Difference between revisions of "Notes:Richard Sharp question"
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(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}...") |
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==Problem== | ==Problem== | ||
I wish to show: | I wish to show: | ||
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# {{MM|1=\sum^n_{k\eq 0}(k-nx)^2\ {}^nC_kx^k(1-x)^{n-1} \eq nx(1-x) }} | # {{MM|1=\sum^n_{k\eq 0}(k-nx)^2\ {}^nC_kx^k(1-x)^{n-1} \eq nx(1-x) }} | ||
# {{MM|1=\sum^n_{k\eq 0}k\ {}^nC_kx^k(1-x)^{n-1}\eq nx}} - probably not important | # {{MM|1=\sum^n_{k\eq 0}k\ {}^nC_kx^k(1-x)^{n-1}\eq nx}} - probably not important | ||
| + | ==Possible solution== | ||
| + | Starting from: | ||
| + | * {{MM|1=\sum_{\begin{array}{c}0\le k\le n\\k:\vert x-\frac{k}{n}\vert\ge\delta\end{array} } \vert f(x)-f(\tfrac{k}{n})\vert\ {}^nC_kx^k(1-x)^{n-k} }} | ||
| + | *: {{MM|1=\le 2\Vert f\Vert_\infty \sum_{\begin{array}{c}0\le k\le n\\k:\vert x-\frac{k}{n}\vert\ge\delta\end{array} }{}^nC_kx^k(1-x)^{n-k} }} - by definition of the {{M|\Vert\cdot\Vert_\infty}} norm | ||
| + | *: {{MM|\eq2\Vert f\Vert_\infty \sum_{\begin{array}{c}0\le k\le n\\k:(k-nx)^2\ge \delta^2n^2\end{array} }{}^nC_kx^k(1-x)^{n-k} }} - by faffing about with absolute values | ||
| + | *: {{MM|1=\le 2\Vert f\Vert_\infty \sum_{\begin{array}{c}0\le k\le n\\k:(k-nx)^2\ge \delta^2n^2\end{array} }\frac{(k-nx)^2}{\delta^2n^2}\ {}^nC_kx^k(1-x)^{n-k} }} - as for such {{M|k}} we see {{MM|\frac{(k-nx)^2}{\delta^2n^2}\ge 1}} | ||
| + | *: {{MM|1=\le 2\Vert f\Vert_\infty \sum^n_{k\eq 0}\frac{(k-nx)^2}{\delta^2n^2}\ {}^nC_kx^k(1-x)^{n-k} }} - because for the {{M|k}} terms added we see {{MM|\frac{(k-nx)^2}{\delta^2n^2}\ {}^nC_kx^k(1-x)^{n-k}\ge 0}} - so it can only increase the summation's value | ||
| + | *: {{MM|1=\eq \frac{2\Vert f\Vert_\infty}{\delta^2n^2} \sum^n_{k\eq 0}(k-nx)^2\ {}^nC_kx^k(1-x)^{n-k} }} - factoring | ||
| + | *: {{MM|1=\eq \frac{2\Vert f\Vert_\infty}{\delta^2n^2}\cdot nx(1-x) }} (using result 2) | ||
| + | *: {{MM|1=\le \frac{1}{4}n\frac{2\Vert f\Vert_\infty}{\delta^2n^2} }} - as {{M|x(1-x)\le\frac{1}{4} }} | ||
| + | *: {{MM|1=\eq \frac{\Vert f\Vert_\infty}{2n\delta^2} }} | ||
Latest revision as of 21:03, 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
Possible solution
Starting from:
- [math]\sum_{\begin{array}{c}0\le k\le n\\k:\vert x-\frac{k}{n}\vert\ge\delta\end{array} } \vert f(x)-f(\tfrac{k}{n})\vert\ {}^nC_kx^k(1-x)^{n-k}[/math]
- [math]\le 2\Vert f\Vert_\infty \sum_{\begin{array}{c}0\le k\le n\\k:\vert x-\frac{k}{n}\vert\ge\delta\end{array} }{}^nC_kx^k(1-x)^{n-k}[/math] - by definition of the [ilmath]\Vert\cdot\Vert_\infty[/ilmath] norm
- [math]\eq2\Vert f\Vert_\infty \sum_{\begin{array}{c}0\le k\le n\\k:(k-nx)^2\ge \delta^2n^2\end{array} }{}^nC_kx^k(1-x)^{n-k} [/math] - by faffing about with absolute values
- [math]\le 2\Vert f\Vert_\infty \sum_{\begin{array}{c}0\le k\le n\\k:(k-nx)^2\ge \delta^2n^2\end{array} }\frac{(k-nx)^2}{\delta^2n^2}\ {}^nC_kx^k(1-x)^{n-k}[/math] - as for such [ilmath]k[/ilmath] we see [math]\frac{(k-nx)^2}{\delta^2n^2}\ge 1[/math]
- [math]\le 2\Vert f\Vert_\infty \sum^n_{k\eq 0}\frac{(k-nx)^2}{\delta^2n^2}\ {}^nC_kx^k(1-x)^{n-k}[/math] - because for the [ilmath]k[/ilmath] terms added we see [math]\frac{(k-nx)^2}{\delta^2n^2}\ {}^nC_kx^k(1-x)^{n-k}\ge 0[/math] - so it can only increase the summation's value
- [math]\eq \frac{2\Vert f\Vert_\infty}{\delta^2n^2} \sum^n_{k\eq 0}(k-nx)^2\ {}^nC_kx^k(1-x)^{n-k}[/math] - factoring
- [math]\eq \frac{2\Vert f\Vert_\infty}{\delta^2n^2}\cdot nx(1-x)[/math] (using result 2)
- [math]\le \frac{1}{4}n\frac{2\Vert f\Vert_\infty}{\delta^2n^2}[/math] - as [ilmath]x(1-x)\le\frac{1}{4} [/ilmath]
- [math]\eq \frac{\Vert f\Vert_\infty}{2n\delta^2}[/math]
