amoodarya
  • amoodarya
What strategies are typically used to solve these recurrences?
Mathematics
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SOLVED
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katieb
  • katieb
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amoodarya
  • amoodarya
\[f(0)=1\\f(1)=1\\f(n)=\sum_{i=1}^{n}f(i-1)f(n-i)\\\]
amoodarya
  • amoodarya
close form of answer is \[\frac{ (2n)! }{n!(n+1)! }\] but I would like to know how to arrive at this solution
campbell_st
  • campbell_st
what about guess and check

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amoodarya
  • amoodarya
it is too complicated for check and : I do not want to use " induction "
ParthKohli
  • ParthKohli
@mukushla
anonymous
  • anonymous
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 \(n\), \[\large\sum_{i=1}^nf(i-1)f(n-i)=2\sum_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}f(i-1)f(n-i)+f\left(\left\lfloor\frac{n}{2}\right\rfloor\right)^2\]
anonymous
  • anonymous
Suggestive of what though, not sure yet :P
anonymous
  • anonymous
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
ParthKohli
  • ParthKohli
Goddammit, I knew I'd seen that closed form before.
anonymous
  • anonymous
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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