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I had a problem with solving the following;

$f(x-2)+f(x+2)=f(x)$ I have tried to make it look like $f(x+T)=f(x)$ but I failed. I mostly have difficulty in solving these type of problems. What do you suggest I should do?

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2 Answers 2

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Hint. We have that $$f(x+2)+f(x+6)=f(x+4) $$ and $$ f(x)+f(x+4)=f(x+2) .$$ By adding them, we get $$ f(x)+f(x+6)=0\implies f(x+6)=-f(x).$$
Therefore $$f(x+12)=f((x+6)+6)=-f(x+6)=f(x)$$ for any real $x$ and we may conclude that a period of the function is $12$.

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  • $\begingroup$ Thank you:) How did you get to the last part where $-f(x+6)=f(x+12)$? $\endgroup$ Aug 26, 2017 at 9:25
  • $\begingroup$ @DenizTunaYalçın Substituting $x + 6$ for $x$ in the equation $f(x + 6) = -f(x)$ yields $f(x + 12) = f(x + 6 + 6) = -f(x + 6)$. $\endgroup$ Aug 26, 2017 at 9:33
  • $\begingroup$ May I ask what is the final conclusion ? That the set of solutions is exactly the set of periodical functions $\mathbb R \to \mathbb R$ with period 12 ? $\endgroup$
    – Jean Marie
    Aug 26, 2017 at 9:44
  • $\begingroup$ @Jean Marie Yes, If $f$ is not identically zero, try to show that the period can't be a proper divisor of $12$. $\endgroup$
    – Robert Z
    Aug 26, 2017 at 9:54
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Using this, the characteristic equation of $$f(x+2)-f(x)+f(x-2)=0$$

$t^4-t^2+1=0, t=e^{(2m+1)\pi i/6}$ where $m=0,2,3,5$

$$\implies f(x)=\sum_{m=\{0,2,3,5\}}A_me^{(2m+1)\pi ix/6}$$

Now if $f(x)=f(x+T)\forall x,$ we need $$e^{(2m+1)\pi ix/6}=e^{(2m+1)\pi i(x+T)/6}$$

$\implies\dfrac{(2m+1)\pi T}6\equiv0\pmod{2\pi}$

$\iff(2m+1)T\equiv0\pmod{12}$

As $(2m+1,12)=1$ for $m=0,2,3,5,$ we need $$T\equiv0\pmod{12}$$

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