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@FishFace@ioc.exchange
Post #4008962
2026-07-22 09:45 UTC
@bemmesr@mathstodon.xyz you don't want to pay attention to @SmartmanApps@dotnet.social on this. Check out any of his "#MathsMonday" pinned posts that have replies from others and you'll see what I mean.
In this case, you are right that you *could* just use a single variable that is equal to a polynomial expression, but this is missing the point of the technique: when we create the placeholder variables A, B, etc, there is in fact an assumption that these are not dependent on x (or whatever the unknown is), and this assumption makes the technique work, together with the assumption that the equation you set up holds for all values of x. Because they don't vary with x, if you know that, for example, Ax + B = 3x + 5, it must be that A = 3, B = 5. For particular values of x, you could pick other values of A,B, but these are the only ones which work for all values of x.
Unless, that is, A and B may themselves depend on x! For example, A=5/x, B=3x would work if they could.
In general it is better to make the dependencies of variables on other variables explicit, by using functional notation, so that "A" always stands for an independent variable, with "A(x)" meaning some quantity which depends on x. But this is not how mathematics is taught in practice, so you end up with these situations where you may not realise what all the assumptions are.
Sorry you had to read the charlatan's answer before getting a proper explanation.
Replies (1)
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@FishFace@ioc.exchange that makes sense, thank you. It's a choice to limit the placeholders to constant values. I still don't quite understand why the technique doesn't work if we don't make this restriction though. After all, there ought to be a way to discover which polynomial A(x) would satisfy the equation between the original polynomial fraction and our decomposed sum of fractions, wherein A(x) is one of the numerators.
I could appreciate that in fact the technique is easier if we do make this restriction but I still would like to know where it fails otherwise. Surely we could still achieve those nice constant numerators in the case of linear denominators if we approached it this way, since a constant is also a polynomial, just of order zero.
Open ##4009860