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Babynini

  • one year ago

Squeeze Theorem greatest integer function

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  1. Babynini
    • one year ago
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  2. anonymous
    • one year ago
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    what about them?

  3. anonymous
    • one year ago
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    from \(-\pi \) to \(-\frac{\pi}{2}\) cosiene goes from \(-1\) to \(0\) so it would be identically \(-1\) on that interval

  4. anonymous
    • one year ago
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    then on \(-\frac{\pi}{2}\) to \(0\) it goes from \(0\) to \(1\) so on that interval it would be \(0\)

  5. anonymous
    • one year ago
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    etc

  6. Empty
    • one year ago
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    Might be fun to play here, the greatest integer function is sometimes called the ceiling function. https://www.desmos.com/calculator/fdipg03oxv

  7. anonymous
    • one year ago
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    Do those funky brackets mean the least integer?

  8. anonymous
    • one year ago
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    on the interval \(0\) to \(\frac{\pi}{2}\) it goes from \(1\) to \(0\) so it is 0 there as well

  9. Babynini
    • one year ago
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    So if I were to draw one cosine it would be just one mountain

  10. anonymous
    • one year ago
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    @empty i though greatest integer was floor

  11. anonymous
    • one year ago
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    greatest integer less than like \(\lfloor 2.4\rfloor=2\)

  12. Babynini
    • one year ago
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    |dw:1444265853694:dw| like that!..? haha i'm such an artist.

  13. anonymous
    • one year ago
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    Why can't they use the standard \[\lceil \cos x \rceil\] for ceiling and \[\lfloor \cos x \rfloor\]for floor?

  14. anonymous
    • one year ago
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    who said art was dead that is cosine though not \(\lfloor \cos(x)\rfloor\)

  15. anonymous
    • one year ago
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    "greatest integer" to me means "greatest integer less than" so "floor" whereas "least integer" means "least integer greater than i.e. "ceiling"

  16. Empty
    • one year ago
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    Oh floor ceil, whatever haha yeah I agree with @ospreytriple I wouldn't have messed up if they just wrote \[\left\lfloor \frac{1}{\sec x} \right\rfloor\]

  17. Babynini
    • one year ago
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    Wait, so how is cos(x) looking then? o.o

  18. Empty
    • one year ago
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    Almost identical but shifted https://www.desmos.com/calculator/cowtkonbwe

  19. Babynini
    • one year ago
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    hmm..shifted how? Like the only difference seems to be that there's lines across the bottom and top.

  20. anonymous
    • one year ago
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    it is identically \(-1\) on \((-\pi,-\frac{\pi}{2})\)

  21. anonymous
    • one year ago
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    then it is \(0\) on \(-\frac{\pi}{2},\frac{\pi}{2}\) except it is 1 at \(0\)

  22. anonymous
    • one year ago
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    then \(-1\) again

  23. anonymous
    • one year ago
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    http://www.wolframalpha.com/input/?i=floor%28cos%28x%29%29%2C+-pi..pi

  24. anonymous
    • one year ago
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    you can't see the \((0,1)\) on wolfram, but it is there

  25. anonymous
    • one year ago
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    There seems to be some confusion above somewhere. The greatest integer function is also called the floor function, the greatest integer that is less than or equal to the argument.

  26. anonymous
    • one year ago
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    or here https://www.desmos.com/calculator/cowtkonbwe

  27. anonymous
    • one year ago
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    @ospreytriple that is what i think as well

  28. Babynini
    • one year ago
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    So there's no..normal curve. Ah yeah @Empty showed me the desmos one but I don't get it really haha it looks like a cosine but with lines on the hills/indents

  29. Babynini
    • one year ago
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    So the graph is nor a normal curve?

  30. Babynini
    • one year ago
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    Is it just a point at (0,1)?

  31. anonymous
    • one year ago
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    That's definitely it @satellite73 . The ceiling function mentioned above is also called the least integer function, the smallest integer that is greater than or equal to the argument.

  32. Babynini
    • one year ago
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    |dw:1444266645781:dw|

  33. Babynini
    • one year ago
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    Is what ..i'm imagining. By what yall are saying 0.0

  34. Empty
    • one year ago
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    Yeah that's right, it's what we're saying. :)

  35. Babynini
    • one year ago
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    |dw:1444266755110:dw| like dis?

  36. Babynini
    • one year ago
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    The line on 0 being all but the point (0,0)

  37. Empty
    • one year ago
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    Yeah, since the greatest integer function says [[0]]=0 Looks good to me.

  38. Babynini
    • one year ago
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    Fabulous! phew. part b!

  39. Babynini
    • one year ago
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    i) = 0 ?

  40. Babynini
    • one year ago
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    So for this one, the rules we used for the previous problem (ie: a, or a-1) do not apply? or do they?

  41. Babynini
    • one year ago
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    @SolomonZelman

  42. anonymous
    • one year ago
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    the reason that desmos thing threw you off is that they graphed both the floor function and cosine on the same graph

  43. anonymous
    • one year ago
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    floor is a step function

  44. Babynini
    • one year ago
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    Right right.

  45. anonymous
    • one year ago
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    what is left to do?

  46. Babynini
    • one year ago
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    I'm wondering about the part b now. Because I know the rules, which you stated for the last problem. But I wasn't sure if those applied here.

  47. Babynini
    • one year ago
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    Part b and c.

  48. anonymous
    • one year ago
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    \[\lim_{x\to 0}f(x)=0\] for sure, since it is identically 0 to the right and to the left

  49. Babynini
    • one year ago
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    b: i) 0 ii) if the rules apply then.. 0.57 ?

  50. anonymous
    • one year ago
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    the fact that \(f(0)=1\) does not effect the limit

  51. anonymous
    • one year ago
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    what does "rules apply" mean? this thing never approaches 0.57 it is either 0 or -1

  52. Babynini
    • one year ago
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    the rules of if it approaches from the left it = a - 1 and if it approaches from the right it = a but for this one I guess we just use the graph.

  53. anonymous
    • one year ago
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    \[\lim_{x\to \frac{\pi}{2}}f(x)=0\]

  54. anonymous
    • one year ago
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    ok typo there hold on

  55. anonymous
    • one year ago
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    \[\huge \lim_{x\to \frac{\pi}{2}^-}f(x)=0\]

  56. anonymous
    • one year ago
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    since it is identically 0 on the interval to the left

  57. Babynini
    • one year ago
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    They all look like they're ultimately approaching 0

  58. anonymous
    • one year ago
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    no, from the right it is \(-1\)

  59. anonymous
    • one year ago
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    |dw:1444267958342:dw|

  60. anonymous
    • one year ago
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    \[\huge \lim_{x\to \frac{\pi}{2}^+}f(x)=-1\]

  61. anonymous
    • one year ago
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    and since the right and left hand limits are not the same, the limit does not exist

  62. Babynini
    • one year ago
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    Righto, gotcha.

  63. anonymous
    • one year ago
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    k good to answer the last question, the limit exists everywhere except at those two jumps

  64. Babynini
    • one year ago
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    So I should probably write that in some form of interval haha the two jumps meaning.

  65. anonymous
    • one year ago
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    no limit at \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)

  66. anonymous
    • one year ago
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    at those to points it jumps form -1 to 0 and from 0 to -1 receptively

  67. anonymous
    • one year ago
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    |dw:1444268282111:dw|

  68. Babynini
    • one year ago
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    hahah awesome. Art lives!!

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