Callisto
  • Callisto
Prove \[\frac{sinx}{x} > \frac{2}{\pi} \ \ , \forall x \in (0, \frac{ \pi }{ 2 })\] How to start?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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bahrom7893
  • bahrom7893
For some reason I'm thinking limits but it's probably because of Sinx / x and has nothing to do with this
Callisto
  • Callisto
I think we need to take limit.. but... we have something to do before taking limit?
bahrom7893
  • bahrom7893
Integrate from 0 to pi/2? And something about the area must tell you something? Idk, lol just rambling on.

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Callisto
  • Callisto
Well, I don't think we need integration here.
Callisto
  • Callisto
It's because the prof. hasn't taught integration in the lessons yet (even though I have learnt integration before)
bahrom7893
  • bahrom7893
Ohh Ok. There are some proof of function being greater than the other.. hmm
Callisto
  • Callisto
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 :|
anonymous
  • anonymous
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)
Callisto
  • Callisto
Well, that's the problem.. How are you going to show it? Finding its derivative?
anonymous
  • anonymous
you cannot just say that the roots are at n*pi?
Callisto
  • Callisto
How so?
anonymous
  • anonymous
You want a proof of that as well...
Callisto
  • Callisto
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.
anonymous
  • anonymous
I mean x= Tan x is a pain.....
Callisto
  • Callisto
Yes!
anonymous
  • anonymous
You just have do a graph y=x and Tan x and look or use a CAS
Callisto
  • Callisto
Graphing... is not that good...
Callisto
  • Callisto
If graphing works, why not just graph the function sinx / x to show it's a decreasing function?
anonymous
  • anonymous
Well, quite...:-)
anonymous
  • anonymous
What are you doing, Fourier?
Callisto
  • Callisto
No.. limit and derivative.
anonymous
  • anonymous
That's a bit sneaky, hitting you with sinc function...
anonymous
  • anonymous
Let's see , all you have to do is show that the first positive root is at pi....
anonymous
  • anonymous
Sin pi/pi = 0
anonymous
  • anonymous
And anywhere from 0 to pi is not 0
anonymous
  • anonymous
That's good enough to show the first root is at pi, right?
Callisto
  • Callisto
Oh... so you're solving sinx/x=0? If so, yes!
anonymous
  • anonymous
So now you are done, I think (assuming usual smoothness principles).
TuringTest
  • TuringTest
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
Callisto
  • Callisto
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..
TuringTest
  • TuringTest
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
Callisto
  • Callisto
Wait.. how.. why... xcosx - sinx = 0 xcotx -1 = 0 xcotx = 1 Am I doing anything wrong here?
TuringTest
  • TuringTest
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....
Callisto
  • Callisto
Don't think while you eat!
TuringTest
  • TuringTest
I know I have seen a proof that tanx>x in (0,pi/2) ... that would work if I could remember it
Callisto
  • Callisto
This is complicated...
TuringTest
  • TuringTest
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?
Callisto
  • Callisto
Yes!
TuringTest
  • TuringTest
okay, we can prove this using the mean value theorem...
TuringTest
  • TuringTest
the mean value theorem states that f(x)=f(a)+f'(c)(x-a) for some a
TuringTest
  • TuringTest
getting what we need from the formula f(0)=0, f'(c)=sec^2(c) where 0
TuringTest
  • TuringTest
what do you get then by using this in the formula?
Callisto
  • Callisto
f(x)=f(a)+f'(c)(x-a) tanx = 0 + sec^2 (c) (x) tanx = [sec^2 (c)] x Doesn't look good.
TuringTest
  • TuringTest
it is though, because what do we know about sec(c) for 0
Callisto
  • Callisto
0
TuringTest
  • TuringTest
yeah, but the point is that sec(c)>1 for that interval
TuringTest
  • TuringTest
so tanx=sec^2(c)x 0
Callisto
  • Callisto
1< sec^2(c) x < sec^2 (c) x x< tanx Wow!!
TuringTest
  • TuringTest
tadah! :) so that completes your proof
Callisto
  • Callisto
Uh-huh! Why did you think of mean value theorem for this question????
TuringTest
  • TuringTest
It's the only way I know to prove tanx>x in (0,pi/2) there may be other ways
experimentX
  • experimentX
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)
TuringTest
  • TuringTest
very similar argument^
experimentX
  • experimentX
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
TuringTest
  • TuringTest
no apologies necessary, always good to see more than one explanation :)
experimentX
  • experimentX
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.
Callisto
  • Callisto
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!!
experimentX
  • experimentX
Night!!
TuringTest
  • TuringTest
See ya !
anonymous
  • anonymous
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.

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