6 ms·

Logarithms yearning to be free

75 points by _Microft 5y ago

_Microft 5y ago
You might want to appeal to Alexandra, patron saint of the paywalled scientist, for she might hear you.
- CodesInChaos 5y ago
  Searching for the title of the paper, DDG shows scihub as the first result, while google only has a libgen hit on the second page (and it's the author's list of papers, not the paper itself).
westurner 5y ago
> The author opens with the example of finding the antiderivative of xn. When n ≠ -1 the antiderivative is another power function, but when n = -1 it’s a logarithm. What a neat limit. Probably best to leave the powerfn/logfn() as a dumb symbolic symbol until the end (until after later parameter substitution)?
- nh23423fefe 5y ago
  I don't follow. The antiderivative fails because of division by zero. What limit? And what does symbolic manip do?
  - gus_massa 5y ago
    The idea is that you want to calculate integral{from t=0, to t=x} t^{-1+0} dx but you calculate instead f(x) = lim_{ε->0} [ integral{from t=0, to t=x} t^{-1+ε} dx ] that is just = lim_{ε->0} [ (t^ε - 1) / ε ] replacing t^ε = lim_{ε->0} [ (e^(ln(t)*ε) - 1) / ε ] by https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule = lim_{ε->0} [ ln(t)*e^(ln(t)*ε) / 1 ] that is easy to calculate = ln(t)*e^(ln(t)*0) / 1 = ln(t)*1 / 1 = ln(t) So f(t)=ln(t)
    - _Microft 5y ago
      Just for the sake of correctness: the differential has to be dt instead of dx here.
      - gus_massa 5y ago
        I agree, thanks. But I saw your comment just now and it's too late to edit. :(
    - westurner 5y ago
      So the type of the return value changes at asymptotes/limits (already) and thus that's not a pure function in terms of math. If a [math] function returns a more complex type signature instead of throwing a ZeroDivisionError (as Python core does) what is that then called? Is it differentiable or no, etc? We throw ZeroDivisionError instead of axiomatically defining a ranking for scalar*parameter*inf if x > 0: 2*x*inf > x*inf # because 2 > 1 But basically every CAS just prematurely throws away all terms next to infinity (by replacing the information in that expression with just infinity)? And nothing yet implements e.g. Conway's Surreal numbers infinities? Is negative infinity to the infinity greater or lesser than infinity? assert (-1*math.inf)**math.inf == math.inf assert (-1*sympy.oo)**sympy.oo == sympy.oo Here's a dumb Real/Function instead of prematurely discarding information that could be useful: from sympy import symbol from sympy.abc import x Infinity = symbol('Infinity', real=True) # * # from sympy.symbols import Wild All the axioms just change there. limit(-x**-1) An uphill battle for certain. "[Python-ideas] Re: 'Infinity' constant in Python" https://mail.python.org/archives/list/python-ideas@python.org/message/JQPNUQXELOYKP25XJ77SX3TVVG24QS72/ https://mail.python.org/archives/list/python-ideas@python.or...
      - hither_shores 5y ago
        > So the type of the return value changes at asymptotes/limits (already) and thus that's not a pure function in terms of math. Plenty of functions are discontinuous. Almost all of them, in fact. (This isn't true constructively, but GP is clearly doing classical analysis.) > If a [math] function returns a more complex type signature instead of throwing a ZeroDivisionError (as Python core does) what is that then called? Is it differentiable or no, etc? Depends on the smooth structures you've imposed on the domain and codomain, which requires much more powerful types to represent than are available in any mainstream language I'm aware of. (You might be able to contort Haskell into something sort of close, but it wouldn't be simple.)
CyberRabbi 5y ago
A link between zeroth powers and the logarithm function is such a simple idea but potentially fruitful. Exponentials and logarithms often translate between additive groups and multiplicative groups. Maybe there are interesting isomorphisms where the logarithm can be generalized as a zeroth power.
- hither_shores 5y ago
  This is the starting point of "tropical" analysis
codeflo 5y ago
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 5y 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 5y ago
  Infinitesimal math is completely coherent, it just doesn't have one theory that is "thee algebra" like the real numbers do.
- hansvm 5y 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).
ewewr 5y ago
integral{from t=0, to t=x} t^{-1+0} dx but you calculate instead f(x) = lim_{ε->0} [ integral{from t=0, to t=x} t^{-1+ε} dx ] that is just = lim_{ε->0} [ (t^ε - 1) / ε ] replacing t^ε = lim_{ε->0} [ (e^(ln(t)ε) - 1) / ε ] by https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule = lim_{ε->0} [ ln(t)e^(ln(t)ε) / 1 ] that is easy to calculate = ln(t)e^(ln(t)0) / 1 = ln(t)1 / 1 = ln(t) So f(t)=ln(t) reply

75 points by _Microft 5y ago

_Microft 5y ago

You might want to appeal to Alexandra, patron saint of the paywalled scientist, for she might hear you.

CodesInChaos 5y ago
Searching for the title of the paper, DDG shows scihub as the first result, while google only has a libgen hit on the second page (and it's the author's list of papers, not the paper itself).

westurner 5y ago

> The author opens with the example of finding the antiderivative of xn. When n ≠ -1 the antiderivative is another power function, but when n = -1 it’s a logarithm. What a neat limit. Probably best to leave the powerfn/logfn() as a dumb symbolic symbol until the end (until after later parameter substitution)?

nh23423fefe 5y ago
I don't follow. The antiderivative fails because of division by zero. What limit? And what does symbolic manip do?
- gus_massa 5y ago
  The idea is that you want to calculate integral{from t=0, to t=x} t^{-1+0} dx but you calculate instead f(x) = lim_{ε->0} [ integral{from t=0, to t=x} t^{-1+ε} dx ] that is just = lim_{ε->0} [ (t^ε - 1) / ε ] replacing t^ε = lim_{ε->0} [ (e^(ln(t)*ε) - 1) / ε ] by https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule = lim_{ε->0} [ ln(t)*e^(ln(t)*ε) / 1 ] that is easy to calculate = ln(t)*e^(ln(t)*0) / 1 = ln(t)*1 / 1 = ln(t) So f(t)=ln(t)
  - _Microft 5y ago
    Just for the sake of correctness: the differential has to be dt instead of dx here.
    - gus_massa 5y ago
      I agree, thanks. But I saw your comment just now and it's too late to edit. :(
  - westurner 5y ago
    So the type of the return value changes at asymptotes/limits (already) and thus that's not a pure function in terms of math. If a [math] function returns a more complex type signature instead of throwing a ZeroDivisionError (as Python core does) what is that then called? Is it differentiable or no, etc? We throw ZeroDivisionError instead of axiomatically defining a ranking for scalar*parameter*inf if x > 0: 2*x*inf > x*inf # because 2 > 1 But basically every CAS just prematurely throws away all terms next to infinity (by replacing the information in that expression with just infinity)? And nothing yet implements e.g. Conway's Surreal numbers infinities? Is negative infinity to the infinity greater or lesser than infinity? assert (-1*math.inf)**math.inf == math.inf assert (-1*sympy.oo)**sympy.oo == sympy.oo Here's a dumb Real/Function instead of prematurely discarding information that could be useful: from sympy import symbol from sympy.abc import x Infinity = symbol('Infinity', real=True) # * # from sympy.symbols import Wild All the axioms just change there. limit(-x**-1) An uphill battle for certain. "[Python-ideas] Re: 'Infinity' constant in Python" https://mail.python.org/archives/list/python-ideas@python.org/message/JQPNUQXELOYKP25XJ77SX3TVVG24QS72/ https://mail.python.org/archives/list/python-ideas@python.or...
    - hither_shores 5y ago
      > So the type of the return value changes at asymptotes/limits (already) and thus that's not a pure function in terms of math. Plenty of functions are discontinuous. Almost all of them, in fact. (This isn't true constructively, but GP is clearly doing classical analysis.) > If a [math] function returns a more complex type signature instead of throwing a ZeroDivisionError (as Python core does) what is that then called? Is it differentiable or no, etc? Depends on the smooth structures you've imposed on the domain and codomain, which requires much more powerful types to represent than are available in any mainstream language I'm aware of. (You might be able to contort Haskell into something sort of close, but it wouldn't be simple.)

CyberRabbi 5y ago

A link between zeroth powers and the logarithm function is such a simple idea but potentially fruitful. Exponentials and logarithms often translate between additive groups and multiplicative groups. Maybe there are interesting isomorphisms where the logarithm can be generalized as a zeroth power.

hither_shores 5y ago
This is the starting point of "tropical" analysis

codeflo 5y ago

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 5y 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 5y ago
Infinitesimal math is completely coherent, it just doesn't have one theory that is "thee algebra" like the real numbers do.
hansvm 5y 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).

ewewr 5y ago

integral{from t=0, to t=x} t^{-1+0} dx but you calculate instead f(x) = lim_{ε->0} [ integral{from t=0, to t=x} t^{-1+ε} dx ] that is just = lim_{ε->0} [ (t^ε - 1) / ε ] replacing t^ε = lim_{ε->0} [ (e^(ln(t)ε) - 1) / ε ] by https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule = lim_{ε->0} [ ln(t)e^(ln(t)ε) / 1 ] that is easy to calculate = ln(t)e^(ln(t)0) / 1 = ln(t)1 / 1 = ln(t) So f(t)=ln(t) reply