I tried this code below but it does not execute and I couldn't find out what is wrong in the equation. What did I do wrong?

$MDAC(t_{i,j}) = max_{ti}\epsilon pred(t_{j})(C_{i,j)$
  • 7
    I think we're all dying to know what MDAC is? :P – Au101 Dec 2 '15 at 20:41

Your primary problem is with (C_{i,j), which should be (C_{i,j}), in other words, you have not paired all your delimeters, you need a closing brace. You should have received the error:

! Missing } inserted.

More generally I would recommend:

$MDAC(t_{j}) = \max_{t_{i} \in pred(t_{j})}(C_{i,j})$

Or perhaps you want?


$MDAC(t_{j}) = \max_{t_{i} \in \pred(t_{j})}(C_{i,j})$

enter image description here

According to others, MDAC is a single unit. It is probably better, therefore, to treat it as an operator and set it upright, but suppose you want it set in italics, you can have:

$\mathit{MDAC}(t_{j}) = \max_{t_{i} \in \pred(t_{j})}(C_{i,j})$

Which will improve the spacing between the letters, treating MDAC as one unit, rather than the product of four separate variables M, D, A and C. For upright typesetting follow Bernard's advice or what I have done with \pred:


$\MDAC(t_{j}) = \max_{t_{i} \in \pred(t_{j})}(C_{i,j})$

enter image description here

  • 2
    "MDAC" should be entered either as \mathrm{MDAC} or as \mathit{MDAC}. Entering it as MDAC induces TeX to typeset it as if were a collection of four separate variables named M, D, A, and C. – Mico Dec 2 '15 at 20:34
  • @Mico thanks, fair enough, I'm genuinely unfamiliar with MDAC so I left it alone, for all I knew, it is the product of the three variables M, D and A with C(t_{j}). But I'll edit, although it makes my answer functionally indistinguishable from Bernard's ... – Au101 Dec 2 '15 at 20:35
  • Well, you can choose the \mathit method to make your solution look different from Bernard's. :-) – Mico Dec 2 '15 at 20:37

I don't think MDAC and prec should be typed in italics. I suggest considering them as maths operators, like log, sin, &c., to ensure a correct spacing w. r. t. possible exponents, indices or delimiters.



 $ \MDAC(t_{i,j}) = \max_{t_i\in \pred(t_j)}(C_{i,j}) $ %


enter image description here

  • 1
    Is "MDAC" an operator or the name of a variable? If it's the latter, it's probably better to define \newcommand\MDAC{\mathrm{MDAC}}. – Mico Dec 2 '15 at 20:35
  • 3
    @Mico: I wondered. I opted for an operator because it was followed by a pair of parentheses. Maybe the O.P. will tell us. – Bernard Dec 2 '15 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.