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Are fractional polynomials linearly independent?

  1. Apr 21, 2014 #1
    i.e., does the set of functions of the form,

    [itex] \{ x^{\frac{n}{m}}\}_{n=0}^{\infty} [/itex] for some fixed [itex] m [/itex] produce a linearly independent set? Either way, can you give a brief argument why or why not?

    Just curious :)
     
  2. jcsd
  3. Apr 21, 2014 #2

    disregardthat

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    If you have any linear combination between a finite number of these elements, this relation can be written as

    [tex]a_ny^n+a_{n-1}y^{n-1}+...+a_1y = 0[/tex]

    where [itex]y = x^{\frac{1}{m}}[/itex] and a_n is non-zero. However, a polynomial of degree n has exactly n zeroes, which means that this is impossible.
     
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