Is the Definition for Fractional Derivative Correct?

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The definition for fractional derivative presented in the discussion is confirmed as correct. By applying the change of variable r=-p, the expression for the fractional derivative is derived from the fractional integral, maintaining mathematical validity. The resulting formula, D^{p}=\frac{1}{\Gamma(-p)}\int_c^xf(t)(x-t)^{-(p+1)}, aligns with established forms of fractional derivatives, incorporating the gamma function appropriately. This conclusion is supported by references to educational resources that explain fractional derivatives in detail.

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from the expression for a Fractional integral of arbitrary order:

[tex]D^{-r}=\frac{1}{\Gamma(r)}\int_c^xf(t)(x-t)^{r-1}[/tex]

if we set r=-p then we would have for the Fractional derivative:

[tex]D^{p}=\frac{1}{\Gamma(-p)}\int_c^xf(t)(x-t)^{-(p+1)}[/tex]

is my definition correct?..i mean if its correct to introduce the change of variable r=-p to obtain fractional derivative form fractional integral...

where [tex]\pi/{\Gamma(-p)}=\Gamma(p+1)sen(p+1)\pi[/tex]
 
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The idea of *a* fractional derivative is well known. If you look over the examples sheets for the first year calculus course at www.maths.bris.ac.uk/~madve[/URL] you'll find it explained on there somewhere I think.
 
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Based on the given expression, the definition for fractional derivative appears to be correct. The change of variable r=-p is a valid mathematical operation and does not change the validity of the definition. The resulting expression for the fractional derivative also follows the general form for fractional derivatives, with the integration being over a specific interval and involving the function f(t) and its derivatives. Furthermore, the use of the gamma function in the denominator is also consistent with the definition of fractional derivatives. Therefore, it can be concluded that the given definition for fractional derivative is correct.
 

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