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July 16

Roots

Template:Hat Whate are the roots of zero? ie square root, cube root etc. Are they imaginarry?--213.205.242.206 (talk) 00:20, 16 July 2018 (UTC)

Since a field can have no nontrivial zero divisors, the only roots of zero are zero itself.--Jasper Deng (talk) 01:18, 16 July 2018 (UTC)

According to nth root, the square root, cube root, etc. of zero are also zero. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:45, 16 July 2018 (UTC)

Note that if one works over rings more general than fields, then one can have things like a nilpotent matrix.--Jasper Deng (talk) 02:07, 16 July 2018 (UTC)

Variation on a problem by Hilbert

The article on Hilbert matrix states a problem originally solved by Hilbert (1894): "Assume that Template:Nowrap, is a real interval. Is it then possible to find a non-zero polynomial P with integral coefficients, such that the integral

$\underset{a}{\overset{b}{\int }}P(x{)}^{2}dx$

is smaller than any given bound ε > 0, taken arbitrarily small?" What happens if you replace the word 'polynomial' with 'trigonometric polynomial'? It's clear from Fourier analysis that if b−a≥2π then the answer is no. --RDBury (talk) 09:32, 16 July 2018 (UTC)

Let a = 0 and b = 1 and P(x) = xⁿ where n > 2^-1(ε^-1-1).

Then $\underset{a}{\overset{b}{\int }}P(x{)}^{2}dx=\underset{0}{\overset{1}{\int }}(x^n{)}^{2}dx$ = (1+2n)^-1 < ε

Is it really this trivial? Bo Jacoby (talk) 12:48, 16 July 2018 (UTC).

(1) That "solves" the original problem, not the question asked, and (2) In the original problem, you don't get to pick a and b, the question is to find the condition on a and b such that there is one such polynomial. Your polynomial works for any a,b where there is at most one integer between them (if that integer is k, use (X-k)^n), but Hilbert's answer is (from OP's link) that b-a<4 makes such polynomial exist. Tigraan^{Click here to contact me} 13:32, 16 July 2018 (UTC)

The article wasn't too clear on Hilbert's method other that it involved solving the Hilbert matrix, and while the original paper is freely available on-line, my German is pretty rusty so any additional insight on what he did is helpful. I assume modern approximation theory supersedes it though, but it's not something I know a lot about. In any case you could apply the xⁿ idea to the trigonometric case to get a partial result, since (2cos x)ⁿ is trigonometric polynomial. --RDBury (talk) 02:07, 17 July 2018 (UTC)