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

  1. Mar 26, 2016 #1
    We know that in the continous 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 ?

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  3. Mar 26, 2016 #2


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    You do not have derivatives in the same sense in discrete mathematics so you will need to specify what you mean by f'(n).
  4. Mar 26, 2016 #3


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    If, by f', you mean the finite difference then f'(x)= (f(x+ h)- f(x))/h. In particular, with [itex]f(x)= a^x[/itex], [itex]f'(x)= (a^{x+ h}- a^x)/h= a^x(a^h- 1)/h= f(x)= a^x[/itex] so that [itex]a^h- 1= h[/itex] and [itex]a= (h+ 1)^{1/h}[/itex]. 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: [itex]a= \lim_{h\to 0} (h+ 1)^{1/h}= e[/itex].
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