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Math Help - null sequence

  1. #1
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    null sequence

    Is sequence a_n = (325*n^2)/(n^3 -2) or b_n = (n^2 +12)/(4000n^2 +144n) null? Please explain.

    I'm sorry if i've upset anyone by not using latex... it was my first time using this forum...and i had no idea of the house rules... please forgive.

    Also, i hope you all can still help me

    Thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tsal15 View Post
    Is sequence a_n = (325*n^2)/(n^3 -2) or b_n = (n^2 +12)/(4000n^2 +144n) null? Please explain.

    I'm sorry if i've upset anyone by not using latex... it was my first time using this forum...and i had no idea of the house rules... please forgive.

    Also, i hope you all can still help me

    Thanks
    "null" means the sequence converges to zero. can you tell which of these sequences, if any, converges to zero? each sequence is a ratio of polynomials. what do you know about sequences of that form?
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    if we look at them as ratio polynomials, then neither will converge....I think they will diverge from zero... am I correct?

    Thanks
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    Quote Originally Posted by tsal15 View Post
    Is sequence a_n = (325*n^2)/(n^3 -2) or b_n = (n^2 +12)/(4000n^2 +144n) null? Please explain.
    In the first one you can write,
    a_n = \frac{\frac{325}{n}}{1 - \frac{2}{n^3}} by dividing numerator and denominator by n^3.

    Now the limit of this \frac{0}{1-0} = 0

    Similarly with b_n
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    Red face

    Quote Originally Posted by ThePerfectHacker View Post
    In the first one you can write,
    a_n = \frac{\frac{325}{n}}{1 - \frac{2}{n^3}} by dividing numerator and denominator by n^3.

    Now the limit of this \frac{0}{1-0} = 0

    Similarly with b_n
    I applied a similar method to b_n = \frac{n^2 +12}{4000n^2 +144n}
    Here is my working tell me if i went wrong some where please.

    b_n = \frac{\frac{n^2 +12}{n^2}}{\frac{4000n^2}{n^2} + \frac{144n}{n^2}}

    therefore, b_n = \frac{1 + \frac{12}{n^2}}{4000 + \frac{144}{n}}

    and when n \rightarrow \infty

    b_n = \frac{1 + 0}{4000 + 0}

    Leaving,
    b_n = \frac{1}{4000}

    which is 0.00025 or 2.5 * 10^{-4}

    THANKS HEAPS!!!
    Last edited by tsal15; November 2nd 2008 at 02:18 AM.
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    Yes.
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  7. #7
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    so does that mean a_n was the null sequence or do we assume b_n is almost zero therefore we will considerate zero and thus both a_n and b_nare null sequences?


    Thanks so much Mr. PerfectHacker
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tsal15 View Post
    so does that mean a_n was the null sequence or do we assume b_n is almost zero therefore we will considerate zero and thus both a_n and b_nare null sequences?


    Thanks so much Mr. PerfectHacker
    the limit has to be zero for it to be a null sequence. so b_n isn't a null sequence

    Quote Originally Posted by tsal15 View Post
    if we look at them as ratio polynomials, then neither will converge....I think they will diverge from zero... am I correct?

    Thanks
    going back to what i said about the ratio of polynomials.

    here are some rules you should get familiar with:

    lets say you are given a ratio of polynomials and you want to find the infinite limit.

    if the highest degree of the numerator is the same of the denominator, then the limit as x \to \pm \infty to the ratio of the coefficients. (i hope you noted that when you got 1/4000)

    the the degree in the denominator is greater than that of the numerator, then the ratio goes to zero as x \to \pm \infty

    if the degree of the numerator is greater than that of the denominator, then the limit as x \to \pm \infty goes to \pm \infty, depending on the sign of the polynomials as x gets large
    Last edited by ThePerfectHacker; October 28th 2008 at 08:18 AM.
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  9. #9
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    Smile

    Quote Originally Posted by Jhevon View Post
    if the highest degree of the numerator is the same of the denominator, then the limit as x \to \pm \infty to the ratio of the coefficients. (i hope you noted that when you got 1/4000)
    This doesn't seem to make sense to me...

    Are you trying to say that because the degrees (or powers) are the same \frac{1^1}{4000^1} the limit equals the ratio/fraction as ? If I've understood it wrong, correct me please?

    Everything else makes sense. thanks heaps Jhevon
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tsal15 View Post
    This doesn't seem to make sense to me...

    Are you trying to say that because the degrees (or powers) are the same \frac{1^1}{4000^1} the limit equals the ratio/fraction as ? If I've understood it wrong, correct me please?

    Everything else makes sense. thanks heaps Jhevon
    the degrees of the original function, of course.

    i am talking about this: \frac {n^{\color{red}2} + 12}{4000n^{\color{red}2} + 144n}

    the highest power in the numerator and denominator is 2 for both. so the infinite limits, as n \to \pm \infty, goes to the ratio of the coefficients of the highest power. namely 1 in the numerator, and 4000 in the denominator. hence you get 1/4000 as your limit
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  11. #11
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    Quote Originally Posted by Jhevon View Post
    the degrees of the original function, of course.

    i am talking about this: \frac {n^{\color{red}2} + 12}{4000n^{\color{red}2} + 144n}

    the highest power in the numerator and denominator is 2 for both. so the infinite limits, as n \to \pm \infty, goes to the ratio of the coefficients of the highest power. namely 1 in the numerator, and 4000 in the denominator. hence you get 1/4000 as your limit
    YES YES YES I GET IT NOW!!!!!!!!!!

    THANK YOU, THANK YOU, THANK YOU, THANK YOU, THANK YOU
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