Problem with amsmath's `gather` environment

Question

I have a lot of math to write so I tried to use align inside gather. When the column was over I was getting badboxes. So I used two gather environments and the output is weird. What is happening and why? How can I fix that?

\documentclass[12pt]{article}
\usepackage[top=0.3in, bottom=1.2in, left=0.8in, right=0.8in]{geometry}

\usepackage{multicol}

\usepackage[utf8]{inputenc}

\setlength{\parindent}{0cm}

\usepackage{setspace}

\usepackage{xltxtra}
\usepackage{xgreek}
\setmainfont[Mapping=tex-text]{GFSDidot.otf}
\setsansfont[Mapping=tex-text]{GFSDidot.otf}

\usepackage[fleqn]{amsmath}
\usepackage{unicode-math}

\setlength{\mathindent}{0cm}



\newcommand{\3}{\vspace{0.3cm}}

\title{}
\author{}
\date{}

\begin{document}

\begin{multicols*}{2}

\begin{gather*}
\begin{aligned}
&\text{70}\\
&e^{jz}=\cos z+j\sin z\\
&\cos z=(1/2)(2\cos z)=\\
&=(1/2)(2\cos z+j\sin z-j\sin z)=\\
&=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\
&=(1/2)(e^{jz}+e^{-jz})
\end{aligned}\\
\begin{aligned}
&\text{207}\\
&x(t)=A_{c}\cos \theta(t)\\
&\theta(t)=2\pi f_{c}t+\phi(t)\\
&\phi(t)=K_{p}m(t)\\
&x(t)=A_{c}\cos[2\pi f_{c}t+K_{p}m(t)]
\end{aligned}\\
\begin{aligned}
&\text{208}\\
&x(t)=A_{c}\cos \theta(t)\\
&\theta(t)=2\pi f_{c}t+\phi(t)\\
&\frac{d\phi(t)}{dt}=2\pi K_{f}m(t)\Rightarrow\\
&\Rightarrow \phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\\
&x(t)=A_{c}\cos\left[2\pi f_{c}t+2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\right]
\end{aligned}\\
\begin{aligned}
&\text{208}\\
&x(t)=A_{c}\cos\theta(t)=A_{c}\cos[2\pi f_{c}t+\phi(t)]\\
&\cos(a+b)=\cos a \cos b-\sin a \sin b\\
&\phi(t)=K_{p}m(t)\\
&\phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau
\end{aligned}\\
\begin{aligned}
&\text{265}\\
&g(t)=f(t)\ast h(t)=\int\limits_{-\infty}^{\infty}f(\tau)h(t-\tau)d\tau\\
&j(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_{s})\Rightarrow\\
&\Rightarrow \mathcal{F}[j(t)]=\sum\limits_{k=-\infty}^{\infty}\mathcal{F}[\delta(t)]e^{j2\pi fkT_{s}}\\
&=\sum\limits_{k=-\infty}^{\infty}e^{j2\pi fkT_{s}}
\end{aligned}
\end{gather*}

\end{multicols*}

\end{document}

\documentclass[12pt]{article}
\usepackage[top=0.3in, bottom=1.2in, left=0.8in, right=0.8in]{geometry}

\usepackage{multicol}

\usepackage[utf8]{inputenc}

\setlength{\parindent}{0cm}

\usepackage{setspace}

\usepackage{xltxtra}
\usepackage{xgreek}
\setmainfont[Mapping=tex-text]{GFSDidot.otf}
\setsansfont[Mapping=tex-text]{GFSDidot.otf}

\usepackage[fleqn]{amsmath}
\usepackage{unicode-math}

\setlength{\mathindent}{0cm}



\newcommand{\3}{\vspace{0.3cm}}

\title{}
\author{}
\date{}

\begin{document}

\begin{multicols*}{2}

\begin{gather*}
\begin{aligned}
&\text{70}\\
&e^{jz}=\cos z+j\sin z\\
&\cos z=(1/2)(2\cos z)=\\
&=(1/2)(2\cos z+j\sin z-j\sin z)=\\
&=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\
&=(1/2)(e^{jz}+e^{-jz})
\end{aligned}\\
\begin{aligned}
&\text{207}\\
&x(t)=A_{c}\cos \theta(t)\\
&\theta(t)=2\pi f_{c}t+\phi(t)\\
&\phi(t)=K_{p}m(t)\\
&x(t)=A_{c}\cos[2\pi f_{c}t+K_{p}m(t)]
\end{aligned}\\
\begin{aligned}
&\text{208}\\
&x(t)=A_{c}\cos \theta(t)\\
&\theta(t)=2\pi f_{c}t+\phi(t)\\
&\frac{d\phi(t)}{dt}=2\pi K_{f}m(t)\Rightarrow\\
&\Rightarrow \phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\\
&x(t)=A_{c}\cos\left[2\pi f_{c}t+2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\right]
\end{aligned}\\
\begin{aligned}
&\text{208}\\
&x(t)=A_{c}\cos\theta(t)=A_{c}\cos[2\pi f_{c}t+\phi(t)]\\
&\cos(a+b)=\cos a \cos b-\sin a \sin b\\
&\phi(t)=K_{p}m(t)\\
&\phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau
\end{aligned}\\
\end{gather*}
\begin{gather*}
\begin{aligned}
&\text{265}\\
&g(t)=f(t)\ast h(t)=\int\limits_{-\infty}^{\infty}f(\tau)h(t-\tau)d\tau\\
&j(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_{s})\Rightarrow\\
&\Rightarrow \mathcal{F}[j(t)]=\sum\limits_{k=-\infty}^{\infty}\mathcal{F}[\delta(t)]e^{j2\pi fkT_{s}}\\
&=\sum\limits_{k=-\infty}^{\infty}e^{j2\pi fkT_{s}}
\end{aligned}
\end{gather*}

\end{multicols*}

\end{document}

Edit I:

This is the output with \raggedcolumns as proposed in the comments and obviously there are problems concerning the space left in the first column and the height difference between the first and the second column. Also I don't understand why is there a problem in the first place with gather environments.

\documentclass[12pt]{article}
\usepackage[top=0.3in, bottom=1.2in, left=0.8in, right=0.8in]{geometry}

\usepackage{multicol}

\usepackage[utf8]{inputenc}

\setlength{\parindent}{0cm}

\usepackage{setspace}

\usepackage{xltxtra}
\usepackage{xgreek}
\setmainfont[Mapping=tex-text]{GFSDidot.otf}
\setsansfont[Mapping=tex-text]{GFSDidot.otf}

\usepackage[fleqn]{amsmath}
\usepackage{unicode-math}

\setlength{\mathindent}{0cm}

\newcommand{\3}{\vspace{0.3cm}}

\title{}
\author{}
\date{}

\begin{document}
\raggedcolumns

\begin{multicols*}{2}

\begin{gather*}
\begin{aligned}
&\text{70}\\
&e^{jz}=\cos z+j\sin z\\
&\cos z=(1/2)(2\cos z)=\\
&=(1/2)(2\cos z+j\sin z-j\sin z)=\\
&=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\
&=(1/2)(e^{jz}+e^{-jz})
\end{aligned}\\
\begin{aligned}
&\text{207}\\
&x(t)=A_{c}\cos \theta(t)\\
&\theta(t)=2\pi f_{c}t+\phi(t)\\
&\phi(t)=K_{p}m(t)\\
&x(t)=A_{c}\cos[2\pi f_{c}t+K_{p}m(t)]
\end{aligned}\\
\begin{aligned}
&\text{208}\\
&x(t)=A_{c}\cos \theta(t)\\
&\theta(t)=2\pi f_{c}t+\phi(t)\\
&\frac{d\phi(t)}{dt}=2\pi K_{f}m(t)\Rightarrow\\
&\Rightarrow \phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\\
&x(t)=A_{c}\cos\left[2\pi f_{c}t+2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\right]
\end{aligned}\\
\begin{aligned}
&\text{208}\\
&x(t)=A_{c}\cos\theta(t)=A_{c}\cos[2\pi f_{c}t+\phi(t)]\\
&\cos(a+b)=\cos a \cos b-\sin a \sin b\\
&\phi(t)=K_{p}m(t)\\
&\phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau
\end{aligned}\\
\end{gather*}
\begin{gather*}
\begin{aligned}
&\text{265}\\
&g(t)=f(t)\ast h(t)=\int\limits_{-\infty}^{\infty}f(\tau)h(t-\tau)d\tau\\
&j(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_{s})\Rightarrow\\
&\Rightarrow \mathcal{F}[j(t)]=\sum\limits_{k=-\infty}^{\infty}\mathcal{F}[\delta(t)]e^{j2\pi fkT_{s}}\\
&=\sum\limits_{k=-\infty}^{\infty}e^{j2\pi fkT_{s}}
\end{aligned}
\end{gather*}

\end{multicols*}

\end{document}

Answer 1

With this approach, I build each block as a stack (with a 3pt buffer above and below), and then insert a \vspace{2ex}\par\vfil after each one. The 2ex is the minimum gap between blocks, and the \vfil will smooth out the whole look of the column (see alternative approach without this \vfil in 2nd part of the answer). After the final item in your multipage list, you can add a final\vfil so as to avoid excessive gapping from a partially filled column.

Within a block, the inter-item gap is specified in advance by \setstackgap{S}{6pt}. A block will not break in two across a column or page boundary, but the \vfil solution makes that look not so objectionable.

\documentclass[12pt]{article}
\usepackage[top=0.3in, bottom=1.2in, left=0.8in, right=0.8in]{geometry}

\usepackage{multicol}

\usepackage[utf8]{inputenc}

\setlength{\parindent}{0cm}

\usepackage{setspace}

\usepackage{xltxtra}
\usepackage{xgreek}
\setmainfont[Mapping=tex-text]{GFSDidot.otf}
\setsansfont[Mapping=tex-text]{GFSDidot.otf}

%\usepackage[fleqn]{amsmath}
\usepackage{amsmath}
\usepackage{unicode-math}

%\setlength{\mathindent}{0cm}

\newcommand{\3}{\vspace{0.3cm}}

\title{}
\author{}
\date{}
\usepackage[usestackEOL]{stackengine}
\stackMath
\def\stackalignment{l}
\setstackgap{S}{6pt}
\def\SS#1{\noindent\addstackgap[3pt]{\Shortstack{#1}}\vspace{2ex}\par\vfil}
\begin{document}

\begin{multicols*}{2}
Preceding text\par
\SS{
\text{70}\\
e^{jz}=\cos z+j\sin z\\
\cos z=(1/2)(2\cos z)=\\
=(1/2)(2\cos z+j\sin z-j\sin z)=\\
=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\
=(1/2)(e^{jz}+e^{-jz})
}
\SS{
\text{70}\\
e^{jz}=\cos z+j\sin z\\
\cos z=(1/2)(2\cos z)=\\
=(1/2)(2\cos z+j\sin z-j\sin z)=\\
=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\
=(1/2)(e^{jz}+e^{-jz})
}
\SS{
\text{207}\\
x(t)=A_{c}\cos \theta(t)\\
\theta(t)=2\pi f_{c}t+\phi(t)\\
\phi(t)=K_{p}m(t)\\
x(t)=A_{c}\cos[2\pi f_{c}t+K_{p}m(t)]
}
\SS{
\text{208}\\
x(t)=A_{c}\cos \theta(t)\\
\theta(t)=2\pi f_{c}t+\phi(t)\\
\dfrac{d\phi(t)}{dt}=2\pi K_{f}m(t)\Rightarrow\\
\Rightarrow \phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\\
x(t)=A_{c}\cos\left[2\pi f_{c}t+2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\right]
}
\SS{
\text{208}\\
x(t)=A_{c}\cos\theta(t)=A_{c}\cos[2\pi f_{c}t+\phi(t)]\\
\cos(a+b)=\cos a \cos b-\sin a \sin b\\
\phi(t)=K_{p}m(t)\\
\phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau
}
\SS{
\text{208}\\
x(t)=A_{c}\cos\theta(t)=A_{c}\cos[2\pi f_{c}t+\phi(t)]\\
\cos(a+b)=\cos a \cos b-\sin a \sin b\\
\phi(t)=K_{p}m(t)\\
\phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau
}
\SS{
\text{208}\\
x(t)=A_{c}\cos\theta(t)=A_{c}\cos[2\pi f_{c}t+\phi(t)]\\
\cos(a+b)=\cos a \cos b-\sin a \sin b\\
\phi(t)=K_{p}m(t)\\
\phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau
}
\SS{
\text{265}\\
g(t)=f(t)\ast h(t)=\int\limits_{-\infty}^{\infty}f(\tau)h(t-\tau)d\tau\\
j(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_{s})\Rightarrow\\
\Rightarrow \mathcal{F}[j(t)]=\sum\limits_{k=-\infty}^{\infty}\mathcal{F}[\delta(t)]e^{j2\pi fkT_{s}}\\
=\sum\limits_{k=-\infty}^{\infty}e^{j2\pi fkT_{s}}
}
\SS{
\text{265}\\
g(t)=f(t)\ast h(t)=\int\limits_{-\infty}^{\infty}f(\tau)h(t-\tau)d\tau\\
j(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_{s})\Rightarrow\\
\Rightarrow \mathcal{F}[j(t)]=\sum\limits_{k=-\infty}^{\infty}\mathcal{F}[\delta(t)]e^{j2\pi fkT_{s}}\\
=\sum\limits_{k=-\infty}^{\infty}e^{j2\pi fkT_{s}}
}
\vfil
\end{multicols*}

\end{document}

---

ALTERNATIVE APPROACH (without \vfil between blocks)

If one does not like the evenly spaced vertical column, and prefers a wide gap at column bottom, then defining

\def\SS#1{\noindent\addstackgap[3pt]{\Shortstack{#1}}\\}

seems to resolve that (and also adds a 3pt buffer above/below each stack), as in

\documentclass[12pt]{article}
\usepackage[top=0.3in, bottom=1.2in, left=0.8in, right=0.8in]{geometry}

\usepackage{multicol}

\usepackage[utf8]{inputenc}

\setlength{\parindent}{0cm}

\usepackage{setspace}

\usepackage{xltxtra}
\usepackage{xgreek}
\setmainfont[Mapping=tex-text]{GFSDidot.otf}
\setsansfont[Mapping=tex-text]{GFSDidot.otf}

%\usepackage[fleqn]{amsmath}
\usepackage{amsmath}
\usepackage{unicode-math}

%\setlength{\mathindent}{0cm}

\newcommand{\3}{\vspace{0.3cm}}

\title{}
\author{}
\date{}
\usepackage[usestackEOL]{stackengine}
\stackMath
\def\stackalignment{l}
\setstackgap{S}{6pt}
\def\SS#1{\noindent\addstackgap[3pt]{\Shortstack{#1}}\\}
\begin{document}

\begin{multicols*}{2}
\SS{
\text{70}\\
e^{jz}=\cos z+j\sin z\\
\cos z=(1/2)(2\cos z)=\\
=(1/2)(2\cos z+j\sin z-j\sin z)=\\
=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\
=(1/2)(e^{jz}+e^{-jz})
}
\SS{
\text{70}\\
e^{jz}=\cos z+j\sin z\\
\cos z=(1/2)(2\cos z)=\\
=(1/2)(2\cos z+j\sin z-j\sin z)=\\
=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\
=(1/2)(e^{jz}+e^{-jz})
}
\SS{
\text{207}\\
x(t)=A_{c}\cos \theta(t)\\
\theta(t)=2\pi f_{c}t+\phi(t)\\
\phi(t)=K_{p}m(t)\\
x(t)=A_{c}\cos[2\pi f_{c}t+K_{p}m(t)]
}
\SS{
\text{208}\\
x(t)=A_{c}\cos \theta(t)\\
\theta(t)=2\pi f_{c}t+\phi(t)\\
\dfrac{d\phi(t)}{dt}=2\pi K_{f}m(t)\Rightarrow\\
\Rightarrow \phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\\
x(t)=A_{c}\cos\left[2\pi f_{c}t+2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\right]
}
\SS{
\text{265}\\
g(t)=f(t)\ast h(t)=\int\limits_{-\infty}^{\infty}f(\tau)h(t-\tau)d\tau\\
j(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_{s})\Rightarrow\\
\Rightarrow \mathcal{F}[j(t)]=\sum\limits_{k=-\infty}^{\infty}\mathcal{F}[\delta(t)]e^{j2\pi fkT_{s}}\\
=\sum\limits_{k=-\infty}^{\infty}e^{j2\pi fkT_{s}}
}
\SS{
\text{265}\\
g(t)=f(t)\ast h(t)=\int\limits_{-\infty}^{\infty}f(\tau)h(t-\tau)d\tau\\
j(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_{s})\Rightarrow\\
\Rightarrow \mathcal{F}[j(t)]=\sum\limits_{k=-\infty}^{\infty}\mathcal{F}[\delta(t)]e^{j2\pi fkT_{s}}\\
=\sum\limits_{k=-\infty}^{\infty}e^{j2\pi fkT_{s}}
}
\SS{
\text{265}\\
g(t)=f(t)\ast h(t)=\int\limits_{-\infty}^{\infty}f(\tau)h(t-\tau)d\tau\\
j(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_{s})\Rightarrow\\
\Rightarrow \mathcal{F}[j(t)]=\sum\limits_{k=-\infty}^{\infty}\mathcal{F}[\delta(t)]e^{j2\pi fkT_{s}}\\
=\sum\limits_{k=-\infty}^{\infty}e^{j2\pi fkT_{s}}
}
\vfil
\end{multicols*}

\end{document}

Answer 2

You aren't allowing LaTeX any place to split the columns; if you really have such gigantic displays, then tell

\allowdisplaybreaks

in your document preamble.

The initial vertical space is due to beginning a paragraph with a display environment, which is wrong.

Here's a more minimalistic example:

\documentclass[12pt]{article}
\usepackage[top=0.3in, bottom=1.2in, left=0.8in, right=0.8in]{geometry}

\usepackage{multicol}

\setlength{\parindent}{0cm}

\usepackage[fleqn]{amsmath}
\setlength{\mathindent}{0cm}

\newcommand{\3}{\vspace{0.3cm}}

\title{}
\author{}
\date{}

\allowdisplaybreaks

\begin{document}
\raggedcolumns

\begin{multicols}{2}
70
\begin{gather*}
\begin{aligned}
&e^{jz}=\cos z+j\sin z\\
&\cos z=(1/2)(2\cos z)=\\
&=(1/2)(2\cos z+j\sin z-j\sin z)=\\
&=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\
&=(1/2)(e^{jz}+e^{-jz})
\end{aligned}\\
\begin{aligned}
&\text{207}\\
&x(t)=A_{c}\cos \theta(t)\\
&\theta(t)=2\pi f_{c}t+\phi(t)\\
&\phi(t)=K_{p}m(t)\\
&x(t)=A_{c}\cos[2\pi f_{c}t+K_{p}m(t)]
\end{aligned}\\
\begin{aligned}
&\text{208}\\
&x(t)=A_{c}\cos \theta(t)\\
&\theta(t)=2\pi f_{c}t+\phi(t)\\
&\frac{d\phi(t)}{dt}=2\pi K_{f}m(t)\Rightarrow\\
&\Rightarrow \phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\\
&x(t)=A_{c}\cos\left[2\pi f_{c}t+2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau\right]
\end{aligned}\\
\begin{aligned}
&\text{208}\\
&x(t)=A_{c}\cos\theta(t)=A_{c}\cos[2\pi f_{c}t+\phi(t)]\\
&\cos(a+b)=\cos a \cos b-\sin a \sin b\\
&\phi(t)=K_{p}m(t)\\
&\phi(t)=2\pi K_{f}\int\limits_{-\infty}^{t}m(\tau)d\tau
\end{aligned}\\
\begin{aligned}
&\text{265}\\
&g(t)=f(t)\ast h(t)=\int\limits_{-\infty}^{\infty}f(\tau)h(t-\tau)d\tau\\
&j(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_{s})\Rightarrow\\
&\Rightarrow \mathcal{F}[j(t)]=\sum\limits_{k=-\infty}^{\infty}\mathcal{F}[\delta(t)]e^{j2\pi fkT_{s}}\\
&=\sum\limits_{k=-\infty}^{\infty}e^{j2\pi fkT_{s}}
\end{aligned}
\end{gather*}

\end{multicols}

\end{document}

Of course, balancing the columns is impossible because of the many aligned.

---

As an aside, note that inputenc should never be used when compiling with XeLaTeX or LuaLaTeX.