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