and also
In[8]:= Integrate[DiracDelta[-1 + Sqrt[x^2 + y^2 + z^2]], {x, -Infinity,
Infinity}, {y, -Infinity, Infinity}, {z, -Infinity, Infinity}]
Out[8]= 4 Pi
On 30 December 2011 12:39, [email protected] <
[email protected]> wrote:
> I vaguely remember the trajectory of a relativistic particle presented in
> jackson's electrodynamics. I suppose it's convoluted enough for use in
> tests.
>
> A simpler one will be some surface density over a strange surface. For
> example the total charge of a unitary surface charge distribution on a
> spherical surface (of radius = 1) will be:
>
> <delta(sqrt(x**2+y**2+z**2)-1) | 1> = 4 pi
>
> Well this was just the surface of the sphere so it was not that
> interesting, but one can calculate for example a field using such integral.
>
> I'll try to dig some nice examples.
>
> PS Mathematica can calculate those (at least in 1D and 2D)
>
> In[6]:= Integrate[DiracDelta[-1 + Sqrt[x^2]], {x, -Infinity, Infinity}]
> Out[6]= 2
>
> In[7]:= Integrate[DiracDelta[-1 + Sqrt[x^2 + y^2]], {x, -Infinity,
> Infinity}, {y, -Infinity, Infinity}]
> Out[7]= 2 Pi
>
>
>
> On 30 December 2011 01:22, Aaron Meurer <[email protected]> wrote:
>
>> Does someone know of an example integral containing a delta function
>> with a non-linear argument, with the correct answer? That would help
>> with trying to find the algorithm.
>>
>> Perhaps it can be done by a change of variable, to make the argument
>> linear. You probably have to be careful about some things, though,
>> since delta functions don't follow all the "rules" that normal
>> functions do.
>>
>> Aaron Meurer
>>
>> On Tue, Dec 20, 2011 at 12:51 PM, Matthew Rocklin <[email protected]>
>> wrote:
>> > @Tom - I agree that it makes sense to separate out the common easy case
>> from
>> > the very rare very difficult case.
>> > @Stephan - In the multivariate case the zeros are likely to consist of
>> > continuous lines/surfaces/manifolds of co-dimension 1. We'll need to do
>> an
>> > integral over complex surfaces. This could be challenging.
>> >
>> > From my experience deltaintegrate has some issues, even in the linear
>> case.
>> > The code also seems like it might be more complex than necessary,
>> allowing
>> > odd bugs to filter through. Wolfram handles deltas more robustly than
>> SymPy
>> > does so at least we know that there exists a nicer solution. I wonder
>> if a
>> > system something like what Stephan outlined would be cleaner/more
>> robust.
>> >
>> >
>> > On Tue, Dec 20, 2011 at 11:26 AM, [email protected]
>> > <[email protected]> wrote:
>> >>
>> >> @Tom, the operation is well defined for all sufficiently nice
>> functions.
>> >> @Matthew, The notation that is used there is a bit unclear (for me).
>> The
>> >> domain is all zeros of g and the sigma is just some measure theory
>> notation
>> >> meaning that you should actually take the sum.
>> >>
>> >> To solve this in sympy you can try the following thing (it depends on
>> how
>> >> good 'solve' is in sympy):
>> >>
>> >> 0) to solve integrate(delta(f(x,y,...)*g(x,y,...)), (x,...), (y,...),
>> ...)
>> >> 1) take out the g(x,y,...) function from the delta function
>> >> 2) find the zeros of g(...) (presumably using 'solve')
>> >> 3) evaluate the derivatives of g(...) at those zeros
>> >> 4) your integral is equal to the sum of f(...) / |grad(g(...))|
>> evaluated
>> >> at those zeros.
>> >>
>> >> I think that a nice explanation can be found in Appel. I can send a pdf
>> >> with the book to you if you are interested.
>> >>
>> >>
>> >> On 20 December 2011 19:45, Tom Bachmann <[email protected]> wrote:
>> >>>
>> >>> I think there are two basic cases to consider:
>> >>>
>> >>> 1) delta functions with *linear* arguments.
>> >>>
>> >>> This is (I believe) by a large margin the most common case. For this
>> >>> deltaintegrate should simply be fixed; one only needs to correctly
>> implement
>> >>> integral(f(x) delta(a*x+b), (x, c, d)) and this shouldn't be hard.
>> (Note
>> >>> that this will also yield the multi-dimensional cases, by repeated
>> >>> integration.)
>> >>>
>> >>> 2) delta functions with more complicated arguments
>> >>>
>> >>> Getting this to work is (I believe) much more work. It is non-trivial
>> to
>> >>> even define this.
>> >>>
>> >>> Tom
>> >>>
>> >>>
>> >>> On 20.12.2011 16:21, Matthew Rocklin wrote:
>> >>>>
>> >>>> Hi Everyone,
>> >>>>
>> >>>> I'd like to compute multivariate integrals that contain Dirac
>> >>>> Deltafunctions. I.e. expressions like
>> >>>>
>> >>>> integrate(exp(-(x**2+y**2))/pi * delta(2*x+3*y), (x,-oo, oo), (y,-oo,
>> >>>> oo))
>> >>>>
>> >>>> The deltaintegrate function inside sympy fails to compute these
>> >>>> correctly, see issue 2630.
>> >>>> <http://code.google.com/p/sympy/issues/detail?id=2630>
>> >>>>
>> >>>> Wikipedia says
>> >>>>
>> >>>> <
>> http://en.wikipedia.org/wiki/Dirac_delta_functions#Properties_in_n_dimensions
>> >
>> >>>>
>> >>>> that you can compute general expressions of this form as follows:
>> >>>>
>> >>>> \int_{\mathbb{R}^n} f(\mathbf{x}) \, \delta(g(\mathbf{x})) \,
>> >>>> d\mathbf{x} =
>> >>>>
>> >>>>
>> \int_{g^{-1}(0)}\frac{f(\mathbf{x})}{|\mathbf{\nabla}g|}\,d\sigma(\mathbf{x})
>> >>>>
>> >>>> (hopefully the above image makes it through, if not go here)
>> >>>>
>> >>>>
>> http://upload.wikimedia.org/wikipedia/en/math/3/3/f/33fbbed28ec715257d268faefc9e0e9f.png
>> >>>>
>> >>>> How hard would it be to compute the right hand side in sympy? In
>> >>>> particular I'm confused by how to express the domain and what they
>> mean
>> >>>> by \delta\sigma(x)
>> >>>>
>> >>>> Or, if there is a better way of going about this I'm happy to hear
>> it.
>> >>>>
>> >>>> -Matt
>> >>>>
>> >>>> --
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>> >>>
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