A general formula for the half-derivative?

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Discussion Overview

The discussion revolves around the concept of fractional derivatives, specifically the half-derivative operator, and its general solutions. Participants also explore the meaning of closed loop integrals in relation to fractional calculus.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant inquires about a general solution for the half-derivative operator, H, applied to a function f(x).
  • Another participant suggests searching for "fractional derivatives" or "fractional calculus" for more information.
  • There is a discussion about the notation of closed loop integrals, with some participants explaining that it represents integration over a closed path or surface without limits.
  • One participant references a specific formula for the fractional derivative of a simple function from Wikipedia, questioning its applicability beyond certain conditions.
  • A participant asks about the notation for the inverse of the fractional derivative and the opposite of closed loop integrals.
  • Some participants express uncertainty regarding the notation and concepts related to fractional derivatives and closed loop integrals.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding fractional derivatives and closed loop integrals. There is no consensus on a general formula for the half-derivative, and multiple viewpoints are presented without resolution.

Contextual Notes

Some participants note that the general formula for fractional derivatives may depend on specific conditions or definitions, and there is uncertainty about the notation for the inverse of fractional derivatives.

dimension10
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[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.
 
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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" :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:
 
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tiny-tim said:
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" :wink:)
:

But they have not specified a general formula

tiny-tim said:
∫ with a circle means over a closed path or surface, ie without limits :smile:

Thanks.
 
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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.
 
sorry, don't know :redface:
 
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.
 
HallsofIvy said:
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.

Thanks a lot.
 

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