achild’sgardenoffractionalderivatives-外文文献(编辑修改稿)

achild’sgardenoffractionalderivatives-外文文献(编辑修改稿)内容摘要：

er offunctions. Given any function that can be expanded in a Taylor’s series in powers of x, fx axnnnb g ==∞∑0,assuming we can differentiate term by term we get (6) Dfx aDxnnnααb g ==∞∑0=+−+=∞−∑annxnnnΓΓ110b gb gαα.This final expression presents itself as a possible candidate for the definition of thefractional derivative for the wide variety of functions that can be expanded in a Taylor’sseries in powers of x . However, we will soon see that it leads to contradictions.Questions Is there a meaning for Dfxαbgin geometric terms?65. A mysterious contradiction. In Section 2, we wrote the fractional derivative of exas (7) De exxα= .Let us now pare this expression with (6) to see if they agree. From the Taylor Series,enxxnn==∞∑10!, (6) gives (8) Dexnxnnααα=−+−=∞∑Γ()10. But (7) and (8) do not match unless α is a whole number! When α is a whole number,the right side of (8) will be the series for ex, with different indexing. But when α is nota whole number, we have two entirely different functions. We have discovered acontradiction that historically has caused great problems. It appears as though ourexpression (1) for the fractional derivative of the exponential is inconsistent with ourformula (6) for the fractional derivative of a power. This inconsistency is one reason the fractional calculus is not found in elementarytexts. In the traditional calculus, where α is a whole number, the derivative of anelementary function is an elementary function. Unfortunately, in the fractional calculusthis is not true. The fractional derivative of an elementary function is usually a highertranscendental function. For a table of fractional derivatives see [3]. At this point you may be asking what is going on? The mystery will be solved in latersections. Stay tuned... .6. Iterated integrals. We have been talking about repeated derivatives. Integrals canalso be repeated. We could write Dfx fxdx−=z1b g b g , but the right hand side is7indefinite. We will instead write Dfx ftdtx−=z10bg bg . The second integral will thenbe Dfx ftdtdttx−=zz2100122b g b g . The region of integration is the triangle in figure 1.Figure 1If we interchange the order of integration, the righthand diagram in Figure 1 shows that Dfx ftdtdttxx−=zz210211bg bg .Since ft1b g is not a function of t2, it can be moved outside the inner integral, soDfx ft dtdt ft xtdtxtxx−==−zz z21021 101bg bg bgb gorDfx ftxtdtx−=−z20bg bgb g .Using the same procedure we can show thatDfx ftxtdtx−=−z30212bg bgb g , Dfx ftxtdtx−=⋅−z403123bg bgb g ,8and in general, Dfxnft x t dtnxn−−=−−zbg bgb g1101()!. Now, as we have previously done, let us replace the −n with arbitrary α and thefactorial with the gamma function to get(9)。