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Math Help - Proving Inequality

  1. #1
    Member mohammadfawaz's Avatar
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    Proving Inequality

    how can we prove that \int_0^1(1-x^2)^ndx \leq \frac{\pi}{2\sqrt{n}}

    I think the proof starts by letting x=cos(u) but I didn't reach anything
    Thanks
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by mohammadfawaz View Post
    how can we prove that \int_0^1(1-x^2)^ndx \leq \frac{\pi}{2\sqrt{n}}

    I think the proof starts by letting x=cos(u) but I didn't reach anything
    I don't know where the \pi/2 comes from, but there is a fairly simple way to prove a stronger inequality.

    Let I_n = \int_0^1(1-x^2)^n\,dx. Integrate by parts to get I_n = \Bigl[x(1-x^2)^n\Bigr]_0^1 + 2n\int_0^1x^2(1-x^2)^{n-1}dx = 2n(I_{n-1}-I_n). Therefore I_n = \frac{2n}{2n+1}I_{n-1}.

    Using that recurrence relation, you can show by induction that I_n\leqslant \frac1{\sqrt{n+1}}. In fact, the base case n=0 is obvious. The inductive step amounts to the inequality \frac{2n+2}{2n+3}\,\frac1{\sqrt{n+1}}\leqslant \frac1{\sqrt{n+2}}, which you can check by squaring both sides.

    Since \frac1{\sqrt{n+1}} is clearly less than \frac\pi{2\sqrt n} , that is a better inequality than what was asked for.
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