Text Formatting ⇒ Equation numbering

[psw1937]

How do you get equations numbers of the form "page number.equation number”. Is there a way to set things up so the equation command numbers this way automatically?

Thanks

[rais]

How do you get equations numbers of the form "page number.equation number”.

Isn't that what amsmath's \numberwithin command is for? See `texdoc amsldoc', section `Equation numbering'.

KR
Rainer

[psw1937]

Hi Rainer

Yes it does do what I wanted. I was told this just after having posted my question. But you have to be careful with \numberwithin because it doesn't always number equations that are at the top of a page correctly. Sometimes it carries the numbering over from the previous page and other times it starts to number from zero rather than 1.

[Stefan Kottwitz]

Hi,

Sometimes it carries the numbering over from the previous page and other times it starts to number from zero rather than 1.

interesting, do you have examples for this?

Stefan

[psw1937]

Yes. I have had both occur in the same document. Unfortunately its a 200 page document and finding the bit of code that causes the problem is difficult. I have code that numbers the first equation on the top of the page 0. The equation at the top of page 2 is numbered 2.0, or at least it was when I ran this code. I'll keep looking for the example of other problem.

Code: Select all

Code, edit and compile here:

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\documentclass[UKenglish,11pt,a4paper]{book}
\usepackage{babel,pstricks,pst-plot,longtable,url,graphicx,setspace}
\usepackage{anyfontsize,fancyhdr,amssymb,amsmath}
\pagestyle{fancy}
\fancyhead{}
\fancyfoot{}
\renewcommand{\headrulewidth}{0pt}
\renewcommand{\headheight}{13.6pt}
%\usepackage{makeidx}
\newcounter{pic}[page]
\newcounter{fig}[page]
\setcounter{secnumdepth}{4}
\newtheorem{ption}{Proposition}
\newtheorem{assum}{Assumption}
\newtheorem{them}{Theorem}
\newtheorem{lem}{Lemma}
\newcounter{thm}[page]
\numberwithin{equation}{page}
\begin{document}
substituting in the value for $\pi^{\star\star\star}$ gives,\footnote{This follows from the fact that\begin{eqnarray*}\Delta S^{\star\star\star}&=&\pi^{\star\star\star} (v-c)-\frac{(\pi^{\star\star\star})^2}{2}\\
&=&\pi^{\star\star\star}\left(v-c-\frac{1}{2}\pi^{\star\star\star}\right)\label{eqn1}\\
&=&\pi^{\star\star\star}\left(v-c-\frac{1}{2}v+\frac{1}{4}c\right)\\
&=&\pi^{\star\star\star}\left(\frac{1}{2}v-\frac{3}{4}c\right)\\
&=&\left(v-\frac{1}{2}c\right)\left(\frac{1}{2}v-\frac{3}{4}c\right)\\
&=&\frac{1}{2}v^2-\frac{3}{4}vc-\frac{1}{4}vc+\frac{3}{8}c^2\\
&=&\frac{1}{2}v^2-vc+\frac{3}{8}c^2.
\end{eqnarray*}$\Delta S^\star$ can be written\begin{eqnarray*}\Delta S^\star&=&\frac{1}{2}(v-c)^2\\
&=&\frac{1}{2}(v^2-2vc+c^2)\\
&=&\frac{1}{2}v^2-vc+\frac{1}{2}c^2~~~~~~>\Delta S^{\star\star\star}\end{eqnarray*}}
\begin{eqnarray*}\Delta S^{\star\star\star}&=&\frac{1}{2}v^2-vc+\frac{3}{8}c^2~~~~~~<\Delta S^\star\end{eqnarray*}
Note that if $v^2\geq 2c$ then $\Delta S^{\star\star\star}\geq \Delta S^{\star\star}$.\footnote{To see this note that\begin{eqnarray*}\Delta S^{\star\star\star}&\geq&\Delta S^{\star\star}\\
\Rightarrow \frac{1}{2}v^2-vc+\frac{3}{8}c^2&\geq&\frac{3}{8}(v-c)^2\\
\Rightarrow \frac{1}{2}v^2-vc+\frac{3}{8}c^2&\geq&\frac{3}{8}(v^2-2vc+c^2)\\
\Rightarrow \frac{1}{2}v^2-vc+\frac{3}{8}c^2&\geq&\frac{3}{8}v^2-\frac{6}{8}vc+\frac{3}{8}c^2\\
\Rightarrow \frac{1}{2}v^2-vc&\geq&\frac{3}{8}v^2-\frac{6}{8}vc\\
\Rightarrow \frac{1}{2}(v^2-2vc)&\geq&\frac{3}{8}(v^2-2vc)\\
\mathrm{which~is~true~when~v^2}&\geq&2vc.\end{eqnarray*}}
This means that if the gains that result from the improvement in quality are large enough Buyer control is preferred to both Seller control or having separate firms. The advantage of Buyer control is that it encourages investment by the Buyer. But as he doesn't take into account the costs, $c$, the Buyer overinvests relative to the first best outcome.%\newpage\setcounter{equation}{1}
In the special case of $v^2\geq 2vc$ the Buyer will be willing, and able, to pay the Seller a fixed amount at time A to obtain the right to make the decision about the improvement. That amounts to the Buyer buying the Seller's firm. The Seller will be able to demand amount equal to the difference in his expected profits, $\Delta S_S^{\star\star}-\Delta S_S^{\star\star\star}$, in return for letting the Buyer make the improvement decision.
 
 
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[psw1937]

Hi Stefan

I have found the case where the top equation on the second page is numbered following on from the numbering of the first page. In this case the top equation on page 2 is numbered 1.3 while the second equation on page 2 is numbered 2.1.

Code: Select all

Code, edit and compile here:

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\documentclass[UKenglish,11pt,a4paper]{book}
%\usepackage[colorlinks,urlcolor=blue]{hyperref}
\usepackage{babel,pstricks,pst-plot,longtable,url,graphicx,setspace}
\usepackage{anyfontsize,fancyhdr,amssymb,amsmath}
\pagestyle{fancy}
\fancyhead{}
\fancyfoot{}
\renewcommand{\headrulewidth}{0pt}
\renewcommand{\headheight}{13.6pt}
%\usepackage{makeidx}
\newcounter{pic}[page]
\newcounter{fig}[page]
\setcounter{secnumdepth}{4}
\newtheorem{ption}{Proposition}
\newtheorem{assum}{Assumption}
\newtheorem{them}{Theorem}
\newtheorem{lem}{Lemma}
%\newcounter{eqnn}[page]
\newcounter{thm}[page]
\numberwithin{equation}{page}
%\makeindex
\begin{document}
Now let $i$ be an/the member for whom the gain in utility (reduction in disutility) from changing his effort level is the smallest, that is, \[v_i(a_i^\star)-v_i(a_i(\varepsilon))\leq v_j(a^\star_j)-v_j(a_j(\varepsilon))~~~~~~~~~\mathrm{for~all~}j.\]Summing over $j$ gives\begin{eqnarray}n(v_i(a_i^\star)-v_i(a_i(\varepsilon)))&\leq&\sum_{j=1}^n v_j(a^\star_j)-v_j(a_j(\varepsilon))\leq x^\star- (x^\star-\varepsilon)\nonumber\\
\Rightarrow \label{eqn8}x^\star -(x^\star -\varepsilon)&\geq &n(v_i(a_i^\star)-v_i(a_i(\varepsilon)))> 0.\end{eqnarray}(Note that $i$ may depend on $\varepsilon$ but the notion used here does not reflect this.)
Using Taylor's Theorem\footnote{Young's Form of Taylor’s Theorem: Let $f:(\alpha,\beta) \rightarrow \Re$ be $n-1$ times continuously differentiable on $(\alpha,\beta)$ and assume that $f$ has an $n^{th}$ derivative at the point $x$ in $(\alpha,\beta)$. For any $v$ such that $x + v$ belongs to $(\alpha,\beta)$,
\[f(x+v)=f(x)+\sum^n_{i=1}\frac{f^k(x)}{k!}v^k +\frac{r(v)}{n!}v^n\]
where the remainder term $r(v)$ satisfies \[\lim_{v\rightarrow 0}r(v)=0.\] Here we have $f(\cdot)=v(\cdot), k=1,x+v=x_i(\varepsilon), v=(a_i(\varepsilon)-a^\star_i)<0$ and $x=a^\star_i$.
Substituting in to the formula above we get
\begin{eqnarray*}v_i(a_i(\varepsilon))&=&v_i(a^\star_i)+\frac{v^{'}_i(a^\star_i)}{1!}(a_i(\varepsilon)-a^\star_i) +\hat{r_i}(\varepsilon)\\
\Rightarrow -v^{'}_i(a^\star_i)(a_i(\varepsilon)-a^\star_i)-\hat{r_i}(\varepsilon)&=&v_i(a^\star_i)-v_i(a_i(\varepsilon))\\
\Rightarrow v_i(a^\star_i)-v_i(a_i(\varepsilon))&=&v^{'}_i(a^\star_i)(a^\star_i-a_i(\varepsilon))-\hat{r_i}(\varepsilon)
\end{eqnarray*}
} (or the definition of a derivative) we get
\begin{eqnarray}\label{eqn9}v_i(a^\star_i)-v_i(a_i(\varepsilon))&=&v^{'}_i(a^\star_i)(a^\star_i-a_i(\varepsilon))-\hat{r_i}(\varepsilon)\\
&&~~~~~\mathrm{where}~~\frac{\hat{r}_i(\varepsilon)}{(a^\star -a_i(\varepsilon))}\rightarrow 0~\mathrm{as~}\varepsilon \rightarrow 0.\nonumber\end{eqnarray}%\newpage\noindent
Similarly,\footnote{
Here we have $f(\cdot)=v(\cdot), k=1,x+v=x(a_{-i}^\star,a_i(\varepsilon)), v=(a_i(\varepsilon)-a^\star_i)<0$ and $x=a^\star$.
Substituting in to the formula above we get
\begin{eqnarray*}x(a_{-i}^\star,a_i(\varepsilon))&=&x(a^\star_i)+\frac{\frac{\partial x(a^\star)}{\partial a_i}}{1!}(a_i(\varepsilon)-a^\star_i) +\tilde{r_i}(\varepsilon)\\
\Rightarrow -\frac{\partial x(a^\star)}{\partial a_i}(a_i(\varepsilon)-a^\star_i)-\tilde{r_i}(\varepsilon)&=&x(a^\star_i)-x(a_{-i}^\star,a_i(\varepsilon))\\
 
 
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[rais]

I have found the case where the top equation on the second page is numbered following on from the numbering of the first page. In this case the top equation on page 2 is numbered 1.3 while the second equation on page 2 is numbered 2.1.

Sorry if I misled you, I didn't keep in mind that the page content is still under construction when \thepage is called...
Heiko's zref package might just do the trick:

Code: Select all

Code, edit and compile here:

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\documentclass[UKenglish,11pt,a4paper]{book}
\usepackage{babel,amsmath}
\usepackage{blindtext}
\usepackage{zref-perpage}
%\numberwithin{equation}{page}
\zmakeperpage{equation}
\renewcommand*\theequation{\thezpage.\arabic{equation}}
\begin{document}
\blindtext[5]
\begin{equation}
y=f(x)
\end{equation}
\end{document}
 
 
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Run LaTeX here

KR
Rainer

[psw1937]

I have found the case where the top equation on the second page is numbered following on from the numbering of the first page. In this case the top equation on page 2 is numbered 1.3 while the second equation on page 2 is numbered 2.1.

Sorry if I misled you, I didn't keep in mind that the page content is still under construction when \thepage is called...
Heiko's zref package might just do the trick:

Thanks Rainer. The way I am doing it now is with the "perpage" package and the commands

\MakePerPage{equation}
\renewcommand{\theequation}{\theperpage.\arabic{equation}}

I guess these two packages will be doing the same thing.

Paul.