How to enclose theorems and definition in rectangles?












1















The following code



documentclass{article}


usepackage{amsthm}
usepackage{amsmath}
usepackage{mathtools}

usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}



newtheorem{definition}{Definition}
newtheorem{theorem}{Theorem}


begin{document}
title{Extra Credit}
maketitle

begin{definition}
If f is analytic at $z_0$, then the series

begin{equation}
f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
end{equation}

is called the Taylor series for f around $z_0$.
end{definition}

begin{theorem}
If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
begin{equation}
f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
end{equation}
end{theorem}

begin{theorem}
(Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
begin{equation}
f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
end{equation}
end{theorem}
noindent hrulefill

begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
end{theorem}


produces the following image
enter image description here



How can I enclose Definition 1, Theorem 1, and Theorem 2 in separate rectangles. And have these rectangles separated by a space?










share|improve this question























  • Do you want all theorems/definition to be enclosed in a frame, or only some?

    – Bernard
    Apr 2 at 21:55













  • I would like all theorems/definitions to be enclosed in a frame except for Theorem 3

    – K.M
    Apr 2 at 21:57











  • In this case you should take a look at the newframedtheorem command in ntheorem.

    – Bernard
    Apr 2 at 22:06
















1















The following code



documentclass{article}


usepackage{amsthm}
usepackage{amsmath}
usepackage{mathtools}

usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}



newtheorem{definition}{Definition}
newtheorem{theorem}{Theorem}


begin{document}
title{Extra Credit}
maketitle

begin{definition}
If f is analytic at $z_0$, then the series

begin{equation}
f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
end{equation}

is called the Taylor series for f around $z_0$.
end{definition}

begin{theorem}
If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
begin{equation}
f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
end{equation}
end{theorem}

begin{theorem}
(Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
begin{equation}
f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
end{equation}
end{theorem}
noindent hrulefill

begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
end{theorem}


produces the following image
enter image description here



How can I enclose Definition 1, Theorem 1, and Theorem 2 in separate rectangles. And have these rectangles separated by a space?










share|improve this question























  • Do you want all theorems/definition to be enclosed in a frame, or only some?

    – Bernard
    Apr 2 at 21:55













  • I would like all theorems/definitions to be enclosed in a frame except for Theorem 3

    – K.M
    Apr 2 at 21:57











  • In this case you should take a look at the newframedtheorem command in ntheorem.

    – Bernard
    Apr 2 at 22:06














1












1








1








The following code



documentclass{article}


usepackage{amsthm}
usepackage{amsmath}
usepackage{mathtools}

usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}



newtheorem{definition}{Definition}
newtheorem{theorem}{Theorem}


begin{document}
title{Extra Credit}
maketitle

begin{definition}
If f is analytic at $z_0$, then the series

begin{equation}
f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
end{equation}

is called the Taylor series for f around $z_0$.
end{definition}

begin{theorem}
If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
begin{equation}
f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
end{equation}
end{theorem}

begin{theorem}
(Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
begin{equation}
f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
end{equation}
end{theorem}
noindent hrulefill

begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
end{theorem}


produces the following image
enter image description here



How can I enclose Definition 1, Theorem 1, and Theorem 2 in separate rectangles. And have these rectangles separated by a space?










share|improve this question














The following code



documentclass{article}


usepackage{amsthm}
usepackage{amsmath}
usepackage{mathtools}

usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}



newtheorem{definition}{Definition}
newtheorem{theorem}{Theorem}


begin{document}
title{Extra Credit}
maketitle

begin{definition}
If f is analytic at $z_0$, then the series

begin{equation}
f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
end{equation}

is called the Taylor series for f around $z_0$.
end{definition}

begin{theorem}
If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
begin{equation}
f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
end{equation}
end{theorem}

begin{theorem}
(Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
begin{equation}
f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
end{equation}
end{theorem}
noindent hrulefill

begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
end{theorem}


produces the following image
enter image description here



How can I enclose Definition 1, Theorem 1, and Theorem 2 in separate rectangles. And have these rectangles separated by a space?







spacing






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Apr 2 at 21:38









K.MK.M

1676




1676













  • Do you want all theorems/definition to be enclosed in a frame, or only some?

    – Bernard
    Apr 2 at 21:55













  • I would like all theorems/definitions to be enclosed in a frame except for Theorem 3

    – K.M
    Apr 2 at 21:57











  • In this case you should take a look at the newframedtheorem command in ntheorem.

    – Bernard
    Apr 2 at 22:06



















  • Do you want all theorems/definition to be enclosed in a frame, or only some?

    – Bernard
    Apr 2 at 21:55













  • I would like all theorems/definitions to be enclosed in a frame except for Theorem 3

    – K.M
    Apr 2 at 21:57











  • In this case you should take a look at the newframedtheorem command in ntheorem.

    – Bernard
    Apr 2 at 22:06

















Do you want all theorems/definition to be enclosed in a frame, or only some?

– Bernard
Apr 2 at 21:55







Do you want all theorems/definition to be enclosed in a frame, or only some?

– Bernard
Apr 2 at 21:55















I would like all theorems/definitions to be enclosed in a frame except for Theorem 3

– K.M
Apr 2 at 21:57





I would like all theorems/definitions to be enclosed in a frame except for Theorem 3

– K.M
Apr 2 at 21:57













In this case you should take a look at the newframedtheorem command in ntheorem.

– Bernard
Apr 2 at 22:06





In this case you should take a look at the newframedtheorem command in ntheorem.

– Bernard
Apr 2 at 22:06










2 Answers
2






active

oldest

votes


















1














You can try with shadethm package, it can do all you want and many more. In you example what you need is:



documentclass{article}
usepackage{shadethm}
usepackage{mathtools}

newshadetheorem{boxdef}{Definition}[section]
newshadetheorem{boxtheorem}[boxdef]{Theorem}
newtheorem{theorem}[boxdef]{Theorem}

setlength{shadeboxsep}{2pt}
setlength{shadeboxrule}{.4pt}
setlength{shadedtextwidth}{textwidth}
addtolength{shadedtextwidth}{-2shadeboxsep}
addtolength{shadedtextwidth}{-2shadeboxrule}
setlength{shadeleftshift}{0pt}
setlength{shaderightshift}{0pt}
definecolor{shadethmcolor}{cmyk}{0,0,0,0}
definecolor{shaderulecolor}{cmyk}{0,0,0,1}

begin{document}

section{Boxed theorems}

begin{boxdef}
If f is analytic at $z_0$, then the series

begin{equation}
f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
end{equation}

is called the Taylor series for f around $z_0$.
end{boxdef}

begin{boxtheorem}
If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
begin{equation}
f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
end{equation}
end{boxtheorem}

begin{boxtheorem}
(Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
begin{equation}
f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
end{equation}
end{boxtheorem}
noindent hrulefill

begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
end{theorem}

end{document}


which produces the following:



enter image description here






share|improve this answer
























  • For newshadetheorem{boxdef}{Definition}[section] newshadetheorem{boxtheorem}[boxdef]{Theorem} newtheorem{theorem}[boxdef]{Theorem}, why is boxdef in brackets?

    – K.M
    Apr 2 at 22:37








  • 1





    In the first box, the space above the equation is larger than that below the equation The reason for this is the blank line above begin{equation}. Blank lines in that position should be avoided.

    – barbara beeton
    Apr 3 at 21:06






  • 1





    @K.M the brackets [boxdef] is to enumerate all different kind of theorems with the same enumeration

    – Luis Turcio
    Apr 3 at 21:50











  • @barbarabeeton the spacing is due to the original code written by K.M, it has a blank line before begin{equation} and one after end{equation}. Removing or commenting this blank lines should be enough to correct spacing.

    – Luis Turcio
    Apr 3 at 21:54











  • @LuisTurcio -- Indeed, commenting or removing the blank line is what is recommended. I really should have posted this comment to the original question.

    – barbara beeton
    Apr 4 at 0:12



















2














Here is a solution with thmtools, which cooperates wit amsthm. Unrelated: you don't have to load amsmath if you load mathtools, as the latter does it for you:



documentclass{article}
usepackage{amsthm, thmtools}
usepackage{mathtools}

usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}

newtheorem{definition}{Definition}
newtheorem{theorem}{Theorem}

declaretheorem[sibling=definition, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Definition]{boxeddef}
declaretheorem[sibling=theorem, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Theorem]{boxedthm}

begin{document}
title{Extra Credit}
author{}
maketitle

begin{boxeddef}
If f is analytic at $z_0$, then the series

begin{equation}
f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
end{equation}

is called the Taylor series for f around $z_0$.
end{boxeddef}

begin{boxedthm}
If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
begin{equation}
f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
end{equation}
end{boxedthm}

begin{boxedthm}
(Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
begin{equation}
f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
end{equation}
end{boxedthm}

noindent hrulefill

begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
end{theorem}

end{document}


enter image description here






share|improve this answer
























    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "85"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f482860%2fhow-to-enclose-theorems-and-definition-in-rectangles%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    You can try with shadethm package, it can do all you want and many more. In you example what you need is:



    documentclass{article}
    usepackage{shadethm}
    usepackage{mathtools}

    newshadetheorem{boxdef}{Definition}[section]
    newshadetheorem{boxtheorem}[boxdef]{Theorem}
    newtheorem{theorem}[boxdef]{Theorem}

    setlength{shadeboxsep}{2pt}
    setlength{shadeboxrule}{.4pt}
    setlength{shadedtextwidth}{textwidth}
    addtolength{shadedtextwidth}{-2shadeboxsep}
    addtolength{shadedtextwidth}{-2shadeboxrule}
    setlength{shadeleftshift}{0pt}
    setlength{shaderightshift}{0pt}
    definecolor{shadethmcolor}{cmyk}{0,0,0,0}
    definecolor{shaderulecolor}{cmyk}{0,0,0,1}

    begin{document}

    section{Boxed theorems}

    begin{boxdef}
    If f is analytic at $z_0$, then the series

    begin{equation}
    f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
    end{equation}

    is called the Taylor series for f around $z_0$.
    end{boxdef}

    begin{boxtheorem}
    If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
    end{equation}
    end{boxtheorem}

    begin{boxtheorem}
    (Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
    end{equation}
    end{boxtheorem}
    noindent hrulefill

    begin{theorem}
    If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
    end{theorem}

    end{document}


    which produces the following:



    enter image description here






    share|improve this answer
























    • For newshadetheorem{boxdef}{Definition}[section] newshadetheorem{boxtheorem}[boxdef]{Theorem} newtheorem{theorem}[boxdef]{Theorem}, why is boxdef in brackets?

      – K.M
      Apr 2 at 22:37








    • 1





      In the first box, the space above the equation is larger than that below the equation The reason for this is the blank line above begin{equation}. Blank lines in that position should be avoided.

      – barbara beeton
      Apr 3 at 21:06






    • 1





      @K.M the brackets [boxdef] is to enumerate all different kind of theorems with the same enumeration

      – Luis Turcio
      Apr 3 at 21:50











    • @barbarabeeton the spacing is due to the original code written by K.M, it has a blank line before begin{equation} and one after end{equation}. Removing or commenting this blank lines should be enough to correct spacing.

      – Luis Turcio
      Apr 3 at 21:54











    • @LuisTurcio -- Indeed, commenting or removing the blank line is what is recommended. I really should have posted this comment to the original question.

      – barbara beeton
      Apr 4 at 0:12
















    1














    You can try with shadethm package, it can do all you want and many more. In you example what you need is:



    documentclass{article}
    usepackage{shadethm}
    usepackage{mathtools}

    newshadetheorem{boxdef}{Definition}[section]
    newshadetheorem{boxtheorem}[boxdef]{Theorem}
    newtheorem{theorem}[boxdef]{Theorem}

    setlength{shadeboxsep}{2pt}
    setlength{shadeboxrule}{.4pt}
    setlength{shadedtextwidth}{textwidth}
    addtolength{shadedtextwidth}{-2shadeboxsep}
    addtolength{shadedtextwidth}{-2shadeboxrule}
    setlength{shadeleftshift}{0pt}
    setlength{shaderightshift}{0pt}
    definecolor{shadethmcolor}{cmyk}{0,0,0,0}
    definecolor{shaderulecolor}{cmyk}{0,0,0,1}

    begin{document}

    section{Boxed theorems}

    begin{boxdef}
    If f is analytic at $z_0$, then the series

    begin{equation}
    f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
    end{equation}

    is called the Taylor series for f around $z_0$.
    end{boxdef}

    begin{boxtheorem}
    If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
    end{equation}
    end{boxtheorem}

    begin{boxtheorem}
    (Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
    end{equation}
    end{boxtheorem}
    noindent hrulefill

    begin{theorem}
    If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
    end{theorem}

    end{document}


    which produces the following:



    enter image description here






    share|improve this answer
























    • For newshadetheorem{boxdef}{Definition}[section] newshadetheorem{boxtheorem}[boxdef]{Theorem} newtheorem{theorem}[boxdef]{Theorem}, why is boxdef in brackets?

      – K.M
      Apr 2 at 22:37








    • 1





      In the first box, the space above the equation is larger than that below the equation The reason for this is the blank line above begin{equation}. Blank lines in that position should be avoided.

      – barbara beeton
      Apr 3 at 21:06






    • 1





      @K.M the brackets [boxdef] is to enumerate all different kind of theorems with the same enumeration

      – Luis Turcio
      Apr 3 at 21:50











    • @barbarabeeton the spacing is due to the original code written by K.M, it has a blank line before begin{equation} and one after end{equation}. Removing or commenting this blank lines should be enough to correct spacing.

      – Luis Turcio
      Apr 3 at 21:54











    • @LuisTurcio -- Indeed, commenting or removing the blank line is what is recommended. I really should have posted this comment to the original question.

      – barbara beeton
      Apr 4 at 0:12














    1












    1








    1







    You can try with shadethm package, it can do all you want and many more. In you example what you need is:



    documentclass{article}
    usepackage{shadethm}
    usepackage{mathtools}

    newshadetheorem{boxdef}{Definition}[section]
    newshadetheorem{boxtheorem}[boxdef]{Theorem}
    newtheorem{theorem}[boxdef]{Theorem}

    setlength{shadeboxsep}{2pt}
    setlength{shadeboxrule}{.4pt}
    setlength{shadedtextwidth}{textwidth}
    addtolength{shadedtextwidth}{-2shadeboxsep}
    addtolength{shadedtextwidth}{-2shadeboxrule}
    setlength{shadeleftshift}{0pt}
    setlength{shaderightshift}{0pt}
    definecolor{shadethmcolor}{cmyk}{0,0,0,0}
    definecolor{shaderulecolor}{cmyk}{0,0,0,1}

    begin{document}

    section{Boxed theorems}

    begin{boxdef}
    If f is analytic at $z_0$, then the series

    begin{equation}
    f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
    end{equation}

    is called the Taylor series for f around $z_0$.
    end{boxdef}

    begin{boxtheorem}
    If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
    end{equation}
    end{boxtheorem}

    begin{boxtheorem}
    (Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
    end{equation}
    end{boxtheorem}
    noindent hrulefill

    begin{theorem}
    If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
    end{theorem}

    end{document}


    which produces the following:



    enter image description here






    share|improve this answer













    You can try with shadethm package, it can do all you want and many more. In you example what you need is:



    documentclass{article}
    usepackage{shadethm}
    usepackage{mathtools}

    newshadetheorem{boxdef}{Definition}[section]
    newshadetheorem{boxtheorem}[boxdef]{Theorem}
    newtheorem{theorem}[boxdef]{Theorem}

    setlength{shadeboxsep}{2pt}
    setlength{shadeboxrule}{.4pt}
    setlength{shadedtextwidth}{textwidth}
    addtolength{shadedtextwidth}{-2shadeboxsep}
    addtolength{shadedtextwidth}{-2shadeboxrule}
    setlength{shadeleftshift}{0pt}
    setlength{shaderightshift}{0pt}
    definecolor{shadethmcolor}{cmyk}{0,0,0,0}
    definecolor{shaderulecolor}{cmyk}{0,0,0,1}

    begin{document}

    section{Boxed theorems}

    begin{boxdef}
    If f is analytic at $z_0$, then the series

    begin{equation}
    f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
    end{equation}

    is called the Taylor series for f around $z_0$.
    end{boxdef}

    begin{boxtheorem}
    If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
    end{equation}
    end{boxtheorem}

    begin{boxtheorem}
    (Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
    end{equation}
    end{boxtheorem}
    noindent hrulefill

    begin{theorem}
    If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
    end{theorem}

    end{document}


    which produces the following:



    enter image description here







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Apr 2 at 22:27









    Luis TurcioLuis Turcio

    1359




    1359













    • For newshadetheorem{boxdef}{Definition}[section] newshadetheorem{boxtheorem}[boxdef]{Theorem} newtheorem{theorem}[boxdef]{Theorem}, why is boxdef in brackets?

      – K.M
      Apr 2 at 22:37








    • 1





      In the first box, the space above the equation is larger than that below the equation The reason for this is the blank line above begin{equation}. Blank lines in that position should be avoided.

      – barbara beeton
      Apr 3 at 21:06






    • 1





      @K.M the brackets [boxdef] is to enumerate all different kind of theorems with the same enumeration

      – Luis Turcio
      Apr 3 at 21:50











    • @barbarabeeton the spacing is due to the original code written by K.M, it has a blank line before begin{equation} and one after end{equation}. Removing or commenting this blank lines should be enough to correct spacing.

      – Luis Turcio
      Apr 3 at 21:54











    • @LuisTurcio -- Indeed, commenting or removing the blank line is what is recommended. I really should have posted this comment to the original question.

      – barbara beeton
      Apr 4 at 0:12



















    • For newshadetheorem{boxdef}{Definition}[section] newshadetheorem{boxtheorem}[boxdef]{Theorem} newtheorem{theorem}[boxdef]{Theorem}, why is boxdef in brackets?

      – K.M
      Apr 2 at 22:37








    • 1





      In the first box, the space above the equation is larger than that below the equation The reason for this is the blank line above begin{equation}. Blank lines in that position should be avoided.

      – barbara beeton
      Apr 3 at 21:06






    • 1





      @K.M the brackets [boxdef] is to enumerate all different kind of theorems with the same enumeration

      – Luis Turcio
      Apr 3 at 21:50











    • @barbarabeeton the spacing is due to the original code written by K.M, it has a blank line before begin{equation} and one after end{equation}. Removing or commenting this blank lines should be enough to correct spacing.

      – Luis Turcio
      Apr 3 at 21:54











    • @LuisTurcio -- Indeed, commenting or removing the blank line is what is recommended. I really should have posted this comment to the original question.

      – barbara beeton
      Apr 4 at 0:12

















    For newshadetheorem{boxdef}{Definition}[section] newshadetheorem{boxtheorem}[boxdef]{Theorem} newtheorem{theorem}[boxdef]{Theorem}, why is boxdef in brackets?

    – K.M
    Apr 2 at 22:37







    For newshadetheorem{boxdef}{Definition}[section] newshadetheorem{boxtheorem}[boxdef]{Theorem} newtheorem{theorem}[boxdef]{Theorem}, why is boxdef in brackets?

    – K.M
    Apr 2 at 22:37






    1




    1





    In the first box, the space above the equation is larger than that below the equation The reason for this is the blank line above begin{equation}. Blank lines in that position should be avoided.

    – barbara beeton
    Apr 3 at 21:06





    In the first box, the space above the equation is larger than that below the equation The reason for this is the blank line above begin{equation}. Blank lines in that position should be avoided.

    – barbara beeton
    Apr 3 at 21:06




    1




    1





    @K.M the brackets [boxdef] is to enumerate all different kind of theorems with the same enumeration

    – Luis Turcio
    Apr 3 at 21:50





    @K.M the brackets [boxdef] is to enumerate all different kind of theorems with the same enumeration

    – Luis Turcio
    Apr 3 at 21:50













    @barbarabeeton the spacing is due to the original code written by K.M, it has a blank line before begin{equation} and one after end{equation}. Removing or commenting this blank lines should be enough to correct spacing.

    – Luis Turcio
    Apr 3 at 21:54





    @barbarabeeton the spacing is due to the original code written by K.M, it has a blank line before begin{equation} and one after end{equation}. Removing or commenting this blank lines should be enough to correct spacing.

    – Luis Turcio
    Apr 3 at 21:54













    @LuisTurcio -- Indeed, commenting or removing the blank line is what is recommended. I really should have posted this comment to the original question.

    – barbara beeton
    Apr 4 at 0:12





    @LuisTurcio -- Indeed, commenting or removing the blank line is what is recommended. I really should have posted this comment to the original question.

    – barbara beeton
    Apr 4 at 0:12











    2














    Here is a solution with thmtools, which cooperates wit amsthm. Unrelated: you don't have to load amsmath if you load mathtools, as the latter does it for you:



    documentclass{article}
    usepackage{amsthm, thmtools}
    usepackage{mathtools}

    usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}

    newtheorem{definition}{Definition}
    newtheorem{theorem}{Theorem}

    declaretheorem[sibling=definition, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Definition]{boxeddef}
    declaretheorem[sibling=theorem, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Theorem]{boxedthm}

    begin{document}
    title{Extra Credit}
    author{}
    maketitle

    begin{boxeddef}
    If f is analytic at $z_0$, then the series

    begin{equation}
    f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
    end{equation}

    is called the Taylor series for f around $z_0$.
    end{boxeddef}

    begin{boxedthm}
    If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
    end{equation}
    end{boxedthm}

    begin{boxedthm}
    (Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
    begin{equation}
    f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
    end{equation}
    end{boxedthm}

    noindent hrulefill

    begin{theorem}
    If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
    end{theorem}

    end{document}


    enter image description here






    share|improve this answer




























      2














      Here is a solution with thmtools, which cooperates wit amsthm. Unrelated: you don't have to load amsmath if you load mathtools, as the latter does it for you:



      documentclass{article}
      usepackage{amsthm, thmtools}
      usepackage{mathtools}

      usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}

      newtheorem{definition}{Definition}
      newtheorem{theorem}{Theorem}

      declaretheorem[sibling=definition, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Definition]{boxeddef}
      declaretheorem[sibling=theorem, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Theorem]{boxedthm}

      begin{document}
      title{Extra Credit}
      author{}
      maketitle

      begin{boxeddef}
      If f is analytic at $z_0$, then the series

      begin{equation}
      f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
      end{equation}

      is called the Taylor series for f around $z_0$.
      end{boxeddef}

      begin{boxedthm}
      If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
      begin{equation}
      f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
      end{equation}
      end{boxedthm}

      begin{boxedthm}
      (Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
      begin{equation}
      f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
      end{equation}
      end{boxedthm}

      noindent hrulefill

      begin{theorem}
      If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
      end{theorem}

      end{document}


      enter image description here






      share|improve this answer


























        2












        2








        2







        Here is a solution with thmtools, which cooperates wit amsthm. Unrelated: you don't have to load amsmath if you load mathtools, as the latter does it for you:



        documentclass{article}
        usepackage{amsthm, thmtools}
        usepackage{mathtools}

        usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}

        newtheorem{definition}{Definition}
        newtheorem{theorem}{Theorem}

        declaretheorem[sibling=definition, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Definition]{boxeddef}
        declaretheorem[sibling=theorem, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Theorem]{boxedthm}

        begin{document}
        title{Extra Credit}
        author{}
        maketitle

        begin{boxeddef}
        If f is analytic at $z_0$, then the series

        begin{equation}
        f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
        end{equation}

        is called the Taylor series for f around $z_0$.
        end{boxeddef}

        begin{boxedthm}
        If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
        begin{equation}
        f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
        end{equation}
        end{boxedthm}

        begin{boxedthm}
        (Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
        begin{equation}
        f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
        end{equation}
        end{boxedthm}

        noindent hrulefill

        begin{theorem}
        If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
        end{theorem}

        end{document}


        enter image description here






        share|improve this answer













        Here is a solution with thmtools, which cooperates wit amsthm. Unrelated: you don't have to load amsmath if you load mathtools, as the latter does it for you:



        documentclass{article}
        usepackage{amsthm, thmtools}
        usepackage{mathtools}

        usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}

        newtheorem{definition}{Definition}
        newtheorem{theorem}{Theorem}

        declaretheorem[sibling=definition, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Definition]{boxeddef}
        declaretheorem[sibling=theorem, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Theorem]{boxedthm}

        begin{document}
        title{Extra Credit}
        author{}
        maketitle

        begin{boxeddef}
        If f is analytic at $z_0$, then the series

        begin{equation}
        f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n
        end{equation}

        is called the Taylor series for f around $z_0$.
        end{boxeddef}

        begin{boxedthm}
        If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
        begin{equation}
        f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )
        end{equation}
        end{boxedthm}

        begin{boxedthm}
        (Cauchy's Integral Formula) Let $Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $Gamma$ and $z_0$ is any point inside $Gamma$, then
        begin{equation}
        f(z_0)= frac{1}{2pi i} int_{Gamma} frac{f(z)}{z-z_0} dz
        end{equation}
        end{boxedthm}

        noindent hrulefill

        begin{theorem}
        If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
        end{theorem}

        end{document}


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Apr 2 at 22:40









        BernardBernard

        175k778209




        175k778209






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to TeX - LaTeX Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f482860%2fhow-to-enclose-theorems-and-definition-in-rectangles%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Plaza Victoria

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...