notice that in the previous post, c is in between 0 and 1, but not 0 or 1. Therefore you know that c, c^2, c^3, c^4... are distinct roots, and so the equation has n-1 roots in total. So you can confirm this by doing (2002b)

Suppose that f: R --> R is a differentiable function with k real roots a(1) < a(2) < ... a(n), and thus f(a(i)) = 0 for each i. Show that f'(x) has at least k-1 real roots.

Again, apply Rolle's theorem n-1 times on each of the intervals a(i), a(i+1), since f(a(i)) are all equal to 0. This implies that there is a c in each of the n-1 intervals such that f'(c) = 0. So this implies that it has at least n-1 real roots.

Now the next part stumps me.

Show that f'(x)sin(x) + f(x)cos(x) has infinitely many real roots, whenever f:R--->R is differentiable.

Ok I notice that this is the derivative of f(x)sin(x), by the product rule. So, surprise, surprise, use Rolle's theorem on each of the intervals ((n-1)*pi, n*pi). So we know that there is a c in between (n-1)*pi and n*pi such that f(c) which is the function in question is a root. So because sin(x) has an infinite number of real roots space at intervals of pi, the function has an infinite number of roots as well.

sin(x) has an infinite number of real roots, but i'm not sure if i'm supposed to take this as given or prove it. I am not sure if the fundamental theorem of algebra applies, but anyway, since it can be expanded as an infinite series with polynomial up to infinite power, does the FTA imply that sin(x) has an infinite number of roots? well, we know this from intuition anyway.

Infinite series, the Basel problem is at the bottom of the page. Also, apparently an open question is the sum of inverse cubes.


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