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Math Help - Inequality

  1. #1
    Senior Member pankaj's Avatar
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    Inequality

    Prove that for -1<x\leq0

    log_{e}(1+x)\geq\frac{tan^{-1}x}{1+x}
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Inequality

    Quote Originally Posted by pankaj View Post
    Prove that for -1<x\leq0

    log_{e}(1+x)\geq\frac{tan^{-1}x}{1+x}

    Define: f(x)=ln(1+x) and g(x)=1/{1+x^2} and use Mean value theorem - Wikipedia, the free encyclopedia on  [x,0] \subset (-1,0] where x>-1.

    Not sure... trying it...

    Edit:

    It's works!
    Last edited by Also sprach Zarathustra; July 1st 2011 at 08:54 AM.
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  3. #3
    Senior Member pankaj's Avatar
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    Re: Inequality

    What do you do for f(-1).
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  4. #4
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    Re: Inequality

    1st let f(x) = (1+x) ln(1+x) - Tan ' x

    and then using calculus show that f(x) is increasing function in interval ]-1,0]
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  5. #5
    Senior Member pankaj's Avatar
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    Re: Inequality

    Appears to be easier said than done. I have been trying this approach but am not able to reach a definite conclusion
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Inequality

    Quote Originally Posted by pankaj View Post
    Appears to be easier said than done. I have been trying this approach but am not able to reach a definite conclusion
    Read my post again(read the Cauchy mean value theorem).

    c\in (-1,0], [x,0]\subset (-1,0] x>-1

    \frac{f(x)-f(0)}{g(x)-g(0)}=\frac{\ln(1+x)-0}{\arctan(x)-0}=\frac{f'(c)}{g'(c)}=\frac{\frac{1}{c+1}}{\frac{  1}{1+c^2}}=\frac{1+c^2}{c+1}>\frac{1}{1+x}


    Prove the last inequality.
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  7. #7
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    Re: Inequality

    the said inequality does hold in the given interval eg if x = -0.5 ln( 0.5) < Tan ' (0.5) / 0.5
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  8. #8
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    Re: Inequality

    Quote Originally Posted by pankaj View Post
    Prove that for -1<x\leq0

    \log_{e}(1+x)\geq\frac{\tan^{-1}x}{1+x}
    Here is one way to do it. Multiply by 1+x (which doesn't change the inequality since 1+x>0). Then we want to show that

    f(x) \overset{\text{d{e}f}}{=} (1+x)\ln(1+x) - \tan^{-1}x \geqslant 0\quad(-1<x\leqslant0).

    Since f(0) = 0 it will be sufficient to show that f(x) is decreasing in that interval. So we want to show that its derivative is negative. But

    f'(x) = \ln(1+x) + 1 - \frac1{1+x^2}.

    We want to show that f'(x)\leqslant0 for -1<x\leqslant0. To do that, repeat the previous argument. Since f'(0)=0 it is sufficient to show that f'(x) is increasing in that interval. So we want to show that its derivative is positive. But

    f''(x) = \frac1{1+x} + \frac{2x}{(1+x^2)^2} = \frac{(1+x)^2 + 3x^2+x^4}{(1+x)(1+x^2)^2},

    which is clearly positive for x>1.
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