How can I invert the product rule nicely [where there are extra numerical factors]?

Question

I am trying to invert the product rule in the following situation:

ex := \partial_{a}{A} B + 2 A \partial_{a}{B}

(update: here I stress the extra factor of 2 in front of the last term)

and I tried the command

substitute(ex, $\partial_{a?}{A??} B?? + A?? \partial_{a?}{B??}
-> \partial_{a?}{A?? B??}$);

but it does not work. I don't want to do this one by one as there are many terms like this in my calculation.

I wonder if there's a method that I can unwrap

2 A \partial_{a}{B} -> A \partial_{a}{B} + A \partial_{a}{B}

so that these can be tackled individually?

Thanks very much!

Answer 1

I could factor the terms as you wanted.

The behaviour of your session might be due to the properties of your objects.

I took your (incomplete) example, and re-ensembled it like this:

{a,b}::Indices.
\partial{#}::PartialDerivative.
{A,B,C,D}::Depends(\partial{#}).

ex := \partial_{a}{A} B + A \partial_{a}{B};

rule := \partial_{b?}{C??} D?? + C?? \partial_{b?}{D??} 
    -> \partial_{b?}{C?? D??};

substitute(ex, rule);

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Update

Physics Cat thank you for highlighting that the extra factor of 2 was not a typo.

In your case, I'd follow Kasper's (usual) advise, and use the substitution the other way around,

rule2 := \partial_{b?}{C??} D?? -> - C?? \partial_{b?}{D??} 
    + \partial_{b?}{C?? D??};

The whole notebook (MWE) would be:

{a,b}::Indices.
\partial{#}::PartialDerivative.
{A,B,C,D}::NonCommuting.
{A,B,C,D}::Depends(\partial{#}).

ex1 := \partial_{a}{A} B + A \partial_{a}{B};
rule := \partial_{b?}{C??} D?? + C?? \partial_{b?}{D??} 
    -> \partial_{b?}{C?? D??};
substitute(ex1, rule);

ex2 := \partial_{a}{A} B + 2 A \partial_{a}{B};
rule2 := \partial_{b?}{C??} D?? -> - C?? \partial_{b?}{D??} 
    + \partial_{b?}{C?? D??};
substitute(ex2, rule2);

Hope this could be of use!

Comments

What is the purpose or meaning of the question marks (and double question marks) in your rule?

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Those question marks are for pattern matching (like a kind of regular expression).

Please check the section in the book: https://cadabra.science/notebooks/ref_patterns.html

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Thanks for that.

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This does not work for my specific question... The main subtlety of my question is the extra factor of 2 in front of one of the terms

A \partial_{a}{B} + 2 \partial_{a}{A} B

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Hi Physics Cat I've updated the answer to your question. Hope this could be useful.

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Thanks very much! It works perfectly!

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Excellent! Nice to hear that.