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
    3 hours ago













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

    – K.M
    3 hours ago











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

    – Bernard
    3 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
    3 hours ago













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

    – K.M
    3 hours ago











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

    – Bernard
    3 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 3 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
    3 hours ago













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

    – K.M
    3 hours ago











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

    – Bernard
    3 hours ago



















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

    – Bernard
    3 hours ago













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

    – K.M
    3 hours ago











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

    – Bernard
    3 hours ago

















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

– Bernard
3 hours ago







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

– Bernard
3 hours ago















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

– K.M
3 hours ago





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

– K.M
3 hours ago













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

– Bernard
3 hours ago





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

– Bernard
3 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
    3 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






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    2 Answers
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    oldest

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    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
      3 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
      3 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 3 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
      3 hours ago





















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

      – K.M
      3 hours ago



















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

    – K.M
    3 hours ago







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

    – K.M
    3 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 2 hours ago









        BernardBernard

        175k776207




        175k776207






















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










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