[HN Gopher] A cute proof that makes e natural
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A cute proof that makes e natural
Author : georgecmu
Score : 36 points
Date : 2025-04-17 16:19 UTC (6 hours ago)
(HTM) web link (www.poshenloh.com)
(TXT) w3m dump (www.poshenloh.com)
| analog31 wrote:
| It also makes f flat.
| johnp314 wrote:
| At first your comment fell flat with me but then I realized it
| was pretty sharp. You are a natural.
|
| But the cute proof was pretty cute. I recommend calc teachers
| try to work it into their lecture on e.
| btilly wrote:
| While this may convince students, you haven't actually prove that
| any exponential function has a slope. The usual presentation
| doesn't even demonstrate that such functions are defined at
| irrational numbers.
|
| That said, it is worthwhile to go through the algebra exercise to
| convince yourself that, for large n, (1+x/n)^n expands out to
| approximately 1 + x + x^2/2 + x^3/6 + ...
|
| Hint. The x^k terms come out to (x/n)^k (n choose k). This will
| turn out to be x^k/k! + O(x^k/n). As n goes to infinity, the
| error term drops out, and we're just left with the series that we
| want.
| ogogmad wrote:
| > for large n, (1+x/n)^n expands out to approximately 1 + x +
| x^2/2 + x^3/6 + ...
|
| The rigorous version of this argument uses the Dominated
| Convergence Theorem in the special case of infinite series.
| btilly wrote:
| There are several ways to make this rigorous.
|
| An explicit epsilon-delta style proof is not that hard to
| produce, it's just a little messy. What you have to do is,
| for a given x and 0<e, pick N large enough that you can bound
| the tail after x^N/N! onwards with a geometric series adding
| up to at most e/2. Now pick n large enough enough that the
| sum of errors in the terms up to N is also bounded by e/2.
| From that n on, (1+x/n)^n is within e of the power series for
| e^x.
| dawnofdusk wrote:
| The arxiv preprint linked in this is _really_ good. I 'm American
| so I got my education on e from the compound interest limit which
| isn't natural at all, as Loh points out. Why should it matter how
| many times I "split up" my compounding?
|
| IMO exponentials should just not be taught at all without basic
| notions of calculus (slopes of tangent lines suffice, as Po Shen
| Loh does here). The geometric intuition matters more than how to
| algebraically manipulate derivatives. The differential equation
| is by far the most natural approach, and it deserves to be taught
| earlier to students as is done apparently in France and Russia.
| munchler wrote:
| > Why should it matter how many times I "split up" my
| compounding
|
| It doesn't, but the limit as the number of splits approaches
| infinity is obviously an interesting (i.e. "natural") result.
|
| The perimeter of a polygon with an infinite number of sides is
| also interesting for the same reason.
| ogogmad wrote:
| Looking at it in terms of compound interest seems random.
| That said, I think the expression lim((1+x/n)^n) is better
| motivated within Lie theory, since every Lie group admits a
| faithful linear representation in which the expression
| lim((1+x/n)^n) makes sense (even if infinite-dimensional
| Hilbert spaces might be needed, as in the metaplectic group).
| Then the subexpression 1+x/n approximates a tangent vector to
| 1.
| LegionMammal978 wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| 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.
| LegionMammal978 wrote:
| I suppose the method would be to derive ln(xy) = ln(x) +
| ln(y) and the corresponding exp(x + y) = exp(x)exp(y), then
| see how this lets exp(y ln(x)) coincide with repeated
| multiplication for integer y. Connecting this to the series
| definition of exp(x) would also take some work, but my
| textbook wasn't very big on series definitions in general.
| ogogmad wrote:
| Tangential fact (har har): The Taylor series for e^x, combined
| with the uniqueness of representing a real number in base-
| factorial, immediately shows that e is irrational.
| nathan_douglas wrote:
| That was lovely; I really enjoyed it. Thank you.
| russellbeattie wrote:
| [delayed]
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