Good typesetting practice for long equations
19
I have the following (quite horrible) equation in my thesis:
begin{align*}
frac{partial^2}{partial t_1^2} f(t_0,t_1) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot bigl(
frac{partial^2}{partial t_1^2}alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +\
alpha'(w-t_0+t_1) cdot 2 cdot log ( delta+2t_0+2t_1)+
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot frac{2}{delta+2t_0+2t_1} +\
2 frac{partial}{partial t_1} alpha( w-t_0+t_1 ) bigr) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-2}cdot\
bigl( frac{partial}{partial t_1} alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) + (alpha (w-t_0+t_1) -2) bigr) cdot \
bigl( alpha'(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +
2alpha( w-t_0+t_1)bigr) = \
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot Bigl(
frac{partial^2}{partial t_1^2}alpha(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +\
2 cdot alpha'(w-t_0+t_1) cdot bigl( 2 + log ( delta+2t_0+2t_1) bigr) Bigr) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1)-2} cdot Bigl( \
alpha '(w-t_0+t_1) cdot
(delta + 2t_0+2t_1) cdot log (delta + 2t_0+2t_1) +
bigl(alpha (w-t_0+t_1) -2) bigr) cdot
bigl( \
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +2alpha( w-t_0+t_1)bigr) Bigr) < 0
end{align*}
Using this exact piece of code, without any special formatting commands such as & or [2mm] the resulting mathematical text is quite unreadable:
How would you format such equations in LaTeX and what would you say is good practice when typesetting such large equations?
math-mode equations typography best-practices
edited Jun 15 '17 at 18:02
Moriambar
7,88731846
asked Apr 14 '13 at 11:08
malinmalin
1,59731730
add a comment |
19
I have the following (quite horrible) equation in my thesis:
begin{align*}
frac{partial^2}{partial t_1^2} f(t_0,t_1) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot bigl(
frac{partial^2}{partial t_1^2}alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +\
alpha'(w-t_0+t_1) cdot 2 cdot log ( delta+2t_0+2t_1)+
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot frac{2}{delta+2t_0+2t_1} +\
2 frac{partial}{partial t_1} alpha( w-t_0+t_1 ) bigr) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-2}cdot\
bigl( frac{partial}{partial t_1} alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) + (alpha (w-t_0+t_1) -2) bigr) cdot \
bigl( alpha'(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +
2alpha( w-t_0+t_1)bigr) = \
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot Bigl(
frac{partial^2}{partial t_1^2}alpha(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +\
2 cdot alpha'(w-t_0+t_1) cdot bigl( 2 + log ( delta+2t_0+2t_1) bigr) Bigr) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1)-2} cdot Bigl( \
alpha '(w-t_0+t_1) cdot
(delta + 2t_0+2t_1) cdot log (delta + 2t_0+2t_1) +
bigl(alpha (w-t_0+t_1) -2) bigr) cdot
bigl( \
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +2alpha( w-t_0+t_1)bigr) Bigr) < 0
end{align*}
Using this exact piece of code, without any special formatting commands such as & or [2mm] the resulting mathematical text is quite unreadable:
How would you format such equations in LaTeX and what would you say is good practice when typesetting such large equations?
math-mode equations typography best-practices
edited Jun 15 '17 at 18:02
Moriambar
7,88731846
asked Apr 14 '13 at 11:08
malinmalin
1,59731730
Either keep youralign*or use asplitinside a display-math environment, but add breaks in places unlikely to throw your reader off. Break lines before plus signs, but after multiplication signs. For the latter, I thinktimesis easier to parse thancdot, here. Also, useleft(,right)for an automatic hierarchy in delimiter size; that will help your reader parse your equation.
– jubobs
Apr 14 '13 at 11:30
1
You want to prove that this second derivative is negative, right? I guess there's no good typesetting answer to your question, but my mathematical answer is: try and give the proof more structure.
– Hendrik Vogt
Apr 15 '13 at 12:12
add a comment |
19
19
19
3
I have the following (quite horrible) equation in my thesis:
begin{align*}
frac{partial^2}{partial t_1^2} f(t_0,t_1) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot bigl(
frac{partial^2}{partial t_1^2}alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +\
alpha'(w-t_0+t_1) cdot 2 cdot log ( delta+2t_0+2t_1)+
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot frac{2}{delta+2t_0+2t_1} +\
2 frac{partial}{partial t_1} alpha( w-t_0+t_1 ) bigr) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-2}cdot\
bigl( frac{partial}{partial t_1} alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) + (alpha (w-t_0+t_1) -2) bigr) cdot \
bigl( alpha'(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +
2alpha( w-t_0+t_1)bigr) = \
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot Bigl(
frac{partial^2}{partial t_1^2}alpha(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +\
2 cdot alpha'(w-t_0+t_1) cdot bigl( 2 + log ( delta+2t_0+2t_1) bigr) Bigr) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1)-2} cdot Bigl( \
alpha '(w-t_0+t_1) cdot
(delta + 2t_0+2t_1) cdot log (delta + 2t_0+2t_1) +
bigl(alpha (w-t_0+t_1) -2) bigr) cdot
bigl( \
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +2alpha( w-t_0+t_1)bigr) Bigr) < 0
end{align*}
Using this exact piece of code, without any special formatting commands such as & or [2mm] the resulting mathematical text is quite unreadable:
How would you format such equations in LaTeX and what would you say is good practice when typesetting such large equations?
math-mode equations typography best-practices
edited Jun 15 '17 at 18:02
Moriambar
7,88731846
asked Apr 14 '13 at 11:08
malinmalin
1,59731730
I have the following (quite horrible) equation in my thesis:
begin{align*}
frac{partial^2}{partial t_1^2} f(t_0,t_1) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot bigl(
frac{partial^2}{partial t_1^2}alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +\
alpha'(w-t_0+t_1) cdot 2 cdot log ( delta+2t_0+2t_1)+
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot frac{2}{delta+2t_0+2t_1} +\
2 frac{partial}{partial t_1} alpha( w-t_0+t_1 ) bigr) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-2}cdot\
bigl( frac{partial}{partial t_1} alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) + (alpha (w-t_0+t_1) -2) bigr) cdot \
bigl( alpha'(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +
2alpha( w-t_0+t_1)bigr) = \
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot Bigl(
frac{partial^2}{partial t_1^2}alpha(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +\
2 cdot alpha'(w-t_0+t_1) cdot bigl( 2 + log ( delta+2t_0+2t_1) bigr) Bigr) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1)-2} cdot Bigl( \
alpha '(w-t_0+t_1) cdot
(delta + 2t_0+2t_1) cdot log (delta + 2t_0+2t_1) +
bigl(alpha (w-t_0+t_1) -2) bigr) cdot
bigl( \
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +2alpha( w-t_0+t_1)bigr) Bigr) < 0
end{align*}
Using this exact piece of code, without any special formatting commands such as & or [2mm] the resulting mathematical text is quite unreadable:
How would you format such equations in LaTeX and what would you say is good practice when typesetting such large equations?
math-mode equations typography best-practices
math-mode equations typography best-practices
edited Jun 15 '17 at 18:02
Moriambar
7,88731846
asked Apr 14 '13 at 11:08
malinmalin
1,59731730
edited Jun 15 '17 at 18:02
Moriambar
7,88731846
asked Apr 14 '13 at 11:08
malinmalin
1,59731730
edited Jun 15 '17 at 18:02
Moriambar
7,88731846
edited Jun 15 '17 at 18:02
Moriambar
7,88731846
edited Jun 15 '17 at 18:02
Moriambar
7,88731846
7,88731846
asked Apr 14 '13 at 11:08
malinmalin
1,59731730
asked Apr 14 '13 at 11:08
malinmalin
1,59731730
asked Apr 14 '13 at 11:08
malinmalin
1,59731730
1,59731730
Either keep youralign*or use asplitinside a display-math environment, but add breaks in places unlikely to throw your reader off. Break lines before plus signs, but after multiplication signs. For the latter, I thinktimesis easier to parse thancdot, here. Also, useleft(,right)for an automatic hierarchy in delimiter size; that will help your reader parse your equation.
– jubobs
Apr 14 '13 at 11:30
1
You want to prove that this second derivative is negative, right? I guess there's no good typesetting answer to your question, but my mathematical answer is: try and give the proof more structure.
– Hendrik Vogt
Apr 15 '13 at 12:12
add a comment |
Either keep youralign*or use asplitinside a display-math environment, but add breaks in places unlikely to throw your reader off. Break lines before plus signs, but after multiplication signs. For the latter, I thinktimesis easier to parse thancdot, here. Also, useleft(,right)for an automatic hierarchy in delimiter size; that will help your reader parse your equation.
– jubobs
Apr 14 '13 at 11:30
1
You want to prove that this second derivative is negative, right? I guess there's no good typesetting answer to your question, but my mathematical answer is: try and give the proof more structure.
– Hendrik Vogt
Apr 15 '13 at 12:12
Either keep your
align* or use a split inside a display-math environment, but add breaks in places unlikely to throw your reader off. Break lines before plus signs, but after multiplication signs. For the latter, I think times is easier to parse than cdot, here. Also, use left(, right) for an automatic hierarchy in delimiter size; that will help your reader parse your equation.– jubobs
Apr 14 '13 at 11:30
Either keep your
align* or use a split inside a display-math environment, but add breaks in places unlikely to throw your reader off. Break lines before plus signs, but after multiplication signs. For the latter, I think times is easier to parse than cdot, here. Also, use left(, right) for an automatic hierarchy in delimiter size; that will help your reader parse your equation.– jubobs
Apr 14 '13 at 11:30
1
1
You want to prove that this second derivative is negative, right? I guess there's no good typesetting answer to your question, but my mathematical answer is: try and give the proof more structure.
– Hendrik Vogt
Apr 15 '13 at 12:12
You want to prove that this second derivative is negative, right? I guess there's no good typesetting answer to your question, but my mathematical answer is: try and give the proof more structure.
– Hendrik Vogt
Apr 15 '13 at 12:12
add a comment |
4 Answers
4
active
oldest
votes
14
I'd try to make the equation smaller by grouping parts:
- Don't use
cdotwhere it's not necessary. I use it only for scalar products of vectors and for numbers, but not for symbolic factors or before parentheses. - Derivatives are often written as
partial_{t_1}instead offrac{partial}{partial t_1}. This can save some space. - Introducing substitutions can be helpful. In your code
(delta+2t_0+2t_1)appears quite often and it could be replaced by a new symbol which will be defined before or after the equation - Align the equation at least on all equal signs:
&=
- Other line breaks may be before
+signs to "group" summands (this shows that the equation consists of similar parts that are added together)
edited Apr 15 '13 at 9:59
egreg
717k8719023197
answered Apr 14 '13 at 11:29
Foo BarFoo Bar
5,99494092
add a comment |
37
breaking before not after operators and defining names for the subterms
documentclass{article}
usepackage{amsmath}
begin{document}
begin{align*}
frac{partial^2}{partial t_1^2} f(t_0,t_1)
&=
b^{a-1} cdot bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
a' cdot 2 cdot log ( b)+
a' cdot b cdot frac{2}{b} +
2 frac{partial}{partial t_1} a bigr) \
&quad+
b^{a-2}cdot
bigl( frac{partial}{partial t_1}a cdot b cdot log ( b) + (a -2) bigr) cdot
bigl( a' cdot b cdot log ( b) + 2abigr)\
& =
b^{a-1} cdot Bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
2 cdot a' cdot bigl( 2 + log ( b) bigr) Bigr)\
&quad +
b^{a-2} cdot bigl(a' cdot
c cdot log (c) +
bigl(a -2) bigr) cdot
bigl(a' cdot b cdot log ( b) +2a)bigr)bigr)\
&< 0
end{align*}
where:\
$a=alpha( w-t_0+t_1 )$\
$a'=alpha'(w-t_0+t_1)$\
$b=delta+2t_0+2t_1$\
$c=delta + 2t_0+2t_1$
end{document}
answered Apr 14 '13 at 11:33
David CarlisleDavid Carlisle
488k4111291877
4
I think adding aqquadafter the first and third line breaks would help the parsing.
– jubobs
Apr 14 '13 at 11:35
1
@FooBar'is equivalent to^prime.
– jubobs
Apr 14 '13 at 11:36
1
@Jubobs always?
– Foo Bar
Apr 14 '13 at 11:37
2
@FooBar yes (unless you define it not to be)
– David Carlisle
Apr 14 '13 at 11:38
1
@David Carlisle, Thanks, didn't knew that.
– Foo Bar
Apr 14 '13 at 11:39
|
show 11 more comments
12
Actually, I would like to start answering with a question: Is it very informative to display an equation that long?
I would try to identify parts in your equation, and write something like
[a (A + B + C) < 0]
where
[a = ... ]
and
begin{align}
A &= ... \
B &= ... \
C &= ...
end{align}
this makes it much easier to read it, and you can maybe give also an exlanation to every term.
edited Jul 15 '17 at 16:41
ShreevatsaR
27.6k87199
answered Apr 14 '13 at 11:24
jjdbjjdb
1,51811033
add a comment |
2
Try using the breqn package. Begin with usepackage{breqn}, then replace the align* environment with dmath*. Then remove all the manual linebreaks \, because breqn does the line-breaking and aligning automatically. Also you can replace bigl and bigr with left and right, because breqn allows line breaks within a left-right pair.
documentclass{article}
usepackage{breqn} % from the "mh" bundle
begin{document}
begin{dmath*}
frac{partial^2}{partial t_1^2} f(t_0,t_1) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
alpha'(w-t_0+t_1) cdot 2 cdot log ( delta+2t_0+2t_1)+
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot frac{2}{delta+2t_0+2t_1} +
2 frac{partial}{partial t_1} alpha( w-t_0+t_1 ) right) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-2}cdot
left( frac{partial}{partial t_1} alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1)
cdot log ( delta+2t_0+2t_1) + (alpha (w-t_0+t_1) -2) right) cdot
left( alpha'(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +
2alpha( w-t_0+t_1)right) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
2 cdot alpha'(w-t_0+t_1) cdot left( 2 + log ( delta+2t_0+2t_1) right) right)
+ ( delta+2t_0+2t_1)^{alpha( w-t_0+t_1)-2} cdot Bigl(
alpha '(w-t_0+t_1) cdot
(delta + 2t_0+2t_1) cdot log (delta + 2t_0+2t_1) +
left(alpha (w-t_0+t_1) -2 right) cdot
left(
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +2alpha(
w-t_0+t_1)right) Bigr) < 0
end{dmath*}
end{document}
answered Apr 15 '13 at 9:51
Ari BrodskyAri Brodsky
1,4181230
add a comment |
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4 Answers
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4 Answers
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14
I'd try to make the equation smaller by grouping parts:
- Don't use
cdotwhere it's not necessary. I use it only for scalar products of vectors and for numbers, but not for symbolic factors or before parentheses. - Derivatives are often written as
partial_{t_1}instead offrac{partial}{partial t_1}. This can save some space. - Introducing substitutions can be helpful. In your code
(delta+2t_0+2t_1)appears quite often and it could be replaced by a new symbol which will be defined before or after the equation - Align the equation at least on all equal signs:
&=
- Other line breaks may be before
+signs to "group" summands (this shows that the equation consists of similar parts that are added together)
edited Apr 15 '13 at 9:59
egreg
717k8719023197
answered Apr 14 '13 at 11:29
Foo BarFoo Bar
5,99494092
add a comment |
14
I'd try to make the equation smaller by grouping parts:
- Don't use
cdotwhere it's not necessary. I use it only for scalar products of vectors and for numbers, but not for symbolic factors or before parentheses. - Derivatives are often written as
partial_{t_1}instead offrac{partial}{partial t_1}. This can save some space. - Introducing substitutions can be helpful. In your code
(delta+2t_0+2t_1)appears quite often and it could be replaced by a new symbol which will be defined before or after the equation - Align the equation at least on all equal signs:
&=
- Other line breaks may be before
+signs to "group" summands (this shows that the equation consists of similar parts that are added together)
edited Apr 15 '13 at 9:59
egreg
717k8719023197
answered Apr 14 '13 at 11:29
Foo BarFoo Bar
5,99494092
add a comment |
14
14
14
I'd try to make the equation smaller by grouping parts:
- Don't use
cdotwhere it's not necessary. I use it only for scalar products of vectors and for numbers, but not for symbolic factors or before parentheses. - Derivatives are often written as
partial_{t_1}instead offrac{partial}{partial t_1}. This can save some space. - Introducing substitutions can be helpful. In your code
(delta+2t_0+2t_1)appears quite often and it could be replaced by a new symbol which will be defined before or after the equation - Align the equation at least on all equal signs:
&=
- Other line breaks may be before
+signs to "group" summands (this shows that the equation consists of similar parts that are added together)
edited Apr 15 '13 at 9:59
egreg
717k8719023197
answered Apr 14 '13 at 11:29
Foo BarFoo Bar
5,99494092
I'd try to make the equation smaller by grouping parts:
- Don't use
cdotwhere it's not necessary. I use it only for scalar products of vectors and for numbers, but not for symbolic factors or before parentheses. - Derivatives are often written as
partial_{t_1}instead offrac{partial}{partial t_1}. This can save some space. - Introducing substitutions can be helpful. In your code
(delta+2t_0+2t_1)appears quite often and it could be replaced by a new symbol which will be defined before or after the equation - Align the equation at least on all equal signs:
&=
- Other line breaks may be before
+signs to "group" summands (this shows that the equation consists of similar parts that are added together)
edited Apr 15 '13 at 9:59
egreg
717k8719023197
answered Apr 14 '13 at 11:29
Foo BarFoo Bar
5,99494092
edited Apr 15 '13 at 9:59
egreg
717k8719023197
edited Apr 15 '13 at 9:59
egreg
717k8719023197
edited Apr 15 '13 at 9:59
egreg
717k8719023197
717k8719023197
answered Apr 14 '13 at 11:29
Foo BarFoo Bar
5,99494092
answered Apr 14 '13 at 11:29
Foo BarFoo Bar
5,99494092
answered Apr 14 '13 at 11:29
Foo BarFoo Bar
5,99494092
5,99494092
add a comment |
add a comment |
37
breaking before not after operators and defining names for the subterms
documentclass{article}
usepackage{amsmath}
begin{document}
begin{align*}
frac{partial^2}{partial t_1^2} f(t_0,t_1)
&=
b^{a-1} cdot bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
a' cdot 2 cdot log ( b)+
a' cdot b cdot frac{2}{b} +
2 frac{partial}{partial t_1} a bigr) \
&quad+
b^{a-2}cdot
bigl( frac{partial}{partial t_1}a cdot b cdot log ( b) + (a -2) bigr) cdot
bigl( a' cdot b cdot log ( b) + 2abigr)\
& =
b^{a-1} cdot Bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
2 cdot a' cdot bigl( 2 + log ( b) bigr) Bigr)\
&quad +
b^{a-2} cdot bigl(a' cdot
c cdot log (c) +
bigl(a -2) bigr) cdot
bigl(a' cdot b cdot log ( b) +2a)bigr)bigr)\
&< 0
end{align*}
where:\
$a=alpha( w-t_0+t_1 )$\
$a'=alpha'(w-t_0+t_1)$\
$b=delta+2t_0+2t_1$\
$c=delta + 2t_0+2t_1$
end{document}
answered Apr 14 '13 at 11:33
David CarlisleDavid Carlisle
488k4111291877
4
I think adding aqquadafter the first and third line breaks would help the parsing.
– jubobs
Apr 14 '13 at 11:35
1
@FooBar'is equivalent to^prime.
– jubobs
Apr 14 '13 at 11:36
1
@Jubobs always?
– Foo Bar
Apr 14 '13 at 11:37
2
@FooBar yes (unless you define it not to be)
– David Carlisle
Apr 14 '13 at 11:38
1
@David Carlisle, Thanks, didn't knew that.
– Foo Bar
Apr 14 '13 at 11:39
|
show 11 more comments
37
breaking before not after operators and defining names for the subterms
documentclass{article}
usepackage{amsmath}
begin{document}
begin{align*}
frac{partial^2}{partial t_1^2} f(t_0,t_1)
&=
b^{a-1} cdot bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
a' cdot 2 cdot log ( b)+
a' cdot b cdot frac{2}{b} +
2 frac{partial}{partial t_1} a bigr) \
&quad+
b^{a-2}cdot
bigl( frac{partial}{partial t_1}a cdot b cdot log ( b) + (a -2) bigr) cdot
bigl( a' cdot b cdot log ( b) + 2abigr)\
& =
b^{a-1} cdot Bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
2 cdot a' cdot bigl( 2 + log ( b) bigr) Bigr)\
&quad +
b^{a-2} cdot bigl(a' cdot
c cdot log (c) +
bigl(a -2) bigr) cdot
bigl(a' cdot b cdot log ( b) +2a)bigr)bigr)\
&< 0
end{align*}
where:\
$a=alpha( w-t_0+t_1 )$\
$a'=alpha'(w-t_0+t_1)$\
$b=delta+2t_0+2t_1$\
$c=delta + 2t_0+2t_1$
end{document}
answered Apr 14 '13 at 11:33
David CarlisleDavid Carlisle
488k4111291877
4
I think adding aqquadafter the first and third line breaks would help the parsing.
– jubobs
Apr 14 '13 at 11:35
1
@FooBar'is equivalent to^prime.
– jubobs
Apr 14 '13 at 11:36
1
@Jubobs always?
– Foo Bar
Apr 14 '13 at 11:37
2
@FooBar yes (unless you define it not to be)
– David Carlisle
Apr 14 '13 at 11:38
1
@David Carlisle, Thanks, didn't knew that.
– Foo Bar
Apr 14 '13 at 11:39
|
show 11 more comments
37
37
37
breaking before not after operators and defining names for the subterms
documentclass{article}
usepackage{amsmath}
begin{document}
begin{align*}
frac{partial^2}{partial t_1^2} f(t_0,t_1)
&=
b^{a-1} cdot bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
a' cdot 2 cdot log ( b)+
a' cdot b cdot frac{2}{b} +
2 frac{partial}{partial t_1} a bigr) \
&quad+
b^{a-2}cdot
bigl( frac{partial}{partial t_1}a cdot b cdot log ( b) + (a -2) bigr) cdot
bigl( a' cdot b cdot log ( b) + 2abigr)\
& =
b^{a-1} cdot Bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
2 cdot a' cdot bigl( 2 + log ( b) bigr) Bigr)\
&quad +
b^{a-2} cdot bigl(a' cdot
c cdot log (c) +
bigl(a -2) bigr) cdot
bigl(a' cdot b cdot log ( b) +2a)bigr)bigr)\
&< 0
end{align*}
where:\
$a=alpha( w-t_0+t_1 )$\
$a'=alpha'(w-t_0+t_1)$\
$b=delta+2t_0+2t_1$\
$c=delta + 2t_0+2t_1$
end{document}
answered Apr 14 '13 at 11:33
David CarlisleDavid Carlisle
488k4111291877
breaking before not after operators and defining names for the subterms
documentclass{article}
usepackage{amsmath}
begin{document}
begin{align*}
frac{partial^2}{partial t_1^2} f(t_0,t_1)
&=
b^{a-1} cdot bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
a' cdot 2 cdot log ( b)+
a' cdot b cdot frac{2}{b} +
2 frac{partial}{partial t_1} a bigr) \
&quad+
b^{a-2}cdot
bigl( frac{partial}{partial t_1}a cdot b cdot log ( b) + (a -2) bigr) cdot
bigl( a' cdot b cdot log ( b) + 2abigr)\
& =
b^{a-1} cdot Bigl(
frac{partial^2}{partial t_1^2}a cdot b cdot log ( b) +
2 cdot a' cdot bigl( 2 + log ( b) bigr) Bigr)\
&quad +
b^{a-2} cdot bigl(a' cdot
c cdot log (c) +
bigl(a -2) bigr) cdot
bigl(a' cdot b cdot log ( b) +2a)bigr)bigr)\
&< 0
end{align*}
where:\
$a=alpha( w-t_0+t_1 )$\
$a'=alpha'(w-t_0+t_1)$\
$b=delta+2t_0+2t_1$\
$c=delta + 2t_0+2t_1$
end{document}
answered Apr 14 '13 at 11:33
David CarlisleDavid Carlisle
488k4111291877
edited Mar 21 '17 at 21:45
answered Apr 14 '13 at 11:33
David CarlisleDavid Carlisle
488k4111291877
answered Apr 14 '13 at 11:33
David CarlisleDavid Carlisle
488k4111291877
answered Apr 14 '13 at 11:33
David CarlisleDavid Carlisle
488k4111291877
488k4111291877
4
I think adding aqquadafter the first and third line breaks would help the parsing.
– jubobs
Apr 14 '13 at 11:35
1
@FooBar'is equivalent to^prime.
– jubobs
Apr 14 '13 at 11:36
1
@Jubobs always?
– Foo Bar
Apr 14 '13 at 11:37
2
@FooBar yes (unless you define it not to be)
– David Carlisle
Apr 14 '13 at 11:38
1
@David Carlisle, Thanks, didn't knew that.
– Foo Bar
Apr 14 '13 at 11:39
|
show 11 more comments
4
I think adding aqquadafter the first and third line breaks would help the parsing.
– jubobs
Apr 14 '13 at 11:35
1
@FooBar'is equivalent to^prime.
– jubobs
Apr 14 '13 at 11:36
1
@Jubobs always?
– Foo Bar
Apr 14 '13 at 11:37
2
@FooBar yes (unless you define it not to be)
– David Carlisle
Apr 14 '13 at 11:38
1
@David Carlisle, Thanks, didn't knew that.
– Foo Bar
Apr 14 '13 at 11:39
4
4
I think adding a
qquad after the first and third line breaks would help the parsing.– jubobs
Apr 14 '13 at 11:35
I think adding a
qquad after the first and third line breaks would help the parsing.– jubobs
Apr 14 '13 at 11:35
1
1
@FooBar
' is equivalent to ^prime.– jubobs
Apr 14 '13 at 11:36
@FooBar
' is equivalent to ^prime.– jubobs
Apr 14 '13 at 11:36
1
1
@Jubobs always?
– Foo Bar
Apr 14 '13 at 11:37
@Jubobs always?
– Foo Bar
Apr 14 '13 at 11:37
2
2
@FooBar yes (unless you define it not to be)
– David Carlisle
Apr 14 '13 at 11:38
@FooBar yes (unless you define it not to be)
– David Carlisle
Apr 14 '13 at 11:38
1
1
@David Carlisle, Thanks, didn't knew that.
– Foo Bar
Apr 14 '13 at 11:39
@David Carlisle, Thanks, didn't knew that.
– Foo Bar
Apr 14 '13 at 11:39
|
show 11 more comments
12
Actually, I would like to start answering with a question: Is it very informative to display an equation that long?
I would try to identify parts in your equation, and write something like
[a (A + B + C) < 0]
where
[a = ... ]
and
begin{align}
A &= ... \
B &= ... \
C &= ...
end{align}
this makes it much easier to read it, and you can maybe give also an exlanation to every term.
edited Jul 15 '17 at 16:41
ShreevatsaR
27.6k87199
answered Apr 14 '13 at 11:24
jjdbjjdb
1,51811033
add a comment |
12
Actually, I would like to start answering with a question: Is it very informative to display an equation that long?
I would try to identify parts in your equation, and write something like
[a (A + B + C) < 0]
where
[a = ... ]
and
begin{align}
A &= ... \
B &= ... \
C &= ...
end{align}
this makes it much easier to read it, and you can maybe give also an exlanation to every term.
edited Jul 15 '17 at 16:41
ShreevatsaR
27.6k87199
answered Apr 14 '13 at 11:24
jjdbjjdb
1,51811033
add a comment |
12
12
12
Actually, I would like to start answering with a question: Is it very informative to display an equation that long?
I would try to identify parts in your equation, and write something like
[a (A + B + C) < 0]
where
[a = ... ]
and
begin{align}
A &= ... \
B &= ... \
C &= ...
end{align}
this makes it much easier to read it, and you can maybe give also an exlanation to every term.
edited Jul 15 '17 at 16:41
ShreevatsaR
27.6k87199
answered Apr 14 '13 at 11:24
jjdbjjdb
1,51811033
Actually, I would like to start answering with a question: Is it very informative to display an equation that long?
I would try to identify parts in your equation, and write something like
[a (A + B + C) < 0]
where
[a = ... ]
and
begin{align}
A &= ... \
B &= ... \
C &= ...
end{align}
this makes it much easier to read it, and you can maybe give also an exlanation to every term.
edited Jul 15 '17 at 16:41
ShreevatsaR
27.6k87199
answered Apr 14 '13 at 11:24
jjdbjjdb
1,51811033
edited Jul 15 '17 at 16:41
ShreevatsaR
27.6k87199
edited Jul 15 '17 at 16:41
ShreevatsaR
27.6k87199
edited Jul 15 '17 at 16:41
ShreevatsaR
27.6k87199
27.6k87199
answered Apr 14 '13 at 11:24
jjdbjjdb
1,51811033
answered Apr 14 '13 at 11:24
jjdbjjdb
1,51811033
answered Apr 14 '13 at 11:24
jjdbjjdb
1,51811033
1,51811033
add a comment |
add a comment |
2
Try using the breqn package. Begin with usepackage{breqn}, then replace the align* environment with dmath*. Then remove all the manual linebreaks \, because breqn does the line-breaking and aligning automatically. Also you can replace bigl and bigr with left and right, because breqn allows line breaks within a left-right pair.
documentclass{article}
usepackage{breqn} % from the "mh" bundle
begin{document}
begin{dmath*}
frac{partial^2}{partial t_1^2} f(t_0,t_1) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
alpha'(w-t_0+t_1) cdot 2 cdot log ( delta+2t_0+2t_1)+
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot frac{2}{delta+2t_0+2t_1} +
2 frac{partial}{partial t_1} alpha( w-t_0+t_1 ) right) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-2}cdot
left( frac{partial}{partial t_1} alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1)
cdot log ( delta+2t_0+2t_1) + (alpha (w-t_0+t_1) -2) right) cdot
left( alpha'(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +
2alpha( w-t_0+t_1)right) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
2 cdot alpha'(w-t_0+t_1) cdot left( 2 + log ( delta+2t_0+2t_1) right) right)
+ ( delta+2t_0+2t_1)^{alpha( w-t_0+t_1)-2} cdot Bigl(
alpha '(w-t_0+t_1) cdot
(delta + 2t_0+2t_1) cdot log (delta + 2t_0+2t_1) +
left(alpha (w-t_0+t_1) -2 right) cdot
left(
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +2alpha(
w-t_0+t_1)right) Bigr) < 0
end{dmath*}
end{document}
answered Apr 15 '13 at 9:51
Ari BrodskyAri Brodsky
1,4181230
add a comment |
2
Try using the breqn package. Begin with usepackage{breqn}, then replace the align* environment with dmath*. Then remove all the manual linebreaks \, because breqn does the line-breaking and aligning automatically. Also you can replace bigl and bigr with left and right, because breqn allows line breaks within a left-right pair.
documentclass{article}
usepackage{breqn} % from the "mh" bundle
begin{document}
begin{dmath*}
frac{partial^2}{partial t_1^2} f(t_0,t_1) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
alpha'(w-t_0+t_1) cdot 2 cdot log ( delta+2t_0+2t_1)+
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot frac{2}{delta+2t_0+2t_1} +
2 frac{partial}{partial t_1} alpha( w-t_0+t_1 ) right) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-2}cdot
left( frac{partial}{partial t_1} alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1)
cdot log ( delta+2t_0+2t_1) + (alpha (w-t_0+t_1) -2) right) cdot
left( alpha'(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +
2alpha( w-t_0+t_1)right) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
2 cdot alpha'(w-t_0+t_1) cdot left( 2 + log ( delta+2t_0+2t_1) right) right)
+ ( delta+2t_0+2t_1)^{alpha( w-t_0+t_1)-2} cdot Bigl(
alpha '(w-t_0+t_1) cdot
(delta + 2t_0+2t_1) cdot log (delta + 2t_0+2t_1) +
left(alpha (w-t_0+t_1) -2 right) cdot
left(
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +2alpha(
w-t_0+t_1)right) Bigr) < 0
end{dmath*}
end{document}
answered Apr 15 '13 at 9:51
Ari BrodskyAri Brodsky
1,4181230
add a comment |
2
2
2
Try using the breqn package. Begin with usepackage{breqn}, then replace the align* environment with dmath*. Then remove all the manual linebreaks \, because breqn does the line-breaking and aligning automatically. Also you can replace bigl and bigr with left and right, because breqn allows line breaks within a left-right pair.
documentclass{article}
usepackage{breqn} % from the "mh" bundle
begin{document}
begin{dmath*}
frac{partial^2}{partial t_1^2} f(t_0,t_1) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
alpha'(w-t_0+t_1) cdot 2 cdot log ( delta+2t_0+2t_1)+
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot frac{2}{delta+2t_0+2t_1} +
2 frac{partial}{partial t_1} alpha( w-t_0+t_1 ) right) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-2}cdot
left( frac{partial}{partial t_1} alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1)
cdot log ( delta+2t_0+2t_1) + (alpha (w-t_0+t_1) -2) right) cdot
left( alpha'(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +
2alpha( w-t_0+t_1)right) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
2 cdot alpha'(w-t_0+t_1) cdot left( 2 + log ( delta+2t_0+2t_1) right) right)
+ ( delta+2t_0+2t_1)^{alpha( w-t_0+t_1)-2} cdot Bigl(
alpha '(w-t_0+t_1) cdot
(delta + 2t_0+2t_1) cdot log (delta + 2t_0+2t_1) +
left(alpha (w-t_0+t_1) -2 right) cdot
left(
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +2alpha(
w-t_0+t_1)right) Bigr) < 0
end{dmath*}
end{document}
answered Apr 15 '13 at 9:51
Ari BrodskyAri Brodsky
1,4181230
Try using the breqn package. Begin with usepackage{breqn}, then replace the align* environment with dmath*. Then remove all the manual linebreaks \, because breqn does the line-breaking and aligning automatically. Also you can replace bigl and bigr with left and right, because breqn allows line breaks within a left-right pair.
documentclass{article}
usepackage{breqn} % from the "mh" bundle
begin{document}
begin{dmath*}
frac{partial^2}{partial t_1^2} f(t_0,t_1) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
alpha'(w-t_0+t_1) cdot 2 cdot log ( delta+2t_0+2t_1)+
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot frac{2}{delta+2t_0+2t_1} +
2 frac{partial}{partial t_1} alpha( w-t_0+t_1 ) right) +
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-2}cdot
left( frac{partial}{partial t_1} alpha(w-t_0+t_1) cdot ( delta+2t_0+2t_1)
cdot log ( delta+2t_0+2t_1) + (alpha (w-t_0+t_1) -2) right) cdot
left( alpha'(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +
2alpha( w-t_0+t_1)right) =
( delta+2t_0+2t_1)^{alpha( w-t_0+t_1 )-1} cdot left(
frac{partial^2}{partial t_1^2}alpha(w -t_0+t_1) cdot ( delta+2t_0+2t_1) cdot
log ( delta+2t_0+2t_1) +
2 cdot alpha'(w-t_0+t_1) cdot left( 2 + log ( delta+2t_0+2t_1) right) right)
+ ( delta+2t_0+2t_1)^{alpha( w-t_0+t_1)-2} cdot Bigl(
alpha '(w-t_0+t_1) cdot
(delta + 2t_0+2t_1) cdot log (delta + 2t_0+2t_1) +
left(alpha (w-t_0+t_1) -2 right) cdot
left(
alpha'(w-t_0+t_1) cdot ( delta+2t_0+2t_1) cdot log ( delta+2t_0+2t_1) +2alpha(
w-t_0+t_1)right) Bigr) < 0
end{dmath*}
end{document}
answered Apr 15 '13 at 9:51
Ari BrodskyAri Brodsky
1,4181230
answered Apr 15 '13 at 9:51
Ari BrodskyAri Brodsky
1,4181230
answered Apr 15 '13 at 9:51
Ari BrodskyAri Brodsky
1,4181230
answered Apr 15 '13 at 9:51
Ari BrodskyAri Brodsky
1,4181230
1,4181230
add a comment |
add a comment |
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Either keep your
align*or use asplitinside a display-math environment, but add breaks in places unlikely to throw your reader off. Break lines before plus signs, but after multiplication signs. For the latter, I thinktimesis easier to parse thancdot, here. Also, useleft(,right)for an automatic hierarchy in delimiter size; that will help your reader parse your equation.– jubobs
Apr 14 '13 at 11:30
1
You want to prove that this second derivative is negative, right? I guess there's no good typesetting answer to your question, but my mathematical answer is: try and give the proof more structure.
– Hendrik Vogt
Apr 15 '13 at 12:12