and the award for completely sidestepping valid points goes to........
you. not me.
hint: how many algebraic terms do you think one would need to construct a polynomial function that estimates price?
That's because I agree with your points about calculus, volume, and momentum.
It depends on how much information you have. If you knew everything you could likely reduce it to a very small number of variables (ie big players). If you are trying to tease it out of a bunch of noisy signals, quite a few. Your example was just a quadratic shifted linearly. I've seen those before.
i'm glad we agree on something
im not sure if your formulation would work though. i'm talking about a function that traces the actual path of the price, which could be estimated by a function of only one variable with a really, really large taylor expansion (c1 x^n + c2 x^(n-1) + c3 x^(n-3) ... + cn x + C) rather than one which solves for price given multiple parameters f(n, V, N) where n= number of large players, V=daily volume, n=good news or something like that.
the beauty of taylor approximations is that they're possible. any differentiable function of one variable can be estimated by a polynomial to any arbitrary precision.
but that's besides the point. the original example was a mixed cubic. its derivative was a mixed quadratic. good on you that you know that it has a basic cup-shaped graph but you wouldn't be able to, for instance, immediately tell me what its intercepts and minimum value were without a little bit of math. if you have the graph in front of you, these things are immediately obvious.
but this explanation is mostly for other readers anyway, as it seems we are in agreement as to why representing the data graphically is advantageous.
If you're going past x^4 and you don't have a damn good reason you are overfitting. Why would you try to model price like that?