3

For the following two approaches or sets of equations, I would like to

1- (in the first set) make the first for align with the two fors below

2- (in the second set) keep the font and interspaces the same as the first set

enter image description here

\documentclass{article}
\usepackage{mathtools,nccmath}
\begin{document}
\begin{fleqn}
    \begin{gather}
    t_k = \left\{
    \begin{alignedat}{2}
    &{\frac{k}{N}}^\nu & \quad &\text{for } \nu > 1\\
    &\left\{\begin{alignedat}{2}
    &t_0 \left( \frac{k}{N_0} \right)^\nu & \quad &\text{for } 0 \leq k \leq N_0\\
    %
    &t_0 + \frac{k-N_0}{N-N_0} \left(t - t_0\right) & &\text{for} \enspace N_0 < k \leq N, t_0 < t
    \end{alignedat}\right.
    \end{alignedat}\right.\\[\baselineskip]
    %%
    % https://tex.stackexchange.com/a/84072/2288
    t_k = 
    {\left\{\begin{array}{@{}l@{\quad}l@{}}
        {\frac{k}{N}}^\nu & \text{for } \nu > 1 \\
        \left\{\begin{array}{@{}l@{}}
        t_0 \left( \frac{k}{N_0} \right)^\nu \\
        t_0 + \frac{k-N_0}{N-N_0} \left(t - t_0\right)
        \end{array}\right.\kern-\nulldelimiterspace
        & \begin{array}{@{}l@{}}
        \text{for } 0 \leq k \leq N_0 \\ 
        \text{for } N_0 < k \leq N, t_0 < t
        \end{array}
        \end{array}\right.}
    \end{gather}
\end{fleqn}
\end{document}
  • 1
    I tried understanding the definition of t_k, but I can't. In the second set there is no n, for instance. You're overcomplicating things, in my opinion. – egreg Jul 28 at 13:02
  • @egreg This question is more of a practice to understand the nesting than a real math problem. Therefore, I summed up all my thoughts in a single question instead of asking many questions on the site. – Diaa Jul 28 at 13:21
  • 1
    I'm not sure that's sensible notation in any case. – egreg Jul 28 at 13:31
  • 1
    You wrote in a comment, "I summed up all my thoughts in a single question". That's actually not a clever strategy; if the issues are not self-evidently related to each other, it's almost certainly better to post separate queries. – Mico Jul 28 at 13:41
  • 1
    @Mico I understand what you both mean, so I happily edit the question to make it mathematically consistent :) – Diaa Jul 28 at 13:42
3

Here's a solution to your second question. AFAICT, the solution answers also your first question implicitly.

Some comments:

  • I added a \left( ... \right) "wrapper" around the first fraction term, as otherwise it's not obvious whether the \nu exponent term pertains to the entire fraction or to just the numerator.

  • The solution also inserts typographic struts in the second "inner" array to assure that the rows' heights are the same as in the first "inner" array.

  • Finally, \addlinespace directives (provided by the booktabs package) are inserted to increase the vertical separation between rows.

enter image description here

That said, I don't believe that this layout is all that clear. For instance, shouldn't there be an indication somewhere that rows 2 and 3 pertain to the case $\nu \le 1?

\documentclass{article} 
\usepackage{nccmath}   % for 'fleqn' env.
\usepackage{array}    % for \newcolumntype macro
\usepackage{booktabs} % for \addlinespace macrdo
\newcolumntype{L}{>{\displaystyle}l}
% Define two (typographical) struts (to be used in math mode only):
\newcommand\strutA{\vphantom{%
   \displaystyle\left( \frac{k}{N_0} \right)^{\nu}}}
\newcommand\strutB{\vphantom{%
   \displaystyle\frac{k-N_0}{N-N_0}}}
   
\begin{document}
\begin{fleqn}
\begin{equation}
t_k = 
    \left\{
    \begin{array}{@{} L @{\quad} L @{}}
        \left(\frac{k}{N}\right)^{\!\nu} 
        & \text{for $\nu > 1$} \\
        \addlinespace
        \left\{ 
        \begin{array}{@{} L @{}}
           t_0 \left(\frac{k}{N_0}\right)^{\!\nu} \\
           \addlinespace
           t_0 + \frac{k-N_0}{N-N_0} (t - t_0)
        \end{array}
        \right.\kern-\nulldelimiterspace
        &
        \begin{array}{@{} L @{}}
           \text{for $0 \leq k \leq N_0$}\strutA \\ 
           \addlinespace
           \text{for $N_0 < k \leq N$, $t_0 < t$}\strutB
        \end{array}
    \end{array}
    \right.
\end{equation}
\end{fleqn}
\end{document}
| improve this answer | |
  • Thanks for your answer. Actually, \nu >1 is a general condition for every single case. So, if you could propose a better place for ot, I would be grateful. – Diaa Jul 28 at 14:24
  • @Diaa - If \nu>1 is always true, then only the first of the three rows can ever be true, right? In that case, why bother with showing rows 2 and 3? – Mico Jul 28 at 14:42
  • @Diaa - Please indicate whether \nu>1 and t_0<t are both always true. – Mico Jul 28 at 15:49
  • Only \nu > 1 is the real conditions here. t0 is a mark over the timeline. So, t can be less than t0. Sorry for late reply, I was away from PC – Diaa Jul 28 at 16:18
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    @Diaa - Honestly, I am still quite unsure about what it is you're trying to express. Hence, I am in no position to suggest anything one might label "the right way". But I can say confidently that the three-row cases-like expression you've presented is not the right way to go. For sure, if the information in row 1 is always true, you should not place that piece of information in a cases-like structure along with two other rows. Readers have every right to expect that rows in cases-like structure are mutually exclusive and exhaustive. – Mico Jul 28 at 17:50

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