8 points | by davedx 7 hours ago ago
3 comments
let e^2 = 0.
the dual numbers are D = {x + ye | x,y in R}.
let f(z) be some polynomial.
Then by binomial theorem f(z + e) - f(z) = z^n + n*z^(n-1)*e + O(e^2) - z^n = f'(z)*e.
And so df = df/dz * e
Really let down by these short sentences.
“Here’s where it gets interesting.”
“That’s not a ____. It’s a _____.”
I’ll gladly read this if the author does an editing pass and makes it read more naturally.
Neato introduction, but the text reeks of LLMs so I stopped reading.
let e^2 = 0.
the dual numbers are D = {x + ye | x,y in R}.
let f(z) be some polynomial.
Then by binomial theorem f(z + e) - f(z) = z^n + n*z^(n-1)*e + O(e^2) - z^n = f'(z)*e.
And so df = df/dz * e
Really let down by these short sentences.
“Here’s where it gets interesting.”
“That’s not a ____. It’s a _____.”
I’ll gladly read this if the author does an editing pass and makes it read more naturally.
Neato introduction, but the text reeks of LLMs so I stopped reading.