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anonymous

  • one year ago

\[a_{n+2}-(2^{n+1}+2^{-n}) a_{n+1}+a_n=0\]\(a_1=1, a_2=3\). Find \(a_n\).

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  1. anonymous
    • one year ago
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    What have you tried?

  2. ganeshie8
    • one year ago
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    I saw this problem before and know a solution which is somewhat complicated. I'm going to wait for other solutions :) I think this problem is composed by mukushla, he wants us to try...

  3. anonymous
    • one year ago
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    thanks gane :)

  4. Empty
    • one year ago
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    Here's my thoughts so far: The ordinary power series generating function satisfies this equation: \[f(x)-\frac{2}{x} f(\frac{x}{2})-2f(2x)-\frac{1}{x}=0\]

  5. anonymous
    • one year ago
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    @Empty you study physics? That seems like a physicist approach!

  6. Empty
    • one year ago
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    I study everything I can get my hands on lol

  7. anonymous
    • one year ago
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    good :) btw, how you got that?

  8. Empty
    • one year ago
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    Sorry that wasn't quite the link I wanted, this is the book I'm reading currently generatingfunctionology: https://www.math.upenn.edu/~wilf/DownldGF.html

  9. anonymous
    • one year ago
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    thanks for the link

  10. anonymous
    • one year ago
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    \[\begin{split} a_n &= \left(2^{n+1}+2^{-n}\right)a_{n-1}-a_{n-2} =ba_{n-1}-a_{n-2}\\ &= b\left(ba_{n-2}-a_{n-3}\right)-a_{n-2} \\ &= \left(b^2-1\right)a_{n-2}-a_{n-3} \\ &= \left(b^3-b-1\right)a_{n-3}-a_{n-4} \\ &= \left(b^{k+1}-\sum_{m=0}^{k-1}b^m\right)a_{n-k-1}-a_{n-k-2} \\ &= \left(b^{k+1}-\sum_{m=0}^{k-1}b^m\right)a_{n-k-1}-a_{n-k-2} \\ &= \left(b^{k+1}-\frac{1-b^{k}}{1-b}\right)a_{n-k-1}-a_{n-k-2} \\ \end{split}\]For our base case, we plug in \(k=n-3\): \[ a_n = \left(b^{n-2}-\frac{1-b^{n-3}}{1-b}\right)a_2+a_1 = 3\left(b^{n-2}-\frac{1-b^{n-3}}{1-b}\right)+1 \]Then substitute the \(b=2^{n+1}+2^{-n}\).

  11. anonymous
    • one year ago
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    It looks good, maybe you have some little mistakes, find \(a_3\) from your formula and once again directly from the relation and compare, see if the formula is right or not :)

  12. anonymous
    • one year ago
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    \[\begin{split} a_n &= \left(2^{n+1}+2^{-n}\right)a_{n-1}-a_{n-2} =ba_{n-1}-a_{n-2}\\ &= b\left(ba_{n-2}-a_{n-3}\right)-a_{n-2} \\ &= \left(b^2-1\right)a_{n-2}-a_{n-3} \\ &= \left(b^3-b-1\right)a_{n-3}-a_{n-4} \\ &= \left(b^{k}-\sum_{m=0}^{k-2}b^m\right)a_{n-k}-a_{n-k-1} \\ &= \left(b^{k}-\sum_{m=0}^{k-2}b^m\right)a_{n-k}-a_{n-k-1} \\ &= \left(b^{k}-\frac{1-b^{k-1}}{1-b}\right)a_{n-k}-a_{n-k-1} \\ \end{split}\]For our base case, we plug in \(k=n-2\): \[ a_n = \left(b^{n-2}-\frac{1-b^{n-3}}{1-b}\right)a_2+a_1 = 3\left(b^{n-2}-\frac{1-b^{n-3}}{1-b}\right)+1 \]Then substitute the \(b=2^{n+1}+2^{-n}\).

  13. anonymous
    • one year ago
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    There is a special case at \(n=3\) because for that case, we have the coefficient \(b\) rather than \(b-\frac{1}{1-b}\)

  14. anonymous
    • one year ago
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    In fact, all the \(n<4\) cases probably will not hold, because that geometric sum assumes higher values for \(n\).

  15. anonymous
    • one year ago
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    thanks

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