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Math Help - sum, difference and product of two null-sequence (need some help)

  1. #1
    Senior Member yeKciM's Avatar
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    Smile [SOLVED] sum, difference and product of two null-sequence (need some help)

    It's like this.
    I didn't have any problems to prove some of definitions of sequences like :

    lim_{n->+\infty}{(ax_n+by_n)}=ax+by

    Let
    lim_{n->+\infty}{x_n}=x
    and
    lim_{n->+\infty}{y_n}=y
    for any \xi>0. Starting from some index n_1 every member of sequence (x_n) are in \xi-region of point x. Same, from some index n_2 every member of sequence (y_n) are in \xi-region of point y.

    If we put n_0=max {\{n_1  ,n_2 \}} then for every n >= n_0 these inequality are true :

    |x_n-x|<\xi and |y_n-y|<\xi

    so from inequality of triangle for n >= n_0 we have:

    |ax_n+by_n-(ax+by) |=|ax_n-ax+by_n-by|\le |a||x_n-x|+|b||y_n-y| \le (|a|+|b| )\xi


    Because (|a|+|b|) is fix real number and \xi any small number so it's (|a|+|b| )\xi any small number so there we have that is:

    lim_{n->+\infty}{(ax_n+by_n)}=ax+by

    true.

    This and some another I got it OK

    But I can't, or don't see how to prove (or show) that sum of two null-sequences are again null-sequence (and difference of two sequences) becose it's to obviously. And for the product of two null-sequences is null-sequence and how we can relax conditions for the product of two null-sequences?

    any help will be most appreciated
    Thanks very much
    Last edited by yeKciM; July 20th 2010 at 11:15 PM.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by yeKciM View Post
    It's like this.
    I didn't have any problems to prove some of definitions of sequences like :

    lim_{n->+\infty}{(ax_n+by_n)}=ax+by

    Let
    lim_{n->+\infty}{x_n}=x
    and
    lim_{n->+\infty}{y_n}=y
    for any \xi>0. Starting from some index n_1 every member of sequence (x_n) are in \xi-region of point x. Same, from some index n_2 every member of sequence (y_n) are in \xi-region of point y.

    If we put n_0=max {\{n_1  ,n_2 \}} then for every n >= n_0 these inequality are true :

    |x_n-x|<\xi and |y_n-y|<\xi

    so from inequality of triangle for n >= n_0 we have:

    |ax_n+by_n-(ax+by) |=|ax_n-ax+by_n-by|\le |a||x_n-x|+|b||y_n-y| \le (|a|+|b| )\xi


    Because (|a|+|b|) is fix real number and \xi any small number so it's (|a|+|b| )\xi any small number so there we have that is:

    lim_{n->+\infty}{(ax_n+by_n)}=ax+by

    true.

    This and some another I got it OK

    But I can't, or don't see how to prove (or show) that sum of two null-sequences are again null-sequence (and difference of two sequences) becose it's to obviously.
    Well, that's funny: you just wrote that you have no trouble believing (even proving) that

    \lim_{n\to\infty}(ax_n+by_n)=ax+by

    is true, provided x_n\to x, y_n\to y for n\to \infty.

    Now, if (x_n), (y_n) are null-sequences, then what you know is that any linear combination z_n := ax_n+by_n is, again a null sequence, because in that case you have z_n\to z=a\cdot 0+b\cdot 0=0.

    If you know that from x_n\to x and y_n\to y it follows that x_n y_n\to x y, then it follows again, that the product of two null-sequences (which jus means that x=0 and y=0), is again a null-sequence z_n := x_n\cdot y_n\to 0\cdot 0=0.

    And for the product of two null-sequences is null-sequence and how we can relax conditions for the product of two null-sequences?
    As to relaxing the restrictions on the product of null-sequences: if you multiply a null-sequence x_n\to 0 with a bounded sequence y_n (which means that there exists a constant M such that for all n |y_n|\leq M), you get that the product z_n := x_n \cdot y_n is also a null- sequence.
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  3. #3
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456

    Talking

    Thanks very very much !!!

    It seems to me that's not enough... that's why i didn't post any my attempts of proofing thath one

    It just seems to simple i don't like obviously proofs

    THANKS!!!!
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