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Math Help - A differential equation

  1. #1
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    A differential equation

    I won't be posting new problems for a couple of weeks. This one is nice:


    Suppose n \geq 1 and f: (a,b) \longrightarrow \mathbb{R} is a C^n function (see here for the definition) and f(x)f'(x)f''(x) \cdots f^{(n)}(x) = 0, for all a < x < b. Show that f is a polynomial of degree at most n-1.
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  2. #2
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    Quote Originally Posted by NonCommAlg View Post
    I won't be posting new problems for a couple of weeks. This one is nice:


    Suppose n \geq 1 and f: (a,b) \longrightarrow \mathbb{R} is a C^n function (see here for the definition) and f(x)f'(x)f''(x) \cdots f^{(n)}(x) = 0, for all a < x < b. Show that f is a polynomial of degree at most n-1.
    if we assume that f(x) is a polynomial of the degree n-1,

    we get,



    If f(x) has the degree n-2,

    then the (n-1)th derivative of f becomes zero.

    The same holds for all degrees <(or equal to) (n-1)

    hence, the given product always becomes vanishes.


    However, if we now start making the degrees > (n-1)

    Lets assume that the function is a polynomial of degree n.

    I'll take the simplest case here. assuming



    Then, the nth derivative of the function:



    And all the other derivatives would be of the form:




    it follows that

    that none of the derivatives will now be zero hence, the product cannot be zero.
    Last edited by Sine; May 3rd 2009 at 07:29 PM. Reason: added the working
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