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Prove \[\frac{sinx}{x} > \frac{2}{\pi} \ \ , \forall x \in (0, \frac{ \pi }{ 2 })\] How to start?
 one year ago
 one year ago
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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bahrom7893Best ResponseYou've already chosen the best response.0
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

CallistoBest ResponseYou've already chosen the best response.1
I think we need to take limit.. but... we have something to do before taking limit?
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.0
Integrate from 0 to pi/2? And something about the area must tell you something? Idk, lol just rambling on.
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Well, I don't think we need integration here.
 one year ago

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

bahrom7893Best ResponseYou've already chosen the best response.0
Ohh Ok. There are some proof of function being greater than the other.. hmm
 one year ago

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

estudierBest ResponseYou've already chosen the best response.0
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

CallistoBest ResponseYou've already chosen the best response.1
Well, that's the problem.. How are you going to show it? Finding its derivative?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
you cannot just say that the roots are at n*pi?
 one year ago

estudierBest ResponseYou've already chosen the best response.0
You want a proof of that as well...
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Just 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.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
I mean x= Tan x is a pain.....
 one year ago

estudierBest ResponseYou've already chosen the best response.0
You just have do a graph y=x and Tan x and look or use a CAS
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Graphing... is not that good...
 one year ago

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

estudierBest ResponseYou've already chosen the best response.0
What are you doing, Fourier?
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
No.. limit and derivative.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
That's a bit sneaky, hitting you with sinc function...
 one year ago

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

estudierBest ResponseYou've already chosen the best response.0
And anywhere from 0 to pi is not 0
 one year ago

estudierBest ResponseYou've already chosen the best response.0
That's good enough to show the first root is at pi, right?
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Oh... so you're solving sinx/x=0? If so, yes!
 one year ago

estudierBest ResponseYou've already chosen the best response.0
So now you are done, I think (assuming usual smoothness principles).
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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

CallistoBest ResponseYou've already chosen the best response.1
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

TuringTestBest ResponseYou've already chosen the best response.1
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

CallistoBest ResponseYou've already chosen the best response.1
Wait.. how.. why... xcosx  sinx = 0 xcotx 1 = 0 xcotx = 1 Am I doing anything wrong here?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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

CallistoBest ResponseYou've already chosen the best response.1
Don't think while you eat!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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

CallistoBest ResponseYou've already chosen the best response.1
This is complicated...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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

TuringTestBest ResponseYou've already chosen the best response.1
okay, we can prove this using the mean value theorem...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the 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
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.1
what do you get then by using this in the formula?
 one year ago

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

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

CallistoBest ResponseYou've already chosen the best response.1
0<c<pi/2 1<sec (c)< undefined (+infinity?!)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yeah, but the point is that sec(c)>1 for that interval
 one year ago

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

CallistoBest ResponseYou've already chosen the best response.1
1< sec^2(c) x < sec^2 (c) x x< tanx Wow!!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
tadah! :) so that completes your proof
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Uhhuh! Why did you think of mean value theorem for this question????
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
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

TuringTestBest ResponseYou've already chosen the best response.1
very similar argument^
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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

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

experimentXBest ResponseYou've already chosen the best response.0
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

CallistoBest ResponseYou've already chosen the best response.1
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

estudierBest ResponseYou've already chosen the best response.0
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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