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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.

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.

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.

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,

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

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

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.

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.

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.