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A general formula for the half-derivative?

  1. Jul 15, 2011 #1
    [tex]{H}^{2}x=\frac{dy}{dx}[/tex]

    Where

    H is the Half-derivative operator.

    My question is:

    Is there a general solution for:

    [tex]\frac{{d}^{\frac{1}{2}}}{{d}^{\frac{1}{2}}x}f(x)[/tex]

    or alternatively

    [tex]H\; f(x)[/tex]


    I have another question. What is the meaning of the symbol:

    [tex]\oint_{\alpha}^{\Omega}f'(x)\; dx[/tex]

    or

    [tex]\oint f'(x)\; dx[/tex]

    I don't get what it means when you have that circle in the centre.

    Thanks.
     
  2. jcsd
  3. Jul 15, 2011 #2

    tiny-tim

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    hi dimension10! :smile:

    the phrase to google is "fractional derivatives" or "fractional calculus" (or just go straight to http://en.wikipedia.org/wiki/Fractional_calculus" [Broken] :wink:)

    and ∫ on its own means between limits (which may be ±∞),

    but ∫ with a circle means over a closed path or surface, ie without limits :smile:
     
    Last edited by a moderator: May 5, 2017
  4. Jul 15, 2011 #3
    But they have not specified a general formula

    Thanks.
     
    Last edited by a moderator: May 5, 2017
  5. Jul 15, 2011 #4

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  6. Jul 15, 2011 #5
    Last edited by a moderator: Apr 26, 2017
  7. Jul 23, 2011 #6
    What is the notation for the opposite of the fractional derivative? As in,

    [tex]\frac{d}{dx}f(x)=f'(x) \Rightarrow \int f'(x) dx =f (x)[/tex]

    What about the fractional derivative?

    [tex]\frac{{d}^{\frac{1}{2}}}{{dx}^{\frac{1}{2}}}f(x)=f'(x)\Rightarrow Statement(P)[/tex]

    Then what is the notation for the statement P?

    Also, what is the opposite of the closed loop-integral thing?

    Thanks.
     
  8. Jul 23, 2011 #7

    tiny-tim

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    sorry, don't know :redface:
     
  9. Jul 23, 2011 #8

    HallsofIvy

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    That same site tiny-tim links to gives another site:
    http://en.wikipedia.org/wiki/Riemann–Liouville_integral
    that gives
    [tex]I^\alpha f(x)= \frac{1}{\gamma(\alpha)}\int_a^x f(t)(x- t)^{\alpha- 1} dt[/tex]
    as the "[itex]\alpha[/itex] integral" for any real [itex]\alpha[/itex].

    The "closed loop integral" is like a definite integral- a number. It does not take one function to another and so has no inverse.
     
  10. Jul 23, 2011 #9
    Thanks a lot.
     
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