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Does there exist a function such that

  1. Nov 16, 2009 #1
    If we define a finite difference operator as [tex]\Delta a_n = a_{n+1}-a_n[/tex]

    Can we prove or disprove the existence of a function F, [tex]F:\mathbb{Z}\rightarrow\mathbb{Z}[/tex], such that [tex]\Delta F(g_n)=\frac{\Delta g_n}{ g_n}[/tex], where g is some arbitrary function?



    Edit: fixed Big typo
     
    Last edited: Nov 16, 2009
  2. jcsd
  3. Nov 16, 2009 #2

    CRGreathouse

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    Is that [itex]\exists g\exists F[/itex], [itex]\forall g\exists F[/itex], or [itex]\exists F\forall g[/itex]?
     
  4. Nov 16, 2009 #3
    [tex]\exists F\forall g[/tex]
     
    Last edited: Nov 16, 2009
  5. Nov 16, 2009 #4
    suppose [itex] g_0 = 1, g_1 = 2 [/itex] and [itex]g_2 = 1 [/itex]

    Then [tex] f(2) - f(1) = \frac {g_2 - g_1} {g_1} = 1 [/tex] and

    [tex] f(1) - f(2) = \frac {g_3 - g_2} {g_2} = -1/2 [/tex]

    so there can't be any F for this g
     
  6. Nov 16, 2009 #5
    Can you generalize a non-piecewise function for g that has the values [itex]g_0 = 1, g_1 = 2[/itex], and [itex]g_2 = 1[/itex]?
     
  7. Nov 16, 2009 #6
    Why? Of course there is a quadratic going through these points.

    Is the range of F really [itex]\mathbb{Z} [/itex]?. That is a problem with the division by [itex] g_n [/itex]
     
  8. Nov 17, 2009 #7
    ]

    Now that I think about it, it shouldn't be.
     
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