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Math Help - Find the the number of multiple roots of a equation

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    Find the the number of multiple roots of a equation

    Let f(x)= 1+ x + \frac{x^2}{2!} + \frac{x^3}{3!}  +...+ \frac{x^{2n}}{(2n)!}.
    How many multiple roots does the equation f(x)=0 have?

    Can someone pls explain how to solve this? Your help is very much appreciated.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Find the the number of multiple roots of a equation

    Quote Originally Posted by shiny718 View Post
    Let f(x)= 1+ x + \frac{x^2}{2!} + \frac{x^3}{3!}  +...+ \frac{x^{2n}}{(2n)!}.
    How many multiple roots does the equation f(x)=0 have?
    Suppose a is a multiple root of f(x)=0, then f(a)=0 and f'(a)=0. This implies f(a)-f'(a)=a^{2n}/(2n)!=0 or equivalently a=0. But f(0)\neq 0, that is, f(x)=0 has no multiple roots.
    Thanks from shiny718 and awkward
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