From 4f670b750af5c11e1eac16d9cd8556455f89f46a Mon Sep 17 00:00:00 2001 From: Juan Marín Noguera Date: Fri, 16 May 2025 22:18:44 +0200 Subject: Changed layout for more manageable volumes --- 1.2.3.lyx | 1732 ------------------------------------------------------------- 1 file changed, 1732 deletions(-) delete mode 100644 1.2.3.lyx (limited to '1.2.3.lyx') diff --git a/1.2.3.lyx b/1.2.3.lyx deleted file mode 100644 index 0648fc2..0000000 --- a/1.2.3.lyx +++ /dev/null @@ -1,1732 +0,0 @@ -#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 -\begin_preamble -\input defs -\end_preamble -\use_default_options true -\maintain_unincluded_children no -\language english -\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 false -\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 ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc1[10] -\end_layout - -\end_inset - -The text says that -\begin_inset Formula $a_{1}+a_{2}+\dots+a_{0}=0$ -\end_inset - -. - What then, - is -\begin_inset Formula $a_{2}+\dots+a_{0}$ -\end_inset - -? -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -We could set that to be -\begin_inset Formula $-a_{1}$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc2[01] -\end_layout - -\end_inset - -What does the notation -\begin_inset Formula $\sum_{1\leq j\leq n}a_{j}$ -\end_inset - - mean, - if -\begin_inset Formula $n=3.14$ -\end_inset - -? -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - - -\begin_inset Formula $a_{1}+a_{2}+a_{3}$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc3[13] -\end_layout - -\end_inset - -Without using the -\begin_inset Formula $\sum$ -\end_inset - --notation, - write out the equivalent of -\begin_inset Formula -\[ -\sum_{0\leq n\leq5}\frac{1}{2n+1}, -\] - -\end_inset - -and also the equivalent of -\begin_inset Formula -\[ -\sum_{0\leq n^{2}\leq5}\frac{1}{2n^{2}+1}. -\] - -\end_inset - -Explain why the two results are different, - in spite of rule (b). -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - - -\begin_inset Formula -\begin{eqnarray*} -\sum_{0\leq n\leq5}\frac{1}{2n+1} & = & \frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11},\\ -\sum_{0\leq n^{2}\leq5}\frac{1}{2n^{2}+1} & = & \frac{1}{9}+\frac{1}{3}+\frac{1}{1}+\frac{1}{3}+\frac{1}{9}, -\end{eqnarray*} - -\end_inset - -as the integers with -\begin_inset Formula $0\leq n^{2}\leq5$ -\end_inset - - are -\begin_inset Formula $\{-2,-1,0,1,2\}$ -\end_inset - -. - Rule (b) states that a permutation of the integers satisfying the condition doesn't alter the result, - but because -\begin_inset Formula $n\mapsto n^{2}$ -\end_inset - - is not a permutation of such -\emph on -integers -\emph default -, - the rule doesn't apply. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc4[10] -\end_layout - -\end_inset - -Without using the -\begin_inset Formula $\sum$ -\end_inset - --notation, - write out the equivalent of each side of Eq. - (10) as a sum of sums for the case -\begin_inset Formula $n=3$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - - -\begin_inset Formula $a_{11}+a_{21}+a_{22}+a_{31}+a_{32}+a_{33}=a_{11}+a_{21}+a_{31}+a_{22}+a_{32}+a_{33}$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc5[HM20] -\end_layout - -\end_inset - -Prove that rule (a) is valid for arbitrary infinite series, - provided that the series converge. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -Since infinite series are defined for cases where the total number of nonzero elements is at most countable, - we might assume that the series are indexed by positive integers. - Thus, - we'd have to prove that -\begin_inset Formula -\[ -\left(\sum_{i=1}^{\infty}a_{i}\right)\left(\sum_{j=1}^{\infty}b_{j}\right)=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i}b_{j}, -\] - -\end_inset - -assuming that the relevant series converge. - We have -\end_layout - -\begin_layout Standard -\begin_inset Formula -\[ -\left(\sum_{i=0}^{\infty}a_{i}\right)\left(\sum_{j=0}^{\infty}b_{j}\right)=\sum_{i=0}^{\infty}\left(a_{i}\sum_{j=0}^{\infty}b_{j}\right)=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}b_{j}, -\] - -\end_inset - -by entering the second series as a constant into the first series (because it converges) and then entering the element of the first series, - considered as a constant, - into the inner series. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc9[05] -\end_layout - -\end_inset - -Is the derivation of Eq. - (14) valid even if -\begin_inset Formula $n=-1$ -\end_inset - -? -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -The result is -\begin_inset Formula -\[ -\sum_{0\leq j\leq-1}ax^{j}=0=a\left(\frac{1-x^{0}}{1-x}\right), -\] - -\end_inset - -which is correct. - However, - the derivation is not correct as the rule (d) cannot be applied, - because the applications in the derivation assume -\begin_inset Formula $n\geq0$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc10[05] -\end_layout - -\end_inset - -Is the derivation of Eq. - (14) valid even if -\begin_inset Formula $n=-2$ -\end_inset - -? -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -No (see previous exercise). -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc11[03] -\end_layout - -\end_inset - -What should the right-hand side of Eq. - (14) be if -\begin_inset Formula $x=1$ -\end_inset - -? -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -Mere substitution in the original formula yields an undefined result, - but it's clear that -\begin_inset Formula -\[ -\sum_{0\leq j\leq n}a\cdot1^{j}=\sum_{0\leq j\leq n}a=(n+1)a. -\] - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc12[10] -\end_layout - -\end_inset - -What is -\begin_inset Formula $1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}+\dots+\left(\frac{1}{7}\right)^{n}$ -\end_inset - -? -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -By Eq. - (14), - this is -\begin_inset Formula -\[ -\sum_{k=0}^{n}\left(\frac{1}{7}\right)^{k}=\left(\frac{1-\left(\frac{1}{7}\right)^{n+1}}{1-\frac{1}{7}}\right)=\frac{1-\left(\frac{1}{7}\right)^{n+1}}{\frac{6}{7}}=\frac{7}{6}\left(1-\frac{1}{7^{n+1}}\right). -\] - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc13[10] -\end_layout - -\end_inset - -Using Eq. - (15) and assuming that -\begin_inset Formula $m\leq n$ -\end_inset - -, - evaluate -\begin_inset Formula $\sum_{j=m}^{n}j$ -\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 Standard -\begin_inset Formula -\begin{multline*} -\sum_{m\leq j\leq n}j=\sum_{m\leq j+m\leq n}(j+m)=\sum_{0\leq j\leq n-m}(m+j)=\\ -=m(n-m+1)+\frac{1}{2}(n-m)(n-m+1)=\frac{1}{2}(n(n+1)-m(m-1)). -\end{multline*} - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc14[11] -\end_layout - -\end_inset - -Using the result of the previous exercise, - evaluate -\begin_inset Formula $\sum_{j=m}^{n}\sum_{k=r}^{s}jk$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - - -\begin_inset Formula -\begin{multline*} -\sum_{j=m}^{n}\sum_{k=r}^{s}jk=\left(\sum_{j=m}^{n}j\right)\left(\sum_{k=r}^{s}k\right)=\\ -=\frac{1}{4}(n(n+1)-m(m-1))(s(s+1)-r(r-1)). -\end{multline*} - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc15[M22] -\end_layout - -\end_inset - -Compute the sum -\begin_inset Formula $1\times2+2\times2^{2}+3\times2^{3}+\dots+n\times2^{n}$ -\end_inset - - for small values of -\begin_inset Formula $n$ -\end_inset - -. - Do you see the pattern developing in these numbers? - If not, - discover it by manipulations similar to those leading up to Eq. - (14). -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -Let -\begin_inset Formula $\tau(n)$ -\end_inset - - be such a such sum -\begin_inset Formula -\begin{align*} -\tau(1) & =2, & \tau(2) & =2+8=10, & \tau(3) & =10+24=34,\\ -\tau(4) & =34+64=98, & \tau(5) & =98+160=258, & & \dots -\end{align*} - -\end_inset - - -\end_layout - -\begin_layout Standard -Now, -\begin_inset Formula -\begin{align*} -\tau(n) & =\sum_{1\leq k\leq n}k2^{k} & & \text{by definition}\\ - & =\sum_{0\leq k\leq n-1}(k+1)2^{k+1} & & \text{by rule (b)}\\ - & =2\sum_{0\leq k\leq n-1}(k+1)2^{k} & & \text{by a special case of (a)}\\ - & =2\left(\sum_{0\leq k\leq n-1}k2^{k}+\sum_{0\leq k\leq n-1}2^{k}\right) & & \text{by rule (c)}\\ - & =2\left(\sum_{0\leq k\leq n}k2^{k}-n2^{n}+(2^{n}-1)\right) & & \text{by rule (d) and Eq. (14)}\\ - & =2\left(\sum_{1\leq k\leq n}k2^{k}-n2^{n}+(2^{n}-1)\right) & & \text{by rule (d)} -\end{align*} - -\end_inset - - -\end_layout - -\begin_layout Standard -This means -\begin_inset Formula $\tau(n)=2(\tau(n)-n2^{n}+2^{n}-1)=2\tau(n)-n2^{n+1}+2^{n+1}-2$ -\end_inset - -, - so -\begin_inset Formula -\[ -\tau(n)=n2^{n+1}-2^{n+1}+2=(n-1)2^{n+1}+2. -\] - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc17[M00] -\end_layout - -\end_inset - -Let -\begin_inset Formula $S$ -\end_inset - - be a set of integers. - What is -\begin_inset Formula $\sum_{j\in S}1$ -\end_inset - -? -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - - -\begin_inset Formula $|S|$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc20[25] -\end_layout - -\end_inset - -Dr. - I. - J. - Matrix has observed a remarkable sequence of formulas: -\begin_inset Formula -\begin{align*} -9\times1+2 & =11, & 9\times12+3 & =111,\\ -9\times123+4 & =1111, & 9\times1234+5 & =11111. -\end{align*} - -\end_inset - - -\end_layout - -\begin_layout Enumerate -Write the good doctor's great discovery in terms of the -\begin_inset Formula $\sum$ -\end_inset - --notation. -\end_layout - -\begin_layout Enumerate -Your answer to part (a) undoubtedly involves the number 10 as base of the decimal system; - generalize this formula so that you get a formula that will perhaps work in any base -\begin_inset Formula $b$ -\end_inset - -. -\end_layout - -\begin_layout Enumerate -Prove your formula from part (b) by using formulas derived in the text or in exercise 16 above. -\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 -For -\begin_inset Formula $n\geq1$ -\end_inset - -, - -\begin_inset Formula -\[ -9\sum_{j=0}^{n-1}10^{j}(n-j)+n+1=\sum_{j=0}^{n}10^{j}. -\] - -\end_inset - - -\end_layout - -\begin_layout Enumerate -\begin_inset Formula -\[ -(b-1)\sum_{j=0}^{n-1}b^{j}(n-j)+n+1=\sum_{j=0}^{n}b^{j}. -\] - -\end_inset - - -\end_layout - -\begin_layout Enumerate -We have -\begin_inset Formula -\begin{eqnarray*} -\sum_{j=0}^{n-1}b^{j}(n-j) & = & n\sum_{j=0}^{n-1}b^{j}-\sum_{j=0}^{n-1}jb^{j}\\ - & = & n\frac{1-b^{n}}{1-b}-\frac{(n-1)b^{n+1}-nb^{n}+b}{(b-1)^{2}}\\ - & = & n\frac{b^{n}-1}{b-1}-\frac{(n-1)b^{n+1}-nb^{n}+b}{(b-1)^{2}}, -\end{eqnarray*} - -\end_inset - -so -\begin_inset Formula -\begin{align*} - & (b-1)\sum_{j=0}^{n-1}b^{j}(n-j)+n+1\\ - & =n(b^{n}-1)-\frac{(n-1)b^{n+1}-nb^{n}+b}{b-1}+n+1\\ - & =\frac{n(b^{n}-1)(b-1)-(n-1)b^{n+1}+nb^{n}-b+n(b-1)+b-1}{b-1}\\ - & =\frac{nb^{n+1}-nb^{n}-nb+n-nb^{n+1}+b^{n+1}+nb^{n}-b+nb-n+b-1}{b-1}\\ - & =\frac{b^{n+1}-1}{b-1}=\frac{1-b^{n+1}}{b-1}=\sum_{j=0}^{n}b^{j}. -\end{align*} - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc21[M25] -\end_layout - -\end_inset - -Derive rule (d) from (8) and (17). -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - - -\begin_inset Formula -\begin{multline*} -\sum_{R(j)}a_{j}+\sum_{S(j)}a_{j}=\sum_{j}a_{j}[R(j)]+\sum_{j}a_{j}[S(j)]=\sum_{j}a_{j}([R(j)]+[S(j)])=\\ -\sum_{j}a_{j}([R(j)\lor S(j)]+[R(j)\land S(j)])=\sum_{R(j)\lor S(j)}a_{j}+\sum_{R(j)\land S(j)}a_{j}. -\end{multline*} - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc22[20] -\end_layout - -\end_inset - -State the appropriate analogs of Eqs. - (5), - (7), - (8), - and (11) for -\emph on -products -\emph default - instead of sums. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - - -\begin_inset Formula -\begin{align*} -\prod_{R(i)}a_{i} & =\prod_{R(j)}a_{j}=\prod_{R(p(j))}a_{p(j)}, & \prod_{R(i)}\prod_{S(j)}a_{ij} & =\prod_{S(j)}\prod_{R(i)}a_{ij},\\ -\prod_{R(i)}b_{i}c_{i} & =\prod_{R(i)}b_{i}+\prod_{R(i)}c_{i}, & \prod_{R(j)}a_{j}\prod_{S(j)}a_{j} & =\prod_{R(j)\lor S(j)}a_{j}\prod_{R(j)\land S(j)}a_{j}. -\end{align*} - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc23[10] -\end_layout - -\end_inset - -Explain why it is a good idea to define -\begin_inset Formula $\sum_{R(j)}a_{j}$ -\end_inset - - and -\begin_inset Formula $\prod_{R(j)}a_{j}$ -\end_inset - - as zero and one, - respectively, - when no integers satisfy -\begin_inset Formula $R(j)$ -\end_inset - -. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -Zero and one are the neutral terms for sum and product, - respectively, - so by defining it this way, - rule (d) (among others) applies independently of whether there are integers satisfying the properties or not. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc25[15] -\end_layout - -\end_inset - -Consider the following derivation; - is anything amiss? -\begin_inset Formula -\[ -\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{j=1}^{n}\frac{1}{a_{j}}\right)=\sum_{1\leq i\leq n}\sum_{1\leq j\leq n}\frac{a_{i}}{a_{j}}=\sum_{1\leq i\leq n}\sum_{1\leq i\leq n}\frac{a_{i}}{a_{i}}=\sum_{i=1}^{n}1=n. -\] - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -answer -\end_layout - -\end_inset - -First, - the change of variable on the second equality is invalid since it changes to a bound variable, - not to a free variable. - Second, - it then converts the double summation into a single summation, - which is invalid too. -\end_layout - -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -rexerc29[M30] -\end_layout - -\end_inset - - -\end_layout - -\begin_layout Enumerate -Express -\begin_inset Formula $\sum_{i=0}^{n}\sum_{j=0}^{i}\sum_{k=0}^{j}a_{i}a_{j}a_{k}$ -\end_inset - - in terms of the multiple-sum notation explained at the end of the section. -\end_layout - -\begin_layout Enumerate -Express the same sum in terms of -\begin_inset Formula $\sum_{i=0}^{n}a_{i}$ -\end_inset - -, - -\begin_inset Formula $\sum_{i=0}^{n}a_{i}^{2}$ -\end_inset - -, - and -\begin_inset Formula $\sum_{i=0}^{n}a_{i}^{3}$ -\end_inset - - [see Eq. - (13)]. -\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 -\begin_inset Formula -\[ -\sum_{0\leq k\leq j\leq i\leq n}a_{i}a_{j}a_{k}. -\] - -\end_inset - - -\end_layout - -\begin_deeper -\begin_layout Standard -\noindent -We have -\begin_inset Formula -\begin{align*} -S & :=\sum_{i=0}^{n}\sum_{j=0}^{i}\sum_{k=0}^{j}a_{i}a_{j}a_{k}=\sum_{i=0}^{n}\sum_{j=0}^{i}\sum_{k=j}^{i}a_{i}a_{j}a_{k}\\ - & =\sum_{i=0}^{n}\sum_{j=i}^{n}\sum_{k=0}^{i}a_{i}a_{j}a_{k}=\sum_{i=0}^{n}\sum_{j=0}^{i}\sum_{k=i}^{n}a_{i}a_{j}a_{k}\\ - & =\sum_{i=0}^{n}\sum_{j=i}^{n}\sum_{k=i}^{j}a_{i}a_{j}a_{k}=\sum_{i=0}^{n}\sum_{j=i}^{n}\sum_{k=j}^{n}a_{i}a_{j}a_{k}. -\end{align*} - -\end_inset - -Thus, -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -bgroup -\backslash -small -\end_layout - -\end_inset - - -\begin_inset Formula -\begin{eqnarray*} -6S & = & \sum_{i=0}^{n}\sum_{j=0}^{i}\left(\sum_{k=0}^{j}a_{i}a_{j}a_{k}+\sum_{k=j}^{i}a_{i}a_{j}a_{k}+\sum_{k=i}^{n}a_{i}a_{j}a_{k}\right)\\ - & & +\sum_{i=0}^{n}\sum_{j=i}^{n}\left(\sum_{k=0}^{i}a_{i}a_{j}a_{k}+\sum_{k=i}^{j}a_{i}a_{j}a_{k}+\sum_{i=j}^{n}a_{i}a_{j}a_{k}\right)\\ - & = & \sum_{i=0}^{n}\left(\sum_{j=0}^{i}\left(\sum_{k=0}^{n}a_{i}a_{j}a_{k}+a_{i}a_{j}^{2}+a_{i}^{2}a_{j}\right)+\sum_{j=i}^{n}\left(\sum_{k=0}^{n}a_{i}a_{j}a_{k}+a_{i}a_{j}^{2}+a_{j}^{2}a_{i}\right)\right)\\ - & = & \sum_{i=0}^{n}\left(\sum_{j=0}^{n}\left(\sum_{k=0}^{n}a_{i}a_{j}a_{k}+a_{i}a_{j}^{2}+a_{i}^{2}a_{j}\right)+\sum_{k=0}^{n}a_{i}^{2}a_{k}+a_{i}^{3}+a_{i}^{3}\right)\\ - & = & \sum_{i=0}^{n}\sum_{j=0}^{n}\sum_{k=0}^{n}a_{i}a_{j}a_{k}+\sum_{i=0}^{n}\sum_{j=0}^{n}a_{i}a_{j}^{2}+\sum_{i=0}^{n}\sum_{j=0}^{n}a_{i}^{2}a_{j}+\sum_{i=0}^{n}\sum_{k=0}^{n}a_{i}^{2}a_{k}+2\sum_{i=0}^{n}a_{i}^{3}\\ - & = & \left(\sum_{i=0}^{n}a_{i}\right)^{3}+\left(\sum_{i=0}^{n}a_{i}\right)\left(\sum_{i=0}^{n}a_{i}^{2}\right)+\left(\sum_{i=0}^{n}a_{i}^{2}\right)\left(\sum_{i=0}^{n}a_{i}\right)\\ - & & +\left(\sum_{i=0}^{n}a_{i}^{2}\right)\left(\sum_{i=0}^{n}a_{i}\right)+2\sum_{i=0}^{n}a_{i}^{3}\\ - & = & \left(\sum_{i=0}^{n}a_{i}\right)^{3}+3\left(\sum_{i=0}^{n}a_{i}\right)\left(\sum_{i=0}^{n}a_{i}^{2}\right)+2\left(\sum_{i=0}^{n}a_{i}^{3}\right). -\end{eqnarray*} - -\end_inset - - -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -egroup -\end_layout - -\end_inset - -This means -\begin_inset Formula -\[ -S=\frac{1}{6}\left(\sum_{i=0}^{n}a_{i}\right)^{3}+\frac{1}{2}\left(\sum_{i=0}^{n}a_{i}\right)\left(\sum_{i=0}^{n}a_{i}^{2}\right)+\frac{1}{3}\left(\sum_{i=0}^{n}a_{i}^{3}\right). -\] - -\end_inset - - -\end_layout - -\end_deeper -\begin_layout Standard -\begin_inset ERT -status open - -\begin_layout Plain Layout - - -\backslash -exerc30[M23] -\end_layout - -\end_inset - -(J. - Binet, - 1812.) Without using induction, - prove the identity -\begin_inset Formula -\[ -\left(\sum_{j=1}^{n}a_{j}x_{j}\right)\left(\sum_{j=1}^{n}b_{j}y_{j}\right)=\left(\sum_{j=1}^{n}a_{j}y_{j}\right)\left(\sum_{j=1}^{n}b_{j}x_{j}\right)+\sum_{1\leq j1$ -\end_inset - -, - assuming this holds for -\begin_inset Formula $n-1$ -\end_inset - -, - subtracting from each row the previous one multiplied by -\begin_inset Formula $x_{1}$ -\end_inset - -, - expanding by cofactors, - factoring out, - and applying the induction hypothesis, - we have -\begin_inset Formula -\begin{eqnarray*} -\left|\begin{array}{cccc} -x_{1} & x_{2} & \cdots & x_{n}\\ -x_{1}^{2} & x_{2}^{2} & \cdots & x_{n}^{2}\\ -\vdots & \vdots & & \vdots\\ -x_{1}^{n} & x_{2}^{n} & \cdots & x_{n}^{n} -\end{array}\right| & = & \left|\begin{array}{cccc} -x_{1} & x_{2} & \cdots & x_{n}\\ -0 & x_{2}(x_{2}-x_{1}) & \cdots & x_{n}(x_{n}-x_{1})\\ -\vdots & \vdots & & \vdots\\ -0 & x_{2}^{n-1}(x_{2}-x_{1}) & \cdots & x_{n}^{n-1}(x_{n}-x_{1}) -\end{array}\right|\\ - & = & x_{1}\left|\begin{array}{ccc} -x_{2}(x_{2}-x_{1}) & \cdots & x_{n}(x_{n}-x_{1})\\ -\vdots & & \vdots\\ -x_{2}^{n-1}(x_{2}-x_{1}) & \cdots & x_{n}^{n-1}(x_{n}-x_{1}) -\end{array}\right|\\ - & = & x_{1}(x_{2}-x_{1})\cdots(x_{n}-x_{1})\left|\begin{array}{ccc} -x_{2} & \cdots & x_{n}\\ -\vdots & & \vdots\\ -x_{2}^{n-1} & \cdots & x_{n}^{n-1} -\end{array}\right|\\ - & = & x_{1}\prod_{2\leq j\leq n}(x_{j}-x_{1})\prod_{2\leq j\leq n}x_{j}\prod_{2\leq i1$ -\end_inset - - and that the hypothesis holds for -\begin_inset Formula $n-1$ -\end_inset - -. - Multiplying each row -\begin_inset Formula $i$ -\end_inset - - by -\begin_inset Formula $\frac{x_{1}+y_{1}}{x_{i}+y_{1}}$ -\end_inset - - times the first row, - we get -\begin_inset Formula -\begin{multline*} -\left|\begin{array}{cccc} -\frac{1}{x_{1}+y_{1}} & \frac{1}{x_{1}+y_{2}} & \cdots & \frac{1}{x_{1}+y_{n}}\\ -\frac{1}{x_{2}+y_{1}} & \frac{1}{x_{2}+y_{2}} & \cdots & \frac{1}{x_{2}+y_{n}}\\ -\vdots & \vdots & & \vdots\\ -\frac{1}{x_{n}+y_{1}} & \frac{1}{x_{n}+y_{2}} & \cdots & \frac{1}{x_{n}+y_{n}} -\end{array}\right|=\\ -=\left|\begin{array}{cccc} -\frac{1}{x_{1}+y_{1}} & \frac{1}{x_{1}+y_{2}} & \cdots & \frac{1}{x_{1}+y_{n}}\\ -0 & \frac{1}{x_{2}+y_{2}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{2})(x_{2}+y_{1})} & \cdots & \frac{1}{x_{2}+y_{n}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{n})(x_{2}+y_{1})}\\ -\vdots & \vdots & & \vdots\\ -0 & \frac{1}{x_{n}+y_{2}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{2})(x_{n}+y_{1})} & \cdots & \frac{1}{x_{n}+y_{n}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{n})(x_{n}+y_{1})} -\end{array}\right|=\\ -=\frac{1}{x_{1}+y_{1}}\left|\begin{array}{ccc} -\frac{1}{x_{2}+y_{2}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{2})(x_{2}+y_{1})} & \cdots & \frac{1}{x_{2}+y_{n}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{n})(x_{2}+y_{1})}\\ -\vdots & & \vdots\\ -\frac{1}{x_{n}+y_{2}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{2})(x_{n}+y_{1})} & \cdots & \frac{1}{x_{n}+y_{n}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{n})(x_{n}+y_{1})} -\end{array}\right|. -\end{multline*} - -\end_inset - -Now, -\begin_inset Formula -\begin{multline*} -\frac{1}{x_{i}+y_{j}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{j})(x_{i}+y_{1})}=\frac{(x_{1}+y_{j})(x_{i}+y_{1})-(x_{1}+y_{1})(x_{i}+y_{j})}{(x_{i}+y_{j})(x_{1}+y_{j})(x_{i}+y_{1})}=\\ -=\frac{1}{x_{i}+y_{j}}\frac{x_{1}y_{1}+x_{i}y_{j}-x_{1}y_{j}-x_{i}y_{1}}{(x_{1}+y_{j})(x_{i}+y_{1})}=\frac{1}{x_{i}+y_{j}}\frac{(x_{i}-x_{1})(y_{j}-y_{1})}{(x_{1}+y_{j})(x_{i}+y_{1})}, -\end{multline*} - -\end_inset - -so -\begin_inset Formula -\begin{multline*} -\left|\begin{array}{cccc} -\frac{1}{x_{1}+y_{1}} & \frac{1}{x_{1}+y_{2}} & \cdots & \frac{1}{x_{1}+y_{n}}\\ -\frac{1}{x_{2}+y_{1}} & \frac{1}{x_{2}+y_{2}} & \cdots & \frac{1}{x_{2}+y_{n}}\\ -\vdots & \vdots & & \vdots\\ -\frac{1}{x_{n}+y_{1}} & \frac{1}{x_{n}+y_{2}} & \cdots & \frac{1}{x_{n}+y_{n}} -\end{array}\right|=\\ -=\frac{1}{x_{1}+y_{1}}\left|\begin{array}{ccc} -\frac{1}{x_{2}+y_{2}}\frac{(x_{2}-x_{1})(y_{2}-y_{1})}{(x_{1}+y_{2})(x_{2}+y_{1})} & \cdots & \frac{1}{x_{2}+y_{n}}\frac{(x_{2}-x_{1})(y_{n}-y_{1})}{(x_{1}+y_{n})(x_{2}+y_{1})}\\ -\vdots & & \vdots\\ -\frac{1}{x_{n}+y_{2}}\frac{(x_{n}-x_{1})(y_{2}-y_{1})}{(x_{1}+y_{2})(x_{n}+y_{1})} & \cdots & \frac{1}{x_{n}+y_{n}}\frac{(x_{n}-x_{1})(y_{n}-y_{1})}{(x_{1}+y_{n})(x_{n}+y_{1})} -\end{array}\right|=\\ -=\frac{1}{x_{1}+y_{1}}\prod_{i=2}^{n}\frac{x_{i}-x_{1}}{x_{i}+y_{1}}\prod_{j=2}^{n}\frac{y_{j}-y_{1}}{x_{1}+y_{j}}\left|\begin{array}{ccc} -\frac{1}{x_{2}+y_{2}} & \cdots & \frac{1}{x_{2}+y_{n}}\\ -\vdots & & \vdots\\ -\frac{1}{x_{n}+y_{2}} & \cdots & \frac{1}{x_{n}+y_{n}} -\end{array}\right|=\\ -=\frac{\prod_{j=2}^{n}(x_{j}-x_{1})(y_{j}-y_{1})}{(x_{1}+y_{1})\prod_{i=2}^{n}(x_{j}+y_{1})\prod_{j=2}^{n}(x_{1}+y_{j})}\frac{\prod_{2\leq i