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Nonstandard analysis is one of those things that sounds nice in the abstract, but even though I first learned about it more than ten years ago, I have yet to se
by scythe 3y ago
Nonstandard analysis is one of those things that sounds nice in the abstract, but even though I first learned about it more than ten years ago, I have yet to see a worked didactic-level example of how you would, for example, obtain the derivative of sine using nonstandard analysis, or derive the natural logarithm.
- ducttapecrown 3y agoI 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.
- yuppiemephisto 3y agohttps://alok.github.io/2022/09/12/y539043/ https://alok.github.io/2022/09/12/y539043/