# I have message that is missing in line of this code \begin{equation} \begin{aligned} \f(\kappa,\rho,\alpha,\beta_i,X_i;x_i,y_i) & =\dfrac{\exp \big(k cos(x_i-X_i)\big)}{2\pi I_0}\dfrac{(1-\rho^2)}{2\pi\big[1+\rho^2-2\rho \cos(y_i-\alpha-\beta_i X_i)\big]}\\ & =\dfrac{1}{(2\pi)^2 I_0(\lambda \rho)}\exp \big(k \cos(x_i-X_i)\big)\dfrac{(1-\rho^2)}{(1+\rho^2)}\dfrac{1}{\big[1-\dfrac{2\rho}{(1+\rho^2)} \cos(y_i-\alpha-\beta_i X_i)\big]}\\ & =\dfrac{1}{(2\pi)^2I_0(\lambda \rho)}\exp \big(k \cos(x_i-X_i)\big)\dfrac{(1-\rho^2)}{(1+\rho^2)}\big[1-\dfrac{2\rho}{(1+\rho^2)}\cos(y_i-\alpha-\beta_i X_i)\big]^{-1}\\ & =\dfrac{1}{(2\pi)^2I_0(\lambda \rho)} \exp \big(\cos(x_i-X_i)\big)\dfrac{(1-\rho^2)}{(1+\rho^2)}\big[1-\dfrac{2\rho}{(1+\rho^2)}{\cos(y_i-\beta_i X_i) \cos\alpha+\sin(y_i-\beta_i X_i \sin\alpha)}\big]^{-1} \end{aligned} \end{equation}  Where should I put the ?

• remove the blank lines, you are not allowed paragraph breaks in math Feb 17 '18 at 10:57
• unrelated but use \bigl(...\bigr) never use \big which is the internal command used to define \bigl and \bigr Feb 17 '18 at 10:58
• what mean by blank line@D Feb 17 '18 at 10:59
• The blank line is an empty line, with only spaces. Feb 17 '18 at 11:02
• @DoāaY.El-borai: The empty line between \begin{aligned} and \f: you should remove it, as well as the empty line before \end{equation}.
– GuM
Feb 17 '18 at 11:03

You haven't defined the macro \f; in the code below, I've set \f equal to just f.

Since the term \frac{\exp\big(k\cos(x_i-X_i)\big)}{(2\pi)^2 I_0} occurs repeatedly, I'd like to suggest that you give it a new, compact name, say \zeta_i.

The following may then be what you're looking for; note the extra linebreak in the final row. \documentclass{article}
\usepackage{amsmath}
\newcommand\f{f} % ??
\begin{document}
Put $\zeta_i= \dfrac{\exp \bigl(k\cos(x_i-X_i)\bigr)}{(2\pi)^2 I_0}$.
\begin{equation}
\begin{aligned}[b]
\f(\kappa,\rho,\alpha,\beta_i,X_i;x_i,y_i)
&= \zeta_i\,\frac{(1-\rho^2)}{\bigl[1+\rho^2-2\rho \cos(y_i-\alpha-\beta_i X_i)\bigr]}\\
&= \frac{\zeta_i}{\lambda\rho}\,\frac{(1-\rho^2)}{(1+\rho^2)}
\frac{1}{\bigl[1-\frac{2\rho}{(1+\rho^2)} \cos(y_i-\alpha-\beta_i X_i)\bigr]}\\
&= \frac{\zeta_i}{\lambda\rho}\,\frac{(1-\rho^2)}{(1+\rho^2)}
\bigl[1-\frac{2\rho}{(1+\rho^2)}\cos(y_i-\alpha-\beta_i X_i)\bigr]^{-1}\\
&= \frac{\zeta_i}{\lambda\rho}\,\frac{(1-\rho^2)}{(1+\rho^2)}
\bigl[1-\frac{2\rho}{(1+\rho^2)}\cos(y_i-\beta_i X_i) \cos\alpha\\
\end{aligned}
\end{equation}
\end{document}


I changed your code a bit, to something that could compile:

\documentclass{standalone}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{mathtools}
\begin{document}
\begin{equation}
\begin{aligned}
f(\kappa,\rho,\alpha,\beta_i,X_i;x_i,y_i) & =\dfrac{\exp \big(k cos(x_i-X_i)\big)}{2\pi I_0}\dfrac{(1-\rho^2)}{2\pi\big[1+\rho^2-2\rho \cos(y_i-\alpha-\beta_i X_i)\big]}\\
& =\dfrac{1}{(2\pi)^2 I_0(\lambda \rho)}\exp \big(k \cos(x_i-X_i)\big)\dfrac{(1-\rho^2)}{(1+\rho^2)}\dfrac{1}{\big[1-\dfrac{2\rho}{(1+\rho^2)} \cos(y_i-\alpha-\beta_i X_i)\big]}\\
& =\dfrac{1}{(2\pi)^2I_0(\lambda \rho)}\exp \big(k \cos(x_i-X_i)\big)\dfrac{(1-\rho^2)}{(1+\rho^2)}\big[1-\dfrac{2\rho}{(1+\rho^2)}\cos(y_i-\alpha-\beta_i X_i)\big]^{-1}\\
& =\dfrac{1}{(2\pi)^2I_0(\lambda \rho)} \exp \big(\cos(x_i-X_i)\big)\dfrac{(1-\rho^2)}{(1+\rho^2)}\big[1-\dfrac{2\rho}{(1+\rho^2)}{\cos(y_i-\beta_i X_i) \cos\alpha+\sin(y_i-\beta_i X_i \sin\alpha)}\big]^{-1} \end{aligned}
\end{equation}
\end{document}


This produces: I have removed the blank lines you had inside the aligned environment, and the \f

If you really want blank lines, you need to mark them with percentage signs:

\documentclass{standalone}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{mathtools}
\begin{document}
\begin{equation}
\begin{aligned}
%
f(\kappa,\rho,\alpha,\beta_i,X_i;x_i,y_i) & =\dfrac{\exp \big(k cos(x_i-X_i)\big)}{2\pi I_0}\dfrac{(1-\rho^2)}{2\pi\big[1+\rho^2-2\rho \cos(y_i-\alpha-\beta_i X_i)\big]}\\
& =\dfrac{1}{(2\pi)^2 I_0(\lambda \rho)}\exp \big(k \cos(x_i-X_i)\big)\dfrac{(1-\rho^2)}{(1+\rho^2)}\dfrac{1}{\big[1-\dfrac{2\rho}{(1+\rho^2)} \cos(y_i-\alpha-\beta_i X_i)\big]}\\
& =\dfrac{1}{(2\pi)^2I_0(\lambda \rho)}\exp \big(k \cos(x_i-X_i)\big)\dfrac{(1-\rho^2)}{(1+\rho^2)}\big[1-\dfrac{2\rho}{(1+\rho^2)}\cos(y_i-\alpha-\beta_i X_i)\big]^{-1}\\
& =\dfrac{1}{(2\pi)^2I_0(\lambda \rho)} \exp \big(\cos(x_i-X_i)\big)\dfrac{(1-\rho^2)}{(1+\rho^2)}\big[1-\dfrac{2\rho}{(1+\rho^2)}{\cos(y_i-\beta_i X_i) \cos\alpha+\sin(y_i-\beta_i X_i \sin\alpha)}\big]^{-1}
%
\end{aligned}
\end{equation}
\end{document}


# Edit:

I originally compiled the code above in the memoir class, but the math was too wide, which is why i changed to standalone. In the standalone class I do get an error, but this error is not present in memoir, and the document still compiles to the shown picture

• i solve it by write f without \.... thanx alot Feb 17 '18 at 11:17
• If this worked @DoāaY.El-borai then please press the tick mark under the arrows, so the question is closed :-) Feb 17 '18 at 11:22
• cos and sin are not variable, so you should write \cos and \sin ... Feb 17 '18 at 16:27