# Problem with compiling formula by LaTeX

• LaTeX
linderox
I read a lot literature how to provide dividing of the formulas to new lines,but it's work just on a short one good like 1st one and doesn't on main (2nd one)
Code:
$$\begin{split} \Psi = & \cos kz + i\sin kz + {} \\ & {} + \frac{f(\theta)}{r} (\cos kr + i\sin kr) \end{split}$$

I try different ways used {split} but everything stopped on the 2nd "\\". Can anybody say me why?

Code:
$$\begin{split} U = U_0\cos[c_1 \cos(\omega t) + c_2 cos(\omega_1 t) + \phi] \\ =U_0 \cos\phi\left[|J_0(c_1)J_0(c_2)| + \sum_n |2J_0(c_2 )J_{2n}(c_1)|\cos 2n\omega t \\ +\sum_k|2J_0(c_1 )J_{2k}(c_2)|\cos 2k\omega_1 t + \sum_n\sum_k |2J_2n(c_1)J_{2k}(c_2)| \cos [2(k\omega_1 — n\omega)t] \\ +\sum_n \sum_k|2J_{2n}(c_1)J_{2k}(c_2)|\cos [2(k\omega_1 + n\omega)t] \\ +\sum_n \sum_k|J_{2n-1}(c_1)J_{2k-1}(c_2)|\cos[(2k —1)\omega_1—(2n — 1)\omega]t \\ +\sum_n \sum_k|J_{2n-1}(c_1)J_{2k-1}(c_2)|\cos[(2k —1)\omega_1+(2n — 1)\omega]t\right] \\ +U_0\sin\phi\left[\sum_n |2J_0(c_2)J_{2n-1}(c_1)|\cos[(2n—1)\omega t] \\ +\sum_k | 2J_0(c_1)J_{2k-1}(c_2)|\cos[(2k — 1)\omega_1 t] \\ +\sum_n \sum_k|2J_{2n-1}(c_1)J_{2k}(c_2)|\cos[2k\omega_1—(2n—1)\omega]t \\ +\sum_n \sum_k|2J_{2n-1}(c_1)J_{2k}(c_2)|\cos[2k\omega_1+(2n—1)\omega]t \\ +\sum_n \sum_k|J_{2n}(c_1)J_{2k-1}(c_2)|\cos [(2k—1)\omega_1—2n\omega]t \\ +\sum_n \sum_k|J_{2n}(c_1)J_{2k-1}(c_2)|\cos [(2k—1)\omega_1+2n\omega]t\right] \end{split}$$

Last edited:

Staff Emeritus
Gold Member
Well, the most obvious problem is you forgot the alignment character (&). I could believe that would result in the bottom part of your equation sitting a few feet past the right margin.

linderox
can you write right version here?
this version for example doesn't work too!
Code:
\begin{eqnarray*}
U=U_0\cos[c_1 \cos(\omega t) + c_2 cos(\omega_1 t) + \phi] \\
& {} U=U_0 \cos\phi\left[|J_0(c_1)J_0(c_2)| + \sum_n |2J_0(c_2 )J_{2n}(c_1)|\cos 2n\omega t \\
& {} +\sum_k|2J_0(c_1 )J_{2k}(c_2)|\cos 2k\omega_1 t + \sum_n\sum_k |2J_2n(c_1)J_{2k}(c_2)| \cos [2(k\omega_1 — n\omega)t] \right]
\end{eqnarray*}

Staff Emeritus
I would imagine that your problem is that you are trying to split equations before you have closed the brackets (i.e. you have a \left[ but no \right] before the \\). A way to remedy this is to use phantom delimiters (\left. and \right.) , so your input would read (with additions in red):

Code:
$$\begin{split} U = U_0\cos[c_1 \cos(\omega t) + c_2 cos(\omega_1 t) + \phi] \\ =U_0 \cos\phi\left[|J_0(c_1)J_0(c_2)| + \sum_n |2J_0(c_2 )J_{2n}(c_1)|\cos 2n\omega t [color=red]\right.[/color]\\ +\sum_k|2J_0(c_1 )J_{2k}(c_2)|\cos 2k\omega_1 t + \sum_n\sum_k |2J_2n(c_1)J_{2k}(c_2)| \cos [2(k\omega_1 — n\omega)t] \\ +\sum_n \sum_k|2J_{2n}(c_1)J_{2k}(c_2)|\cos [2(k\omega_1 + n\omega)t] \\ +\sum_n \sum_k|J_{2n-1}(c_1)J_{2k-1}(c_2)|\cos[(2k —1)\omega_1—(2n — 1)\omega]t \\ +\sum_n \sum_k|J_{2n-1}(c_1)J_{2k-1}(c_2)|\cos[(2k —1)\omega_1+(2n — 1)\omega]t\right] \\ +U_0\sin\phi\left[\sum_n |2J_0(c_2)J_{2n-1}(c_1)|\cos[(2n—1)\omega t] \\ +\sum_k | 2J_0(c_1)J_{2k-1}(c_2)|\cos[(2k — 1)\omega_1 t] \\ +\sum_n \sum_k|2J_{2n-1}(c_1)J_{2k}(c_2)|\cos[2k\omega_1—(2n—1)\omega]t \\ +\sum_n \sum_k|2J_{2n-1}(c_1)J_{2k}(c_2)|\cos[2k\omega_1+(2n—1)\omega]t \\ +\sum_n \sum_k|J_{2n}(c_1)J_{2k-1}(c_2)|\cos [(2k—1)\omega_1—2n\omega]t \\ [color=red]\left.[/color]+\sum_n \sum_k|J_{2n}(c_1)J_{2k-1}(c_2)|\cos [(2k—1)\omega_1+2n\omega]t\right] \end{split}$$

You should also correct the mistake that Hurkyl has pointed out.

linderox
but it's it very long brackets!

Staff Emeritus
but it's it very long brackets!

Did I miss something? Is there only one set of brackets in your equation, or are there several? You should repeat the \right. \\ left. procedure for all sets of brackets. Note that this won't do anything to the appearance of the equation, but will just make it work.

linderox
Yes, Cristo! It works with your version,but What I have write - it is my long formula with a beginning bracket \left[ at the 1st line and
ended bracket \right] at the end of the whole formula

i don't understand what you meant... "You should repeat the \right. \\ left. procedure for all sets of brackets..." . maybe you can give some example?
You should repeat the \right. \\ left. procedure for all sets of brackets. Note that this won't do anything to the appearance of the equation, but will just make it work.

2) How better present this formula?
I want to see all lines connected to the left side of the page,but not like now...

Last edited:
Staff Emeritus
This is how I'd typset it (I'm not really that familiar with "split")

Code:
\begin{eqnarray}
%\begin{split}
U&=& U_0\cos[c_1 \cos(\omega t) + c_2 cos(\omega_1 t) + \phi] \\
&=&U_0 \cos\phi\left[|J_0(c_1)J_0(c_2)| + \sum_n |2J_0(c_2 )J_{2n}(c_1)|\cos 2n\omega t \right.\\
&&+\sum_k|2J_0(c_1 )J_{2k}(c_2)|\cos 2k\omega_1 t + \sum_n\sum_k |2J_2n(c_1)J_{2k}(c_2)| \cos [2(k\omega_1 — n\omega)t] \\
&&+\sum_n \sum_k|2J_{2n}(c_1)J_{2k}(c_2)|\cos [2(k\omega_1 + n\omega)t] \\
&&+\sum_n \sum_k|J_{2n-1}(c_1)J_{2k-1}(c_2)|\cos[(2k —1)\omega_1—(2n — 1)\omega]t \\
&&\left.+\sum_n \sum_k|J_{2n-1}(c_1)J_{2k-1}(c_2)|\cos[(2k —1)\omega_1+(2n — 1)\omega]t\right] \\
&&+U_0\sin\phi\left[\sum_n |2J_0(c_2)J_{2n-1}(c_1)|\cos[(2n—1)\omega t\right. \\
&&+\sum_k | 2J_0(c_1)J_{2k-1}(c_2)|\cos[(2k — 1)\omega_1 t] \\
&&+\sum_n \sum_k|2J_{2n-1}(c_1)J_{2k}(c_2)|\cos[2k\omega_1—(2n—1)\omega]t \\
&&+\sum_n \sum_k|2J_{2n-1}(c_1)J_{2k}(c_2)|\cos[2k\omega_1+(2n—1)\omega]t \\
&&+\sum_n \sum_k|J_{2n}(c_1)J_{2k-1}(c_2)|\cos [(2k—1)\omega_1—2n\omega]t \\
&&\left.+\sum_n \sum_k|J_{2n}(c_1)J_{2k-1}(c_2)|\cos [(2k—1)\omega_1+2n\omega]t\right]
%\end{split}
\end{eqnarray}

i don't understand what you meant... "You should repeat the \right. \\ left. procedure for all sets of brackets..." . maybe you can give some example?
I just meant that you've got more than one set of brackets, so need to put more that one \left. and \right. in (I've done it above)

linderox
Wow! great Thank! but now i have a problem counting each line... how to remove everything except the label to the formula?

Staff Emeritus
Code:
\begin{eqnarray}
%\begin{split}
U&=& U_0\cos[c_1 \cos(\omega t) + c_2 cos(\omega_1 t) + \phi] \\
&=&U_0 \cos\phi\left[|J_0(c_1)J_0(c_2)| + \sum_n |2J_0(c_2 )J_{2n}(c_1)|\cos 2n\omega t \right.\nonumber \\
&&+\sum_k|2J_0(c_1 )J_{2k}(c_2)|\cos 2k\omega_1 t + \sum_n\sum_k |2J_2n(c_1)J_{2k}(c_2)| \cos [2(k\omega_1 — n\omega)t] \nonumber \\
&&+\sum_n \sum_k|2J_{2n}(c_1)J_{2k}(c_2)|\cos [2(k\omega_1 + n\omega)t]\nonumber \\
&&+\sum_n \sum_k|J_{2n-1}(c_1)J_{2k-1}(c_2)|\cos[(2k —1)\omega_1—(2n — 1)\omega]t \nonumber \\
&&\left.+\sum_n \sum_k|J_{2n-1}(c_1)J_{2k-1}(c_2)|\cos[(2k —1)\omega_1+(2n — 1)\omega]t\right] \nonumber \\
&&+U_0\sin\phi\left[\sum_n |2J_0(c_2)J_{2n-1}(c_1)|\cos[(2n—1)\omega t\right. \nonumber \\
&&+\sum_k | 2J_0(c_1)J_{2k-1}(c_2)|\cos[(2k — 1)\omega_1 t] \nonumber \\
&&+\sum_n \sum_k|2J_{2n-1}(c_1)J_{2k}(c_2)|\cos[2k\omega_1—(2n—1)\omega]t \nonumber \\
&&+\sum_n \sum_k|2J_{2n-1}(c_1)J_{2k}(c_2)|\cos[2k\omega_1+(2n—1)\omega]t \nonumber \\
&&+\sum_n \sum_k|J_{2n}(c_1)J_{2k-1}(c_2)|\cos [(2k—1)\omega_1—2n\omega]t \nonumber \\
&&\left.+\sum_n \sum_k|J_{2n}(c_1)J_{2k-1}(c_2)|\cos [(2k—1)\omega_1+2n\omega]t\right] \nonumber
%\end{split}
\end{eqnarray}

linderox
Great thanks

Staff Emeritus
Great thanks

You're welcome

linderox
hm... but why it does not work?
\be
\begin{split}
P(t)=cos\theta J_0(\alpha) + 2 \cos \theta \sum_{n=1}^\infty J_{2n}(\alpha)\cos[2n f_0 t +l_0] \\
&&-2\sin\theta \sum_{n=1}^\infty J_{2n-1}(\alpha)cos[(2n-1) f_0 t +l_0]
\end{split}
\label{equ:P_t_bessel}
\ee

Staff Emeritus
I'm not sure; like I said, I don't really know too much about the split command. I can see one problem, namely that you have no && in your first line. Here's another way to write it, along with the output:

Code:
\be
\begin{array}{ccc}
P(t)&=&cos\theta J_0(\alpha) + 2 \cos \theta \sum_{n=1}^\infty J_{2n}(\alpha)\cos[2n f_0 t +l_0] \\
&&-2\sin\theta \sum_{n=1}^\infty J_{2n-1}(\alpha)cos[(2n-1) f_0 t +l_0]
\end{array}
\ee

$$\begin{array}{ccc} P(t)&=&cos\theta J_0(\alpha) + 2 \cos \theta \sum_{n=1}^\infty J_{2n}(\alpha)\cos[2n f_0 t +l_0] \\ &&-2\sin\theta \sum_{n=1}^\infty J_{2n-1}(\alpha)cos[(2n-1) f_0 t +l_0] \end{array}$$

Last edited:
imanius
I have a problem,
I have \left{ and a \righ} in two lines, I want them to be the same size, both big, but only one that has the summation infront of it is big

thank you

\begin{eqnarray*}
\log p(\mathbf{x}_n,\mathbf{z}_n| \bm{\mu},\bm{\pi}) &=& \log \prod_{k=1}^K \left[\pi_k \, p(\mathbf{x}_n|\bm{\mu_k}) \right]^{z_{nk}} \\
&=& \sum_{k=1}^K z_{nk} \{ \log \pi_k + \log p(\mathbf{x}_n|\bm{\mu_k}) \} \\
&=& \sum_{k=1}^K z_{nk} \left\{ \log \pi_k \right. \\
&+& \left. \sum_{i=1}^D [x_{ni}\, \log \mu_{ki}+(1-x_{ni})\, \log (1-\mu_{ki})] \right\}
\end{eqnarray*}

Staff Emeritus