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Thread: Sequence Proof (Hard)

  1. #1
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    Sequence Proof (Hard)

    Hi! I need some help with this question:



    (a)
    Statement reads: For every $\displaystyle \epsilon$ there exists a point in the sequence $\displaystyle N \in N$ after which all terms lie between the lines $\displaystyle y= L-\epsilon$ and $\displaystyle y= L+ \epsilon$.

    Now if we interchange the quantifiers it reads: There exists $\displaystyle N \in N $for every $\displaystyle \epsilon > 0$ such that for every point greater than or equal to N, all terms are between the lines $\displaystyle y= L-\epsilon$ and $\displaystyle y= L+ \epsilon$.

    (sigh), it's the same thing! What's the difference if we interchange the two quantifiers? And how do we go about proving it? I tried everything, even drew a diagram. But man it’s hard...

    So any help here is appreciated.
    Last edited by Roam; Oct 5th 2009 at 12:35 AM.
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  2. #2
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    Hello Roam
    Quote Originally Posted by Roam View Post
    Hi! I need some help with this question:



    (a)
    Statement reads: For every $\displaystyle \epsilon$ there exists a point in the sequence $\displaystyle N \in N$ after which all terms lie between the lines $\displaystyle y= L-\epsilon$ and $\displaystyle y= L+ \epsilon$.

    Now if we interchange the quantifiers it reads: There exists $\displaystyle N \in N $for every $\displaystyle \epsilon > 0$ such that for every point greater than or equal to N, all terms are between the lines $\displaystyle y= L-\epsilon$ and $\displaystyle y= L+ \epsilon$.

    (sigh), it's the same thing! What's the difference if we interchange the two quantifiers? And how do we go about proving it? I tried everything, even drew a diagram. But man itís hard...

    So any help here is appreciated.
    No, it isn't the same thing at all. In the first statement, we choose $\displaystyle \epsilon$ first, and then find the $\displaystyle N$ for which $\displaystyle |a_n -L|<\epsilon$ whenever $\displaystyle n\ge N$. In other words, the value of $\displaystyle N$ depends upon the value of $\displaystyle \epsilon$.

    In the second statement, there is a fixed value of $\displaystyle N$, such that whenever $\displaystyle n\ge N, |a_n-L|$ can be made arbitrarily small.

    Grandad
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    Thanks Grandad, I see the difference now.

    But I still don't know what method to use in order to prove the second statement. Could you give me some clues?
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    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Roam View Post
    Thanks Grandad, I see the difference now.

    But I still don't know what method to use in order to prove the second statement. Could you give me some clues?
    Notice: $\displaystyle |a_{n}|=|(a_{n}-a)+a|\leq\\|a_{n}-a|+|a|$. If we subtract $\displaystyle |a|$ form both sides, we have that, $\displaystyle |a_{n}|-|a|\leq\\|a_{n}-a|$. Thus, $\displaystyle |a_{n}-a|\geq\\||a_{n}|-|a||$.

    Now for the proof...

    Pick and arbitrary $\displaystyle \epsilon>0$. Since $\displaystyle (a_{n})$ converges to some limit a, there exists a natrual number $\displaystyle N$ such that, whenever $\displaystyle n\geq\\N$ implies $\displaystyle |a_{n}-a|<\epsilon$. However, $\displaystyle ||a_{n}|-|a||\leq\\|a_{n}-a|<\epsilon$ also. Therefore, $\displaystyle ||a_{n}|-|a||<\epsilon$.
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  5. #5
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    Quote Originally Posted by Danneedshelp View Post
    Notice: $\displaystyle |a_{n}|=|(a_{n}-a)+a|\leq\\|a_{n}-a|+|a|$. If we subtract $\displaystyle |a|$ form both sides, we have that, $\displaystyle |a_{n}|-|a|\leq\\|a_{n}-a|$. Thus, $\displaystyle \color{red}|a_{n}-a|\geq\\||a_{n}|-|a||$.
    You have skipped a vital step there
    See this proof.
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  6. #6
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Plato View Post
    You have skipped a vital step there
    See this proof.
    Thanks,

    I just skipped to the result needed for the proof, but thanks for pointing that out.
    Last edited by Danneedshelp; Oct 5th 2009 at 07:58 PM.
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  7. #7
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    Quote Originally Posted by Danneedshelp View Post
    Notice: $\displaystyle |a_{n}|=|(a_{n}-a)+a|\leq\\|a_{n}-a|+|a|$. If we subtract $\displaystyle |a|$ form both sides, we have that, $\displaystyle |a_{n}|-|a|\leq\\|a_{n}-a|$. Thus, $\displaystyle |a_{n}-a|\geq\\||a_{n}|-|a||$.

    Now for the proof...

    Pick and arbitrary $\displaystyle \epsilon>0$. Since $\displaystyle (a_{n})$ converges to some limit a, there exists a natrual number $\displaystyle N$ such that, whenever $\displaystyle n\geq\\N$ implies $\displaystyle |a_{n}-a|<\epsilon$. However, $\displaystyle ||a_{n}|-|a||\leq\\|a_{n}-a|<\epsilon$ also. Therefore, $\displaystyle ||a_{n}|-|a||<\epsilon$.
    Hi!

    I was thinking of using a proof by contradiction. I'm not sure how to do it but something like:

    Suppose $\displaystyle a_n \neq L$ for any $\displaystyle n \geq N$. And then somehow try to prove that this assumption violates the condition:

    $\displaystyle
    \exists N \in \mathbb{N} \forall \epsilon > 0 \forall n \geq N |a_n -L|< \epsilon
    $

    Is this a good strategy?
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