I know this may sound peculiar, but I was wondering if any mathematicians have found a method to finding roots purely through calculus without iteration. I can't imagine that such a method exists for all the roots of a polynomial, but I can imagine we can find a couple of roots given that they are either stationary points or points of inflection which can quite easily be found having once found the stationary points and points of inflection. However, beyond this, I can't possibly imagine there exists method to determine roots which are not stationary points or points if inflection. So, my question is: are there methods to find the a particular root of a polynomial, where the particular root is not a stationary point or point of inflection, purely through calculus and not iteration, including iterations using calculus (e.g. Newton's method)?
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It's not clear what you're asking. Can you give an example of a problem where a root can be found "purely by calculus"? – Ben Grossmann Jul 20 '15 at 17:30
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The answer is no. See: Abel–Ruffini_theorem. – user251257 Jul 20 '15 at 17:30
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1@ReinhildVanResenu note also that the only reason finding stationary/inflection points is usually "easy" is that they require solving a lower degree polynomial. For a polynomial of high enough degree, any of the "usually easy" tasks become difficult. – Ben Grossmann Jul 20 '15 at 17:32
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Also, what do you mean by "find the root"? As in, find an arbitrary decimal expansion, or find an exact algebraic form (that is, something like 1.41421356, or something like $\sqrt 2$)? – Ben Grossmann Jul 20 '15 at 17:33
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Are you referring, for instance, to series representations of Bring radicals, based on Lagrange's inversion theorem? – Lucian Jul 20 '15 at 17:38
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By the root I mean the actual solution. – Reinhild Van Rosenú Jul 20 '15 at 17:42
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@Omnomnomnom By "easy" I meant "doable". – Reinhild Van Rosenú Jul 20 '15 at 17:43
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@ReinhildVanRosenú so, what is the "actual (positive) solution" to $x^2 = 2$? Is $x = \sqrt 2$ the "actual solution", or is $x = 1.41421356...$ the "actual solution". – Ben Grossmann Jul 20 '15 at 17:45
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@user251257 I have acquainted with the AR Theorem before, but I believe it proves that no polynomials of degree higher than 5 can be solved through radicals, which I believe does not consider methods used by calculus. – Reinhild Van Rosenú Jul 20 '15 at 17:45
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@Omnomnomnom I mean the root in the format $$(x-a)$$ – Reinhild Van Rosenú Jul 20 '15 at 17:47
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1Well, do you want to express the roots as radicals or as series? But series are just iterations... so.... – user251257 Jul 20 '15 at 17:47
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@user251257 I think that made things clear. – Reinhild Van Rosenú Jul 20 '15 at 17:48
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@ReinhildVanRosenú that would be a factor, not a root. – Ben Grossmann Jul 20 '15 at 17:49
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@Omnomnomnom But is there not an interchangeability between the two, ugh that if w know one factor, we know one root vice versa? – Reinhild Van Rosenú Jul 20 '15 at 17:50
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1In any case, my question was what is the "actual" positive "root" to $x^2 - 2 = 0$. I suppose I was trying to get to the point that user251257 made. – Ben Grossmann Jul 20 '15 at 17:53
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@Lucian I believe the series representations for a Bring radicals ae only for quintic equations, and I'm not certain if the methods can be applied to higher degree polynomial equations. – Reinhild Van Rosenú Jul 20 '15 at 17:53
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@Omnomnomnom I suppose I meant its radical form. – Reinhild Van Rosenú Jul 20 '15 at 17:54
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2@ReinhildVanRosenú then we get stuck with the AR theorem. There exist polynomials where, in whatever capacity you happen to "know" the solution (via calculus, perhaps), there is no way to "express" that exact solution using finitely many radicals (or $n$th roots). – Ben Grossmann Jul 20 '15 at 17:56
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1@ReinhildVanRosenú: I believe you will find this post to be very interesting. – Lucian Jul 20 '15 at 20:16