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Feynman vs. Computer

31 pointsby cgdltoday at 4:03 PM16 commentsview on HN

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JKCalhountoday at 5:25 PM

As a hobbyist, I'm playing with analog computer circuits right now. If you can match your curve with a similar voltage profile, a simple analog integrator (an op-amp with a capacitor connected in feedback) will also give you the area under the curve (also as a voltage of course).

Analog circuits (and op-amps just generally) are surprising cool. I know, kind of off on a tangent here but I have integration on the brain lately. You say "4 lines of Python", and I say "1 op-amp".)

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bananaflagtoday at 5:32 PM

> I hear that in electronics and quantum dynamics, there are sometimes integrals whose value is not a number, but a function, and knowing that function is important in order to know how the thing it’s modeling behaves in interactions with other things.

I'd be interested in this. So finding classical closed form solutions is the actual thing desired there?

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Animatstoday at 7:01 PM

Good numerical integration is easy, because summing smooths out noise. Good numerical differentiation is hard, because noise is amplified.

Conversely, good symbolic integration is hard, because you can get stuck and have to try another route through a combinatoric maze. Good symbolic differentiation is easy, because just applying the next obvious operation usually converges.

Huh.

Mandatory XKCD: [1]

[1] https://xkcd.com/2117/

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ogogmadtoday at 7:31 PM

The usage of confidence intervals here reminds me of the clearest way to see that integration is a computable operator, in exactly the same way that a function like sin() or sqrt() is computable. It's true thanks to a natural combination of (i) interval arithmetic and (ii) the "Darboux integral" approach to defining integration. So, intervals can do magic.

messetoday at 5:42 PM

An integral trick I picked up from a lecturer at university: if you know the result has to be of the form ax^n for some a that's probably rational and some integer n but you're feeling really lazy and/or it's annoying to simplify (even for mathematica), just plug in a transcendental value for x like Zeta[3].

Then just divide by powers of that irrational number until you have something that looks rational. That'll give you a and n. It's more or less numerical dimensional analysis.

It's not that useful for complicated integrals, but when you're feeling lazy it's a fucking godsend to know what the answer should be before you've proven it.

EDIT: s/irrational/transcendental/

eigtoday at 5:04 PM

What is the advantage of this Monte Carlo approach over a typical numerical integration method (like Runge-Kutta)?

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