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Loser66

  • one year ago

Explain me, please.

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  1. dan815
    • one year ago
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    for a decreasing sequence bn that summation converges

  2. dan815
    • one year ago
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    is it dirichlet?

  3. Loser66
    • one year ago
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    Those are what my prof did in class.

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  4. Loser66
    • one year ago
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    The red parts are my explanation on the lecture so that I remember what is going on when reviewing for test. :)

  5. dan815
    • one year ago
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    it is still downloading, its going very slow

  6. Loser66
    • one year ago
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    dan!!

  7. dan815
    • one year ago
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    ok hmm i see

  8. dan815
    • one year ago
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    lets take cos(ntheta)= Real part of {e^(itheta)}

  9. dan815
    • one year ago
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    what they are using as their argument is that hey look just like how an alternating sequence of (-1)^n multiplied by a decreasing series goes to 0, this also does the same thing

  10. dan815
    • one year ago
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    because for any thetea u pick, for varying n, you keep getting some number of positive and some number of negatives , both equal to each other, tho the addition of this decreasing bn sequences works slightly differently

  11. dan815
    • one year ago
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    but you can still definately show that you will be subtracting greater part than what u are adding in the next set

  12. dan815
    • one year ago
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    here is a picture to help you see

  13. dan815
    • one year ago
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    |dw:1441645035166:dw|

  14. dan815
    • one year ago
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    varying theta is just like varying the period you sample your dots

  15. Loser66
    • one year ago
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    next?

  16. dan815
    • one year ago
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    |dw:1441645239534:dw|

  17. dan815
    • one year ago
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    so in fact this is not much different from alternating series at all, its just put in a different way

  18. dan815
    • one year ago
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    let f(n) be some decreasing function for the first 3 values of n let cosntheta be positive for the next 3 values of n let cos n*theta be negative can you see why now, this has to be convergin

  19. dan815
    • one year ago
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    liket lets take a random decreasing function

  20. dan815
    • one year ago
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    |dw:1441645504019:dw|

  21. dan815
    • one year ago
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    |dw:1441645527838:dw|

  22. Loser66
    • one year ago
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    |dw:1441645615856:dw|

  23. dan815
    • one year ago
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    thats wht u are doing with chos (ntheta) you are taking samples in each part of that curve adding and subtracting

  24. Loser66
    • one year ago
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    dan, slow down on let f(n) be some decreasing function for the first 3 values of n let cosntheta be positive for the next 3 values of n let cos n*theta be negative can you see why now, this has to be convergin

  25. anonymous
    • one year ago
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    wassup dan

  26. dan815
    • one year ago
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    |dw:1441645682902:dw|

  27. Loser66
    • one year ago
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    f(n) = cos (ntheta) 3 values of n let cos n theta > 0

  28. dan815
    • one year ago
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    these could be your sambled points when you multiply by cos n theta

  29. dan815
    • one year ago
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    yeah like

  30. dan815
    • one year ago
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    |dw:1441645810574:dw|

  31. Loser66
    • one year ago
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    |dw:1441645965886:dw|

  32. Loser66
    • one year ago
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    They are the value of cos (n theta) when n = 1, 2, 3... right?

  33. dan815
    • one year ago
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    |dw:1441646040644:dw|

  34. Loser66
    • one year ago
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    yes

  35. dan815
    • one year ago
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    yeah

  36. dan815
    • one year ago
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    now you say the sum of all these samples values in between is bounded by that formula they gave u up at the top

  37. Loser66
    • one year ago
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    suppose 0<theta < pi, then it limits on |dw:1441646174573:dw|

  38. Loser66
    • one year ago
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    now what?

  39. Loser66
    • one year ago
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    sum cos (n theta) = cos (0theta) + cos (1 theta) + cos (2 theta) + ..... cos (ntheta) , right? then?

  40. Loser66
    • one year ago
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    is it not that it is =0?

  41. dan815
    • one year ago
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    |dw:1441646218632:dw|

  42. dan815
    • one year ago
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    now u can use that bound

  43. anonymous
    • one year ago
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    @Adamlevine1986

  44. dan815
    • one year ago
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    notice how if theta -->0 its going to finity as u will be sampling an infinite points in each of the interval

  45. Loser66
    • one year ago
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    then?

  46. dan815
    • one year ago
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    @empty how do u show this series converges

  47. Loser66
    • one year ago
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    Is it not that if it is bounded and continuous, it converges?

  48. dan815
    • one year ago
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    its kind of weird to think about actually

  49. dan815
    • one year ago
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    like say you take an finite number of points on the highest parto of your decreasing function

  50. dan815
    • one year ago
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    |dw:1441646971233:dw|

  51. dan815
    • one year ago
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    now lets say u subtract an infinite set of points but lesser value.. the difference between these 2 is still infinity is it notq

  52. dan815
    • one year ago
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    |dw:1441647101921:dw|

  53. dan815
    • one year ago
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    but then u subtraced with some other difference an infinite set of points when compared to the next block of addition

  54. Loser66
    • one year ago
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    I gotta go. You guys PPPPPPLLLLLEASE, freely discuss. I' ll take it later. Thanks in advance.

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