Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Lets define new differintegral formula as

$$\mathbb{D}^s_xf(x)= \sum_{m=0}^{\infty} \binom {s}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$

or, equivalently,

$$\mathbb{D}^s_xf(x)= \lim_{t\to s} \lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f^{(k)}(x)}{(t-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(t-k) k!(n-k)!}}$$

with the conditions that the above expression converges, and $(\mathbb{D}^{-s}_xf(x))^{(s)}=f(x)$ for each x and natural s.

Will this formula where the conditions are met give the same results as Riemann–Liouville differintegral, Grunwald–Letnikov differintegral and Weyl differintegral?

Update

I have started a bounty.

share|improve this question
    
@Gerald Edgar do you have something to say here? –  Anixx May 17 '13 at 11:24
1  
Why in the second formula there is the limit $t\to x$? What will change if to take in the rhs $f^{(k)}(x)$ instead of $f^{(k)}(t)$? –  Andrew May 25 '13 at 6:56
    
@Andrew indeed it is typo. I have corrected. –  Anixx May 25 '13 at 16:05
    
One of the ways to do fractional derivatives and integrals is to do them as multiplications in Laplace transform space. A drawback might be the difficulty of computing the inverse Laplace transform to get back. If you are using all the derivatives of the function, as here, then maybe you can do the inverse Laplace transform using Post's inversion formula en.wikipedia.org/wiki/Post%27s_inversion_formula –  Gerald Edgar May 31 '13 at 13:28

1 Answer 1

Your formula is based on the identity for $s\in\mathbb N$, $$ \partial^s=(1+(-1)+\partial)^s=\sum_{m\ge 0}C_s^m((-1)+\partial)^m=\sum_{m\ge 0\atop 0\le k\le m}C_s^mC_m^k(-1)^{m-k}\partial^k. \tag{$\flat$}$$ It is interesting to compare it to the Fourier multiplier method $$ \partial =(2iπ\xi)^s=(2π)^s\exp{(s\text{Log}(i\xi))} $$ with the principal determination of the logarithm ($i\xi\in i\mathbb R$). We have $$ \widehat{\partial^s u}=(2iπ\xi)^s\hat u(\xi),\quad (\partial^s u)(x)=\int e^{2iπ x\xi}(2iπ\xi)^s\hat u(\xi)d\xi. \tag{$\sharp$} $$ Of course, using here the notation $D=\frac{\partial}{2iπ\partial x}$, we can also write as you did $$ \partial^s=(1+(-1)+2iπ D)^s $$ but the radius of convergence of $(1+z)^s$ is 1 and you will be forced to drastic assumptions on the decay of the derivatives to have something convergent. Note however that the coefficients in the expansion are the same as in $(\flat)$. A very natural requirement is that $\partial^s$ sends the homogeneous Sobolev space $\dot H^m$ into $\dot H^{m-s}$: we can define in $n$ dimensions for $m>-n/2$ $$\dot H^m=\{u\in\mathscr S'(\mathbb R^n),\hat u\in L^2_{loc}, \vert\xi\vert ^m \hat u\in L^2(\mathbb R^n)\}. $$ In one dimension, for $s\in\mathbb R$, $m>s+\frac12$, Definition $(\sharp)$ provides the continuity of $\partial^s$ from $\dot H^m$ into $\dot H^{m-s}$. I believe that with Formula $(\flat)$, since the homogeneity is not obvious, it should not be so easy to prove that continuity property. On the other hand, for $s\in \mathbb R$, the property $$ \partial ^s\partial^{-s} u=u $$ is obvious with $(\sharp),$ provided we assume that $u$ belongs to $\dot H^{m}$ with $m>\frac12+\vert s\vert$, which is a mild requirement compared to the decay estimates on derivatives that you will need to have $(\flat)$ convergent.

share|improve this answer
    
Where is (♯) and (♭)? –  Anixx May 31 '13 at 21:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.