assert diophantine(x**2 + y**2 +3*x- 5, permute=True) == \
{(-1, 1), (-4, -1), (1, -1), (1, 1), (-4, 1), (-1, -1), (4, 1), (4, -1)}
#test issue 18186
assert diophantine(y**4 + x**4 - 2**4 - 3**4, syms=(x, y), permute=True) == \
{(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)}
assert diophantine(y**4 + x**4 - 2**4 - 3**4, syms=(y, x), permute=True) == \
{(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)}
# issue 18122
assert check_solutions(x**2 - y)
assert check_solutions(y**2 - x)
assert diophantine((x**2 - y), t) == {(t, t**2)}
assert diophantine((y**2 - x), t) == {(t**2, -t)}
f = Poly("-x**3 + 3 * x**2 + 10 * x - 24")
vec_x = np.arange(-5, 5, 0.1)
vec_y = np.vectorize(f)(vec_x)
pprint(vec_y)
r = list(roots(f).keys())
print(r)
plt.scatter(r, len(r) * [0])
plot(vec_x, vec_y)
header("[1.1/2]")
x, y = symbols("x, y")
eq_1 = Eq(x**2 + 3 * y, 10)
eq_2 = Eq(4 * x - y**2, -2)
r = solve([eq_1, eq_2], (x, y))
print("solved")
header("[1.1/3]")
pprint(r)
print(f"len(r)={len(r)}")
header("[1.1/4]")
for sol in r:
print(f"x, y = {[sym.evalf() for sym in sol]}")
header("[1.1/5]")
def solve(eq, x):
res = sympy.solve(eq, x)
if len(res) == 1:
return res[0]
else:
return res
half = sympy.Rational(1, 2)
third = sympy.Rational(1, 3)
Kf = sympy.IndexedBase('Kf', shape=(sympy.oo, ))
K_fac = Symbol(
'Kf_1'
) # currently sympy bugs just using Kf[1], so treat as separate symbol
K_fac_sub = 1 - 3 * K / k**2
K_sub = Eq(K, solve(K_fac_sub - K_fac, K))
H_t = diff(a, t) / a
dH = -a**2 / 6 * kappa * (rho + 3 * P)
Friedmann = Eq(H**2, a**2 * kappa * rho / 3 - K)
Friedmann_subs = [Friedmann, Eq(diff(H, t), dH), Eq(diff(a, t), a * H)]
Friedmann_Kfac_subs = subs(K_sub, Friedmann_subs)
delta_frame = Function('uprime')(t)
def LinearPerturbation(name,
species=None,
camb_var=None,
camb_sub=None,
frame_dependence=None,
def test_equality_subs1():
f = Function("f")
x = abc.x
eq = Eq(f(x)**2, x)
res = Eq(Integer(16), x)
assert eq.subs(f(x), 4) == res
def check(pdf, set):
x = Dummy('x')
val = Sum(pdf(x), (x, set._inf, set._sup)).doit()
_value_check(Eq(val, 1) != S.false, "The pdf is incorrect on the given set.")
)
globals().update(get_symbs()) # see common.py: x, Y, Z, k_f, ...
subs = {}
eqs = []
# rate of x
rate_expr = k_f*Y*(Z-x) - k_b*x
rate_eq = Eq(Derivative(x,t), rate_expr)
eqs.append(rate_eq)
integrand = 1/rate_expr
inte_eq_lhs = Integral(integrand.subs({x: chi}), (chi,0,x))
inte_eq_rhs = Integral(1, (tau,0,t))
inte_eq = Eq(inte_eq_lhs, inte_eq_rhs)
eqs.append(inte_eq)
expl_in_x_eq = inte_eq.doit().simplify()
eqs.append(expl_in_x_eq)
expl_in_t_eq = Eq(x, solve(expl_in_x_eq, x)[0])
eqs.append(expl_in_t_eq)
alt_expl_in_t = Z*k_f*Y/(k_f*Y+k_b)*(1-exp(-t*(k_f*Y+k_b)))
assert (expl_in_t_eq.rhs - alt_expl_in_t).simplify() == 0
alt_expl_in_t_eq = Eq(x, alt_expl_in_t)
eqs.append(alt_expl_in_t_eq)
###################################
## 4. Adiabatic compression
###################################
p_i = Symbol("p_i", positive=True)
V_i = Symbol("V_i", positive=True)
tau_i = Symbol("tau_i", positive=True)
N_i = Symbol("N_i", positive=True)
idealgaslaw_i = Eq( p_i*V_i, N_i*tau_i)
p_f = Symbol("p_f", positive=True)
V_f = Symbol("V_f", positive=True)
tau_f = Symbol("tau_f", positive=True)
idealgaslaw_f = Eq( p_f*V_f, N_i*tau_f)
adia_tV = Eq( tau_i*V_i**(gamma-1) , tau_f*V_f**(gamma-1) )
##############################
# (a)
##############################
roomtemp_K = KCconv.subs(T_C,20).rhs # room temperature in Kelvin
Prob0104ans = adia_tV.subs(gamma,1.4).subs(V_f,1).subs(V_i,15).subs(tau_i, roomtemp_K) # answer to Problem 4 of Chapter 1
Prob0104ans = N( Prob0104ans.lhs) # 866.016969686253 K
Prob0104ansC = solve( KCconv.subs( T_K, Prob0104ans), T_C )[0] # 592.866969686253 C
solve( FCconv.subs( T_C, Prob0104ansC ), T_F)[0] # 1099.16054543526 F
##############################
# (b)