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2
This is the following code that I have so far:
documentclass{article}
usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
usepackage{amsmath}
usepackage{mathtools}
usepackage{amsthm}
usepackage{graphicx}
usepackage{setspace}
onehalfspacing
DeclareMathOperator{Log}{Log}
begin{document}
title{Homework Chapter 5}
author{}
maketitle
section*{Section 5.2: 4, 5, and 7}
paragraph{4.}
Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken
as $e^{alpha Log(1+z)}$, then for $|z|< 1$
begin{equation*}
(1+z)^{alpha} = 1 + frac{alpha}{1}z +
frac{alpha(alpha -1)}{1cdot2}z^{2} +
frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb
end{equation*}
In general,
begin{equation}
frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
end{equation}
We can prove result (1) by induction. For our base case, we let $j=1$. Hence,
begin{equation}
frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}
end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
begin{equation}
frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}
end{equation}
And so,
begin{equation}
frac{d}{dz}Big[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Big] = frac{d}{dz}Big[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Big] = frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!} = frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}
end{equation}
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
Expanding (1), we have
begin{equation}
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!} = frac{alpha (alpha-1) dots (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!} = alpha (alpha - 1) dots (alpha - j + 1) (1+z_0)^{alpha - j}.
end{equation}
The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus
begin{equation}
1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dots (alpha - j + 1)}{j!}z^{j}
end{equation}
end{document}
How do I create vertical space between
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
and
Expanding (1), we have
spacing
asked 10 hours ago
K.MK.M
1675
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.
add a comment |
2
This is the following code that I have so far:
documentclass{article}
usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
usepackage{amsmath}
usepackage{mathtools}
usepackage{amsthm}
usepackage{graphicx}
usepackage{setspace}
onehalfspacing
DeclareMathOperator{Log}{Log}
begin{document}
title{Homework Chapter 5}
author{}
maketitle
section*{Section 5.2: 4, 5, and 7}
paragraph{4.}
Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken
as $e^{alpha Log(1+z)}$, then for $|z|< 1$
begin{equation*}
(1+z)^{alpha} = 1 + frac{alpha}{1}z +
frac{alpha(alpha -1)}{1cdot2}z^{2} +
frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb
end{equation*}
In general,
begin{equation}
frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
end{equation}
We can prove result (1) by induction. For our base case, we let $j=1$. Hence,
begin{equation}
frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}
end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
begin{equation}
frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}
end{equation}
And so,
begin{equation}
frac{d}{dz}Big[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Big] = frac{d}{dz}Big[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Big] = frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!} = frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}
end{equation}
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
Expanding (1), we have
begin{equation}
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!} = frac{alpha (alpha-1) dots (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!} = alpha (alpha - 1) dots (alpha - j + 1) (1+z_0)^{alpha - j}.
end{equation}
The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus
begin{equation}
1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dots (alpha - j + 1)}{j!}z^{j}
end{equation}
end{document}
How do I create vertical space between
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
and
Expanding (1), we have
spacing
asked 10 hours ago
K.MK.M
1675
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.
You could use something likemedskiporbigskip.
– leandriis
10 hours ago
1
paragraphs should be separated by a blank line in the source. Also note thatparagraphis a 4th level heading designed to follow sussubsection
– David Carlisle
10 hours ago
log with a capital L is called the principal log. It's used in complex analysis to specify that the log function is continuous everywhere except the negative real axis.
– K.M
9 hours ago
@DavidCarlisle: Where specifically would I add the blank line?
– K.M
9 hours ago
If you wantExpanding...to start a new paragraph then it should have a blank line before it, otherwise it is a continuation of the previous sentence and will appear on the same line.
– David Carlisle
9 hours ago
add a comment |
2
2
2
This is the following code that I have so far:
documentclass{article}
usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
usepackage{amsmath}
usepackage{mathtools}
usepackage{amsthm}
usepackage{graphicx}
usepackage{setspace}
onehalfspacing
DeclareMathOperator{Log}{Log}
begin{document}
title{Homework Chapter 5}
author{}
maketitle
section*{Section 5.2: 4, 5, and 7}
paragraph{4.}
Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken
as $e^{alpha Log(1+z)}$, then for $|z|< 1$
begin{equation*}
(1+z)^{alpha} = 1 + frac{alpha}{1}z +
frac{alpha(alpha -1)}{1cdot2}z^{2} +
frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb
end{equation*}
In general,
begin{equation}
frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
end{equation}
We can prove result (1) by induction. For our base case, we let $j=1$. Hence,
begin{equation}
frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}
end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
begin{equation}
frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}
end{equation}
And so,
begin{equation}
frac{d}{dz}Big[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Big] = frac{d}{dz}Big[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Big] = frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!} = frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}
end{equation}
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
Expanding (1), we have
begin{equation}
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!} = frac{alpha (alpha-1) dots (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!} = alpha (alpha - 1) dots (alpha - j + 1) (1+z_0)^{alpha - j}.
end{equation}
The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus
begin{equation}
1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dots (alpha - j + 1)}{j!}z^{j}
end{equation}
end{document}
How do I create vertical space between
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
and
Expanding (1), we have
spacing
asked 10 hours ago
K.MK.M
1675
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.
This is the following code that I have so far:
documentclass{article}
usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
usepackage{amsmath}
usepackage{mathtools}
usepackage{amsthm}
usepackage{graphicx}
usepackage{setspace}
onehalfspacing
DeclareMathOperator{Log}{Log}
begin{document}
title{Homework Chapter 5}
author{}
maketitle
section*{Section 5.2: 4, 5, and 7}
paragraph{4.}
Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken
as $e^{alpha Log(1+z)}$, then for $|z|< 1$
begin{equation*}
(1+z)^{alpha} = 1 + frac{alpha}{1}z +
frac{alpha(alpha -1)}{1cdot2}z^{2} +
frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb
end{equation*}
In general,
begin{equation}
frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
end{equation}
We can prove result (1) by induction. For our base case, we let $j=1$. Hence,
begin{equation}
frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}
end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
begin{equation}
frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}
end{equation}
And so,
begin{equation}
frac{d}{dz}Big[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Big] = frac{d}{dz}Big[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Big] = frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!} = frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}
end{equation}
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
Expanding (1), we have
begin{equation}
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!} = frac{alpha (alpha-1) dots (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!} = alpha (alpha - 1) dots (alpha - j + 1) (1+z_0)^{alpha - j}.
end{equation}
The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus
begin{equation}
1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dots (alpha - j + 1)}{j!}z^{j}
end{equation}
end{document}
How do I create vertical space between
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
and
Expanding (1), we have
spacing
spacing
asked 10 hours ago
K.MK.M
1675
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 10 hours ago
K.MK.M
1675
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 10 hours ago
K.MK.M
1675
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 10 hours ago
K.MK.M
1675
asked 10 hours ago
K.MK.M
1675
1675
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.
You could use something likemedskiporbigskip.
– leandriis
10 hours ago
1
paragraphs should be separated by a blank line in the source. Also note thatparagraphis a 4th level heading designed to follow sussubsection
– David Carlisle
10 hours ago
log with a capital L is called the principal log. It's used in complex analysis to specify that the log function is continuous everywhere except the negative real axis.
– K.M
9 hours ago
@DavidCarlisle: Where specifically would I add the blank line?
– K.M
9 hours ago
If you wantExpanding...to start a new paragraph then it should have a blank line before it, otherwise it is a continuation of the previous sentence and will appear on the same line.
– David Carlisle
9 hours ago
add a comment |
You could use something likemedskiporbigskip.
– leandriis
10 hours ago
1
paragraphs should be separated by a blank line in the source. Also note thatparagraphis a 4th level heading designed to follow sussubsection
– David Carlisle
10 hours ago
log with a capital L is called the principal log. It's used in complex analysis to specify that the log function is continuous everywhere except the negative real axis.
– K.M
9 hours ago
@DavidCarlisle: Where specifically would I add the blank line?
– K.M
9 hours ago
If you wantExpanding...to start a new paragraph then it should have a blank line before it, otherwise it is a continuation of the previous sentence and will appear on the same line.
– David Carlisle
9 hours ago
You could use something like
medskip or bigskip.– leandriis
10 hours ago
You could use something like
medskip or bigskip.– leandriis
10 hours ago
1
1
paragraphs should be separated by a blank line in the source. Also note that
paragraph is a 4th level heading designed to follow sussubsection– David Carlisle
10 hours ago
paragraphs should be separated by a blank line in the source. Also note that
paragraph is a 4th level heading designed to follow sussubsection– David Carlisle
10 hours ago
log with a capital L is called the principal log. It's used in complex analysis to specify that the log function is continuous everywhere except the negative real axis.
– K.M
9 hours ago
log with a capital L is called the principal log. It's used in complex analysis to specify that the log function is continuous everywhere except the negative real axis.
– K.M
9 hours ago
@DavidCarlisle: Where specifically would I add the blank line?
– K.M
9 hours ago
@DavidCarlisle: Where specifically would I add the blank line?
– K.M
9 hours ago
If you want
Expanding... to start a new paragraph then it should have a blank line before it, otherwise it is a continuation of the previous sentence and will appear on the same line.– David Carlisle
9 hours ago
If you want
Expanding... to start a new paragraph then it should have a blank line before it, otherwise it is a continuation of the previous sentence and will appear on the same line.– David Carlisle
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
2
I don't think it's necessary to create extra vertical whitespace above "Expanding (1), we have". A paragraph break (with indentation of the string "Expanding (1), we have") should be enough for your readers to note a "break" in the action.
Do replace the two instances of dots with either dotsb or cdots, though.
documentclass{article}
usepackage[hmargin=1.5in, top=0.5in]{geometry}
%%usepackage{amsmath}
usepackage{mathtools} % loads "amsmath" automatically
DeclareMathOperator{Log}{Log}
usepackage{amsthm,graphicx}
usepackage{setspace}
onehalfspacing
begin{document}
title{Homework Chapter 5}
author{}
maketitle
section*{Section 5.2: 4, 5, and 7}
paragraph{4.}
Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken
as $e^{alpha Log(1+z)}$, then for $|z|< 1$,
begin{equation*}
(1+z)^{alpha} = 1 + frac{alpha}{1}z +
frac{alpha(alpha -1)}{1cdot2}z^{2} +
frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb
end{equation*}
In general,
begin{equation} label{eq:1}
frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
end{equation}
% A blank line creates a paragraph break:
We can prove result eqref{eq:1} by induction. For our base case,
we let $j=1$. Hence,
begin{equation}
frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}
end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
begin{equation}label{eq:3}
frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!},.
end{equation}
And so,
begin{equation}
frac{d}{dz}Bigl[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Bigr]
= frac{d}{dz}Bigl[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Bigr]
= frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!}
= frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}
end{equation}
Since eqref{eq:3} holds for $j = k + 1$, it follows that eqref{eq:1}
holds for all positive integers~$j$. QED
% A blank line creates a paragraph break:
Expanding eqref{eq:1}, we have
begin{equation}
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
= frac{alpha (alpha-1) dotsb (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!}
= alpha (alpha - 1) dotsb (alpha - j + 1) (1+z_0)^{alpha - j}.
end{equation}
The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus
begin{equation}
1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dotsb (alpha - j + 1)}{j!}z^{j}
end{equation}
end{document}
answered 9 hours ago
MicoMico
285k31388778
Incidentally, I think it wouldn't hurt to remind your readers that the final result was derived under the assumption that|z|<1.
– Mico
8 hours ago
add a comment |
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1 Answer
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1 Answer
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I don't think it's necessary to create extra vertical whitespace above "Expanding (1), we have". A paragraph break (with indentation of the string "Expanding (1), we have") should be enough for your readers to note a "break" in the action.
Do replace the two instances of dots with either dotsb or cdots, though.
documentclass{article}
usepackage[hmargin=1.5in, top=0.5in]{geometry}
%%usepackage{amsmath}
usepackage{mathtools} % loads "amsmath" automatically
DeclareMathOperator{Log}{Log}
usepackage{amsthm,graphicx}
usepackage{setspace}
onehalfspacing
begin{document}
title{Homework Chapter 5}
author{}
maketitle
section*{Section 5.2: 4, 5, and 7}
paragraph{4.}
Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken
as $e^{alpha Log(1+z)}$, then for $|z|< 1$,
begin{equation*}
(1+z)^{alpha} = 1 + frac{alpha}{1}z +
frac{alpha(alpha -1)}{1cdot2}z^{2} +
frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb
end{equation*}
In general,
begin{equation} label{eq:1}
frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
end{equation}
% A blank line creates a paragraph break:
We can prove result eqref{eq:1} by induction. For our base case,
we let $j=1$. Hence,
begin{equation}
frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}
end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
begin{equation}label{eq:3}
frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!},.
end{equation}
And so,
begin{equation}
frac{d}{dz}Bigl[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Bigr]
= frac{d}{dz}Bigl[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Bigr]
= frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!}
= frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}
end{equation}
Since eqref{eq:3} holds for $j = k + 1$, it follows that eqref{eq:1}
holds for all positive integers~$j$. QED
% A blank line creates a paragraph break:
Expanding eqref{eq:1}, we have
begin{equation}
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
= frac{alpha (alpha-1) dotsb (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!}
= alpha (alpha - 1) dotsb (alpha - j + 1) (1+z_0)^{alpha - j}.
end{equation}
The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus
begin{equation}
1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dotsb (alpha - j + 1)}{j!}z^{j}
end{equation}
end{document}
answered 9 hours ago
MicoMico
285k31388778
Incidentally, I think it wouldn't hurt to remind your readers that the final result was derived under the assumption that|z|<1.
– Mico
8 hours ago
add a comment |
2
I don't think it's necessary to create extra vertical whitespace above "Expanding (1), we have". A paragraph break (with indentation of the string "Expanding (1), we have") should be enough for your readers to note a "break" in the action.
Do replace the two instances of dots with either dotsb or cdots, though.
documentclass{article}
usepackage[hmargin=1.5in, top=0.5in]{geometry}
%%usepackage{amsmath}
usepackage{mathtools} % loads "amsmath" automatically
DeclareMathOperator{Log}{Log}
usepackage{amsthm,graphicx}
usepackage{setspace}
onehalfspacing
begin{document}
title{Homework Chapter 5}
author{}
maketitle
section*{Section 5.2: 4, 5, and 7}
paragraph{4.}
Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken
as $e^{alpha Log(1+z)}$, then for $|z|< 1$,
begin{equation*}
(1+z)^{alpha} = 1 + frac{alpha}{1}z +
frac{alpha(alpha -1)}{1cdot2}z^{2} +
frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb
end{equation*}
In general,
begin{equation} label{eq:1}
frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
end{equation}
% A blank line creates a paragraph break:
We can prove result eqref{eq:1} by induction. For our base case,
we let $j=1$. Hence,
begin{equation}
frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}
end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
begin{equation}label{eq:3}
frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!},.
end{equation}
And so,
begin{equation}
frac{d}{dz}Bigl[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Bigr]
= frac{d}{dz}Bigl[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Bigr]
= frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!}
= frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}
end{equation}
Since eqref{eq:3} holds for $j = k + 1$, it follows that eqref{eq:1}
holds for all positive integers~$j$. QED
% A blank line creates a paragraph break:
Expanding eqref{eq:1}, we have
begin{equation}
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
= frac{alpha (alpha-1) dotsb (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!}
= alpha (alpha - 1) dotsb (alpha - j + 1) (1+z_0)^{alpha - j}.
end{equation}
The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus
begin{equation}
1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dotsb (alpha - j + 1)}{j!}z^{j}
end{equation}
end{document}
answered 9 hours ago
MicoMico
285k31388778
Incidentally, I think it wouldn't hurt to remind your readers that the final result was derived under the assumption that|z|<1.
– Mico
8 hours ago
add a comment |
2
2
2
I don't think it's necessary to create extra vertical whitespace above "Expanding (1), we have". A paragraph break (with indentation of the string "Expanding (1), we have") should be enough for your readers to note a "break" in the action.
Do replace the two instances of dots with either dotsb or cdots, though.
documentclass{article}
usepackage[hmargin=1.5in, top=0.5in]{geometry}
%%usepackage{amsmath}
usepackage{mathtools} % loads "amsmath" automatically
DeclareMathOperator{Log}{Log}
usepackage{amsthm,graphicx}
usepackage{setspace}
onehalfspacing
begin{document}
title{Homework Chapter 5}
author{}
maketitle
section*{Section 5.2: 4, 5, and 7}
paragraph{4.}
Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken
as $e^{alpha Log(1+z)}$, then for $|z|< 1$,
begin{equation*}
(1+z)^{alpha} = 1 + frac{alpha}{1}z +
frac{alpha(alpha -1)}{1cdot2}z^{2} +
frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb
end{equation*}
In general,
begin{equation} label{eq:1}
frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
end{equation}
% A blank line creates a paragraph break:
We can prove result eqref{eq:1} by induction. For our base case,
we let $j=1$. Hence,
begin{equation}
frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}
end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
begin{equation}label{eq:3}
frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!},.
end{equation}
And so,
begin{equation}
frac{d}{dz}Bigl[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Bigr]
= frac{d}{dz}Bigl[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Bigr]
= frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!}
= frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}
end{equation}
Since eqref{eq:3} holds for $j = k + 1$, it follows that eqref{eq:1}
holds for all positive integers~$j$. QED
% A blank line creates a paragraph break:
Expanding eqref{eq:1}, we have
begin{equation}
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
= frac{alpha (alpha-1) dotsb (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!}
= alpha (alpha - 1) dotsb (alpha - j + 1) (1+z_0)^{alpha - j}.
end{equation}
The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus
begin{equation}
1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dotsb (alpha - j + 1)}{j!}z^{j}
end{equation}
end{document}
answered 9 hours ago
MicoMico
285k31388778
I don't think it's necessary to create extra vertical whitespace above "Expanding (1), we have". A paragraph break (with indentation of the string "Expanding (1), we have") should be enough for your readers to note a "break" in the action.
Do replace the two instances of dots with either dotsb or cdots, though.
documentclass{article}
usepackage[hmargin=1.5in, top=0.5in]{geometry}
%%usepackage{amsmath}
usepackage{mathtools} % loads "amsmath" automatically
DeclareMathOperator{Log}{Log}
usepackage{amsthm,graphicx}
usepackage{setspace}
onehalfspacing
begin{document}
title{Homework Chapter 5}
author{}
maketitle
section*{Section 5.2: 4, 5, and 7}
paragraph{4.}
Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken
as $e^{alpha Log(1+z)}$, then for $|z|< 1$,
begin{equation*}
(1+z)^{alpha} = 1 + frac{alpha}{1}z +
frac{alpha(alpha -1)}{1cdot2}z^{2} +
frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb
end{equation*}
In general,
begin{equation} label{eq:1}
frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
end{equation}
% A blank line creates a paragraph break:
We can prove result eqref{eq:1} by induction. For our base case,
we let $j=1$. Hence,
begin{equation}
frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}
end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
begin{equation}label{eq:3}
frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=
frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!},.
end{equation}
And so,
begin{equation}
frac{d}{dz}Bigl[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Bigr]
= frac{d}{dz}Bigl[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Bigr]
= frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!}
= frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}
end{equation}
Since eqref{eq:3} holds for $j = k + 1$, it follows that eqref{eq:1}
holds for all positive integers~$j$. QED
% A blank line creates a paragraph break:
Expanding eqref{eq:1}, we have
begin{equation}
frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}
= frac{alpha (alpha-1) dotsb (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!}
= alpha (alpha - 1) dotsb (alpha - j + 1) (1+z_0)^{alpha - j}.
end{equation}
The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus
begin{equation}
1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dotsb (alpha - j + 1)}{j!}z^{j}
end{equation}
end{document}
answered 9 hours ago
MicoMico
285k31388778
answered 9 hours ago
MicoMico
285k31388778
answered 9 hours ago
MicoMico
285k31388778
answered 9 hours ago
MicoMico
285k31388778
285k31388778
Incidentally, I think it wouldn't hurt to remind your readers that the final result was derived under the assumption that|z|<1.
– Mico
8 hours ago
add a comment |
Incidentally, I think it wouldn't hurt to remind your readers that the final result was derived under the assumption that|z|<1.
– Mico
8 hours ago
Incidentally, I think it wouldn't hurt to remind your readers that the final result was derived under the assumption that
|z|<1.– Mico
8 hours ago
Incidentally, I think it wouldn't hurt to remind your readers that the final result was derived under the assumption that
|z|<1.– Mico
8 hours ago
add a comment |
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You could use something like
medskiporbigskip.– leandriis
10 hours ago
1
paragraphs should be separated by a blank line in the source. Also note that
paragraphis a 4th level heading designed to follow sussubsection– David Carlisle
10 hours ago
log with a capital L is called the principal log. It's used in complex analysis to specify that the log function is continuous everywhere except the negative real axis.
– K.M
9 hours ago
@DavidCarlisle: Where specifically would I add the blank line?
– K.M
9 hours ago
If you want
Expanding...to start a new paragraph then it should have a blank line before it, otherwise it is a continuation of the previous sentence and will appear on the same line.– David Carlisle
9 hours ago