| 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.) |