Wikipedia:Reference desk/Archives/Mathematics/2018 July 16

= July 16 =

Roots
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 , is a real interval.  Is it then possible to find a non-zero polynomial P with integral coefficients, such that the integral
 * $$\int_{a}^b P(x)^2 dx$$

is smaller than any given bound &epsilon; > 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(&epsilon;-1-1).
 * Then $$\int_{a}^b P(x)^2 dx=\int_0^1 (x^n)^2 dx$$ = (1+2n)-1 < &epsilon;
 * 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)