3
\begin{align}
\varphi_m(v_i)&= \left\lceil \frac{2i+3m}{4m}\right\rceil, &&\text{for $i=1,2,\dots, n$,}\\
$$\varphi_m(u_i)&= \begin {cases}
\left\lceil \frac{i}{2m}\right\rceil, &&\text{for $i &\equiv 0, \frac {m} {2} \mod 2m \  $,}\\
\left\lceil \frac{i}{m+1}\right\rceil, &&\text{for $i &\not\equiv 0, \frac {m} {2} \mod 2m \  $,}\\
\end{cases}$$
$$\dim(G_{4,6}^{(k)})=\dim(K_4^{(k)}) = \begin{cases}
2 & \text{for $2\le k\equiv 0 \ ({\rm{mod}}\ 2)$,}\\
3 & \text{for $k=0$ or $1\le k\equiv 1\ ({\rm{mod}}\ 2)$.}
\end{cases}$$
\varphi_4(v_iv_{i+1})&= \left\lceil  \frac{i+1}{8}\right\rceil, &&\text{for $i=1,2,\dots, n-1$,}\\
\varphi_4(u_iu_{i+1})&= \left\lceil  \frac{i}{8}\right\rceil, &&\text{for $i=1,2,\dots, n-1$,}\\
\varphi_4(v_iu_i)&= \left\lceil  \frac{i+4}{8}\right\rceil, &&\text{for $i=1,2,\dots, n$,}
\end{align}
1
  • 3
    You shouldn't use an alignment character inside \text, nor should you use $$...$$ inside align (or anywhere...).
    – Werner
    Feb 7 '17 at 7:01
8

You asked,

This code gives the error “missing } inserted”. What is this error?

As a general rule, once sufficiently many syntax errors have piled up, and especially if the syntax errors interact with each other, TeX's error messages can become uninformative. Unfortunately, your code contains a disturbing number of syntax errors. For sure, don't ever use $$ when already inside a math environment. Your code also features several instances of either surplus or missing & (alignment character) and $ (inline math initiator/terminator) characters.

After fixing these issues, this is what I came up with -- is this what you're looking for?

enter image description here

\documentclass{article}
\usepackage[letterpaper,margin=1in]{geometry} % choose page settings
\usepackage{mathtools}
\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}
\begin{document} 

\begin{align}
\varphi_m(v_i)  &= \ceil*{\frac{2i+3m}{4m}}, &&\text{for $i=1,2,\dots, n$,}\\
\varphi_m(u_i)  &= 
   \begin {dcases}
      \ceil*{\frac{i}{2m}},  & \text{for $i\equiv 0,\frac{m}{2} \mod 2m$,}\\
      \ceil*{\frac{i}{m+1}}, & \text{for $i\not\equiv 0,\frac{m}{2} \mod 2m$,}
   \end{dcases} \\
\dim(G_{4,6}^{(k)})  &=\dim(K_4^{(k)}) = 
   \begin{cases}
      2 & \text{for $2\le k\equiv 0 \ ({\rm{mod}}\ 2)$,}\\
      3 & \text{for $k=0$ or $1\le k\equiv 1\ ({\rm{mod}}\ 2)$.}
   \end{cases} \\
\varphi_4(v_iv_{i+1})&= \ceil*{\frac{i+1}{8}}, &&\text{for $i=1,2,\dots, n-1$,}\\
\varphi_4(u_iu_{i+1})&= \ceil*{\frac{i}{8}},   &&\text{for $i=1,2,\dots, n-1$,}\\
\varphi_4(v_iu_i)    &= \ceil*{\frac{i+4}{8}}, &&\text{for $i=1,2,\dots, n$,}
\end{align}

\end{document}

Addendum: You may also want to replace all instances of the dreadful-looking \ ({\rm{mod}}\ 2) with \pmod{2}. For the sake of consistency, you should also write \pmod{2m} in one of the other equations. For a more compact and probably more readable look, consider replacing all instances of && spacers with \qquad. Reducing the vertical size of the "ceiling" brackets may also improve the look of the equations.

enter image description here

\documentclass{article}
\usepackage[letterpaper,margin=1in]{geometry} % text block settings
\usepackage{mathtools}
\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}
\begin{document} 
\begin{align}
\varphi_m(v_i)&= \ceil[\Big]{\frac{2i+3m}{4m}}, \qquad \text{for $i=1,2,\dots, n$,}\\
\varphi_m(u_i)&= 
   \begin {dcases}
      \ceil[\Big]{\frac{i}{2m}},  & \text{for $i\equiv 0,m/2\pmod{2m}$,}\\
      \ceil[\Big]{\frac{i}{m+1}}, & \text{for $i\not\equiv 0,m/2\pmod{2m}$,}
   \end{dcases} \\
\dim(G_{4,6}^{(k)})=\dim(K_4^{(k)}) &= 
   \begin{cases}
      2, & \text{for $2\le k\equiv 0 \pmod{2}$,}\\
      3, & \text{for $k=0$ or $1\le k\equiv 1 \pmod{2}$.}
   \end{cases} \\
\varphi_4(v_iv_{i+1})&= \ceil[\Big]{\frac{i+1}{8}}, \qquad\text{for $i=1,2,\dots,n-1$,}\\
\varphi_4(u_iu_{i+1})&= \ceil[\Big]{\frac{i}{8}},   \qquad\text{for $i=1,2,\dots,n-1$,}\\
\varphi_4(v_iu_i)    &= \ceil[\Big]{\frac{i+4}{8}}, \qquad\text{for $i=1,2,\dots,n$.}
\end{align}
\end{document}
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  • 1
    It should probably be \pmod{2m} and also \pmod{2} for consistency. If it is to be interpreted as a binary operation symbol, it should be \bmod.
    – egreg
    Feb 7 '17 at 9:11
  • @egreg - Thanks for this. (I am more than happy to defer to you on issues of mathematical notation!) I already corrected the \pmod2 business in the addendum; I'll apply the same treatment to the 2m items.
    – Mico
    Feb 7 '17 at 9:22
  • 1
    I insist: teaching \pmod 2 is wrong, because it doesn't underline that \pmod takes an argument. This can lead astray the beginner who's trying to typeset congruences modulo 11.
    – egreg
    Feb 7 '17 at 9:34
  • @egreg -- Sorry, I completely missed that. I'll update the code right away.
    – Mico
    Feb 7 '17 at 9:35

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