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Math Help - asymptotic sequences

  1. #1
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    asymptotic sequences

    which of the following sequences is asymptotic to the sequence

    [(e^-n)+sinn-n^(1/3)]/[2cosn-n^(5/6)]?

    a) [-e^(-n)]/n^(5/6)]

    b) e^-n

    c) sinn/2cosn

    d) 1/n^(1/2)

    I was thinking that the part of the given sequence that "has more weight" is e^-n so..well,...no..uhmm...WHY????
    Last edited by CaptainBlack; October 6th 2006 at 03:19 PM.
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  2. #2
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    Quote Originally Posted by Aglaia View Post
    which of the following sequences is asymptotic to the sequence [(e^-n)+sinn-n^(1/3)]/[2cosn-n^(5/6)]? a) [-e^(-n)]/n^(5/6)] ; b) e^-n; c) sinn/2cosn; d) 1/n^(1/2) i was thinking that the part of the given sequence that "has more weight" is e^-n so..well,...no..uhmm...WHY????
    For large n:
    [e^(-n) + sin(n) - n^(1/3)]/[2cos(n) - n^(5/6)] -> 1/n^(1/2).

    Why?
    e^(-n) -> 0 and sin(n) and cos(n) become indeterminate, but vary between -1 and 1. However both n^(1/3) and n^(5/6) get very large for large n, so these will be the dominant terms. Thus

    [e^(-n) + sin(n) - n^(1/3)]/[2cos(n) - n^(5/6)] -> -n^(1/3) / -n^(5/6) = 1/n^(1/2)

    -Dan
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Aglaia View Post
    which of the following sequences is asymptotic to the sequence

    [(e^-n)+sinn-n^(1/3)]/[2cosn-n^(5/6)]?
    multiply top and bottom by n^{-1/3} to get:

    [n^{-1/3} e^{-n}+n^{-1/3}sin(n)-1]/[2 n^{-1/3} cos(n) - n^{1/2}]

    which behaves like 1/[n^{1/2}] for large n (because all the other terms at the top
    and on the bottom go to 0 as n becomes large), which is answer d).

    a) [-e^(-n)]/n^(5/6)]

    b) e^-n

    c) sinn/2cosn

    d) 1/n^(1/2)

    I was thinking that the part of the given sequence that "has more weight" is e^-n so..well,...no..uhmm...WHY????
    RonL
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