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A Freshman Dream Quotient Rule

  1. Jun 27, 2016 #1
    I have been interested in this idea of the FDQR. This idea states the following. upload_2016-6-27_9-12-0.png

    I have been trying to see if there is some 2 functions which make this true, but have not found it on research or with trying functions. Does anyone have any insight on this. I think it is just neat and want to know more about what people know about it.
     
  2. jcsd
  3. Jun 27, 2016 #2
    What ever satisifies:

    [tex] g' = g^2 [/tex]

    and

    [tex] f' = f \frac{g^2}{g-1} [/tex]
     
  4. Jun 27, 2016 #3

    micromass

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    There are infinitely many solutions. Given any function ##g##, the associated function can be found by solving the following ODE:

    [tex]f' g (g - g') = (g')^2 f[/tex]

    So for example, let ##g(x) = e^x##. Then this reduces to ##f(x) = 0##.
    If ##g(x) = x^2##, then
    [tex] f'(x) x^2 (x^2 - 2x) = 4x^2 f(x)[/tex]
    or
    [tex]\frac{f'(x)}{f(x)} = \frac{4}{x(x-2)}[/tex]
    Integrating yields
    [tex]\log(f(x)) = 2\log(2-x) - 2\log(x)[/tex]
    or
    [tex]f(x) = \frac{(2-x)^2}{x^2}[/tex]
    In general, given any function ##g## not satisfying ##g=g'## or ##g=0##, we can find an associated function ##f## by
    [tex]f(x) = g(x) + e^{\int \frac{g'}{g-g'}}[/tex]
     
  5. Jun 29, 2016 #4

    Ssnow

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    I think there a minus ##f'g(g-g')=-g'^{2}f## ...
     
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