How do I center a \Large equation in display mode?

Question

I have an equation in display mode and would like to make it bigger using \Large, but the equation runs off the right side of the page while there is much room on the left side of the page.

The equation would fit if I could just center it.

How can I do this?

\documentclass[12pt, landscape]{article}

\usepackage{amsmath}
\begin{document}

 {\Large$$\sum_{t_1=1}^{n}\binom{n}{t_1}+
 \sum_{t_1=1}^{n-1}\sum_{t_2=1}^{n-1}\binom{n}{t_1,t_2}+\cdots+
 \sum_{t_1=1}^{1}\sum_{t_2=1}^{1}\cdots\sum_{t_n=1}^{1}
  \binom{n}{t_1,t_2, \dots, t_n}$$}

\end{document}

Answer 1

I advise against using \Large equations, particularly if this creates too wide an object.

However, here's a possible solution that doesn't mess up the vertical spacing:

\documentclass[12pt]{article}
\usepackage{amsmath}

\usepackage{lipsum}

\begin{document}

\lipsum*[2]
\[
\noindent\begin{minipage}{\textwidth}
\Large
\[
\hidewidth
\sum_{t_1=1}^{n}\binom{n}{t_1}+
 \sum_{t_1=1}^{n-1}\sum_{t_2=1}^{n-1}\binom{n}{t_1,t_2}+\dots+
 \sum_{t_1=1}^{1}\sum_{t_2=1}^{1}\cdots\sum_{t_n=1}^{1}
  \binom{n}{t_1,t_2, \dots, t_n}
\hidewidth
\]
\end{minipage}
\]
\lipsum[3]

\end{document}

Answer 2

Equations centre by default:

\documentclass{article}  


\begin{document}

\noindent X\dotfill X

aaaaa
%
{\large
\[
a+b+c+d=
a+b+c+d=
a+b+c+d=
a+b+c+d=
a+b+c+d
\]}


\end{document}

Answer 3

A small suggestion to start with: Don't use \cdots; since you're using the amsmath package, just use \dots and let the software work out which type of typographic ellipses are best.

If you want to typeset the equation in portrait mode at 12pt and a \Large relative font size, and if you're working with an "A4" paper size, you'll need to set the margins to 2cm.

\documentclass[12pt]{article}
\usepackage[a4paper,margin=2cm]{geometry}
\usepackage{amsmath}
\begin{document}
{\Large
\[
\sum_{t_1=1}^{n}\binom{n}{t_1}+
\sum_{t_1=1}^{n-1}\sum_{t_2=1}^{n-1}
  \binom{n}{t_1,t_2}+\dots+
\sum_{t_1=1}^{1}\sum_{t_2=1}^{1}\dots\sum_{t_n=1}^{1}
  \binom{n}{t_1,t_2, \dots, t_n}
\]
}
\end{document}

Answer 4

You have more possibilities:

• if equation is not numbered, than try:

\begin{center}\Large $\displaystyle a^2 + b^2 = c^2 \end{center}

• if equation is numbered, than try:

\begin{equation}\Large \hspace*{-< x em>} a^2 + b^2 = c^2 \end{equation}

where < x em> you select so, that equation will be centered. Both suggestion are not tested, since you doesn't provide minimal working example, which would show your problem.

Answer 5

Alright.

Displayed equations are always centred. The exact middle of the equation shall be exactly half way between the margins if the equation is smaller than the margins. So, you can imagine, a centred equation will expand outwards until it comes up against the margins (I doubt this is precisely how TeX places its centred material, but obviously when you centre something, the centre of that thing should be mid-way between the two margins and the longer the equation, the closer the left- and rightmost ends of the equation shall be to the margins.) This is what TeX does. Thus:

\documentclass[12pt, landscape]{article}
\usepackage[showframe,margin=2in]{geometry}
\geometry{a4paper}
\usepackage{amsmath}

\begin{document}

\[
  x
\]
\[
  x + y
\]
\[
  \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} \psi -
  \nabla^{2} \psi + \frac{m^{2}c^{2}}{\hbar^{2}} \psi = 0
\]
\[
  \overbrace{\int_{y_{1}}^{y_{2}} x(y) \, dy}^{A_{1}} +
  \overbrace{\int_{x_{1}}^{x_{2}} y(x) \, dx}^{A_{2}} = x.y(x)
  \biggr|_{x_{1}}^{x_{2}} = y.x(y) \biggr|_{y_{1}}^{y_{2}}
\]

\end{document}

If your equation is as wide as the space between the margins, obviously, it will fill up the entire space. If your equation is wider than the margins, then the leftmost end of the equation will be forced up against the margin and TeX will have no choice but to print the rightmost end of the equation in the right hand margin and eventually off the page.

Adding:

\[
  \int \frac{dx}{x^{2^{n}} + 1} = \sum_{k = 1}^{2^{n - 1}}
  \left\{\frac{1}{2^{n - 1}} \left[\sin\left(\frac{(2k -
          1)\pi}{2^{n}}\right) \arctan \left[\left(x -
          \cos\left(\frac{(2k - 1)\pi}{2^{n}}\right)\right) \csc
        \left( \frac{(2k - 1)\pi}{2^{n}}\right)\right]\right] -
    \frac{1}{2^{n}} \left[\cos \left(\frac{(2k - 1)\pi}{2^{n}}\right)
      \ln \left| x^{2} - 2x \cos \left(\frac{(2k -
            1)\pi}{2^{n}}\right) + 1\right| \right]\right\}
\]

TeX doesn't centre content on the page, it centres it between the margins, between the margins is where the text goes. So, if your equation is longer than the space between the margins, you have to either modify the margins, or the equation.

It's not that the final equation isn't centred. The left-margin is the limit. Because the equation is too long, TeX has been forced to print the excess in the margin and off the page. If I have a 5 metre long garage and a 4.2 metre long car, I can park my car in the centre of my garage. The back of the car will be 40 cm from the back of the garage and the front of the car will be 40 cm from the front of the garage. If I have a 5 metre long car, the back of the car will be up against the back wall and the front of the car will be up against the door. If I have a 6 metre long car, the front of the car is going to stick out of the front of the garage. That's what's happened to you.

So:

---

Consider

\usepackage[margin=2in]{geometry}
\geometry{a4paper}

With \usepackage{amsmath} and \begin{document} & \end{document} added to your MWE:

\documentclass[12pt, landscape]{article}
\usepackage[margin=2in]{geometry}
\geometry{a4paper}
\usepackage{amsmath}

\begin{document}

{\Large
  $$\sum_{t_{1} = 1}^{n} \binom{n}{t_{1}} + \sum_{t_{1} = 1}^{n-1}
  \sum_{t_{2} = 1}^{n-1} \binom{n}{t_{1},t_{2}} + \cdots + \sum_{t_{1} =
    1}^{1} \sum_{t_{2} = 1}^{1} \cdots \sum_{t_{n} = 1}^{1}
  \binom{n}{t_{1},t_{2}, \dots, t_{n}}$$
}

\end{document}

Also try \[ ... \] instead of $$ ... $$, see: Why is \[ ... \] preferable to $$ ... $$?