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How my high-school calculus textbook did it was to first define ln(x) so that ln(1) = 0 and d/dx ln(x) = 1/x, then take exp(x) as the inverse function of ln(x),

How my high-school calculus textbook did it was to first define ln(x) so that ln(1) = 0 and d/dx ln(x) = 1/x, then take exp(x) as the inverse function of ln(x), and finally set e = exp(1). It's definitely a bit different from the exp-first formulation, but it does do a good job connecting the natural logarithm to a natural definition. (It's an interesting exercise to show, using only limit identities and algebraic manipulation, that this is equivalent to the usual compound-interest version of e.)

ogogmad 2y ago
I think this approach is the most logically "efficient". You can phrase it as defining ln(x) to be the integral of 1/t from 1 to x. Maybe not the most intuitive, though. Interestingly, a similar approach gives the shortest proof that exp(x) and ln(x) are computable functions (since integration is a computable functional, thanks to interval arithmetic), and therefore that e = exp(1) is a computable real number.
- LegionMammal978 2y ago
  Yeah, the hairiest part is probably the existence and uniqueness of the antiderivative, followed by the existence of an inverse for exp(1). In fact, I can't quite recall whether the book defined it as a Riemann integral or an antiderivative, but of course it had a statement of the FTC which would connect the two. (It was just a high-school textbook, so it tended to gloss over the finer points of existence and uniqueness.)
jcranmer 2y ago
That's how my textbook did it as well (well, it defined e as ln(e) = 1, but only because it introduced e before exp). The problem with this approach is that, since we were already introduced to exponents and logarithms in algebra but via different definitions, it always left this unanswered question in my head about how we knew these two definitions were the same, since everyone quickly glossed over that fact.
- ogogmad 2y ago
  Did you eventually realise that the expression a^b should be understood to "really" mean exp(b * ln(a)), at least in the case that b might not be an integer? I think even in complex analysis, the above definition a^b := exp(b ln(a)) makes sense, since the function ln() admits a Riemann surface as its natural domain and the usual complex numbers as its codomain. [EDIT] Addressing your response: > Calculus glosses over the case when a is negative The Riemann surface approach mostly rescues this. When "a" is negative, and b is 1/3 (for instance), choose "a" = (r, theta) = (|a|, 3 pi). This gives ln(a) = ln |a| + i (3 pi). Then a^b = exp((|a| + i 3 pi) / 3) = exp(ln |a|/3 + i pi) = -|a|^(1/3), as desired. Notice though that I chose to represent "a" using theta=3pi, instead of let's say 5pi.
  - jcranmer 2y ago
    The problem is a^b := exp(b ln(a)) sort of breaks down when a is negative, which is a case that is covered in algebra class but glossed over in calculus.
    - deleted 2y ago
      [deleted]
    - dawnofdusk 2y ago
      It doesn't break down, one just needs the complex logarithm. If you ignore complex numbers it breaks down in both cases. If you allow complex numbers it works in both cases.
      - ogogmad 2y ago
        No, using complex numbers alone DOES NOT work. To really allow complex numbers, you also need Riemann surfaces. The function "ln" has type ln: R -> CC where "R" denotes the Riemann surface corresponding to the natural domain of ln, and "CC" denotes the complex numbers. See here for details: https://en.wikipedia.org/wiki/Complex_logarithm#The_associated_Riemann_surface https://en.wikipedia.org/wiki/Complex_logarithm#The_associat...

ogogmad 2y ago

I think this approach is the most logically "efficient". You can phrase it as defining ln(x) to be the integral of 1/t from 1 to x. Maybe not the most intuitive, though. Interestingly, a similar approach gives the shortest proof that exp(x) and ln(x) are computable functions (since integration is a computable functional, thanks to interval arithmetic), and therefore that e = exp(1) is a computable real number.

LegionMammal978 2y ago
Yeah, the hairiest part is probably the existence and uniqueness of the antiderivative, followed by the existence of an inverse for exp(1). In fact, I can't quite recall whether the book defined it as a Riemann integral or an antiderivative, but of course it had a statement of the FTC which would connect the two. (It was just a high-school textbook, so it tended to gloss over the finer points of existence and uniqueness.)

jcranmer 2y ago

That's how my textbook did it as well (well, it defined e as ln(e) = 1, but only because it introduced e before exp). The problem with this approach is that, since we were already introduced to exponents and logarithms in algebra but via different definitions, it always left this unanswered question in my head about how we knew these two definitions were the same, since everyone quickly glossed over that fact.

ogogmad 2y ago
Did you eventually realise that the expression a^b should be understood to "really" mean exp(b * ln(a)), at least in the case that b might not be an integer? I think even in complex analysis, the above definition a^b := exp(b ln(a)) makes sense, since the function ln() admits a Riemann surface as its natural domain and the usual complex numbers as its codomain. [EDIT] Addressing your response: > Calculus glosses over the case when a is negative The Riemann surface approach mostly rescues this. When "a" is negative, and b is 1/3 (for instance), choose "a" = (r, theta) = (|a|, 3 pi). This gives ln(a) = ln |a| + i (3 pi). Then a^b = exp((|a| + i 3 pi) / 3) = exp(ln |a|/3 + i pi) = -|a|^(1/3), as desired. Notice though that I chose to represent "a" using theta=3pi, instead of let's say 5pi.
- jcranmer 2y ago
  The problem is a^b := exp(b ln(a)) sort of breaks down when a is negative, which is a case that is covered in algebra class but glossed over in calculus.
  - deleted 2y ago
    [deleted]
  - dawnofdusk 2y ago
    It doesn't break down, one just needs the complex logarithm. If you ignore complex numbers it breaks down in both cases. If you allow complex numbers it works in both cases.
    - ogogmad 2y ago
      No, using complex numbers alone DOES NOT work. To really allow complex numbers, you also need Riemann surfaces. The function "ln" has type ln: R -> CC where "R" denotes the Riemann surface corresponding to the natural domain of ln, and "CC" denotes the complex numbers. See here for details: https://en.wikipedia.org/wiki/Complex_logarithm#The_associated_Riemann_surface https://en.wikipedia.org/wiki/Complex_logarithm#The_associat...