Revision as of 03:36, 9 June 2024 by Bot (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> Show that the Taylor series expansion for <math>\sqrt{1 - x}</math> is <math display="block"> \sqrt{1 - x} = \sum_{n = 0}^\infty {{1/2} \choose n} x^n\ , </math> where the binomial coefficient <math>{1/2} \choose n</math> is <math display="bloc...")
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BBy Bot
Jun 09'24

Exercise

[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]

Show that the Taylor series expansion for [math]\sqrt{1 - x}[/math] is

[[math]] \sqrt{1 - x} = \sum_{n = 0}^\infty {{1/2} \choose n} x^n\ , [[/math]]

where the binomial coefficient [math]{1/2} \choose n[/math] is

[[math]] {{1/2} \choose n} = \frac{(1/2)(1/2 - 1) \cdots (1/2 - n + 1)}{n!}\ . [[/math]]

Using this and the result of Exercise Exercise, show that the probability that the gambler is ruined on the [math]n[/math]th step is

[[math]] p_T(n) = \left \{ \begin{array}{ll} \frac{(-1)^{k - 1}}{2p} {{1/2} \choose k} (4pq)^k, & \mbox{if $n = 2k - 1$,} \\ 0, & \mbox{if \ltmath\gtn = 2k[[/math]]

.}

                \end{array}
       \right.   

</math>