3 ms·
Its interesting to see the discrete case. Let D_n be a difference operator like D_n(F) = F(x+n) - F(x) Solving by induction for the equation D_n(F)/D_n(x) = F
by huachimingo 5y ago
Its interesting to see the discrete case.
Let D_n be a difference operator like
D_n(F) = F(x+n) - F(x)
Solving by induction for the equation D_n(F)/D_n(x) = F(x), F(0)=1, it gives (1+n)^(x/n).
So now you have a "family" of functions that when "derived" by its difference, gives the same constant ^ x. Euler is the case when n→0.
This aplies for every n, including complex numbers.