Exempt portion of equation line from aligning?
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
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asked 38 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
18 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
15 mins ago
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
just now
add a comment |
eqparbox
allows you to store the lengths of boxes via a <tag>
. Boxes with the same <tag>
are set with the maximum width across all content. 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}
Since eqparbox
uses TeX's label
-ref
system, you need to compile twice for every change in the content of the maximum width.
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 28 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
18 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
15 mins ago
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
just now
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
18 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
15 mins ago
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
just now
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 15 mins ago
answered 21 mins ago
marmotmarmot
107k5129243
107k5129243
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
18 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
15 mins ago
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
just now
add a comment |
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
18 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
15 mins ago
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
just now
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
18 mins ago
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
18 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
15 mins ago
@PGmath Very good catch! My bad. I updated.
– marmot
15 mins ago
Thanks. Can you explain a little what
[t]
does? I've never done much involved stuff with arrays before.– PGmath
just now
Thanks. Can you explain a little what
[t]
does? I've never done much involved stuff with arrays before.– PGmath
just now
add a comment |
eqparbox
allows you to store the lengths of boxes via a <tag>
. Boxes with the same <tag>
are set with the maximum width across all content. 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}
Since eqparbox
uses TeX's label
-ref
system, you need to compile twice for every change in the content of the maximum width.
add a comment |
eqparbox
allows you to store the lengths of boxes via a <tag>
. Boxes with the same <tag>
are set with the maximum width across all content. 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}
Since eqparbox
uses TeX's label
-ref
system, you need to compile twice for every change in the content of the maximum width.
add a comment |
eqparbox
allows you to store the lengths of boxes via a <tag>
. Boxes with the same <tag>
are set with the maximum width across all content. 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}
Since eqparbox
uses TeX's label
-ref
system, you need to compile twice for every change in the content of the maximum width.
eqparbox
allows you to store the lengths of boxes via a <tag>
. Boxes with the same <tag>
are set with the maximum width across all content. 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}
Since eqparbox
uses TeX's label
-ref
system, you need to compile twice for every change in the content of the maximum width.
edited 3 mins ago
answered 15 mins ago
WernerWerner
446k699871692
446k699871692
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