4 ms·

It's not immediately clear (though I'm sure it is true) from reading the article that F(\pi) + F(0) != 0 (assuming \pi = \frac{a}{b}). Given that F(x) is an alt

• ykonstant 2y ago
  The article shows in (1) that F(\pi) + F(0) is the integral of f(x)sin(x) from 0 to pi; the integrand is non-negative there and thus so is the integral.
  • red_trumpet 2y ago
    Even stronger, the integrand is positive, hence the integral is positive as well, so it cannot be zero.
• csense 2y ago
  F(\pi) + F(0) is the integral of f(x) sin(x) from 0 to \pi. f(x) sin(x) is the product of four factors (not necessarily integers): f(x) sin(x) = (1/n!) (x^n) (a-bx)^n sin(x) Examine the signs of these terms on the interval 0 to \pi. All four factors are non-negative everywhere, so the integral has to be non-negative. Could the integral be zero? 1/n! is a nonzero constant. x^n = 0 only when x = 0. (a-bx)^n = 0 only when x = a/b = \pi. And sin(x) = 0 only when x = 0 or x = \pi. The four factors are all positive inside the interval, and nice continuous functions, so there's definitely a rectangle of positive area underneath their product (and the product can never be negative so you can never get a negative area to get the integral back down to 0).