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Proof that lim sup cos(n) is 1

  1. Jul 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that the lim sup cos(n)=1

    2. Relevant equations

    given any [itex]α\in\mathbb{R}-\mathbb{Q} , \exists P_n,Q_n\in\mathbb{N}.\ni.\frac{P_n}{Q_n} \rightarrow {a} [/itex] and by definition, [itex] \frac{1}{Q_n} \rightarrow 0[/itex]

    Also, [itex] cos^{2}(x)=\frac{1+cos(2x)}{2}[/itex]

    My attempt

    I first consider [itex]lim\:sup\:cos^{2}(n) [/itex]
    We know that [itex]\exists P_n,Q_n\in\mathbb{N}.\ni.\frac{P_n}{Q_n} \rightarrow {\pi}[/itex] which means [itex]\forall \epsilon>0 \: \exists{N} \: .\ni. \forall{n}>N, \: |\frac{P_n}{Q_n}-\pi|<\epsilon[/itex]
    pick [itex]N.\ni.\:\epsilon<\frac{1}{Q_{n}^2}[/itex]
    So it follows that [itex]|\frac{P_n}{Q_n}-\pi|<\frac{1}{Q_{n}^2}[/itex]
    and therefore [itex]|P_n-Q_n\pi|<\frac{1}{Q_n} [/itex], am I allowed to make assume by continuity that [itex]|cos^{2}(P_n)-cos^{2}(Q_n\pi)|=|cos^{2}(P_n)-1|<ε [/itex] and since [itex] P_k = a_{n_{k}} [/itex] where [itex]a_n=n[/itex]?
     
  2. jcsd
  3. Jul 3, 2012 #2

    HallsofIvy

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    Why would you do all of that? You know that [itex]cos(x)\le 1[/itex] for all x and that cos(0)= 1. Therefore the maximum value of cos(x) is 1 and so is "lim sup".

    (Strictly speaking "lim sup" is defined for sets, not functions, but I assume that by "lim sup cos(x)" you mean "lim sup {y| y= cos(x) for some x}".)
     
  4. Jul 3, 2012 #3
    It's not for a function.... but the sequence cos(n)
    where n=1,2,3... and so on
    obviously I know that cosine is max at 1, but i have to prove that the lim sup is 1 over natural numbers
     
  5. Jul 3, 2012 #4
    to clarify, I need to show that
    [itex] \stackrel{limit}{N \rightarrow{\infty}}\:sup\left\{cos(n):n\in\mathbb{N} \wedge n>N\right\}=1[/itex]
     
  6. Jul 3, 2012 #5
    also.... having a maximum value of 1 at 0 is not the same as having the lim sup being 1...
    max value of 1/n is 1 (n=1)... but the lim sup {1/n:n is a natural number and n>N}=0
     
  7. Jul 3, 2012 #6

    HallsofIvy

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    Yes, now that I see that it is the limit of hte sequence, not the function, you are correct.
     
  8. Jul 3, 2012 #7
    but as far as my proof, is it a correct assumption?
     
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