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If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, then the period of $f(x)$ is?

I tried replacing $x$ with $x-1$ and stuff, but didn't lead me to anything...

  • 0
    $$f(x) = \sin{\left (\frac{\pi}{4} x \right)}$$ – 2017-01-08
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    @RonGordon: this is far from being the only solution. – 2017-01-08
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    @RonGordon How did you figure it out though? – 2017-01-08

3 Answers 3

4

Make $x=x-1$

$$f(x)+f(x-2)=\sqrt{2}f(x-1) \tag {1}$$

Make $x=x+1$

$$f(x+2)+f(x)=\sqrt{2}f(x+1) \tag {2}$$

Now sum $(1)$ and $(2)$:

$$2f(x)+f(x+2)+f(x-2)=\sqrt{2}(f(x+1)+f(x-1))=2f(x) $$

$$f(x+2)+f(x-2)=0$$

Now make $x=x+2$ then $f(x+4)=-f(x)$.

Make $x+4$ and get $f(x+8)=-f(x+4)=f(x)$, then the period is $8$.

  • 0
    $f(x)=-f(x+4)$ proves that $4$ is not the period (the antiperiod, maybe ?) – 2017-01-08
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    How did you conclude it is $8$? – 2017-01-08
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    This shows that $8$ is a multiple of the period. Now use e.g. the comment of Ron Gordon to show that $8$ is the smallest period. – 2017-01-08
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    @WimC: the particular solution given by RonGordon doesn't prove the smallest period. – 2017-01-08
2

Note that $\sqrt{2} = \frac{2}{\sqrt{2}}$. Hence,

$$f(x + 1) + f(x - 1) = \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}f(x)$$ $$\begin{align}f(x + 1) &= \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(f(x) - \sqrt{2}f(x - 1)\right)\\ &=\frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(-f(x - 2)\right)\\ &= \frac{1}{\sqrt{2}}\left(f(x) - f(x - 2)\right)\end{align}$$

$$\sqrt{2}f(x + 1) = f(x) - f(x - 2)$$ $$\sqrt{2}f(x) = f(x - 1) - f(x - 3)$$

Hence,

$$f(x - 1) - f(x - 3) = f(x + 1) + f(x - 1)$$ $$f(x + 1) = -f(x - 3)$$ $$f(x) = -f(x - 4)$$

Therefore,

$$f(x) = -f(x - 4) = -(-f(x - 8)) = f(x - 8)$$

thus the period is $8$.

  • 1
    Or $8$ is a multiple of the period. – 2017-01-08
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    "the" period... More like "a" period, unless you prove $8$ is the smallest period. – 2017-01-08
1

This is an ordinary linear recurrence, which you can solve using the characteristic equation

$$r+\frac1r=\sqrt 2.$$

The roots are $r=\dfrac{1\pm i}{\sqrt2}$, and the general solution is

$$f(n)=C\left(\dfrac{1+i}{\sqrt2}\right)^n+C^*\left(\dfrac{1-i}{\sqrt2}\right)^n=Ce^{in\pi/4}+C^*e^{-in\pi/4}=A\cos\frac{n\pi}4+B\sin\frac{n\pi}4.$$

The constants $C$ or $A,B$ can be obtained from some given conditions such as $f(0)$ and $f(1)$. This will only solve for integer arguments. For real arguments, you can specify the initial conditions by assigning $f(x)$ arbitrary values for $0\le x < 2$.

This proves that the period is $8$.