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aef321

  • 2 years ago

how do i find the limit as an goes to infinity of the following? (the equation's in the comments)

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  1. alffer1
    • 2 years ago
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    \[\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2} -4(\frac{ 4a }{ n }))\] This is the equation in question

  2. alffer1
    • 2 years ago
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    \[\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2}) - \frac{ 4 }{ n }\sum_{a=1}^{n}(4(\frac{ 4a }{ n}))\]

  3. alffer1
    • 2 years ago
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    we're not sure where to go from here

  4. anonymous
    • 2 years ago
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    looks like you are trying to compute some sort or reimann sum

  5. alffer1
    • 2 years ago
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    exactly. a right riemann sum of -x^2 -4x to be exact.

  6. anonymous
    • 2 years ago
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    don't let the \(n\) through you off in the summation you are summing over \(a\) (never seen an \(a\) used like that before, but no matter) pull it right out front of the sum then use the formula for \[\sum_{a=1}^na^2\] and also \[\sum_{a=1}^na\]

  7. tkhunny
    • 2 years ago
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    After a little algebra. \(-\dfrac{16}{n^{2}}\left[\dfrac{1}{n}\sum\limits_{a=1}^{n}a^{2} + \sum\limits_{a=1}^{n}a\right]\)

  8. primeralph
    • 2 years ago
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    @satellite73 Got this?

  9. anonymous
    • 2 years ago
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    \[\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2})\] \[=-\frac{64}{n^3}\sum_{a=1}^na^2\] for example @primeralph maybe, i am tired and all this latex is wearing me out

  10. anonymous
    • 2 years ago
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    @tkhunny i think the \(4\) is being squared as well

  11. tkhunny
    • 2 years ago
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    Whoops. Make that 64, instead of 16.

  12. anonymous
    • 2 years ago
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    not really sure the sum is correct, but assuming it is for the moment \[-\frac{64}{n^3}\sum_{k=1}^nk^2=-\frac{64}{n^3}\frac{n(n+1)(2n+1)}{6}\] for the first one

  13. alffer1
    • 2 years ago
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    Wait but how does this help us get the limit as n -> infinity?

  14. anonymous
    • 2 years ago
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    taking the limit as \(n\to \infty\) is very easy because the numerator is a polynomial of degree 3 will leading coefficient -128 and the denominator is a polynomial of degree 3 with leading coefficent 6

  15. anonymous
    • 2 years ago
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    as you remember from some pre calc course about "horizontal asymptotes" that makes the limit \(-\frac{128}{6}\)

  16. alffer1
    • 2 years ago
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    oh oops duh that makes sense.

  17. anonymous
    • 2 years ago
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    easy if you think about it that way instead of the way they do it in the text by dividing top and bottom by \(n^3\) and all that other nonsense

  18. aef321
    • 2 years ago
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    Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

  19. alffer1
    • 2 years ago
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    @satellite73 where did -128 come from?

  20. alffer1
    • 2 years ago
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    sorry probably a dumb question but...

  21. alffer1
    • 2 years ago
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    oh waitn.....nevermind

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