How do I align the equations to the left?

Question

The following is the full code that produces the image shown below:

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}

\usepackage{amsthm}
\usepackage{amsmath}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}


\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}


\begin{document}
    \title{Extra Credit}
    \maketitle

    \begin{definition}
        If f is analytic at $z_0$, then the series

        \begin{equation}
            f(z_0) + f'(z_0)(z-z_0) + \frac{f''(z_0)}{2!}(z-z_0)^2 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
        \end{equation}

        is called the Taylor series for f around $z_0$.
    \end{definition}

    \begin{theorem}
        If f is analytic inside and on the simple closed positively oriented contour $\Gamma$ and if $z_0$ is any point inside $\Gamma$, then
        \begin{equation}
            f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta \hspace{1cm} (n=1,2,3, \cdots )
        \end{equation}
    \end{theorem}
    \hrulefill



\begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk. 
\end{theorem}

\begin{proof}
Suppose that the function \textit{f} is analytic in the disk $|z-z_0|<R'$. We can define a positively oriented contour $C$ as $$ C:=\Big\{z:|z-z_0|=\frac{R + R'}{2}, 0<R< R'  \Big\}.$$ Letting $\zeta$ be an arbitrary point on $C$ and applying $(2)$ to $(1)$, we get 

\begin{equation}
\sum_{n=0}^{\infty} \frac{(z-z_0)^{n}}{2\pi i} \int_{C} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta.
\end{equation}\\ 

Or equivalently, we have that


\begin{align*}
  \sum_{n=0}^{\infty} \frac{(z-z_0)^{n}}{2\pi i} \int_{C} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta
  &= \frac{1}{2\pi i} \sum_{n=0}^{\infty} \int_{C} \frac{(z-z_0)^{n}f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta 
  \\ &= \frac{1}{2\pi i}\int_{C}\sum_{n=0}^{\infty} \frac{1}{\zeta - z_0}\frac{f(\zeta)}{\displaystyle\frac{(\zeta - z_0)^n}{(z-z_0)^n}}d\zeta                   
  \\ &= \frac{1}{2\pi i} \int_{C} \frac{f(\zeta)}{\zeta - z_0}\sum_{n=0}^{\infty}\left( \frac{z-z_0}{\zeta - z_0}\right)^{n} d\zeta
\                   
\end{align*}


Now since 
\begin{equation}
\frac{|z-z_0|}{|\zeta - z_0|}<\frac{R}{\displaystyle\frac{R'+R}{2}} < \frac{\displaystyle\frac{R'+R}{2}}{\displaystyle\frac{R'+R}{2}}=1,
\end{equation}
it follows that 
\begin{equation}
\sum_{n=0}^{\infty}\left( \frac{z-z_0}{\zeta - z_0}\right)^{n}=\frac{1}{1-\displaystyle\frac{z-z_0}{\zeta-z_0}}. 
\end{equation}
The last result comes from the observation that 
\begin{flalign}
&\Big[1-\left(\frac{z-z_0}{\zeta - z_0}\right)\Big]\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]  
\\&= 1 + \Big(\frac{z-z_0}{\zeta-z_0}\Big)+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}+ \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}-\Big(\frac{z-z_0}{\zeta-z_0}\Big)-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}- \cdots -\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\\&=1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\
\end{flalign}
\\
And so,
\begin{flalign}
&\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]
\\&=\frac{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\\&=\frac{1}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}-\frac{\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\ 
\end{flalign}

where specifically,

Now since 
\begin{equation}
\frac{|z-z_0|}{|\zeta - z_0|}<\frac{R}{\displaystyle\frac{R'+R}{2}} < \frac{\displaystyle\frac{R'+R}{2}}{\displaystyle\frac{R'+R}{2}}=1,
\end{equation}
it follows that 
\begin{equation}
\sum_{n=0}^{\infty}\left( \frac{z-z_0}{\zeta - z_0}\right)^{n}=\frac{1}{1-\displaystyle\frac{z-z_0}{\zeta-z_0}}. 
\end{equation}
The last result comes from the observation that 
\begin{flalign}
&\Big[1-\left(\frac{z-z_0}{\zeta - z_0}\right)\Big]\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]  
\\&= 1 + \Big(\frac{z-z_0}{\zeta-z_0}\Big)+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}+ \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}-\Big(\frac{z-z_0}{\zeta-z_0}\Big)-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}- \cdots -\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\\&=1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\
\end{flalign}
\\
And so,
\begin{flalign}
&\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]
\\&=\frac{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\\&=\frac{1}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}-\frac{\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\ 
\end{flalign}
\\

produces the image above. How can I get the lines after And so in the image to align beneath And so. And how can I only keep the numbering for equation (8) and equation (11) as shown in the image?

the fleqn documentclass option makes equations left aligned, but you also need to remove misplaced \\ such as \end{flalign} \\ — David Carlisle, Mar 31 '19 at 17:04
probably a duplicate of https://tex.stackexchange.com/questions/54478/how-to-left-align-a-set-of-equations — David Carlisle, Mar 31 '19 at 17:53
I'm not able to get it to work using the link you've posted. — Skm, Mar 31 '19 at 18:18

John Kormylo · Accepted Answer · 2019-03-31T19:13:12.303

Is this what you want?

Basically, you want to overlap left aligned equations and centered (more or less) equations. This uses \mathrlap (mathtools package) to prevent the left aligned equations from hogging the space. The use of &&& skips over the left aligned portion.

\documentclass{article}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
%\usepackage[utf8]{inputenc}% all they do for me is slow down the build.
%\usepackage[english]{babel}
\usepackage{amsthm}
\usepackage{mathtools}

\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}

\begin{document}
Now since 
\begin{equation}
\frac{|z-z_0|}{|\zeta - z_0|}<\frac{R}{\displaystyle\frac{R'+R}{2}} < \frac{\displaystyle\frac{R'+R}{2}}{\displaystyle\frac{R'+R}{2}}=1,
\end{equation}
it follows that 
\begin{equation}
\sum_{n=0}^{\infty}\left( \frac{z-z_0}{\zeta - z_0}\right)^{n}=\frac{1}{1-\displaystyle\frac{z-z_0}{\zeta-z_0}}. 
\end{equation}
The last result comes from the observation that 
\begin{flalign}
\mathrlap{\Big[1-\left(\frac{z-z_0}{\zeta - z_0}\right)\Big]\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}
  + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1} + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]} &&&&
\\&&&= 1 + \Big(\frac{z-z_0}{\zeta-z_0}\Big)+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}+ \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n} \notag
\\&&&\quad-\Big(\frac{z-z_0}{\zeta-z_0}\Big)-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2}- \cdots -\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}
  -\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\\&&&=1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}
\intertext{And so,}
\mathrlap{\Big[1 + \frac{z-z_0}{\zeta-z_0}+ \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{2} + \cdots + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n-1}
  + \Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n}\Big]} &&&
\\&&&=\frac{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\\&&&=\frac{1}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}-\frac{\Big(\frac{z-z_0}{\zeta-z_0}\Big)^{n+1}}{1-\Big(\frac{z-z_0}{\zeta-z_0}\Big)}
\ 
\end{flalign}
\end{document}

How do I align the equations to the left?

1 Answers1