I said I hate doing hackish things instead of understanding what's happening, so of course eventually I did go figure it out...
The problem is that TeX, when placing things, has an idea of a baseline, and things have a height (how far do I extend above the baseline?) and a depth (how far do I ...
Search found 4 matches
- Fri Sep 11, 2009 1:41 am
- Forum: Math & Science
- Topic: getting spacing right in complicated continued fraction
- Replies: 4
- Views: 5942
- Thu Jul 09, 2009 3:54 am
- Forum: Math & Science
- Topic: getting spacing right in complicated continued fraction
- Replies: 4
- Views: 5942
getting spacing right in complicated continued fraction
Thanks to localghost's suggestion, I have something that looks pretty good now, so I'll share in case anyone else finds this: I used a savebox. I took the z^{t+2} fraction part and separated it out:
\newsavebox{\foo}
\savebox{\foo}{$1 - \dfrac{z^{t+2}}
{\bigg(1 - \dfrac{z^{t+3}}
{(1 - \cdots ...
\newsavebox{\foo}
\savebox{\foo}{$1 - \dfrac{z^{t+2}}
{\bigg(1 - \dfrac{z^{t+3}}
{(1 - \cdots ...
- Wed Jul 08, 2009 2:46 pm
- Forum: Math & Science
- Topic: getting spacing right in complicated continued fraction
- Replies: 4
- Views: 5942
Re: getting spacing right in complicated continued fraction
Huh, I didn't think it was the delimiters that were making the fractions so big. That's a very good start, I'll keep experimenting and see if I can improve it. Thanks for the help.
- Wed Jul 08, 2009 6:51 am
- Forum: Math & Science
- Topic: getting spacing right in complicated continued fraction
- Replies: 4
- Views: 5942
getting spacing right in complicated continued fraction
I have a somewhat unusual continued fraction and the vertical spacing is just not working properly. Here's my fraction:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\cfrac{1}
{1 - \cfrac{z^{t+1}}
{\left(1 - \cfrac{z^{t+2}}
{\left(1 - \cfrac{z^{t+3}}
{(1 - \cdots)^{t}}\right ...
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\cfrac{1}
{1 - \cfrac{z^{t+1}}
{\left(1 - \cfrac{z^{t+2}}
{\left(1 - \cfrac{z^{t+3}}
{(1 - \cdots)^{t}}\right ...