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A cute proof that makes e natural
- analog31 1y agoIt also makes f flat.
- johnp314 1y agoAt 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.
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- nayuki 1y agoExplaining the joke: In standard Western music's 12-tone equal temperament scale, the pitch class E-natural is exactly the same as the pitch class F-flat. Putting it another way, look at an F key on a piano, flatten it by one semitone by moving left, and you get an E key.
- btilly 1y agoWhile 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 1y ago> 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 1y agoThere 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<ε, pick N large enough that you can bound the tail after x^N/N! onwards with a geometric series adding up to at most ε/2. Now pick n large enough enough that the sum of errors in the terms up to N is also bounded by ε/2. From that n on, (1+x/n)^n is within ε of the power series for e^x.
- dawnofdusk 1y agoThe 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 1y ago> 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 1y agoLooking 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.
- dawnofdusk 1y agoA nice approach (which is also in Po Shen Loh's preprint) is that this limit can be seen as the Euler scheme for solving the differential equation f' = f. This is related to your viewpoint about Lie theory... maybe slightly more digestible for high school students.
- thaumasiotes 1y ago>> 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. Except that the limit as the number of splits approaches infinity is just the declared rate of interest. The computation that ultimately yields e is a mistake, not a natural quantity to calculate.
- LegionMammal978 1y agoHow 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 1y agoI 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 1y agoYeah, 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 1y agoThat'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 1y ago
- ogogmad 1y agoTangential 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.
- jeremyscanvic 1y agoI know that the irrationality of e can be proven using the theory of continued fractions or by showing that finite Taylor sums for e give out rational approximations that are too good for e to be rational. What you're saying does not ring a bell though. Can you elaborate on that? This sounds really interesting!
- ogogmad 1y agoBase-factorial is a representation for real numbers between 0 and 1. The representation has the form A/2! + B/3! + C/4! +..., where A is in {0,1}, B is in {0,1,2}, C is in {0,1,2,3}, etc. Every irrational number in [0,1) has only 1 such representation, while every rational number in [0,1) has exactly 2 such representations (where one of those is finite). Notice that e-2 is in [0,1), and the Taylor series of e^x produces a representation of e-2 in base-factorial which corresponds to neither possible forms that a rational number can take. It follows that e is irrational. Some details here: https://en.wikipedia.org/wiki/Factorial_number_system#Fractional_values https://en.wikipedia.org/wiki/Factorial_number_system#Fracti... See also: https://math.stackexchange.com/questions/425963/is-there-a-simple-proof-that-e2-is-irrational-using-a-positional-numeral-syst https://math.stackexchange.com/questions/425963/is-there-a-s... Ultimately, the above argument resembles Fourier's proof that e is irrational. But I find it more intuitive. A lot of other special constants can be proved irrational using similar techniques.
- jeremyscanvic 1y agoWhat a wild number system! That's very cool. Thanks!
- nathan_douglas 1y agoThat was lovely; I really enjoyed it. Thank you.
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- pwdisswordfishz 1y agoAnd here I thought it was irrational.