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How to formally prove that $f(n)=\Theta f(n+1)$?

It's supposed to be easy, but I still can't get it. Thank you very much.

  • 1
    Beware the abuse of notation there.2012-05-19

1 Answers 1

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This depends on the sequence $(f(n))$. For example:

  • If $f(n)=n!$ then $f(n)\ll f(n+1)$ hence $f(n)\notin\Theta(f(n+1))$.
  • If $f(n)=\frac1{n!}$ then $f(n)\gg f(n+1)$ hence $f(n)\notin\Theta(f(n+1))$.
  • If $f(n)=a^n$ with $a\ne0$ then $f(n)=\frac1af(n+1)$ hence $f(n)\in\Theta(f(n+1))$.
  • 1
    This (abuse of) notation suggests that numbers can be in Landau relations. I thin$k$ you should properly defi$n$e the two seque$n$ces behind the scenes as to not confuse anybody.2012-05-19