If $f(x+1)= -f(x-1)$, prove that the function f is a periodic function and its period is $4$

Question

I sure do know that I should arrive to the point where $f(x)= f(x+T)$. I've tried replacing $x+1=a$ but the problem seems to get more and more complicated.

Answer 1

Hint: fix any $y$, and apply your hypothesis to both $x\stackrel{\rm def}{=} y+1$ and $x'\stackrel{\rm def}{=} y+3$.

In detail: (place your mouse over the grey areas to reveal the contents)

Fix any $y\in\mathbb{R}$, and let $x\stackrel{\rm def}{=} y+1$, $x'\stackrel{\rm def}{=} y+3$.

$$ f(y) = f(x-1) \stackrel{\rm (assumption)}{=} -f(x+1) = -f(y+2) $$

while

$$ f(y+2) = f(x'-1) \stackrel{\rm (assumption)}{=} -f(x'+1) = -f(y+4)$$

so combining the two,

$$ f(y) = -f(y+2) = f(y+4) $$

and since this holds for any $y$, $4$ is a period of $f$. (I.e., $f$ is $4$-periodic. There may be other periods.)

Answer 2

Hint: Start by trying to evaluate $f(x+4)$ and using the functional equation multiple times:

Full solution:

$$\begin{align*}f(x+4) &= f(x+ 3 + 1)\\&=-f(x+3-1)\\&= -f(x+2)\\&=-f(x+1+1)\\&=-(-f(x+1-1))\\&= f(x)\end{align*}$$

Answer 3

Substituting $y:=x-1$ shows that $$f(y+2)=-f(y)=-(-f(y-2))=f(y-2),$$ holds for all $y$, or equivalently $f(y+4)=f(y)$, so 'its period' is at most $4$.

As others have noted, it can be rather ambigious to speak of the period of a function. Such a function $f$ may have period $4$, but it may also have other (shorter) periods. And for some functions $f$ it is not clear what is meant by 'its' period, for example $f=0$. So I would rather say that we have shown that $4$ is a period of $f$.