3 ms·

I my naive mind, if you can differentiate a number than you should be able to integrate it. So what then is the integral of a number? if I(n) is the integral o

I my naive mind, if you can differentiate a number than you should be able to integrate it. So what then is the integral of a number? if I(n) is the integral of n, then shouldn't I(D(n)) == n? But if D(prime) = 1, then what prime is the answer to I(1) ?? If this can't be done, then what were doing here isn't differentiation as I understand it. So why call this the derivative of a number? Why not call it something else?

yau8edq12i 2y ago
It is, indeed, not differentiation in the sense you're used to. Here it just means a map that satisfies the Leibniz rule, (fg)' = f' g + f g' or D(fg) = D(f) g + f D(g). Maps with this property are usually called "derivations". Although you might also want to consider that "integration" (really, indefinite integrals, AKA antiderivative) is only defined up to a constant. So why couldn't it be the same for this "number derivative"? Perhaps the "antiderivative" is only defined up to something. It'd be a fun exercise, if you're interested. Can you figure out under what conditions do you get D(a) = D(b)? Put differently, given an integer c, what are the solutions to D(x) = c?
- plank 2y ago
  Solutions to D(x)=C is probably something like p_a^a * p_b^b * … * p_n^n with a+b+d+…+n (yes, I am skipping c) equal to your constant C, and p_a, p_b etc all prime.
  - yau8edq12i 2y ago
    No, that looks wrong. Take c = 2, for example. You seem to be saying that p^2 ought to be a solution. But D(p^2) = 2p (because D(p^2) = D(p) p + p D(p) = p + p = 2p).
GrantMoyer 2y ago
Starting from my discussion in [1], you can find a sort of antiderivative of natural numbers by finding which natural number (if any) ∫F_n maps to. Note however that D is not injective and has no inverse. [1]: https://news.ycombinator.com/item?id=40327885 https://news.ycombinator.com/item?id=40327885
rm445 2y ago
Heh, the "integral" I(1) = 2 + C, where C is a constant from the set {0, 1, 3, 5, 9,...} (has to add up to another prime number).

yau8edq12i 2y ago

It is, indeed, not differentiation in the sense you're used to. Here it just means a map that satisfies the Leibniz rule, (fg)' = f' g + f g' or D(fg) = D(f) g + f D(g). Maps with this property are usually called "derivations". Although you might also want to consider that "integration" (really, indefinite integrals, AKA antiderivative) is only defined up to a constant. So why couldn't it be the same for this "number derivative"? Perhaps the "antiderivative" is only defined up to something. It'd be a fun exercise, if you're interested. Can you figure out under what conditions do you get D(a) = D(b)? Put differently, given an integer c, what are the solutions to D(x) = c?

plank 2y ago
Solutions to D(x)=C is probably something like p_a^a * p_b^b * … * p_n^n with a+b+d+…+n (yes, I am skipping c) equal to your constant C, and p_a, p_b etc all prime.
- yau8edq12i 2y ago
  No, that looks wrong. Take c = 2, for example. You seem to be saying that p^2 ought to be a solution. But D(p^2) = 2p (because D(p^2) = D(p) p + p D(p) = p + p = 2p).

GrantMoyer 2y ago

Starting from my discussion in [1], you can find a sort of antiderivative of natural numbers by finding which natural number (if any) ∫F_n maps to. Note however that D is not injective and has no inverse. [1]: https://news.ycombinator.com/item?id=40327885 https://news.ycombinator.com/item?id=40327885

rm445 2y ago

Heh, the "integral" I(1) = 2 + C, where C is a constant from the set {0, 1, 3, 5, 9,...} (has to add up to another prime number).