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Thread: 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 $\displaystyle \{ x_n \}$, $\displaystyle \{ y_n \}$ and $\displaystyle \{ \alpha _n \}$ are sequences of real number then the following is true:
    if $\displaystyle x_n=o( \alpha _n)$ then $\displaystyle x_n=O(\alpha _n )$.

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

    But to me, any constant in R that I'd take would make $\displaystyle \lim _{n \to \infty} \frac{x_n}{\alpha _n} \neq 0$. For instance if the sequence $\displaystyle \alpha_n =x_n$, then $\displaystyle \lim _{n\to \infty} \frac{x_n}{\alpha _n}=\frac{1}{C} \neq 0$. 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 $\displaystyle \alpha_n=x_n$, we don't have $\displaystyle x_n=o(\alpha_n)$ 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 $\displaystyle \lim _{n\to \infty} \left | \frac{x_n}{\alpha _n} \right | \leq $C. I take $\displaystyle C=0$ 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 $\displaystyle \lim _{n\to \infty} \left | \frac{x_n}{\alpha _n} \right | \leq $C. I take $\displaystyle C=0$ and I'm done?
    This looks odd, why don't you work with the basic definition of limits: $\displaystyle |x_n|<\varepsilon |\alpha _n|$ if $\displaystyle n>N$. Now pick a c such that $\displaystyle |x_m|<c|\alpha _m|$ if $\displaystyle 0<m<N$ 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 $\displaystyle |x_n| <C|\alpha _n|$ for n>r is equivalent to write $\displaystyle \lim _{n\to \infty} \left | \frac{x_n}{\alpha _n} \right |<C$.
    But I start with knowing that $\displaystyle \lim _{n\to \infty} \frac{x_n}{\alpha _n} =0$ and since $\displaystyle C \geq 0$ 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 $\displaystyle n \to \infty$, 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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