If f is a continuous function from R to R having a periodic antiderivative ( primitive ), then the set of its zeros is at least countable and contains an unbounded sequence (xn) such that f(xn/n) converges to f(0).
If T is the principal period of F then on each interval (nT, (n + 1)T) f has a zero (by Rolle). Here n is an integer. That shows the part about at least a countable set of zeros. Let u_n be the root in the interval above. The sequence x_n is like this
u_1, u_2, u_2, u_3, u_3, u_3, ....
so
x_(n(n+1)/2 + k) = u_(n + 1) where 1 \leq k \leq n + 1
Because u_(n + 1) is in the interval (n T, (n + 1) T) we get that x_n/n has limit 0
If T is the principal period of F then on each interval (nT, (n + 1)T) f has a zero (by Rolle). Here n is an integer. That shows the part about at least a countable set of zeros. Let u_n be the root in the interval above. The sequence x_n is like this
u_1, u_2, u_2, u_3, u_3, u_3, ....
so
x_(n(n+1)/2 + k) = u_(n + 1) where 1 \leq k \leq n + 1
Because u_(n + 1) is in the interval (n T, (n + 1) T) we get that x_n/n has limit 0
I think we should add to the assumptions of the problem that f is continuous, since the existence of an antiderivative of f does not imply that f is continuous, and continuity of f is necessary for the second question; in particular, to show that f(xn/n) converges to f(0). Since xn is in (nT,(n+1)T) for every positive integer n, we have T
IVT is a fundamental property of continuous functions and has many applications in real life. in numerical methods for finding roots of the functions. We then have that f (x n/n) converges to f(T). Finally, since the antiderivative off is periodic with period T, so is f, and thus f(T)= f (0).
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