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amoodarya

  • one year ago

What strategies are typically used to solve these recurrences?

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  1. amoodarya
    • one year ago
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    \[f(0)=1\\f(1)=1\\f(n)=\sum_{i=1}^{n}f(i-1)f(n-i)\\\]

  2. amoodarya
    • one year ago
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    close form of answer is \[\frac{ (2n)! }{n!(n+1)! }\] but I would like to know how to arrive at this solution

  3. campbell_st
    • one year ago
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    what about guess and check

  4. amoodarya
    • one year ago
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    it is too complicated for check and : I do not want to use " induction "

  5. ParthKohli
    • one year ago
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    @mukushla

  6. anonymous
    • one year ago
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    The symmetry of the summation is a bit suggestive. For even \(n\), \[\large\sum_{i=1}^nf(i-1)f(n-i)=2\sum_{i=1}^\frac{n}{2}f(i-1)f(n-i)\] while for odd \(n\), \[\large\sum_{i=1}^nf(i-1)f(n-i)=2\sum_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}f(i-1)f(n-i)+f\left(\left\lfloor\frac{n}{2}\right\rfloor\right)^2\]

  7. anonymous
    • one year ago
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    Suggestive of what though, not sure yet :P

  8. anonymous
    • one year ago
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    Some rewriting of the closed form: \[\frac{(2n)!}{n!(n+1)!}=\frac{1}{n+1}\frac{(2n)!}{n!n!}=\frac{1}{n+1}\binom{2n}n\] which is the closed form for the sequence of Catalan numbers: https://en.wikipedia.org/wiki/Catalan_number

  9. ParthKohli
    • one year ago
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    Goddammit, I knew I'd seen that closed form before.

  10. anonymous
    • one year ago
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    Admittedly, we're working backwards here so it's not much help, but there's at least one proof on the wiki page.

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