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$f(x+1/2)+f(x-1/2)= f(x)$. Then the period of $f(x)$ is:

a)$1$

b)$2$

c)$3$

d)$4$?

Attempt:

I substituted $x= x \pm1/2$ but the equations I got didn't help at all.

How do I go about solving such a question? I am just looking for a hint and not the entire solution.

Archer
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    Small hint: take the functional equation $f(x)=f(x-1/2)+f(x+1/2)$, and expand one of the terms using the same equation. – Mike Earnest Apr 18 '18 at 17:09

6 Answers6

4

We may conclude that $f(x)=f(x+3)$ holds for all $x$.

Following @Mike Earnest's suggestion, \begin{align} f(x)&=f(x+1/2)+f(x-1/2),\\ f(x-1/2)&=f(x)+f(x-1). \end{align} Add up these two equations, and you will have $$ f(x+1/2)+f(x-1)=0, $$ or, due to the arbitrariness of $x$, $$ f(x+3/2)+f(x)=0. $$ Now let $x\to x+3/2$, and $$ f(x+3)+f(x+3/2)=0. $$ The difference of the last two equations gives $$ f(x)=f(x+3). $$

Of course, as @Przemysław Scherwentke has mentioned, this might not mean that $3$ is the period of $f$, as we do not know if this $3$ is the smallest non-negative value $T$ that ensures $f(x)=f(x+T)$. All in all, this is what we can get from the given relation.

hypernova
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4

Let $L$ be the operator on functions on $\mathbb{R}$ such that $(Lg)(x) = g(x+1/2)$. In terms of $L$, the equation at hand $f(x+1/2) + f(x-1/2) = f(x)$ is equivalent to $(L^2-L+1)f = 0$.

Notice the factorization of polynomial $$(t^6-1) = (t^3-1)(t^3+1) = (t^3-1)(t+1)(t^2-t+1)$$ Any solution of $(L^2-L+1)f = 0$ automatically satisfy $(L^6-1)f = 0$ or equivalently $f(x+3) = f(x)$. This means $3$ is a period (through not necessary the smallest period) of $f$.

For non-trivial examples, take $f(x) = \sin\left(\frac{2\pi}{3}(6k+1 )x\right)$ where $k \in \mathbb{Z}$, the smallest period of $f(x)$ is $\frac{3}{|6k+1|}$ and you can verify they are solutions to the equation at hand.

achille hui
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$f(x+1) + f(x) = f(x + \frac 12)= f(x) - f(x - \frac 12)$

$f(x+1) = -f(x - \frac 12)$

So $f(x - \frac 12) = -f(x - \frac 12 - 1\frac 12) = -f(x-2)$

So $f(x+1) = -f(x-\frac 12) = f(x-2)$.

So period is at most $3$.

Not sure off the top of my head how to show it doesn't necessarily need to be smaller.

fleablood
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The period may not exist. For example a constant function $f\equiv0$ is an example with no (minimal) period.

Przemysław Scherwentke
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    While there is no shortest or "fundamental" period of a constant function, most authors would include constant functions in the collection of functions have specified period $L$ (rather than say it has no period). – hardmath Apr 18 '18 at 18:12
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    @hardmath Yes, but in this notation every positive constant is a period, so any answer to the question in the present formulation may be, in a special case, correct. The question probably should be reformulated: "for every $f$...". – Przemysław Scherwentke Apr 18 '18 at 18:19
  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Henrik supports the community Apr 18 '18 at 20:28
  • @Henrik In the present (and especially previous) formulation of the question it does provide an answer to it. – Przemysław Scherwentke Apr 18 '18 at 20:44
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hint: $$ \eqalign{ & f(x - 1/2) + f(x + 1/2) = f(x) \cr & x = y/2\quad \Rightarrow \cr & f\left( {\left( {y - 1} \right)/2} \right) + f\left( {\left( {y + 1} \right)/2} \right) = f\left( {y/2} \right) \cr & f\left( {y/2} \right) = g\left( y \right)\quad \Rightarrow \cr & g\left( {y + 1} \right) = g\left( y \right) - g\left( {y - 1} \right) \cr} $$

which is a Finite Difference equation of 2nd degree, whose solution depends on two starting conditions ...

G Cab
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$f(x+1/2)=f(x+1)+f(x)$; $f(x-1/2)=f(x)+f(x-1)$; substituting in the original equation: $2f(x)+f(x+1)+f(x-1)=f(x)$, $f(x)=-f(x-1)-f(x+1)$.
Using the last identity, $f(x+1)=-f(x)-f(x+2)$ and substituting in the last equation $f(x+1)$, we obtain $f(x)=-f(x-1)+f(x)+f(x+2)$ or $f(x-1)=f(x+2)$ and the period is $3$

Vasili
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