Math & Science ⇒ beamer | Reducing horizontal Space in Equations

[latexhelp1]

I am having trouble with the horizontal spacing in equations. I would like to reduce the spacing within the equation so that it looks more natural.

Here's the first one:

Code: Select all

\begin{equation*}
  \int_s^\infty x(t) e^{-bp}\,dp 
\end{equation*}

There is too much space between the integral sign and the x in x(t).

Here's the second one:

Code: Select all

\begin{equation*}
  ln(z) =
  \tikz[baseline]{
    \node[anchor=base](t1){$g$};
  } 
  \tikz[baseline]{
    \node[anchor=base](t2){$(t(b), b)$};
  } 
\end{equation*}

There is too much space between the h and the ln(z) and the h and the (t(b, b)).

For a minimum working example that embeds these 2:

Code: Select all

    \documentclass[professionalfonts, smaller]{beamer} 
    \usetheme{CambridgeUS}
    \usepackage[latin1]{inputenc}
    \usefonttheme{professionalfonts}
    \usepackage{tikz}
    \usepackage{iwona}
    \usepackage{amsmath}
    \usepackage{verbatim}
    \usepackage[T1]{fontenc}
    \usepackage{mathtools}
    \usepackage{fix-cm}
    \usetikzlibrary{arrows,shapes}
    \DeclareMathOperator*{\Max}{Max}
    \newcommand\deriv{^{\mkern4mu\prime}}

    \begin{document}

    \tikzstyle{every picture}+=[remember picture]

    \everymath{\displaystyle}

    \begin{frame}
    \frametitle{Frame Title}

    \tikzstyle{na} = [baseline=-.5ex]

\begin{equation*}
    \int_s^\infty x(t) e^{-bp}\,dp 
     \end{equation*}

    \vspace{0.2cm}

     \begin{equation*}
           ln(z) =
            \tikz[baseline]{
             \node[anchor=base](t1){$g$};
            } 
            \tikz[baseline]{
               \node[anchor=base](t2){$(t(b), b)$};
          } 
    \end{equation*}

[Stefan Kottwitz]

1. You could use limits above and below, instead of right of the expression:
   

   Code: Select all

   \int\limits_s^\infty x(t) e^{-bp}\,dp

2. Try

   Code: Select all

   \begin{equation*}
        \ln(z) =
         \tikz[overlay, remember picture, baseline=(t1.base)]{
           \node(t1){$g$};
         }
         \tikz[overlay, remember picture, baseline=(t2.base)]{
            \node[anchor=west, right of=t1, node distance=20pt](t2){$(t(b), b)$};
         }
   \end{equation*}

equations.png (6 KiB) Viewed 6531 times

Stefan

[latexhelp1]

Stefan,

That works very well. I had to make a slight change though to the code. Otherwise, it was changing the position of the equation on the slide with every time I compiled it.

Code: Select all

        \documentclass[professionalfonts, smaller]{beamer}
        \usetheme{CambridgeUS}
        \usepackage[latin1]{inputenc}
        \usefonttheme{professionalfonts}
        \usepackage{tikz}
        \usepackage{iwona}
        \usepackage{amsmath}
        \usepackage{verbatim}
        \usepackage[T1]{fontenc}
        \usepackage{mathtools}
        \usepackage{fix-cm}
        \usetikzlibrary{arrows,shapes}
        \DeclareMathOperator*{\Max}{Max}
        \newcommand\deriv{^{\mkern4mu\prime}}

        \begin{document}

        \tikzstyle{every picture}+=[remember picture]

        \everymath{\displaystyle}

        \begin{frame}
        \frametitle{Frame Title}

        \tikzstyle{na} = [baseline=-.5ex]

    \begin{equation*}
        \int_s^\infty x(t) e^{-bp}\,dp
         \end{equation*}

        \vspace{0.2cm}

         \begin{equation*}
               ln(z) =
                \tikz[overlay, remember picture, baseline]{
                 \node[anchor=base](t1){$g$};
                }
                \tikz[overlay, remember picture, baseline]{
                   \node[anchor=base, right of=t1, node distance20pt](t2){$(t(b), b)$};
              }
        \end{equation*}

However, when I try that to a different slide where I use the \Max command that you wrote for me earlier, the \Max command ends up not being even with the rest of the equation:

Code: Select all

 \begin{equation*}
            \hspace{-4.5cm} \vspace{2cm}
            \tikz[overlay, remember picture, baseline]{
                 \node[anchor=base](t1){$\Max\limits_x$};
             } \vspace{-2cm}
            \tikz[overlay, remember picture, baseline]{
             \node[anchor=base, right of=t1, node distance=30pt](t2){$L(x) =$};
            } 
            \tikz[overlay, remember picture, baseline]{
                \node[anchor=base, right of=t2, node distance=35pt](t3){$p(x) +$};
            } 
            \tikz[overlay, remember picture, baseline]{
                \node[anchor=base, right of=t3, node distance=29pt](t4){$\gamma d(x)$};
            }
    \end{equation*}

I would very much appreciate any help you can provide in straightening out the Max with respect to the rest of the equation.

1. You could use limits above and below, instead of right of the expression:
   

   Code: Select all

   \int\limits_s^\infty x(t) e^{-bp}\,dp

2. Try

   Code: Select all

   \begin{equation*}
        \ln(z) =
         \tikz[overlay, remember picture, baseline=(t1.base)]{
           \node(t1){$g$};
         }
         \tikz[overlay, remember picture, baseline=(t2.base)]{
            \node[anchor=west, right of=t1, node distance=20pt](t2){$(t(b), b)$};
         }
   \end{equation*}

equations.png

Stefan

[Stefan Kottwitz]

Specify the base line of the tikzpicture environment to be the baseline of the corresponding node:

Code: Select all

\begin{equation*}
            \hspace{-4.5cm} \vspace{2cm}
            \tikz[overlay, remember picture, baseline=(t1.base)]{
                 \node(t1){$\Max\limits_x$};
             } %\vspace{-2cm}
            \tikz[overlay, remember picture, baseline=(t2.base)]{
             \node[anchor=base, right of=t1, node distance=30pt](t2){$L(x) =$};
            }
            \tikz[overlay, remember picture, baseline=(t3.base))]{
                \node[anchor=base, right of=t2, node distance=35pt](t3){$p(x) +$};
            }
            \tikz[overlay, remember picture, baseline=(t4.base)]{
                \node[anchor=base, right of=t3, node distance=29pt](t4){$\gamma d(x)$};
            }
    \end{equation*}

baselines.png (2.49 KiB) Viewed 6518 times

Stefan

[latexhelp1]

Adding baseline = (t1.base) for each line instead of just baseline fixes the spacing issue. However, every time I compile the file, it shifts the position of the entire equation. So, if I compile it say 40 times, it looks grossly different than when I first compiled it. This problem goes away if it is just baseline.

Specify the base line of the tikzpicture environment to be the baseline of the corresponding node:

Code: Select all

\begin{equation*}
            \hspace{-4.5cm} \vspace{2cm}
            \tikz[overlay, remember picture, baseline=(t1.base)]{
                 \node(t1){$\Max\limits_x$};
             } %\vspace{-2cm}
            \tikz[overlay, remember picture, baseline=(t2.base)]{
             \node[anchor=base, right of=t1, node distance=30pt](t2){$L(x) =$};
            }
            \tikz[overlay, remember picture, baseline=(t3.base))]{
                \node[anchor=base, right of=t2, node distance=35pt](t3){$p(x) +$};
            }
            \tikz[overlay, remember picture, baseline=(t4.base)]{
                \node[anchor=base, right of=t3, node distance=29pt](t4){$\gamma d(x)$};
            }
    \end{equation*}

baselines.png

Stefan

[Stefan Kottwitz]

However, every time I compile the file, it shifts the position of the entire equation. So, if I compile it say 40 times, it looks grossly different than when I first compiled it.

You can fix the position of the equation by absolutely positioning the reference node, the first one, referring to the current page node:

Code: Select all

\node(t1) at ([xshift=.4\textwidth, yshift=-.5\textheight]
    current page.north west) {$\Max\limits_x$};

Complete example with your code:

Code: Select all

\documentclass[professionalfonts, smaller]{beamer}
\usetheme{CambridgeUS}
\usepackage[latin1]{inputenc}
\usefonttheme{professionalfonts}
\usepackage{tikz}
\usepackage{iwona}
\usepackage{amsmath}
\usepackage{verbatim}
\usepackage[T1]{fontenc}
\usepackage{mathtools}
\usepackage{fix-cm}
\usetikzlibrary{arrows,shapes}
\DeclareMathOperator*{\Max}{Max}
\newcommand\deriv{^{\mkern4mu\prime}}
\begin{document}
\begin{frame}
    \begin{equation*}
            \hspace{-4.5cm} \vspace{2cm}
            \tikz[overlay, remember picture, baseline=(t1.base)]{
                 \node(t1) at ([xshift=.4\textwidth, yshift=-.5\textheight]
                    current page.north west) {$\Max\limits_x$};
             } %\vspace{-2cm}
            \tikz[overlay, remember picture, baseline=(t2.base)]{
             \node[anchor=base, right of=t1, node distance=30pt](t2){$L(x) =$};
            }
            \tikz[overlay, remember picture, baseline=(t3.base))]{
                \node[anchor=base, right of=t2, node distance=35pt](t3){$p(x) +$};
            }
            \tikz[overlay, remember picture, baseline=(t4.base)]{
                \node[anchor=base, right of=t3, node distance=29pt](t4){$\gamma d(x)$};
            }
    \end{equation*}
\end{frame}
\end{document}

Stefan