#LyX 2.4 created this file. For more info see https://www.lyx.org/
\lyxformat 620
\begin_document
\begin_header
\save_transient_properties true
\origin unavailable
\textclass book
\use_default_options true
\maintain_unincluded_children no
\language american
\language_package default
\inputencoding utf8
\fontencoding auto
\font_roman "default" "default"
\font_sans "default" "default"
\font_typewriter "default" "default"
\font_math "auto" "auto"
\font_default_family default
\use_non_tex_fonts false
\font_sc false
\font_roman_osf false
\font_sans_osf false
\font_typewriter_osf false
\font_sf_scale 100 100
\font_tt_scale 100 100
\use_microtype false
\use_dash_ligatures true
\graphics default
\default_output_format default
\output_sync 0
\bibtex_command default
\index_command default
\float_placement class
\float_alignment class
\paperfontsize default
\spacing single
\use_hyperref false
\papersize default
\use_geometry false
\use_package amsmath 1
\use_package amssymb 1
\use_package cancel 1
\use_package esint 1
\use_package mathdots 1
\use_package mathtools 1
\use_package mhchem 1
\use_package stackrel 1
\use_package stmaryrd 1
\use_package undertilde 1
\cite_engine basic
\cite_engine_type default
\biblio_style plain
\use_bibtopic false
\use_indices false
\paperorientation portrait
\suppress_date false
\justification true
\use_refstyle 1
\use_formatted_ref 0
\use_minted 0
\use_lineno 0
\index Index
\shortcut idx
\color #008000
\end_index
\secnumdepth 3
\tocdepth 3
\paragraph_separation indent
\paragraph_indentation default
\is_math_indent 0
\math_numbering_side default
\quotes_style english
\dynamic_quotes 0
\papercolumns 1
\papersides 1
\paperpagestyle default
\tablestyle default
\tracking_changes false
\output_changes false
\change_bars false
\postpone_fragile_content true
\html_math_output 0
\html_css_as_file 0
\html_be_strict false
\docbook_table_output 0
\docbook_mathml_prefix 1
\end_header

\begin_body

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
TODO 4,
 6 (1 p.) (est.
 0:13:30)
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc4[HM10]
\end_layout

\end_inset

Prove Eq.
 (13).
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

Since 
\begin_inset Formula $u\geq0$
\end_inset

,
 
\begin_inset Formula $\ln(1+u)\geq0$
\end_inset

 and 
\begin_inset Formula $u=v+\ln(1+u)\geq v$
\end_inset

.
 On the other hand,
 
\begin_inset Formula $x$
\end_inset

 is fixed with respect to 
\begin_inset Formula $v$
\end_inset

,
 so by monotonicity of the integral,
 if 
\begin_inset Formula $v\geq1$
\end_inset

,
\begin_inset Formula 
\begin{multline*}
0\leq\int_{r}^{\infty}x\text{e}^{-xv}\left(1+\frac{1}{u}\right)\text{d}v\leq\int_{r}^{\infty}x\text{e}^{-xv}\left(1+\frac{1}{v}\right)\text{d}v\leq\int_{r}^{\infty}2x\text{e}^{-xv}=\\
=\left.-2\text{e}^{-xv}\right|_{v=r}^{\infty}=2\text{e}^{-rx}.
\end{multline*}

\end_inset

This means 
\begin_inset Formula $\int_{r}^{\infty}x\text{e}^{-xv}\left(1+\frac{1}{u}\right)\text{d}v=O(\text{e}^{-rx})$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc6[HM20]
\end_layout

\end_inset

Prove Eq.
 (23).
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset


\begin_inset Formula 
\[
(n+\alpha)^{n+\beta}=n^{n+\beta}\left(1+\frac{\alpha}{n}\right)^{n+\beta}.
\]

\end_inset

For the second factor,
\begin_inset Formula 
\begin{multline*}
\ln\left(1+\frac{\alpha}{n}\right)^{n+\beta}=(n+\beta)\ln\left(1+\frac{\alpha}{n}\right)=(n+\beta)\left(\frac{\alpha}{n}-\frac{\alpha^{2}}{2n^{2}}+O(n^{-3})\right)=\\
=\alpha+\frac{1}{n}\left(\alpha\beta-\frac{\alpha^{2}}{2}\right)+O(n^{-2}),
\end{multline*}

\end_inset

so
\begin_inset Formula 
\begin{multline*}
n^{n+\beta}\left(1+\frac{\alpha}{n}\right)^{n+\beta}=n^{n+\beta}\text{e}^{\alpha}\exp\left(\frac{1}{n}\left(\alpha\beta-\frac{\alpha^{2}}{2}\right)+O(n^{-2})\right)=\\
n^{n+\beta}\text{e}^{\alpha}\left(1+\frac{\alpha}{n}\left(\beta-\frac{\alpha}{2}\right)+O(n^{-2})\right).
\end{multline*}

\end_inset


\end_layout

\end_body
\end_document