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Maybe a stupid question but I've been brushing up on my calculus in preparation to go back to school for physics. Is this equivalent to the tangent to the equa

Maybe a stupid question but I've been brushing up on my calculus in preparation to go back to school for physics. Is this equivalent to the tangent to the equation at the root? I mean, obviously with exponents higher than 2 you get polynomials not lines, but the given derivation involves just taking the derivative so shouldn't that work? If so, this massively simplifies a bunch of the practice problems I'm doing right now. I should probably just try to prove it

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.

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.