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Thread: An inequality

  1. #1
    MHF Contributor

    May 2008

    An inequality

    The upper bound given in this problem is a good approximation for the integral:

    Problem: Prove that \int_0^1 e^{\cos x} \ dx \leq e \sqrt{2} \tan^{-1}(\sqrt{2}/2).
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  2. #2
    Senior Member Sampras's Avatar
    May 2009
    By the Rectangle Rule,  \int_{0}^{1} e^{\cos x} \ dx \approx e^{\cos \frac{1}{2}} . Or we know that  e^{\cos x} has a bounded derivative over  [0,1] . Now by the MVT,  xf'(y_x) = f(x)-f(0) for some  y_x \in [a,x] . Also,  x<1 . Then

     \left|\int_{0}^{1} e^{\cos x} \ dx -e^{\cos 1} \right| = \left|\int_{0}^{1} x f'(y_x) \ dx \right|

     \left|\int_{0}^{1} e^{\cos x} \ dx -e^{\cos 1} \right| \leq \frac{1}{2} \sup_{0 \leq x \leq 1} |-e^{\cos x} \sin x|
    Last edited by Sampras; Jul 6th 2009 at 12:06 PM.
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  3. #3
    Mar 2009
    OK, I have come up with an incredibly contrived solution suggested by the answer. Clearly it is equivalent to

    \int_0^1\mathrm e^{\cos x-1}\mathrm dx\leq\int_0^1\frac 2{2+x^2}\mathrm dx, so this suggests the inequality \mathrm e^{\cos x-1}\leq\frac 2{2+x^2}. If there's a better way...

    Here we go. Start with \sin x\leq x and \sin^2 x+\cos x=1+\cos x(1-\cos x)\geq 1 on [0,1]. Therefore \sin x\leq x(\sin^2 x+\cos x).

    This rearranges to \frac{\sin x-x\cos x}{\sin^2 x}\leq x. The LHS is the derivative of \frac x{\sin x}, so integrate both sides from 0 to t, where t\in[0,1].

    Thus \frac t{\sin t}-1\leq \frac{t^2}2 and so \sin t\geq\frac t{1+t^2/2}. Integrate this from 0 to x and 1-\cos x\geq \ln(1+x^2/2) pops up.

    Guess what? \cos x-1\leq\ln\left(\frac2{2+x^2}\right), and we arrive at our destination. Where's my hat? Rabbit, anyone?
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