exercise:8b7f1ebd97: Difference between revisions
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(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> <ul><li> Show that for <math>m \ge 1</math>, <math display="block"> f_{2m} = u_{2m-2} - u_{2m}\ . </math> </li> <li> Using part (a), find a closed-form expression for the sum <math display="block"> f_2 + f_4 + \cdots + f_{2m}\ . </math> </l...") |
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Latest revision as of 00:29, 15 June 2024
[math]
\newcommand{\NA}{{\rm NA}}
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\newcommand{\mathds}{\mathbb}[/math]
- Show that for [math]m \ge 1[/math],
[[math]] f_{2m} = u_{2m-2} - u_{2m}\ . [[/math]]
- Using part (a), find a closed-form expression for the sum
[[math]] f_2 + f_4 + \cdots + f_{2m}\ . [[/math]]
- Using part (b), show that
[[math]] \sum_{m = 1}^\infty f_{2m} = 1\ . [[/math]](One can also obtain this statement from the fact that[[math]] F(x) = 1 - (1-x)^{1/2}\ .) [[/math]]
- Using parts (a) and (b), show that the probability of no equalization in the first [math]2m[/math] outcomes equals the probability of an equalization at time [math]2m[/math].