numerical integration methods question python
I am trying to integrate the following formula:
Below is my attempt to perform this integration using a homemade scheme for f(a) = sin(a).
def func(x):
return math.sin(x)
def integration (f, n, r, a, dtheta ):
summation = 0
theta = 0
while theta <= 2*np.pi:
f_arg = a + r*np.exp(1j*theta)
second = np.exp(-1j*theta*n)
summation += f(f_arg) * second * dtheta
theta += dtheta
return math.factorial(n)*summation / (2*np.pi*r**n)
integration(func, n=1, r=1, a=0, dtheta=2*np.pi/10000)
The first derivative (n=1) of f(a) = sin(a) is f'(a) = cos(a). When evaluated at a = 0, this should give cos(0) = 1, however, it does not. Where am I going wrong?
python numerical-methods
edited Nov 12 '18 at 23:07
Matthias Ossadnik
57427
asked Nov 12 '18 at 19:57
RedPenRedPen
14317
add a comment |
I am trying to integrate the following formula:
Below is my attempt to perform this integration using a homemade scheme for f(a) = sin(a).
def func(x):
return math.sin(x)
def integration (f, n, r, a, dtheta ):
summation = 0
theta = 0
while theta <= 2*np.pi:
f_arg = a + r*np.exp(1j*theta)
second = np.exp(-1j*theta*n)
summation += f(f_arg) * second * dtheta
theta += dtheta
return math.factorial(n)*summation / (2*np.pi*r**n)
integration(func, n=1, r=1, a=0, dtheta=2*np.pi/10000)
The first derivative (n=1) of f(a) = sin(a) is f'(a) = cos(a). When evaluated at a = 0, this should give cos(0) = 1, however, it does not. Where am I going wrong?
python numerical-methods
edited Nov 12 '18 at 23:07
Matthias Ossadnik
57427
asked Nov 12 '18 at 19:57
RedPenRedPen
14317
return math.factorial(n) / (2*np.pi*r**n)doesn't seem to involvef,theta,dtheta, or anything you did in thewhileloop.
– user2357112
Nov 12 '18 at 20:00
@user2357112, thank you for attention, updating question (as that was a typo), still didn't fix.
– RedPen
Nov 12 '18 at 20:01
add a comment |
I am trying to integrate the following formula:
Below is my attempt to perform this integration using a homemade scheme for f(a) = sin(a).
def func(x):
return math.sin(x)
def integration (f, n, r, a, dtheta ):
summation = 0
theta = 0
while theta <= 2*np.pi:
f_arg = a + r*np.exp(1j*theta)
second = np.exp(-1j*theta*n)
summation += f(f_arg) * second * dtheta
theta += dtheta
return math.factorial(n)*summation / (2*np.pi*r**n)
integration(func, n=1, r=1, a=0, dtheta=2*np.pi/10000)
The first derivative (n=1) of f(a) = sin(a) is f'(a) = cos(a). When evaluated at a = 0, this should give cos(0) = 1, however, it does not. Where am I going wrong?
python numerical-methods
edited Nov 12 '18 at 23:07
Matthias Ossadnik
57427
asked Nov 12 '18 at 19:57
RedPenRedPen
14317
I am trying to integrate the following formula:
Below is my attempt to perform this integration using a homemade scheme for f(a) = sin(a).
def func(x):
return math.sin(x)
def integration (f, n, r, a, dtheta ):
summation = 0
theta = 0
while theta <= 2*np.pi:
f_arg = a + r*np.exp(1j*theta)
second = np.exp(-1j*theta*n)
summation += f(f_arg) * second * dtheta
theta += dtheta
return math.factorial(n)*summation / (2*np.pi*r**n)
integration(func, n=1, r=1, a=0, dtheta=2*np.pi/10000)
The first derivative (n=1) of f(a) = sin(a) is f'(a) = cos(a). When evaluated at a = 0, this should give cos(0) = 1, however, it does not. Where am I going wrong?
python numerical-methods
python numerical-methods
edited Nov 12 '18 at 23:07
Matthias Ossadnik
57427
asked Nov 12 '18 at 19:57
RedPenRedPen
14317
edited Nov 12 '18 at 23:07
Matthias Ossadnik
57427
asked Nov 12 '18 at 19:57
RedPenRedPen
14317
edited Nov 12 '18 at 23:07
Matthias Ossadnik
57427
edited Nov 12 '18 at 23:07
Matthias Ossadnik
57427
edited Nov 12 '18 at 23:07
Matthias Ossadnik
57427
57427
asked Nov 12 '18 at 19:57
RedPenRedPen
14317
asked Nov 12 '18 at 19:57
RedPenRedPen
14317
asked Nov 12 '18 at 19:57
RedPenRedPen
14317
14317
return math.factorial(n) / (2*np.pi*r**n)doesn't seem to involvef,theta,dtheta, or anything you did in thewhileloop.
– user2357112
Nov 12 '18 at 20:00
@user2357112, thank you for attention, updating question (as that was a typo), still didn't fix.
– RedPen
Nov 12 '18 at 20:01
add a comment |
return math.factorial(n) / (2*np.pi*r**n)doesn't seem to involvef,theta,dtheta, or anything you did in thewhileloop.
– user2357112
Nov 12 '18 at 20:00
@user2357112, thank you for attention, updating question (as that was a typo), still didn't fix.
– RedPen
Nov 12 '18 at 20:01
return math.factorial(n) / (2*np.pi*r**n) doesn't seem to involve f, theta, dtheta, or anything you did in the while loop.– user2357112
Nov 12 '18 at 20:00
return math.factorial(n) / (2*np.pi*r**n) doesn't seem to involve f, theta, dtheta, or anything you did in the while loop.– user2357112
Nov 12 '18 at 20:00
@user2357112, thank you for attention, updating question (as that was a typo), still didn't fix.
– RedPen
Nov 12 '18 at 20:01
@user2357112, thank you for attention, updating question (as that was a typo), still didn't fix.
– RedPen
Nov 12 '18 at 20:01
add a comment |
1 Answer
1
active
oldest
votes
It seems that your problem is the math.sin function, which doesn't support complex arguments:
i = np.exp(.5j * np.pi)
math.sin(i), np.sin(i)
(6.123233995736766e-17, (9.44864380126377e-17+1.1752011936438014j))
It also throws a warning (but not an error...):
ComplexWarning: Casting complex values to real discards the imaginary part
Using np.sin instead fixes the problem.
In general, the implementation might be simpler to express (and easier to debug) with more use of numpy, like
def integration(func, a, n, r, n_steps):
z = r * np.exp(2j * np.pi * np.arange(0, 1, 1. / n_steps))
return math.factorial(n) * np.mean(func(a + z) / z**n)
np.allclose(1., integration(np.sin, a=0., n=1, r=1., n_steps=100))
True
answered Nov 12 '18 at 20:22
Matthias OssadnikMatthias Ossadnik
57427
1
Replacingmath.sinwithnp.sinisn't enough to produce a result of 1 when I try it.
– user2357112
Nov 12 '18 at 22:45
1
@user2357112 in the original example there is also n=2 used instead of n=1 in the function call. If you change this it is OK. Have edited question.
– Matthias Ossadnik
Nov 12 '18 at 22:49
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It seems that your problem is the math.sin function, which doesn't support complex arguments:
i = np.exp(.5j * np.pi)
math.sin(i), np.sin(i)
(6.123233995736766e-17, (9.44864380126377e-17+1.1752011936438014j))
It also throws a warning (but not an error...):
ComplexWarning: Casting complex values to real discards the imaginary part
Using np.sin instead fixes the problem.
In general, the implementation might be simpler to express (and easier to debug) with more use of numpy, like
def integration(func, a, n, r, n_steps):
z = r * np.exp(2j * np.pi * np.arange(0, 1, 1. / n_steps))
return math.factorial(n) * np.mean(func(a + z) / z**n)
np.allclose(1., integration(np.sin, a=0., n=1, r=1., n_steps=100))
True
answered Nov 12 '18 at 20:22
Matthias OssadnikMatthias Ossadnik
57427
1
Replacingmath.sinwithnp.sinisn't enough to produce a result of 1 when I try it.
– user2357112
Nov 12 '18 at 22:45
1
@user2357112 in the original example there is also n=2 used instead of n=1 in the function call. If you change this it is OK. Have edited question.
– Matthias Ossadnik
Nov 12 '18 at 22:49
add a comment |
It seems that your problem is the math.sin function, which doesn't support complex arguments:
i = np.exp(.5j * np.pi)
math.sin(i), np.sin(i)
(6.123233995736766e-17, (9.44864380126377e-17+1.1752011936438014j))
It also throws a warning (but not an error...):
ComplexWarning: Casting complex values to real discards the imaginary part
Using np.sin instead fixes the problem.
In general, the implementation might be simpler to express (and easier to debug) with more use of numpy, like
def integration(func, a, n, r, n_steps):
z = r * np.exp(2j * np.pi * np.arange(0, 1, 1. / n_steps))
return math.factorial(n) * np.mean(func(a + z) / z**n)
np.allclose(1., integration(np.sin, a=0., n=1, r=1., n_steps=100))
True
answered Nov 12 '18 at 20:22
Matthias OssadnikMatthias Ossadnik
57427
1
Replacingmath.sinwithnp.sinisn't enough to produce a result of 1 when I try it.
– user2357112
Nov 12 '18 at 22:45
1
@user2357112 in the original example there is also n=2 used instead of n=1 in the function call. If you change this it is OK. Have edited question.
– Matthias Ossadnik
Nov 12 '18 at 22:49
add a comment |
It seems that your problem is the math.sin function, which doesn't support complex arguments:
i = np.exp(.5j * np.pi)
math.sin(i), np.sin(i)
(6.123233995736766e-17, (9.44864380126377e-17+1.1752011936438014j))
It also throws a warning (but not an error...):
ComplexWarning: Casting complex values to real discards the imaginary part
Using np.sin instead fixes the problem.
In general, the implementation might be simpler to express (and easier to debug) with more use of numpy, like
def integration(func, a, n, r, n_steps):
z = r * np.exp(2j * np.pi * np.arange(0, 1, 1. / n_steps))
return math.factorial(n) * np.mean(func(a + z) / z**n)
np.allclose(1., integration(np.sin, a=0., n=1, r=1., n_steps=100))
True
answered Nov 12 '18 at 20:22
Matthias OssadnikMatthias Ossadnik
57427
It seems that your problem is the math.sin function, which doesn't support complex arguments:
i = np.exp(.5j * np.pi)
math.sin(i), np.sin(i)
(6.123233995736766e-17, (9.44864380126377e-17+1.1752011936438014j))
It also throws a warning (but not an error...):
ComplexWarning: Casting complex values to real discards the imaginary part
Using np.sin instead fixes the problem.
In general, the implementation might be simpler to express (and easier to debug) with more use of numpy, like
def integration(func, a, n, r, n_steps):
z = r * np.exp(2j * np.pi * np.arange(0, 1, 1. / n_steps))
return math.factorial(n) * np.mean(func(a + z) / z**n)
np.allclose(1., integration(np.sin, a=0., n=1, r=1., n_steps=100))
True
answered Nov 12 '18 at 20:22
Matthias OssadnikMatthias Ossadnik
57427
edited Nov 12 '18 at 20:29
answered Nov 12 '18 at 20:22
Matthias OssadnikMatthias Ossadnik
57427
answered Nov 12 '18 at 20:22
Matthias OssadnikMatthias Ossadnik
57427
answered Nov 12 '18 at 20:22
Matthias OssadnikMatthias Ossadnik
57427
57427
1
Replacingmath.sinwithnp.sinisn't enough to produce a result of 1 when I try it.
– user2357112
Nov 12 '18 at 22:45
1
@user2357112 in the original example there is also n=2 used instead of n=1 in the function call. If you change this it is OK. Have edited question.
– Matthias Ossadnik
Nov 12 '18 at 22:49
add a comment |
1
Replacingmath.sinwithnp.sinisn't enough to produce a result of 1 when I try it.
– user2357112
Nov 12 '18 at 22:45
1
@user2357112 in the original example there is also n=2 used instead of n=1 in the function call. If you change this it is OK. Have edited question.
– Matthias Ossadnik
Nov 12 '18 at 22:49
1
1
Replacing
math.sin with np.sin isn't enough to produce a result of 1 when I try it.– user2357112
Nov 12 '18 at 22:45
Replacing
math.sin with np.sin isn't enough to produce a result of 1 when I try it.– user2357112
Nov 12 '18 at 22:45
1
1
@user2357112 in the original example there is also n=2 used instead of n=1 in the function call. If you change this it is OK. Have edited question.
– Matthias Ossadnik
Nov 12 '18 at 22:49
@user2357112 in the original example there is also n=2 used instead of n=1 in the function call. If you change this it is OK. Have edited question.
– Matthias Ossadnik
Nov 12 '18 at 22:49
add a comment |
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return math.factorial(n) / (2*np.pi*r**n)doesn't seem to involvef,theta,dtheta, or anything you did in thewhileloop.– user2357112
Nov 12 '18 at 20:00
@user2357112, thank you for attention, updating question (as that was a typo), still didn't fix.
– RedPen
Nov 12 '18 at 20:01