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Math Help - can someone prove this

  1. #1
    Junior Member miss_lolitta's Avatar
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    can someone prove this

    Hello,
    can someone prove this to me as.
    Any help would help save thier hair I have not torn out as yet.

    If
    a_n,b_n
    are sequences of real number ,n>m then:

    \sum_{k=m}^{n}a_k .b_k= a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k

    Where
    S_n
    is the partial sum of sequence
    \sum_{k=1}^{\infty}b_n

    Thanks for any help
    Last edited by CaptainBlack; April 25th 2006 at 01:08 PM. Reason: To try to correct the TeX
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by miss_lolitta
    Hello,
    can someone prove this to me as.
    Any help would help save thier hair I have not torn out as yet.

    If
    a_n,b_n
    are sequences of real number ,n>m then:

    \sum_{k=m}^{n}a_k .b_k= a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k

    Where
    S_n
    is the partial sum of sequence
    \sum_{k=1}^{\infty}b_n

    Thanks for any help
    Either my correction of the TeX has introduced an error, or this cannot
    be an identity, as the LHS is independent of a_{n+1}, while the RHS is dependent
    on a_{n+1}.

    RonL
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  3. #3
    Junior Member miss_lolitta's Avatar
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    the relation is correct..sure
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  4. #4
    Junior Member miss_lolitta's Avatar
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    Were is the proof??
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by miss_lolitta
    Were is the proof??
    There is no proof because it cant be an identity.

    RonL
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  6. #6
    Junior Member miss_lolitta's Avatar
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    How??
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by miss_lolitta
    How??
    We seek to prove that for arbitary sequences a_i,\ i=1 \dots and b_i,\ i=1 \dots ,
    and S_n the sequence of partial sums of the b_i's:

    \sum_{k=m}^{n}a_k .b_k= a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k\ \ \dots(1).

    Rearranging:

    a_{n+1}=(-\sum_{k=m}^{n}a_k .b_k+ a_{n+1}S_n-a_m S_{m-1}+\sum_{k=m}^{n}( a_k - b_{k+1})S_k\left)/S_n

    Now everything on the RHS is independent of a_{n+1}, so
    a_{n+1} is deternined by the sequence b_i,\ i=1\dots
    and a_i,\ i=1 \dots n, and so a_i,\ i=1 \dots is not an arbitary sequence,
    or if it is then (1) is not an identity.

    RonL
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