[Tex/LaTex] Put a box around an item (homework)

#enumerateframedlists

I am using the enumerate environment to do homework. In the following, first I write out the question, then I write out the answers to the first two parts.

What I'd like to do is have a box around the question. How can I do that?

I tried using tikzmark as suggested here but it did not work (the box shows up in the wrong place when I try to use math mode. See the code at the end of the article

\documentclass{article}    
\usepackage{amsthm, amssymb, amsfonts, amsmath}

    \begin{document}

    \begin{enumerate}
    \item[1.]
    Consider the initial value problem 
    \[
     \frac{dy}{dx} = 1- (y-x)^2, \qquad y=0 \,\, \text{at} \,\, x=0
    \]
    \begin{enumerate}
     \item[(a)] Find a solution to the problem.
     \item[(b)] Regardless of (a), is the solution unique?
     \item[(c)] Write down the Picard iteration scheme for this ODE and find some $r>0$ for which it converges $|x|<r$.
     \item[(d)] Find some $R>0$ such that the Picard iterates \textit{diverge} for $|x|>R$. 


    \end{enumerate}

    \item[1a.]
       $\phi(x) = x$ is a solution, since $$\phi'(x) = 1 = 1 - (x-x)^2.$$

    \item[1b.]
       Note that the function $1-(y-x)^2$ is $C^{\infty}$ in both $x$ and $y$, and therefore given $x_0$, the function is locally Lipschitz in $y$, uniformly with respect to $x$ for $x$ in a neighborhood of $x_0$. Therefore such a solution is unique.
     \end{enumerate}
     \end{document}

Using the tikzmark idea (didn't work)

\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usetikzlibrary{calc,shapes}
\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};}
\begin{document}
\begin{enumerate}
\item[\tikzmark{bl} 1.]
Consider the initial value problem 
\[
 \frac{dy}{dx} = 1- (y-x)^2, \qquad y=0 \,\, \text{at} \,\, x=0
\]
\begin{enumerate}
 \item[(a)] Find a solution to the problem.
 \item[(b)] Regardless of (a), is the solution unique?
 \item[(c)] Write down the Picard iteration scheme for this ODE and find some $r>0$ for which it converges $|x|<r$.
 \item[(d)] Find some $R>0$ such that the Picard iterates \textit{diverge} for $|x|>R$. 
\end{enumerate}
\tikzmark{br}
\item[2.] Something not boxed
\item[3.] Also not boxed.
\end{enumerate}
\tikz[overlay,remember picture]{\draw[red]
  ($(bl)+(-0.2em,0.9em)$) rectangle
  ($(br)+(0.2em,-0.3em)$);}
\end{document}

Edit: A problem came when I tried to apply the suggestion given below. tikzmark always wants to put the box on the last page. So if the document is only one page, it is fine. But if it jumps to two pages, it is wrong.

\documentclass{article}
\usepackage{tikz}
\usepackage{enumitem}
\usepackage{amsmath}
\usetikzlibrary{calc,shapes}
\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};}
\begin{document}
\begin{enumerate}
\item
\leavevmode
\strut
\vadjust{%%
  \noindent
  \raisebox{\dimexpr\dp\strutbox+\ht\strutbox+1ex}[0pt][0pt]{\tikzmark{bl}}}%%
Consider the initial value problem 
\[
  \frac{dy}{dx} = 1- (y-x)^2, \qquad y=0 \,\, \text{at} \,\, x=0
\]
  \begin{enumerate}
   \item Find a solution to the problem.
   \item Regardless of (a), is the solution unique?
   \item Write down the Picard iteration scheme for this ODE and find some $r>0$ for which it converges $|x|<r$.
   \item Find some $R>0$ such that the Picard iterates \textit{diverge} for $|x|>R$. 
  \end{enumerate}

\leavevmode
\vadjust{%
  \noindent
  \hspace*{\dimexpr\textwidth+1ex}\tikzmark{br}}%%

  \item[1a.]
   $\phi(x) = x$ is a solution, since $$\phi'(x) = 1 = 1 - (x-x)^2.$$

\item[1b.]
   Note that the function $1-(y-x)^2$ is $C^{\infty}$ in both $x$ and $y$, and therefore given $x_0$, the function is locally Lipschitz in $y$, uniformly with respect to $x$ for $x$ in a neighborhood of $x_0$. Therefore such a solution is unique.

   \item[1c.]
    For convenience of notation, let us agree to label the constant function $0$ as the first Picard iterate $y_1(x)$ rather than the $0$th Picard iterate.
 \begin{align*}
  y_2(x) &= \int_0^x 1 - (y_1 - \xi)^2 d\xi = x - \frac13 x^3 \\
  y_3(x) &= \int_0^x 1 - (y_2(\xi) - \xi)^2 d\xi = \int_0^x 1 - \left(\frac13 \xi^3\right)^2 d\xi = x - \frac{1}{7\cdot 3^2}x^7 \\
  y_4(x) &= \dotsb = x - \frac{1}{15\cdot 7^2 \cdot 3^4}x^{15}\\
  \vdots \\
  y_n(x) &= x - \frac{1}{\prod_{k=1}^{n} (2^k-1)^{2^{(n-k)}}} x^{2^n - 1}
 \end{align*}
 The iterates will converge pointwise for $|x|<1$ (making no claim that that is the largest such interval). Since the modulus of the second term of $y_n$ is increasing with the modulus of $x$, we can say that the iterates will converge uniformly on compact subsets of the unit interval. (Given $[-a, a]\subseteq (-1,1)$, pick $n$ such that the second term of $y_n(a)$ is less than $\epsilon$; then this $n$ will work for all $x \in [-a, a]$.)

\item[1d.]
 Suppose we find an $x$ for which 
 \begin{align*}
  x-\frac{1}{\prod_{k=1}^{n} (2^k-1)^{2^{(n-k)}}} x^{2^n - 1}
  \end{align*}
 diverges to infinity (of course it will diverge iff the second term diverges). Then by the same considerations as in part (c), we can say that it will diverge for $\xi$ such that $|\xi| > |x|$.  

 Suppose we choose $x=4$. We have 

 \[
 y_n(x) = 4 - \frac{2^{2(2^n-1)}}{\prod_{k=1}^{n} (2^k-1)^{2^{(n-k)}}}.
 \]
 Note that the denominator satisfies
 \[
  (2^n-1)^{2^0}(2^{n-1}-1)^{2^1}\dotsb (2^2-1)^{2^{n-2}}\leq 2^{n2^0 + (n-1)2^1 + \dotsb + (2)2^{n-2}},
 \]
 where we have left out the final factor because it is equal to one. Some algebra tells us that $n2^0 + (n-1)2^1 + \dotsb + (2)2^{n-2} = 2^{n}+2^{n-1}-(n+2)\leq 2^{n+1}-(n+2)$. When we consider 
 \[
  \frac{2^{2(2^n-1)}}{\prod_{k=1}^{n} (2^k-1)^{2^{(n-k)}}} \geq \frac{C2^{2^{n+1}}}{2^{2^{n+1}-(n+2)}}=C2^{n+2},
 \]
 this clearly diverges.

 \item[2.] Let $y=\phi(x)$ and $y=\psi(x)$ be linearly independent solutions of the ODE $y'' + p(x)y' + q(x)y = 0$, where $p$ and $q$ are continuous on an open interval $I$. Prove the following statements:
 \begin{enumerate}
 \item[(a)] Suppose that $x_0 \in I$ is a zero of $\phi$, then $\psi$ cannot have a relative extremum value at $x_0$.
 \item[(b)] Suppose that $x_0$ and $x_1$ are two consecutive zeros of $\phi$ in $I$. Then $\psi$ must have a zero at some $\xi$, $x_0<\xi<x_1$.

 \end{enumerate}

[\textit{Hint:} Think Wronskian.]

\end{enumerate}
\tikz[overlay,remember picture]{\draw[red]
  (bl) rectangle
  (br);}
\end{document}

Best Answer

Based on my answer at How can I add lists and similar in a savebox?, I placed the question in a temporary vbox, and then embraced it in an \fbox, making the appropriate allowances for indentation.

I here define the process as a macro, \boxitem{}:

\def\boxitem#1{\setbox0=\vbox{#1}{\centering\makebox[0pt]{%
  \fboxrule=2pt\color{red}\fbox{\hspace{\leftmargini}\color{black}\box0}}\par}}

And then use it inside the enumerate as \boxitem{\item[]...}. Note: \boxitem was REVISED to make it insensitive to succeeding blank lines.

\documentclass{article}    
\usepackage{amsthm, amssymb, amsfonts, amsmath, xcolor}% ADDED xcolor
\def\boxitem#1{\setbox0=\vbox{#1}{\centering\makebox[0pt]{%
  \fboxrule=2pt\color{red}\fbox{\hspace{\leftmargini}\color{black}\box0}}\par}}
    \begin{document}
    \begin{enumerate}
\boxitem{% THIS LINE ADDED
    \item[1.]
    Consider the initial value problem 
    \[
     \frac{dy}{dx} = 1- (y-x)^2, \qquad y=0 \,\, \text{at} \,\, x=0
    \]
    \begin{enumerate}
     \item[(a)] Find a solution to the problem.
     \item[(b)] Regardless of (a), is the solution unique?
     \item[(c)] Write down the Picard iteration scheme for this ODE and find some $r>0$ for which it converges $|x|<r$.
     \item[(d)] Find some $R>0$ such that the Picard iterates \textit{diverge} for $|x|>R$. 
    \end{enumerate}
}% THIS LINE ADDED
    \item[1a.]
       $\phi(x) = x$ is a solution, since $$\phi'(x) = 1 = 1 - (x-x)^2.$$
    \item[1b.]
       Note that the function $1-(y-x)^2$ is $C^{\infty}$ in both $x$ and $y$, and therefore given $x_0$, the function is locally Lipschitz in $y$, uniformly with respect to $x$ for $x$ in a neighborhood of $x_0$. Therefore such a solution is unique.
     \end{enumerate}
     \end{document}

enter image description here

POSTSCRIPT: From my exchange with the OP in the comments, I will just add that if one wanted to add a fixed vertical space before and after my \fbox, the definition of \boxitem could be defined as follows, where a space of 3ex is added, by way of example.

\def\boxitem#1{\vspace{3ex}\setbox0=\vbox{#1}{\centering\makebox[0pt]{%
  \fboxrule=2pt\color{red}\fbox{\hspace{\leftmargini}\color{black}\box0}}\vspace{3ex}\par}}

Code:

\documentclass{article}
\usepackage{amsmath}
\usepackage{tcolorbox}

\begin{document}
\begin{tcolorbox}

blablabla

\begin{align}
E &= mc^2 & \text{Formula of the universe}
\end{align}


Hoaray

\end{tcolorbox}

\end{document}

[Tex/LaTex] Box around a few items in an itemize environment

One way to do that is to adapt the code from Highlight elements in the matrix. Below, I provide two macros \DrawBox and \DrawBoxWide depending on if you want to include the list markers or not. The default below is to produce a red box, but this can be adjusted to suit, or the options can be specified on a per use basis (as in the second example):

enter image description here

However, since the width of the box is based on the width of the last item, there is a possibility that the box won't be wide enough if the last item is shorter than the wides item within the box. For this case I have defined \DrawBox* and \DrawBoxWide* which use enclose the items in a box as wide as the line:

enter image description here

Note that in the above the width of the box is the full \linewidth. This is not an issue if the text uses the full width as is the case of the blue box above, but for the case where the text is not the full width (as is the case for the red box), one might desire a tight box around the text. In this case I would recommend that the last line be set in a box as wide as the widest text using:

\makebox[\widthof{\WidestText}][l]{<text>}\tikzmark{right}

which yields:

enter image description here

Syntax:

To use this you place \tikzmark{left} to mark the position where the top left of the box is, and \tikzmark{right} to mark the position of the bottom right of the box. After these two positions have been marked you call the \DrawBox or \DrawBoxWide (or the starred variant) macros to draw the box.

Notes:

This does require two runs: the first to compute the positions of the box, and the second to draw it in the correct spot.
Since this is using tikz, you automatically get all the flexibility inherent in tikz, such as line styles, line thickness, line color, fill, etc. These can be passed to the \DrawBox macros to customize each instance or provided as default options to maintain consistency.

Code:

\documentclass{article}
\usepackage{xparse}%  For \NewDocumentCommand
\usepackage{calc}%    For the \widthof macro
\usepackage{tikz}
\usetikzlibrary{calc}

\newcommand{\Text}{Lorem ipsum dolor sit amet, consectetur adipiscing elit. 
    Sed accumsan nulla ac ipsum elementum interdum. }

\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};}

\makeatletter
\NewDocumentCommand{\DrawBox}{s O{}}{%
    \tikz[overlay,remember picture]{
    \IfBooleanTF{#1}{%
        \coordinate (RightPoint) at ($(left |- right)+(\linewidth-\labelsep-\labelwidth,0.0)$);
    }{%
        \coordinate (RightPoint) at (right.east);
    }%
    \draw[red,#2]
      ($(left)+(-0.2em,0.9em)$) rectangle
      ($(RightPoint)+(0.2em,-0.3em)$);}
}

\NewDocumentCommand{\DrawBoxWide}{s O{}}{%
    \tikz[overlay,remember picture]{
    \IfBooleanTF{#1}{%
        \coordinate (RightPoint) at ($(left |- right)+(\linewidth-\labelsep-\labelwidth,0.0)$);
    }{%
        \coordinate (RightPoint) at (right.east);
    }%
    \draw[red,#2]
      ($(left)+(-\labelwidth,0.9em)$) rectangle
      ($(RightPoint)+(0.2em,-0.3em)$);}
}
\makeatother

\begin{document}
Using the \verb|\DrawBox and \DrawBoxWide|:
\medskip\par\noindent
\begin{minipage}{0.4\linewidth}
  \begin{enumerate}
        \item First Item
        \item \tikzmark{left}Second Item
        \item Third Item
        \item Fourth Item\tikzmark{right}
        \item Fifth Item
  \end{enumerate}
    \DrawBox[thick]
\end{minipage}
%
\begin{minipage}{0.4\linewidth}
  \begin{enumerate}
        \item First Item
        \item \tikzmark{left}Second Item
        \item Third item
        \item Fourth Item\tikzmark{right}
        \item Fifth Item
  \end{enumerate}
    \DrawBoxWide[thick, blue, fill=yellow, fill opacity=0.2]
\end{minipage}

Using the \verb|\DrawBox* and \DrawBoxWide*|:
\medskip\par\noindent
\begin{minipage}{0.4\linewidth}
  \begin{enumerate}
        \item First Item
        \item \tikzmark{left}Second Item
        \item Third item.
        \item Fourth Item\tikzmark{right}
        \item Fifth Item
  \end{enumerate}
    \DrawBox*[thick]
\end{minipage}
%
\begin{minipage}{0.4\linewidth}
  \begin{enumerate}
        \item First Item
        \item \tikzmark{left}Second Item
        \item Third item: \Text
        \item Fourth Item\tikzmark{right}
        \item Fifth Item
  \end{enumerate}
    \DrawBoxWide*[thick, blue, fill=yellow, fill opacity=0.2]
\end{minipage}

\noindent
For the case where the last line is not the longest and the lines before do not use the full width of the text, as would be the case below with \verb|\DrawBoxWide*|, you can use \verb|\makebox{\widthof{<widest_text>}{<text>}|

\newcommand*{\WidestText}{Third item which is wider}%
\begin{enumerate}
    \item First Item
    \item \tikzmark{left}Second Item
    \item Third item which is wider
    \item \makebox[\widthof{\WidestText}][l]{Fourth Item}\tikzmark{right}
    \item Fifth Item
\end{enumerate}
\DrawBoxWide[thick, brown]
\end{document}

Best Answer

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