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Callisto
Prove \[\frac{sinx}{x} > \frac{2}{\pi} \ \ , \forall x \in (0, \frac{ \pi }{ 2 })\] How to start?
For some reason I'm thinking limits but it's probably because of Sinx / x and has nothing to do with this
I think we need to take limit.. but... we have something to do before taking limit?
Integrate from 0 to pi/2? And something about the area must tell you something? Idk, lol just rambling on.
Well, I don't think we need integration here.
It's because the prof. hasn't taught integration in the lessons yet (even though I have learnt integration before)
Ohh Ok. There are some proof of function being greater than the other.. hmm
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 :|
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)
Well, that's the problem.. How are you going to show it? Finding its derivative?
you cannot just say that the roots are at n*pi?
You want a proof of that as well...
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.
I mean x= Tan x is a pain.....
You just have do a graph y=x and Tan x and look or use a CAS
Graphing... is not that good...
If graphing works, why not just graph the function sinx / x to show it's a decreasing function?
What are you doing, Fourier?
No.. limit and derivative.
That's a bit sneaky, hitting you with sinc function...
Let's see , all you have to do is show that the first positive root is at pi....
And anywhere from 0 to pi is not 0
That's good enough to show the first root is at pi, right?
Oh... so you're solving sinx/x=0? If so, yes!
So now you are done, I think (assuming usual smoothness principles).
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
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..
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
Wait.. how.. why... xcosx - sinx = 0 xcotx -1 = 0 xcotx = 1 Am I doing anything wrong here?
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....
Don't think while you eat!
I know I have seen a proof that tanx>x in (0,pi/2) ... that would work if I could remember it
This is complicated...
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?
okay, we can prove this using the mean value theorem...
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
getting what we need from the formula f(0)=0, f'(c)=sec^2(c) where 0<c<x<pi/2
what do you get then by using this in the formula?
f(x)=f(a)+f'(c)(x-a) tanx = 0 + sec^2 (c) (x) tanx = [sec^2 (c)] x Doesn't look good.
it is though, because what do we know about sec(c) for 0<c<pi/2 ???
0<c<pi/2 1<sec (c)< undefined (+infinity?!)
yeah, but the point is that sec(c)>1 for that interval
so tanx=sec^2(c)x 0<c<pi/2 implies that...?
1< sec^2(c) x < sec^2 (c) x x< tanx Wow!!
tadah! :) so that completes your proof
Uh-huh! Why did you think of mean value theorem for this question????
It's the only way I know to prove tanx>x in (0,pi/2) there may be other ways
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)
very similar argument^
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
no apologies necessary, always good to see more than one explanation :)
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.
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!!
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.