-2

I want to find the periodicity of the following function

$f(x) = \sqrt{\sin(x)}+\sqrt{\cos(x)}$

my attempt: I've posed $f(x)=f(x+T)$.So $$ \begin{align} f(x+T) &= \sqrt{\sin(x+T)}+ \sqrt{\cos(x+T)}\\ &= \sqrt{\sin(x)\cos(T)+\cos(t)\sin(T)}+\sqrt{\cos(x)\cos(T)-\sin(t)\sin(T)} \tag{1} \end{align} $$ if $T=2\pi$, then Eq.(1) becomes $$ \begin{align} f(x+T) &= \sqrt{\sin(x+T)}+ \sqrt{\cos(x+T)}\\ &= \sqrt{\sin(x)}+\sqrt{\cos(x)} \end{align} $$ this shows that the function is a periodic and $T=2\pi$,

user62498
  • 3,556
  • 5
    It should be easy to test any particular guess you had in mind, right? But...did you forget to add absolute values? As it stands, your function is ill defined for many values of $x$. – lulu Oct 23 '21 at 18:14
  • Analyze the domain of the function, and it will be clear that the $T=2\pi$. – Zaragosa Oct 23 '21 at 18:34
  • It is $\pi/2$, assuming they are in absolute values. I have proof that would only be tangentially (pun intended) accepted by a professor if this is an assignment. Otherwise, domain suggests $2\pi$. – ck1987pd Oct 23 '21 at 18:43
  • But what I know for certainty is that it is not $2/\pi$. If $f(x)$ is periodic with $a$, so is $g(f(x))$, so assuming $\sqrt{.}$ is just any $g(.)$, it can either be $2\pi$ or $\pi$, not anything else. – ck1987pd Oct 23 '21 at 18:45
  • ZaragosaوI edited the answer – user62498 Oct 23 '21 at 19:42
  • This is not a proof of periodicity. You just showed that $2\pi$ is a period. It is not necessarily the period. Besides, since $\sqrt{\sin(x)}$ is obviously periodic with $2\pi$ and so is $\sqrt{\cos(x)}$, it follows that $\sqrt{\sin(x)}+\sqrt{\cos(x)}$ also has a period of $2\pi$. – ck1987pd Oct 23 '21 at 19:51
  • You are still conjecturing that the period should be 2pi. Why do you conjecture that it is 2pi and why is it not pi? – ck1987pd Oct 23 '21 at 22:48

1 Answers1

1

A semi-decent way might be this. The general information about periodicity is given here.

We know that $\sin(x)$ is periodic with $2\pi$. If $f(x)$ is periodic with $a$, so is $g(f(x))$, hence $\sqrt{\sin(x)}$ is also periodic with $2\pi$. For the same reason, $\sqrt{\cos(x)}$ is also periodic with $2\pi$.

When two periodic functions are summed/multiplied the least common multiple of the individual functions is the new period. Therefore, $2\pi$ is a period of the $\sqrt{\sin(x)}+\sqrt{\cos(x)}$. But we do not yet know if it is the fundamental period.

Since it is the summation of two functions of the same period, $\pi$ might also be a period. (I don't have a good proof of this statement, this is why this is semi-decent, but you can think of addition of shifted impulse trains or frequency multiplier).

$$ \sqrt{\sin(x)}+\sqrt{\cos(x)}\stackrel{?}{=}\sqrt{\sin(x+\pi)}+\sqrt{\cos(x+\pi)}\\ \sqrt{\sin(x)}+\sqrt{\cos(x)}\stackrel{?}{=}\sqrt{-\sin(x)}+\sqrt{\cos(x)}, $$ such that $$ \sqrt{\sin(x)}+\sqrt{\cos(x)}\neq\sqrt{\sin(x+\pi)}+\sqrt{\cos(x+\pi)}, $$ hence $\pi$ is not a period. This leaves $2\pi$ as the fundamental period.

ck1987pd
  • 1,104