Freshman Dream Quotient Rule

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  • Thread starter JasMath33
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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
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What ever satisifies:

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

and

[tex] f' = f \frac{g^2}{g-1} [/tex]
 
  • #3
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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]
 
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  • #4
Ssnow
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##f′g(g−g′)=(g′)^2f##
I think there a minus ##f'g(g-g')=-g'^{2}f## ...
 
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