[Tex/LaTex] Aligning Equations & Removing Equation Label

amsmathequationshorizontal alignment

I have the code:

\begin{flalign} 
&f_1(y_i; \boldsymbol{\theta}_1) = \lambda e^{-\lambda y_i} \\
&l_1(\boldsymbol{\theta}_1; \textbf{y}, \textbf{x}) = `r num.not.censored` 
  \log{(\lambda)} - \lambda \sum_{i=1}^{`r num.data`} y_i \\
&f_2(y_i; x_i, \boldsymbol{\theta}_2) = e^{\beta_1 + \beta_2 x_i} 
  \exp{(-e^{\beta_1 + \beta_2 x_i}y_i)} \\
&l_2(\boldsymbol{\theta}_2; \textbf{y}, \textbf{x}) = 
  \sum_{i=1}^{`r num.not.censored`} (\beta_1 + \beta_2 x_i) 
 -\sum_{i=1}^{`r num.data`} (e^{\beta_1 + \beta_2 x_i}y_i) \\
&f_3(y_i; x_i, \boldsymbol{\theta}_3) = \frac{1}{\sqrt{2\pi\sigma^2}} 
   \exp{\bigg(\frac{-(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2\sigma^2}\bigg)} \\
&l_3(\boldsymbol{\theta}_3; \textbf{y}, \textbf{x}) = - 28 
   \log(2 \pi \sigma^2) - \sum_{i=1}^{`r num.not.censored`} 
   \bigg(\frac{(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2 \sigma^2}\bigg)
   \space + \\ 
&\sum_{i=`r num.not.censored + 1`}^{`r num.data`} \log 
   \bigg(\frac{1}{2} - \frac{1}{2}\text{erf}\Big(\frac{y_i - 
   (\gamma_1 + \gamma_2 x_i)}{\sigma \sqrt{2}}\Big)\bigg)
\end{flalign}

It renders as:

enter image description here

How can I shift line 7 to the right, so that it starts at equals sign of line 6?

I also want to remove the equation (7) label, since it is still part of equation (6).

Thanks,

Jack

Best Answer

You seem to want the fleqn option rather than flalign:

\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{bm}

\DeclareMathOperator{\erf}{erf}

\begin{document}

\begin{gather}
f_1(y_i; \bm{\theta}_1) = \lambda e^{-\lambda y_i} \\
l_1(\bm{\theta}_1; \mathbf{y}, \mathbf{x}) = 56 
  \log{(\lambda)} - \lambda \sum_{i=1}^{75} y_i \\
f_2(y_i; x_i, \bm{\theta}_2) = e^{\beta_1 + \beta_2 x_i} 
  \exp{(-e^{\beta_1 + \beta_2 x_i}y_i)} \\
l_2(\bm{\theta}_2; \mathbf{y}, \mathbf{x}) = 
  \sum_{i=1}^{56} (\beta_1 + \beta_2 x_i) 
 -\sum_{i=1}^{75} (e^{\beta_1 + \beta_2 x_i}y_i) \\
f_3(y_i; x_i, \bm{\theta}_3) = \frac{1}{\sqrt{2\pi\sigma^2}} 
   \exp{\biggl(\frac{-(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2\sigma^2}\biggr)} \\
\begin{split}
l_3(\bm{\theta}_3; \mathbf{y}, \mathbf{x}) 
 ={}&{-}28\log(2 \pi \sigma^2) - \sum_{i=1}^{56} 
   \biggl(\frac{(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2 \sigma^2}\biggr)
   \\ 
 &+\sum_{i=57}^{75} \log 
   \biggl(\frac{1}{2} - \frac{1}{2}\erf\Bigl(\frac{y_i - 
   (\gamma_1 + \gamma_2 x_i)}{\sigma \sqrt{2}}\Bigr)\biggr)
\end{split}
\end{gather}

\end{document}

enter image description here

You can easily change to alignment to the equals signs:

\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{bm}

\DeclareMathOperator{\erf}{erf}

\begin{document}

\begin{align}
f_1(y_i; \bm{\theta}_1) &= \lambda e^{-\lambda y_i} \\
l_1(\bm{\theta}_1; \mathbf{y}, \mathbf{x}) &= 56 
  \log{(\lambda)} - \lambda \sum_{i=1}^{75} y_i \\
f_2(y_i; x_i, \bm{\theta}_2) &= e^{\beta_1 + \beta_2 x_i} 
  \exp{(-e^{\beta_1 + \beta_2 x_i}y_i)} \\
l_2(\bm{\theta}_2; \mathbf{y}, \mathbf{x}) &= 
  \sum_{i=1}^{56} (\beta_1 + \beta_2 x_i) 
 -\sum_{i=1}^{75} (e^{\beta_1 + \beta_2 x_i}y_i) \\
f_3(y_i; x_i, \bm{\theta}_3) &= \frac{1}{\sqrt{2\pi\sigma^2}} 
   \exp{\biggl(\frac{-(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2\sigma^2}\biggr)} \\
\begin{split}
l_3(\bm{\theta}_3; \mathbf{y}, \mathbf{x}) 
 &=-28\log(2 \pi \sigma^2) - \sum_{i=1}^{56} 
   \biggl(\frac{(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2 \sigma^2}\biggr)
   \\ 
 &\mathrel{\hphantom{=}}+\sum_{i=57}^{75} \log 
   \biggl(\frac{1}{2} - \frac{1}{2}\erf\Bigl(\frac{y_i - 
   (\gamma_1 + \gamma_2 x_i)}{\sigma \sqrt{2}}\Bigr)\biggr)
\end{split}
\end{align}

\end{document}

enter image description here

A few points to note.

  1. \boldsymbol is deprecated and \bm (with the bm package) is to be preferred.

  2. All \textbf commands should be \mathbf (in an italic context, such as a theorem statement, the letters would be typeset in italic).

  3. Instead of \text{erf} one should define a suitable operator with \DeclareMathOperator (same reason as before).

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