numerical integration methods question python

I am trying to integrate the following formula:

enter image description here

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?

edited Nov 12 '18 at 23:07

Matthias Ossadnik

57427

asked Nov 12 '18 at 19:57

RedPen

14317

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

add a comment |

I am trying to integrate the following formula:

enter image description here

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?

edited Nov 12 '18 at 23:07

Matthias Ossadnik

57427

asked Nov 12 '18 at 19:57

RedPen

14317

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

add a comment |

I am trying to integrate the following formula:

enter image description here

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?

edited Nov 12 '18 at 23:07

Matthias Ossadnik

57427

asked Nov 12 '18 at 19:57

RedPen

14317

I am trying to integrate the following formula:

enter image description here

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

RedPen

14317

edited Nov 12 '18 at 23:07

Matthias Ossadnik

57427

asked Nov 12 '18 at 19:57

RedPen

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

asked Nov 12 '18 at 19:57

RedPen

14317

asked Nov 12 '18 at 19:57

RedPen

14317

asked Nov 12 '18 at 19:57

RedPen

14317

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

add a comment |

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

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

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

edited Nov 12 '18 at 20:29

answered Nov 12 '18 at 20:22

Matthias Ossadnik

57427

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

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

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

edited Nov 12 '18 at 20:29

answered Nov 12 '18 at 20:22

Matthias Ossadnik

57427

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

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

edited Nov 12 '18 at 20:29

answered Nov 12 '18 at 20:22

Matthias Ossadnik

57427

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

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

edited Nov 12 '18 at 20:29

answered Nov 12 '18 at 20:22

Matthias 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

edited Nov 12 '18 at 20:29

answered Nov 12 '18 at 20:22

Matthias Ossadnik

57427

edited Nov 12 '18 at 20:29

answered Nov 12 '18 at 20:22

Matthias Ossadnik

57427

answered Nov 12 '18 at 20:22

Matthias Ossadnik

57427

answered Nov 12 '18 at 20:22

Matthias Ossadnik

57427

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

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

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

@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

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

@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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