Revision as of 02:30, 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 if <math display="block"> h(z) = \frac{1 - \sqrt{1 - 4pqz^2}}{2qz}\ , </math> then <math display="block"> h(1) = \left \{ \begin{array}{ll} p/q, & \mbox{if $p \leq q,$} \\ 1, & \mbox{if <math>p \geq q,<...")
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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 if

[[math]] h(z) = \frac{1 - \sqrt{1 - 4pqz^2}}{2qz}\ , [[/math]]

then

[[math]] h(1) = \left \{ \begin{array}{ll} p/q, & \mbox{if $p \leq q,$} \\ 1, & \mbox{if \ltmath\gtp \geq q,[[/math]]

} \end{array}\right. </math> and

[[math]] h'(1) = \left \{ \begin{array}{ll} 1/(p - q), & \mbox{if $p \gt q,$}\\ \infty, & \mbox{if \ltmath\gtp = q.[[/math]]

} \end{array}\right. </math>