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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)[reply]

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)[reply]

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?

Both Y and its complement are dense.
Both Y and its complement have empty interiors.
Y is dense and has an empty interior.
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)[reply]

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

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)[reply]

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