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Callisto

Prove \[\frac{sinx}{x} > \frac{2}{\pi} \ \ , \forall x \in (0, \frac{ \pi }{ 2 })\] How to start?

  • one year ago
  • one year ago

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  1. bahrom7893
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    For some reason I'm thinking limits but it's probably because of Sinx / x and has nothing to do with this

    • one year ago
  2. Callisto
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    I think we need to take limit.. but... we have something to do before taking limit?

    • one year ago
  3. bahrom7893
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    Integrate from 0 to pi/2? And something about the area must tell you something? Idk, lol just rambling on.

    • one year ago
  4. Callisto
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    Well, I don't think we need integration here.

    • one year ago
  5. Callisto
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    It's because the prof. hasn't taught integration in the lessons yet (even though I have learnt integration before)

    • one year ago
  6. bahrom7893
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    Ohh Ok. There are some proof of function being greater than the other.. hmm

    • one year ago
  7. Callisto
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    1. it's a decreasing function. 2. \[\lim_{x \rightarrow 0} \frac{sinx}{x}=1\] 3. \[\lim_{x \rightarrow \frac{\pi}{2}} \frac{sin\frac{\pi }{2}}{\frac{\pi}{2}}=\frac{2}{\pi}\] Uh-oh! I'm not on the right track :|

    • one year ago
  8. estudier
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    Why can't you just show that it is uniformly decreasing over the interval and calculate the least value? (Once you have shown that the limit is 1 at start)

    • one year ago
  9. Callisto
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    Well, that's the problem.. How are you going to show it? Finding its derivative?

    • one year ago
  10. estudier
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    you cannot just say that the roots are at n*pi?

    • one year ago
  11. Callisto
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    How so?

    • one year ago
  12. estudier
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    You want a proof of that as well...

    • one year ago
  13. Callisto
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    Just a minute \[\frac{d}{dx}\frac{sinx}{x} = \frac{xcosx-sinx}{x^2}\]If I put it =0, I'll get xcosx - sinx = 0 That's where I stuck at. And I haven't learnt the general solution (hmm, for that n*pi) yet, so I would like to know how you get that.

    • one year ago
  14. estudier
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    I mean x= Tan x is a pain.....

    • one year ago
  15. Callisto
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    Yes!

    • one year ago
  16. estudier
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    You just have do a graph y=x and Tan x and look or use a CAS

    • one year ago
  17. Callisto
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    Graphing... is not that good...

    • one year ago
  18. Callisto
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    If graphing works, why not just graph the function sinx / x to show it's a decreasing function?

    • one year ago
  19. estudier
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    Well, quite...:-)

    • one year ago
  20. estudier
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    What are you doing, Fourier?

    • one year ago
  21. Callisto
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    No.. limit and derivative.

    • one year ago
  22. estudier
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    That's a bit sneaky, hitting you with sinc function...

    • one year ago
  23. estudier
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    Let's see , all you have to do is show that the first positive root is at pi....

    • one year ago
  24. estudier
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    Sin pi/pi = 0

    • one year ago
  25. estudier
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    And anywhere from 0 to pi is not 0

    • one year ago
  26. estudier
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    That's good enough to show the first root is at pi, right?

    • one year ago
  27. Callisto
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    Oh... so you're solving sinx/x=0? If so, yes!

    • one year ago
  28. estudier
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    So now you are done, I think (assuming usual smoothness principles).

    • one year ago
  29. TuringTest
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    I don't see the problem by saying that the limit at 0 is 1>pi/2, and that sinx/x is uniformly decreasing over (0,pi/2), and the limit approaching pi/2=2/pi that shows that sinx/x>2/pi in that region

    • one year ago
  30. Callisto
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    So, step 1, 2, 3 are correct. But how to show step 1 is the point. I'm not quite sure if I can get estudier's idea..

    • one year ago
  31. TuringTest
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    how to show it's decreasing? find critical points\[f'=\frac{x\cos x-\sin x}{x^2}=0\implies x\cot x=0\]so the critical points are\[x=\{0,\frac\pi2\}\]hence you just need to test a point in that region, say \(x=\pi/4\), and that tells you whether the function is increasing or decreasing on that interval

    • one year ago
  32. Callisto
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    Wait.. how.. why... xcosx - sinx = 0 xcotx -1 = 0 xcotx = 1 Am I doing anything wrong here?

    • one year ago
  33. TuringTest
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    oh shnap I am out to lunch, but I still say we find that critical point and just test the interval let me think how....

    • one year ago
  34. Callisto
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    Don't think while you eat!

    • one year ago
  35. TuringTest
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    I know I have seen a proof that tanx>x in (0,pi/2) ... that would work if I could remember it

    • one year ago
  36. Callisto
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    This is complicated...

    • one year ago
  37. TuringTest
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    I got it :) so are you with me up to the idea that tanx>x in (0,pi/2) would prove that this is decreasing?

    • one year ago
  38. Callisto
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    Yes!

    • one year ago
  39. TuringTest
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    okay, we can prove this using the mean value theorem...

    • one year ago
  40. TuringTest
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    the mean value theorem states that f(x)=f(a)+f'(c)(x-a) for some a<c<x our f(x)=tanx, and our interval is (0,pi/2) so a=0

    • one year ago
  41. TuringTest
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    getting what we need from the formula f(0)=0, f'(c)=sec^2(c) where 0<c<x<pi/2

    • one year ago
  42. TuringTest
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    what do you get then by using this in the formula?

    • one year ago
  43. Callisto
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    f(x)=f(a)+f'(c)(x-a) tanx = 0 + sec^2 (c) (x) tanx = [sec^2 (c)] x Doesn't look good.

    • one year ago
  44. TuringTest
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    it is though, because what do we know about sec(c) for 0<c<pi/2 ???

    • one year ago
  45. Callisto
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    0<c<pi/2 1<sec (c)< undefined (+infinity?!)

    • one year ago
  46. TuringTest
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    yeah, but the point is that sec(c)>1 for that interval

    • one year ago
  47. TuringTest
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    so tanx=sec^2(c)x 0<c<pi/2 implies that...?

    • one year ago
  48. Callisto
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    1< sec^2(c) x < sec^2 (c) x x< tanx Wow!!

    • one year ago
  49. TuringTest
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    tadah! :) so that completes your proof

    • one year ago
  50. Callisto
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    Uh-huh! Why did you think of mean value theorem for this question????

    • one year ago
  51. TuringTest
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    It's the only way I know to prove tanx>x in (0,pi/2) there may be other ways

    • one year ago
  52. experimentX
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    Given the interval is \( (0, {\pi \over 2}) \) to find the maximum or minimum, we have \[ x = \tan x\] the first part is a line y=x passing through origin while the latter one is y=tan(x) |dw:1349628691638:dw| we can see that tan(x) > x since tan(x) > slope is sec^2(x) > slope of (x) = 1 so next time tan(x) meets x in the interval pi/2, pi Hence we do not have any critical point in the interval (0, pi/2)

    • one year ago
  53. TuringTest
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    very similar argument^

    • one year ago
  54. experimentX
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    sorry I didn't follow the post For Global extremum, we check the for max or min value in the endpoints of interval as well as in the boundary. The minimum value will be the minimum value among those points. So, you don't have critical points on the interval. the min value will be among the min value of function at the boundary which is pi/2. since your interval is open, f(x) > pi/2

    • one year ago
  55. TuringTest
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    no apologies necessary, always good to see more than one explanation :)

    • one year ago
  56. experimentX
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    although it is enough to show that you have -ve slope in the interval ... x1 > x2, f(x1) > f(x2) usually it's customary to deal problem with Global extremum.

    • one year ago
  57. Callisto
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    Thanks everyone for you help, but I think I need some time to think over it again. I hope you don't mind if I come back few hours later as it's actually quite late here. I'm sorry.. and thank you so much!!

    • one year ago
  58. experimentX
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    Night!!

    • one year ago
  59. TuringTest
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    See ya !

    • one year ago
  60. estudier
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    Going back to the usual method, maybe u can just do like this, you have sin x < x < tan x so for x in 0 to pi/2 1 < x/sinx < 1/cos x -> x cos x < sin x so therefore the derivative is negative and you are done.

    • one year ago
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