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Math Help - Big O, little o demonstration

  1. #1
    MHF Contributor arbolis's Avatar
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    Big O, little o demonstration

    I must demonstrate that if \{ x_n \}, \{ y_n \} and \{ \alpha _n \} are sequences of real number then the following is true:
    if x_n=o( \alpha _n) then x_n=O(\alpha _n ).

    My attempt: I must show that knowing that \lim _{n \to \infty} \frac{x_n}{\alpha _n}=0 then there exist a constant C and an integer r such that |x_n|<C|\alpha _n| for n>r.

    But to me, any constant in R that I'd take would make \lim _{n \to \infty} \frac{x_n}{\alpha _n} \neq  0. For instance if the sequence \alpha_n =x_n, then \lim _{n\to \infty} \frac{x_n}{\alpha _n}=\frac{1}{C} \neq 0. In fact C would have to be infinite for the relation to be true.

    Am I missing something or the problem asks me to show a wrong result?
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  2. #2
    Super Member girdav's Avatar
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    If you take \alpha_n=x_n, we don't have x_n=o(\alpha_n) so we can't deduce the conclusion of the result is true.
    The result is surely true because a convergent sequence is bounded.
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  3. #3
    MHF Contributor arbolis's Avatar
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    You're right, I cannot choose my example.
    To prove the afirmation, is my work valid?
    What I must prove is equivalent to prove that \lim _{n\to \infty} \left | \frac{x_n}{\alpha _n} \right | \leq C. I take C=0 and I'm done?
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  4. #4
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    Quote Originally Posted by arbolis View Post
    You're right, I cannot choose my example.
    To prove the afirmation, is my work valid?
    What I must prove is equivalent to prove that \lim _{n\to \infty} \left | \frac{x_n}{\alpha _n} \right | \leq C. I take C=0 and I'm done?
    This looks odd, why don't you work with the basic definition of limits: |x_n|<\varepsilon |\alpha _n| if n>N. Now pick a c such that |x_m|<c|\alpha _m| if 0<m<N and just take C the maximum of both.
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  5. #5
    MHF Contributor arbolis's Avatar
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    Thanks Jose. Actually I think it would be overkill.
    I want to correct my previous post into this one: The existance of a constant C and an integer r such that |x_n| <C|\alpha _n| for n>r is equivalent to write \lim _{n\to \infty} \left | \frac{x_n}{\alpha _n}  \right |<C.
    But I start with knowing that \lim _{n\to \infty} \frac{x_n}{\alpha _n} =0 and since C \geq 0 then the implication is obvious.

    I said overkill in your base because from what I understand you want to prove it for all n while I think they ask for n \to \infty, unless I'm misunderstanding once again the exercise.
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  6. #6
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    You're right, I missed the part where it said that only for sufficiently large n.
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  7. #7
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Jose27 View Post
    You're right, I missed the part where it said that only for sufficiently large n.
    They do not state "for sufficiently large n", but this is what I believe from the definitions I have for the little and big o notations. The problem is the one I wrote in the first post in the lines before "My attempt".

    By the way I didn't know the result would hold for any n, I think this is a nice result and I thank you for pointing it out.
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