3 ms·
I literally saw the sin one yesterday, so I'll rewrite it!! Let e be an infinitesimal. We write st(x) for the function dropping the infinitesimal part of a nu
by ducttapecrown 3y ago
I literally saw the sin one yesterday, so I'll rewrite it!!
Let e be an infinitesimal.
We write st(x) for the function dropping the infinitesimal part of a number.
Then f'(x) = st(1/e (f(x+e)-f(x)))
Now use the angle sum identity and cos(e) = 1 - e^2, sin(e) = e. I don't know how to justify these values other than the power series identities for sin and cos...
So st(1/e (sin(x+e)-sin(x))) = st(1/e (sin(x)cos(e)+cos(x)sin(e)-sin(x))
= st(1/e (sin(x)(1-e^2)+cos(x)e - sin(x)))
= st(1/e (sin(x) - sin(x) - e^2 sin(x) + cos(x)e))
= st(e sin(x) + cos(x))
= cos(x)
- ianai 3y ago"cos(e) = 1 - e^2, sin(e) = e. I don't know how to justify these values other than the power series identities for sin and cos..." Isn't that from the definition of cos and sin, even geometrically?
- karatinversion 3y agoThe version of rigorous infinitesimals I've seen didn't let you have these identities, because e^2 != 0 (because it still maintains the field axiom xy = 0 => x=0 V y=0); but the argument still goes through by carrying the whole power series in place of these simplifications – you can write the higher order terms as e*f(x, e) and drop them when you evaluate st(...).
- scythe 3y agoYou can get the relationship: 1 - x^2 < cos(x)^2 < 1 / (1 + x^2) from ordinary trigonometry, although you can't just do this with nilsquare infinitesimals; you need a more sophisticated setup.