hi,
is there a way to let ConTeXt make nested parentheses bigger automatically (based on level of nesting)? if you have like (((((..).)..))..), it is very hard to read if all parentheses are the same size.
tobias
On Tue, Dec 01, 2009 at 01:47:17PM +0100, commb07@googlemail.com wrote:
hi,
is there a way to let ConTeXt make nested parentheses bigger automatically (based on level of nesting)? if you have like (((((..).)..))..), it is very hard to read if all parentheses are the same size.
What about:
$\Bigg(\bigg(\Big(\big((..).\big)..\Big)\bigg)..\Bigg)$
Regards, Khaled
What about:
$\Bigg(\bigg(\Big(\big((..).\big)..\Big)\bigg)..\Bigg)$
Regards, Khaled
thanks for your reply, but i would like to have it done automatically, not by hand; something like a switch for parentheses mode. also i´m not sure if those \big( parentheses vary their size according to what they surround, so i should have asked for parentheses with sizing based on level of nesting AND what´s inside.
tobias
commb07@googlemail.com wrote:
What about:
$\Bigg(\bigg(\Big(\big((..).\big)..\Big)\bigg)..\Bigg)$
Regards, Khaled
thanks for your reply, but i would like to have it done automatically, not by hand; something like a switch for parentheses mode. also i´m not sure if those \big( parentheses vary their size according to what they surround, so i should have asked for parentheses with sizing based on level of nesting AND what´s inside.
Like with \left?
$\left(\left( ... \right)\right)$
Best wishes, Taco
Like with \left?
$\left(\left( ... \right)\right)$
Best wishes, Taco
maybe im doing something wrong here. i have this expression
\startformula \sum\limits_{i = 1}^n {\left( {\left( {\left( {\alpha { \odot _{C}}{{\tilde k}_i}\left( y \right)} \right){ \oplus _{C}}\left( {\beta { \odot _{C}}{{\tilde k}_i}\left( z \right)} \right)} \right){ \odot _Y}{u_i}} \right)} \stopformula
and the first 3 parentheses have the same size. i would like the outer to be bigger than the inner, like in the attached image (i hope attachments work).
tobias
commb07@googlemail.com wrote:
Like with \left?
$\left(\left( ... \right)\right)$
Best wishes, Taco
maybe im doing something wrong here. i have this expression
\startformula \sum\limits_{i = 1}^n {\left( {\left( {\left( {\alpha { \odot _{C}}{{\tilde k}_i}\left( y \right)} \right){ \oplus _{C}}\left( {\beta { \odot _{C}}{{\tilde k}_i}\left( z \right)} \right)} \right){ \odot _Y}{u_i}} \right)} \stopformula
and the first 3 parentheses have the same size. i would like the outer to be bigger than the inner, like in the attached image (i hope attachments work).
I see. You can add \delimiterfactor=1001 in the preamble somewhere.
Best wishes, Taco
commb07@googlemail.com wrote:
Like with \left?
$\left(\left( ... \right)\right)$
Best wishes, Taco
maybe im doing something wrong here. i have this expression
\startformula \sum\limits_{i = 1}^n {\left( {\left( {\left( {\alpha { \odot _{C}}{{\tilde k}_i}\left( y \right)} \right){ \oplus _{C}}\left( {\beta { \odot _{C}}{{\tilde k}_i}\left( z \right)} \right)} \right){ \odot _Y}{u_i}} \right)} \stopformula
and the first 3 parentheses have the same size. i would like the outer to be bigger than the inner, like in the attached image (i hope attachments work).
I see. You can add \delimiterfactor=1001 in the preamble somewhere.
Best wishes, Taco
yes, that´s what i was searching for! thank you so much for your fantastic help!
\delimitershortfall=-2pt seems to be another alternative.
tobias
commb07@googlemail.com wrote:
yes, that´s what i was searching for! thank you so much for your fantastic help!
\delimitershortfall=-2pt seems to be another alternative.
Yes. There is a small bit of danger in the value -2pt, though: at smaller pointsizes you may skip over an intermediate size. Any negative value will do, so \delimitershortfall=-1sp is better than -2pt.
Best wishes, Taco
On Tue, Dec 1, 2009 at 3:30 PM, commb07@googlemail.com wrote:
Like with \left?
$\left(\left( ... \right)\right)$
Best wishes, Taco
maybe im doing something wrong here. i have this expression
\startformula \sum\limits_{i = 1}^n {\left( {\left( {\left( {\alpha { \odot _{C}}{{\tilde k}_i}\left( y \right)} \right){ \oplus _{C}}\left( {\beta { \odot _{C}}{{\tilde k}_i}\left( z \right)} \right)} \right){ \odot _Y}{u_i}} \right)} \stopformula
and the first 3 parentheses have the same size. i would like the outer to be bigger than the inner, like in the attached image (i hope attachments work).
a bit ot.. It sounds to me like a hint: is the third parentheses necessary ?
is the third parentheses necessary ?
well, i guess you´re right, it´s not absolutely necessary because all the expressions have indizes, so one could guess all belong to the sum. but this way it´s clearly visible and formally precise that all expressions belong to the sum.
tobias
On Wed, Dec 2, 2009 at 6:33 PM, commb07@googlemail.com wrote:
is the third parentheses necessary ?
well, i guess you´re right, it´s not absolutely necessary because all the expressions have indizes, so one could guess all belong to the sum. but this way it´s clearly visible and formally precise that all expressions belong to the sum.
Yes, true but a sum sign smaller than first parenthesis looks a bit strange . Anyway it' really my opinion.