Tuesday, 18 December 2012

calculus - Proving frac2pixlesinxlex for xin[0,fracpi2]




Prove 2πxsinxx for x[0,π2].





I tried to do this in two ways, I'm not sure about CMVT and I have a problem with the other way.






Using Cauchy's MVT:



RHS:
sinxxsinxx1
So define:
f(x)=sinx, g(x)=x then from CMVT:

f(π2)g(π2)=f(c)g(c)=cosc
and from the fact that c is between 0 and π/2cosc1.



LHS: In the same manner but here I run into some trouble:
2πxsinx2xπsinx1
So:
f(π2)g(π2)=f(c)g(c)1sinπ2=2πcosc
Here actually
1sinπ2=1 so it's also 1




Is it correct to use CMVT like this ?






The other way:



We want to show: f(x)=sinxx<0 and g(x)=2xπsinx<0 by deriving both it's easy to show that the inequality stands for f but for g it isn't so obvious that g(x)=2πcosx is negative. In fact for x=π2 it's positive. Please help figure this out.







This is the same The sine inequality 2πxsinxx for $0 but all the answers there are partial or hints and I want to avoid convexity.



Note: I can't use integrals.


Answer



To show that sinxx you can apply the Cauchy mean value theorem. (Note that you want to show the inequality for any x[0,π2]. )Consider, as you have done, f(x)=sinx and g(x)=x. Apply the theorem in the interval [0,x] and you will get the inequality, as a consequence of cosc1. Indeed, there exists c(0,x) such that sinx=g(c)(f(x)f(0))=f(c)(g(x)g(0))=(cosc)xx.



To show the other inequality consider f(x)=sinx2πx. We have that f(0)=f(π/2)=0. Since f is continuous and [0,π/2] is compact it attains a global minimum. If the minimum is not achieved at the extrema of the interval then it belongs to the open interval (0,π/2). Let c be the point where f achieves its global minimum. Then f(c)0, but f(c)=sinc<0 for any c(0,π/2). So the minimum value is f(0)=f(π/2)=0, from where 0f(x)=sinx2πx, which shows the inequality.


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