Peter Pfannenschmid <peter.pfannenschmid at guido-kuebler-gmbh.de> writes:
> I have tried subst, but of course it did not work (i*k is not a
> complete subexpression in the example expression). Then I read a hint
> regarding ratsubst, but this does the wrong thing as well. Below is a
> transcription showing the problem:
>
> (%i7) test: (i*k*l + i*k*h)^2;
> 2
> (%o7) (i k l + h i k)
> (%i8) subst(s, (i*k), test);
> 2
> (%o8) (i k l + h i k)
> (%i9) ratsubst(s, (i*k), test);
> 2 2 2 2 2
> (%o9) l s + 2 h l s + h s
>
> Is there any way to achieve what I want?
Well, the lazy answer is that you can tell Maxima to try and factor the
result (which ratsubst multiplied out):
(%i27) factor(ratsubst(s, (i*k), test));
2 2
(%o27) (l + h) s
But I admit that's cheating! Here, another option is to notice that
ratsubst wouldn't have anything to expand if test wasn't squared. With
that in mind, you can do something like
(%i28) map(lambda([x], ratsubst(s, i*k, x)), test);
2
(%o28) (l s + h s)
That basically applies the ratsubst step to the two arguments of test
(ikl+hik and 2). This is also rather hacky, I admit!
One last trick, which I have used quite a lot is the following: Notice
that s = i*k is the same as i = s/k (well, as long as k is nonzero). So
you can do this:
(%i30) subst(s/k, i, test);
2
(%o30) (l s + h s)
Obviously, this doesn't work so well when you're trying to substitute
for a more complex expression.
I'm hoping that some other users can come up with more "high-brow"
solutions!
Rupert
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