Homework Statement
A function f(x) has periodic derivative. In other words, f'(x +p) = f'(x) for some real value of p. Is f(x) necessarily periodic? Prove or give a counterexample.
I believe it is true simply because of trigonometric functions. However, I do not know how to prove it. I want to claim if the derivative is periodic than the anti-derivative is also periodic. However, I don't think that is good enough. Any hints?
Of course that isn't good enough; you are just stating the conclusion with no proof.
Hint: Call G(x) = f(x+p)-f(x). Then calculate G'(x) and see if that leads you anywhere. (Keep an open mind)
taking G'(x)= would only give f'(x+p)-f'(x) correct? How is this proof of the claim? Don't you have to prove it for all cases of the situation?
Don't try to prove it's true. It's not. Try to find a counterexample. Sure f'(x+p)-f'(x) is zero. The derivative of a function being zero doesn't show that the function is zero.
I don't understand how it could not be true. All the functions that were periodic had a derivative that was also periodic that we had taken up so far in class. I can only think that sin(x^-1) = e(x) that e'(x) would not be periodic simply because e(x) has not period but is periodic.
Try and follow LCKurtz's hint. If g(x)=f(x+p)-f(x) then g'(x)=f'(x+p)-f'(x)=0. Does that mean g(x) is zero? What else could it be?
This is where I am at. f(x+p) = f(x)
Let g(x) = f(x+p)-f(x)
g'(x) = f'(x+p)-f'(x).
g'(x) = f'(x+p)-f'(x) because the constant p'=0. Therefore, it becomes
g'(x) = f'(x+p)-f'(x). The only point that g'(x) = 0 is when p=0 and thereby, making f'(x+o)=f'(x) => that g'(x)=0. From here no idea where to go.
Ok, I'm going be blunt. The hint may not be all that obvious. Suppose f(x)=sin(x)+x. Is f(x) periodic? Is f'(x) periodic?
Thank you very much. I think I understand. F'(x) is periodic because it is cos(x)+1, but f(x) is not periodic because it does not have a phase point in which is does not repeat itself. It increases at a certain period time, but never repeats itself, correct?
Right, that's what LCKurtz's hint was supposed to suggest. You can add a function that increases at a constant rate to a periodic function and get one whose derivative is periodic while the function itself is not.
This is assuming that p is 0 right?
Thank you for your help. To show this it is as follows:
let g'(x) = f'(x+p) - f'(x)
Assume that f'(x+p) = cosx and f'(x) = 1
Therefore, we can conclude that g'(x) is periodic because it is cosx-1 but g(x) = sinx-x.Sinx-x is not periodic because it has no period and/or oscillation in which the function repeats itself.
Now why would you say that? Don't shake my confidence in you. The period of f'(x) is p=2pi, isn't it? f(x) has no period.
sin(x)+x isn't periodic because it's always increasing since its derivative is nonnegative. Show that. I think you might be getting tired.
Thanks for grabbing the ball and following up Dick. I had to go to choir practice with the wife.
Oh, I wasn't doing much else anyway. If I hadn't somebody else would have. The g'(x)=0 hint doesn't seem be working as well as I'd hoped, though. There's another thread on this same problem around. Try and talk that one through better than I did.