## Callisto Group Title Prove $\frac{sinx}{x} > \frac{2}{\pi} \ \ , \forall x \in (0, \frac{ \pi }{ 2 })$ How to start? one year ago one year ago

1. bahrom7893 Group Title

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

2. Callisto Group Title

I think we need to take limit.. but... we have something to do before taking limit?

3. bahrom7893 Group Title

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

4. Callisto Group Title

Well, I don't think we need integration here.

5. Callisto Group Title

It's because the prof. hasn't taught integration in the lessons yet (even though I have learnt integration before)

6. bahrom7893 Group Title

Ohh Ok. There are some proof of function being greater than the other.. hmm

7. Callisto Group Title

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 Group Title

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 Group Title

Well, that's the problem.. How are you going to show it? Finding its derivative?

10. estudier Group Title

you cannot just say that the roots are at n*pi?

11. Callisto Group Title

How so?

12. estudier Group Title

You want a proof of that as well...

13. Callisto Group Title

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 Group Title

I mean x= Tan x is a pain.....

15. Callisto Group Title

Yes!

16. estudier Group Title

You just have do a graph y=x and Tan x and look or use a CAS

17. Callisto Group Title

Graphing... is not that good...

18. Callisto Group Title

If graphing works, why not just graph the function sinx / x to show it's a decreasing function?

19. estudier Group Title

Well, quite...:-)

20. estudier Group Title

What are you doing, Fourier?

21. Callisto Group Title

No.. limit and derivative.

22. estudier Group Title

That's a bit sneaky, hitting you with sinc function...

23. estudier Group Title

Let's see , all you have to do is show that the first positive root is at pi....

24. estudier Group Title

Sin pi/pi = 0

25. estudier Group Title

And anywhere from 0 to pi is not 0

26. estudier Group Title

That's good enough to show the first root is at pi, right?

27. Callisto Group Title

Oh... so you're solving sinx/x=0? If so, yes!

28. estudier Group Title

So now you are done, I think (assuming usual smoothness principles).

29. TuringTest Group Title

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 Group Title

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 Group Title

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 Group Title

Wait.. how.. why... xcosx - sinx = 0 xcotx -1 = 0 xcotx = 1 Am I doing anything wrong here?

33. TuringTest Group Title

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 Group Title

Don't think while you eat!

35. TuringTest Group Title

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

36. Callisto Group Title

This is complicated...

37. TuringTest Group Title

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 Group Title

Yes!

39. TuringTest Group Title

okay, we can prove this using the mean value theorem...

40. TuringTest Group Title

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 Group Title

getting what we need from the formula f(0)=0, f'(c)=sec^2(c) where 0<c<x<pi/2

42. TuringTest Group Title

what do you get then by using this in the formula?

43. Callisto Group Title

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 Group Title

it is though, because what do we know about sec(c) for 0<c<pi/2 ???

45. Callisto Group Title

0<c<pi/2 1<sec (c)< undefined (+infinity?!)

46. TuringTest Group Title

yeah, but the point is that sec(c)>1 for that interval

47. TuringTest Group Title

so tanx=sec^2(c)x 0<c<pi/2 implies that...?

48. Callisto Group Title

1< sec^2(c) x < sec^2 (c) x x< tanx Wow!!

49. TuringTest Group Title

50. Callisto Group Title

Uh-huh! Why did you think of mean value theorem for this question????

51. TuringTest Group Title

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

52. experimentX Group Title

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 Group Title

very similar argument^

54. experimentX Group Title

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 Group Title

no apologies necessary, always good to see more than one explanation :)

56. experimentX Group Title

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 Group Title

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 Group Title

Night!!

59. TuringTest Group Title

See ya !

60. estudier Group Title

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.