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

- Callisto

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

- schrodinger

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- bahrom7893

For some reason I'm thinking limits but it's probably because of Sinx / x and has nothing to do with this

- Callisto

I think we need to take limit.. but... we have something to do before taking limit?

- 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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## More answers

- Callisto

Well, I don't think we need integration here.

- Callisto

It's because the prof. hasn't taught integration in the lessons yet (even though I have learnt integration before)

- bahrom7893

Ohh Ok. There are some proof of function being greater than the other.. hmm

- 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

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

Well, that's the problem.. How are you going to show it? Finding its derivative?

- anonymous

you cannot just say that the roots are at n*pi?

- Callisto

How so?

- anonymous

You want a proof of that as well...

- 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

I mean x= Tan x is a pain.....

- Callisto

Yes!

- anonymous

You just have do a graph y=x and Tan x and look or use a CAS

- Callisto

Graphing... is not that good...

- Callisto

If graphing works, why not just graph the function sinx / x to show it's a decreasing function?

- anonymous

Well, quite...:-)

- anonymous

What are you doing, Fourier?

- Callisto

No.. limit and derivative.

- anonymous

That's a bit sneaky, hitting you with sinc function...

- anonymous

Let's see , all you have to do is show that the first positive root is at pi....

- anonymous

Sin pi/pi = 0

- anonymous

And anywhere from 0 to pi is not 0

- anonymous

That's good enough to show the first root is at pi, right?

- Callisto

Oh... so you're solving sinx/x=0? If so, yes!

- anonymous

So now you are done, I think (assuming usual smoothness principles).

- 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

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

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

Wait.. how.. why...
xcosx - sinx = 0
xcotx -1 = 0
xcotx = 1
Am I doing anything wrong here?

- 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

Don't think while you eat!

- TuringTest

I know I have seen a proof that tanx>x in (0,pi/2) ... that would work if I could remember it

- Callisto

This is complicated...

- 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

Yes!

- TuringTest

okay, we can prove this using the mean value theorem...

- TuringTest

the mean value theorem states that f(x)=f(a)+f'(c)(x-a) for some a

- TuringTest

getting what we need from the formula
f(0)=0, f'(c)=sec^2(c) where 0

- TuringTest

what do you get then by using this in the formula?

- 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

it is though, because what do we know about sec(c) for 0

- Callisto

0

- TuringTest

yeah, but the point is that sec(c)>1 for that interval

- TuringTest

so
tanx=sec^2(c)x 0

- Callisto

1< sec^2(c)
x < sec^2 (c) x
x< tanx
Wow!!

- TuringTest

tadah! :)
so that completes your proof

- Callisto

Uh-huh! Why did you think of mean value theorem for this question????

- TuringTest

It's the only way I know to prove tanx>x in (0,pi/2)
there may be other ways

- 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

very similar argument^

- 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

no apologies necessary, always good to see more than one explanation :)

- 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

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

Night!!

- TuringTest

See ya !

- 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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