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Exempt portion of equation line from aligning?
The mysteries of mathpaletteAligning plain `align` and `cases`?Alignment across nested aligned environmentsHow to match left alignment of equations in math mode outside and inside an array?How to modify eqnarray?Alignment of two equations on LaTeXHow center align two equations separated by a lineequation custom horizontal alignment & numbering each rowFlushed-left and flushed-right text in align or alignat environmentHow can I align this equation in the center?breqn not aligning first two lines
I am using an array environment to get aligned portions of a series of equations to center (instead of left-justify), as shown below:
usepackage{array,amsmath}
[
begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
sumlimits_{r=0}^{n+1} binom{n+1}{r} & binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
end{array}
]
The array environment (I believe) is necessary here to get each of the columns to center instead of justifying left.
Now my problem is that these two lines are part of a greater series of equations, where the others do not follow this pattern to be aligned. However, I need the equals signs to line up across all lines.
My current approach is follow the array with a normal align
environment, having one equation line mirroring the longest line above but enclosed in phantom{}
to get the align spacing right. But this leaves a single empty line with an equals in it.
...
begin{align*}
&= 2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right] \
phantom{sumlimits_{r=0}^{n+1} binom{n+1}{r}} &= phantom{ binom{n+1}{0} + binom{n+1}{1} + ldots + binom{n+1}{n} + binom{n+1}{n+1}}
end{align*}
How can I get this result, but without the extraneous equals line at the end? Preferable a more elegant one, as this idea relies on several iffy factors such as none of the following equations exceeding the size of the one governing the special alignment.
math-mode horizontal-alignment align arrays
New contributor
add a comment |
I am using an array environment to get aligned portions of a series of equations to center (instead of left-justify), as shown below:
usepackage{array,amsmath}
[
begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
sumlimits_{r=0}^{n+1} binom{n+1}{r} & binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
end{array}
]
The array environment (I believe) is necessary here to get each of the columns to center instead of justifying left.
Now my problem is that these two lines are part of a greater series of equations, where the others do not follow this pattern to be aligned. However, I need the equals signs to line up across all lines.
My current approach is follow the array with a normal align
environment, having one equation line mirroring the longest line above but enclosed in phantom{}
to get the align spacing right. But this leaves a single empty line with an equals in it.
...
begin{align*}
&= 2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right] \
phantom{sumlimits_{r=0}^{n+1} binom{n+1}{r}} &= phantom{ binom{n+1}{0} + binom{n+1}{1} + ldots + binom{n+1}{n} + binom{n+1}{n+1}}
end{align*}
How can I get this result, but without the extraneous equals line at the end? Preferable a more elegant one, as this idea relies on several iffy factors such as none of the following equations exceeding the size of the one governing the special alignment.
math-mode horizontal-alignment align arrays
New contributor
add a comment |
I am using an array environment to get aligned portions of a series of equations to center (instead of left-justify), as shown below:
usepackage{array,amsmath}
[
begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
sumlimits_{r=0}^{n+1} binom{n+1}{r} & binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
end{array}
]
The array environment (I believe) is necessary here to get each of the columns to center instead of justifying left.
Now my problem is that these two lines are part of a greater series of equations, where the others do not follow this pattern to be aligned. However, I need the equals signs to line up across all lines.
My current approach is follow the array with a normal align
environment, having one equation line mirroring the longest line above but enclosed in phantom{}
to get the align spacing right. But this leaves a single empty line with an equals in it.
...
begin{align*}
&= 2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right] \
phantom{sumlimits_{r=0}^{n+1} binom{n+1}{r}} &= phantom{ binom{n+1}{0} + binom{n+1}{1} + ldots + binom{n+1}{n} + binom{n+1}{n+1}}
end{align*}
How can I get this result, but without the extraneous equals line at the end? Preferable a more elegant one, as this idea relies on several iffy factors such as none of the following equations exceeding the size of the one governing the special alignment.
math-mode horizontal-alignment align arrays
New contributor
I am using an array environment to get aligned portions of a series of equations to center (instead of left-justify), as shown below:
usepackage{array,amsmath}
[
begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
sumlimits_{r=0}^{n+1} binom{n+1}{r} & binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
end{array}
]
The array environment (I believe) is necessary here to get each of the columns to center instead of justifying left.
Now my problem is that these two lines are part of a greater series of equations, where the others do not follow this pattern to be aligned. However, I need the equals signs to line up across all lines.
My current approach is follow the array with a normal align
environment, having one equation line mirroring the longest line above but enclosed in phantom{}
to get the align spacing right. But this leaves a single empty line with an equals in it.
...
begin{align*}
&= 2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right] \
phantom{sumlimits_{r=0}^{n+1} binom{n+1}{r}} &= phantom{ binom{n+1}{0} + binom{n+1}{1} + ldots + binom{n+1}{n} + binom{n+1}{n+1}}
end{align*}
How can I get this result, but without the extraneous equals line at the end? Preferable a more elegant one, as this idea relies on several iffy factors such as none of the following equations exceeding the size of the one governing the special alignment.
math-mode horizontal-alignment align arrays
math-mode horizontal-alignment align arrays
New contributor
New contributor
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asked 29 mins ago
PGmathPGmath
1162
1162
New contributor
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add a comment |
3 Answers
3
active
oldest
votes
try
documentclass{article}
usepackage{array,amsmath}
begin{document}
[
begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
sum_{r=0}^{n+1} binom{n+1}{r}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
& multicolumn{3}{>{displaystyle}l}{
2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
}
end{array}
]
end{document}
add a comment |
Use the [t]
option. Then you do not need to use multicolumn
many times if you have many subsequent lines.
documentclass{article}
usepackage{array,amsmath}
begin{document}
begin{align*}
sumlimits_{r=0}^{n+1} binom{n+1}{r}
&begin{array}[t]{@{}>{displaystyle}c @{{}={}}@{}>{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
end{array}\
&=2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
end{align*}
end{document}
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
8 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
5 mins ago
add a comment |
eqparbox
allows you to store the lengths of boxes via a <tag>
. Below I use this approach with a newly-defined eqmathbox[<tag>][<align>]
(default for <align>
is to c
entre the content) to add content to three different <tag>
ged boxes:
documentclass{article}
usepackage{eqparbox,xparse,amsmath}
% https://tex.stackexchange.com/a/34412/5764
makeatletter
NewDocumentCommand{eqmathbox}{o O{c} m}{%
IfValueTF{#1}
{defeqmathbox@##1##2{eqmakebox[#1][#2]{$##1##2$}}}
{defeqmathbox@##1##2{eqmakebox{$##1##2$}}}
mathpaletteeqmathbox@{#3}
}
makeatother
begin{document}
begin{align*}
sum_{r = 0}^{n + 1} binom{n + 1}{r}
&= eqmathbox[LEFT]{binom{n + 1}{0}} + eqmathbox[CENTRE]{binom{n + 1}{1} + dots + binom{n + 1}{n}} + eqmathbox[RIGHT]{binom{n + 1}{n + 1}} \
&= eqmathbox[LEFT]{1} + eqmathbox[CENTRE]{sum_{r = 1}^n binom{n + 1}{r}} + eqmathbox[RIGHT]{1} \
&= 2 + sum_{r = 1}^n biggl[ binom{n}{r} + binom{n}{r - 1} biggr]
end{align*}
end{document}
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
try
documentclass{article}
usepackage{array,amsmath}
begin{document}
[
begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
sum_{r=0}^{n+1} binom{n+1}{r}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
& multicolumn{3}{>{displaystyle}l}{
2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
}
end{array}
]
end{document}
add a comment |
try
documentclass{article}
usepackage{array,amsmath}
begin{document}
[
begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
sum_{r=0}^{n+1} binom{n+1}{r}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
& multicolumn{3}{>{displaystyle}l}{
2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
}
end{array}
]
end{document}
add a comment |
try
documentclass{article}
usepackage{array,amsmath}
begin{document}
[
begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
sum_{r=0}^{n+1} binom{n+1}{r}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
& multicolumn{3}{>{displaystyle}l}{
2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
}
end{array}
]
end{document}
try
documentclass{article}
usepackage{array,amsmath}
begin{document}
[
begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
sum_{r=0}^{n+1} binom{n+1}{r}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
& multicolumn{3}{>{displaystyle}l}{
2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
}
end{array}
]
end{document}
answered 18 mins ago
ZarkoZarko
126k868165
126k868165
add a comment |
add a comment |
Use the [t]
option. Then you do not need to use multicolumn
many times if you have many subsequent lines.
documentclass{article}
usepackage{array,amsmath}
begin{document}
begin{align*}
sumlimits_{r=0}^{n+1} binom{n+1}{r}
&begin{array}[t]{@{}>{displaystyle}c @{{}={}}@{}>{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
end{array}\
&=2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
end{align*}
end{document}
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
8 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
5 mins ago
add a comment |
Use the [t]
option. Then you do not need to use multicolumn
many times if you have many subsequent lines.
documentclass{article}
usepackage{array,amsmath}
begin{document}
begin{align*}
sumlimits_{r=0}^{n+1} binom{n+1}{r}
&begin{array}[t]{@{}>{displaystyle}c @{{}={}}@{}>{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
end{array}\
&=2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
end{align*}
end{document}
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
8 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
5 mins ago
add a comment |
Use the [t]
option. Then you do not need to use multicolumn
many times if you have many subsequent lines.
documentclass{article}
usepackage{array,amsmath}
begin{document}
begin{align*}
sumlimits_{r=0}^{n+1} binom{n+1}{r}
&begin{array}[t]{@{}>{displaystyle}c @{{}={}}@{}>{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
end{array}\
&=2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
end{align*}
end{document}
Use the [t]
option. Then you do not need to use multicolumn
many times if you have many subsequent lines.
documentclass{article}
usepackage{array,amsmath}
begin{document}
begin{align*}
sumlimits_{r=0}^{n+1} binom{n+1}{r}
&begin{array}[t]{@{}>{displaystyle}c @{{}={}}@{}>{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}
& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \
& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \
end{array}\
&=2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]
end{align*}
end{document}
edited 5 mins ago
answered 12 mins ago
marmotmarmot
107k5129243
107k5129243
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
8 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
5 mins ago
add a comment |
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
8 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
5 mins ago
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
8 mins ago
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
8 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
5 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
5 mins ago
add a comment |
eqparbox
allows you to store the lengths of boxes via a <tag>
. Below I use this approach with a newly-defined eqmathbox[<tag>][<align>]
(default for <align>
is to c
entre the content) to add content to three different <tag>
ged boxes:
documentclass{article}
usepackage{eqparbox,xparse,amsmath}
% https://tex.stackexchange.com/a/34412/5764
makeatletter
NewDocumentCommand{eqmathbox}{o O{c} m}{%
IfValueTF{#1}
{defeqmathbox@##1##2{eqmakebox[#1][#2]{$##1##2$}}}
{defeqmathbox@##1##2{eqmakebox{$##1##2$}}}
mathpaletteeqmathbox@{#3}
}
makeatother
begin{document}
begin{align*}
sum_{r = 0}^{n + 1} binom{n + 1}{r}
&= eqmathbox[LEFT]{binom{n + 1}{0}} + eqmathbox[CENTRE]{binom{n + 1}{1} + dots + binom{n + 1}{n}} + eqmathbox[RIGHT]{binom{n + 1}{n + 1}} \
&= eqmathbox[LEFT]{1} + eqmathbox[CENTRE]{sum_{r = 1}^n binom{n + 1}{r}} + eqmathbox[RIGHT]{1} \
&= 2 + sum_{r = 1}^n biggl[ binom{n}{r} + binom{n}{r - 1} biggr]
end{align*}
end{document}
add a comment |
eqparbox
allows you to store the lengths of boxes via a <tag>
. Below I use this approach with a newly-defined eqmathbox[<tag>][<align>]
(default for <align>
is to c
entre the content) to add content to three different <tag>
ged boxes:
documentclass{article}
usepackage{eqparbox,xparse,amsmath}
% https://tex.stackexchange.com/a/34412/5764
makeatletter
NewDocumentCommand{eqmathbox}{o O{c} m}{%
IfValueTF{#1}
{defeqmathbox@##1##2{eqmakebox[#1][#2]{$##1##2$}}}
{defeqmathbox@##1##2{eqmakebox{$##1##2$}}}
mathpaletteeqmathbox@{#3}
}
makeatother
begin{document}
begin{align*}
sum_{r = 0}^{n + 1} binom{n + 1}{r}
&= eqmathbox[LEFT]{binom{n + 1}{0}} + eqmathbox[CENTRE]{binom{n + 1}{1} + dots + binom{n + 1}{n}} + eqmathbox[RIGHT]{binom{n + 1}{n + 1}} \
&= eqmathbox[LEFT]{1} + eqmathbox[CENTRE]{sum_{r = 1}^n binom{n + 1}{r}} + eqmathbox[RIGHT]{1} \
&= 2 + sum_{r = 1}^n biggl[ binom{n}{r} + binom{n}{r - 1} biggr]
end{align*}
end{document}
add a comment |
eqparbox
allows you to store the lengths of boxes via a <tag>
. Below I use this approach with a newly-defined eqmathbox[<tag>][<align>]
(default for <align>
is to c
entre the content) to add content to three different <tag>
ged boxes:
documentclass{article}
usepackage{eqparbox,xparse,amsmath}
% https://tex.stackexchange.com/a/34412/5764
makeatletter
NewDocumentCommand{eqmathbox}{o O{c} m}{%
IfValueTF{#1}
{defeqmathbox@##1##2{eqmakebox[#1][#2]{$##1##2$}}}
{defeqmathbox@##1##2{eqmakebox{$##1##2$}}}
mathpaletteeqmathbox@{#3}
}
makeatother
begin{document}
begin{align*}
sum_{r = 0}^{n + 1} binom{n + 1}{r}
&= eqmathbox[LEFT]{binom{n + 1}{0}} + eqmathbox[CENTRE]{binom{n + 1}{1} + dots + binom{n + 1}{n}} + eqmathbox[RIGHT]{binom{n + 1}{n + 1}} \
&= eqmathbox[LEFT]{1} + eqmathbox[CENTRE]{sum_{r = 1}^n binom{n + 1}{r}} + eqmathbox[RIGHT]{1} \
&= 2 + sum_{r = 1}^n biggl[ binom{n}{r} + binom{n}{r - 1} biggr]
end{align*}
end{document}
eqparbox
allows you to store the lengths of boxes via a <tag>
. Below I use this approach with a newly-defined eqmathbox[<tag>][<align>]
(default for <align>
is to c
entre the content) to add content to three different <tag>
ged boxes:
documentclass{article}
usepackage{eqparbox,xparse,amsmath}
% https://tex.stackexchange.com/a/34412/5764
makeatletter
NewDocumentCommand{eqmathbox}{o O{c} m}{%
IfValueTF{#1}
{defeqmathbox@##1##2{eqmakebox[#1][#2]{$##1##2$}}}
{defeqmathbox@##1##2{eqmakebox{$##1##2$}}}
mathpaletteeqmathbox@{#3}
}
makeatother
begin{document}
begin{align*}
sum_{r = 0}^{n + 1} binom{n + 1}{r}
&= eqmathbox[LEFT]{binom{n + 1}{0}} + eqmathbox[CENTRE]{binom{n + 1}{1} + dots + binom{n + 1}{n}} + eqmathbox[RIGHT]{binom{n + 1}{n + 1}} \
&= eqmathbox[LEFT]{1} + eqmathbox[CENTRE]{sum_{r = 1}^n binom{n + 1}{r}} + eqmathbox[RIGHT]{1} \
&= 2 + sum_{r = 1}^n biggl[ binom{n}{r} + binom{n}{r - 1} biggr]
end{align*}
end{document}
answered 5 mins ago
WernerWerner
446k699871692
446k699871692
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