3 ms·
We hit comment cooldown but it's not foolery exactly, more like reflecting the futility of trying to make these kinds of arguments. The problem with this whole
by hooloovoo_zoo 3y ago
We hit comment cooldown but it's not foolery exactly, more like reflecting the futility of trying to make these kinds of arguments. The problem with this whole chain is we haven't actually defined what a neuron is. Take the purported bound above. Suppose I move an associated potassium ion farther away. Have I not changed the state? If yes, what's the limit to how far away I can move that ion and have it still count as part of the neuron? That's just one of the limits of physical laws, you can make very precise statements, but only under very narrow constraints.
- SideburnsOfDoom 3y agoHere are some bounds, from most theoretical to most practical: The size of the earth's biosphere The size of the largest animal. The size of a human skull cavity. The size of a human skull cavity, divided by the number of neurons therein ( a few Billion). Is this not obvious? There's nothing futile about stating these obvious facts. Do not fool around with "if I move a potassium ion farther away, isn't that a different bound" ? correct, but moving the bound slightly, is irrelevant to the obvious statement that there are still bounds, and they are practical. Stating that the neuron is not well-defined is similar irrelevance, if you want to precisely measure the bound, you will need to know precisely what is bounded, but I did not, merely stated the obvious truth that every such definition will come with a bound. One can say "definitely smaller than this" in a wide range of ways, without defining the exact edge. What's the alternate argument: "I won't define it, therefor it might be larger than the solar system". This is not a serious thing to say. And to think that such bounded systems "have infinite states would thereby be capable of storing infinite information" is pseudo-mystical foolery. Large, or even "vast", is not infinite. And as sibling comment has mentioned, there is a "noise floor" below which analog resolution means nothing.