Why is the math output hard to read sometimes?

[squidsoft]

May I suggest improving the format of the math output in the forum.

Consider the following code:

$$\mathop\textnormal{Res}\limits_{z=-n}\left\{\frac{\pi}{x^s\sin(\pi s)}\right\}=(-x)^n,\quad n=0,-1,-2,\cdots$$

The equal sign is not well displayed under the Res symbol and the "s" in sine is broken up. I've noticed other problems like this in general. I think PF would look more polished if the math output was nicer looking.

 

[phreak]

If I recall correctly, it used to be better. I'm not sure when or why the change occurred.

 

[DrGreg]

I suspect the problem might be that the LaTeX renderer (which generates the equation images) may work on the assumption that the equations will be displayed on a white background. On a grey background, some of the pixels are too faint. Is it possible to tweak the LaTeX renderer to take account of the grey background?

 

[chroot]

Hey all,

A year ago or so, something changed in the fonts included in the normal LaTeX distributions that come with most Linux distributions. Along with it were a number of other changes that broke PF's latex system. I rewrote some of it, but never really figured out the problem with the fonts.

I will look into it more. I don't actually think it has anything to do with anti-aliasing. The images are currently anti-aliased to white, and then white is dropped out as transparent. If the strokes look correct when anti-aliased to white, it seems that changing the surrounding white pixels to transparent would not affect them. It's worth a shot, though.

- Warren

 

[DrGreg]

Hey all,

A year ago or so, something changed in the fonts included in the normal LaTeX distributions that come with most Linux distributions. Along with it were a number of other changes that broke PF's latex system. I rewrote some of it, but never really figured out the problem with the fonts.

I will look into it more. I don't actually think it has anything to do with anti-aliasing. The images are currently anti-aliased to white, and then white is dropped out as transparent. If the strokes look correct when anti-aliased to white, it seems that changing the surrounding white pixels to transparent would not affect them. It's worth a shot, though.

- Warren

For what it's worth, I took the PNG image in post #1, on its default white background, and decreased the brightness until its background matched this thread's grey background. I think the result (attached) is therefore what you'd get if anti-aliased to grey. Slightly more legible, I think, but still not great, and I guess that's down to a poor choice of font. Or something.

 

[Moonbear]

Can the font be made bold, either in a default setting or when typed by the user (I never use LaTex, so don't know the ins and outs of this)? It just looks like the font is a bit thin and loses something, so if there's a way to make it bold, that might be enough to improve readability.

 

[DrGreg]

Can the font be made bold, either in a default setting or when typed by the user (I never use LaTex, so don't know the ins and outs of this)? It just looks like the font is a bit thin and loses something, so if there's a way to make it bold, that might be enough to improve readability.

That wouldn't be a solution as such, because some equations use both bold and plain font, e.g.

$$\mathbf{z} = a\mathbf{x} + b\mathbf{y}$$​

although personally I prefer

$$\textbf{z} = a\textbf{x} + b\textbf{y}$$​

However, if you have a greater choice of font weights than just "plain" and "bold", then some slightly heavier fonts might help.

 

[chroot]

Okay, guys... I changed some of the antialiasing behavior in Ghostscript (I turned it down!), and I think the output looks a little better now. If you could, post some troublesome LaTeX here and see if it renders better now.

- Warren

 

[chroot]

$$<br /> \mathop\textnormal{Res}\limits_{z=-n}\left\{\frac{\pi}{x^s\sin(\pi s)}\right\}=(-x)^n,\quad n=0,-1,-2,\cdots<br />$$

 

[CRGreathouse]

$$\sum_{n=a}^bf(n)$$ has a very strong summation symbol.

 

[chroot]

This is how it looked with the old antialiasing options:

$$\sum_{n=a}^bf(n)$$

- Warren

 

[chroot]

And now the new:

$$\sum_{n=a}^bf(n)$$

It's really strange that antialiasing options could even cause this in the first place...

- Warren

 

[chroot]

And with no anti-aliasing at all:

$$\sum_{n=a}^bf(n)$$

- Warren

 

[chroot]

Fooling around some more:

$$\sum_{n=a}^bf(n)$$

 

[chroot]

Hmmm...

$$\sum_{n=a}^bf(n)$$

 

[chroot]

Try try again:

$$\sum_{n=a}^bf(n)$$

 

[chroot]

$$<br /> \mathop\textnormal{Res}\limits_{z=-n}\left\{\frac{\pi}{x^s\sin(\pi s)}\right\}=(-x)^n,\quad n=0,-1,-2,\cdots<br />$$

 

[chroot]

I'm not really sure I've found a solution. I'll have to keep hunting.

$$<br /> \mathop\textnormal{Res}\limits_{z=-n}\left\{\frac{\pi}{x^s\sin(\pi s)}\right\}=(-x)^n,\quad n=0,-1,-2,\cdots<br />$$

- Warren

 

[Moonbear]

Some of those versions looked better...not perfect, but certainly better.

 

[Fredrik]

$$\begin{pmatrix}1 & 0 & 0\\ 0 & \frac{u_x}{u} & \frac{u_y}{u}\\ 0 & -\frac{u_y}{u} & \frac{u_x}{u} \end{pmatrix}$$

Hm, both the parentheses and the zeroes look better than they did here. They used to look like the pixel size was bigger in the LaTeX font. I'm not a big fan of the new $$\sum$$ though, and x and y are still just barely legible. Have you tried a slightly bigger font size?

It would also be nice if the \dot code would make a slightly bigger dot: $$\dot{\vec r}$$ (but I realize of course that you can't do anything that changes only that symbol).

 

[robphy]

Could it be an issue with the number of colors in the rendered image?

It seems that many of the old LaTeX images were 8-bit images (up to 256 colors) but the new ones are 4-bit (up to 16 colors) [which have smaller file sizes]. (To save, I right-click on the image then save to my desktop.)

When counting colors in some of the new images, I get 4 colors.

 

[chroot]

I'm playing with it again, so expect a little weirdness.

robphy, the images are being generated explicitly as 8-bit. If they're somehow being down-converted, I'll have to figure out where...

- Warren

 

[tiny-tim]

Strange brackets (I'm seeing the right bracket much thicker at the top ):

$$\frac{dr}{d\tau}\ =\ \pm\sqrt{E^2\ -\ \left(1 - \frac{2M}{r}\right)\left(m^2\ +\ \frac{L^2}{r^2}\right)}$$

and don't the r and the tau look very similar?

 

[lisab]

Strange brackets (I'm seeing the right bracket much thicker at the top ):

$$\frac{dr}{d\tau}\ =\ \pm\sqrt{E^2\ -\ \left(1 - \frac{2M}{r}\right)\left(m^2\ +\ \frac{L^2}{r^2}\right)}$$

and don't the r and the tau look very similar?

Yes, the r and tau are too similar, IMO. But the brackets have a nice calligraphy look to them.

 

[chroot]

Argh I'm just making it worse!

- Warren

 

[qntty]

I don't know how they did it but mathlinks has very readable equations without cranking up the resolution higher than PF

 

[Hans de Vries]

I don't know how they did it but mathlinks has very readable equations without cranking up the resolution higher than PF

Impressive indeed:

http://alt2.mathlinks.ro/latexrender/pictures/e/6/f/e6f7873e4868755812ca60c34dd13a22e4ea785b.gif

Gif images with alpha anti-aliasing, so I presume they map to 32 bit ARGB colors...

Maybe one of these?

http://packages.debian.org/search?keywords=mimetex

Regards, Hans.

 

[CRGreathouse]

They say they're using LatexRender:
http://www.mayer.dial.pipex.com/tex.htm

 

[robphy]

Impressive indeed:

http://alt2.mathlinks.ro/latexrender/pictures/e/6/f/e6f7873e4868755812ca60c34dd13a22e4ea785b.gif

Gif images with alpha anti-aliasing, so I presume they map to 32 bit ARGB colors...

That image is 8-bit with about 27 indexed-colors...with transparency.
I suspect one can do fine with 4-bit grey images if it uses up to 16 indexed-colors.
Some of the images I found have been 4-bit with only 4 colors.

When PF first supported $$\LaTeX$$, it also was and has been impressive.
It's just recently that something seems to have changed.

 

[Hans de Vries]

It's indeed not mimetex which is like this:

http://www.sciencechatforum.com/documents/#preview

using:

\sum_{n=1}^{\infty}\frac{\chi(n) }{n^{s}}=\prod_{p\in\mathbb{P}}\left(\frac{1}{1-\frac{\chi(p)}{p^{s}}}\right)


Regards, Hans