def test_odefun_harmonic():
mp.dps = 15
# Harmonic oscillator
f = odefun(lambda x, y: [-y[1], y[0]], 0, [1, 0])
for x in [0, 1, 2.5, 8, 3.7]: # we go back to 3.7 to check caching
c, s = f(x)
assert c.ae(cos(x))
assert s.ae(sin(x))
def test_odefun_harmonic():
mp.dps = 15
# Harmonic oscillator
f = odefun(lambda x, y: [-y[1], y[0]], 0, [1, 0])
for x in [0, 1, 2.5, 8, 3.7]: # we go back to 3.7 to check caching
c, s = f(x)
assert c.ae(cos(x))
assert s.ae(sin(x))
def test_odefun_sinc_large():
mp.dps = 15
# Sinc function; test for large x
f = sinc
g = odefun(lambda x, y: [(cos(x) - y[0]) / x],
1, [f(1)],
tol=0.01,
degree=5)
assert abs(f(100) - g(100)[0]) / f(100) < 0.01
plt.plot(space, fapprox)
legendList.append('series, N = ' + str(n))
plt.xlabel('x')
plt.ylabel('f')
plt.xlim(space[0], space[-1])
plt.ylim(-2, 2)
#plt.title("f = %s, together with Taylor approximations of f" % f)
plt.legend(legendList, loc=3)
#plt.savefig('../figs/sympy_exercise2b.png')
#### Problem 3a ####
from sympy.mpmath import odefun
theta0 = 0.1 # initial value of thetea
theta = odefun(lambda t, y: [y[1], -y[0]],
0, [theta0, 0],
tol=0.5,
degree=2,
method='taylor')
"""odefun solves an ODE initial value problem using Taylors method.
odefun(lambda t, y: [y[1], -y[0]], 0, [theta0, 0] ) means that t is the independent variable to perform series expansions around.
y: [[y[1], -y[0]]], is the derivates of y written in the standard form of higher order ODE's written as a set of 1st order ODE's:
[y1', y2']
0, means that the initial values are taken at t = 0. this is also the t that odefun will evaluate the taylor series around
[theta0, 0] are the initial values of [y1, y2]"""
time = np.linspace(0, 5 * 2 * np.pi, 10)
y = np.zeros_like(time)
for k, t in enumerate(time):
y[k] = theta(
t
)[0] # evaluate the solution at time t, theta(t) is a vector of [theta, theta']. [0] means assigning Y[k] to theta
def test_odefun_sinc_large():
mp.dps = 15
# Sinc function; test for large x
f = sinc
g = odefun(lambda x, y: [(cos(x)-y[0])/x], 1, [f(1)], tol=0.01, degree=5)
assert abs(f(100) - g(100)[0])/f(100) < 0.01
def test_odefun_rational():
mp.dps = 15
# A rational function
f = lambda t: 1/(1+mpf(t)**2)
g = odefun(lambda x, y: [-2*x*y[0]**2], 0, [f(0)])
assert f(2).ae(g(2)[0])
def test_odefun_rational():
mp.dps = 15
# A rational function
f = lambda t: 1 / (1 + mpf(t)**2)
g = odefun(lambda x, y: [-2 * x * y[0]**2], 0, [f(0)])
assert f(2).ae(g(2)[0])
fapprox = fTaylorfunc(space) # calculate from 0 to 2pi
plt.plot(space, fapprox)
legendList.append('series, N = ' + str(n))
plt.xlabel('x')
plt.ylabel('f')
plt.xlim(space[0], space[-1])
plt.ylim(-2, 2)
#plt.title("f = %s, together with Taylor approximations of f" % f)
plt.legend(legendList, loc=3)
#plt.savefig('../figs/sympy_exercise2b.png')
#### Problem 3a ####
from sympy.mpmath import odefun
theta0 = 0.1 # initial value of thetea
theta = odefun(lambda t, y: [y[1], -y[0]], 0, [theta0, 0], tol = 0.5, degree=2, method='taylor')
"""odefun solves an ODE initial value problem using Taylors method.
odefun(lambda t, y: [y[1], -y[0]], 0, [theta0, 0] ) means that t is the independent variable to perform series expansions around.
y: [[y[1], -y[0]]], is the derivates of y written in the standard form of higher order ODE's written as a set of 1st order ODE's:
[y1', y2']
0, means that the initial values are taken at t = 0. this is also the t that odefun will evaluate the taylor series around
[theta0, 0] are the initial values of [y1, y2]"""
time = np.linspace(0,5*2*np.pi, 10)
y = np.zeros_like(time)
for k, t in enumerate(time):
y[k] = theta(t)[0] # evaluate the solution at time t, theta(t) is a vector of [theta, theta']. [0] means assigning Y[k] to theta
sampletimes = np.linspace(0, 1, 4)
print "analytic: Taylors solution:"
for sampletime in sampletimes:
