Difference between revisions of "Notes:Richard Sharp question"
From Maths
(Created page with "==Problem== I wish to show: * <span style="font-size:1.1em;">{{MM|1=2\Vert f\Vert_\infty \sum_{k:(k-nx)^2\ge n^2\delta^2}{}^nC_kx^k(1-x)^{n-k}\le 2\Vert f\Vert_\infty \frac{1}...") |
(No difference)
|
Revision as of 20:07, 25 November 2016
Problem
I wish to show:
- [math]2\Vert f\Vert_\infty \sum_{k:(k-nx)^2\ge n^2\delta^2}{}^nC_kx^k(1-x)^{n-k}\le 2\Vert f\Vert_\infty \frac{1}{n^2\delta^2}\sum_{k\eq 0}^n(k-nx)^2\ {}^nC_kx^k(1-x)^{n-k}[/math]
I am completely happy with the LHS, I understand the RHS is probably best written as:
- [math]2\Vert f\Vert_\infty \sum_{k\eq 0}^n\frac{(k-nx)^2}{n^2\delta^2}\ {}^nC_kx^k(1-x)^{n-k} [/math]
By hypothesis, for the summation on the LHS, [ilmath]k[/ilmath] is such that:
- [ilmath](k-nx)^2\ge n^2\delta^2\implies\frac{(k-nx)^2}{n^2\delta^2}\ge 1[/ilmath]
I cannot see how to get to the RHS
Supporting equations
Note that:
- [math]\sum^n_{k\eq 0}{}^nC_kx^k(1-x)^{n-x}\eq 1[/math]
- [math]\sum^n_{k\eq 0}(k-nx)^2\ {}^nC_kx^k(1-x)^{n-1} \eq nx(1-x)[/math]
- [math]\sum^n_{k\eq 0}k\ {}^nC_kx^k(1-x)^{n-1}\eq nx[/math] - probably not important