# Thread: Periodic functions - prove that p is a rational number?

1. ## Periodic functions - prove that p is a rational number?

If the function given by $\displaystyle f(x) = \sin x + \cos px$ is periodic, prove that p is a rational number.

2. Originally Posted by fardeen_gen
If the function given by $\displaystyle f(x) = \sin x + \cos px$ is periodic, prove that p is a rational number.
suppose $\displaystyle T \neq 0$ is a period of $\displaystyle f.$ then $\displaystyle \sin(x+T)+\cos(px+pT)=\sin x + \cos px.$ call this (1). put $\displaystyle x=0$ in (1) to get: $\displaystyle \sin T + \cos pT = 1.$ call this one (2). now differentiate (1) twice to get:

$\displaystyle \sin(x+T)+p^2 \cos(px+pT)=\sin x + p^2\cos px,$ and put $\displaystyle x=0$ to get: $\displaystyle \sin T + p^2 \cos pT=p^2.$ call this (3). now (2) and (3) give us: $\displaystyle (p^2-1)(\cos pT - 1)=0.$ therefore either $\displaystyle p^2=1$ and so

$\displaystyle p=\pm 1 \in \mathbb{Q},$ or $\displaystyle \cos pT = 1$ and thus by (2): $\displaystyle \sin T=0.$ hence $\displaystyle pT=2k \pi$ and $\displaystyle T=\ell \pi,$ for some integers $\displaystyle k$ and $\displaystyle \ell \neq 0.$ thus: $\displaystyle p=\frac{2k}{\ell} \in \mathbb{Q}.$

3. Period of $\displaystyle \sin x=2\pi$

Period of $\displaystyle \cos px=\frac{2\pi}{p}$

Period of $\displaystyle f(x)=L.C.M\left\{2\pi,\frac{2\pi}{p}\right\}$=$\displaystyle 2\pi L.C.M\left\{1,\frac{1}{p}\right\}$.

If f(x) is to be periodic then the L.C.M must exist.This L.C.M will exist only if p is a rational number and $\displaystyle 2\pi L.C.M\left\{1,\frac{1}{p}\right\}$also may not be the fundamental period.

4. Originally Posted by pankaj
If f(x) is to be periodic then the L.C.M must exist.
How do you know that the period of the sum must be a period of both?
Assuming this, of course, the problem is easy; but you should supply proofs of your statements!