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







New contributor




K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





















  • 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







New contributor




K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





















  • 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







New contributor




K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












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







New contributor




K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Apr 2 at 21:38









K.MK.M

1555




1555




New contributor




K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.













  • 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
    2 days ago






  • 1





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

    – Luis Turcio
    2 days ago











  • @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
    2 days ago











  • @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
    2 days ago



















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
    });


    }
    });






    K.M is a new contributor. Be nice, and check out our Code of Conduct.










    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
      2 days ago






    • 1





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

      – Luis Turcio
      2 days ago











    • @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
      2 days ago











    • @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
      2 days ago
















    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
      2 days ago






    • 1





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

      – Luis Turcio
      2 days ago











    • @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
      2 days ago











    • @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
      2 days ago














    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

    1259




    1259













    • 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
      2 days ago






    • 1





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

      – Luis Turcio
      2 days ago











    • @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
      2 days ago











    • @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
      2 days ago



















    • 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
      2 days ago






    • 1





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

      – Luis Turcio
      2 days ago











    • @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
      2 days ago











    • @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
      2 days ago

















    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
    2 days ago





    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
    2 days ago




    1




    1





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

    – Luis Turcio
    2 days ago





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

    – Luis Turcio
    2 days ago













    @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
    2 days ago





    @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
    2 days ago













    @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
    2 days ago





    @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
    2 days ago











    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

        175k777207




        175k777207






















            K.M is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            K.M is a new contributor. Be nice, and check out our Code of Conduct.













            K.M is a new contributor. Be nice, and check out our Code of Conduct.












            K.M is a new contributor. Be nice, and check out our Code of Conduct.
















            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

            數位音樂下載

            格利澤436b

            When can things happen in Etherscan, such as the picture below?