F'(n) = f(n) in discrete calculus

[Bruno Tolentino]

We know that in the continuous math, e is special number because if f(x) = e^x, so f'(x) = f(x). But in discrete math, what's the constante base that satisfies this condition? Is not the 2? I. e. f(n) = 2^n ?

Thanks,

 

[Orodruin]

You do not have derivatives in the same sense in discrete mathematics so you will need to specify what you mean by f'(n).

 

[HallsofIvy]

If, by f', you mean the finite difference then f'(x)= (f(x+ h)- f(x))/h. In particular, with $f(x)= a^x$, $f'(x)= (a^{x+ h}- a^x)/h= a^x(a^h- 1)/h= f(x)= a^x$ so that $a^h- 1= h$ and $a= (h+ 1)^{1/h}$. In the case that h= 1, which is the most common case, a= 2.

The "Calculus" case can be recovered by taking the limit as h goes to 0: $a= \lim_{h\to 0} (h+ 1)^{1/h}= e$.