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

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February 5

Coefficients of a Legendre polynomial

Starting from the definition $P_{n}(x)={\frac {1}{2^{n}n!}}{\frac {d^{n}}{dx^{n}}}(x^{2}-1)^{n}$ 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 $2^{n}n!.$ 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 $2^{n}z^{n}+c_{n{+}1}z^{n+1}+...~.$ The $n$-fold derivative is then $2^{n}n!+c_{n{+}1}(n{+}1)!z+...~.$ 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. pma 00:06, 6 February 2021 (UTC)