[jordigh]

Usually people who don't understand stuff like this are unable to even begin to explain what they don't understand.

I don't really understand people like this. I have tried to guess many times what the problem is. I think they may be too embarrassed to admit they don't know what an integral or factorial sign means, or perhaps something like why f is a polynomial, or what a polynomial is.

[jameshart]

I'm not the OP, but your attitude here is unnecessarily dismissive. Not everybody has the calculus skills to understand this, let alone an awareness of mid-twentieth century calculus notation conventions.

For my part, I do know what polynomials are, what a Taylor series is, and so on, and in theory I can trace through the steps here and agree that yes, one follows from the other. Yet I find this proof unsatisfying because it doesn't demonstrate clearly, to me, which particular properties of pi it is using that bring about the contradiction. When the proof makes use of pi, it doesn't explain why the statements it is making are specifically true for pi, and not for some other number.

Take this same proof, substitute the number 3 for every occurrence of pi. Now pinpoint for me the place in the proof where it is clear and obvious that the proof makes an invalid claim about the number three (but where for pi, it was clearly and obviously correct). If you can't find it, then this proof structure equally serves as a convincing argument that three is irrational. That's quite unsatisfying - though of course, a proof doesn't have to be convincing, it just has to be right. Nevertheless, to qualify as a 'simple' proof, I think it does have to appeal to intuitions and concepts in a way that simply convinces you of its truth.

This proof is short, but it is not simple. It doesn't satisfy because it doesn't show me how the ratio of a circumference to a diameter has to be irrational - only how a number called pi which has particular relationships (not specified clearly in the proof) to the sin and cos functions, has to be irrational.

[jordigh]

I didn't mean my attitude to be dismissive. It's just that truly a lot of people respond this way. "I don't understand it at all." And when you ask what don't they understand, they're unable to say it. Witness for example how exacube seems to have vanished and will probably never tell us what she or he did not understand.

I think what happens is that people are so overwhelmed with unfamiliar ideas when they encounter a proof like this that they just grind to a halt, curl up into a ball, and scream how much they hate it all and don't understand a bit of it. We have at least a couple of other people in this thread who have expressed their hatred of calculus. Starting from that it seems pretty hopeless to try to explain to them this proof.

    Yet I find this proof unsatisfying because it doesn't demonstrate
    clearly, to me, which particular properties of pi it is using that
    bring about the contradiction.

Only one: that it's a root of sin(x). The proof actually works for any nonzero root of sin(x).

In fact, that's a great definition of pi: the least positive root of sin. It's a much easier definition to work with than ratio of circumference to diameter (how do you define cirumference? What is length? What is a curve?)

[jameshart]

well sine has to come from lengths of curves, doesn't it? Because the geometric definition of sine is a function from angles to numbers, and to get pi in there you need to introduce radians as a way to measure angles - otherwise I could argue that the roots of sin are 0, 180, 360, etc. - and none of them are particularly irrational.

[jordigh]

No. There's the analytic definition of sin that does not mention curves or lengths at all: the sine is the unique function that satisfies the following differential equation and initial value problem:

    s''(x) + s(x) = 0
    s(0)  = 0
    s'(0) = 1

It's a nifty way to define sine purely by its differential properties. Of course, it requires some work to show that differential equations have a solution and that this particular solution is sine (i.e. has the properties you want a sine to have), but once you do that work, it's pretty easy to prove things such as sin^2(x) + cos^2(x) = 1.

On the other hand, starting from the geometric definitions (and building the framework for that, such as arclength, which really requires calculus), it takes a longer route to get to the calculus of sine. Historically this was the route, but we have found shortcuts since then.

[retbull]

Dude I needed you when I was working through my Calc 2 stuff. I didn't ever see that definition as a route to use. Way nicer than memorizing the unit circle which eventually translated into "knowing" the answers to the derivatives and integrals.

[jameshart]

Interesting - I was sort of wondering if you could get sine just from simple harmonic motion, and that's basically what those equations do - the first is equivalent to saying s''(x) = -s(x), so acceleration is opposite to and linearly proportional to (well, equal to in this simple case) displacement. That's a good point.

[vortico]

Just curious: What mid-twentieth century calculus notation is being used here?

[jordigh]

The same we use today.

I suppose the complaint was to hint that mathematics that is older than 50 years is too old for modern tastes, but most of our modern notation was already established at the start of the 19th century. Even Euler uses almost completely modern notation, about 250 years ago.

[jameshart]

One slightly obscure notation is for the evaluated definite integral - square brackets suffixed with the integral range.

[jordigh]

Oh, I didn't even notice that. I encounter that notation frequently enough, but it wasn't the one I was taught in high school. Kind of hard to do a web search for its origins. I tried and failed.

[retbull]

I recognized it immediately but each math teacher I had in college used different notation for the definite integral. I really did think it was common. I need to study more math.

[madprof]

But once they get all that they get stuck on what a Taylor series is, what the integrals of trigonometric functions are or what the squeeze rule is.

[mauricioc]

This is not directed at you specifically (I get your point), but here's an attempt at an explanation for those parts:

1) Verifying that f^(j) = 0 is 0 for all j doesn't require Taylor series (though, as 'dnautics pointed out, it does motivate the construction):

1.1) f^(j)(0) is zero for 0 <= j < n because every term of the polynomial f has degree at least n (and therefore you won't get a nonzero constant coefficient if you derive fewer than n times).

1.2) f^(j)(0) is zero for j >= n because once you derive n times you will get a factor of n! in each term, thus cancelling the only source of "non-integerness".

2) Point taken, you have to know how to differentiate a product and what the derivative of the sine and cosine functions is. With this in mind, checking the equation before equation (1) is routine, though. You then apply the fundamental theorem of calculus to get equation (1).

3) He is not applying the squeeze rule here, as it would not produce a contradiction. The squeeze rule would say that the limit of f(x) sin(x) (there's an implicit dependence on n here) is zero, which would say that F(pi) - F(0) is zero, which is not a contradiction.

The argument requires less machinery: For large enough n (not in the limit), f(x) sin(x) is strictly between zero and 1 (and is thus not an integer), because pi^n * a^n / n! is smaller than 1 if n is large enough. The simplest way of explaining this is that n! >= (n/2)^(n/2), as there are n/2 terms each larger than n/2 in the definition of n!. Thus the expression is at most ((pi a)^2/(n/2))^(n/2), and thus taking taking any n such that n/2 >= (pi a)^2 works for making the right side less than 1.

(If you know the definition of the Euler constant e as a series, you'll see that (pi a)^n / n! appears in the expansion of e^(pi a). Since the series converges, this means that (pi a)^n / n! is less than 1 if n is large enough. But this requires more previous knowledge.)

[hasenj]

I don't understand calculus and have no problem admitting to it.

I hated calculus throughout highschool and college.