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Math Help - proving an inequality

  1. #1
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    proving an inequality

    proving an inequality-p5.jpg

    i tried differentiatin both sides, i did that 4 times and i still couldnt find the pattern that gets me to sin any idea how to solve this ??
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  2. #2
    Super Member girdav's Avatar
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    Did you compute \dfrac{d^n}{dx^n}\left(\dfrac 1{x-i}\right) and \dfrac{d^n}{dx^n}\left(\dfrac 1{x+i}\right) ?
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  3. #3
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    I think we are using the same book! The last exercise of Chapter IX, right?
    Quote Originally Posted by PeaceSoul View Post
    i tried differentiatin both sides, i did that 4 times and i still couldnt find the pattern that gets me to sin any idea how to solve this ?
    Show that  \displaystyle \frac{d^n}{dx^n} \left(\frac{1}{x^2+1} \right) = \frac{i(-1)^nn!}{2(x+i)^{n+1}}+\frac{i(-1)^{n+1}n!}{2(x-i)^{n+1}}.

    I avoid the differentiation mess by using the following formula:

     \displaystyle \frac{d^n}{dx^n} (ax+b)^m =  a^n\left(ax+b\right)^{m-n}\prod_{1 \le k \le n}\left(m-k+1\right).
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    yes i did that, but i got no where,,please guide me im really lost :S
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  5. #5
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     \displaystyle  \frac{d^n}{dx^n} \left(\frac{1}{x^2+1} \right) = \frac{i(-1)^nn!}{2(x+i)^{n+1}}+\frac{i(-1)^{n+1}n!}{2(x-i)^{n+1}}  = (-1)^nn!\bigg(\frac{i}{2(x+i)^{n+1}}-\frac{i}{2(x-i)^{n+1}}\bigg).

    Denote the stuff under the brackets by S. Now, let x = \cot{\theta}, then we have:

     \begin{aligned} \displaystyle   S & =  \frac{i}{2(\cot{\theta}+i)^{n+1}}-\frac{i}{2(\cot{\theta}-i)^{n+1}} = \frac{i}{2\left(\frac{\cos{\theta}}{\sin{\theta}}+  i\right)^{n+1}}-\frac{i}{2\left(\frac{\cos{\theta}}{ \sin{\theta}}-i\right)^{n+1}} \\ & = \frac{i}{\frac{2}{ \sin^{n+1}{\theta}}\left(\cos{\theta}+i\sin{\theta  }\right)^{n+1}}-\frac{i}{\frac{2}{ \sin^{n+1}{\theta}}\left(\cos{\theta}-i\sin{\theta}\right)^{n+1}}\\ & = \frac{i\sin^{n+1}\left(\cos{\theta}+i\sin{\theta}\  right)^{-n-1}}{2}-\frac{i\sin^{n+1}\left(\cos{\theta}-i\sin{\theta}\right)^{-n-1}}{2} \end{aligned}

    Recall that \left(\cos{\varphi}+i\sin{\varphi}\right)^{t} = \cos{t\varphi} +i\sin{t\varphi} (De Moivre's theorem), then:

    \displaystyle \begin{aligned} S & = \frac{i\sin^{n+1}\left(\cos[(-n-1)\theta]+i\sin[(-n-1)\theta]\right)}{2}-\frac{i\sin^{n+1}\left(\cos[(-n-1)\theta]-i\sin[(-n-1)\theta]\right)}{2} \\ & = \frac{i\sin^{n+1}\left(\cos[(n+1)\theta]-i\sin[(n+1)\theta]\right)}{2}-\frac{i\sin^{n+1}\left(\cos[(n+1)\theta]+i\sin[(n+1)\theta]\right)}{2} \\&  = \frac{i\sin^{n+1}{\theta}\cos[(n+1)\theta)-i^2sin[(n+1)\theta]-i\sin^{n+1}\cos[(n+1)\theta]-i^2\sin^{n+1}{\theta}\sin[(n+1)\theta]}{2} \\ & = \frac{-2i^2\sin^{n+1}\sin[(n+1)\theta]}{2}= -(-1)\sin^{n+1}{\theta}\sin[(n+1)\theta] = \sin^{n+1}{\theta}\sin[(n+1){\theta}]. \end{aligned}

    Thus S = \sin^{n+1}{\theta}\sin[(n+1)\theta], therefore  \displaystyle \frac{d^n}{dx^n} \left(\frac{1}{x^2+1} \right) = (-1)^nn!\sin^{n+1}{\theta}\sin[(n+1)\theta].
    Last edited by TheCoffeeMachine; March 6th 2011 at 02:32 PM.
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  6. #6
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    Thank you for an amazinggggggggggggg answer !!!!!! You rock
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