amsmathboldmath-mode
\begin{equation}
\begin{align}
\mathbf{x_{t}=f_{t}(x_{t-1},u_{t})}\\
\mathbf{y_{t}=g_{t}(x_{t},v_{t})}
\label{eq:state-space&obs-equ}
\end{align}
\end{equation}
I want that y(t) equation should come on next line but it comes on same line? how to solve this problem?
I suggest using the multlined environment from mathtools to break the radicand . You also can choose not to break it, but to type in medsize (~80 % of \displaystyle) with the \mediummath command from nccmath. I took the opportunity to clean your code:
\documentclass[11pt,a4paper,twocolumn]{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools, nccmath}
\begin{document}
\begin{equation*}
\bigl| \dot{r} \bigr| =
\sqrt{\begin{multlined}[b] -\!\bigl(e^{-t} ( \cos t + \sin t )\bigr)^2 + \\ \bigl(e^{-t} (\cos t + \sin t)\bigr)^2 + \bigl(-e^{-t}\bigr)^2 \end{multlined}}
\end{equation*}
\begin{align*}
\bigl| \dot{r} \bigr| & =
\sqrt{\medmath{ -\!\bigl(e^{-t} ( \cos t + \sin t )\bigr)^2 + \bigl(e^{-t} (\cos t + \sin t)\bigr)^2 + \bigl(-e^{-t}\bigr)^2}} \\
& = \sqrt{ -\!\bigl(e^{-t} ( \cos t + \sin t )\bigr)^2 + \bigl(e^{-t} (\cos t + \sin t)\bigr)^2 + \bigl(-e^{-t}\bigr)^2}
\end{align*}
\end{document}
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}
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}
A few points to note.
\boldsymbol is deprecated and \bm (with the bm package) is to be preferred.
All \textbf commands should be \mathbf (in an italic context, such as a theorem statement, the letters would be typeset in italic).
Instead of \text{erf} one should define a suitable operator with \DeclareMathOperator (same reason as before).
Best Answer