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Tom Copeland
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Extensions of the Leibnitz rule to fractional calculus of a type similar to that Gessel depicts were investigated by Osler in the 1970s. See, e.g., "Fractional derivatives and special functions" by Lavoie, Osler, and Tremblay.

One can even make expansions of the type

$$D^{-1} f(x)g(x) = (D_L+D_R)^{-1} f(x)g(x) = D_L^{-1}(1+D_R/D_L)^{-1}f(x)g(x)$$

$$ = \sum_{n \geq 0} \binom{-1}{n} [D^{-n-1}f(x)][D^ng(x)]= \sum_{n \geq 0} (-1)^n [D^{-n-1}f(x)][D^ng(x)]$$

for some classes of functions with, e.g., $D^{-1}f(x) = \int_0^x f(t)dt$ and the indicessubscripts on $D$ designating action on either the Left or Right factor. Try

For example, let $f(x) = x^{q-r}$ and $g(x) = x^{p+r}$ . Then

$$D^{-1}f(x)g(x) = D^{-1}x^{p+q} = \frac{x^{p+q+1}}{p+q+1}$$

and

$$\sum_{n \geq 0} (-1)^n [D^{-n-1}f(x)][D^ng(x)] = \sum_{n \geq 0} (-1)^n [D^{-n-1}x^{q-r}][D^nx^{p+r}]$$

$$ =x^{p+q+1} \sum_{n \geq 0} (-1)^n \frac{(q-r)!}{(q-r+n+1)!} \frac{(p+r)!}{(p+r-n)!} , $$

so the generalized Leibnitz rule holds in this case with the generalized integration $D^{-1} \frac{x^s}{s!} = \frac{x^{s+1}}{(s+1)!}$ for $x >0$ when

$$\frac{1}{p+q+1} =\sum_{n \geq 0} (-1)^n \frac{(q-r)!}{(q-r+n+1)!} \frac{(p+r)!}{(p+r-n)!} $$

which is apparently valid for $p+q +1 > 0.$$p+q +1 > 0$

(see p. 268 in Section 105 of Die Gammafunktion by Niels Nielsen for convergence of this sum).

Extensions of the Leibnitz rule to fractional calculus of a type similar to that Gessel depicts were investigated by Osler in the 1970s. See, e.g., "Fractional derivatives and special functions" by Lavoie, Osler, and Tremblay.

One can even make expansions of the type

$$D^{-1} f(x)g(x) = (D_L+D_R)^{-1} f(x)g(x) = D_L^{-1}(1+D_R/D_L)^{-1}f(x)g(x)$$

$$ = \sum_{n \geq 0} \binom{-1}{n} [D^{-n-1}f(x)][D^ng(x)]= \sum_{n \geq 0} (-1)^n [D^{-n-1}f(x)][D^ng(x)]$$

for some classes of functions with, e.g., $D^{-1}f(x) = \int_0^x f(t)dt$ and the indices on $D$ designating action on either the Left or Right factor. Try $f(x) = x^{q-r}$ and $g(x) = x^{p+r}$ . Then

$$D^{-1}f(x)g(x) = D^{-1}x^{p+q} = \frac{x^{p+q+1}}{p+q+1}$$

and

$$\sum_{n \geq 0} (-1)^n [D^{-n-1}f(x)][D^ng(x)] = \sum_{n \geq 0} (-1)^n [D^{-n-1}x^{q-r}][D^nx^{p+r}]$$

$$ =x^{p+q+1} \sum_{n \geq 0} (-1)^n \frac{(q-r)!}{(q-r+n+1)!} \frac{(p+r)!}{(p+r-n)!} , $$

so the generalized Leibnitz rule holds in this case when

$$\frac{1}{p+q+1} =\sum_{n \geq 0} (-1)^n \frac{(q-r)!}{(q-r+n+1)!} \frac{(p+r)!}{(p+r-n)!} $$

which is apparently valid for $p+q +1 > 0.$

Extensions of the Leibnitz rule to fractional calculus were investigated by Osler in the 1970s. See, e.g., "Fractional derivatives and special functions" by Lavoie, Osler, and Tremblay.

One can even make expansions of the type

$$D^{-1} f(x)g(x) = (D_L+D_R)^{-1} f(x)g(x) = D_L^{-1}(1+D_R/D_L)^{-1}f(x)g(x)$$

$$ = \sum_{n \geq 0} \binom{-1}{n} [D^{-n-1}f(x)][D^ng(x)]= \sum_{n \geq 0} (-1)^n [D^{-n-1}f(x)][D^ng(x)]$$

for some classes of functions with, e.g., $D^{-1}f(x) = \int_0^x f(t)dt$ and the subscripts on $D$ designating action on either the Left or Right factor.

For example, let $f(x) = x^{q-r}$ and $g(x) = x^{p+r}$ . Then

$$D^{-1}f(x)g(x) = D^{-1}x^{p+q} = \frac{x^{p+q+1}}{p+q+1}$$

and

$$\sum_{n \geq 0} (-1)^n [D^{-n-1}f(x)][D^ng(x)] = \sum_{n \geq 0} (-1)^n [D^{-n-1}x^{q-r}][D^nx^{p+r}]$$

$$ =x^{p+q+1} \sum_{n \geq 0} (-1)^n \frac{(q-r)!}{(q-r+n+1)!} \frac{(p+r)!}{(p+r-n)!} , $$

so the generalized Leibnitz rule holds in this case with the generalized integration $D^{-1} \frac{x^s}{s!} = \frac{x^{s+1}}{(s+1)!}$ for $x >0$ when

$$\frac{1}{p+q+1} =\sum_{n \geq 0} (-1)^n \frac{(q-r)!}{(q-r+n+1)!} \frac{(p+r)!}{(p+r-n)!} $$

which is valid for $p+q +1 > 0$

(see p. 268 in Section 105 of Die Gammafunktion by Niels Nielsen for convergence of this sum).

Source Link
Tom Copeland
  • 10.5k
  • 3
  • 57
  • 84

Extensions of the Leibnitz rule to fractional calculus of a type similar to that Gessel depicts were investigated by Osler in the 1970s. See, e.g., "Fractional derivatives and special functions" by Lavoie, Osler, and Tremblay.

One can even make expansions of the type

$$D^{-1} f(x)g(x) = (D_L+D_R)^{-1} f(x)g(x) = D_L^{-1}(1+D_R/D_L)^{-1}f(x)g(x)$$

$$ = \sum_{n \geq 0} \binom{-1}{n} [D^{-n-1}f(x)][D^ng(x)]= \sum_{n \geq 0} (-1)^n [D^{-n-1}f(x)][D^ng(x)]$$

for some classes of functions with, e.g., $D^{-1}f(x) = \int_0^x f(t)dt$ and the indices on $D$ designating action on either the Left or Right factor. Try $f(x) = x^{q-r}$ and $g(x) = x^{p+r}$ . Then

$$D^{-1}f(x)g(x) = D^{-1}x^{p+q} = \frac{x^{p+q+1}}{p+q+1}$$

and

$$\sum_{n \geq 0} (-1)^n [D^{-n-1}f(x)][D^ng(x)] = \sum_{n \geq 0} (-1)^n [D^{-n-1}x^{q-r}][D^nx^{p+r}]$$

$$ =x^{p+q+1} \sum_{n \geq 0} (-1)^n \frac{(q-r)!}{(q-r+n+1)!} \frac{(p+r)!}{(p+r-n)!} , $$

so the generalized Leibnitz rule holds in this case when

$$\frac{1}{p+q+1} =\sum_{n \geq 0} (-1)^n \frac{(q-r)!}{(q-r+n+1)!} \frac{(p+r)!}{(p+r-n)!} $$

which is apparently valid for $p+q +1 > 0.$