Prove that
I thought of integrating the inequality from 0 to 1 but the inequality isn't strict. I'm not sure what I could do to make it strict; I think a different approach is needed.
Any ideas? Thanks.
Hello,
I don't know how you got the inequality... But I think it's a good start.
You can prove that it is a strict inequality... The way I found looks a bit messy ><
Prove that there isn't such that (we can exlude 1, because it's not equal when x=1)
Consider

But since x is in [0,1), 2x
~~~~~~~~~~
Now what's the sign of ?
Differentiate and you'll get :
, which is always positive, for any x in [0,1)
Thus is increasing in [0,1)
So and we have
So there exists a unique value such that (by the intermediate value theorem, or something like that)
and for any and for any
But . So they have the same sign.
~~~~~~~~~~
Hence :
So , for any
Now you're up to prove that :
We know that is such that , that is to say :
By , we have :
, where
Note that << MISTAKE
Hence
We can easily see that , because .
And similarly, we can see that , because
Thus
I hope this ... erm helps ^^
Edit : okay, because of the mistake pointed out above, the conclusion is false... I'm too tired to think again about it, sorry ><
Hiya Moo,
pi/2 is the area of the semicircle with radius 1, which we can write as the integral from 1 to 1 of sqrt(1x^2), and since the function is even, we can write this as the integral from 0 to 1 of sqrt(1x^2). I think a function like 2/(x^2 + 1) would work, too.
I haven't checked your working yet (I will now), but can't we simply substitute x = 0 into the equation which results in equality? Hence the inequality can't be strict over [0,1]. I might be making a silly mistake, though.
Thanks.
Ah, you noticed. It happens to the best of us!
Quick followup: it turns out that even though the inequality isn't strict, we can make the integral inequality strict! The following theorem is what was needed:
Suppose f => 0 for all x in [a,b] and f is continuous at one point c in [a,b] such that f(c) > 0. Then