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laurie214

Turing Test or Zarkon..another question to my Reimann sum problem....

  • one year ago
  • one year ago

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  1. laurie214
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    when I simplify that long formula and take it's limit, I get a -1/4 and that's not possible is it??

    • one year ago
  2. laurie214
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    And how can you do what zarcon suggested, just look at the exponents for the leading n values and know you'll get their product in the numerator? can you do the same to the constants (+1, +1, and -1 to know you'll get a -1 for the final constant in the numerator?

    • one year ago
  3. Murph79
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    It's quite possible. What is the equation?

    • one year ago
  4. freckles
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    Is this what you are evaluating? \[\int\limits_0^1x^4dx=\lim_{n\to\infty}\sum_{j=1}^nf(\Delta x_j^*)\Delta x=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^4}\cdot\frac1n=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^5} \]

    • one year ago
  5. laurie214
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    yep...

    • one year ago
  6. freckles
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    \[\lim_{n\to\infty}\frac1{n^5}\sum_{j=1}^\ni^4=\lim_{n\to\infty}\frac1{n^5}{n(n+1)(2n+1)(3n^2 + 3n -1)\over30} \] Ok so if we multiply the top out we only care about the coefficient of n^5 If we multiply all the crap out what would be the coefficient of n^5 well what is 1(1)(2)(3) ?

    • one year ago
  7. laurie214
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    and then is became 1/n^5 n(n+1)(2n+1)(3n^2+3n-1) all /3

    • one year ago
  8. freckles
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    the the coefficient of n^5 on bottom is 30

    • one year ago
  9. freckles
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    so we have 6/30 right? :)

    • one year ago
  10. laurie214
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    yeah, see that and that's what I get when I DO multiply it all out, but is that one of the shortcut things you can just use or KNOW...was there a proof on it somewhere?

    • one year ago
  11. laurie214
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    no...it didn't come out for me that simply!

    • one year ago
  12. laurie214
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    OH...I forgot to keep bringng down the 30!!

    • one year ago
  13. freckles
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    |dw:1336266690533:dw| Do you remember how to find horizontal asypmtotes for polynomials/polynomials? if the degrees on top and bottom are the same then just take the coeffiicent of the term with the highest exponent on top /over/ the coefficient of the term with the highest exponent on bottom and that would be the limit

    • one year ago
  14. freckles
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    |dw:1336266798515:dw|

    • one year ago
  15. laurie214
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    but sstilldon't see how the 30 can cancel.....the n^5 does...eventually down to ...I can't get it to factor so IT will cancel out or simplify..:-(

    • one year ago
  16. freckles
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    \[\lim_{n \rightarrow \infty}\frac{n(n)(2n)(3n^2)}{30n^5}=\lim_{n \rightarrow \infty}\frac{6n^5}{30n^5}\]

    • one year ago
  17. laurie214
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    now how did you simplify that so easily??? how can I just drop out the linear terms and constant?

    • one year ago
  18. freckles
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    Ok did you see the paragraph i wrote above?

    • one year ago
  19. freckles
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    Do you remember how to find horizontal asypmtotes for polynomials/polynomials? if the degrees on top and bottom are the same then just take the coeffiicent of the term with the highest exponent on top /over/ the coefficient of the term with the highest exponent on bottom and that would be the limit

    • one year ago
  20. freckles
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    That one ^

    • one year ago
  21. laurie214
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    yes, I remember that now from when I tutor precal....duh....thought about that. :-)

    • one year ago
  22. freckles
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    The degree of the top was 5 The degree of the bottom was 5

    • one year ago
  23. freckles
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    Since the degrees are the same we only care about the coefficient of the n^5 on top and the coefficient of n^5 on bottom

    • one year ago
  24. laurie214
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    But I need to calculate the upper reimann sumof all that mess....so what does it simplify down to??

    • one year ago
  25. laurie214
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    I have 6n^4 + 15n^3 -10n^2 -1/ 30n^4 and can get no further...what am I doing wriong??

    • one year ago
  26. freckles
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    \[\lim_{n \rightarrow \infty}\frac{n(n)(2n)(3n^2)}{30n^5}=\lim_{n \rightarrow \infty}\frac{6n^5}{30n^5} \] so are you asking me to evaluate this limit for you? It is just 6/30 since the n^5 's cancel

    • one year ago
  27. laurie214
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    that's what I thought I should get, but see what I got above by multiplying it all out and not enough canceled....

    • one year ago
  28. freckles
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    Ok remember what you have on top and bottom the degrees are the same just take the coeifficent of n^4 on top and on bottom

    • one year ago
  29. freckles
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    you don't have to multiply all of that out and nor do i recommend it

    • one year ago
  30. bmp
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    @laurie214 Think about it this way. Let 15n^3 - 10n^2 - 1 = m. Then: \[ \lim_{n \rightarrow \infty} \frac{6n^4 + m}{30n^4} \implies \lim_{n \rightarrow \infty} \frac{6n^4}{30n^4} + \lim_{n \rightarrow \infty} \frac{m}{30n^4} \]Now, the highest deegre term for m is n^3. Do you agree that n^4 goes to infinity much quicker than n^4? That implies that the second limit goes to zero and we are left with\[ \lim_{n \rightarrow \infty} \frac{6n^4}{30n^4} \]as wanted

    • one year ago
  31. freckles
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    \[\lim_{n \rightarrow \infty}\frac{\frac{6n^4}{n^4}+\frac{15n^3}{n^4}-\frac{10^2}{n^4}-\frac{1}{n^4}}{\frac{n^4}{n^4}}\]

    • one year ago
  32. freckles
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    \[\lim_{n \rightarrow \infty}\frac{6+0-0-0}{30}\] type-o above

    • one year ago
  33. freckles
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    i left out coefficient of n^4/n^4 on bottom

    • one year ago
  34. laurie214
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    when the prof did the problem for x^3, he got down to (n+1)^2/4n^2 for the evaluation of the summation....shouldn't ok...see what you're doing above here....

    • one year ago
  35. laurie214
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    WE only need to evaluate that summation for X^4 the Reimann way....how they did it before Leibnitz came on the scene. :-)

    • one year ago
  36. laurie214
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    so was my 6n^4 etc technically correct for the evaluation of the sum??

    • one year ago
  37. freckles
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    6/30

    • one year ago
  38. laurie214
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    And to get the limit, you wouldn't need to multiply it all out....I see that, but to get a formula or an evaluation, you would, right?

    • one year ago
  39. bmp
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    If the amount of "rectangles" that we want is infinity (as the limit is saying), it should equal the defined integral, that is, it should equal 1/5. The evaluation is the limit. :-)

    • one year ago
  40. freckles
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    you don't need to multiply it all out to evaluate the limit

    • one year ago
  41. freckles
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    i don't know what you mean looking for a formula for definite integral

    • one year ago
  42. freckles
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    definite integrals you only get a number

    • one year ago
  43. bmp
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    Or using a better wording, the limit is the evaluation of the sum, because, in your other question, you have already set up the rectangles and f(x*). Now, to evaluate it, we have to take the limit, that should be equal to the definite integral.

    • one year ago
  44. laurie214
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    ok, so you're saying the Reimann sum LPLUS the limit IS the same as the integral!

    • one year ago
  45. laurie214
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    duh...I was trying to equate the RS ONLY to the integral....left out the limit step. :)

    • one year ago
  46. bmp
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    I am saying that after setting up a Riemann sum, if we evaluate it to infinity, that is the Riemann integral. :-)

    • one year ago
  47. laurie214
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    vs the leibnitz integral...the short cut. right?

    • one year ago
  48. bmp
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    If we consider a finite amount of rectangles, we always have an error. Because n -> infinity, the error is minimal, and we consider it equal to the area under the curve, i.e., the Riemann integral in the case of f(x) = x^4.

    • one year ago
  49. laurie214
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    limit = definite integral in other words, like you said above...got it! so this integral = 1/5 either way,,,duh, so the sum of the first n squares is 1/3, the first n cubes is 1/4 (we found on last hmwk) and the sum of the first n^4 is 1/5....denominator is always just one more that the exponent

    • one year ago
  50. bmp
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    Maybe this will make you have more intuition: http://en.wikipedia.org/wiki/File:Riemann.gif We are still dealing with a finite amount of rectangles, but you can see that if we grow and grow the number of rectangles, the area approximation has a smaller and smaller error.

    • one year ago
  51. laurie214
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    new I was supposed to get that, but the variable expression fo rthe summation threw me off..I HAD to go another step to get it's limit...

    • one year ago
  52. bmp
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    And, yes, if we are dealing with the interval [0,1] for polynomial functions like those you said, x^5 = 1/6, x^7 = 1/8 :-)

    • one year ago
  53. laurie214
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    AND if the interval value changes, say to n=2? then calculate that sum by plugging 2 into the formula ?

    • one year ago
  54. bmp
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    I think the \( \Delta x \) changes in the Riemann sum, or, if you want to integrate, it will be (2^n)/n. It will be, for instance, (2^4)/4, and so on. I will double check the Riemann sum part.

    • one year ago
  55. laurie214
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    were supposed to verify the formula fo rn=2, 3, 5, 8

    • one year ago
  56. bmp
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    Ah, yes. That is correct. For instance, if you have [0,1], you will want to cut the interval in n pieces: 1/n. If it's from [0,2], you will want to cut the interval in n pieces again: 2/n and so on.

    • one year ago
  57. bmp
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    Remember the formula TuringTest showed you? Pick the size of the interval and do it like that: 2/n, 3/n, etc. Albeit, I think this n means the limit. So we aren't really taking it to infinity.

    • one year ago
  58. laurie214
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    yes, each time....

    • one year ago
  59. bmp
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    Then, all that changes is that the interval has to be 2/n, 3/n and so on :-)

    • one year ago
  60. laurie214
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    so do the reimann sum method separately for each n, just like I did for n=1?

    • one year ago
  61. bmp
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    Yup. It's a bit of a torture, I agree. But always remember that, if we are evaluating the sum to infinity, the result should be the same as the integral.

    • one year ago
  62. laurie214
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    ok, that's what it means to evaluate these....but the limit will be 1/5 each time, no matter what the interval is (or width of the rectangle, in other words)

    • one year ago
  63. laurie214
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    that makes sense....

    • one year ago
  64. bmp
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    No. The limit does change. Now, instead of the definite integral being : \[\int_{0}^{1} x^4 dx \]we are dealing now with: \[ \int_{0}^{2} x^4dx \]Or more generally: \[\Large\int_{0}^{\text{end of interval}}x^4dx \]

    • one year ago
  65. laurie214
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    ok....

    • one year ago
  66. bmp
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    The limit should be : \[ \frac{(\text{end of interval})^{5}}{5} \]

    • one year ago
  67. laurie214
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    gotcha...thanks SO much !! Sorry I took you away from other people with questions..

    • one year ago
  68. bmp
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    No problem. Glad to be helpful :-) and sorry about the other question, I started answering it but got distracted.

    • one year ago
  69. laurie214
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    that's ok....is there anyway to priint all of this out???

    • one year ago
  70. bmp
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    To print the text, I don't think so. Only printing the page, but that is not that much helpful, I think.

    • one year ago
  71. laurie214
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    ok :-)

    • one year ago
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