exercise:5a5fe2185b: 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 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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Show that if | |||
<math display="block"> | <math display="block"> | ||
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h(1) = \left \{ \begin{array}{ll} | h(1) = \left \{ \begin{array}{ll} | ||
p/q, & \mbox{if $p \leq q,$} \\ | p/q, & \mbox{if $p \leq q,$} \\ | ||
1, & \mbox{if | 1, & \mbox{if $p \geq q,$} \end{array}\right. | ||
</math> | </math> | ||
and | and | ||
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h'(1) = \left \{ \begin{array}{ll} | h'(1) = \left \{ \begin{array}{ll} | ||
1/(p - q), & \mbox{if $p > q,$}\\ | 1/(p - q), & \mbox{if $p > q,$}\\ | ||
\infty, & \mbox{if | \infty, & \mbox{if $p = q.$} \end{array}\right. | ||
</math> | </math> | ||
Latest revision as of 23:46, 14 June 2024
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 $p \geq q,$} \end{array}\right.
[[/math]]
and
[[math]]
h'(1) = \left \{ \begin{array}{ll}
1/(p - q), & \mbox{if $p \gt q,$}\\
\infty, & \mbox{if $p = q.$} \end{array}\right.
[[/math]]