# pylab, integral of sinc function

Discussion in 'Python' started by =?ISO-8859-1?Q?Sch=FCle_Daniel?=, Feb 19, 2007.

1. ### =?ISO-8859-1?Q?Sch=FCle_Daniel?=Guest

Hello,

In [19]: def simple_integral(func,a,b,dx = 0.001):
....: return sum(map(lambda x:dx*x, func(arange(a,b,dx))))
....:

In [20]: simple_integral(sin, 0, 2*pi)
Out[20]: -7.5484213527594133e-08

ok, can be thought as zero

In [21]: simple_integral(sinc, -1000, 1000)
Out[21]: 0.99979735786416357

hmm, it should be something around pi
it is a way too far from it, even with a=-10000,b=10000

In [22]: def ppp(x):
....: return sin(x)/x
....:

In [23]: simple_integral(ppp, -1000, 1000)
Out[23]: 3.1404662440661117

nice

is my sinc function in pylab broken?
is there a better way to do numerical integration in pylab?

Regards, Daniel

=?ISO-8859-1?Q?Sch=FCle_Daniel?=, Feb 19, 2007

2. ### =?ISO-8859-1?Q?Sch=FCle_Daniel?=Guest

my fault

In [31]: simple_integral(lambda x:sinc(x/pi), -1000, 1000)
Out[31]: 3.14046624406611

=?ISO-8859-1?Q?Sch=FCle_Daniel?=, Feb 19, 2007

3. ### =?ISO-8859-1?Q?Sch=FCle_Daniel?=Guest

[...]

>> In [19]: def simple_integral(func,a,b,dx = 0.001):
>> ....: return sum(map(lambda x:dx*x, func(arange(a,b,dx))))

>
> Do you mean
>
> def simple_integral(func,a,b,dx = 0.001):
> return dx * sum(map(func, arange(a,b,dx)))
>

yes, this should be faster

=?ISO-8859-1?Q?Sch=FCle_Daniel?=, Feb 19, 2007
4. ### Paul RubinGuest

Schüle Daniel <-karlsruhe.de> writes:
> In [19]: def simple_integral(func,a,b,dx = 0.001):
> ....: return sum(map(lambda x:dx*x, func(arange(a,b,dx))))

Do you mean

def simple_integral(func,a,b,dx = 0.001):
return dx * sum(map(func, arange(a,b,dx)))

Paul Rubin, Feb 20, 2007
5. ### Paul RubinGuest

Schüle Daniel <-karlsruhe.de> writes:
> > return dx * sum(map(func, arange(a,b,dx)))

> yes, this should be faster

You should actually use itertools.imap instead of map, to avoid
creating a big intermediate list. However I was mainly concerned that
the original version might be incorrect. I don't use pylab and don't
know what happens if you pass the output of arange to a function.
I only guessed at what arange does.

Paul Rubin, Feb 20, 2007
6. ### Robert KernGuest

SchÃ¼le Daniel wrote:
> Hello,
>
> In [19]: def simple_integral(func,a,b,dx = 0.001):
> ....: return sum(map(lambda x:dx*x, func(arange(a,b,dx))))
> ....:
>
> In [20]: simple_integral(sin, 0, 2*pi)
> Out[20]: -7.5484213527594133e-08
>
> ok, can be thought as zero
>
> In [21]: simple_integral(sinc, -1000, 1000)
> Out[21]: 0.99979735786416357
>
> hmm, it should be something around pi
> it is a way too far from it, even with a=-10000,b=10000
>
> In [22]: def ppp(x):
> ....: return sin(x)/x
> ....:
>
> In [23]: simple_integral(ppp, -1000, 1000)
> Out[23]: 3.1404662440661117
>
> nice
>
> is my sinc function in pylab broken?

A couple things:

1) The function is not from pylab, it is from numpy.

2) Look at the docstring of the function, and you will notice that the
convention that sinc() uses is different than what you think it is.

In [3]: numpy.sinc?
Type: function
Base Class: <type 'function'>
Namespace: Interactive
File:
/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/numpy-1.0.2.dev3521-py2.5-macosx-10.3-fat.egg/numpy/lib/function_base.py
Definition: numpy.sinc(x)
Docstring:
sinc(x) returns sin(pi*x)/(pi*x) at all points of array x.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
an underlying truth."
-- Umberto Eco

Robert Kern, Feb 20, 2007
7. ### Fernando PerezGuest

SchÃ¼le Daniel wrote:

> Hello,
>
> In [19]: def simple_integral(func,a,b,dx = 0.001):
> ....: return sum(map(lambda x:dx*x, func(arange(a,b,dx))))
> ....:
>
> In [20]: simple_integral(sin, 0, 2*pi)
> Out[20]: -7.5484213527594133e-08
>
> ok, can be thought as zero
>
> In [21]: simple_integral(sinc, -1000, 1000)
> Out[21]: 0.99979735786416357
>
> hmm, it should be something around pi
> it is a way too far from it, even with a=-10000,b=10000
>
> In [22]: def ppp(x):
> ....: return sin(x)/x
> ....:
>
> In [23]: simple_integral(ppp, -1000, 1000)
> Out[23]: 3.1404662440661117
>
> nice
>
> is my sinc function in pylab broken?
> is there a better way to do numerical integration in pylab?

Pylab is mostly a plotting library, which happens (for historical reasons I
won't go into) to expose a small set of numerical algorithms, most of them
actually residing in Numpy. For a more extensive collection of scientific
and numerical algorithms, you should look into using SciPy:

In [34]: import scipy.integrate

In [35]: import scipy as S

In [36]: import scipy.integrate

In [37]: S.integrate.
S.integrate.Inf S.integrate.composite
S.integrate.NumpyTest S.integrate.cumtrapz
S.integrate.__delattr__ S.integrate.inf
S.integrate.__dict__ S.integrate.newton_cotes
S.integrate.__doc__ S.integrate.ode
S.integrate.__file__ S.integrate.odeint
S.integrate.__getattribute__ S.integrate.odepack
S.integrate.__path__ S.integrate.romb
S.integrate.__reduce__ S.integrate.romberg
S.integrate.__reduce_ex__ S.integrate.simps
S.integrate.__repr__ S.integrate.test
S.integrate.__str__ S.integrate.trapz
S.integrate._odepack S.integrate.vode

These will provide dramatically faster performance, and far better
algorithmic control, than the simple_integral:

In [4]: time simple_integral(lambda x:sinc(x/pi), -100, 100)
CPU times: user 7.08 s, sys: 0.42 s, total: 7.50 s
Wall time: 7.58
Out[4]: 3.1244509352

In [40]: time S.integrate.quad(lambda x:sinc(x/pi), -100, 100)
CPU times: user 0.05 s, sys: 0.00 s, total: 0.05 s
Wall time: 0.06
Out[40]: (3.124450933778113, 6.8429604895257158e-10)

Note that I used only -100,100 as the limits so I didn't have to wait
forever for simple_integral to finish.

As you know, this is a nasty, highly oscillatory integral for which almost
any 'black box' method will have problems, but at least scipy is nice
enough to let you know:

In [41]: S.integrate.quad(lambda x:sinc(x/pi), -1000, 1000)
Warning: The maximum number of subdivisions (50) has been achieved.
If increasing the limit yields no improvement it is advised to analyze
the integrand in order to determine the difficulties. If the position of
a
local difficulty can be determined (singularity, discontinuity) one will
probably gain from splitting up the interval and calling the integrator
on the subranges. Perhaps a special-purpose integrator should be used.
Out[41]: (3.5354545588973298, 1.4922039610659907)

In [42]: S.integrate.quad(lambda x:sinc(x/pi), -1000, 1000,limit=1000)
Out[42]: (3.1404662439375475, 4.5659823144674379e-08)

Cheers,

f

ps - the 2nd number is the error estimate.

Fernando Perez, Feb 23, 2007