Paper 4, Section II, 7D

Numbers and Sets | Part IA, 2017

(a) For positive integers n,m,kn, m, k with knk \leqslant n, show that

(nk)(kn)m=(n1k1)=0m1an,m,(k1n1)m1\left(\begin{array}{l} n \\ k \end{array}\right)\left(\frac{k}{n}\right)^{m}=\left(\begin{array}{l} n-1 \\ k-1 \end{array}\right) \sum_{\ell=0}^{m-1} a_{n, m, \ell}\left(\frac{k-1}{n-1}\right)^{m-1-\ell}

giving an explicit formula for an,m,a_{n, m, \ell}. [You may wish to consider the expansion of (k1n1+1n1)m1.]\left.\left(\frac{k-1}{n-1}+\frac{1}{n-1}\right)^{m-1} .\right]

(b) For a function f:[0,1]Rf:[0,1] \rightarrow \mathbb{R} and each integer n1n \geqslant 1, the function Bn(f):[0,1]RB_{n}(f):[0,1] \rightarrow \mathbb{R} is defined by

Bn(f)(x)=k=0nf(kn)(nk)xk(1x)nkB_{n}(f)(x)=\sum_{k=0}^{n} f\left(\frac{k}{n}\right)\left(\begin{array}{l} n \\ k \end{array}\right) x^{k}(1-x)^{n-k}

For any integer m0m \geqslant 0 let fm(x)=xmf_{m}(x)=x^{m}. Show that Bn(f0)(x)=1B_{n}\left(f_{0}\right)(x)=1 and Bn(f1)(x)=xB_{n}\left(f_{1}\right)(x)=x for all n1n \geqslant 1 and x[0,1]x \in[0,1].

Show that for each integer m0m \geqslant 0 and each x[0,1]x \in[0,1],

Bn(fm)(x)fm(x) as nB_{n}\left(f_{m}\right)(x) \rightarrow f_{m}(x) \text { as } n \rightarrow \infty

Deduce that for each integer m0m \geqslant 0,

limn14nk=02n(kn)m(2nk)=1\lim _{n \rightarrow \infty} \frac{1}{4^{n}} \sum_{k=0}^{2 n}\left(\frac{k}{n}\right)^{m}\left(\begin{array}{c} 2 n \\ k \end{array}\right)=1

Typos? Please submit corrections to this page on GitHub.