3 ms·

If you want to make a statement like that, then logarithms might not be the zeroth power, but the epsilonth power, where ε is an “infinitesimal”. Bear with me

If you want to make a statement like that, then logarithms might not be the zeroth power, but the epsilonth power, where ε is an “infinitesimal”. Bear with me here, I know infinitesimal math isn’t a fully coherent thing. But there’s a reason why Newton used it, it sometimes works surprisingly well to make intuitive analogies. Maybe there’s a way to make it work here. For example: The function ln(x) grows strictly slower than x^a for any positive real a, but faster than x^0. Hence, it’s x^ε, where 0 < ε < a for any positive real a.

gradschool 4y ago
Where does that leave superpolylogarithmic subexponential functions (sung to the tune from Mary Poppins)? https://www.csee.umbc.edu/~sherman/Papers/superpoly.ps https://www.csee.umbc.edu/~sherman/Papers/superpoly.ps
voxl 4y ago
Infinitesimal math is completely coherent, it just doesn't have one theory that is "thee algebra" like the real numbers do.
hansvm 4y ago
That's actually close to what happens with any well-founded infinitesimal theory. Bear with me a moment. Note that lots of functions have rates of growth between x^0 and all x^a, so the rationale isn't great. Log(x) doesn't actually have to behave like any exponential. Worse yet, it's similarly pretty easy to prove that x^ε must be closer to 1 than any other real (for positive x). Log(x) seems curvier than a constant, so x^ε doesn't look like quite the right representation. That said, (x^ε-1)/ε _is_ actually closer to log(x) than any other real. The easiest way to get an intuitive feel for that is to rewrite x^ε as e^(ε log(x)) and examine the maclaurin series. There are some details in proving that works appropriately, but everything does actually fall into place. In some view then, x^ε really does have the same asymptomatic structure as log(x). They just differ by infinite constant factors rather than finite ones, so they definitely don't grow at the same rate in any classical sense (Big O, Big Theta, ...). Incidentally, exactly that same idea plays into one of the examples from TFA. Integrate x^(ε-1) with the ordinary power rule, and you'll find (1/ε)x^ε, or with that same maclaurin expansion (1/ε) + log(x) + O(ε). Admittedly the infinite constant looks a little odd (and I haven't taken care to prove that the results of that particular integration procedure are anything more than a happy accident), but that bears a striking resemblance to C + log(x).
- codeflo 4y ago
  Yeah, I hadn't thought this fully through when posting, and was almost right more by accident than anything. To further show your point, if you actually take the (real) limit, you get lim t->0 (x^t-1)/t = ln(x) using L'Hopital's rule, so ln(x) = (x^ε-1)/ε makes some sense in that regard.
  - hansvm 4y ago
    Yep. That's actually one of the main points of infinitesimal theories -- all the standard theorems have infinitesimal analogues and vice versa. Rounding to the closest real number in an infinitesimal theory is the same as taking a limit in real theory. It is definitely nice to have multiple routes to the same answer though to help verify that some stupid mistake didn't infect the results.

gradschool 4y ago

Where does that leave superpolylogarithmic subexponential functions (sung to the tune from Mary Poppins)? https://www.csee.umbc.edu/~sherman/Papers/superpoly.ps https://www.csee.umbc.edu/~sherman/Papers/superpoly.ps

voxl 4y ago

Infinitesimal math is completely coherent, it just doesn't have one theory that is "thee algebra" like the real numbers do.

hansvm 4y ago

That's actually close to what happens with any well-founded infinitesimal theory. Bear with me a moment. Note that lots of functions have rates of growth between x^0 and all x^a, so the rationale isn't great. Log(x) doesn't actually have to behave like any exponential. Worse yet, it's similarly pretty easy to prove that x^ε must be closer to 1 than any other real (for positive x). Log(x) seems curvier than a constant, so x^ε doesn't look like quite the right representation. That said, (x^ε-1)/ε _is_ actually closer to log(x) than any other real. The easiest way to get an intuitive feel for that is to rewrite x^ε as e^(ε log(x)) and examine the maclaurin series. There are some details in proving that works appropriately, but everything does actually fall into place. In some view then, x^ε really does have the same asymptomatic structure as log(x). They just differ by infinite constant factors rather than finite ones, so they definitely don't grow at the same rate in any classical sense (Big O, Big Theta, ...). Incidentally, exactly that same idea plays into one of the examples from TFA. Integrate x^(ε-1) with the ordinary power rule, and you'll find (1/ε)x^ε, or with that same maclaurin expansion (1/ε) + log(x) + O(ε). Admittedly the infinite constant looks a little odd (and I haven't taken care to prove that the results of that particular integration procedure are anything more than a happy accident), but that bears a striking resemblance to C + log(x).

codeflo 4y ago
Yeah, I hadn't thought this fully through when posting, and was almost right more by accident than anything. To further show your point, if you actually take the (real) limit, you get lim t->0 (x^t-1)/t = ln(x) using L'Hopital's rule, so ln(x) = (x^ε-1)/ε makes some sense in that regard.
- hansvm 4y ago
  Yep. That's actually one of the main points of infinitesimal theories -- all the standard theorems have infinitesimal analogues and vice versa. Rounding to the closest real number in an infinitesimal theory is the same as taking a limit in real theory. It is definitely nice to have multiple routes to the same answer though to help verify that some stupid mistake didn't infect the results.