Suppose I had a function that satisfied the property $f(x)=f(x-1)+g(x)$. For any $x\in\mathbb N$, it is easy enough to see that this boils down to the statement


If we return to our functional equation and differentiate it $n$ times, we get


Once again, for any $x\in\mathbb N$, we have


One could then integrate both sides. For example, this asserts that


If the sum inside the intagral is generalized to any upper bound.

How would I justify this for discrete functions?

An example of this is letting $f_p(x)=\sum_{k=1}^xk^p$, as it seems to follow the above result. We can see from the above that



which appears to be a true recursive formula.

Under what conditions can I extend discrete functions $f(x)$ and $g(x)$ into continuous differentiable functions and apply this method?

  • 2
    $\begingroup$ What is exactly is the method? This is clearly true whenever $f$ and $g$ are differentiable, but the derivative of an extension of a discrete function to one defined on all of $\mathbb R$, being a local property, seems rather meaningless without further constraints. $\endgroup$ – Milo Brandt Nov 27 '16 at 23:54
  • $\begingroup$ @MiloBrandt I've tried to be more clear on this. If it is rather meaningless, when can I determine if it is meaningful? $\endgroup$ – Simply Beautiful Art Nov 28 '16 at 0:00
  • $\begingroup$ @MiloBrandt I've added what this lets me do for general functions. $\endgroup$ – Simply Beautiful Art Nov 28 '16 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.