I have this question. I would like to help me with this problem please . If $f'(x)$ is a periodic function, with period $a$, prove that $f(x)$ is a periodic function, if and only if $f(a)=f(0)$. I appreciate your help.

  • $\begingroup$ It is given that $f'(x)$ is periodic. $\endgroup$ – Doug M Jul 7 '16 at 22:23
  • 2
    $\begingroup$ if $f(x)$ is $a$-periodic then $f(x)+C$ is $a$-periodic, so the integration constant isn't the problem. An example of the problem is $f'(x) = 1$, to be compared with $f'(x) = \cos(2\pi x / a)$ $\endgroup$ – reuns Jul 7 '16 at 22:25
  • $\begingroup$ and I prefer a more intuitive and explicit statement : if $g(x)$ is $a$-periodic, then there is a unique constant $\bar{g}$ such that $f(x) = \int_{x_0}^x (g(x)-\bar{g}) dx$ is $a$-periodic $\endgroup$ – reuns Jul 7 '16 at 22:29

if $f(x)$ is periodic with period $a$ then $f(x) = f(x+a)$ for all x. $f(0) = f(a)$ and $f'(x) = \lim_\limits{x\to 0} \frac {f(x+h) + f(x)}{h} = \lim_\limits{x\to 0} \frac {f(x+a + h) + f(x+a)}{h} = f'(x+a)$

To go the other direction.

It is a necessary condition that $f(0) = f(a)$ for $f(x)$ to be periodic.

but is $f'(x)$ periodic and $f(0) = f(a)$ sufficient?

$f(x+a) - f(x) = \int_x^{x+a} f'(x) dx$

Given that $f'(x)$ is periodic $\int_x^{x+a} f'(x) dx$ is constant.

$f(0) - f(a) \implies\int_0^{a} f'(x) dx = 0\\ f(x+a) = f(x)$


\begin{equation} \int_0^xf^\prime(t)\,dt=\int_a^{a+x}f^\prime(t)\,dt \end{equation}

\begin{equation} f(x)-f(0)=f(a+x)-f(a) \end{equation}


\begin{equation} f(x)=f(x+a)+f(0)-f(a) \end{equation}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.