# A method for evaluating sums/discrete functions by assuming they can be made continuous and differentiable?

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

$$f(x)=f(0)+\sum_{k=1}^xg(k)$$

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

$$f^{(n)}(x)=f^{(n)}(x-1)+g^{(n)}(x)$$

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

$$f^{(n)}(x)=f^{(n)}(0)+\sum_{k=1}^xg^{(n)}(k)$$

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

$$\sum_{k=1}^xg(k)=xf'(0)+\int_0^x\sum_{k=1}^tg'(k)dt$$

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

$$f_p(x)=a_px+p\int_0^xf_{p-1}(t)dt$$

$$a_p=1-p\int_0^1f_{p-1}(t)dt$$

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?

• 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. – Milo Brandt Nov 27 '16 at 23:54
• @MiloBrandt I've tried to be more clear on this. If it is rather meaningless, when can I determine if it is meaningful? – Simply Beautiful Art Nov 28 '16 at 0:00
• @MiloBrandt I've added what this lets me do for general functions. – Simply Beautiful Art Nov 28 '16 at 18:06