diff --git a/books/bookheader.tex b/books/bookheader.tex index 865b863..026549f 100644 --- a/books/bookheader.tex +++ b/books/bookheader.tex @@ -1,4 +1,6 @@ \usepackage{hyperref} +\hypersetup{colorlinks=true,linkcolor=blue,pdfborderstyle={/S/U/W 1}, +citecolor=red} \usepackage{amssymb} \usepackage{axiom} \usepackage{makeidx} diff --git a/books/bookvol10.1.pamphlet b/books/bookvol10.1.pamphlet index e63cdd5..b86e43c 100644 --- a/books/bookvol10.1.pamphlet +++ b/books/bookvol10.1.pamphlet @@ -256,8 +256,8 @@ an exception. If it contains a negative part, the implementation will crop it to only its non-negative part to allow that computations such as $\sqrt{0}$ ca be carried out in exact real arithmetic. +\chapter{Integration \cite{Bro98b}} -\chapter{Integration} An {\sl elementary function} \index{elementary function} of a variable $x$ is a function that can @@ -267,7 +267,8 @@ numbers or functions. Since $\sqrt{-1}$ is elementary, the trigonometric functions and their inverses are also elementary (when they are rewritten using complex exponentials and logarithms) as well as all the ``usual'' functions of calculus. For example, -\begin{equation} + +\begin{equation}\label{Int1} \sin(x+\tan(x^3-\sqrt{x^3-x+1})) \end{equation} is elementary when rewritten as @@ -300,7 +301,7 @@ correctness of the algorithms presented here can be found in several of the references, and are generally too long and too detailed to be described in this tutorial. -{\bf Notations}: we write $x$ for the variable of integration, and ' +{\bf Notations}: we write $x$ for the variable of integration, and $\prime$ for the derivation $d/dx$. $\mathbb{Z}$,$\mathbb{Q}$,$\mathbb{R}$,and $\mathbb{C}$ denote respectively the integers, rational, real and complex numbers. All fields are commutative and, except when mentioned @@ -313,7 +314,7 @@ By a {\sl rational function}, we mean a quotient of polynomials in the integration variable $x$. This means that other functions can appear in the integrand, provided they do not involve $x$, hence that the coefficients of our polynomials in $x$ lie in an arbitrary field $K$ -satisfying: $\forall{a} \in K,\ a^{'}=0$. +satisfying: $\forall{a} \in K,\ a^{\prime}=0$. \subsection{The full partial-fraction algorithm} This method, which dates back to Newton, Leibniz, and Bernoulli, @@ -323,8 +324,8 @@ factorization of the denominator of the integrand over the real or complex numbers. We outline it because it provides the theoretical foundations for all the subsequent algorithms. Let $f \in \mathbb{R}(x)$ be our integrand, and write -$f=P+A/D$ where $P, A, D \in \mathbb{R}[x]$, $gcd(A,D)=1$, and -$deg(A) < deg(D)$. Let +$f=P+A/D$ where $P, A, D \in \mathbb{R}[x]$, gcd$(A,D)=1$, and +deg$(A) < $deg$(D)$. Let \[ D=c\prod_{i=1}^n(x-a_i)^{e_i}\prod_{j=1}^m(x^2+b_jx+c_j)^{f_j} \] @@ -344,7 +345,7 @@ $\mathbb{R}$. Hence, \] Computing $\int{P}$ poses no problem (it will for any other class of functions), and for the other terms we have -\begin{equation} +\begin{equation}\label{Int2} \int{\frac{A_{ik}}{(x-a_i)^k}}=\left\{ \begin{array}{lc} A_{ik}(x-a_i)^{1-k}/(1-k)&{\rm if\ } k > 1\\ @@ -374,8 +375,8 @@ and for $k > 1$, This last formula is then used recursively until $k=1$. An alternative is to factor $D$ linearly over $\mathbb{C}$: -$D=\prod_{i=1}^q(x-\alpha_i)^{e_i}$, and then use (2) on each term of -\begin{equation} +$D=\prod_{i=1}^q(x-\alpha_i)^{e_i}$, and then use \ref{Int2} on each term of +\begin{equation}\label{Int3} f=P+\sum_{i=1}^q\sum_{j=1}^{e_i}\frac{A_{ij}}{(x-\alpha_i)^j} \end{equation} Note that this alternative is applicable to coefficients in any field @@ -390,21 +391,21 @@ f=\frac{A_{ie_i}}{(x-\alpha_i)^{e_i}} +\frac{A_{i1}}{(x-\alpha_i)} +\cdots \] -where the $A_{ij}$'s are the same as those in (3). Thus, this approach +where the $A_{ij}$'s are the same as those in \ref{Int3}. Thus, this approach can be seen as expanding the integrand into series around all the poles (including $\infty$), then integrating the series termwise, and then interpolating for the answer, by summing all the polar terms, -obtaining the integral of (3). In addition, this alternative shows +obtaining the integral of \ref{Int3}. In addition, this alternative shows that any rational function $f \in K(x)$ has an elementary integral of the form -\begin{equation} +\begin{equation}\label{Int4} \int{f}=v+c_1\log(u_1)+\cdots+c_m\log(u_m) \end{equation} where $v,u_1,\ldots,u_m \in \overline{K}(x)$ are the rational functions, and $c_1,\ldots,c_m \in \overline{K}$ are constants. The original Risch algorithm is essentially a generalization of this approach that searches for integrals of arbitrary elementary functions -in a form similar to (4). +in a form similar to \ref{Int4}. \subsection{The Hermite reduction} The major computational inconvenience of the full partial fraction @@ -422,59 +423,64 @@ calculations being done in $K(x)$, and express the integral \end{itemize} The first rational algorithms for integration date back to the -$19^{{\rm th}}$ century, when both Hermite\cite{Her1872} and -Ostrogradsky\cite{Ost1845} invented methods for computing the $v$ of (4) +$19^{{\rm th}}$ century, when both Hermite \cite{Her1872} and +Ostrogradsky \cite{Ost1845} invented methods for +computing the $v$ of \ref{Int4} entirely within $K(x)$. We describe here only Hermite's method, since it is the one that has been generalized to arbitrary elementary functions. The basic idea is that if an irreducible $p \in K[x]$ appears with multiplicity $k > 1$ in the factorization of the -denominator of the integrand, then (2) implies that it appears with +denominator of the integrand, then \ref{Int2} implies that it appears with multiplicity $k-1$ in the denominator of the integral. Furthermore, it is possible to compute the product of all such irreducibles for each $k$ without factoring the denominator into irreducibles by computing its {\sl squarefree factorization}, {\sl i.e} a factorization $D=D_1D_2^2\cdots D_m^m$, where each $D_i$ is squarefree and -$gcd(D_i,D_j)=1$ for $i \ne j$. A straightforward way to compute it is -as follows: let $R=gcd(D,D^{'})$, then $R=D_2D_2^3\cdots D_m^{m-1}$, so -$D/R=D_1D_2\cdots D_m$ and $gcd(R,D/R)=D_2\cdots D_m$, which implies +gcd$(D_i,D_j)=1$ for $i \ne j$. A straightforward way to compute it is +as follows: let $R={\rm gcd}(D,D^{\prime})$, +then $R=D_2D_2^3\cdots D_m^{m-1}$, so +$D/R=D_1D_2\cdots D_m$ and gcd$(R,D/R)=D_2\cdots D_m$, which implies finally that \[ -D_1=\frac{D/R}{gcd(R,D/R)} +D_1=\frac{D/R}{{\rm gcd}(R,D/R)} \] Computing recursively a squarefree factorization of $R$ completes the one for $D$. Note that \cite{Yu76} presents a more efficient method for this decomposition. Let now $f \in K(x)$ be our integrand, and write -$f=P+A/D$ where $P,A,D \in K[x]$, $gcd(A,D)=1$, and $deg(A) 1$, and that solution always has a denominator coprime with V. Furthermore, the denominator of each -$w_i^{'}$ must be squarefree, implying that the denominator of $h$ is +$w_i^{\prime}$ must be squarefree, implying that the denominator of $h$ is a factor of $FUV^{m-1}$ where $F \in K[x]$ is squarefree and coprime with $UV$. He also described an algorithm for computing an integral basis, a necessary preprocessing for his Hermite reduction. The main problem with that approach is that computing the integral basis, whether by the method of \cite{Tr84} or the local alternative \cite{vH94}, can be in general more expansive than the rest of the reduction -process. We describe here the lazy Hermite reduction \cite{Bro98}, which +process. We describe here the lazy Hermite reduction \cite{REF-Bro98}, which avoids the precomputation of an integral basis. It is based on the -observation that if $m > 1$ and (8) does not have a solution allowing +observation that if $m > 1$ and \ref{Int8} does not have a solution allowing us to perform the reduction, then either \begin{itemize} \item the $S_i$'s are linearly dependent over $K(x)$, or -\item (8) has a unique solution in $K(x)$ whose denominator has a +\item \ref{Int8} has a unique solution in $K(x)$ whose denominator has a nontrivial common factor with $V$, or \item the denominator of some $w_i$ is not squarefree \end{itemize} @@ -735,8 +745,8 @@ also made up of integral elements, so that that $K[x]$-module generated by the new basis strictly contains the one generated by $w$: \noindent -{\bf Theorem 1 (\cite{Bro98})} {\sl Suppose that $m \ge 2$ and that -$\{S_1,\ldots,S_n\}$ as given by (9) are linearly dependent over $K(x)$, +{\bf Theorem 1 (\cite{REF-Bro98})} {\sl Suppose that $m \ge 2$ and that +$\{S_1,\ldots,S_n\}$ as given by \ref{Int9} are linearly dependent over $K(x)$, and let $T_1,\ldots,T_n \in K[x]$ be not all 0 and such that $\sum_{i=1}^n T_iS_i=0$. Then, \[ @@ -746,8 +756,8 @@ Furthermore, if $\gcd(T_1,\ldots,T_n)=1$ then $w_0 \notin K[x]w_1+\cdots+K[x]w_n$.} \noindent -{\bf Theorem 2 (\cite{Bro98})} {\sl Suppose that $m \ge 2$ and that -$\{S_1,\ldots,S_n\}$ as given by (9) are linearly independent over +{\bf Theorem 2 (\cite{REF-Bro98})} {\sl Suppose that $m \ge 2$ and that +$\{S_1,\ldots,S_n\}$ as given by \ref{Int9} are linearly independent over $K(x)$, and let $Q,T_1,\ldots,T_n \in K[x]$ be such that \[ \sum_{i=1}^n A_iw_i = \frac{1}{Q}\sum_{i=1}^n T_iS_i @@ -761,15 +771,15 @@ Furthermore, if $\gcd(Q,T_1,\ldots,T_n)=1$ and $\deg(\gcd(V,Q)) \ge 1$, then $w_0 \notin K[x]w_1+\cdots+K[x]w_n$.} -{\bf Theorem 3 (\cite{Bro98})} {\sl Suppose that the denominator $F$ of +{\bf Theorem 3 (\cite{REF-Bro98})} {\sl Suppose that the denominator $F$ of some $w_i$ is not squarefree, and let $F=F_1F_2^2\cdots F_k^k$ be its squarefree factorization. Then,} \[ -w_0=F_1\cdots F_kw_i^{'} \in {\bf O}_{K[x]} \backslash +w_0=F_1\cdots F_kw_i^{\prime} \in {\bf O}_{K[x]} \backslash (K[x]w_1+\cdots+K[x]w_n). \] -The lazy Hermite reduction proceeds by solving the system (8) in +The lazy Hermite reduction proceeds by solving the system \ref{Int8} in $K(x)$. Either the reduction will succeed, or one of the above theorems produces an element $w_0 \in {\bf O}_{K[x]} \backslash (K[x]w_1+\cdots+K[x]w_n).$ Let then @@ -845,7 +855,7 @@ x& & & \\ \right) \] so the new basis is $\overline{w}=(1,y,y^2,y^3/x)$, and the -denominator of $f$ with respect to $\overline{w}$ is 1, which is +denominator of $f$ with respect to $\overline{w}$ is $x$, which is squarefree. \subsection{Simple radical extensions} @@ -863,31 +873,31 @@ z^n=\left(y\frac{D}{F}\right)^n=\frac{y^nD^n}{F^n}=\frac{AD^{n-1}}{F} Since $r_i < n$ for each $i$, the squarefree factorization of $H$ is of the form $H=H_1H_2^2\cdots H_m^m$ with $m 0$, we have +{\sl special} $S \in K[t]$ with ${\rm deg}_t(S) > 0$, we have \[ -R=\frac{1}{deg_t(S)}\frac{r_{d-1}}{c_d}\frac{S'}{S}+\overline{R} +R=\frac{1}{{\rm deg}_t(S)}\frac{r_{d-1}}{c_d}\frac{S'}{S}+\overline{R} \] where $\overline{R} \in K[t]$ is such that $\overline{R}=0$ or -$deg_t(\overline{R}) < e-1$. Furthermore, it can be proven \cite{Bro97} +${\rm deg}_t(\overline{R}) < e-1$. Furthermore, it can be proven \cite{Bro97} that if $R+A/D$ has an elementary integral over $K(t)$, then $r_{d-1}/{c_d}$ is a constant, which implies that \[ -\int{R}=\frac{1}{deg_t(S)}\frac{r_{d-1}}{c_d}\log(S) +\int{R}=\frac{1}{{\rm deg}_t(S)}\frac{r_{d-1}}{c_d}\log(S) +\int\left(\overline{R}+\frac{A}{D}\right) \] so we are left with an integrand whose polynomial part has degree at -most $deg_t(t')-2$. In this case $t=\tan(b)$ for $b \in K$, then +most ${\rm deg}_t(t')-2$. In this case $t=\tan(b)$ for $b \in K$, then $t'=b't^2+b'$, so $\overline{R} \in K$. {\bf Example 10} {\sl @@ -1437,7 +1451,7 @@ The integrand is f=1+xt+t^2 \in K(t)\quad{\rm where\ }K=\mathbb{Q}(x) {\rm\ and\ }t'=t^2+1 \] -Using (15), we get $\overline{P}=f-t'=f-(t^2+1)=xt$ so +Using \ref{Int15}, we get $\overline{P}=f-t'=f-(t^2+1)=xt$ so \[ \int(1+x\tan(x)+\tan(x)^2)~dx=\tan(x)+\int{x\tan(x)}~dx \] @@ -1449,10 +1463,10 @@ Similarly to the Hermite reduction, the Rothstein-Trager and Lazard-Rioboo-Trager algorithms are easy to generalize to the transcendental case $E=K(t)$ for arbitrary monomials $t$: let $f\in K(t)$ be our integrand and write $f=P+A/D+B/S$ where -$P,A,D,B,S \in K[t]$, $deg(A) < deg(D)$, $S$ is special and, following +$P,A,D,B,S \in K[t]$, ${\rm deg}(A) < {\rm deg}(D)$, $S$ is special and, following the Hermite reduction, $D$ is normal. Let then $z$ be a new indeterminate, $\kappa : K[z] \rightarrow K[z]$ be give by -$\kappa(\sum_i a_iz^i)=\sum_i a_i^{'}z^i$, +$\kappa(\sum_i a_iz^i)=\sum_i a_i^{\prime}z^i$, \[ R={\rm resultant_t}(D,A-zD') \in K[z] \] @@ -1471,14 +1485,14 @@ case. It can then be proven \cite{Bro97} that \item $f-g'$ is always ``simpler'' than $f$ \item the splitting field of $Q_1\cdots Q_k$ over $K$ is the minimal algebraic extension of $K$ needed in order to express $\int f$ in the -form (4) +form \ref{Int4} \item if $f$ has an elementary integral over $K(t)$, then $R | \kappa(R)$ in $K[z]$ and the denominator of $f-q'$ is special \end{itemize} Thus, while in the pure rational function case the remaining integrand is a polynomial, in this case the remaining integrand has a special denominator. In that case we have additionally that if its integral is -elementary, then (13) has a solution such that $v\in K(t)$ has a +elementary, then \ref{Int13} has a solution such that $v\in K(t)$ has a special denominator, and each $u_i \in K(c_1,\ldots,c_k)[t]$ is special. @@ -1525,11 +1539,11 @@ Computing $f-g'$ we find =\frac{1}{2}\log\left(\frac{\log(x)+x}{\log(x)-x}\right) +\int{\frac{dx}{\log(x)}} \] -and since $deg_z(Q_1) < deg_z(R)$, it follows that the remaining +and since ${\rm deg}_z(Q_1) < {\rm deg}_z(R)$, it follows that the remaining integral is not an elementary function (it is in fact the logarithmic integral $Li(x)$).} -In the most general case, when $E=K(t)(j)$ is algebraic over $K(t)$ and +In the most general case, when $E=K(t)(y)$ is algebraic over $K(t)$ and $y$ is integral over $K[t]$, the criterion part of the above result remains valid: let $w=(w_1,\ldots,w_n)$ be an integral basis for $E$ over $K(t)$ and write the integrand $f \in E$ as @@ -1538,22 +1552,22 @@ and, following the Hermite reduction, $D$ is normal. Write $\sum_{i=1}^n A_iw_i=G/H$, where $G \in K[t,y]$ and $H \in K[t]$, let $F \in K[t,y]$ be the (monic) minimum polynomial for $y$ over $K(t)$, $z$ be a new indeterminante and compute -\begin{equation} +\begin{equation}\label{Int16} R(z)={\rm resultant_t}({\rm pp_z}({\rm resultant_y}(G-tHD',F)),D) \in K[t] \end{equation} -It can then be proven \cite{Bro90} that if $f$ has an elementary integral +It can then be proven \cite{Bro90c} that if $f$ has an elementary integral over $E$, then $R|\kappa(R)$ in $K[z]$. {\bf Example 12} {\sl Consider -\begin{equation} +\begin{equation}\label{Int17} \int{\frac{\log(1+e^x)^{(1/3)}}{1+\log(1+e^x)}}~dx \end{equation} The integrand is \[ f=\frac{y}{t+1} \in E = K(t)[y]/(y^3-t) \] -where $K=\mathbb{Q}(x)(t_1)$,$t_1=e^x$ and $t=\log(1+t_1)$. Its +where $K=\mathbb{Q}(x)(t_1)$, $t_1=e^x$ and $t=\log(1+t_1)$. Its denominator with respect to the integral basis $w=(1,y,y^2)$ is $D=t+1$, which is normal, and the resultant is \[ @@ -1564,7 +1578,7 @@ We have \[ \kappa(R)=-\frac{3t_1^3}{(1+t_1)^4}z^3 \] -which is coprime with $R$ in $K[z]$, implying that the integral (17) +which is coprime with $R$ in $K[z]$, implying that the integral \ref{Int17} is not an elementary function. } @@ -1572,10 +1586,10 @@ is not an elementary function. Suppose now that $t=\log(b)$ for some $b \in K^{*}$, and that $E=K(t)$. Then, every special polynomial must be in $K$, so, following the residue criterion, we must look for a solution $v \in K[t]$, -$u_1,\ldots,u_k \in K(c_1,\ldots,c_n)^{*}$ of (13). Furthermore, the +$u_1,\ldots,u_k \in K(c_1,\ldots,c_n)^{*}$ of \ref{Int13}. Furthermore, the integrand $f$ is also in $K[t]$, so write $f=\sum_{i=0}^d f_it^i$ where $f_0,\ldots,f_d \in K$ and $f_d \ne 0$. We -must have $deg{}_t(v) \le d_1$, so writing $v=\sum_{i=0}^{d+1} v_it^i$, +must have ${\rm deg}{}_t(v) \le d+1$, so writing $v=\sum_{i=0}^{d+1} v_it^i$, we get \[ \int f_dt^d+\cdots+f_1t+f_0=v_{d+1}t^{d+1}+\cdots+v_1t+v_0 @@ -1585,22 +1599,22 @@ If $d=0$, then the above is simply an integration problem for $f_0 \in K$, which can be solved recursively. Otherwise, differentiating both sides and equating the coefficients of $t^d$, we get ${v_{d+1}}'=0$ and -\begin{equation} +\begin{equation}\label{Int18} f_d=v_d'+(d+1)v_{d+1}\frac{b'}{b} \end{equation} Since $f_d \in K$, we can recursively apply the integration algorithm -to $f_d$, either proving that (18) has no solution, in which case $f$ +to $f_d$, either proving that \ref{Int18} has no solution, in which case $f$ has no elementary integral, or obtaining the constant $v_{d+1}$, and $v_d$ up to an additive constant (in fact, we apply recursively a specialized version of the integration algorithm to equations of the -form (18), see \cite{Bro97} for details). Write then +form \ref{Int18}, see \cite{Bro97} for details). Write then $v_d=\overline{v_d}+c_d$ where $\overline{v_d} \in K$ is known and $c_d \in {\rm Const}(K)$ is undetermined. Equating the coefficients of $t^{d-1}$ yields \[ f_{d-1}-d\overline{v_d}\frac{b'}{b}={v_{d-1}}'+dc_d\frac{b'}{b} \] -which is an equation of the form (18), so we again recursively compute +which is an equation of the form \ref{Int18}, so we again recursively compute $c_d$ and $v_{d-1}$ up to an additive constant. We repeat this process until either one of the recursive integrations fails, in which case $f$ has no elementary integral, or we reduce our integrand to an element @@ -1617,7 +1631,7 @@ $at^m$ for $a \in K^{*}$ and $m \in \mathbb{N}$. Since \frac{(at^m)'}{at^m}=\frac{a'}{a}+m\frac{t'}{t}=\frac{a'}{a}+mb' \] we must then look for a solution $v\in K[t,t^{-1}]$, -$u_1,\ldots,u_k \in K(c_1,\ldots,c_n)^{*}$ of (13). Furthermore, the +$u_1,\ldots,u_k \in K(c_1,\ldots,c_n)^{*}$ of \ref{Int13}. Furthermore, the integrand $f$ is also in $K[t,t^{-1}]$, so write $f=\sum_{i=e}^d f_it^i$ where $f_e,\ldots,f_d \in K$ and $e,d\in \mathbb{Z}$. Since $(at^{m})'=(a'+mb')t^m$ for any @@ -1629,18 +1643,18 @@ integer $M$, hence Differentiating both sides and equating the coefficients of each power to $t^d$, we get \[ -f_0=(v_0+Mb)'+\sum_{i=1}^k c_i\frac{u_i^{'}}{u_i} +f_0=(v_0+Mb)'+\sum_{i=1}^k c_i\frac{u_i^{\prime}}{u_i} \] which is simply an integration problem for $f_0 \in K$, and \[ -f_i=v_i^{'}+ib'v_i\quad{\rm for\ }e \le i \le d, i \ne 0 +f_i=v_i^{\prime}+ib'v_i\quad{\rm for\ }e \le i \le d, i \ne 0 \] -The above problem is called a {\sl Risch differential equation overK}. +The above problem is called a {\sl Risch differential equation over K}. Although solving it seems more complicated than solving $g'=f$, it is actually simpler than an integration problem because we look for the solutions $v_i$ in $K$ only rather than in an extension of -$K$. Bronstein \cite{Bro90,Bro91,Bro97} and Risch +$K$. Bronstein \cite{Bro90c,Bro91a,Bro97} and Risch \cite{Ris68,Ris69a,Ris69b} describe algorithms for solving this type of equation when $K$ is an elementary extension of the rational function field. @@ -1660,12 +1674,12 @@ $m_1,\ldots,m_k \in \mathbb{N}$, constants $c_1,\ldots,c_k \in \overline{K}$ and $u_1,\ldots,u_k \in K(c_1,\ldots,c_k)^{*}$ such that \[ -f=v'+2b't\sum_{i=1}^k c_im_i + \sum_{i=1}^k c_i\frac{u_i^{'}}{u_i} +f=v'+2b't\sum_{i=1}^k c_im_i + \sum_{i=1}^k c_i\frac{u_i^{\prime}}{u_i} \] Furthermore, the integrand $f \in K(t)$ following the residue criterion must be of the form $f=A/(t^2+1)^M$ where $A \in K[t]$ and $M \ge 0$. If $M > 0$, it can be shown that $m=M$ and that -\begin{equation} +\begin{equation}\label{Int19} \left( \begin{array}{c} c'\\ @@ -1711,7 +1725,7 @@ The integrand is f=\frac{2t/x}{t^2+1} \in K(t)\quad{\rm where\ }K=\mathbb{Q}(x) {\rm\ and\ }t=\tan\left(\frac{x}{2}\right) \] -Its denominator is $D=t^2+1$, which is special, and the system (19) +Its denominator is $D=t^2+1$, which is special, and the system \ref{Int19} becomes \[ \left( @@ -1747,10 +1761,10 @@ The transcendental logarithmic case method also generalizes to the case when $E=K(t)(y)$ is algebraic over $K(t)$, $t=log(b)$ for $b \in K^{*}$ and $y$ is integral over $K[t]$: following the residue criterion, we can assume that $R | \kappa(R)$ where $R$ is given by -(16), hence that all its roots in $\overline{K}$ are constants. The +\ref{Int16}, hence that all its roots in $\overline{K}$ are constants. The polynomial part of the integrand is replace by a family of at most $[E : K(t)]$ Puiseux expansions at infinity, each of the form -\begin{equation} +\begin{equation}\label{Int20} a_{-m}\theta^{-m}+\cdots+a_{-1}\theta^{-1}+\sum_{i \ge 0} a_i\theta^i \end{equation} where $\theta^r=t^{-1}$ for some positive integer $r$. Applying the @@ -1784,17 +1798,17 @@ nonzero integer $N$, then the integral is not elementary, otherwise, let $n_1,\ldots,n_k$ be nonzero integers such that $n_j\delta_j$ is principal for each $j$, and \[ -h=f-\frac{1}{m}\sum_{j=1}^k\frac{q_j}{n_j}\frac{u_j^{'}}{u_j} +h=f-\frac{1}{m}\sum_{j=1}^k\frac{q_j}{n_j}\frac{u_j^{\prime}}{u_j} \] where $f$ is the integrand and $u_j \in E(\alpha_1,\ldots,\alpha_s,\rho_1,\ldots,\rho_q)^{*}$ is such that $n_j\delta_j=(u_j)$. If the integral of $h$ is elementary, then -(13) must have a solution with $v \in {\bf O}_{K[x]}$ and +\ref{Int13} must have a solution with $v \in {\bf O}_{K[x]}$ and $u_1,\ldots,u_k \in \overline{K}$ so we must solve -\begin{equation} +\begin{equation}\label{Int21} h=\frac{\sum_{i=1}^n A_iw_i}{D} -=\sum_{i=1}^n v_i^{'}w_i+\sum_{i=1}^n v_iw_i^{'} -+\sum_{i=1}^k c_i\frac{u_i^{'}}{u_i} +=\sum_{i=1}^n v_i^{\prime}w_i+\sum_{i=1}^n v_iw_i^{\prime} ++\sum_{i=1}^k c_i\frac{u_i^{\prime}}{u_i} \end{equation} for $v_1,\ldots,v_n \in K[t]$, constants $c_1,\ldots,c_n \in \overline{K}$ and @@ -1802,19 +1816,20 @@ $u_1,\ldots,u_k \in \overline{K}^{*}$ where $w=(w_1,\ldots,w_n)$ is an integral basis for $E$ over $K(t)$. If $E$ is a simple radical extension of $K(t)$, and we use the basis -(11) and the notation of that section, then $w_1=1$ and -\begin{equation} -w_i^{'}=\left(\frac{i-1}{n}\frac{H'}{H}-\frac{D_{i-1}^{'}}{D_{i-1}}\right)w_i +\ref{Int11} and the notation of that section, then $w_1=1$ and +\begin{equation}\label{Int22} +w_i^{\prime}=\left(\frac{i-1}{n}\frac{H'}{H}- +\frac{D_{i-1}^{\prime}}{D_{i-1}}\right)w_i \quad{\rm for\ }1 \le i \le n \end{equation} -This implies that (21) becomes -\begin{equation} -\frac{A_1}{D}=v_1^{'}+\sum_{i=1}^k c_i\frac{u_i^{'}}{u_i} +This implies that \ref{Int21} becomes +\begin{equation}\label{Int23} +\frac{A_1}{D}=v_1^{\prime}+\sum_{i=1}^k c_i\frac{u_i^{\prime}}{u_i} \end{equation} which is simply an integration problem for $A_1/D \in K(t)$, and -\begin{equation} -\frac{A_i}{D}=v_i^{'}+\left(\frac{i-1}{n}\frac{H'}{H} --\frac{D_{i-1}^{'}}{D_{i-1}}\right)v_i\quad{\rm for\ }1 < i \le n +\begin{equation}\label{Int24} +\frac{A_i}{D}=v_i^{\prime}+\left(\frac{i-1}{n}\frac{H'}{H} +-\frac{D_{i-1}^{\prime}}{D_{i-1}}\right)v_i\quad{\rm for\ }1 < i \le n \end{equation} which are Risch differential equations over $K(t)$ @@ -1838,7 +1853,7 @@ $D=xt^2-(x^3-2x^2)t-x^4+x^3$, which is normal, and the resultant is R&=&{\rm resultant_t}({\rm pp_z}({\rm resultant_y}(((3x+1)t-x^3+x^2)y\\ &&\\ &&\hbox{\hskip 2.0cm} --(2x^2-x-1)t-2x^3+x^2+x-zD^{'},F)),D)\\ +-(2x^2-x-1)t-2x^3+x^2+x-zD^{\prime},F)),D)\\ &&\\ &=&x^{12}(2x+1)^2(x+1)^2(x-1)^2z^3(z-2)\\ \end{array} @@ -1861,41 +1876,41 @@ $\delta=(u)$ where $u=(x+y)^2 \in E^{*}$, so the new integrand is \[ h=f-\frac{u'}{u}=f-2\frac{(x+y)'}{x+y}=\frac{(x+1)y}{xt+x^2} \] -We have $y^2=t+x$, which is squarefree, so (23) becomes +We have $y^2=t+x$, which is squarefree, so \ref{Int23} becomes \[ -0=v_1^{'}+\sum_{i=1}^k c_i\frac{u_i^{'}}{u_i} +0=v_1^{\prime}+\sum_{i=1}^k c_i\frac{u_i^{\prime}}{u_i} \] -whose solution is $v_1=k=0$ and (24) becomes +whose solution is $v_1=k=0$ and \ref{Int24} becomes \[ -\frac{x+1}{xt+x^2}=v_2^{'}+\frac{x+1}{2xt+2x^2}v_2 +\frac{x+1}{xt+x^2}=v_2^{\prime}+\frac{x+1}{2xt+2x^2}v_2 \] whose solution is $v_2=2$, implying that $h=2y'$, hence that \[ \begin{array}{l} \displaystyle \int{\frac{(x^2+2x+1)\sqrt{x+\log(x)}+(3x+1)\log(x)+3x^2+x} -{(x\log(x)+x^2)\sqrt{x+\log(x)}+x^2\log(x)+x^3}}~dx\\ +{(x\log(x)+x^2)\sqrt{x+\log(x)}+x^2\log(x)+x^3}}~dx =\\ \\ \displaystyle \hbox{\hskip 4.0cm}2\sqrt{x+\log(x)}+2\log\left(x+\sqrt{x+\log(x)}\right) \end{array} \]} In the general case when $E$ is not a radical extension of $K(t)$, -(21) is solved by bounding $deg_t(v_i)$ and comparing the Puiseux +\ref{Int21} is solved by bounding ${\rm deg}_t(v_i)$ and comparing the Puiseux expansions at infinity of $\sum_{i=1}^n v_iw_i$ with those of the form -(20) of $h$, see \cite{Bro90,Ris68} for details. +\ref{Int20} of $h$, see \cite{Bro90c,Ris68} for details. \subsection{The algebraic exponential case} The transcendental exponential case method also generalizes to the case when $E=K(t)(y)$ is algebraic over $K(t)$, $t=e^b$ for $b \in K$ and $y$ is integral over $K[t]$: following the residue criterion, we -can assume that $R|\kappa(R)$ where $R$ is given by (16), hence that +can assume that $R|\kappa(R)$ where $R$ is given by \ref{Int16}, hence that all its roots in $\overline{K}$ are constants. The denominator of the -integrancd must be of the form $D=t^mU$ where $\gcd(U,t)=1$, $U$ is +integrand must be of the form $D=t^mU$ where $\gcd(U,t)=1$, $U$ is squarefree and $m \ge 0$. If $m > 0$, $E$ is a simple radical extension of $K(t)$, and we use the -basis (11), then it is possible to reduce the power of $t$ appearing +basis \ref{Int11}, then it is possible to reduce the power of $t$ appearing in $D$ by a process similar to the Hermite reduction: writing the integrand $f=\sum_{i=1}^n A_iw_i/(t^mU)$, we ask whether we can compute $b_1,\ldots,b_n \in K$ and $C_1,\ldots,C_n \in K[t]$ such that @@ -1907,35 +1922,35 @@ compute $b_1,\ldots,b_n \in K$ and $C_1,\ldots,C_n \in K[t]$ such that Differentiating both sides and multiplying through by $t^m$ we get \[ \frac{\sum_{i=1}^n A_iw_i}{U} -=\sum_{i=1}^n b_i^{'}w_i+\sum_{i=1}^n b_iw_i^{'} +=\sum_{i=1}^n b_i^{\prime}w_i+\sum_{i=1}^n b_iw_i^{\prime} -mb'\sum_{i=1}^n b_iw_i+\frac{t\sum_{i=1}^n C_iw_i}{U} \] -Using (22) and equating the coefficients of $w_i$ on both sides, we +Using \ref{Int22} and equating the coefficients of $w_i$ on both sides, we get -\begin{equation} -\frac{A_i}{U}=b_i^{'}+(\omega_i-mb')b_i+\frac{tC_i}{U} +\begin{equation}\label{Int25} +\frac{A_i}{U}=b_i^{\prime}+(\omega_i-mb')b_i+\frac{tC_i}{U} \quad{\rm for\ }1 \le i \le n \end{equation} where \[ -\omega_i=\frac{i-1}{n}\frac{H'}{H}-\frac{D_{i-1}^{'}}{D_{i-1}} \in K(t) +\omega_i=\frac{i-1}{n}\frac{H'}{H}-\frac{D_{i-1}^{\prime}}{D_{i-1}} \in K(t) \] Since $t'/t=b' \in K$, it follows that the denominator of $\omega_i$ -is not divisible by $t$ in $K[t]$, hence, evaluating (25) at $t=0$, we +is not divisible by $t$ in $K[t]$, hence, evaluating \ref{Int25} at $t=0$, we get -\begin{equation} -\frac{A_i(0)}{U(0)}=b_i^{'}+(\omega_i(0)-mb')b_i +\begin{equation}\label{Int26} +\frac{A_i(0)}{U(0)}=b_i^{\prime}+(\omega_i(0)-mb')b_i \quad{\rm for\ }1 \le i \le n \end{equation} which are Risch differential equations over $K(t)$. If any of them has no solution in $K(t)$, then the integral is not elementary, otherwise we repeat this process until the denominator of the integrand is -normal. We then perform the change of variabl $\overline{t}=t^{-1}$, +normal. We then perform the change of variable $\overline{t}=t^{-1}$, which is also exponential over $K$ since $\overline{t}'=-b'\overline{t}$, and repeat the above process in order to eliminate the power of $\overline{t}$ from the denominator of the integrand. It can be shown that after this process, any solution of -(13) must have $v \in K$. +\ref{Int13} must have $v \in K$. \noindent {\bf Example 15} {\sl @@ -1965,7 +1980,7 @@ We have so $R|\kappa(R)$ in $K[z]$, its only root being 0. Since $D$ is not divisible by $t$, let $\overline{t}=t^{-1}$ and $z=\overline{t}y$. We have $\overline{t}'=-\overline{t}$ and -$z^3-\overline{t}^2-x\overline{t}^3=0$, so the integral basis (11) is +$z^3-\overline{t}^2-x\overline{t}^3=0$, so the integral basis \ref{Int11} is \[ \overline{w}=(\overline{w}_1,\overline{w}_2,\overline{w}_3) =\left(1,z,\frac{z^2}{\overline{t}}\right) @@ -1984,15 +1999,15 @@ $D_2=\overline{t}$, implying that \omega_1=0, \omega_2=\frac{(1-3x)\overline{t}-2}{3x\overline{t}+3}, {\rm\ and\ } \omega_3=\frac{(2-3x)\overline{t}-1}{3x\overline{t}+3} \] -Therefore the equations (26) become +Therefore the equations \ref{Int26} become \[ -0=b_1^{'}+b_1,0=b_2^{'}+\frac{1}{3}b_2,{\rm\ and\ } -2x+3=b_3^{'}+\frac{2}{3}b_3 +0=b_1^{\prime}+b_1,0=b_2^{\prime}+\frac{1}{3}b_2,{\rm\ and\ } +2x+3=b_3^{\prime}+\frac{2}{3}b_3 \] whose solutions are $b_1=b_2=0$ and $b_3=3x$, implying that the new integrand is \[ -h=f-\left(\frac{3x\overline{w}_3}{\overline{t}}\right)^{'}=\frac{3}{x} +h=f-\left(\frac{3x\overline{w}_3}{\overline{t}}\right)^{\prime}=\frac{3}{x} \] hence that \[ @@ -2002,13 +2017,14 @@ hence that } In the general case when $E$ is not a radical extension of $K(t)$, -following the Hermite reduction, any solution of (13) must have +following the Hermite reduction, any solution of \ref{Int13} must have $v=\sum_{i=1}^n v_iw_i/t^m$ where $v_1,\ldots,v_m \in K[t]$. We can -compute $v$ by bounding $deg_t(v_i)$ and comparing the Puiseux +compute $v$ by bounding ${\rm deg}_t(v_i)$ and comparing the Puiseux expansions at $t=0$ and at infinity of $\sum_{i=1}^n v_iw_i/t^m$ with -those of the form (20) of the integrand, see \cite{Bro90,Ris68} for details. +those of the form \ref{Int20} of the integrand, +see \cite{Bro90c,Ris68} for details. -Once we are reduced to solving (13) for $v \in K$, constants +Once we are reduced to solving \ref{Int13} for $v \in K$, constants $c_1,\ldots,c_k \in \overline{K}$ and $u_1,\ldots,u_k \in E(c_1,\ldots,c_k)^{*}$, constants $\rho_1,\ldots,\rho_s \in \overline{K}$ can be determined at all the @@ -2016,9 +2032,9 @@ places above $t=0$ and at infinity in a manner similar to the algebraic logarithmic case, at which point the algorithm proceeds by constructing the divisors $\delta_j$ and the $u_j$'s as in that case. Again, the details are quite technical and can be found in -\cite{Bro90,Ris68,Ris69a}. +\cite{Bro90c,Ris68,Ris69a}. -\chapter{Singular Value Decomposition} +\chapter{Singular Value Decomposition \cite{Pu09}} \section{Singular Value Decomposition Tutorial} When you browse standard web sources like Wikipedia to learn about @@ -2429,7 +2445,7 @@ are the same. We are trying to predict patterns of how words occur in documents instead of trying to predict patterns of how players score on holes. \chapter{Quaternions} -from\cite{Alt05}: +from \cite{Alt05}: \begin{quotation} Quaternions are inextricably linked to rotations. Rotations, however, are an accident of three-dimensional space. @@ -2451,7 +2467,7 @@ The Theory of Quaternions is due to Sir William Rowan Hamilton, Royal Astronomer of Ireland, who presented his first paper on the subject to the Royal Irish Academy in 1843. His Lectures on Quaternions were published in 1853, and his Elements, in 1866, -shortly after his death. The Elements of Quaternions by Tait\cite{Ta1980} is +shortly after his death. The Elements of Quaternions by Tait \cite{Ta1890} is the accepted text-book for advanced students. Large portions of this file are derived from a public domain version @@ -2489,7 +2505,7 @@ are measured. {\bf 4}. In the usual notation of Analytical Geometry of two dimensions, when rectangular axes are employed, this amounts to reckoning each unit of length along $Oy$ as $+\sqrt{-1}$, and on -$Oy^{'}$ as $-\sqrt{-1}$ ; while on $Ox$ each unit is $+1$, and on +$Oy^{\prime}$ as $-\sqrt{-1}$ ; while on $Ox$ each unit is $+1$, and on $Ox$ it is $-1$. If we look at these four lines in circular order, i.e. in the order of @@ -2563,36 +2579,36 @@ $$OP = a + b\sqrt{-1}$$ the line $OP$ considered as that by which we pass from one extremity, $O$, to the other, $P$. In this sense it is called a VECTOR. Considering, in the plane, any other vector, -$$OQ = a^{'}+b^{'}\sqrt{-1}$$ +$$OQ = a^{\prime}+b^{\prime}\sqrt{-1}$$ \boxed{4.6in}{ \vskip 0.1cm In order to created superscripted variables we use the superscript -function from the SYMBOL domain. So we can create $a^{'}$ as ``ap'' -(that is, ``a-prime'') and $b^{'}$ as ``bp'' (``b-prime'') thus +function from the SYMBOL domain. So we can create $a^{\prime}$ as ``ap'' +(that is, ``a-prime'') and $b^{\prime}$ as ``bp'' (``b-prime'') thus (also note that the underscore character is Axiom's escape character which removes any special meaning of the next character, in this case, the quote character): \spadcommand{ap:=superscript(a,[\_'])} -$$a^{'}$$ +$$a^{\prime}$$ \returnType{Type: Symbol} } \boxed{4.6in}{ \vskip 0.1cm \spadcommand{bp:=superscript(b,[\_'])} -$$b^{'}$$ +$$b^{\prime}$$ \returnType{Type: Symbol} } \boxed{4.6in}{ \vskip 0.1cm at this point we can type \spadcommand{ap+bp*\%i} -$$a^{'}+b^{'}\ \%i$$ +$$a^{\prime}+b^{\prime}\ \%i$$ \returnType{Type: Complex Polynomial Integer} } the addition of these two lines obviously gives -$$OR = a + a^{'} + (b + b^{'})\sqrt{-1}$$ +$$OR = a + a^{\prime} + (b + b^{\prime})\sqrt{-1}$$ \boxed{4.6in}{ \vskip 0.1cm @@ -2604,13 +2620,13 @@ $$a + b\ \%i$$ \boxed{4.6in}{ \vskip 0.1cm \spadcommand{oq:=complex(ap,bp)} -$$a^{'} + b^{'}\ \%i$$ +$$a^{\prime} + b^{\prime}\ \%i$$ \returnType{Type: Complex Polynomial Integer} } \boxed{4.6in}{ \vskip 0.1cm \spadcommand{op + oq} -$$a + a^{'} + (b + b^{'})\%i$$ +$$a + a^{\prime} + (b + b^{\prime})\%i$$ \returnType{Type: Complex Polynomial Integer} } @@ -2710,9 +2726,9 @@ proportional to unity and the two lines, thus $$ \begin{array}{lclr} & \textrm{ } & -1 : a+b\sqrt{-1} :: a^{'}+b^{'}\sqrt{-1} : \prod\\ +1 : a+b\sqrt{-1} :: a^{\prime}+b^{\prime}\sqrt{-1} : \prod\\ \textrm{or} & \textrm{ } & -\prod = (aa^{'} - bb^{'})+(a^{'}b+b^{'}a)\sqrt{-1} +\prod = (aa^{\prime} - bb^{\prime})+(a^{\prime}b+b^{\prime}a)\sqrt{-1} \end{array} $$ @@ -2743,13 +2759,13 @@ $$ \begin{array}{lclr} & \textrm{ OP} & = & a + b\sqrt{-1}\\ \textrm{but by} & \textrm{ OP} & = & pa + qb\\ -\textrm{And if, similarly,} & \textrm{ OQ} & = & pa^{'} + qb^{'}\\ +\textrm{And if, similarly,} & \textrm{ OQ} & = & pa^{\prime} + qb^{\prime}\\ \end{array} $$ the addition of these two lines gives for $OR$ (which retains its previous signification) -$$OR = p(a+a^{'} + q(b+b^{'})$$ +$$OR = p(a+a^{\prime} + q(b+b^{\prime})$$ {\bf 12}. Beyond this, few attempts were made, or at least recorded, in earlier times, to extend the principle to space of three dimensions; @@ -3042,12 +3058,14 @@ straight line drawn from the origin parallel to $\beta$ (\S 22). The straight line drawn from $A$, where $\overline{OA} = \alpha$, and parallel to $\beta$, has the equation -$$\rho = \alpha + x\beta\eqno(1)$$ +\begin{equation}\label{Vec1} +\rho = \alpha + x\beta +\end{equation} In words, we may pass directly from $O$ to $P$ by the vector $\overline{OP}$ or $\rho$; or we may pass first to $A$, by means of $\overline{OA}$ or $\alpha$, and then to $P$ along a vector parallel to $\beta$ (\S 16). -Equation (1) is one of the many useful forms into which +Equation \ref{Vec1} is one of the many useful forms into which Quaternions enable us to throw the general equation of a straight line in space. As we have seen (\S 25) it is equivalent to three numerical equations; but, as these involve the indefinite quantity @@ -3171,7 +3189,7 @@ results which are sometimes useful. They may be easily verified by producing $\overline{Aa}$ to twice its length and joining the extremity with $B$. -($b^{'}$) {\sl The bisectors of the sides of a triangle meet in a point, +($b^{\prime}$) {\sl The bisectors of the sides of a triangle meet in a point, which trisects each of them}. Taking $A$ as origin, and putting $\alpha$, $\beta$, $\gamma$ @@ -3241,7 +3259,7 @@ y(e-1)=ze \right\} $$ These give $xy = e$, and the equation of the locus of $Q$ is -$$\rho = e\beta+y^{'}\alpha$$ +$$\rho = e\beta+y^{\prime}\alpha$$ i.e. a straight line parallel to $OA$, drawn through $N$ in $OB$ produced, so that $$ON : OB :: OM : OA$$ @@ -3308,18 +3326,20 @@ $OB$ drawn from the origin, and $\alpha$ is $OA$ the tangent at the origin. In the figure $$\overline{QP}=\alpha t,\hbox{\hskip 1cm}\overline{OQ}=\frac{\beta t^2}{2}$$ -The secant joining the points where $t$ has the values $t$ and $t^{'}$ is +The secant joining the points where $t$ has the values $t$ and $t^{\prime}$ is represented by the equation $$ \begin{array}{rcl} -\rho&=&\alpha t +\frac{\beta t^2}{2}+x\left(\alpha t^{'}+\frac{\beta t^{'2}}{2} +\rho&=&\alpha t +\frac{\beta t^2}{2}+ +x\left(\alpha t^{\prime}+\frac{\beta t^{'2}}{2} -\alpha t-\frac{\beta t^2}{2}\right)\hbox{\hskip 1cm}(\S 30)\\ -&=&\alpha t+\frac{\beta t^2}{2}+x(t^{'}-t)\left\{\alpha+\beta\frac{t^{'}-t}{2} +&=&\alpha t+\frac{\beta t^2}{2}+ +x(t^{\prime}-t)\left\{\alpha+\beta\frac{t^{\prime}-t}{2} \right\} \end{array} $$ -Write $x$ for $x(t^{'}-t)$ [which may have any value], then put -$t^{'}=t$, and the equation of the tangent at the point ($t$) is +Write $x$ for $x(t^{\prime}-t)$ [which may have any value], then put +$t^{\prime}=t$, and the equation of the tangent at the point ($t$) is $$\rho=\alpha t + \frac{\beta t^2}{2}+x(\alpha+\beta t)$$ In this put $x = -t$, and we have $$\rho=-\frac{\beta t^2}{2}$$ @@ -3500,7 +3520,9 @@ fact that, in these equations, $\phi(t)$ or $\phi(t, u)$ is necessarily a vector expression, since it is equated to a vector, $\rho$.] (m) Thus the equation -$$\rho = \alpha\cos t+\beta\sin t + \gamma t\eqno(1)$$ +\begin{equation}\label{Quat1} +\rho = \alpha\cos t+\beta\sin t + \gamma t +\end{equation} belongs to a helix, \boxed{4.6in}{ \vskip 0.1cm @@ -3509,7 +3531,9 @@ In Axiom we can draw this with the commands: tpdhere } while -$$\rho = \alpha\cos t+\beta\sin t + \gamma u\eqno(2)$$ +\begin{equation}\label{Quat2} +\rho = \alpha\cos t+\beta\sin t + \gamma u +\end{equation} represents a cylinder whose generating lines are parallel to $\gamma$, \boxed{4.6in}{ \vskip 0.1cm @@ -3627,7 +3651,7 @@ $$\textrm{Limit}\left( \right)_{\delta t=0} =\frac{d\rho}{dt} =\frac{d\phi(t)}{dt} -=\phi^{'}(t)$$ +=\phi^{\prime}(t)$$ In such a case as this, then, we are permitted to differentiate, or to form the differential coefficient of, a vector, according to the @@ -3692,16 +3716,16 @@ $$ \begin{array}{rcl} d\rho & = & L_{x=\infty}x \left\{ -\phi^{'}(t)\frac{1}{x}dt+ -\phi^{''}(t)\frac{1}{x^2}\frac{(dt)^2}{1\textrm{ . }2}+ +\phi^{\prime}(t)\frac{1}{x}dt+ +\phi^{\prime\prime}(t)\frac{1}{x^2}\frac{(dt)^2}{1\textrm{ . }2}+ \textrm{\&c} \right\}\\ &&\\ -& = & \phi^{'}(t)dt +& = & \phi^{\prime}(t)dt \end{array} $$ And, if we choose, we may now write -$$\frac{d\rho}{dt}=\phi^{'}(t)$$ +$$\frac{d\rho}{dt}=\phi^{\prime}(t)$$ {\bf 37}. But it is to be most particularly remarked that in the whole of this investigation no regard whatever has been paid to @@ -3715,9 +3739,9 @@ of a 'celestial Atwood's machine'. {\bf 38}. If we suppose the variable, in terms of which $\rho$ is expressed, to be the arc, $s$, of the curve measured from some fixed point, we find as before -$$d\rho = \phi^{'}(x)ds$$ +$$d\rho = \phi^{\prime}(x)ds$$ From the very nature of the question it is obvious that the length -of $dp$ must in this case be $ds$, so that $\phi^{'}(s)$ +of $dp$ must in this case be $ds$, so that $\phi^{\prime}(s)$ is necessarily a unit-vector. This remark is of importance, as we shall see later; and it may therefore be useful to obtain afresh the above result without @@ -4221,11 +4245,11 @@ the reciprocal. \includegraphics{ps/quat7.ps} -Thus, if $OA$, $OB$, $OA^{'}$ , lie in one plane, and if -$OA^{'} = OA$, and $\angle A^{'}OB = \angle BOA$, we have +Thus, if $OA$, $OB$, $OA^{\prime}$ , lie in one plane, and if +$OA^{\prime} = OA$, and $\angle A^{\prime}OB = \angle BOA$, we have $$\frac{\overline{OB}}{\overline{OA}}=q$$, and -$$\frac{\overline{OB}}{\overline{OA^{'}}}=\textrm{ congugate of }q=Kq$$ +$$\frac{\overline{OB}}{\overline{OA^{\prime}}}=\textrm{ congugate of }q=Kq$$ By last section we see that $$Kq=(Tq)^2q^{-1}$$ @@ -4660,18 +4684,18 @@ have taken. \includegraphics{ps/quat15.ps} -Let $ABA^{'}$ be a semicircle, whose centre -is $0$, and let $OB$ be perpendicular to $AOA^{'}$. +Let $ABA^{\prime}$ be a semicircle, whose centre +is $0$, and let $OB$ be perpendicular to $AOA^{\prime}$. -Then ${\displaystyle\frac{\overline{OB}}{\overline{OA^{'}}}}=q$ +Then ${\displaystyle\frac{\overline{OB}}{\overline{OA^{\prime}}}}=q$ suppose, is a quadrantal versor, and is evidently equal to -${\displaystyle\frac{\overline{OA^{'}}}{\overline{OB}}}$ ; +${\displaystyle\frac{\overline{OA^{\prime}}}{\overline{OB}}}$ ; \S\S 50, 53. Hence -$$q^2=\frac{\overline{OA^{'}}}{\overline{OB}}. +$$q^2=\frac{\overline{OA^{\prime}}}{\overline{OB}}. \frac{\overline{OB}}{\overline{OA}}= -\frac{\overline{OA^{'}}}{\overline{OA}}=-1]$$ +\frac{\overline{OA^{\prime}}}{\overline{OA}}=-1]$$ {\bf 69}. Having thus found that the squares of {\sl i}, {\sl j}, {\sl k} are each @@ -4961,14 +4985,15 @@ from which the quaternion derives its name, exhibited in the most simple form. And now we see at once that an equation such as -$$q^{'}=q$$ -where\hbox{\hskip 3cm}$q^{'}=w^{'}+x^{'}i+y^{'}j+z^{'}k$\\ +$$q^{\prime}=q$$ +where\hbox{\hskip 3cm}$q^{\prime}= +w^{\prime}+x^{\prime}i+y^{\prime}j+z^{\prime}k$\\ involves, of course, the {\sl four} equations $$ -w^{'}=w\textrm{, } -x^{'}=x\textrm{, } -y^{'}=y\textrm{, } -z^{'}=z +w^{\prime}=w\textrm{, } +x^{\prime}=x\textrm{, } +y^{\prime}=y\textrm{, } +z^{\prime}=z $$ {\bf 81}. We proceed to indicate another mode of proof of the distributive @@ -4985,15 +5010,17 @@ But, writing $\alpha$ for $\alpha^{-1}$, we see that this involves the equality $$(\beta+\gamma)\alpha = \beta\alpha+\gamma\alpha$$ from which, by taking the conjugates of both sides, we derive -$$\alpha^{'}(\beta^{'}+\gamma^{'})=\alpha^{'}\beta^{'}+\alpha^{'}\gamma^{'} +$$\alpha^{\prime}(\beta^{\prime}+\gamma^{\prime})= +\alpha^{\prime}\beta^{\prime}+\alpha^{\prime}\gamma^{\prime} (\S 55)$$ And a combination of these results (putting -$\beta+\gamma$ for $\alpha^{'}$ in the latter, for instance) gives +$\beta+\gamma$ for $\alpha^{\prime}$ in the latter, for instance) gives $$ \begin{array}{lcr} -(\beta+\gamma)(\beta^{'}+\gamma^{'}) & = & -(\beta+\gamma)\beta^{'}+(\beta+\gamma)\gamma^{'}\\ -& = & \beta\beta^{'}+\gamma\beta^{'}+\beta\gamma^{'}+\gamma\gamma^{'} +(\beta+\gamma)(\beta^{\prime}+\gamma^{\prime}) & = & +(\beta+\gamma)\beta^{\prime}+(\beta+\gamma)\gamma^{\prime}\\ +& = & \beta\beta^{\prime}+\gamma\beta^{\prime}+ +\beta\gamma^{\prime}+\gamma\gamma^{\prime} \end{array} $$ by the former. @@ -5026,7 +5053,8 @@ a\frac{\beta}{\delta}+\frac{\gamma}{\beta}.\frac{\beta}{\delta}\\ \end{array} $$ And the conjugate may be written -$$q^{'}(a^{'}+\alpha^{'})=q^{'}a^{'}+q^{'}\alpha^{'} (\S 55)$$ +$$q^{\prime}(a^{\prime}+\alpha^{\prime})= +q^{\prime}a^{\prime}+q^{\prime}\alpha^{\prime} (\S 55)$$ Hence, generally, $$(a+\alpha)(b+\beta)=ab+a\beta+b\alpha+\alpha\beta$$ or, breaking up $a$ and $b$ each into the sum of two scalars, and @@ -5053,11 +5081,12 @@ we have $(p+q)(r+s)=pr+ps+qr+qs$ {\bf 82}. Cayley suggests that the laws of quaternion multiplication may be derived more directly from those of vector multiplication, supposed to be already established. Thus, let $\alpha$ be the unit vector -perpendicular to the vector parts of $q$ and of $q^{'}$. Then let -$$\rho=q.\alpha,\;\;\;\sigma=-\alpha .q^{'}$$ +perpendicular to the vector parts of $q$ and of $q^{\prime}$. Then let +$$\rho=q.\alpha,\;\;\;\sigma=-\alpha .q^{\prime}$$ as is evidently permissible, and we have -$$p\alpha=q.\alpha\alpha=-q;\;\;\;\alpha\sigma=-\alpha\alpha.q^{'}=q^{'}$$ -so that\hbox{\hskip 4cm}$-q.q^{'}=\rho\alpha.\alpha\sigma=-\rho.\sigma$ +$$p\alpha=q.\alpha\alpha=-q;\;\;\;\alpha\sigma= +-\alpha\alpha.q^{\prime}=q^{\prime}$$ +so that\hbox{\hskip 4cm}$-q.q^{\prime}=\rho\alpha.\alpha\sigma=-\rho.\sigma$ The student may easily extend this process. @@ -5071,7 +5100,7 @@ We will commence by proving the result of \S 77 anew. {\bf 83}. Let $$\alpha=xi+yj+zk$$ -$$\beta=x^{'}i+y^{'}j+z^{'}k$$ +$$\beta=x^{\prime}i+y^{\prime}j+z^{\prime}k$$ Then, because by \S 71 every product or quotient of $i$, $j$, $k$ is reducible to one of them or to a number, we are entitled to assume $$q=\frac{\beta}{\alpha}=\omega+\xi i+\eta j +\zeta k$$ @@ -5086,12 +5115,12 @@ where the various sums are to be interpreted as in \S 61. All such things become obvious in view of the properties of $i$, $j$ ,$k$.] {\bf 84}. But it may be interesting to find $\omega$, $\xi$, $\eta$, $\zeta$ -in terms of $x$, $y$, $z$, $x^{'}$, $y^{'}$ , $z^{'}$ . +in terms of $x$, $y$, $z$, $x^{\prime}$, $y^{\prime}$ , $z^{\prime}$ . We have $$\beta=q\alpha$$ or -$$x^{'}i+y^{'}j+z^{'}k=(\omega+\xi i+\eta j+\zeta k)(xi+yj+zk)$$ +$$x^{\prime}i+y^{\prime}j+z^{\prime}k=(\omega+\xi i+\eta j+\zeta k)(xi+yj+zk)$$ $$=-(\xi x+\eta y+\zeta z) +(\omega x+\eta z-\zeta y)i +(\omega y+\zeta x-\xi z)j @@ -5105,21 +5134,21 @@ This (\S 80) resolves itself into the four equations $$ \begin{array}{lllllllll} 0 & = & & & \xi x & + & \eta y & + & \zeta z\\ -x^{'} & = & \omega x & & & + & \eta z & - & \zeta y\\ -y^{'} & = & \omega y & - & \xi z & & & + & \zeta x\\ -z^{'} & = & \omega z & + & \xi y & - & \eta x\\ +x^{\prime} & = & \omega x & & & + & \eta z & - & \zeta y\\ +y^{\prime} & = & \omega y & - & \xi z & & & + & \zeta x\\ +z^{\prime} & = & \omega z & + & \xi y & - & \eta x\\ \end{array} $$ The three last equations give -$$xx^{'}+yy^{'}+zz^{'}=\omega(x^2+y^2+z^2)$$ +$$xx^{\prime}+yy^{\prime}+zz^{\prime}=\omega(x^2+y^2+z^2)$$ which determines $\omega$. Also we have, from the same three, by the help of the first, -$$\xi x^{'}+\eta y^{'}+\zeta z^{'} = 0$$ +$$\xi x^{\prime}+\eta y^{\prime}+\zeta z^{\prime} = 0$$ which, combined with the first, gives -$$\frac{\xi}{yz^{'}-zy^{'}} -=\frac{\eta}{zx^{'}-xz^{'}} -=\frac{\zeta}{xy^{'}-yx^{'}} +$$\frac{\xi}{yz^{\prime}-zy^{\prime}} +=\frac{\eta}{zx^{\prime}-xz^{\prime}} +=\frac{\zeta}{xy^{\prime}-yx^{\prime}} $$ and the common value of these three fractions is then easily seen to be @@ -5134,8 +5163,9 @@ giving it, we shall refer back to this section. by means of the distributive (\S 81). We leave the proof to the student. He has merely to multiply together the factors $$w+xi+yj+zk,\;\;\;\; -w+x^{'}i+y^{'}j+z^{'}k,\;\;\;\;\textrm{ and } -w^{''} + x^{''}i + y^{''}j + z^{''}k$$ +w+x^{\prime}i+y^{\prime}j+z^{\prime}k,\;\;\;\;\textrm{ and } +w^{\prime\prime} + x^{\prime\prime}i + y^{\prime\prime}j + +z^{\prime\prime}k$$ as follows : @@ -5149,19 +5179,19 @@ $i$, $j$, $k$, in these products, respectively equal, each to each. {\bf 86}. With the same expressions for $\alpha$, $\beta$, as in section 83, we have -$$\alpha\beta=(xi+yj+zk)(x^{'}i+y^{'}j+z^{'}k)$$ -$$\;\;=-(xx^{'}+yy^{'}+zz^{'}) -+(yz^{'}-zy^{'})i -+(zx^{'}-xz^{'})j -+(xy^{'}-yx^{'})k +$$\alpha\beta=(xi+yj+zk)(x^{\prime}i+y^{\prime}j+z^{\prime}k)$$ +$$\;\;=-(xx^{\prime}+yy^{\prime}+zz^{\prime}) ++(yz^{\prime}-zy^{\prime})i ++(zx^{\prime}-xz^{\prime})j ++(xy^{\prime}-yx^{\prime})k $$ But we have also $$\beta\alpha= --(xx^{'}+yy^{'}+zz^{'}) --(yz^{'}-zy^{'})i --(zx^{'}-xz^{'})j --(xy^{'}-yx^{'})k +-(xx^{\prime}+yy^{\prime}+zz^{\prime}) +-(yz^{\prime}-zy^{\prime})i +-(zx^{\prime}-xz^{\prime})j +-(xy^{\prime}-yx^{\prime})k $$ The only difference is in the sign of the vector parts. Hence @@ -5172,7 +5202,7 @@ $$\alpha\beta-\beta\alpha=2V\alpha\beta\eqno{(4)}$$ $$\alpha\beta=K.\beta\alpha\eqno{(5)}$$ {\bf 87}. If $\alpha=\beta$ we have of course (\S 25) -$$x=x^{'},\;\;\;\;y=y^{'},\;\;\;\;z=z^{'}$$ +$$x=x^{\prime},\;\;\;\;y=y^{\prime},\;\;\;\;z=z^{\prime}$$ and the formulae of last section become $$\alpha\beta=\beta\alpha=\alpha^2=-(x^2+y^2+z^2)$$ which was anticipated in \S 73, where we proved the formula @@ -5629,28 +5659,28 @@ where $x$ is an undetermined scalar. {\bf 99}. If we write, as in \S\S 83, 84, $$\alpha=ix+jy+kz$$ -$$\beta=ix^{'}+jy^{'}+kz^{'}$$ +$$\beta=ix^{\prime}+jy^{\prime}+kz^{\prime}$$ we have, at once, by \S 86, $$\begin{array}{rcl} -S\alpha\beta&=&-xx^{'}-yy^{'}-zz^{'}\\ -&=&-rr^{'}\left( -\frac{x}{r}\frac{x^{'}}{r^{'}}+ -\frac{y}{r}\frac{y^{'}}{r^{'}}+ -\frac{z}{r}\frac{z^{'}}{r^{'}} +S\alpha\beta&=&-xx^{\prime}-yy^{\prime}-zz^{\prime}\\ +&=&-rr^{\prime}\left( +\frac{x}{r}\frac{x^{\prime}}{r^{\prime}}+ +\frac{y}{r}\frac{y^{\prime}}{r^{\prime}}+ +\frac{z}{r}\frac{z^{\prime}}{r^{\prime}} \right) \end{array} $$ where $$ r=\sqrt{x^2+y^2+z^2},\;\;\;\; -r^{'}=\sqrt{x^{'2}+y^{'2}+z^{'2}} +r^{\prime}=\sqrt{x^{'2}+y^{'2}+z^{'2}} $$ Also $$ -V\alpha\beta=rr^{'}\left\{ -\frac{yz^{'}-zy^{'}}{rr^{'}}i+ -\frac{zx^{'}-xz^{'}}{rr^{'}}j+ -\frac{xy^{'}=yx^{'}}{rr^{'}}k +V\alpha\beta=rr^{\prime}\left\{ +\frac{yz^{\prime}-zy^{\prime}}{rr^{\prime}}i+ +\frac{zx^{\prime}-xz^{\prime}}{rr^{\prime}}j+ +\frac{xy^{\prime}=yx^{\prime}}{rr^{\prime}}k \right\} $$ @@ -5708,17 +5738,17 @@ $$S.\alpha\beta\gamma=S.\alpha\beta(p\alpha+q\beta)=0$$ This property of the expression $S.\alpha\beta\gamma$ prepares us to find that it is a determinant. And, in fact, if we take $\alpha$,$\beta$ as in \S 83, and in addition -$$\gamma=ix^{''}+jy^{''}+kz^{''}$$ +$$\gamma=ix^{\prime\prime}+jy^{\prime\prime}+kz^{\prime\prime}$$ we have at once -$$S.\alpha\beta\gamma=-x^{''}(yz^{'}-zy^{'})- -y^{''}(zx^{'}-xz^{'})- -z^{''}(xy^{'}-yx^{'})$$ +$$S.\alpha\beta\gamma=-x^{\prime\prime}(yz^{\prime}-zy^{\prime})- +y^{\prime\prime}(zx^{\prime}-xz^{\prime})- +z^{\prime\prime}(xy^{\prime}-yx^{\prime})$$ $$ =-\left\vert \begin{array}{ccc} x & y & z\\ -x^{'} & y^{'} & z^{'}\\ -x^{''}&y^{''}&z^{''} +x^{\prime} & y^{\prime} & z^{\prime}\\ +x^{\prime\prime}&y^{\prime\prime}&z^{\prime\prime} \end{array} \right\vert $$ @@ -5729,9 +5759,9 @@ $$S.\alpha\beta\gamma=-S.\beta\alpha\gamma=S.\beta\gamma\alpha \textrm{, \&c}$$ If we take three new vectors -$$\alpha_1=ix+jx^{'}+kx^{''}$$ -$$\beta_1 =iy+jy^{'}+ky^{''}$$ -$$\gamma_1=iz+jz^{'}+kz^{''}$$ +$$\alpha_1=ix+jx^{\prime}+kx^{\prime\prime}$$ +$$\beta_1 =iy+jy^{\prime}+ky^{\prime\prime}$$ +$$\gamma_1=iz+jz^{\prime}+kz^{\prime\prime}$$ we thus see that they are coplanar if $\alpha$, $\beta$, $\gamma$ are so. That is, if $$S.\alpha\beta\gamma=0$$ @@ -5755,8 +5785,8 @@ In Cartesian coordinates this is\\ \vskip 0.1cm $(x^2+y^2+z^z)(x^{'2}+y^{'2}+z^{'2})$ $$ -=(xx^{'}+yy^{'}+zz^{'})^2+(yz^{'}-zy^{'})^2+ -(zx^{'}-xz^{'})^2+(xy^{'}-yx^{'})^2 +=(xx^{\prime}+yy^{\prime}+zz^{\prime})^2+(yz^{\prime}-zy^{\prime})^2+ +(zx^{\prime}-xz^{\prime})^2+(xy^{\prime}-yx^{\prime})^2 $$ More generally we have $$ @@ -5767,11 +5797,13 @@ $$ $$ If we write $$q=w+\alpha=w+ix+jy+kz$$ -$$r=w^{'}+\beta=w^{'}+ix^{'}+jy^{'}+kz^{'}$$ +$$r=w^{\prime}+\beta=w^{\prime}+ix^{\prime}+jy^{\prime}+kz^{\prime}$$ this becomes $$(w^2+x^2+y^2+z^2)(w^{'2}+x^{'2}+y^{'2}+z^{'2})$$ -$$=(ww^{'}-xx^{'}-yy^{'}-zz^{'})^2+(wx^{'}+w^{'}x+yz^{'}-zy^{'})^2$$ -$$=(xy^{'}+w^{'}y+zx^{'}-xz^{'})^2+(wz^{'}+w^{'}z+xy^{'}-yx^{'})^2$$ +$$=(ww^{\prime}-xx^{\prime}-yy^{\prime}-zz^{\prime})^2+ +(wx^{\prime}+w^{\prime}x+yz^{\prime}-zy^{\prime})^2$$ +$$=(xy^{\prime}+w^{\prime}y+zx^{\prime}-xz^{\prime})^2+ +(wz^{\prime}+w^{\prime}z+xy^{\prime}-yx^{\prime})^2$$ a formula of algebra due to Euler. {\bf 104}. We have, of course, by multiplication, @@ -5904,10 +5936,10 @@ Thus $$T(\rho+\alpha)=T(\rho-\alpha)$$ [which expresses that if $$ \overline{OA}=\alpha\;\;\;\; -\overline{OA^{'}}=-\alpha\;\;\;\;\textrm{ and }\;\;\;\; +\overline{OA^{\prime}}=-\alpha\;\;\;\;\textrm{ and }\;\;\;\; \overline{OP}=\rho $$ -we have\hbox{\hskip 4cm}$AP=A^{'}P$\\ +we have\hbox{\hskip 4cm}$AP=A^{\prime}P$\\ and thus that $P$ is any point equidistant from two fixed points,] may be written $$(\rho+\alpha)^2=(\rho-\alpha)^2$$ or\hbox{\hskip 3cm}$\rho^2+2S\alpha\rho+\alpha^2= @@ -6002,16 +6034,22 @@ $$ which, when equated to zero, gives the same relation as before. [See Ex. 10 at the end of this Chapter.] -An additional point, with $\epsilon=x^{'}\alpha+y^{'}\beta+z^{'}\gamma$ +An additional point, with $\epsilon=x^{\prime}\alpha+ +y^{\prime}\beta+z^{\prime}\gamma$ gives six additional equations like (1) ; i. e. $$ \begin{array}{rlll} -S\alpha\epsilon&=x^{'}\alpha^2&+y^{'}S\alpha\beta&+z^{'}S\alpha\gamma\\ -S\beta\epsilon&=x^{'}S\beta\alpha&+y^{'}\beta^2&+z^{'}S\beta\gamma\\ -S\gamma\epsilon&=x^{'}S\gamma\alpha&+y^{'}S\gamma\beta&+z^{'}\gamma^2\\ -S\delta\epsilon&=x^{'}S\delta\alpha&+y^{'}S\delta\beta&+z^{'}S\delta\gamma\\ +S\alpha\epsilon&=x^{\prime}\alpha^2&+ +y^{\prime}S\alpha\beta&+z^{\prime}S\alpha\gamma\\ +S\beta\epsilon&=x^{\prime}S\beta\alpha&+y^{\prime}\beta^2&+ +z^{\prime}S\beta\gamma\\ +S\gamma\epsilon&=x^{\prime}S\gamma\alpha&+y^{\prime}S\gamma\beta&+ +z^{\prime}\gamma^2\\ +S\delta\epsilon&=x^{\prime}S\delta\alpha&+y^{\prime}S\delta\beta&+ +z^{\prime}S\delta\gamma\\ &=xS\epsilon\alpha&+yS\epsilon\beta&+zS\epsilon\gamma\\ -\epsilon^2&=x^{'}S\alpha\epsilon&+y^{'}S\beta\epsilon&+z^{'}S\gamma\epsilon +\epsilon^2&=x^{\prime}S\alpha\epsilon&+y^{\prime}S\beta\epsilon&+ +z^{\prime}S\gamma\epsilon \end{array} $$ from which corresponding conclusions may be drawn. @@ -6377,13 +6415,13 @@ triangles, with a common side, and having their other sides bisected by the same great circle (i.e. having their vertices in a small circle parallel to this great circle) have equal areas, \&c. ] -{\bf 118}. Let $\overline{Oa}=\alpha^{'}$, $\overline{Ob}=\beta^{'}$, -$\overline{Oc}=\gamma^{'}$, and we have +{\bf 118}. Let $\overline{Oa}=\alpha^{\prime}$, $\overline{Ob}=\beta^{\prime}$, +$\overline{Oc}=\gamma^{\prime}$, and we have $$ \begin{array}{rcl} -\left(\frac{\alpha^{'}}{\beta^{'}}\right)^{\frac{1}{2}} -\left(\frac{\beta^{'}}{\gamma^{'}}\right)^{\frac{1}{2}} -\left(\frac{\gamma^{'}}{\alpha^{'}}\right)^{\frac{1}{2}}&=& +\left(\frac{\alpha^{\prime}}{\beta^{\prime}}\right)^{\frac{1}{2}} +\left(\frac{\beta^{\prime}}{\gamma^{\prime}}\right)^{\frac{1}{2}} +\left(\frac{\gamma^{\prime}}{\alpha^{\prime}}\right)^{\frac{1}{2}}&=& {\stackrel{\frown}{Ca}}.{\stackrel{\frown}{cA}}.{\stackrel{\frown}{Bc}}\\ &=&{\stackrel{\frown}{Ca}}.{\stackrel{\frown}{BA}}\\ &=&{\stackrel{\frown}{EG}}.{\stackrel{\frown}{FE}}= @@ -6394,13 +6432,13 @@ $$ But $FG$ is the complement of $DF$. Hence the {\sl angle of the quaternion} $$ -\left(\frac{\alpha^{'}}{\beta^{'}}\right)^{\frac{1}{2}} -\left(\frac{\beta^{'}}{\gamma^{'}}\right)^{\frac{1}{2}} -\left(\frac{\gamma^{'}}{\alpha^{'}}\right)^{\frac{1}{2}} +\left(\frac{\alpha^{\prime}}{\beta^{\prime}}\right)^{\frac{1}{2}} +\left(\frac{\beta^{\prime}}{\gamma^{\prime}}\right)^{\frac{1}{2}} +\left(\frac{\gamma^{\prime}}{\alpha^{\prime}}\right)^{\frac{1}{2}} $$ {\sl is half the spherical excess of the triangle whose angular points are -at the extremities of the unit-vectors} $\alpha^{'}$, $\beta^{'}$, and -$\gamma^{'}$. +at the extremities of the unit-vectors} $\alpha^{\prime}$, $\beta^{\prime}$, +and $\gamma^{\prime}$. [In seeking a purely quaternion proof of the preceding proposi tions, the student may commence by showing that for any three @@ -6417,13 +6455,15 @@ Another easy method is to commence afresh by forming from the vectors of the corners of a spherical triangle three new vectors thus: $$ -\alpha^{'}=\left(\frac{\beta+\gamma}{\alpha}^{2}\right)^2 .\;\alpha,\;\;\;\;\; +\alpha^{\prime}=\left(\frac{\beta+\gamma}{\alpha}^{2}\right)^2 .\; +\alpha,\;\;\;\;\; \textrm{\&c.} $$ -Then the angle between the planes of $\alpha$, $\beta^{'}$ and -$\gamma^{'}$, $\alpha$; or of $\beta$, $\gamma^{'}$ and $\alpha^{'}$, -$\beta$; or of $\gamma$, $\alpha^{'}$ and $\beta^{'}$, $\gamma$ +Then the angle between the planes of $\alpha$, $\beta^{\prime}$ and +$\gamma^{\prime}$, $\alpha$; or of $\beta$, $\gamma^{\prime}$ +and $\alpha^{\prime}$, +$\beta$; or of $\gamma$, $\alpha^{\prime}$ and $\beta^{\prime}$, $\gamma$ is obviously the spherical excess. But a still simpler method of proof is easily derived from the @@ -6448,16 +6488,16 @@ be the pole of $q$. $$ {\stackrel{\frown}{AB}}=q,\;\;\;\; {\stackrel{\frown}{AB^{-1}}}=q^{-1},\;\;\;\; -{\stackrel{\frown}{B^{'}C^{'}}}=r +{\stackrel{\frown}{B^{\prime}C^{\prime}}}=r $$ -Join $C^{'}A$, and make -${\stackrel{\frown}{AC}}={\stackrel{\frown}{C^{'}A}}$. Join $CB$. +Join $C^{\prime}A$, and make +${\stackrel{\frown}{AC}}={\stackrel{\frown}{C^{\prime}A}}$. Join $CB$. Then ${\stackrel{\frown}{CB}}$ is $qrq^{-1}$, its arc $CB$ is evidently equal in length to that -of $r$, $B^{'}C^{'}$; and its plane (making the same angle with -$B^{'}B$ that that of -$B^{'}C^{'}$ does) has evidently been made to revolve about $Q$, the +of $r$, $B^{\prime}C^{\prime}$; and its plane (making the same angle with +$B^{\prime}B$ that that of +$B^{\prime}C^{\prime}$ does) has evidently been made to revolve about $Q$, the pole of $q$, through double the angle of $q$. It is obvious, from the nature of the above proof, that this @@ -6529,19 +6569,19 @@ $$=\lambda\sin d+\mu\cos d$$ where $d$ is its declination. Hence when its hour-angle is $h$, its vector is -$$\delta^{'}=L^{-1}\delta L$$ +$$\delta^{\prime}=L^{-1}\delta L$$ The vertical plane containing it intersects the horizon in -$$iVi\delta^{'}=jSj\delta^{'}+kSk\delta^{'}$$ +$$iVi\delta^{\prime}=jSj\delta^{\prime}+kSk\delta^{\prime}$$ so that -$$\tan(azimuth)=\frac{Sk\delta^{'}}{Sj\delta^{'}}\eqno{(1)}$$ +$$\tan(azimuth)=\frac{Sk\delta^{\prime}}{Sj\delta^{\prime}}\eqno{(1)}$$ [This may also be obtained directly from the last formula (1) of \S 114.] To find its Amplitude, i.e. its azimuth at rising or setting, the hour-angle must be obtained from the condition -$$Si\delta^{'}=0\eqno{(2)}$$ +$$Si\delta^{\prime}=0\eqno{(2)}$$ These relations, with others immediately deducible from them, enable us (at once and for ever) to dispense with the hideous @@ -6551,7 +6591,7 @@ formulae of Spherical Trigonometry. translate the expressions above into the ordinary notation. This is effected at once by means of the expressions for $\lambda$, $\mu$, $L$, and $\delta$ above, which give by inspection -$$\delta^{'}=\lambda\sin d+(\mu\cos h-k\sin h)\cos d$$ +$$\delta^{\prime}=\lambda\sin d+(\mu\cos h-k\sin h)\cos d$$ = x sin d + (fjb cos h k sin h) cos d, and we have from (1) and (2) of last section respectively $$ @@ -6875,10 +6915,10 @@ $$ $$ we obtain, as will be seen in Chapter IV, the following, $$ -S\alpha\rho S\alpha\rho^{'}+ -S\beta\rho S\beta\rho^{'}+ -S\gamma\rho S\gamma\rho^{'}= -\frac{S.(\iota\rho+\rho\kappa)(\kappa\rho^{'}+\rho^{'}\iota)} +S\alpha\rho S\alpha\rho^{\prime}+ +S\beta\rho S\beta\rho^{\prime}+ +S\gamma\rho S\gamma\rho^{\prime}= +\frac{S.(\iota\rho+\rho\kappa)(\kappa\rho^{\prime}+\rho^{\prime}\iota)} {(\kappa^2-\iota^2)^2} $$ where $\rho$ also may be any vector whatever. @@ -6886,7 +6926,7 @@ where $\rho$ also may be any vector whatever. This is another very important formula of transformation ; and it will be a good exercise for the student to prove its truth by processes analogous to those in last section. We may merely -observe, what indeed is obvious, that by putting $\rho^{'}=\rho$ it becomes +observe, what indeed is obvious, that by putting $\rho^{\prime}=\rho$ it becomes the formula of last section. And we see that we may write, with the recent values of $\iota$ and $\kappa$ in terms of $\alpha$, $\beta$, $\gamma$, the identity @@ -6912,12 +6952,13 @@ as in algebra, that any quaternion expression which contains this imaginary can always be broken up into the sum of two parts, one real, the other multiplied by the first power of $\sqrt{-1}$. Such an expression, viz. -$$q=q^{'}+\sqrt{-1}q^{''}$$ -where $q^{'}$ and $q^{''}$ are real quaternions, is called by Hamilton a +$$q=q^{\prime}+\sqrt{-1}q^{\prime\prime}$$ +where $q^{\prime}$ and $q^{\prime\prime}$ are real quaternions, +is called by Hamilton a BIQUATERNION. [The student should be warned that the term Biquaternion has since been employed by other writers in the sense sometimes of a ``set'' of 8 elements, analogous to the -Quaternion 4 ; sometimes for an expression $q^{'} + \theta q^{''}$ +Quaternion 4 ; sometimes for an expression $q^{\prime} + \theta q^{\prime\prime}$ where $\theta$ is not the algebraic imaginary. By them Hamilton s Biquaternion is called simply a quaternion with non-real constituents.] Some @@ -6927,99 +6968,109 @@ that any biquaternion can be divided into a real and an imaginary part, the latter being the product of $\sqrt{-1}$ by a real quaternion; second, that this $\sqrt{-1}$ is commutative with all other quantities in multiplication; third, that if two biquaternions be equal, as -$$q^{'}+\sqrt{-1}\;q^{''}=r^{'}+\sqrt{-1}\;r^{''}$$ +$$q^{\prime}+\sqrt{-1}\;q^{\prime\prime}= +r^{\prime}+\sqrt{-1}\;r^{\prime\prime}$$ we have, as in algebra, -$$q^{'}=r^{'},\;\;\;\;q^{''}=r^{''}$$ +$$q^{\prime}=r^{\prime},\;\;\;\;q^{\prime\prime}=r^{\prime\prime}$$ so that an equation between biquaternions involves in general {\sl eight} equations between scalars. Compare \S 80. {\bf 131}. We have obviously, since $\sqrt{-1}$ is a scalar, -$$S(q^{'}+\sqrt{-1}\;q^{''})=Sq^{'}+\sqrt{-1}\;Sq^{''}$$ -$$V(q^{'}+\sqrt{-1}\;q^{''})=Vq^{'}+\sqrt{-1}\;Vq^{''}$$ +$$S(q^{\prime}+\sqrt{-1}\;q^{\prime\prime})= +Sq^{\prime}+\sqrt{-1}\;Sq^{\prime\prime}$$ +$$V(q^{\prime}+\sqrt{-1}\;q^{\prime\prime})= +Vq^{\prime}+\sqrt{-1}\;Vq^{\prime\prime}$$ Hence (\S 103) -$$\{T(q^{'}+\sqrt{-1}\;q^{''})\}^2$$ +$$\{T(q^{\prime}+\sqrt{-1}\;q^{\prime\prime})\}^2$$ $$ -=(Sq^{'}+\sqrt{-1}\;Sq^{''}+Vq^{'}+\sqrt{-1}\;Vq^{''}) -(Sq^{'}+\sqrt{-1}\;Sq^{''}-Vq^{'}-\sqrt{-1}\;Vq^{''}) +=(Sq^{\prime}+\sqrt{-1}\;Sq^{\prime\prime}+ +Vq^{\prime}+\sqrt{-1}\;Vq^{\prime\prime}) +(Sq^{\prime}+\sqrt{-1}\;Sq^{\prime\prime}-Vq^{\prime}- +\sqrt{-1}\;Vq^{\prime\prime}) $$ -$$=(Sq^{'}+\sqrt{-1}\;Sq^{''})^2-(Vq^{'}+\sqrt{-1}\;Vq^{''})^2$$ -$$=(Tq^{'})^2-(Tq^{''})^2+2\sqrt{-1}\;S.q^{'}Kq^{''}$$ +$$=(Sq^{\prime}+\sqrt{-1}\;Sq^{\prime\prime})^2- +(Vq^{\prime}+\sqrt{-1}\;Vq^{\prime\prime})^2$$ +$$=(Tq^{\prime})^2-(Tq^{\prime\prime})^2+ +2\sqrt{-1}\;S.q^{\prime}Kq^{\prime\prime}$$ The only remark which need be made on such formulae is this, that {\sl the tensor of a biquaternion may vanish while both of the component quaternions are finite}. Thus, if -$$Tq^{'}=Tq^{''}$$ +$$Tq^{\prime}=Tq^{\prime\prime}$$ and -$$S.q^{'}Kq^{''}=0$$ +$$S.q^{\prime}Kq^{\prime\prime}=0$$ the above formula gives -$$T(q^{'}+\sqrt{-1}\;q^{''})=0$$ +$$T(q^{\prime}+\sqrt{-1}\;q^{\prime\prime})=0$$ The condition -$$S.q^{'}Kq^{''}=0$$ +$$S.q^{\prime}Kq^{\prime\prime}=0$$ may be written $$ -Kq^{''}=q^{'-1}\alpha,\;\;\;\textrm{ or }\;\;\; -q^{''}=-\alpha Kq^{'-1}=-\frac{\alpha q^{'}}{(Tq^{'})^2} +Kq^{\prime\prime}=q^{'-1}\alpha,\;\;\;\textrm{ or }\;\;\; +q^{\prime\prime}=-\alpha Kq^{'-1}=-\frac{\alpha q^{\prime}}{(Tq^{\prime})^2} $$ where $\alpha$ is any vector whatever. Hence -$$Tq^{'}=Tq^{''}=TKq^{''}=\frac{T\alpha}{Tq^{''}}$$ +$$Tq^{\prime}=Tq^{\prime\prime}=TKq^{\prime\prime}= +\frac{T\alpha}{Tq^{\prime\prime}}$$ and therefore $$ -Tq^{'}(Uq^{'}-\sqrt{-1}\;U\alpha . Uq^{'})= -(1-\sqrt{-1}\;U\alpha)q^{'} +Tq^{\prime}(Uq^{\prime}-\sqrt{-1}\;U\alpha . Uq^{\prime})= +(1-\sqrt{-1}\;U\alpha)q^{\prime} $$ is the general form of a biquaternion whose tensor is zero. -{\bf 132}. More generally we have, $q$, $r$, $q^{'}$, $r^{'}$ +{\bf 132}. More generally we have, $q$, $r$, $q^{\prime}$, $r^{\prime}$ being any four real and non-evanescent quaternions, $$ -(q+\sqrt{-1}\;q^{'})(r+\sqrt{-1}\;r^{'})= -qr-q^{'}r^{'}+\sqrt{-1}\;(qr^{'}+q^{'}r) +(q+\sqrt{-1}\;q^{\prime})(r+\sqrt{-1}\;r^{\prime})= +qr-q^{\prime}r^{\prime}+\sqrt{-1}\;(qr^{\prime}+q^{\prime}r) $$ That this product may vanish we must have -$$qr=q^{'}r^{'}$$ +$$qr=q^{\prime}r^{\prime}$$ and -$$qr^{'}=-q^{'}r$$ -Eliminating $r^{'}$ we have -$$qq^{'-1}qr=-q^{'}r$$ +$$qr^{\prime}=-q^{\prime}r$$ +Eliminating $r^{\prime}$ we have +$$qq^{'-1}qr=-q^{\prime}r$$ which gives $$(q^{'-1}q)^2=-1$$ i.e. -$$q=q^{'}\alpha$$ +$$q=q^{\prime}\alpha$$ where $\alpha$ is some unit-vector. And the two equations now agree in giving -$$-r=\alpha r^{'}$$ +$$-r=\alpha r^{\prime}$$ so that we have the biquaternion factors in the form -$$q^{'}(\alpha +\sqrt{-1})\;\;\;\textrm{ and }\;\;\;-(\alpha-\sqrt{-1})r^{'}$$ +$$q^{\prime}(\alpha +\sqrt{-1})\;\;\;\textrm{ and } +\;\;\;-(\alpha-\sqrt{-1})r^{\prime}$$ and their product is -$$-q^{'}(\alpha +\sqrt{-1})(\alpha -\sqrt{-1})r^{'}$$ +$$-q^{\prime}(\alpha +\sqrt{-1})(\alpha -\sqrt{-1})r^{\prime}$$ which, of course, vanishes. [A somewhat simpler investigation of the same proposition may be obtained by writing the biquaternions as $$ -q^{'}(q^{'-1}q+\sqrt{-1})\;\;\;\textrm{ and }\;\;\; -(rr^{'-1}+\sqrt{-1})r^{'} +q^{\prime}(q^{'-1}q+\sqrt{-1})\;\;\;\textrm{ and }\;\;\; +(rr^{'-1}+\sqrt{-1})r^{\prime} $$ or $$ -q^{'}(q^{''}+\sqrt{-1})\;\;\;\textrm{ and }\;\;\; -(r^{''}+\sqrt{-1})r^{'} +q^{\prime}(q^{\prime\prime}+\sqrt{-1})\;\;\;\textrm{ and }\;\;\; +(r^{\prime\prime}+\sqrt{-1})r^{\prime} $$ and showing that -$$q^{''}=-r^{''}=\alpha \;\;\;\textrm{ where }\;T\alpha=1]$$ +$$q^{\prime\prime}= +-r^{\prime\prime}=\alpha \;\;\;\textrm{ where }\;T\alpha=1]$$ From this it appears that if the product of two {\sl bivectors} $$ \rho+\sigma\sqrt{-1}\;\;\;\textrm{ and }\;\;\; -\rho^{'}+\sigma^{'}\sqrt{-1} +\rho^{\prime}+\sigma^{\prime}\sqrt{-1} $$ is zero, we must have -$$\sigma^{-1}\rho=-\rho^{'}\sigma^{'-1}=U\alpha$$ +$$\sigma^{-1}\rho=-\rho^{\prime}\sigma^{'-1}=U\alpha$$ where $\alpha$ may be any vector whatever. But this result is still more easily obtained by means of a direct process. @@ -7194,12 +7245,12 @@ $$ Show also that $$ \frac{\alpha+\beta}{\alpha-\beta}= -\frac{V\alpha\beta}{1+S\alpha\beta^{'}} +\frac{V\alpha\beta}{1+S\alpha\beta^{\prime}} $$ and $$ \frac{\alpha-\beta}{\alpha+\beta}= --\frac{V\alpha\beta}{1-S\alpha\beta^{'}} +-\frac{V\alpha\beta}{1-S\alpha\beta^{\prime}} $$ provided $\alpha$ and $\beta$ be unit-vectors. If these conditions are not fulfilled, what are the true values ? @@ -7298,12 +7349,12 @@ perpendicular vectors, can anything be predicated as to $\alpha_1$, $\beta_1$, $\gamma_1$? If $\alpha$, $\beta$, $\gamma$ be rectangular unit-vectors, what of $\alpha_1$, $\beta_1$, $\gamma_1$? -{\bf 12}. If $\alpha$, $\beta$, $\gamma$, $\alpha^{'}$, $\beta^{'}$, -$\gamma^{'}$ be two sets of rectangular unit-vectors, show that +{\bf 12}. If $\alpha$, $\beta$, $\gamma$, $\alpha^{\prime}$, $\beta^{\prime}$, +$\gamma^{\prime}$ be two sets of rectangular unit-vectors, show that $$ -S\alpha\alpha^{'}= -S\gamma\beta^{'}S\beta\gamma^{'}= -S\beta\beta^{'}S\gamma\gamma^{'}\;\;\;\textrm{\&c. \&c.} +S\alpha\alpha^{\prime}= +S\gamma\beta^{\prime}S\beta\gamma^{\prime}= +S\beta\beta^{\prime}S\gamma\gamma^{\prime}\;\;\;\textrm{\&c. \&c.} $$ {\bf 13}. The lines bisecting pairs of opposite sides of a quadrilateral @@ -7441,12 +7492,12 @@ V\beta\gamma S.\alpha\beta\gamma= \gamma(S\alpha\beta S\beta\gamma+S\alpha\gamma) $$ -{\bf 26}. If $i$, $j$, $k$, $i^{'}$, $j^{'}$, $k^{'}$, +{\bf 26}. If $i$, $j$, $k$, $i^{\prime}$, $j^{\prime}$, $k^{\prime}$, be two sets of rectangular unit-vectors, show that $$ \begin{array}{rcl} -S.Vii^{'}Vjj^{'}Vkk^{'}&=&(Sij^{'})^2-(Sji^{'})^2\\ - &=&(Sjk^{'})^2-(Skj^{'})^2=\textrm{\&c.} +S.Vii^{\prime}Vjj^{\prime}Vkk^{\prime}&=&(Sij^{\prime})^2-(Sji^{\prime})^2\\ + &=&(Sjk^{\prime})^2-(Skj^{\prime})^2=\textrm{\&c.} \end{array} $$ and find the values of the vector of the same product. @@ -7600,13 +7651,13 @@ i = \right] $$ -\chapter{Clifford Algebra\cite{Fl09}} +\chapter{Clifford Algebra \cite{Fl09}} -This is quoted from John Fletcher's web page\cite{Fl09} (with permission). +This is quoted from John Fletcher's web page \cite{Fl09} (with permission). The theory of Clifford Algebra includes a statement that each Clifford Algebra is isomorphic to a matrix representation. Several authors -discuss this and in particular Ablamowicz\cite{Ab98} gives examples of +discuss this and in particular Ablamowicz \cite{Ab98} gives examples of derivation of the matrix representation. A matrix will itself satisfy the characteristic polynomial equation obeyed by its own eigenvalues. This relationship can be used to calculate the inverse of @@ -7621,7 +7672,7 @@ Clifford(2), Clifford(3) and Clifford(2,2). Introductory texts on Clifford algebra state that for any chosen Clifford Algebra there is a matrix representation which is equivalent. Several authors discuss this in more detail and in particular, -Ablamowicz\cite{Ab98} shows that the matrices can be derived for each algebra +Ablamowicz \cite{Ab98} shows that the matrices can be derived for each algebra from a choice of idempotent, a member of the algebra which when squared gives itself. The idea of this paper is that any matrix obeys the characteristic equation of its own eigenvalues, and that therefore @@ -7636,7 +7687,7 @@ implementation. This knowledge is not believed to be new, but the theory is distributed in the literature and the purpose of this paper is to make it clear. The examples have been first developed using a system of symbolic algebra described in another paper by this -author\cite{Fl01}. +author \cite{Fl01}. \section{Clifford Basis Matrix Theory} @@ -8078,7 +8129,7 @@ simple cases of wide usefulness. \subsection{Example 3: Clifford (2,2)} -The following basis matrices are given by Ablamowicz\cite{Ab98} +The following basis matrices are given by Ablamowicz \cite{Ab98} \[ \begin{array}{cc} @@ -8328,7 +8379,7 @@ and \[n^{-1}_2 = \frac{n^3_2- 4n^2_2 + 8n_2 - 8}{4}\] This expression can be evaluated easily using a computer algebra -system for Clifford algebra such as described in Fletcher\cite{Fl01}. +system for Clifford algebra such as described in Fletcher \cite{Fl01}. The result is \[ @@ -8372,14 +8423,15 @@ It is well known that the most difficult part in constructing AG-code is the computation of a basis of the vector space ``L(D)'' where D is a divisor of the function field of an irreducible curve. To compute such a basis, PAFF used the Brill-Noether algorithm which was generalized -to any plane curve by D. LeBrigand and J.J. Risler (see [LR88] ). In [Ha96] +to any plane curve by D. LeBrigand and J.J. Risler \cite{LR88}. In +\cite{Ha96} you will find more details about the algorithmic aspect of the Brill-Noether algorithm. Also, if you prefer, as I do, a strictly -algebraic approach, see [Ha95]. This is the approach I used in my thesis -([Ha96]) and of course this is where you will find complete details about +algebraic approach, see \cite{Ha95}. This is the approach I used in my thesis +(\cite{Ha96}) and of course this is where you will find complete details about the implementation of the algorithm. The algebraic approach use the theory of algebraic function field in one variable : you will find in -[St93] a very good introduction to this theory and AG-codes. +\cite{St93} a very good introduction to this theory and AG-codes. It is important to notice that PAFF can be used for most computation related to the function field of an irreducible plane curve. For @@ -8392,7 +8444,7 @@ There is also the package PAFFFF which is especially designed to be used over finite fields. This package is essentially the same as PAFF, except that the computation are done over ``dynamic extensions'' of the ground field. For this, I used a simplify version of the notion of -dynamic algebraic closure as proposed by D. Duval (see [Du95]). +dynamic algebraic closure as proposed by D. Duval \cite{Du95}. Example 1 @@ -8442,187 +8494,176 @@ Gaussian Elimination Diophantine Equations \begin{thebibliography}{99} -\bibitem[Ab98]{Ab98} -Ablamowicz Rafal, ``Spinor Representations of Clifford -Algebras: A Symbolic Approach'', Computer Physics Communications + +\bibitem[Ablamowicz 98]{Ab98} Ablamowicz, Rafal\\ +``Spinor Representations of Clifford Algebras: A Symbolic Approach''\\ +Computer Physics Communications Vol. 115, No. 2-3, December 11, 1998, pages 510-535. -\bibitem[Alt05]{Alt05} -Altmann, Simon L. Rotations, Quaternions, and Double Groups + +\bibitem[Altmann 05]{Alt05} Altmann, Simon L.\\ +``Rotations, Quaternions, and Double Groups''\\ Dover Publications, Inc. 2005 ISBN 0-486-44518-6 -\bibitem[Ber95]{Ber95} -Laurent Bertrand. Computing a hyperelliptic integral using -arithmetic in the jacobian of the curve. {\sl Applicable Algebra in -Engineering, Communication and Computing}, 6:275-298, 1995 -\bibitem[Bro90]{Bro90} -M. 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Clay and Sons, Cambridge University Press Warehouse, Ave Maria Lane 1890 -\bibitem[Tr76]{Tr76} -Trager, Barry ``Algebraic factoring and rational function integration'' + +\bibitem[Trager 76]{Tr76} Trager, Barry\\ +``Algebraic factoring and rational function integration''\\ In {Proceedings of SYMSAC'76} pages 219-226, 1976 -\bibitem[Tr84]{Tr84} -Trager Barry {\sl On the integration of algebraic functions}, + +\bibitem[Trager 84]{Tr84} Trager, Barry\\ +``On the integration of algebraic functions''\\ PhD thesis, MIT, Computer Science, 1984 -\bibitem[vH94]{vH94} -M. van Hoeij. ``An algorithm for computing an integral -basis in an algebraic function field'' {\sl J. Symbolic Computation} -18(4):353-364, October 1994 -\bibitem [Lambov 06]{Lambov06} Lambov, Branimir\\ -``Interval Arithmetic Using SSE-2\\ + +\bibitem[van Hoeij 94]{vH94} van Hoeij, M.\\ +``An algorithm for computing an integral basis in an algebraic +function field''\\ +Journal of Symbolic Computation, 18(4) pp353-363 Oct. 1994 +CODEN JSYCEH ISSN 0747-7171 + +\bibitem[Lambov 06]{Lambov06} Lambov, Branimir\\ +``Interval Arithmetic Using SSE-2''\\ in Lecture Notes in Computer Science, Springer ISBN 978-3-540-85520-0 (2006) pp102-113 -\bibitem[Wa03]{Wa03} -Watt, Stephen, ``Aldor'', \verb|www.aldor.org| -\bibitem[We71]{We71} -Andr\'{e} Weil, {\sl Courbes alg\'{e}briques et -vari\'{e}t\'{e}s Abeliennes} Hermann, Paris, 1971 -\bibitem[Wo09]{Wo09} -Wolfram Research, \verb|mathworld.wolfram.com/Quaternion.html| -\bibitem[Yu76]{Yu76} -D.Y.Y. Yun. ``On square-free decomposition algorithms'' + +\bibitem[Weil 71]{We71} Weil, Andr\'{e}\\ +``Courbes alg\'{e}briques et vari\'{e}t\'{e}s Abeliennes''\\ +Hermann, Paris, 1971 + +\bibitem[Yun 76]{Yu76} Yun, D.Y.Y.\\ +``On square-free decomposition algorithms''\\ {\sl Proceedings of SYMSAC'76} pages 26-35, 1976 + \end{thebibliography} \chapter{Index} \printindex diff --git a/books/bookvolbib.pamphlet b/books/bookvolbib.pamphlet index 7992c6d..fd5d673 100644 --- a/books/bookvolbib.pamphlet +++ b/books/bookvolbib.pamphlet @@ -3176,7 +3176,9 @@ Wiley. (1980) ``An Algorithm for Generalized Matrix Eigenproblems''\\ SIAM J. Numer. Anal. 10 241--256. 1973 -\bibitem[Mulders 97]{Mul97} Mulders. Thom\\``A note on subresultants and a correction to the lazard/rioboo/trager formula in rational function integration''\\ +\bibitem[Mulders 97]{Mul97} Mulders. Thom\\ +``A note on subresultants and a correction to the lazard/rioboo/trager +formula in rational function integration''\\ {\sl Journal of Symbolic Computation}, 24(1):45-50, 1997 \bibitem[Munksgaard 80]{Mun80} Munksgaard N.\\ @@ -3545,7 +3547,7 @@ Mathematical Association of America. (1984) \subsection{T} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\bibitem[Tait 1890]{Ta1980} Tait, P.G.\\ +\bibitem[Tait 1890]{Ta1890} Tait, P.G.\\ ``An Elementary Treatise on Quaternions''\\ C.J. Clay and Sons, Cambridge University Press Warehouse, Ave Maria Lane 1890 diff --git a/changelog b/changelog index 5f980ac..a03c7b0 100644 --- a/changelog +++ b/changelog @@ -1,3 +1,7 @@ +20140719 tpd src/axiom-website/patches.html 20140719.01.tpd.patch +20150719 tpd books/bookvol10.1 clean up mistakes +20150719 tpd books/bookvolbib fix references +20150719 tpd bookheader.tex use underline, colored links 20140718 tpd src/axiom-website/patches.html 20140718.01.tpd.patch 20140718 tpd books/bookvolbib add Jeff97 20140715 tpd src/axiom-website/patches.html 20140715.01.tpd.patch diff --git a/patch b/patch index 80a0955..247adbc 100644 --- a/patch +++ b/patch @@ -1,5 +1,3 @@ -books/bookvolbib add Jeff97 +books/bookvol10.1, bookvolbib, bookheader.tex clean up mistakes -\bibitem[Jeffrey 97]{Jeff97} Jeffrey, D.J.; Rich, A.D.\\ -``Recursive integration of piecewise-continuous functions''\\ -Proc ISSAC'98 pp290-294 +There were several typos, mistakes, and bibliographic noise fixed. \ No newline at end of file diff --git a/src/axiom-website/patches.html b/src/axiom-website/patches.html index 3b64eb1..acbe192 100644 --- a/src/axiom-website/patches.html +++ b/src/axiom-website/patches.html @@ -4552,6 +4552,8 @@ books/bookvol10 absorb src/algebra/Makefile books/bookvolbib add Hoeij04 20140718.01.tpd.patch books/bookvolbib add Jeff97 +20140719.01.tpd.patch +books/bookvol10.1, bookvolbib, bookheader.tex clean up mistakes