CODEWITHSUNDEEP
maple
Rahul P
Rahul P

Reputation: 2663

Is there a way to add multiple conditions while doing the product in Maple?

Note: I've searched for the answer to this already but I can't find what I need. I may have overlooked something or maybe this is referred to in a different way. I'll delete this if someone points out that it is a duplicate, please let me know.

Question: How can I add multiple conditions to the product or mul in Maple?

Example: I am trying to re-create the following, I cannot seem to find a way to add both r != k and r=1 as parameters.

enter image description here

I've gone over the documentation of the product command and maybe I am missing something.

Upvotes: 2

Views: 215

Answers (2)

AmirHosein Sadeghimanesh
AmirHosein Sadeghimanesh

Reputation: 478

If you want the formula that you showed in your question to be displayed symbolically as you have, I don't have any idea at the moment in Maple. As you may have seen in the help page of product, it only accepts ranges of the form idx=n..m or idx=n and that's it. But if you have values in k and j and want to receive the result of your formula, you can use mul, not product. As @Robai did a comparison in her answer between mul and product, the important thing here is that product does not support its second argument (the range) to be in the form idx in set, no "in" for product and that means you can not use product for your case even with the formulation given in @Robai's answer.

Now for mul, other than the set-minus idea of @Robai's answer. Here is another approach (you can read more about it in another answer I posted for another question in this link https://stackoverflow.com/a/72498377/6195473).

j := 5:
k := 2:
idxs := select( x -> x <> k, [ seq( r, r = 1..j-1 ) ] );
(k/(j-k)) * mul( r/(r-k), r in idxs );

Here is a screenshot from the output. enter image description here

Upvotes: 0

Robai
Robai

Reputation: 307

Conditional products can be achieved using sets for index range (in the example below R is a set or can be a list, but set is better in this case since it's easier to remove its members), but that set must not have unknown parameters (so it's a numeric range). For example:

[> j:=6: k:=3:
   R:={seq(s, s in {seq(s,s=1..j-1)} minus {k})};
   mul(r/(r-k), r in R);

The output will be:

        R := {1, 2, 4, 5}
               10

You can do it without additional variable R too:

[> mul(r/(r-k), r in {seq(s, s in {seq(s,s=1..j-1)} minus {k})});

A comparison of mul and product:

  1. mul works only for numeric ranges and there should be no infinity involved, it also works for r in R case.
  2. product can work with symbolic ranges, also range can have infinity, but it doesn't work for r in R case.
  3. product is more powerful, but mul is much faster, so for numeric ranges mul should be always used instead (similarly add should used instead of sum for numeric ranges).

You can use mul with parameters too, but the range should be numeric (R is defined above):

[> mul((a+r)/(b+r-k), r in R);

The output:

Upvotes: 2

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