4 ms·
Not a stupid question, and you're definitely on the right track! It's a local linearization in the neighborhood of a root. Because of how simply polynomials can
by alhirzel 4y ago
Not a stupid question, and you're definitely on the right track! It's a local linearization in the neighborhood of a root. Because of how simply polynomials can be expressed (they can be defined by a scale and a multiset of roots), it happens that you can read this linearization right off the factored form.
If you're interested, I bet you could look at the behavior of the Taylor series expansion of polynomials in the neighborhood of their roots and see something that provides a good foundation for this article.
n.b. this is somewhat akin to a partial derivative of the polynomal, but rather than wrt to a specific variable, wrt a term that dominates a region. By evaluating the rest of the polynomial with the center of that region (the root), you're just killing the factors that are non-dominant. It wouldn't work if you were a bozo and forgot that $(x-1)(x-1) = (x-1)^2$ and so tried to say it looks like the line $x-1$ locally even though it doesn't locally dominate (the effect of the other $x-1$ is large). Also wouldn't work outside of happy polynomial land because who's to say there's not some $sin(x)$ somewhere that has global effects. If you try to go down this rabbit hole with non-localized effects, you'll encounter the Fourier Transform (and Linear Canonical Transform). If you try to force things to be more localized, you'd probably encounter spectral methods like the Short-Time Fourier Transform as well as wavelet methods. The latter are useful for things like compression because they operate using functions that are "semi-local" but non-trivial.