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Calculus with D operator

  1. Nov 20, 2013 #1
    Given that D²f(x) = g(x), one form that eliminate the second derivate is integrating the equation: ∫∫D²f(x)dx² = ∫∫g(x)dx². But, and if I try so:

    [tex]\\ \sqrt{D^2f(x)}=\sqrt{g(x)} \\ D\sqrt{f(x)}=\sqrt{g(x)} \\ PD\sqrt{f(x)}=P\sqrt{g(x)} \\ \sqrt{f(x)}=P\sqrt{g(x)} \\ f(x)=[P\sqrt{g(x)}]^2 \\ f(x)=[\int \sqrt{g(x)}dx]^2[/tex]

    Is it works?
  2. jcsd
  3. Nov 20, 2013 #2


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    Going from the first to the second line you are claiming
    [tex] \sqrt{ f''(x)} = \frac{d}{dx} \sqrt{f(x)} [/tex]

    If we pick a random function, say f(x) = x4, the left hand side of this is
    [tex] \sqrt{12 x^2} = \sqrt{12} |x| [/tex]
    and the right hand side is
    [tex] \frac{d}{dx} x^2 = 2x [/tex]

    so we see they're not equal at all. The D operator is nice for seeing how you are using the linearity of the derivative but don't confuse it for an honest to goodness number that can be manipulated the same in every way.
  4. Nov 20, 2013 #3


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    You can't write ##dx^2## like that. It means integrate with respect to ##x^2## instead of integrating with respect to ##x##.
  5. Nov 21, 2013 #4


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    Every definition of the "differential operators" will include showing which ordinary arithmetic operators work for differential operators as well. Of course you can only use arithmetic operators that have the property that they can be applied to differential operators. Have you seen any proof that the "square root" operator has that property?
  6. Nov 21, 2013 #5
    Not always there is demonstrations for certain relations. I never saw a demo for this but I know it's valid.

    [tex]\\ f^D=f^D \\ log(f^D)=log(f^D) \\ log(f^D)=Dlog(f) \\ f^D=exp(\frac{f'}{f})[/tex]
    PS: f^D = f*(x) = geometric derivate
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