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Math Help - Proof: the following equation has only one real root

  1. #1
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    **solved** Proof: the following equation has only one real root

    Hi. Could somebody show me how to prove that ae^x=1+x+\frac{x^2}{2} has only one real root for a>0 ? All I know so far is that this equation has at least one root because 1+x+\frac{x^2}{2}>0 for all real x.

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    Already solved
    Last edited by wilhelm; December 16th 2012 at 07:00 AM.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Proof: the following equation has only one real root

    I got a solution that i thought was cool so im posting anyways. Now if  ae^x = 1 + x + \frac{x^2}{2} . then  y = ae^x and  y = 1 + x + \frac{x^2}{2} should be solutions to the differential eqn  y' - y = 0 but for  y = 1 + x + \frac{x^2}{2} , we get  y ' - y = \frac{x^2}{2} =0 implies  x = 0 so to have a solution to  ae^x = 1 + x + \frac{x^2}{2} , x = 0 on the right side, so you get  ae^x = 1 which for a > 1, has only 1 real root, namely x = ln(\frac{1}{a})
    Thanks from wilhelm
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