From 4f670b750af5c11e1eac16d9cd8556455f89f46a Mon Sep 17 00:00:00 2001
From: Juan Marín Noguera <juan@mnpi.eu>
Date: Fri, 16 May 2025 22:18:44 +0200
Subject: Changed layout for more manageable volumes

---
 1.2.9.lyx | 846 --------------------------------------------------------------
 1 file changed, 846 deletions(-)
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diff --git a/1.2.9.lyx b/1.2.9.lyx
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--- a/1.2.9.lyx
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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
-\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
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-\spacing single
-\use_hyperref false
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-\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
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-\postpone_fragile_content true
-\html_math_output 0
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-\html_be_strict false
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc2[M13]
-\end_layout
-
-\end_inset
-
-Prove Eq.
- (11).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\[
-\left(\sum_{n\geq0}\frac{a_{n}}{n!}z^{n}\right)\left(\sum_{m\geq0}\frac{b_{m}}{m!}z^{m}\right)=\sum_{n\geq0}\left(\sum_{k}\frac{a_{k}b_{n-k}}{k!(n-k)!}\right)z^{n}=\sum_{n\geq0}\frac{1}{n!}\left(\sum_{k}\binom{n}{k}a_{k}b_{n-k}\right)z^{n}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc4[M01]
-\end_layout
-
-\end_inset
-
-Explain why Eq.
- (19) is a special case of Eq.
- (21).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Just set 
-\begin_inset Formula $t=0$
-\end_inset
-
-,
- then 
-\begin_inset Formula $x=1+z$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc6[HM15]
-\end_layout
-
-\end_inset
-
-Find the generating function for
-\begin_inset Formula 
-\[
-\left\langle \sum_{0<k<n}\frac{1}{k(n-k)}\right\rangle ;
-\]
-
-\end_inset
-
-differentiate it and express the coefficients in terms of harmonic numbers.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The generating function is
-\begin_inset Formula 
-\begin{align*}
-G(z) & \coloneqq\sum_{n\geq0}\sum_{0<k<n}\frac{1}{k(n-k)}z^{n}=\sum_{n,m\geq1}\frac{z^{n+m}}{nm}=\left(\sum_{n\geq1}\frac{z^{n}}{n}\right)\left(\sum_{m\geq1}\frac{z^{m}}{m}\right)=\left(\sum_{n\geq1}\frac{z^{n}}{n}\right)^{2}\\
- & =\left(-\sum_{n\geq1}\frac{(-1)^{n+1}}{n}(-z)^{n}\right)^{2}=(-\ln(1+(-z)))^{2}=\ln(1-z)^{2},
-\end{align*}
-
-\end_inset
-
-by Eq.
- (24).
- Its derivative is
-\begin_inset Formula 
-\[
-\dot{G}(z)=-\frac{2}{1-z}\ln(1-z)=\frac{2}{1-z}\ln\left(\frac{1}{1-z}\right),
-\]
-
-\end_inset
-
-so by Eq.
- (25) with 
-\begin_inset Formula $m=0$
-\end_inset
-
- the coefficients are 
-\begin_inset Formula $[z^{n}]\dot{G}(z)=2(H_{n}-H_{0})\binom{n}{n}=2H_{n}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc10[M25]
-\end_layout
-
-\end_inset
-
-An 
-\emph on
-elementary symmetric function
-\emph default
- is defined by the formula
-\begin_inset Formula 
-\[
-e_{m}=\sum_{1\leq j_{1}<\dots<j_{m}\leq n}x_{j_{1}}\cdots x_{j_{m}}.
-\]
-
-\end_inset
-
-(This is the same as 
-\begin_inset Formula $h_{m}$
-\end_inset
-
- of Eq.
- (33),
- except that equal subscripts are not allowed.) Find the generating function for 
-\begin_inset Formula $e_{m}$
-\end_inset
-
-,
- and express 
-\begin_inset Formula $e_{m}$
-\end_inset
-
- in terms of the 
-\begin_inset Formula $S_{j}$
-\end_inset
-
- in Eq.
- (34).
- Write out the formulas for 
-\begin_inset Formula $e_{1}$
-\end_inset
-
-,
- 
-\begin_inset Formula $e_{2}$
-\end_inset
-
-,
- 
-\begin_inset Formula $e_{3}$
-\end_inset
-
-,
- and 
-\begin_inset Formula $e_{4}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Here 
-\begin_inset Formula $m\leq n$
-\end_inset
-
- or otherwise we have no elements,
- so
-\begin_inset Formula 
-\begin{align*}
-G(z) & \coloneqq\sum_{m\geq0}e_{m}z^{m}=\sum_{m=0}^{n}\sum_{1\leq j_{1}<\dots<j_{k}\leq n}x_{j_{1}}\cdots x_{j_{m}}z^{n}=\sum_{a_{1},\dots,a_{n}\in\{0,1\}}\prod_{i=1}^{n}(x_{i}z)^{a_{i}}\\
- & =\prod_{i=1}^{n}\sum_{a\in\{0,1\}}(x_{i}z)^{a}=\prod_{i=1}^{n}(x_{i}z+1),
-\end{align*}
-
-\end_inset
-
-as we can regard the sum as taking the product of each subset of 
-\begin_inset Formula $\{x_{i}z\}_{i}$
-\end_inset
-
-.
- From here,
-\begin_inset Formula 
-\[
-\ln G(z)=\sum_{i=1}^{n}\ln(1+x_{i}z)=\sum_{i=1}^{n}\sum_{k\geq1}\frac{(-1)^{k+1}}{k}x_{i}^{k}z^{k}=\sum_{k\geq1}\frac{(-1)^{k+1}}{k}S_{k}z^{k},
-\]
-
-\end_inset
-
-so
-\begin_inset Formula 
-\begin{align*}
-G(z) & =\text{e}^{\ln G(z)}=\exp\left(\sum_{k\geq1}\frac{(-1)^{k+1}}{k}S_{k}z^{k}\right)=\prod_{k\geq1}\text{e}^{(-1)^{k+1}S_{k}z^{k}/k}\\
- & =\prod_{k\geq1}\sum_{j\geq0}\frac{1}{j!}\left(\frac{(-1)^{k+1}S_{k}}{k}\right)^{j}z^{jk}=\sum_{m\geq0}\left(\sum_{\begin{subarray}{c}
-j_{1},\dots,j_{m}\geq0\\
-j_{1}+2j_{2}+\dots+mj_{m}=m
-\end{subarray}}\prod_{k=1}^{m}\frac{1}{j_{k}!}\left(\frac{(-1)^{k+1}S_{k}}{k}\right)^{j_{k}}\right)z^{m}.
-\end{align*}
-
-\end_inset
-
-This gives us an explicit formulation for 
-\begin_inset Formula $e_{m}$
-\end_inset
-
-.
- Note that 
-\begin_inset Formula 
-\[
-\prod_{k=1}^{m}(-1)^{(k+1)j_{k}}=(-1)^{\sum_{k=1}^{m}kj_{k}+\sum_{k=1}^{m}j_{k}}=(-1)^{m+\sum_{k}j_{k}},
-\]
-
-\end_inset
-
-so
-\begin_inset Formula 
-\[
-e_{m}=\sum_{\begin{subarray}{c}
-j_{1},\dots,j_{m}\geq0\\
-j_{1}+2j_{2}+\dots+mj_{m}=m
-\end{subarray}}(-1)^{m+j_{1}+\dots+j_{m}}\prod_{k=1}^{m}\frac{S_{k}^{j_{k}}}{k^{j_{k}}j_{k}!}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-For example,
- if we call 
-\begin_inset Formula $p_{kj_{k}}$
-\end_inset
-
- to each of the factors of the product in this latest formula,
- using that each 
-\begin_inset Formula $p_{k0}=1$
-\end_inset
-
- and can therefore be ignored,
-\begin_inset Formula 
-\begin{align*}
-e_{1} & =p_{11}=S_{1},\\
-e_{2} & =p_{12}+p_{21}=\frac{S_{1}^{2}}{2}-\frac{S_{2}}{2}=\frac{1}{2}(S_{1}^{2}-S_{2}),\\
-e_{3} & =p_{13}-p_{11}p_{21}+p_{31}=\frac{S_{1}^{3}}{6}-S_{1}\frac{S_{2}}{2}+\frac{S_{3}}{3}=\frac{1}{6}(S_{1}^{3}+2S_{3}-3S_{1}S_{2}),\\
-e_{4} & =p_{14}-p_{12}p_{21}+p_{11}p_{31}+p_{22}-p_{41}=\frac{S_{1}^{4}}{24}-\frac{S_{1}^{2}}{2}\frac{S_{2}}{2}+S_{1}\frac{S_{3}}{3}+\frac{S_{2}^{2}}{8}-\frac{S_{4}}{4}\\
- & =\frac{1}{24}(S_{1}^{4}-6S_{1}^{2}S_{2}+8S_{1}S_{3}+3S_{2}^{2}-6S_{4}).
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc11[M25]
-\end_layout
-
-\end_inset
-
-Equation (39) can also be used to express the 
-\begin_inset Formula $S$
-\end_inset
-
-'s in terms of the 
-\begin_inset Formula $h$
-\end_inset
-
-'s:
- We find 
-\begin_inset Formula $S_{1}=h_{1}$
-\end_inset
-
-,
- 
-\begin_inset Formula $S_{2}=2h_{2}-h_{1}^{2}$
-\end_inset
-
-,
- 
-\begin_inset Formula $S_{3}=3h_{3}-3h_{1}h_{2}+h_{1}^{3}$
-\end_inset
-
-,
- etc.
- What is the coefficient of 
-\begin_inset Formula $h_{1}^{k_{1}}h_{2}^{k_{2}}\cdots h_{m}^{k_{m}}$
-\end_inset
-
- in this representation of 
-\begin_inset Formula $S_{m}$
-\end_inset
-
-,
- when 
-\begin_inset Formula $k_{1}+2k_{2}+\dots+mk_{m}=m$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Note Greyedout
-status open
-
-\begin_layout Plain Layout
-(I had to look up the solution.)
-\end_layout
-
-\end_inset
-
- We ignore Eq.
- (39) and focus on Eqs.
- (37) and (24).
- There,
-\begin_inset Formula 
-\begin{align*}
-\sum_{k\geq1}\frac{S_{k}z^{k}}{k!} & =\ln G(z)=\ln\left(1+\sum_{j\geq1}h_{j}z^{j}\right)=\sum_{k\geq1}\frac{(-1)^{k+1}}{k}\left(\sum_{j\geq1}h_{j}z^{j}\right)^{k}\\
- & =\sum_{k\geq1}\frac{(-1)^{k+1}}{k}\sum_{j_{1},\dots,j_{k}\geq1}h_{j_{1}}\cdots h_{j_{k}}z^{j_{1}+\dots+j_{k}}\\
- & =\sum_{m\geq1}\sum_{\begin{subarray}{c}
-i_{1},\dots,i_{m}\geq0\\
-i_{1}+2i_{2}+\dots+mi_{m}=m
-\end{subarray}}\frac{(-1)^{i_{1}+\dots+i_{m}+1}}{i_{1}+\dots+i_{m}}\binom{i_{1}+\dots+i_{m}}{i_{1},\dots,i_{m}}h_{1}^{i_{1}}\cdots h_{m}^{i_{m}}z^{m},
-\end{align*}
-
-\end_inset
-
-so the coefficient is
-\begin_inset Formula 
-\[
-\frac{(-1)^{k_{1}+\dots+k_{m}+1}(k_{1}+\dots+k_{m})!}{k_{1}!\cdots k_{m}!}.
-\]
-
-\end_inset
-
-For the last identity,
- we have regrouped all the terms by the value of 
-\begin_inset Formula $m\coloneqq j_{1}+\dots+j_{k}$
-\end_inset
-
- and then by 
-\begin_inset Formula $\{i_{j}\coloneqq|\{t\mid j_{t}=j\}|\}_{j=1}^{m}$
-\end_inset
-
-,
- that is,
- the number of times that each 
-\begin_inset Formula $h_{j}$
-\end_inset
-
- appears rather than the indexes of those that do.
- When doing this,
- we note that 
-\begin_inset Formula $k=i_{1}+\dots+i_{m}$
-\end_inset
-
- and that 
-\begin_inset Formula $h_{j_{1}}\cdots h_{j_{k}}=h_{1}^{i_{1}}\cdots h_{m}^{i_{m}}$
-\end_inset
-
-,
- so all the addends in a given group are the same,
- and the number of addends is the number of orderings of the 
-\begin_inset Formula $k$
-\end_inset
-
- indexes without considering the order of indexes that are equal,
- of which there are 
-\begin_inset Formula $i_{1}$
-\end_inset
-
- indexes equal to 1,
- 
-\begin_inset Formula $i_{2}$
-\end_inset
-
- equal to 2,
- etc.,
- and this is a multinomial coefficient.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc12[M20]
-\end_layout
-
-\end_inset
-
-Suppose we have a doubly subscripted sequence 
-\begin_inset Formula $\langle a_{mn}\rangle$
-\end_inset
-
- for 
-\begin_inset Formula $m,n=0,1,\dots$
-\end_inset
-
-;
- show how this double sequence can be represented by a 
-\emph on
-single
-\emph default
- generating function of two variables,
- and determine the generating function for 
-\begin_inset Formula $\langle\binom{n}{m}\rangle$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-A generating function in this case could be something like
-\begin_inset Formula 
-\[
-G(y,z)\coloneqq\sum_{m,n\geq0}a_{mn}y^{m}z^{n}=\sum_{n\geq0}\left(\sum_{m\geq0}a_{mn}y^{m}\right)z^{n}=\sum_{m\geq0}\left(\sum_{n\geq0}a_{mn}z^{n}\right)y^{m}.
-\]
-
-\end_inset
-
-In our case,
- by the binomial theorem,
-\begin_inset Formula 
-\[
-(1+y)^{n}=\sum_{m\geq0}\binom{n}{m}y^{m},
-\]
-
-\end_inset
-
-so
-\begin_inset Formula 
-\[
-G(y,z)=\sum_{n\geq0}(1+y)^{n}z^{n}=\frac{1}{1-(1+y)z}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc18[M25]
-\end_layout
-
-\end_inset
-
-Given positive integers 
-\begin_inset Formula $n$
-\end_inset
-
- and 
-\begin_inset Formula $r$
-\end_inset
-
-,
- find a simple formula for the value of the following sums:
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\sum_{1\leq k_{1}<k_{2}<\dots<k_{r}\leq n}k_{1}k_{2}\cdots k_{r}$
-\end_inset
-
-;
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\sum_{1\leq k_{1}\leq k_{2}\leq\dots\leq k_{r}\leq n}k_{1}k_{2}\cdots k_{r}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-(For example,
- when 
-\begin_inset Formula $n=3$
-\end_inset
-
- and 
-\begin_inset Formula $r=2$
-\end_inset
-
- the sums are,
- respectively,
- 
-\begin_inset Formula $1\cdot2+1\cdot3+2\cdot3$
-\end_inset
-
- and 
-\begin_inset Formula $1\cdot1+1\cdot2+1\cdot3+2\cdot2+2\cdot3+3\cdot3$
-\end_inset
-
-.)
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-Following Exercise 10 and then applying Equation (27),
-\begin_inset Formula 
-\begin{align*}
-G(z) & \coloneqq\sum_{r\geq0}\sum_{1\leq k_{1}<\dots<k_{r}\leq n}k_{1}\cdots k_{r}=\prod_{k=1}^{n}(1+kz)=z^{n}\prod_{k=1}^{n}\left(\frac{1}{z}+k\right)\\
- & =z^{n+1}\prod_{k=0}^{n}\left(\frac{1}{z}+k\right)=z^{n+1}\sum_{k\geq0}\stirla{n+1}{k}z^{-k}\\
- & =\sum_{k=0}^{n+1}\stirla{n+1}{k}z^{n+1-k}=\sum_{k=0}^{n+1}\stirla{n+1}{n+1-k}z^{k},
-\end{align*}
-
-\end_inset
-
-so the sum is 
-\begin_inset Formula $\stirla{n+1}{n+1-r}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-Doing something similar with Equations (36) and (28),
-\begin_inset Formula 
-\begin{align*}
-G(z) & \coloneqq\sum_{r\geq0}\sum_{1\leq k_{1}\leq\dots\leq k_{r}\leq n}k_{1}\cdots k_{r}=\prod_{k=1}^{n}\frac{1}{1-kz}=\frac{1}{z^{n}}\sum_{k\geq n}\stirlb{k}{n}z^{k}\\
- & =\sum_{k}\stirlb{k+n}{n}z^{k},
-\end{align*}
-
-\end_inset
-
-so the sum is 
-\begin_inset Formula $\stirlb{n+r}{n}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc25[M23]
-\end_layout
-
-\end_inset
-
-Evaluate the sum 
-\begin_inset Formula $\sum_{k}\binom{n}{k}\binom{2n-2k}{n-k}(-2)^{k}$
-\end_inset
-
- by simplifying the equivalent formula 
-\begin_inset Formula $\sum_{k}[w^{k}](1-2w)^{n}[z^{n-k}](1+z)^{2n-2k}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-By the binomial theorem,
-\begin_inset Formula 
-\begin{align*}
-(1-2w)^{n} & =\sum_{k\geq0}\binom{n}{k}(-2)^{k}w^{k}, & (1+z)^{2n-2k} & =\sum_{j\geq0}\binom{2n-2k}{j}z^{j},
-\end{align*}
-
-\end_inset
-
-so the second formula is indeed equivalent to the first,
- assuming of course that 
-\begin_inset Formula $n\in\mathbb{N}$
-\end_inset
-
- and therefore terms with negative 
-\begin_inset Formula $k$
-\end_inset
-
- or 
-\begin_inset Formula $k>n$
-\end_inset
-
- are null.
- Now,
- 
-\begin_inset Formula $[z^{n-k}](1+z)^{2n-2k}=[z^{n}](z^{k}(1+z)^{2n-2k})$
-\end_inset
-
-,
- so this is
-\begin_inset Formula 
-\[
-S\coloneqq[z^{n}](1+z)^{2n}\sum_{k}[w^{k}](1-2w)^{n}\left(\frac{z}{(1+z)^{2}}\right)^{k}=[z^{n}](1+z)^{2n}\left(1-\frac{2z}{(1+z)^{2}}\right)^{n},
-\]
-
-\end_inset
-
-where we evaluating the sum by taking 
-\begin_inset Formula $\frac{z}{(1+z)^{2}}$
-\end_inset
-
- as the argument of the generating function in 
-\begin_inset Formula $w$
-\end_inset
-
-.
- Then,
- we simplify to get
-\begin_inset Formula 
-\[
-S=[z^{n}]\left(\left((1+z)^{2}(1-\nicefrac{2z}{(1+z)^{2}})\right)^{n}\right)=[z^{n}]\left((1+z^{2})^{n}\right)=\binom{n}{\nicefrac{n}{2}}[n\text{ even}].
-\]
-
-\end_inset
-
-
-\end_layout
-
-\end_body
-\end_document
-- 
cgit v1.2.3