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Jun 18, 2014 at 3:01 review Close votes
Jun 18, 2014 at 7:54
Jun 15, 2014 at 18:51 comment added Dirk Polynomials are so useful because they approximate analytic functions so well.
Jun 15, 2014 at 17:44 comment added Sam Hopkins I'm surprised the Nullstellensatz has not been mentioned here (well, the combinatorial one was, but I mean the classical one).
Jun 15, 2014 at 17:00 history made wiki Post Made Community Wiki by Todd Trimble
Jun 15, 2014 at 16:52 answer added Count Iblis timeline score: -7
Jun 15, 2014 at 12:31 comment added user9072 @TomCopeland "Didn't imaginary numbers evolve through their formal utility in solving quadratics?" Not really. But cubics, more specifically the 'casus irreducibilis,' were quite relevant.
Jun 14, 2014 at 14:31 comment added R.P. I am surprised about the number of upvotes this question is garnering. One might as well ask why minor chords are so ubiquitous in music, or how come the word 'and' is used so often in modern literature.
Jun 13, 2014 at 23:49 comment added Joseph O'Rourke @ViditNanda: Your proposed rephrasing is a very interesting, and I think different question. You should consider posing it yourself.
Jun 13, 2014 at 21:57 vote accept Joseph O'Rourke
Jun 13, 2014 at 20:04 answer added Joni Teräväinen timeline score: 9
Jun 13, 2014 at 16:58 answer added PA6OTA timeline score: 2
Jun 13, 2014 at 11:20 comment added Joseph O'Rourke @ManfredWeis: I was more interested in the utility of polynomials within mathematics, as opposed to their myriad applications, such as cubic splines. Although of course there is no clean separation.
Jun 13, 2014 at 6:53 answer added Paul Siegel timeline score: 20
Jun 13, 2014 at 5:53 answer added Włodzimierz Holsztyński timeline score: 2
Jun 13, 2014 at 5:52 answer added Joshua Grochow timeline score: 11
Jun 13, 2014 at 5:44 comment added Manfred Weis @Joseph do you also have in mind the sections of polynoms that constitute to a polynom-spline? Would surely broaden the range of applicability.
Jun 13, 2014 at 2:59 comment added KConrad Yes, over fields in which you can't take limits polynomials are still important (algebraic extensions of fields, algebraic varieties, linear algebra). They're also important just in analysis alone, as you already know (e.g., SW theorem).
Jun 13, 2014 at 2:12 comment added user5794 @KConrad: So if I understand, when we work over fields where we can't do analysis, we still care about polynomials? This makes sense, of course, and I agree. I guess we may be bumping into a divergence in why polynomials are useful. Say I only care about analysis. Aren't polynomials still very important, despite their algebraic origin?
Jun 13, 2014 at 2:07 comment added KConrad @trb456: Yes the SW theorem is very important, but when you work with coefficients that are not the real or complex (or $p$-adic) numbers polynomials are still important and it's not due to anything involving the SW theorem. That is, there are many reasons to care about polynomials other than in their role as approximations to other functions.
Jun 13, 2014 at 2:02 answer added The Masked Avenger timeline score: 21
Jun 13, 2014 at 1:59 comment added user5794 @KConrad: But isn't the fact that SW gives us a link from essentially algebraic objects (polynomials) to analysis (continuous functions) a big deal? Don't we want such connections? I am genuinely interested in a contrary answer.
Jun 13, 2014 at 1:49 answer added Qiaochu Yuan timeline score: 36
Jun 13, 2014 at 1:42 comment added KConrad I am a bit puzzled about why you suggest the Stone-Weierstrass theorem might explain the ubiquity of polynomials, since most uses of polynomials (outside of the functional calculus, say) have no direct link to the SW theorem.
Jun 13, 2014 at 1:41 comment added Vidit Nanda I am usually a big fan of your questions and answers on this site, but I fear that this one might be too broad even for CW. I won't downvote or vote for closure, but would instead suggest rephrasing along the lines of "what are some instances where polynomials appear unexpectedly, or happen to be surprisingly beneficial?".
Jun 13, 2014 at 1:39 comment added Tobias Fritz @GerryMyerson: another way of saying that is that $K[x]$ is the free unital $K$-algebra on one generator. Similarly, $K[x_1,\ldots,x_n]$ is the free $K$-algebra on $n$ generators.
Jun 13, 2014 at 1:38 review Close votes
Jun 13, 2014 at 2:15
Jun 13, 2014 at 1:33 comment added Gerry Myerson $K[x]$ represents the forgetful functor from $K$-algebras to sets, according to math.stackexchange.com/questions/531073/… See also the universal property for polynomials as stated at arbourj.wordpress.com/2012/04/26/…
Jun 13, 2014 at 1:30 history edited Joseph O'Rourke CC BY-SA 3.0
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Jun 13, 2014 at 1:17 comment added user62675 +1. Interesting question! I'm really looking forward to the answers of this question.
Jun 13, 2014 at 1:10 history asked Joseph O'Rourke CC BY-SA 3.0