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Re: Subscripts And Superscripts In Gemtext

In reply to: Curiouser

Curiouser wrote:

Math formulas are especially challenging, not only because they can have so many symbols (such as sigma 'Ξ£'), but those symbols must be rendered in a certain layout (e.g. Ξ£ used to sum a series has the variable and its starting value, as well as the value its going to, arranged in 2 different rows to the right of the Ξ£).

As a further example, sigma notation only behaves that way when it’s inline. β€˜Block-level’ sigma notation places one row above the sigma and the other below, and makes the sigma like twice as large as the rest of the text. This holds true for a lot of other operations as well, such as integrals (∫), sequential products (∏), and sometimes unions (βˆͺ). Limits also follow the same rules, except they’re not extra-large because they don’t have their own symbol.

In fact, text layout of formulas and symbols is such a complicate domain that Donald Knuth literally created TeX, an entire digital typesetting system, while he was writing "The Art of Computer Programming."
Given all this complexity, I didn't even try to represent math formulas in gemtext.

This is what made me reply to this post. You see, I quite like math, so I envision writing some here at times. So as part of working on the back-end, I needed to think about how to typeset math in Gemtext, and I think I’ve come up with a workable solution.

Basically, we can use preformatted blocks to approximate block-level mathematics and put plaintext math using unicode’s extensive set of math symbols in the alt-text as a mostly accessible fallback.

*Note*: This is still not completely accessible, so it’s important to also provide an accessible format. Unfortunately right now, this means HTML for math.

Examples

For these examples, try collapsing the preformatted blocks to see the alt-text if your client supports that.

 ₇
 Ξ£ 𝑖²
ᢀ⁼³

⁂

 Ο€
 ⌠               ⎑         ⎀ Ο€
 ⌑  sin πœƒ dπœƒ  =  ⎒ βˆ’ cos πœƒ βŽ₯
Ο€/6              ⎣         ⎦ Ο€/6

                       √3
              =  1  +  ──
                       2

⁂

     Dβ‚“ cos(π‘₯𝑦)  =  Dβ‚“ (1 + sin 𝑦)
    βˆ’sin(π‘₯𝑦)𝑦𝑦′  =  𝑦′ cos 𝑦
     sin(π‘₯𝑦)𝑦𝑦′
   βˆ’ ──────────  =  1
      𝑦′ cos 𝑦
   sin(π‘₯𝑦)𝑦  𝑦′
βˆ’  ────────  ──  =  1
    cos 𝑦    𝑦′
                       cos 𝑦
    𝑦′ (1 Γ— 1ΒΉ)  =  - ────────
                      sin(π‘₯𝑦)𝑦
                       cos 𝑦
             𝑦′  =  - ────────
                      sin(π‘₯𝑦)𝑦

⁂

Questions, comments, or wrote a reply? Email me.

© DJ Chase, 2023-11-21. Licensed under the Academic Free License (AFL 3.0)