As far as I can tell, the function only supports finite differences of
one variable at a time. My guess is that it wouldn't be too hard to
extend it to do what you want, though.

Aaron Meurer

On Thu, Mar 17, 2016 at 7:07 AM, Mathias Louboutin
<mathias.loubou...@gmail.com> wrote:
> In this example case it would give
>
> .25 * ( p(x,y+h,z,t) - p(x,y,z,t) + p(x-h,y,z,t) - p(x-y,y+h,z,t) ) /h^2
>
> But more generally I would want to be able to take the result of
> as_finite_diff as a new expression I can differentiate :
>
> a = as_finite_diff ( f(.....).diff(x) )
>
> b= as_finite_diff( a.diff(y) )
>
> ....... And so on
>
>
> Le mercredi 16 mars 2016 21:09:53 UTC, Aaron Meurer a écrit :
>>
>> What is the answer that you would expect to get?
>>
>> Aaron Meurer
>>
>> On Wed, Mar 16, 2016 at 12:12 PM, Mathias Louboutin
>> <mathias....@gmail.com> wrote:
>> > Hi
>> >
>> > I am working on PDE's and I am trying to get the symbolic expression for
>> > multivariabe finite differences
>> >
>> > mainly I wan the expression for d/dx(d/dy f), however
>> >
>> > d2pdxdy=as_finite_diff(p(x,y,z,t).diff(x,y),[x,x-h,y,y+h])
>> >
>> > doesn't work, nor any possible combination. Is this just not implemented
>> > or
>> > am I doing it wrong.
>> >
>> > cheers
>> >
>> > mathias
>> >
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