How do I make the characters have the same size?
Th following is the full code that produces the image shown below:
documentclass{article}
usepackage[utf8]{inputenc}
usepackage[english]{babel}
usepackage{amsthm}
usepackage{amsmath}
usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
newtheorem{definition}{Definition}
newtheorem{theorem}{Theorem}
theoremstyle{remark}
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}
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}
begin{proof}
Suppose that the function textit{f} is analytic in the disk $|z-z_0|<R'$. We can define a positively oriented contour $C$ as $$ C:=Big{z:|z-z_0|=frac{R + R'}{2}, 0<R< R' Big}.$$ Letting $zeta$ be an arbitrary point on $C$ and applying Theorem 1 to $(1)$, we get
begin{equation}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta.
end{equation}\
Or equivalently, we have that
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
The above image is created using
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
which is extracted from the full code above. How can I resize the second line of the image, so that all the characters are the same size?
resize
New contributor
add a comment |
Th following is the full code that produces the image shown below:
documentclass{article}
usepackage[utf8]{inputenc}
usepackage[english]{babel}
usepackage{amsthm}
usepackage{amsmath}
usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
newtheorem{definition}{Definition}
newtheorem{theorem}{Theorem}
theoremstyle{remark}
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}
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}
begin{proof}
Suppose that the function textit{f} is analytic in the disk $|z-z_0|<R'$. We can define a positively oriented contour $C$ as $$ C:=Big{z:|z-z_0|=frac{R + R'}{2}, 0<R< R' Big}.$$ Letting $zeta$ be an arbitrary point on $C$ and applying Theorem 1 to $(1)$, we get
begin{equation}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta.
end{equation}\
Or equivalently, we have that
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
The above image is created using
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
which is extracted from the full code above. How can I resize the second line of the image, so that all the characters are the same size?
resize
New contributor
1
Usedisplaystyle
at the beginning of the outer denominator.
– L. F.
5 mins ago
Possible duplicate of equal size numerator and denominator
– L. F.
1 min ago
add a comment |
Th following is the full code that produces the image shown below:
documentclass{article}
usepackage[utf8]{inputenc}
usepackage[english]{babel}
usepackage{amsthm}
usepackage{amsmath}
usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
newtheorem{definition}{Definition}
newtheorem{theorem}{Theorem}
theoremstyle{remark}
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}
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}
begin{proof}
Suppose that the function textit{f} is analytic in the disk $|z-z_0|<R'$. We can define a positively oriented contour $C$ as $$ C:=Big{z:|z-z_0|=frac{R + R'}{2}, 0<R< R' Big}.$$ Letting $zeta$ be an arbitrary point on $C$ and applying Theorem 1 to $(1)$, we get
begin{equation}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta.
end{equation}\
Or equivalently, we have that
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
The above image is created using
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
which is extracted from the full code above. How can I resize the second line of the image, so that all the characters are the same size?
resize
New contributor
Th following is the full code that produces the image shown below:
documentclass{article}
usepackage[utf8]{inputenc}
usepackage[english]{babel}
usepackage{amsthm}
usepackage{amsmath}
usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
newtheorem{definition}{Definition}
newtheorem{theorem}{Theorem}
theoremstyle{remark}
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}
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}
begin{proof}
Suppose that the function textit{f} is analytic in the disk $|z-z_0|<R'$. We can define a positively oriented contour $C$ as $$ C:=Big{z:|z-z_0|=frac{R + R'}{2}, 0<R< R' Big}.$$ Letting $zeta$ be an arbitrary point on $C$ and applying Theorem 1 to $(1)$, we get
begin{equation}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta.
end{equation}\
Or equivalently, we have that
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
The above image is created using
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
which is extracted from the full code above. How can I resize the second line of the image, so that all the characters are the same size?
resize
resize
New contributor
New contributor
New contributor
asked 7 mins ago
K.MK.M
1164
1164
New contributor
New contributor
1
Usedisplaystyle
at the beginning of the outer denominator.
– L. F.
5 mins ago
Possible duplicate of equal size numerator and denominator
– L. F.
1 min ago
add a comment |
1
Usedisplaystyle
at the beginning of the outer denominator.
– L. F.
5 mins ago
Possible duplicate of equal size numerator and denominator
– L. F.
1 min ago
1
1
Use
displaystyle
at the beginning of the outer denominator.– L. F.
5 mins ago
Use
displaystyle
at the beginning of the outer denominator.– L. F.
5 mins ago
Possible duplicate of equal size numerator and denominator
– L. F.
1 min ago
Possible duplicate of equal size numerator and denominator
– L. F.
1 min ago
add a comment |
1 Answer
1
active
oldest
votes
You want to
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\
&= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{displaystylefrac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
or simply dfrac
as you can use amsmath
package.
add a comment |
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You want to
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\
&= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{displaystylefrac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
or simply dfrac
as you can use amsmath
package.
add a comment |
You want to
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\
&= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{displaystylefrac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
or simply dfrac
as you can use amsmath
package.
add a comment |
You want to
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\
&= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{displaystylefrac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
or simply dfrac
as you can use amsmath
package.
You want to
begin{align*}
sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta
&= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta
\
&= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{displaystylefrac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta
\ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta
end{align*}
or simply dfrac
as you can use amsmath
package.
answered 3 mins ago
Przemysław ScherwentkePrzemysław Scherwentke
29.9k54795
29.9k54795
add a comment |
add a comment |
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.
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1
Use
displaystyle
at the beginning of the outer denominator.– L. F.
5 mins ago
Possible duplicate of equal size numerator and denominator
– L. F.
1 min ago