I can't figure out why my code won't work. I need an answer as soon as possible.

\therefore \delta F \left(E_{x} \right )&=\int \left [ \frac{\partial F}{\partial E }\delta E+\frac{\partial F}{\partial\dot{E}}\delta \dot{E}\right ]dx=0\\
&=\int\left {\left[\frac{\partial H_{z}}{\partial y}-\frac{\partial H_{y}}{\partial x}-2i\omega\varepsilon_{0}\varepsilon \\E_{x} \right ]\delta E_{x}-\left[ \frac{\partial}{\partial x}\left ( \frac{\partial F}{\partial E/\partial x}\right )+\frac{\partial}{\partial y}\left ( \frac{\partial F}{\partial E/\partial y}\right )+\frac{\partial}{\partial z}\left ( \frac{\partial F}{\partial E/\partial z}\right )\right ]\delta E_{x}\right } \ dx\\
&=\int\left[\left(\frac{\partial H_{z}}{\partial y}-\frac{\partial H_{y}}{\partial x}-2i\omega\varepsilon_{0}\varepsilon \\E_{x} \right ) - \left(-\frac{\partial H_{z}}{\partial y}+\frac{\partial H_{y}}{\partial x} \right )\right ]\delta E_{x} \ dx\\
&=\int 2 \(\frac{\partial H_{z}}{\partial y}-\frac{\partial H_{y}}{\partial x}-i\omega\varepsilon_{0}\varepsilon \\E_{x}\)\delta E_{x} \ dx &=0.

The error I get is:

! Missing delimiter (. inserted). { l.50 \end{split}

Please help!

  • 1
    cells such as \\E_{x} \right ] have a \right with no matching \left Jul 28, 2020 at 17:15
  • 1
    To elaborate, between \left and \right, you can't have & nor \\ . Also, \left{ won't work. If you want a literal {, you need to type \{, so that you'd have \left\{.
    – Teepeemm
    Jul 28, 2020 at 17:26
  • 1
    To be expand on David's comment, you need to match your \left and \right commands. Note that most of the time you don't need to use these unless you have an inner content that will require larger than usual delimiters. If for some reason you need unbalanced delimiters, you can use \left. or \right. to create an invisible left or right delimiter.
    – Don Hosek
    Jul 28, 2020 at 17:28
  • 1
    i also see a \( on the last light which I imagine was meant to be \left(
    – Don Hosek
    Jul 28, 2020 at 17:28

1 Answer 1


You have some misplaced \\ before E_x

I'd avoid \left and \right for this display, preferring \bigg size. In any case, \left and \right cannot straddle different parts of the alignment.


\newcommand{\pder}[2]{\frac{\partial #1}{\partial #2}}


\therefore \delta F(E_{x})
  &=\int \biggl[ \pder{F}{E}\delta E+\pder{F}{\dot{E}}\delta \dot{E} \biggr]\,dx=0 \\[2ex]
        \pder{H_{z}}{y}-\pder{H_{y}}{x}-2i\omega\varepsilon_{0}\varepsilon E_{x}
      \biggr]\delta E_{x} \\
        \biggl( \pder{F}{E/\partial x} \biggr)+\pder{}{y} \biggl( \pder{F}{E/\partial y} \biggr)
        +\pder{}{z} \biggl( \pder{F}{E/\partial z} \biggr)
      \biggr]\delta E_{x}
    \biggr\}\,dx \\[2ex]
      \biggl(\pder{H_{z}}{y}-\pder{H_{y}}{x}-2i\omega\varepsilon_{0}\varepsilon E_{x} \biggr)
      \biggl(-\pder{H_{z}}{y}+\pder{H_{y}}{x} \biggr)
    \biggr]\delta E_{x} \, dx\\[2ex]
&=\int 2
    \pder{H_{z}}{y}-\pder{H_{y}}{x}-i\omega\varepsilon_{0}\varepsilon E_{x}
  \biggr)\delta E_{x} \, dx\\[2ex]
& =0.


The second integral needs to be split across lines, I added some spacing to make clear that the third line is a continuation of the second one and also added vertical spaces between main lines.

An abbreviation for partial derivatives has been defined, to ease input.

enter image description here

I'd also omit the \therefore symbol: just spell out “therefore” before the display.

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