Timeline for Reasoning on Efficiency (2)

Current License: CC BY-SA 3.0

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Dec 8, 2016 at 11:17	comment	added	David Richerby		Unless you can give an actual reason why looking at any derivative tells you anything relevant about the function, you shouldn't be considering derivatives at all. Anyway, please note that this isn't a discussion site; it's not a suitable place for fleshing out ideas. That sort of thing can be done in Computer Science Chat but it doesn't work in the Q&A part of the site.
Dec 8, 2016 at 10:49	comment	added	David Richerby		Sorry -- replace zero with any other constant. No derivative of $x^{1/2}$ is a constant function. But why do you care about having a constant derivative? What "physical" meaning does that have? As you say, the constant is just a property of the leading coefficient and leading exponent of the polynomial. So why not just compare those directly instead of combining them into an arbitrary function?
Dec 8, 2016 at 8:45	comment	added	David Richerby		It's as if you're saying "$x<y$ is an inadequate way of comparing real numbers. I'm going to use $x-y$ instead." But then restricting only to numbers where $x-y$ is an integer. Why?
Dec 8, 2016 at 8:44	history	edited	David Richerby	CC BY-SA 3.0	added 242 characters in body
Dec 8, 2016 at 8:41	comment	added	David Richerby		@TobiAlafin I ignored the part about the $m$th derivative being zero because, although it's something you state, it doesn't seem relevant to anything. It seems to be just a coincidence: sure, polynomials only have finitely many nonzero derivatives, but so what? In any case, the way you've written it, it's not even true. You assert that $r_i=n^k$ for some real $k$, and then claim that the $\lceil k\rceil$th derivative is zero. Well, take $k=\tfrac12$. The first derivative of $x^{1/2}$ is $\tfrac12x^{-1/2}\neq 0$.
Dec 8, 2016 at 0:54	history	answered	David Richerby	CC BY-SA 3.0