Finding the Period of a Function

[shen07]

Hi guys,

I know that this question may be silly but could you please tell me how do i go about.

if $$\phi(2x+3)+\phi(2x+7)=10,\\\\\\\forall x\in\Re$$,find the period of $\phi(x)$.

 

[chisigma]

Hi guys,

I know that this question may be silly but could you please tell me how do i go about.

if $$\phi(2x+3)+\phi(2x+7)=10,\\\\\\\forall x\in\Re$$,find the period of $\phi(x)$.

Setting $\displaystyle 2\ x + 3 = \xi$ You have that...

$\displaystyle \phi(\xi) = 10 - \phi (\xi + 4)\ (1)$

... but is also...

$\displaystyle \phi(\xi + 4) = 10 - \phi(\xi + 8)\ (2)$

Comparing (1) and (2) You arrive to write...

$\displaystyle \phi(\xi) = \phi(\xi + 8)\ (3)$

... so that the period of $\phi(*)$ is 8...

Kind regards

$\chi$ $\sigma$

 

[johng1]

Hi,
Certainly the previous response, \(\displaystyle \phi(x+8)=\phi(x)\) for all x, is true. However, you asked for the period of the function. Usually when we talk of the period of a function f, we mean the least positive p with f(x + p) = f(x) for all x (sometimes this is called the principal period). For example, the period of sin(x) is \(\displaystyle 2\pi\). Aside: if f is a non-constant continuous function and there is a positive p with f(x + p) = f(x) for all x, then there is a smallest such p; i.e. the period of f exists.

For your specific question, if \(\displaystyle \phi\) is a non-constant continuous function, the period can be an arbitrarily small positive value!
Example:
Let n be a positive integer, \(\displaystyle a={(2n+1)\pi\over4}\) and \(\displaystyle \phi(x)=\text{sin}(ax)+5\). Easily then \(\displaystyle \phi(x)+\phi(x+4)=10\) for all x, and so this function satisfies your equation. The period of this function is \(\displaystyle {2\over 2n+1}\), which can be made as small as desired by taking n sufficiently large.