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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
-\save_transient_properties true
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-\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