Post #4010175

2026-07-22 10:46 UTC

@FishFace@ioc.exchange sort of. I wouldn't say ‘just’ do this, since I'm thinking my approach is actually harder, because we'd be ignoring information we have (as in, we know that the left numerator is a constant and that the right is no more than order one), but out of curiosity, what would happen if we went with this order‐agnostic approach where we simply let the numerators stand in for any polynomial in x? I would expect that we get the same final answer, even if it takes longer. So yeah, let the left numerator be called E(x) and the right D(x), as you say.

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@FishFace@ioc.exchange 2026-07-22 10:57

@bemmesr@mathstodon.xyz so what I would say if we did that is that first of all there is no unique solution for D(x) and E(x) - they could conceivably be all sorts of expressions. There will in general be solutions that aren't even rational functions. As with the example of the 3x + 5 situation above, a complicated expression in D can be cancelled out by a different complicated expression in E. So this way of looking at the problem is not *that* useful. But we can *work out* that one particularly nice solution among the many has the shape "E(x) is just a constant, and D(x) is an order 1 polynomial in x". So the partial fraction step of just writing in the constants of the different polynomial numerators can be seen as doing this all in one go: we know that there is a solution where the numerators *are* be written in this way, so we're going to do that. You might be wondering *how* we can work out that this nice shape of solution works - well, the answer to that is simply that, once you're done you can put your solution over a common denominator and see that it comes out to the same thing you started with, so that proves it! I think you still had another question, but let me pause here in case this isn't clear.

Open ##4010174