πŸ’Ύ Archived View for dj-chase.com β€Ί documents β€Ί re-sub-sup-gemtext.gmi captured on 2022-06-11 at 20:53:24. Gemini links have been rewritten to link to archived content

View Raw

More Information

➑️ Next capture (2023-06-14)

-=-=-=-=-=-=-

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 accessible fallback.

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 or comments? Email me.

© DJ Chase, 2022-06-10. Licensed under the Academic Free License (AFL 3.0)