Kharakov
Quantum Hot Dog
Assume generalized binomial coefficients. Also assume this is not for normal fractional (iterative) derivatives, but for the following operation. I think I should implement it in python or something... just wonder if there is an easy way to do it on a phone (like how do I code it for Wolfram Alpha???) since my computer time is limited.
Latex broken? I specifically prohibit Bilby from replying to this part of the question if no forum user is wearing latex.
\(f^n(x) = \lim_{h \to 0} \frac{ \sum\limits_{k=0}^\infty \binom{\alpha}{k} (-1)^k f(x+ (n-k) h ) } {h^n}\)
So, limits as h-->0:
f0 (x) = [f(x) ]/h^0
f1 (x) = [f(x+1h) - f(x) ]/h^1
f1.5 (x) = [f(x+1.5h) - 1.5 f(x+.5h) + 3/8 f(x-.5h) +1/16 f(x-1.5h) +3/128 f(x-2.5h).... ]/h^1.5
f2 (x) = [f(x+2h) -2 f(x+h) +f(x) ]/h^2
f3 (x) = [f(x+3h) -3 f(x+2h) +3 f(x+h) -f(x) ]/h^3
Latex broken? I specifically prohibit Bilby from replying to this part of the question if no forum user is wearing latex.
\(f^n(x) = \lim_{h \to 0} \frac{ \sum\limits_{k=0}^\infty \binom{\alpha}{k} (-1)^k f(x+ (n-k) h ) } {h^n}\)
So, limits as h-->0:
f0 (x) = [f(x) ]/h^0
f1 (x) = [f(x+1h) - f(x) ]/h^1
f1.5 (x) = [f(x+1.5h) - 1.5 f(x+.5h) + 3/8 f(x-.5h) +1/16 f(x-1.5h) +3/128 f(x-2.5h).... ]/h^1.5
f2 (x) = [f(x+2h) -2 f(x+h) +f(x) ]/h^2
f3 (x) = [f(x+3h) -3 f(x+2h) +3 f(x+h) -f(x) ]/h^3
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