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aef321

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

  • 4 months ago
  • 4 months ago

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  1. alffer1
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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

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

    • 4 months ago
  3. alffer1
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    we're not sure where to go from here

    • 4 months ago
  4. satellite73
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    looks like you are trying to compute some sort or reimann sum

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

    • 4 months ago
  6. satellite73
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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 \[\sum_{a=1}^na\]

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

    • 4 months ago
  8. primeralph
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    @satellite73 Got this?

    • 4 months ago
  9. satellite73
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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

    • 4 months ago
  10. satellite73
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    @tkhunny i think the \(4\) is being squared as well

    • 4 months ago
  11. tkhunny
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    Whoops. Make that 64, instead of 16.

    • 4 months ago
  12. satellite73
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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

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

    • 4 months ago
  14. satellite73
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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

    • 4 months ago
  15. satellite73
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    as you remember from some pre calc course about "horizontal asymptotes" that makes the limit \(-\frac{128}{6}\)

    • 4 months ago
  16. alffer1
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    oh oops duh that makes sense.

    • 4 months ago
  17. satellite73
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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

    • 4 months ago
  18. aef321
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    Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    • 4 months ago
  19. alffer1
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    @satellite73 where did -128 come from?

    • 4 months ago
  20. alffer1
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    sorry probably a dumb question but...

    • 4 months ago
  21. alffer1
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    oh waitn.....nevermind

    • 4 months ago
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