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Use iteration to guess an explicit formula

  1. Oct 11, 2016 #1
    1. The problem statement, all variables and given/known data

    Use iteration to guess an explicit formula for u_k = u_{k−2} * u_{k−1}, for all integers k ≥ 2, u_0 = u_1 = 2 and prove it .

    2. Relevant equations

    Hint: Express the answer using the Fibonacci sequence.

    3. The attempt at a solution

    u_k = u_{k−2} * u_{k−1} and u_0 = u_1 = 2, so

    u_2 = 2^2

    u_3 = 2^3

    u_4 = 2^5

    u_5 = 2^8

    u_6 = 2^13

    u_7 = 2^21

    Then (I think) in general we have ,

    2^{F_k} = 2^{F_k-1 + F_k-2}

    Not finished yet... But does it make sense so far?

    BTW, I tried putting $ signs around expressions, but mathjax(?) doesn't seem to render in preview.
     
  2. jcsd
  3. Oct 11, 2016 #2

    haruspex

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    Yes.
    That was dollar signs in pairs, right? $$latex$$. Not sure if mathjax is same as LaTeX.
     
  4. Oct 11, 2016 #3
    Oh, I never knew you had to put the dollar signs in pairs. Thanks.


    $$2^{F_k} = 2^{F_{k-1} + F_{k-2}}$$ then $$2^{F_n} = 2^ {\frac{\frac {1 + \sqrt 5}{2} - \frac {1 + \sqrt 5}{2}}{\sqrt 5}}$$ which implies $$\log_22^{F_n} = \log_22^ {\frac{\frac {1 + \sqrt 5}{2} - \frac {1 + \sqrt 5}{2}}{\sqrt 5}}$$ meaning $$F_n = {\frac{\frac {1 + \sqrt 5}{2} - \frac {1 + \sqrt 5}{2}}{\sqrt 5}}$$

    Can I claim this to be the solution to the given recurrence relation?
     
  5. Oct 11, 2016 #4

    Charles Link

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    Suggest you google Fibonacci series. I think you might have already solved this one as much as you can in post #1. (You still need to prove that your guess is correct.)
     
    Last edited: Oct 11, 2016
  6. Oct 11, 2016 #5
    I was unsure. Usually when we find an explicit formula for a recurrence, we get rid of indices in the recurrence relation so that we do at most one substitution into the formula. But if you say I am done - I am happy to take a break :) As for the proof, I think induction should take care of it. Thanks.
     
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