A general formula for the half-derivative?

[dimension10]

{H}^{2}x=\frac{dy}{dx}

Where

H is the Half-derivative operator.

My question is:

Is there a general solution for:

\frac{{d}^{\frac{1}{2}}}{{d}^{\frac{1}{2}}x}f(x)

or alternatively

H\; f(x)


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

\oint_{\alpha}^{\Omega}f'(x)\; dx

or

\oint f'(x)\; dx

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

Thanks.

 

[tiny-tim]

hi dimension10!

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

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

but ∫ with a circle means over a closed path or surface, ie without limits

 

[dimension10]

hi dimension10!

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

But they have not specified a general formula

∫ with a circle means over a closed path or surface, ie without limits

Thanks.

 

[tiny-tim]

But they have not specified a general formula

isn't it given at the bottom of http://en.wikipedia.org/wiki/Fractional_calculus#Fractional_derivative_of_a_simple_function" as D^{1/2}f(x) = \frac{1}{\Gamma(1/2)}\frac{d}{dx}\int_0^x\frac{f(t)}{\sqrt(x-t)}dt

 

[dimension10]

isn't it given at the bottom of http://en.wikipedia.org/wiki/Fractional_calculus#Fractional_derivative_of_a_simple_function" as D^{1/2}f(x) = \frac{1}{\Gamma(1/2)}\frac{d}{dx}\int_0^x\frac{f(t)}{\sqrt(x-t)}dt

Oh. I though that was only for f(a)=euler gamma(a)

 

[dimension10]

What is the notation for the opposite of the fractional derivative? As in,

\frac{d}{dx}f(x)=f'(x) \Rightarrow \int f'(x) dx =f (x)

What about the fractional derivative?

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

Then what is the notation for the statement P?

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

Thanks.

 

[tiny-tim]

sorry, don't know

 

[HallsofIvy]

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

The "closed loop integral" is like a definite integral- a number. It does not take one function to another and so has no inverse.

 

[dimension10]

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

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.