Wikipedia:Reference desk/Archives/Mathematics/2021 February 5

= February 5 =

Coefficients of a Legendre polynomial
Starting from the definition $$P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$$ how to show that $$P_n(1)=1$$, i.e. the coefficients of the Legendre polynomials always sum to 1? Thanks Abdul Muhsy (talk) 12:14, 5 February 2021 (UTC)


 * This is equivalent to showing that the value of $$\frac{d^n}{dx^n}(x^2-1)^n$$ at $$x=1$$ equals $$2^nn!.$$ By the substitution $$x=z+1,$$ this is the value of $$\frac{d^n}{dz^n}(2z+z^2)^n$$ at $$z=0$$. Binomial expansion of $$(2z+z^2)^n$$ results in a polynomial in $$z$$, which, presented as a sum of terms of increasing degree, has the form $$2^nz^n+c_{n{+}1}z^{n+1}+...~.$$ The $$n$$-fold derivative is then $$2^nn!+c_{n{+}1}(n{+}1)!z+...~.$$ Putting $$z=0,$$ we see that all but the first term vanish. --Lambiam 13:54, 5 February 2021 (UTC)

Set whose boundary is the entire space
Is there a known name for those subsets Y of a topological space X for which one of the following four equivalent conditions is satisfied?


 * 1) Both Y and its complement are dense.
 * 2) Both Y and its complement have empty interiors.
 * 3) Y is dense and has an empty interior.
 * 4) The boundary of Y is all of X.

Any nonempty proper subset of an indiscrete space satisfies the above four equivalent conditions. Are there any examples of such subsets of the real numbers $$\mathbb{R}$$ with the usual topology? GeoffreyT2000 (talk) 16:29, 5 February 2021 (UTC)


 * Aren't both Q (the rational numbers) and R\Q dense in R? --Lambiam 19:36, 5 February 2021 (UTC)


 * The term I've seen for the first condition is dense/co-dense. The rationals are also the example I would have given. JoelleJay (talk) 21:33, 5 February 2021 (UTC)


 * In fact "dense set with empty interior" is quite common, I'd say the standard term. pm a 00:06, 6 February 2021 (UTC)