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Math Help - Prove f is periodic

  1. #1
    Junior Member mathfun's Avatar
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    Prove f is periodic

    Let \[f:\mathbb{R} \to \mathbb{R}\] such that:
    \[\tfrac{{\sqrt 5  + 1}}{2}f\left( x \right) = f\left( {x + 1} \right) + f\left( {x - 1} \right)\] (1)

    a) Prove that f is periodic
    b) Prove that there are infinitely many functions such that (1) is true
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  2. #2
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    (a)

    Edit major mistake in my working so ive removed it, sorry



    (b) let f(x) be any function satisfying the condition. consider g(x)= kf(x)
    Last edited by SpringFan25; March 31st 2011 at 11:30 AM.
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  3. #3
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    This doesn't explain why such a function has to be periodic, but here is an example of one such function (and sure enough it is periodic, with period 10).

    Let f(x) = \sin(\pi x/5). Then f(x\pm1) = \sin(\pi x/5 \pm \pi/5), and f(x+1)+f(x-1) = 2\sin(\pi x/5)\cos(\pi/5) = \frac{\sqrt5+1}2f(x) since \cos(\pi/5) = \frac{\sqrt5+1}4.
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  4. #4
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    Well, I didn't make all the calculus, but:

    x\leftarrow x+1 \Rightarrow \frac{\sqrt{5}+1}{2}f(x+1)=f(x+2)+f(x)
    \frac{\sqrt{5}+1}{2}f(x+1)=\frac{\sqrt{5}+3}{2}f(x  )-\frac{\sqrt{5}+1}{2}f(x-1)
    \frac{\sqrt{5}+3}{2}f(x)-\frac{\sqrt{5}+1}{2}f(x-1)=f(x+2)+f(x) \Rightarrow f(x+2)=\frac{\sqrt{5}+1}{2}(f(x)-f(x-1))

    Now calculate f(x+3), f(x+4)... and find the period. I am pretty sure this is a way it can be solved.
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  5. #5
    Junior Member mathfun's Avatar
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    Well i have tried many combinations but with no result. With certainty the period is 10 though. If anyone made it, please post a solution
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  6. #6
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    In fact, veileen's method works nicely. Let \tau = \frac{\sqrt5+1}2. Then \tau is the golden ratio, and satisfies the quadratic equation \tau^2-\tau-1=0.

    The function f satisfies f(x+2) = \tau f(x+1) - f(x). Therefore

    \begin{aligned}f(x+3) &= \tau f(x+2) - f(x+1) \\ &= \tau \bigl(\tau f(x+1) - f(x)\bigr) - f(x+1) \\ &= (\tau^2-1)f(x+1) - \tau f(x),\end{aligned}

    \begin{aligned}f(x+4) &= \tau f(x+3) - f(x+2) \\ &= \tau \bigl((\tau^2-1)f(x+1) - \tau f(x)\bigr) - \bigl(\tau f(x+1) - f(x)\bigr)) \\ &= (\tau^3-2\tau)f(x+1) - (\tau^2-1)f(x),\end{aligned}

    \begin{aligned}f(x+5) &= \tau f(x+4) - f(x+3) \\ &= \tau \bigl((\tau^3-2\tau)f(x+1) - (\tau^2-1)f(x)\bigr) - \bigl((\tau^2-1)f(x+1) - \tau f(x)\bigr)) \\ &= (\tau^4-3\tau^2+1)f(x+1) - (\tau^3-2\tau)f(x).\end{aligned}

    Now you can check (using the quadratic equation for \tau) that \tau^4-3\tau^2+1 = 0 and \tau^3-2\tau = 1. Thus f(x+5) = -f(x) and so f(x+10) = f(x).

    Edit. Looking again at veileen's comment, I notice that my solution would be much simpler if I had used the quadratic equation \tau^2-\tau-1=0 at each stage of the calculation, rather than saving it up until the end. In fact,

    f(x+3) = (\tau^2-1)f(x+1) - \tau f(x) = \tau\bigl(f(x+1)-f(x)\bigr) (because \tau^2-1=\tau),

    \begin{aligned}f(x+4) = \tau f(x+3) - f(x+2) &= \tau^2\bigl(f(x+1)-f(x)\bigr) - \bigl(\tau f(x+1) - f(x)\bigr) \\ &= (\tau^2-\tau)f(x+1) - (\tau^2-1)f(x) = f(x+1) - \tau f(x),\end{aligned}

    f(x+5) = \tau f(x+4) - f(x+3) = \tau\bigl( f(x+1) - \tau f(x)\bigr) - \tau\bigl(f(x+1)-f(x)\bigr) = (\tau-\tau^2)f(x) = -f(x).
    Last edited by Opalg; April 1st 2011 at 04:12 AM.
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