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

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bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1For some reason I'm thinking limits but it's probably because of Sinx / x and has nothing to do with this

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1I think we need to take limit.. but... we have something to do before taking limit?

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Integrate from 0 to pi/2? And something about the area must tell you something? Idk, lol just rambling on.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Well, I don't think we need integration here.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1It's because the prof. hasn't taught integration in the lessons yet (even though I have learnt integration before)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Ohh Ok. There are some proof of function being greater than the other.. hmm

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.11. it's a decreasing function. 2. \[\lim_{x \rightarrow 0} \frac{sinx}{x}=1\] 3. \[\lim_{x \rightarrow \frac{\pi}{2}} \frac{sin\frac{\pi }{2}}{\frac{\pi}{2}}=\frac{2}{\pi}\] Uhoh! I'm not on the right track :

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Why 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)

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Well, that's the problem.. How are you going to show it? Finding its derivative?

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0you cannot just say that the roots are at n*pi?

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0You want a proof of that as well...

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Just a minute \[\frac{d}{dx}\frac{sinx}{x} = \frac{xcosxsinx}{x^2}\]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.

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0I mean x= Tan x is a pain.....

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0You just have do a graph y=x and Tan x and look or use a CAS

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Graphing... is not that good...

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1If graphing works, why not just graph the function sinx / x to show it's a decreasing function?

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0What are you doing, Fourier?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1No.. limit and derivative.

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0That's a bit sneaky, hitting you with sinc function...

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Let's see , all you have to do is show that the first positive root is at pi....

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0And anywhere from 0 to pi is not 0

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0That's good enough to show the first root is at pi, right?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Oh... so you're solving sinx/x=0? If so, yes!

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0So now you are done, I think (assuming usual smoothness principles).

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I 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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1So, 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..

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1how 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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Wait.. how.. why... xcosx  sinx = 0 xcotx 1 = 0 xcotx = 1 Am I doing anything wrong here?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1oh shnap I am out to lunch, but I still say we find that critical point and just test the interval let me think how....

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Don't think while you eat!

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I know I have seen a proof that tanx>x in (0,pi/2) ... that would work if I could remember it

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1This is complicated...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I got it :) so are you with me up to the idea that tanx>x in (0,pi/2) would prove that this is decreasing?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1okay, we can prove this using the mean value theorem...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1the mean value theorem states that f(x)=f(a)+f'(c)(xa) for some a<c<x our f(x)=tanx, and our interval is (0,pi/2) so a=0

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1getting what we need from the formula f(0)=0, f'(c)=sec^2(c) where 0<c<x<pi/2

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1what do you get then by using this in the formula?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1f(x)=f(a)+f'(c)(xa) tanx = 0 + sec^2 (c) (x) tanx = [sec^2 (c)] x Doesn't look good.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1it is though, because what do we know about sec(c) for 0<c<pi/2 ???

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.10<c<pi/2 1<sec (c)< undefined (+infinity?!)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yeah, but the point is that sec(c)>1 for that interval

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1so tanx=sec^2(c)x 0<c<pi/2 implies that...?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.11< sec^2(c) x < sec^2 (c) x x< tanx Wow!!

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1tadah! :) so that completes your proof

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Uhhuh! Why did you think of mean value theorem for this question????

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1It's the only way I know to prove tanx>x in (0,pi/2) there may be other ways

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0Given 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)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1very similar argument^

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0sorry 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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1no apologies necessary, always good to see more than one explanation :)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0although 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.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks 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!!

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Going 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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