identify and polyratfun

[jmbrod]

Hi,

I have the following problem: I use polyratfun and then in the end would like to make a substitution of a the polyratfun function. The following is a minimal example:

s x,y;
cf rat,f;

l exp = rat(x-y,y) ;

.sort
polyratfun rat;
id rat(x-y,y) = f(x-y,y) ;

print +s;
.end

This gives rat(x - y,y) while I would expect f( - y + x,y). On the other hand, just not using polyratfun works (but this is not an option in my actual code):

s x,y;
cf rat,f;

l exp = rat(x-y,y) ;

id rat(x-y,y) = f(x-y,y) ;

print +s;
.end

gives the desired result. I tried to "undo" the polyratfun by

s x,y;
cf rat,f;

l exp = rat(x-y,y) ;

.sort
polyratfun rat;
.sort
polyratfun;
id rat(x-y,y) = f(x-y,y) ;

print +s;
.end

but still, rat does not get replaced by f. Probably I'm missing something quite basic...?

Thanks for any help / clarification!

[vermaseren]

The last thing first: If you print the expression with the rat it probably has rat(-y+x,y) and because of that your substitution does not work.

Then the first: the polyratfun acts as the coefficient of the term. And in the same way that you cannot replace the 4/7 in the term 4/7*x you cannot replace the polyratfun.

You could however use the following statement

Multiply replace_(rat,f);

and have a look what you obtain.

On 17 Nov 2023, at 19:14, Joachim Brod @.***> wrote:

Hi,

I have the following problem: I use polyratfun and then in the end would like to make a substitution of a the polyratfun function. The following is a minimal example:

s x,y; cf rat,f;

l exp = rat(x-y,y) ;

.sort polyratfun rat; id rat(x-y,y) = f(x-y,y) ;

print +s; .end This gives rat(x - y,y) while I would expect f( - y + x,y). On the other hand, just not using polyratfun works (but this is not an option in my actual code):

s x,y; cf rat,f;

l exp = rat(x-y,y) ;

id rat(x-y,y) = f(x-y,y) ;

print +s; .end gives the desired result. I tried to "undo" the polyratfun by

s x,y; cf rat,f;

l exp = rat(x-y,y) ;

.sort polyratfun rat; .sort polyratfun; id rat(x-y,y) = f(x-y,y) ;

print +s; .end but still, rat does not get replaced by f. Probably I'm missing something quite basic...?

Thanks for any help / clarification!

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[jodavies]

In your first example you certainly can replace the rat if you use wildcards.

id rat(x?,y?) = f(x,y) ;

will make the replacement that you seem to want.

It will leave a rat(1,1) leftover, since polyratfun is still enabled. If you disable it again as in your last example, you end up with only f(x - y,y).

[jodavies]

Your last example is a bit weird. For me it prints as rat(x - y,y) and yet your id statement still does not match. This is something to do with polyratfun's sort ordering; if you run your script with On highfirst; at the start, your last example works as you would expect.

[vermaseren]

OK, maybe I was to fast with my answer…..

On 17 Nov 2023, at 19:34, jodavies @.***> wrote:

Your last example is a bit weird. For me it prints as rat(x - y,y) and yet your id statement still does not match. This is something to do with polyratfun's sort ordering; if you run your script with On highfirst; at the start, your last example works as you would expect.

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[jmbrod]

Hi, thank you for your quick replies! Yes, I thought the last example was weird :-) The sorting explanation makes sense, and I think in practice I might just use the replace_ function.

[tueda]

This is something to do with polyratfun's sort ordering;

Indeed, inserting argument;endargument; (force it to change the sort ordering) is a workaround (see also #144):

s x,y;
cf rat,f;

l exp = rat(x-y,y) ;

.sort
polyratfun rat;
.sort
polyratfun;
argument;endargument;  * <-- inserted
id rat(x-y,y) = f(x-y,y) ;

print +s;
.end

FORM 4.3.1 (Apr 11 2023, v4.3.1) 64-bits         Run: Mon Nov 20 13:50:00 2023

   exp =
       + f( - y + x,y)
      ;