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Math Help - Functional equeation

  1. #1
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    Functional equeation

    Is it possible to find nonzero solution of functional equation

    <br />
(1+4x)f(2x)=2f(x+2x^2)<br />

    defined and analitic in some neighbourhood of 0
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  2. #2
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    Quote Originally Posted by ns1954 View Post
    Is it possible to find nonzero solution of functional equation

    <br />
(1+4x)f(2x)=2f(x+2x^2)<br />

    defined and analitic in some neighbourhood of 0
    never mind!
    Last edited by NonCommAlg; January 28th 2010 at 07:01 AM.
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  3. #3
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    After taking first derivate in FE, by shain rule, we have relation

    4f(2x)+2(1+4x)[f'(2x)-f'(x+2x^2)]=0 or 4f(2x)+2(1+4x)[g(2x)-g(x+2x^2)]=0, by your notion.

    I can't see how you derived g(2x)=g(x+2x^2) from this.
    Last edited by ns1954; January 24th 2010 at 04:51 AM.
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