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

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

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

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

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

Callisto Group TitleBest 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)
 2 years ago

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

Callisto Group TitleBest 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 :
 2 years ago

estudier Group TitleBest 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)
 2 years ago

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

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

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

Callisto Group TitleBest 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.
 2 years ago

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

estudier Group TitleBest 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
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Graphing... is not that good...
 2 years ago

Callisto Group TitleBest 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?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Well, quite...:)
 2 years ago

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

Callisto Group TitleBest ResponseYou've already chosen the best response.1
No.. limit and derivative.
 2 years ago

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

estudier Group TitleBest 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....
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Sin pi/pi = 0
 2 years ago

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

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

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

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

TuringTest Group TitleBest 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
 2 years ago

Callisto Group TitleBest 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..
 2 years ago

TuringTest Group TitleBest 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
 2 years ago

Callisto Group TitleBest 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?
 2 years ago

TuringTest Group TitleBest 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....
 2 years ago

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

TuringTest Group TitleBest 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
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
This is complicated...
 2 years ago

TuringTest Group TitleBest 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?
 2 years ago

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

TuringTest Group TitleBest 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
 2 years ago

TuringTest Group TitleBest 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
 2 years ago

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

Callisto Group TitleBest 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.
 2 years ago

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

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

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

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

Callisto Group TitleBest ResponseYou've already chosen the best response.1
1< sec^2(c) x < sec^2 (c) x x< tanx Wow!!
 2 years ago

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

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

TuringTest Group TitleBest 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
 2 years ago

experimentX Group TitleBest 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)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
very similar argument^
 2 years ago

experimentX Group TitleBest 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
 2 years ago

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

experimentX Group TitleBest 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.
 2 years ago

Callisto Group TitleBest 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!!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
Night!!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
See ya !
 2 years ago

estudier Group TitleBest 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.
 2 years ago
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