I am trying to write a function to compute 1st derivative with backward difference approximation. $ u'(x_i) = \frac{u(x_i) - u(x_i - \Delta_x)}{\Delta x} \equiv D_- u(x_i).$ And for the first point, I use the forward difference approximation that mirrors the one I just found.
My problem is the function is not accurate, so that I cannot pass the test case. I got error message
AssertionError:
Not equal to tolerance rtol=1e-07, atol=0.001
Mismatched elements: 150 / 150 (100%)
Max absolute difference: 0.0585835752066481
Max relative difference: 0.0184951666851795
x: array([0.519974, 0.519974, 0.538915, 0.557941, 0.577058, 0.596266,
0.61557 , 0.634974, 0.654479, 0.67409 , 0.69381 , 0.713641,
0.733589, 0.753655, 0.773844, 0.794158, 0.814602, 0.835178,...
y: array([0.510532, 0.52943 , 0.548413, 0.567484, 0.586646, 0.605902,
0.625255, 0.644709, 0.664267, 0.683931, 0.703707, 0.723596,
0.743602, 0.763729, 0.78398 , 0.804358, 0.824868, 0.845512,...
So my function itself works, just for some reason the error beyond the tolerance. Is there any logic fault? I really cannot figure our by myself how can I make progress.
Please help with this.
Thank you so much
def compute_prime(x, f):
"""Compute the first derivative"""
N = 150
x_hat = numpy.linspace(x[0], x[-1], N)
delta_x = x_hat[1] - x_hat[0]
f_prime_hat = numpy.empty(x_hat.shape)
for i in range(1, N):
f_prime_hat[i] = (f(x_hat[i]) - f(x_hat[i-1]))/delta_x
#f_prime_hat[0] = (f(x_hat[i+1])-f(x_hat[i]))/delta_x
f_prime_hat[0] = (f(x_hat[1]) - f(x_hat[0])) / delta_x
return f_prime_hat
And the test case is
f = lambda x: x**3 / numpy.sin(x)
f_prime = lambda x: -x**3 * numpy.cos(x) / numpy.sin(x)**2 + 3.0 * x**2 / numpy.sin(x)
x = numpy.linspace(0.25, 0.5 * numpy.pi, 150)
numpy.testing.assert_allclose(compute_prime(x, f), f_prime(x), atol=1e-3)
print("Success!")
Second order method
I am trying to write a function to compute 2nd order 1st derivative with backward difference approximation. $u'(x_n) = \frac{3u(x_n) - 4 u(x_{n-1}) + u(x_{n-2})}{2 \Delta x}$
And for the first point, I use the forward difference approximation $u'(x_n) = \frac{-2u(x_n+2) + 4 u(x_{n+1}) - 3u(x_{n})}{2 \Delta x} $ that mirrors the one I just found.
My problem is the function is not accurate, so that I cannot pass the test case.
I got error message
AssertionError:
Not equal to tolerance rtol=1e-07, atol=0.001
Mismatched elements: 150 / 150 (100%)
Max absolute difference: 7.401621709928578
Max relative difference: 1.004627870636921
x: array([-2.362674e-03, 1.695963e-02, 4.309081e-05, 4.458928e-05,
4.609489e-05, 4.760789e-05, 4.912854e-05, 5.065709e-05,
5.219379e-05, 5.373891e-05, 5.529271e-05, 5.685546e-05,...
y: array([0.510532, 0.52943 , 0.548413, 0.567484, 0.586646, 0.605902,
0.625255, 0.644709, 0.664267, 0.683931, 0.703707, 0.723596,
0.743602, 0.763729, 0.78398 , 0.804358, 0.824868, 0.845512,...
I think how I find the first point is not correct, but really not sure. Please point it out where did I make mistakes.
Please help with this.
Thank you so much
def compute_prime(x, f):
"""Compute the first derivative"""
N = 150
x_hat = numpy.linspace(x[0], x[-1], N)
delta_x = x_hat[1] - x_hat[0]
f_prime_hat = numpy.empty(x_hat.shape)
for i in range(1, N):
f_prime_hat[i] = (3*f(x_hat[i]) - 4*f(x_hat[i-1]) + f(x_hat[i-2]))/2*delta_x
#f_prime_hat[0] = (-f(x_hat[i+2])-4*f(x_hat[i+1])- 3*f(x_hat[i]))/2*delta_x
f_prime_hat[0] = (-f(x_hat[2])-4*f(x_hat[1])- 3*f(x_hat[0]))/2*delta_x
return f_prime_hat
And the test case is
f = lambda x: x**3 / numpy.sin(x)
f_prime = lambda x: -x**3 * numpy.cos(x) / numpy.sin(x)**2 + 3.0 * x**2 / numpy.sin(x)
x = numpy.linspace(0.25, 0.5 * numpy.pi, 150)
numpy.testing.assert_allclose(compute_prime(x, f), f_prime(x), atol=1e-3)
print("Success!")
