exercise:05298cad26: 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> 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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{{1/2} \choose n} = \frac{(1/2)(1/2 - 1) \cdots (1/2 - n + 1)}{n!}\ . | {{1/2} \choose n} = \frac{(1/2)(1/2 - 1) \cdots (1/2 - n + 1)}{n!}\ . | ||
</math> | </math> | ||
Using this and the result of | Using this and the result of [[exercise:85a2ae80e0 |Exercise]], show that the probability | ||
that the gambler is ruined on the <math>n</math>th step is | that the gambler is ruined on the <math>n</math>th step is | ||
| Line 21: | Line 21: | ||
p_T(n) = \left \{ \begin{array}{ll} | p_T(n) = \left \{ \begin{array}{ll} | ||
\frac{(-1)^{k - 1}}{2p} {{1/2} \choose k} (4pq)^k, & \mbox{if $n = 2k - 1$,} \\ | \frac{(-1)^{k - 1}}{2p} {{1/2} \choose k} (4pq)^k, & \mbox{if $n = 2k - 1$,} \\ | ||
0, & \mbox{if | 0, & \mbox{if $n = 2k$.} | ||
\end{array} | \end{array} | ||
\right. | \right. | ||
</math> | </math> | ||
Latest revision as of 01:55, 15 June 2024
[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, 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 $n = 2k$.}
\end{array}
\right.
[[/math]]