Derivative of a sum w.r.t its limits

Question

As an analogue of $$\frac{d}{\,dx} \int_a^x f(y) \,dy = f(x)$$

Could we define something for an expression of the derivatives of partial sums; something like $$\frac{d}{\,dn} \sum_{k=a}^n f(k)$$?

Answer 1

It is well known that:

$$\frac{d}{dx} g(x) = \lim_{h \to 0^+} \frac{g(x+h)-g(x)}{h}.$$

This makes sense when $g : \mathbb{R} \to \mathbb{A} \subseteq \mathbb{R}$. If $g : \mathbb{N} \to \mathbb{A} \subseteq \mathbb{R}$, then the limit used for defining the derivative loses its meaning.

For the discrete case, we can introduce the difference operator, in place of the derivative:

$$D g(k) = g(k+1)-g(k).$$

Specifically, this operator is also known as forward difference operator. Check this link for further details.

It is interesting to notice that: $$\sum_{k=a}^n D f(k) = \sum_{k=a}^n [f(k+1)-f(k)] = \\ = [f(a+1)-f(a)]+[f(a+2)-f(a+1)]+\ldots = f(n+1) - f(a),$$

which is somehow similar to:

$$ \int_a^x \frac{d}{\,dx}f(y) \,dy = f(x)-f(a)$$

Moreover, if we call

$$F(n) = \sum_{k=a}^n f(k),$$ then:

$$D \sum_{k=a}^n f(k) = F(n+1) - F(n) = \sum_{k=a}^{n+1} f(k) - \sum_{k=a}^n f(k) = f(n+1).$$