Prove this inequality via graph

[utkarshakash]

Homework Statement

2x>3sinx-xcosx, 0<x<∏/2

Homework Equations

The Attempt at a Solution

One possible way is to draw the graph of the functions and compare but plotting a graph manually is not easy in this case. I want some other methods.

 

[SteamKing]

You know that sine and cosine vary in amplitude between 0 and 1 on the interval in question. If you substitute the end values of the interval into the inequality, that should show that the inequality does not hold.

 

[utkarshakash]

You know that sine and cosine vary in amplitude between 0 and 1 on the interval in question. If you substitute the end values of the interval into the inequality, that should show that the inequality does not hold.

What you are suggesting to me is just the verification of the result. Proving a result does not mean plugging the values and checking it.

 

[jeppetrost]

Okay this is a bit messy, but it's the best I came up with. Say we define f(x)= 3sinx - xcosx, and we want to show f(x)<2x. Well, we see that at least the endpoints are all right [ f(0)=0 (okay since we don't actually consider 0), f(pi/2)=3 < pi ]. So if we can just show that the function never goes over the line 2x, we're done. Well, if we differentiate both sides, we can equivalently show that f '(x) < 2. Simply doing the differentiation gives us f '(x) = 2cosx-xsinx. We again check that the end points are okay: f '(0)= 2 (okay), f '(pi/2) = pi/2<2 (also okay). Now we need to show that f '(x) doesn't become greater than 2 (this is actually too strict, but we can anyway). So we calculate f ''(x) = x cos x-sinx and want to show that this is always negative. Well, at x=0 we have f ''(0) =0, so that's good. Now we go even further and find f '''(x) = -xsinx. This is obviously negative, which means that f ''(x) is also negative. This in turn means that f '(x) < 2, which means that f(x)<2x.

I hope you follow my train of thoughts. You just go deeper and deeper in the derivatives until you see something you know, and then go back up.

 

[andrien]

you can just easily go with f(x)=2x-3sinx+xcosx,you can see at x=0,it is zero.so just differentiate and try to show that f'(x)>0 for the given interval.

 

[SteamKing]

What you are suggesting to me is just the verification of the result. Proving a result does not mean plugging the values and checking it.

If the inequality doesn't hold, then you are wasting time trying to prove that it does.
Verification that a relation holds at just one point in an interval is insufficient to prove that it holds at all points in the interval. However, showing that a relation fails to hold at just one point in an interval is sufficient to show that the relation does not hold.

 

[jeppetrost]

If the inequality doesn't hold, then you are wasting time trying to prove that it does.
Verification that a relation holds at just one point in an interval is insufficient to prove that it holds at all points in the interval. However, showing that a relation fails to hold at just one point in an interval is sufficient to show that the relation does not hold.

However, it does hold, so I don't exactly see your point.

 

[utkarshakash]

Okay this is a bit messy, but it's the best I came up with. Say we define f(x)= 3sinx - xcosx, and we want to show f(x)<2x. Well, we see that at least the endpoints are all right [ f(0)=0 (okay since we don't actually consider 0), f(pi/2)=3 < pi ]. So if we can just show that the function never goes over the line 2x, we're done. Well, if we differentiate both sides, we can equivalently show that f '(x) < 2. Simply doing the differentiation gives us f '(x) = 2cosx-xsinx. We again check that the end points are okay: f '(0)= 2 (okay), f '(pi/2) = pi/2<2 (also okay). Now we need to show that f '(x) doesn't become greater than 2 (this is actually too strict, but we can anyway). So we calculate f ''(x) = x cos x-sinx and want to show that this is always negative. Well, at x=0 we have f ''(0) =0, so that's good. Now we go even further and find f '''(x) = -xsinx. This is obviously negative, which means that f ''(x) is also negative. This in turn means that f '(x) < 2, which means that f(x)<2x.

I hope you follow my train of thoughts. You just go deeper and deeper in the derivatives until you see something you know, and then go back up.

Thanks. Your method was pretty good.

 

[SteamKing]

Mea culpa.

 

[noc_njz]

Why not use taylor series?