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1st iterated derivatives and the tetration of 0
#1
Hi, everyone!

Let us sign d/dx as just D, let J mean the antiderivative like this way: J f(x) = int from 0 to x f(u) du.
Let n and m be an integer and let z and w be a complex number.
Generally D^n J^n f equals to f while J^n D^n f does not.
There are tries to extend the derivatives, like the Riemann-Liouville definition:
RL: D^z [a;x] f(x) = 1/gamma(ceil(z)-z) D^ceil(z) int from a to x f(u)/(x-u)^(z-ceil(z)+1) du
Or here is the Grünwald-Letnikov definiton:
GL: D^z [a;x] f(x) = lim 1/h^z sum from k=0 to infinity (-1)^k binomial(z,k) f(x+kh)
and if z is not a complex number instead of an integer:
GL: D^n [a;x] f(x) = lim 1/h^n sum from k=0 to n (-1)^k binomial(n,k) f(x+kh)

Here is my question: how to iterate the derivative operator? - My answer is: we just need to iterate the operations of the GL formula. Let us see the first iterated third derivative of a function:

D^3 f(x) = lim 1/h^3 (f(x) - 3 f(x+h) + 3 f(x+2h) - f(x+3h)) = f'''(x)
Its iterated:
D^^3 f(x) = lim h^^3th root of ( (f(x) f(x+2h)^3) ÷ (f(x+h) f(x+3h)^3) )
And because of 0^^3 = 0, therefor
D^^3 f(x) = lim hth root of ( (f(x) f(x+2h)^3) ÷ (f(x+h) f(x+3h)^3) )

E.g.:
D^^3 x = exp(-4/x)
according to Wolfram Alpha.

0^^2 = 1, so the first iterated second derivative is D^^2 f(x) = f(x) as all the eventh derivative.
I am interested in the halfth derivative:

D^^0.5 f(x) = lim 0^^0.5th root of prod(k=0,infintiy,(f(x+kh)^binomial(0.5,k))^((-1)^k))

But what is 0^^0.5?
0^^0 = 1, 0^^1 = 0, 0^^2 = 1, 0^^3 = 0, ...
0^^z = ?
And if you made a pari/gp code, I would be really thankfull.

(And in the lim h approaches to 0.)
Xorter Unizo
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