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Callisto

  • 2 years ago

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

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  1. bahrom7893
    • 2 years ago
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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

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

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

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

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

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

  7. Callisto
    • 2 years ago
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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 :|

  8. estudier
    • 2 years ago
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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)

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

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

  11. Callisto
    • 2 years ago
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    How so?

  12. estudier
    • 2 years ago
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    You want a proof of that as well...

  13. Callisto
    • 2 years ago
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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.

  14. estudier
    • 2 years ago
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    I mean x= Tan x is a pain.....

  15. Callisto
    • 2 years ago
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    Yes!

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

  17. Callisto
    • 2 years ago
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    Graphing... is not that good...

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

  19. estudier
    • 2 years ago
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    Well, quite...:-)

  20. estudier
    • 2 years ago
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    What are you doing, Fourier?

  21. Callisto
    • 2 years ago
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    No.. limit and derivative.

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

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

  24. estudier
    • 2 years ago
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    Sin pi/pi = 0

  25. estudier
    • 2 years ago
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    And anywhere from 0 to pi is not 0

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

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

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

  29. TuringTest
    • 2 years ago
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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

  30. Callisto
    • 2 years ago
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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..

  31. TuringTest
    • 2 years ago
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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

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

  33. TuringTest
    • 2 years ago
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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....

  34. Callisto
    • 2 years ago
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    Don't think while you eat!

  35. TuringTest
    • 2 years ago
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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

  36. Callisto
    • 2 years ago
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    This is complicated...

  37. TuringTest
    • 2 years ago
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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?

  38. Callisto
    • 2 years ago
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    Yes!

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

  40. TuringTest
    • 2 years ago
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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

  41. TuringTest
    • 2 years ago
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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

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

  43. Callisto
    • 2 years ago
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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.

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

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

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

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

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

  49. TuringTest
    • 2 years ago
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    tadah! :) so that completes your proof

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

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

  52. experimentX
    • 2 years ago
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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)

  53. TuringTest
    • 2 years ago
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    very similar argument^

  54. experimentX
    • 2 years ago
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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

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

  56. experimentX
    • 2 years ago
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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.

  57. Callisto
    • 2 years ago
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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!!

  58. experimentX
    • 2 years ago
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    Night!!

  59. TuringTest
    • 2 years ago
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    See ya !

  60. estudier
    • 2 years ago
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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.

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