How to enclose theorems and definition in rectangles?Vertical space around theoremsTheorems and Definitions...

ssTTsSTtRrriinInnnnNNNIiinngg

How can saying a song's name be a copyright violation?

What reasons are there for a Capitalist to oppose a 100% inheritance tax?

files created then deleted at every second in tmp directory

Is this draw by repetition?

Why were 5.25" floppy drives cheaper than 8"?

Do Iron Man suits sport waste management systems?

In the UK, is it possible to get a referendum by a court decision?

How to show a landlord what we have in savings?

One verb to replace 'be a member of' a club

Rotate ASCII Art by 45 Degrees

Sums of two squares in arithmetic progressions

Why do I get negative height?

How could indestructible materials be used in power generation?

Ambiguity in the definition of entropy

Can I hook these wires up to find the connection to a dead outlet?

What is the most common color to indicate the input-field is disabled?

How to coordinate airplane tickets?

Can compressed videos be decoded back to their uncompresed original format?

Did 'Cinema Songs' exist during Hiranyakshipu's time?

Partial fraction expansion confusion

Do creatures with a listed speed of "0 ft., fly 30 ft. (hover)" ever touch the ground?

How to compactly explain secondary and tertiary characters without resorting to stereotypes?

Why didn't Boeing produce its own regional jet?



How to enclose theorems and definition in rectangles?


Vertical space around theoremsTheorems and Definitions as quotesHow to replace all pictures by white rectangles?How to remove line breaks before and after theorems?Horizontal spaces to the left and right of theoremsExtra spacing around restatable theoremsKOMA script and amsthm: Space lost before and after theoremsShrinking spacing around definition environmentTheorems and parskipremove spacing from a definition













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
    5 hours ago













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

    – K.M
    5 hours ago











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

    – Bernard
    4 hours ago
















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
    5 hours ago













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

    – K.M
    5 hours ago











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

    – Bernard
    4 hours ago














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 5 hours ago









K.MK.M

1305




1305




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
    5 hours ago













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

    – K.M
    5 hours ago











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

    – Bernard
    4 hours ago



















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

    – Bernard
    5 hours ago













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

    – K.M
    5 hours ago











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

    – Bernard
    4 hours ago

















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

– Bernard
5 hours ago







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

– Bernard
5 hours ago















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

– K.M
5 hours ago





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

– K.M
5 hours ago













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

– Bernard
4 hours ago





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

– Bernard
4 hours ago










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
    4 hours 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
      4 hours 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
      4 hours 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 4 hours ago









    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
      4 hours ago





















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

      – K.M
      4 hours ago



















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

    – K.M
    4 hours ago







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

    – K.M
    4 hours 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 4 hours ago









        BernardBernard

        175k776207




        175k776207






















            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

            Why not use the yoke to control yaw, as well as pitch and roll? Announcing the arrival of...

            Couldn't open a raw socket. Error: Permission denied (13) (nmap)Is it possible to run networking commands...

            VNC viewer RFB protocol error: bad desktop size 0x0I Cannot Type the Key 'd' (lowercase) in VNC Viewer...