Re: Subscripts And Superscripts In Gemtext

Curiouser wrote:

Math formulas are especially challenging, not only because they can have so many symbols (such as sigma 'Σ'), but those symbols must be rendered in a certain layout (e.g. Σ 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 Σ).

As a further example, sigma notation only behaves that way when it’s inline. ‘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 (∪). 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.

Examples

For these examples, I have put the alt text as the summary and a properly marked-up preformatted block in the details. This more closely resembles the Gemtext expirence.

Σᵢ₌₃⁷ i²

 ₇
 Σ 𝑖²
ᶤ⁼³

∫ _(π/6) ^π sin θ dθ = [−cos θ] _(π/6) ^π = 1 + (√3)/2

 π
 ⌠               ⎡         ⎤ π
 ⌡  sin 𝜃 d𝜃  =  ⎢ − cos 𝜃 ⎥
π/6              ⎣         ⎦ π/6

                       √3
              =  1  +  ──
                       2

Dₓ cos(xy) = Dₓ (1 + sin y)

     Dₓ cos(𝑥𝑦)  =  Dₓ (1 + sin 𝑦)

−sin(xy)yy' = y' cos y

     sin(𝑥𝑦)𝑦𝑦′
   − ──────────  =  1
      𝑦′ cos 𝑦

- ""[sin(xy)y / cos y]"" (y'/y') = 1

   sin(𝑥𝑦)𝑦  𝑦′
−  ────────  ──  =  1
    cos 𝑦    𝑦′

y' (1 × 1¹) = − cos y / [sin(xy)y]

                       cos 𝑦
    𝑦′ (1 × 1¹)  =  - ────────
                      sin(𝑥𝑦)𝑦

y' = − cos y / [sin(xy)y]

                       cos 𝑦
             𝑦′  =  - ────────
                      sin(𝑥𝑦)𝑦